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Qm Math Intervals And Contacts

1 · Prerequisites

2 · Summary

The interval and interface layer of the non-relativistic quantum mechanics mathematical prerequisites, all of it mathematics: one-dimensional Sobolev spaces and their boundary traces, the sine and character bases of $L^{2} (0, L)$ , the Dirichlet, twisted-periodic and separated Robin realisations of $- d^{2} / d x^{2} + V$ , the regular initial value problem and its Prufer angle, the associated Sturm oscillation theory, and the point-interaction realisation of the delta interface together with its rank-one resolvent and unique bound state. Every item carries domain: mathematics and depends only on mathematical suppliers, published or from qm-math-oscillator; nothing here asserts a physical interpretation, measurement model or empirical claim, and the companion page qm-math-intervals-and-contacts-examples stays mathematics as well.

One-dimensional Sobolev spaces on intervals fixes the notation: $H^{1} (I)$ and $H^{2} (I)$ are the classes with one, respectively two, $L^{2}$ weak derivatives on an interval, $H_{0}^{1} (0, L)$ is the closure of the test functions, and the Dirichlet form is $q_{D} [u] = ∥ D u ∥^{2}$ . Absolutely continuous representative of an L2 function with L2 weak derivative then extracts the unique absolutely continuous representative of an $H^{1}$ class with the difference formula $u^{*} (x) - u^{*} (y) = \int_{y}^{x} u^{'}$ and the Holder modulus $\sqrt{∣ x - y ∣} ∥ u^{'} ∥$ , and Endpoint trace bound and decay of H1 functions proves the endpoint bounds $∣ u^{*} (x) ∣ \leq L^{- 1 / 2} ∥ u ∥ + L^{1 / 2} ∥ u^{'} ∥$ , the half-line decay $u^{*} (x) \to 0$ , and the vanishing of the boundary products used for integration by parts. Integration by parts and the endpoint-vanishing space on intervals contains the first- and second-order integration-by-parts identities on bounded intervals, the half-line and line identities, and the identification $H_{0}^{1} (0, L) = {u \in H^{1} : u^{*} (0) = u^{*} (L) = 0}$ by a cutoff-and-mollification argument. Completeness of the Dirichlet sine system on an interval proves that the normalised sines form a complete orthonormal family of $L^{2} (0, L)$ , by odd extension and transport of the torus completeness theorem.

The spectral layer begins with Identification of the Dirichlet Laplacian with the sine diagonal operator, which identifies the diagonal operator of qm-math-oscillator with eigenvalues $λ_{n} = (n π / L)^{2}$ with the Dirichlet Laplacian $A_{D} u = - D^{2} u$ on $D (A_{D}) = {u \in H^{2} (0, L) : u^{*} (0) = u^{*} (L) = 0}$ , proves $A_{D} \geq λ_{1}$ and the form identity $⟨ A_{D} u, u ⟩ = ∥ D u ∥^{2}$ . Self-adjoint operators with compact resolvent: eigenbasis and discrete spectrum abstracts the resulting structure: a self-adjoint operator with compact resolvent has a complete orthonormal eigenfamily, finite-dimensional eigenspaces, spectrum equal to the eigenvalue set and empty essential spectrum. The twisted-periodic momentum operator on an interval proves that $P_{θ} = - i D$ with the twisted-periodic condition $u^{*} (L) = e^{i θ} u^{*} (0)$ is self-adjoint, with the complete orthonormal characters $φ_{n}$ and simple eigenvalues $(2 π n + θ) / L$ . The companion example Dirichlet momentum failure shows that no self-adjoint extension retains both Dirichlet conditions. The free-line resolvent kernel and its inverse operator supplies the free-line kernel $G_{z} (t) = (2 k)^{- 1} e^{- k ∣ t ∣}$ , its two-sided inverse property and the distributional identity $(- d^{2} / d x^{2} - z) G_{z} = δ_{0}$ . Bounded perturbations of the interval Dirichlet Laplacian adds a real continuous potential by Kato-Rellich, proves the compact resolvent and the min-max bounds $∣ E_{n} - λ_{n} ∣ \leq ∥ V ∥_{\infty}$ .

The oscillation layer is Regular interval IVP: existence, a priori bound and entire spectral-parameter dependence, which solves $- u^{''} + V u = z u$ with arbitrary initial data, proves the a priori bound, gives a locally uniform Picard series and entire dependence on the spectral parameter $z$ , and identifies the differentiated equation. Then Prufer angle for the interval equation and strict monotonicity in the energy introduces the Prufer angle $θ$ with $θ^{'} = \cos^{2} θ + (E - V) \sin^{2} θ$ and proves $\partial_{E} θ (L, E) = \int_{0}^{L} u^{2} / (u^{2} + u^{' 2}) > 0$ . The Dirichlet nodal theorem Dirichlet eigenvalues, simplicity and the nodal theorem on an interval then characterises the eigenvalues as the unique roots of $θ (L, E) = k π$ , proves that they are simple and increasing with the bounds $(k π / L)^{2} + V_{\min} \leq E_{k} \leq (k π / L)^{2} + V_{\max}$ , and shows that the $k$ -th eigenfunction has exactly $k - 1$ zeros in $(0, L)$ .

The interface layer is The one-dimensional delta interface operator, the self-adjoint realisation of $- d^{2} / d x^{2}$ with the jump condition $u^{' *} (0 +) - u^{' *} (0 -) = α u^{*} (0)$ ; Rank-one resolvent formula for the delta interface operator computes its resolvent as $R_{0} (z)$ minus the rank-one term $α ℓ_{z} (f) G_{z} / (1 + α G_{z} (0))$ ; and The unique negative eigenvalue of the attractive delta interface operator proves that the sole negative eigenvalue is $- α^{2} / 4$ , present exactly for $α < 0$ , with normalised eigenfunction $\sqrt{- α / 2} e^{α ∣ x ∣ / 2}$ , and records the conditional scaling $H = (ℏ^{2} / 2 m) A_{α}$ , $g = ℏ^{2} α / (2 m)$ , $E_{H} = - m g^{2} / (2 ℏ^{2})$ . Green resolvent for separated Robin boundary conditions builds the Green kernel $G_{z} (x, y) = - u_{a} (\min) u_{b} (\max) / W$ for the separated Robin condition $u^{' *} (0) = α u^{*} (0)$ , $u^{' *} (L) = β u^{*} (L)$ and proves self-adjointness with compact resolvent; Simple complete spectrum of the separated Robin realisation adds the complete simple eigenbasis. Finally Rectangular barrier: transmission coefficients and the thick-barrier error computes the rectangular-barrier transmission coefficient and its controlled thick-barrier error, and Finite square well: root conditions, simplicity and the bound-state count computes the finite-well root conditions, the simplicity of each bound state and the exact count $⌈ 2 a K / π ⌉$ . All items are draft mathematics under construction; physical pages consume them as stated mathematical premises and add their own postulates and interpretation.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

One-dimensional Sobolev spaces on intervals

Definition

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Fix complex scalars. Let $I \subseteq R$ be a nonempty open interval, bounded or unbounded, and write $L^{2} (I; C)$ for the complex Hilbert space of Lebesgue (equivalence classes of) square-integrable functions with the inner product $⟨ u, v ⟩ = \int_{I} u \overline{v}$ linear in the first argument (Complex Lp classes and Euclidean test-function conventions, The space $L^{p} (μ)$ as the quotient by null functions, $L^{2}$ with the integral pairing is a Hilbert space).

First-order space. Define $H^{1} (I) : = {u \in L^{2} (I; C) : the weak derivative D u of the class u exists and D u \in L^{2} (I; C)},$ where $D u$ is the weak derivative of Weak derivative of a locally integrable function: a class $D u \in L_{l o c}^{1} (I; C)$ with $\int_{I} u φ^{'} = (- 1) \int_{I} (D u) φ for every φ \in C_{c}^{\infty} (I),$ an identity that is choice-free (Locally integrable functions as regular distributions). Equip $H^{1} (I)$ with the sesquilinear form $⟨ u, v ⟩_{H^{1}} : = \int_{I} u \overline{v} + \int_{I} (D u) \overline{D v},$ whose associated norm is $∥ u ∥_{H^{1}} = (∥ u ∥_{L^{2}}^{2} + ∥ D u ∥_{L^{2}}^{2})^{1 / 2}$ ; the symbol $u^{'}$ denotes the same weak derivative $D u$ throughout the page.

Second-order space. Define $H^{2} (I) : = {u \in H^{1} (I) : D u \in H^{1} (I)},$ and for $u \in H^{2} (I)$ write $D^{2} u : = D (D u) \in L^{2} (I; C)$ ; the norm is $∥ u ∥_{H^{2}} = (∥ u ∥_{L^{2}}^{2} + ∥ D u ∥_{L^{2}}^{2} + ∥ D^{2} u ∥_{L^{2}}^{2})^{1 / 2}$ . Thus $H^{2} (I)$ consists exactly of the classes in $L^{2} (I)$ whose first and second weak derivatives exist and lie in $L^{2} (I)$ , and $H^{2} (I) \subseteq H^{1} (I)$ .

Endpoint-vanishing first-order space and the Dirichlet form. For the bounded interval $(0, L)$ , $L > 0$ , define $H_{0}^{1} (0, L) : = {\overline{C_{c}^{\infty} (0, L)}}^{∥ \cdot ∥_{H^{1}}},$ the closure of the test functions in $H^{1} (0, L)$ , and define the Dirichlet form on it by $q_{D} [u] : = ∥ D u ∥_{L^{2} (0, L)}^{2} = \int_{0}^{L} ∣ D u ∣^{2}$ .

Elementary consequences, recorded for reuse. Weak derivatives are unique as locally integrable classes under Countable Choice, are additive and $C$ -homogeneous, and classical derivatives of continuously differentiable functions are weak derivatives. The cited published items supply these facts, so $H^{1} (I)$ and $H^{2} (I)$ are well-defined linear subspaces of $L^{2} (I; C)$ . No claim is made here that $H_{0}^{1} (0, L)$ equals the set of classes with vanishing endpoint values; that identification is proved on this page from the trace estimates.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Absolutely continuous representative of an L2 function with L2 weak derivative

Statement

Assume the Axiom of Choice and the Axiom of Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $- \infty < a < b < \infty$ and let $u \in H^{1} (a, b)$ have weak derivative $u^{'} = D u$ (One-dimensional Sobolev spaces on intervals). Then:

(a) there is exactly one representative $u^{*}$ of the class $u$ that is continuous on $[a, b]$ , is absolutely continuous on $[a, b]$ , and satisfies $u^{*} (x) - u^{*} (y) = \int_{y}^{x} u^{'} (t) d t for all x, y \in [a, b],$ where for $x < y$ the right-hand side means $- \int_{x}^{y} u^{'}$ ;

(b) for all $x, y \in [a, b]$ , $∣ u^{*} (x) - u^{*} (y) ∣ \leq \sqrt{∣ x - y ∣} ∥ u^{'} ∥_{L^{2} (a, b)},$ so $u^{*}$ is uniformly continuous, and the assignment $u \mapsto u^{*}$ is linear;

(c) if additionally $u \in H^{2} (a, b)$ , then $(u^{*})^{'}$ is absolutely continuous on $[a, b]$ and $(u^{*})^{'} (x) - (u^{*})^{'} (y) = \int_{y}^{x} D^{2} u$ for all $x, y \in [a, b]$ .

Facts & Assumptions

Given: reals $a < b$ ; a class $u \in H^{1} (a, b) \subseteq L^{2} (a, b; C)$ with weak derivative $u^{'} \in L^{2} (a, b; C)$ .

[F1]

$H^{1} (a, b)$ and $H^{2} (a, b)$ are the spaces of One-dimensional Sobolev spaces on intervals: $u \in H^{1} (a, b)$ means $u \in L^{2} (a, b; C)$ and the weak derivative $u^{'} = D u$ lies in $L^{2} (a, b; C)$ ; $u \in H^{2} (a, b)$ means $u \in H^{1} (a, b)$ and $u^{'} \in H^{1} (a, b)$ , with $D^{2} u = D (u^{'})$ .

[F2]

Assume the Axiom of Choice for the ACL and absolutely continuous fundamental-theorem interfaces. For a nonempty open interval $I \subseteq R$ , $1 \leq p \leq \infty$ and $u \in W^{1, p} (I; C)$ with weak derivative $D u$ , there is exactly one continuous representative $u^{*}$ that is locally absolutely continuous and satisfies $u^{*} (x) - u^{*} (y) = \int_{y}^{x} D u$ ; if $I = (a, b)$ has finite endpoints, then $D u \in L^{1} (a, b)$ , $u^{*}$ extends uniquely to an absolutely continuous function on $[a, b]$ , and $u^{*} (x) = u^{*} (a) + \int_{a}^{x} D u$ for every $x \in [a, b]$ (One-dimensional $W^{1, p}$ functions have unique absolutely continuous representatives).

[F3]

For $1 \leq p < r < \infty$ on a finite-measure space, the real-valued finite-measure inclusion applied to $∣ f ∣$ gives $∥ f ∥_{p} \leq μ (X)^{1 / p - 1 / r} ∥ f ∥_{r}$ also for complex $f$ , by the modulus definition of complex $L^{p}$ (Finite-measure $L^{r}$ includes into $L^{p}$ for $p < r$ ). The complex integral triangle inequality is $∣ \int_{E} f ∣ \leq \int_{E} ∣ f ∣$ (The modulus of an integral is bounded by the integral of the modulus). Applying the real-valued Holder inequality to $∣ f ∣$ and $1$ on a finite-measure set $E$ gives $\int_{E} ∣ f ∣ \leq ∥ f ∥_{L^{2} (E)} μ (E)^{1 / 2}$ (Holder's inequality for integrals, including the endpoint cases).

[F4]

Under [F1], $W^{1, 2} (a, b; C) = H^{1} (a, b)$ as sets with the same weak derivative classes; both conditions are existence of an $L^{2}$ weak derivative, so the identification is immediate from the definitions.

Proof

1.1F1F3F4given

[Reduction to the published corollary] The interval $(a, b)$ has finite Lebesgue measure $b - a$ , and $1 \leq 2 < \infty$ , so [F3] gives $u \in L^{2} (a, b) \subseteq L^{1} (a, b)$ and $u^{'} \in L^{2} (a, b) \subseteq L^{1} (a, b)$ , while $u^{'}$ is the $L^{2}$ weak derivative of $u$ by [F1]. By [F4] the class $u$ lies in $W^{1, 2} (a, b; C)$ with weak derivative $u^{'}$ .

2.1step 1.1F1F2givenalgebra

[Existence and uniqueness of the representative] Apply [F2] with $I = (a, b)$ and $p = 2$ : there is exactly one representative $u^{*}$ of the class $u$ that is continuous, locally absolutely continuous and satisfies the integral formula. Since $I$ has finite endpoints, [F2] gives $u^{'} \in L^{1} (a, b)$ , the unique extension of $u^{*}$ to an absolutely continuous function on $[a, b]$ , and the formula for all $x, y \in [a, b]$ ; the assignment $u \mapsto u^{*}$ is linear because the weak derivative and the integral are linear (One-dimensional Sobolev spaces on intervals records additivity of the weak derivative). This proves (a).

3.1step 2.1F3algebra

[Holder modulus] Let $x, y \in [a, b]$ with $x \leq y$ . If $x = y$ , both sides are zero. If $x < y$ , apply the complex integral triangle inequality in [F3] and then real Holder to $∣ u^{'} ∣$ and $1$ on $(x, y)$ , whose measure is $y - x$ , to obtain $∣ u^{*} (x) - u^{*} (y) ∣ = ∣ \int_{x}^{y} u^{'} ∣ \leq \int_{x}^{y} ∣ u^{'} ∣ \leq ∥ u^{'} ∥_{L^{2} (x, y)} \sqrt{y - x} \leq ∥ u^{'} ∥_{L^{2} (a, b)} \sqrt{y - x};$ the case $x > y$ follows by exchanging $x$ and $y$ . This is (b).

4.1step 2.1F1F2algebra∎

[Second-order representative] If additionally $u \in H^{2} (a, b)$ , then $u^{'} \in H^{1} (a, b)$ with weak derivative $D^{2} u \in L^{2} (a, b)$ by [F1]. Applying the already proved part (a) with $u^{'}$ in place of $u$ supplies a continuous representative $w$ of the class $u^{'}$ , absolutely continuous on $[a, b]$ , with $w (x) - w (y) = \int_{y}^{x} D^{2} u$ for all $x, y \in [a, b]$ . The a.e. derivative of $u^{*}$ equals the class $u^{'}$ , so it equals $w$ almost everywhere. The fundamental theorem for absolutely continuous functions therefore gives $u^{*} (x) = u^{*} (a) + \int_{a}^{x} w (t) d t$ for every $x \in [a, b]$ . Since $w$ is continuous, this identity makes $u^{*}$ continuously differentiable and gives $(u^{*})^{'} (x) = w (x)$ at every $x \in [a, b]$ . Thus $(u^{*})^{'}$ is absolutely continuous and has the displayed difference formula, proving (c).

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Endpoint trace bound and decay of H1 functions

Statement

Assume the Axiom of Choice and Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $u^{*}$ denote the continuous absolutely continuous representative of a class in $H^{1}$ supplied by Absolutely continuous representative of an L2 function with L2 weak derivative, and write $∥ u ∥ = ∥ u ∥_{L^{2}}$ and $∥ u^{'} ∥ = ∥ u^{'} ∥_{L^{2}}$ for the norms of One-dimensional Sobolev spaces on intervals on the interval under consideration.

(a) Bounded-interval trace bound. Let $L > 0$ and $u \in H^{1} (0, L)$ . Then $∣ u^{*} (x) ∣ \leq L^{- 1 / 2} ∥ u ∥_{L^{2} (0, L)} + L^{1 / 2} ∥ u^{'} ∥_{L^{2} (0, L)} for every x \in [0, L] .$ More generally, for $0 \leq c < d \leq L$ and $x \in [c, d]$ , $∣ u^{*} (x) ∣ \leq (d - c)^{- 1 / 2} ∥ u ∥_{L^{2} (c, d)} + (d - c)^{1 / 2} ∥ u^{'} ∥_{L^{2} (c, d)} .$

(b) Half-line. Let $u \in H^{1} (0, \infty)$ , with representative $u^{*}$ defined on $[0, \infty)$ by gluing the representatives on $(0, n)$ . Then $∣ u^{*} (x) ∣ \leq ∥ u ∥_{L^{2} (x, x + 1)} + ∥ u^{'} ∥_{L^{2} (x, x + 1)} (x \geq 0), \sup_{x \geq 0} ∣ u^{*} (x) ∣ \leq ∥ u ∥_{L^{2} (0, \infty)} + ∥ u^{'} ∥_{L^{2} (0, \infty)},$ and $u^{*} (x) \to 0$ as $x \to \infty$ .

(c) Real line and vanishing boundary products. Let $u, v \in H^{1} (R)$ with continuous representatives $u^{*}, v^{*}$ . Then $u^{*}, v^{*}$ are bounded and uniformly continuous with $∣ u^{*} (x) - u^{*} (y) ∣ \leq \sqrt{∣ x - y ∣} ∥ u^{'} ∥_{L^{2} (R)}, \sup_{R} ∣ u^{*} ∣ \leq ∥ u ∥_{L^{2} (R)} + ∥ u^{'} ∥_{L^{2} (R)},$ $u^{*} (x) \to 0$ and $v^{*} (x) \to 0$ as $∣ x ∣ \to \infty$ , and consequently $\lim_{∣ x ∣ \to \infty} u^{*} (x) \overline{v^{*} (x)} = 0$ .

Facts & Assumptions

Given: the spaces $H^{1}$ of One-dimensional Sobolev spaces on intervals; the absolutely continuous representative theorem Absolutely continuous representative of an L2 function with L2 weak derivative and its estimate $∣ u^{*} (x) - u^{*} (y) ∣ \leq \sqrt{∣ x - y ∣} ∥ u^{'} ∥_{L^{2}}$ .

[F1]

For a bounded interval $(c, d)$ and $u \in H^{1} (c, d)$ , the unique absolutely continuous representative $u^{*}$ on $[c, d]$ satisfies $u^{*} (x) - u^{*} (y) = \int_{y}^{x} u^{'}$ and $∣ u^{*} (x) - u^{*} (y) ∣ \leq \sqrt{∣ x - y ∣} ∥ u^{'} ∥_{L^{2} (c, d)}$ (Absolutely continuous representative of an L2 function with L2 weak derivative).

[F2]

On a finite-measure set, $\int_{E} ∣ f ∣ \leq ∥ f ∥_{L^{2}} μ (E)^{1 / 2}$ (Holder's inequality for integrals, including the endpoint cases).

[F3]

The indefinite integral of an integrable function is countably additive over disjoint measurable sets: for pairwise disjoint measurable $E_{n}$ and integrable $f$ , $\int_{⋃_{n} E_{n}} f = \sum_{n} \int_{E_{n}} f$ (The indefinite integral of an integrable function is countably additive on measurable sets).

Proof

1.1F1F2algebra

[Bounded-interval trace bound] Let $u \in H^{1} (0, L)$ and fix $x \in [0, L]$ . By [F1] applied on $(0, L)$ (and on $(c, d)$ for the local form), for every $y \in [0, L]$ $∣ u^{*} (x) ∣ \leq ∣ u^{*} (y) ∣ + ∣ \int_{y}^{x} u^{'} ∣ \leq ∣ u^{*} (y) ∣ + ∥ u^{'} ∥_{L^{2} (0, L)} L^{1 / 2},$ using [F2] for the last estimate. Integrating this inequality in $y$ over $(0, L)$ and using [F2] again, $L ∣ u^{*} (x) ∣ \leq \int_{0}^{L} ∣ u^{*} ∣ + L^{3 / 2} ∥ u^{'} ∥ \leq L^{1 / 2} ∥ u ∥ + L^{3 / 2} ∥ u^{'} ∥,$ so $∣ u^{*} (x) ∣ \leq L^{- 1 / 2} ∥ u ∥ + L^{1 / 2} ∥ u^{'} ∥$ . Replacing $(0, L)$ by any subinterval $(c, d)$ with $x \in [c, d]$ gives the local form by the same computation. This is (a).

2.1step 1.1F1algebra

[Half-line pointwise and supremum bounds] Let $u \in H^{1} (0, \infty)$ . For every $n$ the class lies in $H^{1} (0, n)$ , and the representatives on overlapping intervals agree by uniqueness in [F1], so they glue to a continuous representative $u^{*}$ on $[0, \infty)$ . Step 1.1 on the interval $(x, x + 1)$ , $x \geq 0$ , gives $∣ u^{*} (x) ∣ \leq ∥ u ∥_{L^{2} (x, x + 1)} + ∥ u^{'} ∥_{L^{2} (x, x + 1)}$ . Enlarging each interval to $(0, \infty)$ and using $∥ u ∥_{L^{2} (x, x + 1)} \leq ∥ u ∥_{L^{2} (0, \infty)}$ and likewise for $u^{'}$ gives $\sup_{x \geq 0} ∣ u^{*} (x) ∣ \leq ∥ u ∥_{L^{2} (0, \infty)} + ∥ u^{'} ∥_{L^{2} (0, \infty)}$ .

3.1step 2.1F1F3givenalgebra

[Decay] We show $\lim_{x \to \infty} u^{*} (x) = 0$ . First suppose $∥ u^{'} ∥_{L^{2} (0, \infty)} = : M > 0$ and assume for contradiction that there are $x_{n} \to \infty$ with $∣ u^{*} (x_{n}) ∣ \geq ε$ for some $ε > 0$ . Put $δ : = \min {1, (ε / (2 M))^{2}} > 0$ ; by [F1], for $t \in [x_{n}, x_{n} + δ]$ $∣ u^{*} (t) ∣ \geq ∣ u^{*} (x_{n}) ∣ - \sqrt{∣ t - x_{n} ∣} M \geq ε - \sqrt{δ} M \geq ε / 2.$ Passing to a subsequence, assume $x_{n + 1} \geq x_{n} + δ$ for all $n$ ; then the half-open intervals $E_{n} : = [x_{n}, x_{n} + δ)$ are pairwise disjoint, have measure $δ$ , and satisfy $∣ u^{*} ∣^{2} \geq ε^{2} / 4$ throughout. By [F3], $\int_{0}^{\infty} ∣ u ∣^{2} \geq \int_{⋃_{n} E_{n}} ∣ u ∣^{2} = \sum_{n} \int_{E_{n}} ∣ u ∣^{2} \geq \sum_{n} \frac{ε^{2}}{4} δ = + \infty,$ contradicting $u \in L^{2} (0, \infty)$ . If $M = 0$ , then [F1] gives $∣ u^{*} (x) - u^{*} (y) ∣ \leq 0$ , so $u^{*}$ is a constant whose $L^{2}$ norm over the half-line is finite only if that constant is $0$ . Thus $u^{*} (x) \to 0$ .

4.1step 2.1step 3.1algebra∎

[Real line] For $u \in H^{1} (R)$ restrict to $(0, \infty)$ and $(- \infty, 0)$ ; the representatives glue at $0$ by uniqueness of the continuous representative. Steps 1.1 and 2.1 give the sup bound and uniform continuity, and step 3.1 applied to both half-lines gives $u^{*} (x) \to 0$ as $x \to \pm \infty$ . For $u, v \in H^{1} (R)$ the elementary bound $∣ u^{*} \overline{v^{*}} ∣ \leq \frac{1}{2} (∣ u^{*} ∣^{2} + ∣ v^{*} ∣^{2})$ and $u^{*} (x) \to 0$ , $v^{*} (x) \to 0$ give $u^{*} (x) \overline{v^{*} (x)} \to 0$ as $∣ x ∣ \to \infty$ . This is (c).

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Integration by parts and the endpoint-vanishing space on intervals

Statement

Assume the Axioms of Choice, Countable Choice and Dependent Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ ), The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain). Use the spaces of One-dimensional Sobolev spaces on intervals and the absolutely continuous representatives supplied by Absolutely continuous representative of an L2 function with L2 weak derivative.

(a) Bounded intervals. Let $0 < L < \infty$ and $u, v \in H^{1} (0, L)$ . Then $\int_{0}^{L} u \overline{v^{'}} = u^{*} (L) \overline{v^{*} (L)} - u^{*} (0) \overline{v^{*} (0)} - \int_{0}^{L} u^{'} \overline{v} .$

(b) Endpoint-vanishing space. $H_{0}^{1} (0, L) = {u \in H^{1} (0, L) : u^{*} (0) = u^{*} (L) = 0}$ , with the traces $u^{*} (0), u^{*} (L)$ of the continuous representative.

(c) Second-order identity. For $u, v \in H^{2} (0, L)$ , $\int_{0}^{L} (u^{''} \overline{v} - u \overline{v^{''}}) = [u^{' *} \overline{v^{*}} - u^{*} \overline{v^{' *}}]_{0}^{L},$ where the bracket denotes $u^{' *} (L) \overline{v^{*} (L)} - u^{*} (L) \overline{v^{' *} (L)} - u^{' *} (0) \overline{v^{*} (0)} + u^{*} (0) \overline{v^{' *} (0)}$ .

(d) Half-line and line. If $u, v \in H^{1} (0, \infty)$ , then $\int_{0}^{\infty} u \overline{v^{'}} = - u^{*} (0) \overline{v^{*} (0)} - \int_{0}^{\infty} u^{'} \overline{v}$ , both integrals being absolutely convergent. For $u, v \in H^{1} (R)$ one instead has $\int_{R} u \overline{v^{'}} = - \int_{R} u^{'} \overline{v}$ , with both integrals absolutely convergent.

Facts & Assumptions

Given: the $H^{1}$ , $H^{2}$ and $H_{0}^{1}$ spaces of One-dimensional Sobolev spaces on intervals; the representative theorem Absolutely continuous representative of an L2 function with L2 weak derivative; the trace and decay estimates Endpoint trace bound and decay of H1 functions.

[F1]

For $F, G \in A C [a, b]$ , $\int_{a}^{b} F G^{'} + \int_{a}^{b} F^{'} G = F (b) G (b) - F (a) G (a)$ (Integration by parts for absolutely continuous functions); products of absolutely continuous functions are absolutely continuous (The product of two absolutely continuous functions is absolutely continuous). Applying the real identity to the real and imaginary parts of complex-valued absolutely continuous $F, G$ with $L^{1}$ derivatives gives the complex form $\int_{a}^{b} F \overline{G^{'}} + \int_{a}^{b} F^{'} \overline{G} = F (b) \overline{G (b)} - F (a) \overline{G (a)}$ .

[F2]

An $H^{1}$ class on a bounded interval has an absolutely continuous representative $u^{*}$ on the closed interval with $u^{*} (x) - u^{*} (y) = \int_{y}^{x} u^{'}$ and $∣ u^{*} (x) - u^{*} (y) ∣ \leq \sqrt{∣ x - y ∣} ∥ u^{'} ∥_{L^{2}}$ (Absolutely continuous representative of an L2 function with L2 weak derivative); on $(0, L)$ the trace bound $∣ u^{*} (x) ∣ \leq L^{- 1 / 2} ∥ u ∥_{L^{2} (0, L)} + L^{1 / 2} ∥ u^{'} ∥_{L^{2} (0, L)}$ holds for all $x \in [0, L]$ , and on the half-line $∣ u^{*} (x) ∣ \leq ∥ u ∥_{L^{2} (x, x + 1)} + ∥ u^{'} ∥_{L^{2} (x, x + 1)}$ with $u^{*} (x) \to 0$ as $x \to \infty$ (Endpoint trace bound and decay of H1 functions).

[F3]

The standard smooth step $σ$ of The standard smooth step function is smooth with $σ = 0$ on $(- \infty, 0]$ , $σ = 1$ on $[1, \infty)$ and $0 \leq σ \leq 1$ ; its derivative is bounded with $∥ σ^{'} ∥_{\infty} \leq C_{σ}$ .

[F4]

Fix a real $ρ \in C_{c}^{\infty} (R)$ with $\int ρ = 1$ , for example $ρ = σ^{'}$ from [F3], whose integral is $σ (1) - σ (0) = 1$ by [F1]. Set $ρ_{δ} (t) = δ^{- 1} ρ (t / δ)$ . By Complex translation, convolution, approximate identities, and mollification, $∥ ρ_{δ} * g - g ∥_{L^{2} (R)} \to 0$ for $g \in L^{2} (R)$ , and convolution maps compactly supported locally integrable inputs to smooth compactly supported functions with $(ρ_{δ} * g)^{'} = ρ_{δ}^{'} * g$ . Step 1.2 proves the needed derivative commutation for its particular zero extension by integration by parts; classical derivatives of smooth functions are weak derivatives (Classical derivatives agree with weak derivatives).

[F5]

Dominated convergence (Dominated convergence): if $∣ g_{ε} ∣ \leq h$ with $h$ integrable and $g_{ε} \to g$ pointwise almost everywhere, then $\int g_{ε} \to \int g$ .

Proof

1.1F1F2

[Bounded integration by parts] Let $u, v \in H^{1} (0, L)$ . By [F2], the products $u^{*} \overline{v^{*}}$ and the derivative classes $u^{'} \overline{v}$ and $u \overline{v^{'}}$ are integrable; by [F1] in its complex form applied to $F = u^{*}$ and $G = v^{*}$ (whose real and imaginary parts are absolutely continuous with $L^{1}$ derivatives), $\int_{0}^{L} (u^{'} \overline{v} + u \overline{v^{'}}) = u^{*} (L) \overline{v^{*} (L)} - u^{*} (0) \overline{v^{*} (0)},$ which is exactly (a).

1.2F1F3F4givenconstruct

[Zero traces lie in $H_{0}^{1}$ ] Let $u \in H^{1} (0, L)$ satisfy $u^{*} (0) = u^{*} (L) = 0$ . For $0 < ε < L / 4$ put $χ_{ε} (x) : = σ (2 x / ε - 1) σ (2 (L - x) / ε - 1)$ with $σ$ as in [F3]. Then $χ_{ε}$ is smooth, $0 \leq χ_{ε} \leq 1$ , equals one on $[ε, L - ε]$ , vanishes on $[0, ε / 2] \cup [L - ε / 2, L]$ , and has $∣ χ_{ε}^{'} ∣ \leq 4 C_{σ} / ε$ , supported in $S_{ε} : = [0, ε] \cup [L - ε, L]$ . The AC product rule [F1] gives $u_{ε} : = χ_{ε} u \in H^{1} (0, L)$ with weak derivative $χ_{ε}^{'} u + χ_{ε} u^{'}$ ; its support lies in $[ε / 2, L - ε / 2]$ . Extending by zero to $R$ preserves this weak derivative because the product vanishes near both endpoints. Choose a finite $R \geq 1$ with $supp ρ \subseteq [- R, R]$ . For $0 < δ < ε / (4 R)$ the convolution $u_{ε, δ} : = ρ_{δ} * u_{ε}$ is smooth by [F4] and its support lies in $[ε / 2 - R δ, L - ε / 2 + R δ] \subset (0, L)$ , because the integrand vanishes outside the sum of the two supports. The zero extension of $u_{ε}^{*}$ is absolutely continuous on every bounded interval, with compact support and derivative $u_{ε}^{'}$ . For fixed $x$ , apply [F1] on an interval containing that support to $u_{ε}^{*} (t)$ and $ρ_{δ} (x - t)$ ; the boundary term vanishes and $\frac{d}{d t} ρ_{δ} (x - t) = - ρ_{δ}^{'} (x - t)$ , so $(ρ_{δ} * u_{ε})^{'} = ρ_{δ}^{'} * u_{ε} = ρ_{δ} * u_{ε}^{'}$ . Applying [F4] separately to the function and derivative proves $u_{ε, δ} \to u_{ε}$ in $H^{1}$ as $δ \to 0$ , with every such approximant in $C_{c}^{\infty} (0, L)$ .

2.1step 1.2F2F5algebra

[The cutoff converges to $u$ ] By [F5], $∥ (χ_{ε} - 1) u ∥_{2} \to 0$ and $∥ (χ_{ε} - 1) u^{'} ∥_{2} \to 0$ . The zero traces and the integral representative in [F2] give, for $0 \leq x \leq ε$ , $∣ u^{*} (x) ∣^{2} \leq x \int_{0}^{ε} ∣ u^{'} ∣^{2}$ , and for $L - ε \leq x \leq L$ , $∣ u^{*} (x) ∣^{2} \leq (L - x) \int_{L - ε}^{L} ∣ u^{'} ∣^{2}$ . Thus the derivative term of step 1.2 satisfies $∥ χ_{ε}^{'} u ∥_{2}^{2} \leq (4 C_{σ} / ε)^{2} (ε^{2} / 2) \int_{S_{ε}} ∣ u^{'} ∣^{2} = 8 C_{σ}^{2} \int_{S_{ε}} ∣ u^{'} ∣^{2} \to 0$ by [F5], since $∣ u^{'} ∣^{2}$ is integrable and the strip indicators tend to zero almost everywhere. Hence $u_{ε} \to u$ in $H^{1}$ ; each $u_{ε}$ lies in the $H^{1}$ closure of $C_{c}^{\infty} (0, L)$ by step 1.2, and that closure is closed, so $u \in H_{0}^{1} (0, L)$ .

2.2step 1.1F1F2algebra

[Second-order identity] Let $u, v \in H^{2} (0, L)$ . Applying (a) to the pair $(u^{'}, v)$ and to the pair $(u, v^{'})$ , both in $H^{1} (0, L)$ , gives $\int_{0}^{L} u^{''} \overline{v} = [u^{' *} \overline{v^{*}}]_{0}^{L} - \int_{0}^{L} u^{'} \overline{v^{'}}, \int_{0}^{L} u \overline{v^{''}} = [u^{*} \overline{v^{' *}}]_{0}^{L} - \int_{0}^{L} u^{'} \overline{v^{'}} .$ Subtracting the second identity from the first gives (c).

3.1step 2.1F2

[Traces of $H_{0}^{1}$ classes vanish] Conversely let $u \in H_{0}^{1} (0, L)$ ; by definition there are $u_{n} \in C_{c}^{\infty} (0, L)$ with $u_{n} \to u$ in $H^{1}$ . For each $x \in [0, L]$ the trace bound of [F2] gives $∣ u_{n}^{*} (x) - u^{*} (x) ∣ \leq L^{- 1 / 2} ∥ u_{n} - u ∥_{L^{2}} + L^{1 / 2} ∥ u_{n}^{'} - u^{'} ∥_{L^{2}},$ so $u_{n}^{*} \to u^{*}$ uniformly on $[0, L]$ . Since $u_{n}^{*} (0) = 0 = u_{n}^{*} (L)$ , it follows that $u^{*} (0) = 0 = u^{*} (L)$ . Together with step 2.1 this proves (b).

4.1step 1.1F2F5algebra∎

[Half-line and line] Let $u, v \in H^{1} (0, \infty)$ . Step 1.1 on $(0, R)$ gives $\int_{0}^{R} u \overline{v^{'}} = u^{*} (R) \overline{v^{*} (R)} - u^{*} (0) \overline{v^{*} (0)} - \int_{0}^{R} u^{'} \overline{v}$ . The boundary term at $R$ tends to zero by [F2], while both integrals converge absolutely by Cauchy–Schwarz (or $∣ u \overline{v^{'}} ∣ \leq (∣ u ∣^{2} + ∣ v^{'} ∣^{2}) / 2$ ), so $\int_{0}^{\infty} u \overline{v^{'}} = - u^{*} (0) \overline{v^{*} (0)} - \int_{0}^{\infty} u^{'} \overline{v}$ . For $u, v \in H^{1} (R)$ , apply this identity to their positive-half-line restrictions and to the reflected negative-half-line restrictions $u (- x), v (- x)$ . Reflection reverses the derivative sign, so the negative-half-line identity has the opposite origin term $+ u^{*} (0) \overline{v^{*} (0)}$ . The representatives on the two sides have the same trace at zero by [F2]; adding cancels the origin terms and gives the claimed whole-line identity.

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Completeness of the Dirichlet sine system on an interval

Statement

Assume the Axiom of Choice and Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $0 < L < \infty$ and let $L^{2} (0, L; C)$ be as in One-dimensional Sobolev spaces on intervals. Define, for $n \geq 1$ , $e_{n} (x) : = \sqrt{\frac{2}{L}} \sin ⁣ (\frac{n π x}{L}), x \in [0, L] .$ Then $(e_{n})_{n \geq 1}$ is an orthonormal family and a complete orthonormal family (Hilbert basis) of $L^{2} (0, L; C)$ (Orthonormal families, complete orthonormal systems and Hilbert bases): $⟨ e_{n}, e_{m} ⟩ = δ_{n m}, \overset{}{span} {e_{n} : n \geq 1} = L^{2} (0, L; C) .$

Facts & Assumptions

Given: $0 < L < \infty$ ; the functions $e_{n}$ ; the $2 L$ -periodic characters $ξ_{n} (x) : = e^{i π n x / L}$ for $n \in Z$ .

[F1]

The characters $e_{k}$ , $k \in Z$ , of Fourier coefficients and trigonometric polynomials on the torus form an orthonormal basis of $L^{2} (T; C)$ for the one-dimensional torus $T = R / Z$ with its normalized Haar measure (The one-dimensional torus and its normalized Haar integral, The trigonometric characters are orthonormal in $L^{2}$ of the torus, The trigonometric system is complete in $L^{2}$ of the torus).

[F2]

For an orthonormal family $(e_{i})_{i \in I}$ in a Hilbert space $H$ , the family is complete if and only if the only vector orthogonal to every $e_{i}$ is $0$ ; equivalently the closed span is $H$ and $∥ x ∥^{2} = \sum_{i} ∣ ⟨ x, e_{i} ⟩ ∣^{2}$ for every $x$ (Parseval equivalences for an orthonormal family, Orthonormal families, complete orthonormal systems and Hilbert bases).

[F3]

Sine and cosine addition formulas give $2 \sin a \sin b = \cos (a - b) - \cos (a + b)$ ; the derivatives are $\sin^{'} = \cos$ , $\cos^{'} = - \sin$ ; the second fundamental theorem gives $\int_{c}^{d} \cos (λ t) d t = λ^{- 1} (\sin (λ d) - \sin (λ c))$ for $λ \neq 0$ ; and $\sin (k π) = 0$ , $\cos (k π) = (- 1)^{k}$ for $k \in Z$ (The addition formulas for sine and cosine, The derivatives of sine and cosine are cosine and minus sine, The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , Quarter-turn values and shifts by pi/2 and pi).

[F4]

For $ϕ (t) = 2 L t - L$ , affine scaling and translation invariance of Lebesgue measure give $λ_{1} (ϕ^{- 1} (E)) = (2 L)^{- 1} λ_{1} (E)$ for each Lebesgue-measurable $E \subseteq R$ (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation). The resulting integral identity $\int_{0}^{1} H (2 L t - L) d t = (2 L)^{- 1} \int_{- L}^{L} H (x) d x$ holds first for indicators and simple functions, then for nonnegative measurable $H$ by increasing simple approximation and monotone convergence, and then for integrable complex $H$ by decomposing real and imaginary parts (Monotone convergence for the integral).

[F5]

Lebesgue measure on $R$ is the completion of its Borel restriction, and every real Lebesgue-measurable function has a Borel representative equal to it almost everywhere under Countable Choice ( $L (R^{n})$ is exactly the completion of the restriction of $λ_{n}$ to the Borel sets, A function measurable for a completion is almost everywhere equal to one measurable for the original sigma-algebra). Apply this to both components to choose a Borel representative of a complex class.

[F6]

$L^{2} (0, L; C)$ is a complex Hilbert space with $⟨ f, g ⟩ = \int_{0}^{L} f \overline{g}$ ( $L^{2}$ with the integral pairing is a Hilbert space, One-dimensional Sobolev spaces on intervals); a complex class in $L^{2}$ has real and imaginary parts in $L^{2}$ and $⟨ f, e_{n} ⟩ = ⟨ Re f, e_{n} ⟩ + i ⟨ Im f, e_{n} ⟩$ when $e_{n}$ is real-valued.

Proof

1.1F3F6algebra

[Orthonormality] For $n, m \geq 1$ , [F3] gives $2 \sin (n π x / L) \sin (m π x / L) = \cos ((n - m) π x / L) - \cos ((n + m) π x / L)$ . If $n \neq m$ , then $n - m \neq 0$ and $n + m \neq 0$ , and by [F3] $\int_{0}^{L} \cos ⁣ (\frac{k π x}{L}) d x = \frac{L}{k π} (\sin (k π) - \sin 0) = 0 (k \neq 0),$ so $⟨ e_{n}, e_{m} ⟩ = 0$ . If $n = m$ , the first cosine is $1$ and $\int_{0}^{L} (1 - \cos (2 n π x / L)) d x = L$ , so $⟨ e_{n}, e_{n} ⟩ = (2 / L) \cdot (L / 2) = 1$ . Hence $(e_{n})$ is orthonormal.

1.2F6algebra

[Reduction to real-valued $f$ ] Let $f \in L^{2} (0, L; C)$ satisfy $⟨ f, e_{n} ⟩ = 0$ for all $n \geq 1$ . By [F6] write $f = f_{1} + i f_{2}$ with real-valued $f_{1}, f_{2} \in L^{2}$ ; since each $e_{n}$ is real-valued, $0 = ⟨ f, e_{n} ⟩ = ⟨ f_{1}, e_{n} ⟩ + i ⟨ f_{2}, e_{n} ⟩$ , so both real parts have vanishing sine coefficients. It suffices to prove $f_{1} = f_{2} = 0$ ; hence assume from now on that $f$ is real-valued with $\int_{0}^{L} f (x) \sin (n π x / L) d x = 0$ for all $n \geq 1$ .

2.1step 1.2F3algebra

[Odd extension] Define $g : (- L, L) \to R$ by $g (x) = f (x)$ for $x \in (0, L)$ , $g (x) = - f (- x)$ for $x \in (- L, 0)$ , and $g (0) = 0$ . Then $g$ is measurable, odd, and $\int_{- L}^{L} ∣ g ∣^{2} = 2 \int_{0}^{L} ∣ f ∣^{2} < \infty$ , so $g \in L^{2} (- L, L)$ with exactly the same class as the odd extension. For every $n \in Z$ , $\int_{- L}^{L} g (x) e^{- i π n x / L} d x = 0.$ Indeed, writing $e^{- i π n x / L} = \cos (n π x / L) - i \sin (n π x / L)$ , the product $g \cos (n π x / L)$ is odd, so its integral vanishes, while $g \sin (n π x / L)$ is even, so its integral equals $2 \int_{0}^{L} f (x) \sin (n π x / L) d x$ , which is $0$ for $n \geq 1$ by hypothesis and also for $n \leq - 1$ by oddness of sine; for $n = 0$ the sine factor vanishes.

3.1step 2.1F1F3F4algebra

[Transport to the torus] By [F5], choose a Borel representative of the real-valued class $g$ and assign any values at the endpoints $\pm L$ , which does not change its $L^{2}$ class. Define $G : T \to C$ using the unique representative $t \in [0, 1)$ of each torus point by $G ([t]) : = g (2 L t - L)$ . The representative map $[t] \mapsto t \in [0, 1)$ is Borel: the quotient map restricts to a homeomorphism from $(0, 1)$ onto $T ∖ {[0]}$ , and the remaining singleton is Borel; hence $G$ is measurable. The affine integral identity [F4] gives [step 2.1, F1, F4, F5, algebra] $\int_{T} ∣ G ∣^{2} d m_{T} = \frac{1}{2 L} \int_{- L}^{L} ∣ g (x) ∣^{2} d x < \infty .$ For $k \in Z$ , the substitution $x = 2 L t - L$ gives $e^{- 2 π i k t} = e^{- i π k} e^{- i π k x / L} = (- 1)^{k} e^{- i π k x / L}$ ; hence $\hat{G} (k) = \frac{(- 1)^{k}}{2 L} \int_{- L}^{L} g (x) e^{- i π k x / L} d x = 0$ by step 2.1. The harmless phase factor is retained explicitly.

4.1step 1.1step 3.1F1F2∎

[Conclusion] By [F1] the characters form an orthonormal basis of $L^{2} (T)$ , so by [F2] a vector with all Fourier coefficients zero is zero: $G = 0$ in $L^{2} (T; C)$ . Since $φ (t) = 2 L t - L$ transports the fundamental interval onto $[- L, L)$ and the norm identity in step 3.1 is exact, $g = 0$ almost everywhere on $(- L, L)$ , hence $f = 0$ almost everywhere on $(0, L)$ . Therefore no nonzero class is orthogonal to all $e_{n}$ , and [F2] makes $(e_{n})_{n \geq 1}$ complete. With step 1.1 this proves the statement.

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The free-line resolvent kernel and its inverse operator

Statement

Assume the Axioms of Choice and Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $z \in C ∖ [0, \infty)$ and let $k \in C$ be the unique complex number with $k^{2} = - z$ and $Re k > 0$ . Put $G_{z} (t) : = \frac{1}{2 k} e^{- k ∣ t ∣} (t \in R),$ and let $A_{0}$ be the free one-dimensional operator of Real multiplication operators and free plus bounded-potential Schrödinger operators clause (b) with $d = 1$ , $c = 4 π^{2}$ and $b = 0$ : $D (A_{0}) = H^{2} (R)$ and $A_{0} u = - D^{2} u$ for every $u \in H^{2} (R)$ , where $D^{2}$ is the second weak derivative of One-dimensional Sobolev spaces on intervals. Write $σ : = Re k > 0$ . Then:

(a) $G_{z} \in H^{1} (R) \cap L^{1} (R) \cap L^{2} (R)$ , its weak derivative is the class of $t \mapsto - \frac{1}{2} sgn (t) e^{- k ∣ t ∣}$ , and $∥ G_{z} ∥_{L^{1}} = \frac{1}{∣ k ∣ σ}, ∥ G_{z} ∥_{L^{2}} = \frac{1}{2 ∣ k ∣ \sqrt{σ}}, ∥ D G_{z} ∥_{L^{1}} = \frac{1}{σ};$

(b) for every $f \in L^{2} (R)$ the convolution $u : = G_{z} * f$ , $u (x) = \int_{R} G_{z} (x - y) f (y) d y$ , lies in $H^{2} (R)$ with $∥ u ∥_{L^{2}} \leq ∥ G_{z} ∥_{L^{1}} ∥ f ∥_{L^{2}}, ∥ u ∥_{L^{\infty}} \leq \frac{1}{2 ∣ k ∣ \sqrt{σ}} ∥ f ∥_{L^{2}},$ and $A_{0} u - z u = f$ ;

(c) the linear map $R_{0} (z) : L^{2} (R) \to L^{2} (R)$ , $f \mapsto G_{z} * f$ is bounded with range exactly $H^{2} (R)$ , and $(A_{0} - z) R_{0} (z) = I_{L^{2}}, R_{0} (z) (A_{0} - z) = I_{H^{2}};$ in particular $R_{0} (z) = (A_{0} - z)^{- 1}$ and $A_{0} - z$ is a bijection from $H^{2} (R)$ onto $L^{2} (R)$ ;

(d) $(- \frac{d^{2}}{d x^{2}} - z) G_{z} = δ_{0}$ in the sense of distributions: for every $φ \in C_{c}^{\infty} (R)$ , $\int_{R} G_{z} (x) (- φ^{''} (x) - z φ (x)) d x = φ (0) .$

Facts & Assumptions

Given: $z \notin [0, \infty)$ , the root $k$ with $k^{2} = - z$ , $Re k = σ > 0$ , and the kernel $G_{z}$ .

[F1]

$A_{0}$ of Real multiplication operators and free plus bounded-potential Schrödinger operators clause (b) with $(d, c, b) = (1, 4 π^{2}, 0)$ is self-adjoint with domain $D (A_{0}) = H^{2} (R) = W^{2, 2} (R; C)$ of One-dimensional Sobolev spaces on intervals, and $A_{0} f = - f^{''}$ for Schwartz $f$ . For $u \in H^{2} (R)$ one has $A_{0} u = - D^{2} u$ : for every Schwartz $φ$ , $⟨ A_{0} u, φ ⟩ = ⟨ u, A_{0} φ ⟩ = - ⟨ u, φ^{''} ⟩ = ⟨ - D^{2} u, φ ⟩$ by Integration by parts and the endpoint-vanishing space on intervals clause (d) applied twice, and Schwartz functions are dense in $L^{2}$ .

[F2]

For a class $g \in H^{1} (R)$ , the translate $g_{y} : = g (\cdot - y)$ has weak derivative $D g (\cdot - y)$ : the defining test identity is invariant under the change of variables $x \mapsto x + y$ (One-dimensional Sobolev spaces on intervals).

[F3]

Complex convolution with $K \in L^{1} (R)$ satisfies $∥ K * f ∥_{L^{2}} \leq ∥ K ∥_{L^{1}} ∥ f ∥_{L^{2}}$ , and $K * f$ is defined a.e.; the integral $\int_{R} ∣ K (x - y) ∣ ∣ f (y) ∣ d y$ is finite for almost every $x$ (Complex translation, convolution, approximate identities, and mollification). Also $∣ u (x) ∣ \leq ∥ G_{z} ∥_{L^{2}} ∥ f ∥_{L^{2}}$ by Cauchy-Schwarz (Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors).

[F4]

The complex exponential is differentiable with $\exp^{'} (z) = \exp z$ at every $z \in C$ , by termwise differentiation of its infinite-radius power series (Inside its disc of convergence a complex power series is holomorphic and may be differentiated term by term, The complex exponential by its power series). For fixed $k \neq 0$ and real $t$ , taking the increment $- k h$ in the defining complex difference quotient gives $\frac{d}{d t} e^{- k t} = - k e^{- k t}$ ; the same quotient argument applies to $e^{k t}$ .

[F5]

For real $x, y$ , $∣ \exp (x + i y) ∣ = e^{x}$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ); hence $∣ e^{- k ∣ t ∣} ∣ = e^{- σ ∣ t ∣}$ when $σ = Re k$ .

[F6]

For $σ > 0$ and $λ \in C$ with $Re λ > 0$ , [F4] and the fundamental theorem of calculus give $\int_{0}^{R} e^{- λ t} d t = (1 - e^{- λ R}) / λ$ for finite $R$ ; letting $R \to \infty$ gives $\int_{0}^{\infty} e^{- λ t} d t = λ^{- 1}$ because [F5] gives $∣ e^{- λ R} ∣ = e^{- R Re λ} \to 0$ . Taking $λ = 2 σ$ gives $\int_{0}^{\infty} e^{- 2 σ t} d t = (2 σ)^{- 1}$ .

Proof

1.1F4F5F6givenalgebra

[Elementary properties of $G_{z}$ ] Since $z \notin [0, \infty)$ , $k \neq 0$ . By [F5], $t \mapsto e^{- k ∣ t ∣}$ is differentiable on each half-line, continuous at $0$ , and locally absolutely continuous, with a.e. derivative $- k sgn (t) e^{- k ∣ t ∣}$ . This derivative is in $L^{1} \cap L^{\infty}$ by [F6], so $G_{z} \in H^{1} (R)$ and $D G_{z} (t) = - \frac{1}{2} sgn (t) e^{- k ∣ t ∣}$ ; also $D G_{z} \in L^{1} \cap L^{\infty}$ . Using [F4] and [F6], $∥ G_{z} ∥_{L^{1}} = \frac{1}{2 ∣ k ∣} \int_{R} e^{- σ ∣ t ∣} d t = \frac{1}{∣ k ∣ σ}, ∥ G_{z} ∥_{L^{2}}^{2} = \frac{1}{4 ∣ k ∣^{2}} \int_{R} e^{- 2 σ ∣ t ∣} d t = \frac{1}{4 ∣ k ∣^{2} σ},$ and $∥ D G_{z} ∥_{L^{1}} = \frac{1}{2} \int_{R} e^{- σ ∣ t ∣} d t = σ^{- 1}$ . This is (a).

2.1step 1.1F2F3givenalgebra

[ $G_{z} * f$ solves the resolvent equation] Let $f \in L^{2} (R)$ and $u : = G_{z} * f$ . By [F3], $u \in L^{2}$ with $∥ u ∥_{L^{2}} \leq ∥ G_{z} ∥_{L^{1}} ∥ f ∥_{L^{2}}$ and $∥ u ∥_{\infty} \leq ∥ G_{z} ∥_{L^{2}} ∥ f ∥_{L^{2}}$ . We show that $u$ has weak first derivative $w : = D G_{z} * f$ : for a test function $φ$ , Fubini for the absolutely integrable integrand $∣ G_{z} (x - y) f (y) φ^{'} (x) ∣$ (its integral is at most $∥ G_{z} ∥_{L^{1}} ∥ f ∥_{L^{2}} ∥ φ^{'} ∥_{L^{2}}$ ) gives $\int_{R} u (x) φ^{'} (x) d x = \int_{R} f (y) (\int_{R} G_{z} (x - y) φ^{'} (x) d x) d y .$ For fixed $y$ , the inner integral equals $- \int_{R} D G_{z} (x - y) φ (x) d x$ by [F2] and the integration by parts identity Integration by parts and the endpoint-vanishing space on intervals clause (d) applied to the $H^{1}$ classes $G_{z} (\cdot - y)$ and $φ$ . Substituting and applying Fubini again gives $\int u φ^{'} = - \int w φ$ for every test $φ$ , so $w = D u$ is the weak first derivative of $u$ .

2.2step 1.1F1F4F5givenalgebra

[Injectivity on the free domain] Suppose $v \in H^{2} (R)$ satisfies $A_{0} v = z v$ , i.e. $- D^{2} v = z v$ by [F1]. Since $v$ is continuous with $D^{2} v = - z v$ a continuous class, $v$ is a classical $C^{2}$ solution of $v^{''} = k^{2} v$ . Put $g (x) : = v (x) e^{- k x}$ ; then $g \in C^{2}$ and $g^{''} + 2 k g^{'} = (v^{''} - 2 k v^{'} + k^{2} v) e^{- k x} + 2 k (v^{'} - k v) e^{- k x} = 0$ , so $(g^{'} e^{2 k x})^{'} = 0$ and hence $g^{'} = C e^{- 2 k x}$ , $g = C_{1} + C_{2} e^{- 2 k x}$ ; therefore $v = C_{1} e^{k x} + C_{2} e^{- k x}$ . By [F5], $∣ e^{k x} ∣ = e^{σ x} \to \infty$ as $x \to + \infty$ and $∣ e^{- k x} ∣ = e^{- σ x} \to \infty$ as $x \to - \infty$ ; since $v \in L^{2}$ , necessarily $C_{1} = C_{2} = 0$ , so $v = 0$ .

3.1step 2.1F1givenalgebra

[Second weak derivative] We show that the weak derivative of $w$ is the class $k^{2} u - f$ . Let $φ \in C_{c}^{\infty} (R)$ . As in step 2.1, Fubini reduces $\int w φ^{'}$ to $\int f (y) I (y) d y$ with $I (y) : = \int_{R} D G_{z} (x - y) φ^{'} (x) d x$ . Split $I (y) = \int_{- \infty}^{y} + \int_{y}^{\infty}$ ; on each half-line the integrand $x \mapsto D G_{z} (x - y)$ is $C^{1}$ with derivative $k^{2} G_{z} (x - y)$ (direct differentiation of $- \frac{1}{2} sgn (t) e^{- k ∣ t ∣}$ away from $t = 0$ ), and $D G_{z} (y^{+}) = - \frac{1}{2}$ , $D G_{z} (y^{-}) = \frac{1}{2}$ . Integration by parts on each half-line (the boundary terms at $\pm \infty$ vanish because $φ$ has compact support) gives $I (y) = (D G_{z} (y^{-}) - D G_{z} (y^{+})) φ (y) - \int_{R} k^{2} G_{z} (x - y) φ (x) d x = φ (y) - k^{2} \int_{R} G_{z} (x - y) φ (x) d x .$ Hence $\int w φ^{'} = \int f φ - k^{2} \int u φ = - \int (k^{2} u - f) φ$ for every test $φ$ , so the weak derivative of $w$ is the $L^{2}$ class $k^{2} u - f$ . Thus $u \in H^{2} (R)$ with $D^{2} u = k^{2} u - f$ , and since $k^{2} = - z$ , $- D^{2} u - z u = - (k^{2} u - f) - z u = f - (k^{2} + z) u = f .$ By [F1], $A_{0} u = - D^{2} u$ , so $A_{0} u - z u = f$ . This proves (b).

4.1step 3.1step 2.2F1given

[The two-sided resolvent identity] Step 3.1 shows $(A_{0} - z) R_{0} (z) f = f$ for every $f \in L^{2} (R)$ , so $(A_{0} - z) R_{0} (z) = I_{L^{2}}$ and the range of $R_{0} (z)$ is contained in $D (A_{0}) = H^{2} (R)$ . Conversely let $u \in H^{2} (R)$ and put $f : = A_{0} u - z u \in L^{2}$ . Step 3.1 applied to $f$ gives $(A_{0} - z) R_{0} (z) f = f = (A_{0} - z) u$ , and step 2.2 shows that $A_{0} - z$ is injective on $H^{2} (R)$ ; hence $R_{0} (z) f = u$ , so $R_{0} (z) (A_{0} - z) = I_{H^{2}}$ and the range of $R_{0} (z)$ is exactly $H^{2} (R)$ . This is (c).

5.1step 3.1givenalgebra∎

[Distributional identity] Let $φ \in C_{c}^{\infty} (R)$ . Integrating by parts once ( $G_{z}$ is continuous and $φ^{'}$ has compact support, so the boundary terms vanish at both ends) and using the value of $I (0)$ computed in step 3.1, $\int_{R} G_{z} φ^{''} = - \int_{R} D G_{z} φ^{'} = - φ (0) + k^{2} \int_{R} G_{z} φ = - φ (0) - z \int_{R} G_{z} φ,$ because $D G_{z} (0^{-}) - D G_{z} (0^{+}) = \frac{1}{2} - (- \frac{1}{2}) = 1$ and $k^{2} = - z$ . Rearranging gives $\int_{R} G_{z} (- φ^{''} - z φ) = φ (0)$ , which is (d).

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Identification of the Dirichlet Laplacian with the sine diagonal operator

Statement

Assume the Axiom of Choice and Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $0 < L < \infty$ , let $e_{n} (x) = \sqrt{\frac{2}{L}} \sin ⁣ (\frac{n π x}{L}), λ_{n} : = (\frac{n π}{L})^{2} (n \geq 1),$ let $(e_{n})_{n \geq 1}$ be the complete orthonormal family of Completeness of the Dirichlet sine system on an interval, and let $A_{D}$ be the diagonal operator of Diagonal adjoint and graph-core lemma associated with $(e_{n})$ and $(λ_{n})$ : $A_{D} x = \sum_{n \geq 1} λ_{n} ⟨ x, e_{n} ⟩ e_{n}$ with domain $D (A_{D}) = {x \in L^{2} (0, L; C) : \sum_{n \geq 1} λ_{n}^{2} ∣ ⟨ x, e_{n} ⟩ ∣^{2} < \infty} .$ Let $B$ be the Dirichlet Laplacian $B u : = - D^{2} u, D (B) = {u \in H^{2} (0, L) : u^{*} (0) = u^{*} (L) = 0},$ with $H^{2}$ and the trace $u^{*}$ of One-dimensional Sobolev spaces on intervals and Integration by parts and the endpoint-vanishing space on intervals clause (b). Then:

(a) $D (A_{D}) = D (B)$ and $A_{D} = B$ ; in particular $D (A_{D})$ is a linear subspace of $H^{2} (0, L)$ whose members have vanishing endpoint traces;

(b) $A_{D}$ is self-adjoint (Symmetric, self-adjoint and essentially self-adjoint operators), positive, and its spectrum is $σ (A_{D}) = {λ_{n} : n \geq 1}$ ; each $λ_{n}$ is an eigenvalue with eigenvector $e_{n}$ ; the eigenvalues are simple in the sense that $\ker (A_{D} - λ_{n})$ contains no nonzero vector orthogonal to $e_{n}$ ;

(c) for every $u \in D (A_{D})$ , $⟨ A_{D} u, u ⟩ = ∥ D u ∥_{L^{2} (0, L)}^{2}$ ; consequently $A_{D} \geq λ_{1} = (π / L)^{2}$ in the quadratic-form sense on its domain.

Facts & Assumptions

Given: $L > 0$ ; the complete orthonormal sine family $(e_{n})$ and numbers $λ_{n} = (n π / L)^{2}$ of Completeness of the Dirichlet sine system on an interval; the diagonal operator $A_{D}$ of Diagonal adjoint and graph-core lemma.

[F1]

$(e_{n})_{n \geq 1}$ is an orthonormal family with $⟨ x, e_{n} ⟩$ the Fourier coefficients and $x = \sum_{n} ⟨ x, e_{n} ⟩ e_{n}$ in $L^{2}$ for every $x \in L^{2} (0, L)$ (Completeness of the Dirichlet sine system on an interval, Parseval equivalences for an orthonormal family, Orthonormal families, complete orthonormal systems and Hilbert bases). Also $e_{n} \in H^{2} (0, L)$ , $e_{n} (0) = e_{n} (L) = 0$ , and the weak derivatives satisfy $D e_{n} (x) = \frac{n π}{L} \sqrt{\frac{2}{L}} \cos (\frac{n π x}{L})$ , $D^{2} e_{n} = - λ_{n} e_{n}$ , with $⟨ D e_{n}, D e_{m} ⟩ = λ_{n} δ_{n m}$ (Integration by parts and the endpoint-vanishing space on intervals and the product-to-sum computation of Completeness of the Dirichlet sine system on an interval).

[F2]

The diagonal operator lemma Diagonal adjoint and graph-core lemma states: the finite-support diagonal operator is symmetric, $A_{D}$ is self-adjoint with the displayed domain, $A_{D} e_{n} = λ_{n} e_{n}$ , the finite spans are a graph core, and $⟨ A_{D} x, x ⟩ = \sum_{n} λ_{n} ∣ ⟨ x, e_{n} ⟩ ∣^{2}$ . The companion Diagonal spectrum identifies $σ (A_{D})$ with the closure of the sequence $(λ_{n})$ .

[F3]

For $u, v \in H^{1} (0, L)$ , $\int_{0}^{L} u \overline{v^{'}} = u^{*} (L) \overline{v^{*} (L)} - u^{*} (0) \overline{v^{*} (0)} - \int_{0}^{L} u^{'} \overline{v}$ (Integration by parts and the endpoint-vanishing space on intervals clause (a)); for $u, v \in H^{2} (0, L)$ the second-order identity holds; and $H_{0}^{1} (0, L) = {u \in H^{1} (0, L) : u^{*} (0) = u^{*} (L) = 0}$ (clause (b)).

[F4]

Bessel's inequality for an orthonormal family: for every $x \in L^{2}$ , $\sum_{n} ∣ ⟨ x, e_{n} ⟩ ∣^{2} \leq ∥ x ∥^{2}$ (Parseval equivalences for an orthonormal family).

Proof

1.1F1F2F4givenalgebra

[Series of derivatives] Let $u \in D (A_{D})$ and put $c_{n} : = ⟨ u, e_{n} ⟩$ , so that $\sum_{n} λ_{n}^{2} ∣ c_{n} ∣^{2} < \infty$ by [F2]. By [F4], $\sum_{n} ∣ c_{n} ∣^{2} \leq ∥ u ∥^{2}$ , so Cauchy-Schwarz gives $\sum_{n} λ_{n} ∣ c_{n} ∣^{2} \leq (\sum_{n} λ_{n}^{2} ∣ c_{n} ∣^{2})^{1 / 2} (\sum_{n} ∣ c_{n} ∣^{2})^{1 / 2} < \infty,$ and hence the series $w : = \sum_{n} c_{n} D e_{n}$ converges in $L^{2}$ with $∥ w ∥^{2} = \sum_{n} λ_{n} ∣ c_{n} ∣^{2}$ by [F1].

2.1step 1.1F1F2F3F4algebra

[ $u \in D (B)$ and $A_{D} u = - D^{2} u$ ] We first show $D u = w$ for $u \in D (A_{D})$ . Let $φ \in C_{c}^{\infty} (0, L)$ and put $u_{N} : = \sum_{n \leq N} c_{n} e_{n}$ . Since $u_{N} \to u$ in $L^{2}$ , $\int u_{N} \overline{φ^{'}} \to \int u \overline{φ^{'}}$ . For each $n$ , [F3] clause (a) applied to $e_{n}$ and $φ$ (both in $H^{1} (0, L)$ , and $φ$ compactly supported in the interior) gives $\int e_{n} \overline{φ^{'}} = e_{n}^{*} (L) \overline{φ (L)} - e_{n}^{*} (0) \overline{φ (0)} - \int e_{n}^{'} \overline{φ} = - \int e_{n}^{'} \overline{φ}$ , because $φ$ vanishes near the endpoints and $e_{n} (0) = e_{n} (L) = 0$ . Summing over $n \leq N$ and passing to the limit (justified by absolute convergence: $\sum_{n} ∣ c_{n} ∣ ∣ ⟨ e_{n}^{'}, φ ⟩ ∣ \leq (\sum_{n} λ_{n} ∣ c_{n} ∣^{2})^{1 / 2} (\sum_{n} λ_{n}^{- 1} ∣ ⟨ e_{n}, φ^{'} ⟩ ∣^{2})^{1 / 2} \leq (\sum_{n} λ_{n} ∣ c_{n} ∣^{2})^{1 / 2} λ_{1}^{- 1 / 2} ∥ φ^{'} ∥$ , using $[F 4]$ for $φ^{'}$ and $λ_{n} \geq λ_{1}$ ) gives $\int u \overline{φ^{'}} = - \int w \overline{φ}$ for every test $φ$ ; hence $D u = w \in L^{2}$ and $u \in H^{1} (0, L)$ .

3.1step 1.1step 2.1F1F2F3

[Endpoint traces and second derivative] With $u \in D (A_{D})$ , $D u = w$ , and the partial sums $u_{N}$ of step 2.1, the trace bound of Endpoint trace bound and decay of H1 functions clause (a) gives $∣ u_{N}^{*} (x) - u^{*} (x) ∣ \leq L^{- 1 / 2} ∥ u_{N} - u ∥ + L^{1 / 2} ∥ D u_{N} - D u ∥ (x \in [0, L]),$ and the right side tends to $0$ because $u_{N} \to u$ and $D u_{N} \to D u$ in $L^{2}$ by [F1] and step 1.1; hence $u_{N}^{*} \to u^{*}$ uniformly and, since $u_{N}^{*} (0) = u_{N}^{*} (L) = 0$ , also $u^{*} (0) = u^{*} (L) = 0$ . To identify the weak derivative of $w$ , put $w_{N} : = \sum_{n \leq N} c_{n} D e_{n}$ and $v_{N} : = \sum_{n \leq N} c_{n} D^{2} e_{n} = - \sum_{n \leq N} λ_{n} c_{n} e_{n}$ . By step 1.1, $w_{N} \to w$ in $L^{2}$ ; by [F2] and $\sum_{n} λ_{n}^{2} ∣ c_{n} ∣^{2} < \infty$ , $v_{N} \to - A_{D} u$ in $L^{2}$ . For each $φ \in C_{c}^{\infty} (0, L)$ , integration by parts [F3] applied to $D e_{n}$ and $φ$ gives $\int_{0}^{L} w_{N} \overline{φ^{'}} = - \int_{0}^{L} v_{N} \overline{φ}$ . Passing to the two $L^{2}$ limits proves $D w = - A_{D} u$ . Hence $D u = w \in H^{1} (0, L)$ , $D^{2} u = - A_{D} u$ , and $u \in D (B)$ with $B u = - D^{2} u = A_{D} u$ .

4.1step 3.1F1F3F4algebra

[Reverse inclusion] Let $u \in D (B)$ ; then $u \in H^{2} (0, L)$ and $D^{2} u \in L^{2}$ , and $u^{*} (0) = u^{*} (L) = 0$ . Put $c_{n} : = ⟨ u, e_{n} ⟩$ . For each $n$ , apply the second-order identity [F3] to $u$ and $e_{n}$ : all boundary terms vanish because $u$ and $e_{n}$ vanish at $0$ and $L$ , so $⟨ - D^{2} u, e_{n} ⟩ = \int_{0}^{L} (- D^{2} u) \overline{e_{n}} = - \int_{0}^{L} u \overline{D^{2} e_{n}} = λ_{n} ⟨ u, e_{n} ⟩ = λ_{n} c_{n},$ using $D^{2} e_{n} = - λ_{n} e_{n}$ from [F1]. By Bessel [F4] applied to $- D^{2} u \in L^{2}$ , $\sum_{n} λ_{n}^{2} ∣ c_{n} ∣^{2} = \sum_{n} ∣ ⟨ - D^{2} u, e_{n} ⟩ ∣^{2} \leq ∥ D^{2} u ∥^{2} < \infty,$ so $u \in D (A_{D})$ . Moreover $A_{D} u = \sum_{n} λ_{n} c_{n} e_{n} = \sum_{n} ⟨ - D^{2} u, e_{n} ⟩ e_{n} = - D^{2} u$ by the expansion [F1]. Hence $D (B) \subseteq D (A_{D})$ and $A_{D} u = B u$ .

5.1step 3.1step 4.1

[Conclusion of (a)] Steps 3.1 and 4.1 give $D (A_{D}) = D (B)$ and $A_{D} = B$ .

6.1step 5.1F2

[Self-adjointness, spectrum, positivity] By [F2], $A_{D}$ is self-adjoint with spectrum the closure of ${λ_{n}}$ ; since $λ_{n} = (n π / L)^{2} \to \infty$ and the map $n \mapsto λ_{n}$ is injective, the set ${λ_{n} : n \geq 1}$ has no finite accumulation point and is closed in $R$ , so $σ (A_{D}) = {λ_{n} : n \geq 1}$ , and [F2] gives $A_{D} e_{n} = λ_{n} e_{n}$ . If $x = \sum_{m} c_{m} e_{m} \in D (A_{D})$ lies in $\ker (A_{D} - λ_{n})$ , then $0 = (A_{D} - λ_{n}) x = \sum_{m} (λ_{m} - λ_{n}) c_{m} e_{m}$ , and uniqueness of coefficients in the orthonormal expansion [F1] gives $(λ_{m} - λ_{n}) c_{m} = 0$ for every $m$ , so $c_{m} = 0$ unless $m = n$ and $\ker (A_{D} - λ_{n}) = C e_{n}$ ; this is the simplicity assertion of (b). Finally, for $u \in D (A_{D})$ and $u \neq 0$ , [F2] gives $⟨ A_{D} u, u ⟩ = \sum_{n} λ_{n} ∣ c_{n} ∣^{2} \geq λ_{1} \sum_{n} ∣ c_{n} ∣^{2} = λ_{1} ∥ u ∥^{2} > 0$ , since $λ_{1} > 0$ ; so $A_{D}$ is positive and $A_{D} \geq λ_{1}$ in the quadratic-form sense.

7.1step 5.1F3algebra∎

[Form identity] Let $u \in D (A_{D}) = D (B)$ . Since $u^{*} (0) = u^{*} (L) = 0$ and $A_{D} u = - D^{2} u$ , the first-order identity [F3] applied to the pair $(D u, u)$ gives $⟨ A_{D} u, u ⟩ = \int_{0}^{L} (- D^{2} u) \overline{u} = \int_{0}^{L} ∣ D u ∣^{2} - [D u^{*} \overline{u^{*}}]_{0}^{L} = ∥ D u ∥_{L^{2}}^{2}$ , because the boundary terms contain the factor $u^{*} (0) = u^{*} (L) = 0$ . This is (c).

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Self-adjoint operators with compact resolvent: eigenbasis and discrete spectrum

Statement

Assume the Axiom of Choice and Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $A$ be a self-adjoint operator on an infinite-dimensional complex Hilbert space $H$ (Symmetric, self-adjoint and essentially self-adjoint operators, Hilbert space, Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis), let $c \in R$ lie in the resolvent set of $A$ , and assume that the resolvent $R_{c} : = (A - c)^{- 1} \in B (H)$ is compact (Resolvent and spectrum of an unbounded operator, Compact linear operator). Then:

(a) $H$ has a complete orthonormal family $(e_{n})_{n \in N}$ of eigenvectors of $A$ ; writing $μ_{n}$ for the eigenvalue of $e_{n}$ , one has $∣ μ_{n} ∣ \to \infty$ (each eigenvalue repeated according to its finite multiplicity), and every eigenspace of $A$ is finite dimensional;

(b) $σ (A)$ consists exactly of the eigenvalues $μ_{n}$ of $A$ and has no other points, and the essential spectrum is empty: $σ_{e s s} (A) = \emptyset$ (Discrete and essential spectrum of a self-adjoint operator); thus every spectral point of $A$ is an isolated eigenvalue of finite multiplicity;

Facts & Assumptions

Given: a self-adjoint operator $A$ on a complex Hilbert space $H$ ; $c \in ρ (A)$ with $R_{c} = (A - c)^{- 1}$ compact.

[F1]

$A$ has a spectral PVM; for $c \in ρ (A)$ the inverse $R_{c}$ is bounded with $ran R_{c} = D (A)$ , $(A - c) R_{c} = I_{H}$ and $R_{c} (A - c) = I_{D (A)}$ (Resolvent and spectrum of an unbounded operator, Discrete and essential spectrum of a self-adjoint operator); below $H \neq {0}$ .

[F2]

Compact self-adjoint operators: if $T \in B (H)$ is compact and self-adjoint, then its nonzero eigenvalues are real, each nonzero eigenspace is finite dimensional, for every $ε > 0$ only finitely many nonzero eigenvalues satisfy $∣ λ ∣ \geq ε$ , and the union of orthonormal bases of the nonzero eigenspaces with a Hilbert basis of $\ker T$ is a Hilbert basis of $H$ (Spectral theorem for compact self adjoint operators, including claim 1, Orthonormal eigenbasis for a compact self adjoint operator, Orthonormal families, complete orthonormal systems and Hilbert bases). If $\ker T = {0}$ , the union of orthonormal bases of the nonzero eigenspaces is already a Hilbert basis.

[F3]

Diagonal operators: for a complete orthonormal family $(w_{n})$ and arbitrary real numbers $(ν_{n})$ (including zero), the operator $T x = \sum_{n} ν_{n} ⟨ x, w_{n} ⟩ w_{n}$ with domain ${x : \sum_{n} ν_{n}^{2} ∣ ⟨ x, w_{n} ⟩ ∣^{2} < \infty}$ is self-adjoint with spectrum the closure of ${ν_{n}}$ , and $T w_{n} = ν_{n} w_{n}$ (Diagonal adjoint and graph-core lemma, Diagonal spectrum).

[F4]

Products and composites with compact operators are compact: $K (H) \cdot B (H) \subseteq K (H)$ and $B (H) \cdot K (H) \subseteq K (H)$ (Compositions with a compact operator are compact, Compact linear operator).

Proof

1.1F1F2

[Spectral decomposition of $R_{c}$ ] The resolvent $R_{c}$ is compact by hypothesis and is self-adjoint: for $x, y \in H$ , put $u = R_{c} x$ and $v = R_{c} y$ . Then $u, v \in D (A)$ and $y = (A - c) v$ , $x = (A - c) u$ by [F1], so self-adjointness of $A - c$ gives $⟨ R_{c} x, y ⟩ = ⟨ u, (A - c) v ⟩ = ⟨ (A - c) u, v ⟩ = ⟨ x, R_{c} y ⟩$ . It is injective: if $R_{c} x = 0$ , then $x = (A - c) R_{c} x = 0$ by [F1]. By [F2], with $\ker R_{c} = {0}$ , the nonzero eigenspaces of $R_{c}$ are finite dimensional with real nonzero eigenvalues, and the union of orthonormal bases of those eigenspaces is a complete orthonormal family $(e_{i})_{i \in I}$ of eigenvectors of $R_{c}$ with eigenvalues $ν_{i} \neq 0$ . The eigenvalue set is countable, being the union over $k \geq 1$ of the finite sets ${ν : ∣ ν ∣ \geq 1 / k}$ (Countable unions of at most countable sets, assuming ${A C}_{ω}$ , Finite, countably infinite, countable, uncountable), so $I$ , a union of finitely many bases over a countable set of eigenvalues, is finite or countably infinite; it is infinite because a finite complete orthonormal family would be a finite spanning set of $H$ (its span is finite dimensional, hence closed, and equals its closed span, A finite-dimensional normed subspace is closed), so that $H$ would have a finite basis and be finite dimensional (Every spanning subset of a vector space contains a basis, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis, Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis), contrary to hypothesis. Hence $(e_{i})_{i \in I}$ can be indexed as $(e_{n})_{n \in N}$ , and then $ν_{n} \to 0$ : for every $ε > 0$ only finitely many indices $n$ satisfy $∣ ν_{n} ∣ \geq ε$ by [F2].

2.1step 1.1F1algebra

[From $R_{c}$ -eigenvectors to $A$ -eigenvectors] For each $n$ , $e_{n} = ν_{n}^{- 1} R_{c} e_{n}$ lies in $ran R_{c} = D (A)$ by [F1], and $(A - c) e_{n} = ν_{n}^{- 1} (A - c) R_{c} e_{n} = ν_{n}^{- 1} e_{n},$ so $A e_{n} = (c + ν_{n}^{- 1}) e_{n}$ . Put $μ_{n} : = c + ν_{n}^{- 1}$ . Since $∣ ν_{n} ∣ \to 0$ and $ν_{n} \neq 0$ , $∣ μ_{n} ∣ \geq ∣ ν_{n} ∣^{- 1} - ∣ c ∣ \to \infty$ . The map $λ \mapsto c + λ^{- 1}$ is injective on $R ∖ {0}$ , so distinct $ν_{n}$ give distinct $μ_{n}$ , and every eigenvector $x$ of $A$ with eigenvalue $μ \neq c$ satisfies $R_{c} x = (μ - c)^{- 1} x$ , so $x$ is an $R_{c}$ -eigenvector; conversely every $R_{c}$ -eigenvector is an $A$ -eigenvector with eigenvalue $c + ν^{- 1} \neq c$ . Hence the eigenvalue set of $A$ is ${μ_{n}}$ with the same eigenspaces as those of $R_{c}$ , the associated eigenvalues are exactly $μ_{n}$ , and each eigenspace is finite dimensional; in particular the eigenspace of $A$ at $μ_{n}$ equals the $R_{c}$ -eigenspace at $ν_{n}$ .

3.1step 2.1F1F3algebra

[ $A$ is the diagonal operator in the basis $(e_{n})$ ] We claim $D (A) = {x \in H : \sum_{n} μ_{n}^{2} ∣ ⟨ x, e_{n} ⟩ ∣^{2} < \infty}$ and $A x = \sum_{n} μ_{n} ⟨ x, e_{n} ⟩ e_{n}$ for $x$ in this set. First let $x \in D (A)$ . Then $A x \in H$ and, by self-adjointness of $A$ and step 2.1, $⟨ A x, e_{n} ⟩ = ⟨ x, A e_{n} ⟩ = μ_{n} ⟨ x, e_{n} ⟩$ for every $n$ ; since $(e_{n})$ is a complete orthonormal family, Parseval gives $\sum_{n} μ_{n}^{2} ∣ ⟨ x, e_{n} ⟩ ∣^{2} = ∥ A x ∥^{2} < \infty$ and $A x = \sum_{n} ⟨ A x, e_{n} ⟩ e_{n} = \sum_{n} μ_{n} ⟨ x, e_{n} ⟩ e_{n}$ . Conversely, let $x \in H$ satisfy $\sum_{n} μ_{n}^{2} ∣ ⟨ x, e_{n} ⟩ ∣^{2} < \infty$ and put $y : = \sum_{n} μ_{n} ⟨ x, e_{n} ⟩ e_{n} \in H$ ; the series converges because the coefficient sequence is square summable. We claim $⟨ x, A φ ⟩ = ⟨ y, φ ⟩$ for every $φ \in D (A)$ : the identity holds for every $φ = e_{m}$ , because $⟨ x, A e_{m} ⟩ = μ_{m} ⟨ x, e_{m} ⟩ = ⟨ y, e_{m} ⟩$ by step 2.1 and the definition of $y$ ; moreover the span $Z$ of the $e_{m}$ is contained in $D (A)$ , and it is graph-norm dense in $D (A)$ : for $x^{'} \in D (A)$ the truncations $x_{N}^{'} = \sum_{n \leq N} ⟨ x^{'}, e_{n} ⟩ e_{n}$ satisfy $x_{N}^{'} \to x^{'}$ and $A x_{N}^{'} = \sum_{n \leq N} μ_{n} ⟨ x^{'}, e_{n} ⟩ e_{n} \to A x^{'}$ by the first part of this step. Both sides of the claimed identity are continuous on $(D (A), ∥ \cdot ∥_{A})$ , since $∣ ⟨ x, A φ ⟩ ∣ \leq ∥ x ∥ ∥ φ ∥_{A}$ and $∣ ⟨ y, φ ⟩ ∣ \leq ∥ y ∥ ∥ φ ∥_{A}$ (Unbounded linear operators: domain, graph and extension); hence the identity extends from the dense subspace $Z$ to all of $D (A)$ . Thus the functional $φ \mapsto ⟨ x, A φ ⟩$ equals the bounded functional $φ \mapsto ⟨ y, φ ⟩$ on $D (A)$ , so $x \in D (A^{*}) = D (A)$ with $A^{*} x = y$ , and $A x = A^{*} x = y$ . This proves the claim.

4.1step 2.1step 3.1F3

[Spectrum] By step 3.1, $A$ is unitarily equivalent to the diagonal operator with entries $μ_{n}$ , so [F3] gives $σ (A) = \overline{{μ_{n} : n \in N}} = {μ_{n} : n \in N}$ , the closure being the set itself because $∣ μ_{n} ∣ \to \infty$ (a sequence tending to infinity has no finite accumulation point). Together with step 2.1 this is (a) and the first half of (b): $H$ has a complete orthonormal family of eigenvectors of $A$ with $∣ μ_{n} ∣ \to \infty$ , and the spectrum is exactly ${μ_{n}}$ .

5.1step 2.1step 4.1

[Essential spectrum] Every $μ_{n}$ is an isolated point of $σ (A) = {μ_{n}}$ (the set is discrete in $R$ because $∣ μ_{n} ∣ \to \infty$ ), and its eigenspace is finite dimensional by step 2.1. Hence every $μ_{n}$ belongs to the discrete spectrum in the sense of Discrete and essential spectrum of a self-adjoint operator, and $σ_{e s s} (A) = σ (A) ∖ σ_{d} (A) = \emptyset$ . This completes (b).

6.1F1F4algebra∎

[All resolvents are compact] Let $z \in ρ (A)$ and let $R_{z} : = (A - z)^{- 1}$ . Subtracting the inverse identities for $A - z$ and $A - c$ gives $R_{z} - R_{c} = (z - c) R_{c} R_{z}$ , hence $R_{z} = R_{c} [I + (z - c) R_{z}]$ . The factor $I + (z - c) R_{z}$ is bounded, so [F4] makes $R_{z}$ compact as a product with the compact $R_{c}$ . This is (c).

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The twisted-periodic momentum operator on an interval

Statement

Assume the Axiom of Choice and Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $0 < L < \infty$ and $θ \in [0, 2 π)$ . Put $D_{θ} : = {u \in H^{1} (0, L) : u^{*} (L) = e^{i θ} u^{*} (0)}, P_{θ} u : = - i D u,$ with $H^{1}$ and the trace $u^{*}$ of One-dimensional Sobolev spaces on intervals and Integration by parts and the endpoint-vanishing space on intervals clause (b). Then:

(a) $D_{θ}$ is dense in $L^{2} (0, L; C)$ , $P_{θ}$ is symmetric and self-adjoint;

(b) the functions $φ_{n} (x) : = \frac{1}{\sqrt{L}} e^{i (2 π n + θ) x / L} (n \in Z)$ form a complete orthonormal family of $L^{2} (0, L; C)$ , each $φ_{n} \in D_{θ}$ , and $P_{θ} φ_{n} = κ_{n} φ_{n}$ with $κ_{n} : = \frac{2 π n + θ}{L};$ the eigenvalues $κ_{n}$ are simple and $σ (P_{θ}) = {κ_{n} : n \in Z}$ ;

(c) $P_{θ}$ is the unique self-adjoint extension of the symmetric operator $P_{θ}$ restricted to the finite linear span of the $φ_{n}$ , and its domain is exactly $D_{θ} = {\sum_{n} c_{n} φ_{n} : \sum_{n} κ_{n}^{2} ∣ c_{n} ∣^{2} < \infty} .$

Facts & Assumptions

Given: $0 < L < \infty$ ; $θ \in [0, 2 π)$ ; the space $D_{θ}$ and operator $P_{θ}$ of the statement.

[F1]

$H^{1} (0, L)$ consists of the classes with an $L^{2}$ weak derivative, the trace $u^{*}$ is the continuous absolutely continuous representative, and for $u, v \in H^{1} (0, L)$ , $\int_{0}^{L} u \overline{v^{'}} = u^{*} (L) \overline{v^{*} (L)} - u^{*} (0) \overline{v^{*} (0)} - \int_{0}^{L} u^{'} \overline{v}$ (One-dimensional Sobolev spaces on intervals, Integration by parts and the endpoint-vanishing space on intervals). The normalized sine functions $e_{j} (x) = \sqrt{2 / L} \sin (j π x / L)$ form a complete orthonormal family in $L^{2} (0, L)$ (Completeness of the Dirichlet sine system on an interval); each $e_{j}$ is in $H^{1} (0, L)$ and has zero endpoint traces, hence lies in $D_{θ}$ . Therefore their dense linear span is contained in $D_{θ}$ .

[F2]

The characters $e_{k} (t) = e^{2 π i k t}$ of Fourier coefficients and trigonometric polynomials on the torus form an orthonormal basis of $L^{2} (T; C)$ (The trigonometric system is complete in $L^{2}$ of the torus, The one-dimensional torus and its normalized Haar integral), and for an orthonormal family completeness is equivalent to having no nonzero orthogonal vector, equivalently to the Parseval identity holding for every vector (Parseval equivalences for an orthonormal family, Orthonormal families, complete orthonormal systems and Hilbert bases).

[F3]

Change of variables: for the increasing $C^{1}$ map $φ : t \mapsto L t$ on $[0, 1]$ and continuous $H$ on $[0, L]$ , $\int_{0}^{1} H (L t) d t = L^{- 1} \int_{0}^{L} H (x) d x$ (In one dimension the compact-Jordan formula is substitution over the unoriented image interval with the absolute derivative).

[F4]

Range criterion: a densely defined symmetric operator $T$ is self-adjoint if and only if $ran (T - z) = H$ for every $z \in C ∖ R$ (Range criterion for self-adjointness, Symmetric, self-adjoint and essentially self-adjoint operators, Densely defined, closed and closable operators, and cores).

[F5]

The diagonal operator lemma Diagonal adjoint and graph-core lemma: for a complete orthonormal family $(w_{n})$ and real numbers $(ν_{n})$ , the finite-support diagonal operator is essentially self-adjoint with unique self-adjoint extension $T$ , $T w_{n} = ν_{n} w_{n}$ , and $D (T) = {x : \sum_{n} ν_{n}^{2} ∣ ⟨ x, w_{n} ⟩ ∣^{2} < \infty}$ ; the spectrum is the closure of ${ν_{n}}$ (Diagonal spectrum).

[F6]

The affine map $ϕ (t) = L t$ satisfies $λ_{1} (ϕ^{- 1} (E)) = L^{- 1} λ_{1} (E)$ for Lebesgue-measurable $E$ , by linear scaling of Lebesgue measure (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not). Thus $\int_{0}^{1} Q (L t) d t = L^{- 1} \int_{0}^{L} Q (x) d x$ first for indicators and simple functions, then for nonnegative measurable functions by monotone convergence, and then for integrable complex $Q$ by decomposition (Monotone convergence for the integral).

[F7]

Extend a Lebesgue-measurable function on $(0, L)$ by zero to $R$ ; Lebesgue measure on $R$ is the completion of its Borel restriction, and each real component has a Borel representative equal almost everywhere under Countable Choice ( $L (R^{n})$ is exactly the completion of the restriction of $λ_{n}$ to the Borel sets, A function measurable for a completion is almost everywhere equal to one measurable for the original sigma-algebra). The representative map $[t] \mapsto t \in [0, 1)$ is Borel: the quotient map restricts to a homeomorphism from $(0, 1)$ onto $T ∖ {[0]}$ , and the remaining singleton is Borel; hence the function induced on $T$ from a Borel function on $[0, 1)$ is Borel (The one-dimensional torus and its normalized Haar integral).

[F8]

Termwise differentiation of the complex exponential power series gives $\exp^{'} (z) = \exp z$ on $C$ (Inside its disc of convergence a complex power series is holomorphic and may be differentiated term by term, The complex exponential by its power series). For fixed $c \in C$ , the real-variable difference quotient along $x \mapsto c x$ therefore gives $\frac{d}{d x} e^{c x} = c e^{c x}$ (if $c \neq 0$ , use the increment $c h$ in the complex difference quotient; if $c = 0$ , the function is constant). Classical derivatives of $C^{1}$ functions are weak derivatives (Classical derivatives agree with weak derivatives).

[F9]

The complex exponential satisfies $\exp (z + w) = \exp z \exp w$ , and for real $x, y$ , $∣ \exp (x + i y) ∣ = e^{x}$ and $\exp (i y) = \cos y + i \sin y$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ). In particular $∣ e^{i y} ∣ = 1$ and $\overline{e^{i y}} = e^{- i y}$ : multiplication gives $e^{i y} e^{- i y} = 1$ , and a unit complex number has inverse equal to its conjugate. Sine and cosine have period $2 π$ (The zero sets of sine and cosine and the least positive common period 2 pi); consequently $e^{2 π i n} = 1$ for each $n \in Z$ .

Proof

1.1F1givenalgebra

[Density and symmetry] By [F1], $D_{θ}$ contains the linear span of the complete sine family and is therefore dense in $L^{2} (0, L)$ . For $u, v \in D_{θ}$ , [F1] gives $⟨ P_{θ} u, v ⟩ - ⟨ u, P_{θ} v ⟩ = - i (u^{*} (L) \overline{v^{*} (L)} - u^{*} (0) \overline{v^{*} (0)}) = 0,$ because $u^{*} (L) \overline{v^{*} (L)} = e^{i θ} u^{*} (0) \overline{e^{i θ} v^{*} (0)} = u^{*} (0) \overline{v^{*} (0)}$ . Hence $P_{θ}$ is symmetric.

1.2F2F3F6F7F9algebra

[The twisted characters and their completeness] For $n, m \in Z$ , $⟨ φ_{n}, φ_{m} ⟩ = \frac{1}{L} \int_{0}^{L} e^{i (2 π (n - m)) x / L} d x = δ_{n m},$ since for $n = m$ the integrand is $1$ , and for $n \neq m$ the function $L^{- 1} e^{2 π i (n - m) x / L}$ is the transport under $x = L t$ of the torus character $e_{n - m}$ , whose integral over $T$ vanishes by [F2]; the transport of the continuous integral is [F3]. To see completeness, let $g \in L^{2} (0, L)$ satisfy $⟨ g, φ_{n} ⟩ = 0$ for every $n$ . By [F7] choose a Borel representative of $g$ and set $h (x) : = e^{- i θ x / L} g (x)$ , also Borel. By [F9], $e^{- i θ x / L} φ_{n} (x) = L^{- 1 / 2} e^{2 π i n x / L} = χ_{n} (x)$ , so $⟨ h, χ_{n} ⟩ = \int_{0}^{L} g (x) \overline{φ_{n} (x)} d x = 0$ for every $n$ . Define $H ([t]) : = h (L t)$ for the unique representative $t \in [0, 1)$ ; it is measurable by [F7]. The affine integral identity [F6] gives $\hat{H} (k) = \int_{0}^{1} h (L t) e^{- 2 π i k t} d t = L^{- 1} \int_{0}^{L} h (x) e^{- 2 π i k x / L} d x = L^{- 1 / 2} ⟨ h, χ_{k} ⟩ = 0$ . Completeness of the torus characters [F2] gives $H = 0$ , hence $h = 0$ and $g = 0$ almost everywhere. By [F2] the family $(φ_{n})$ is complete.

1.3F1F4F8F9algebra

[Range of $P_{θ} - z$ ] Let $z \in C ∖ R$ and $f \in L^{2} (0, L)$ . Define $A (x) : = u_{0} + i \int_{0}^{x} e^{- i z t} f (t) d t, u (x) : = e^{i z x} A (x),$ where $u_{0} \in C$ is chosen below. By [F8] and the product rule, $A$ is absolutely continuous with $A^{'} = i e^{- i z x} f \in L^{2}$ and $A (0) = u_{0}$ , so $u \in H^{1} (0, L)$ with $D u = i z u + i f$ ; hence $(P_{θ} - z) u = - i (i z u + i f) - z u = f$ . The boundary condition $u^{*} (L) = e^{i θ} u^{*} (0)$ reads $e^{i z L} (u_{0} + i \int_{0}^{L} e^{- i z t} f (t) d t) = e^{i θ} u_{0}$ , i.e. $u_{0} (e^{i z L} - e^{i θ}) = - i e^{i z L} \int_{0}^{L} e^{- i z t} f (t) d t .$ Since $z \notin R$ , [F9] gives $∣ e^{i z L} ∣ = e^{- L Im z} \neq 1 = ∣ e^{i θ} ∣$ , so $e^{i z L} - e^{i θ} \neq 0$ and the displayed equation has exactly one solution $u_{0} \in C$ ; the resulting $u \in D_{θ}$ satisfies $(P_{θ} - z) u = f$ . Hence $ran (P_{θ} - z) = L^{2} (0, L)$ for every nonreal $z$ .

2.1step 1.2F8F9algebra

[Eigenvalues] Each $φ_{n}$ is $C^{1}$ with $D φ_{n} = i κ_{n} φ_{n}$ by [F8]. By [F9], $φ_{n}^{*} (L) = L^{- 1 / 2} e^{i (2 π n + θ)} = e^{i θ} L^{- 1 / 2} e^{2 π i n} = e^{i θ} φ_{n}^{*} (0)$ , so $φ_{n} \in D_{θ}$ and $P_{θ} φ_{n} = κ_{n} φ_{n}$ . The numbers $κ_{n} = (2 π n + θ) / L$ are pairwise distinct, so the eigenvalues listed are simple provided the eigenvalue set is exactly ${κ_{n}}$ (proved in the final phase).

2.2step 1.1step 1.3F4

[Self-adjointness] Steps 1.1 and 1.3 give a densely defined symmetric operator with $ran (P_{θ} - z) = L^{2}$ for every nonreal $z$ ; by the range criterion [F4], $P_{θ}$ is self-adjoint.

3.1step 1.2step 2.1step 2.2F2F5∎

[Diagonal form] Let $S$ be the restriction of $P_{θ}$ to the linear span $D_{0}$ of the $φ_{n}$ . By step 2.1, $S$ is the finite-support diagonal operator with respect to the complete orthonormal family $(φ_{n})$ and the real numbers $κ_{n}$ ; by [F5] $S$ has a unique self-adjoint extension $T$ with $D (T) = {\sum_{n} c_{n} φ_{n} : \sum_{n} κ_{n}^{2} ∣ c_{n} ∣^{2} < \infty}$ and $T φ_{n} = κ_{n} φ_{n}$ , and $σ (T)$ is the closure of ${κ_{n}}$ . Since $P_{θ}$ is self-adjoint and extends $S$ , uniqueness in [F5] gives $P_{θ} = T$ . Because $κ_{n} \to \pm \infty$ as $n \to \pm \infty$ , the set ${κ_{n}}$ has no finite accumulation point, so $σ (P_{θ}) = {κ_{n} : n \in Z}$ , and the domain is the displayed coefficient domain; the diagonal representation also gives $\ker (P_{θ} - κ_{n}) = C φ_{n}$ by uniqueness of coefficients. This proves (b) and (c).

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Bounded perturbations of the interval Dirichlet Laplacian

Statement

Assume the Axiom of Choice and Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $0 < L < \infty$ , let $A_{D}$ be the Dirichlet Laplacian of Identification of the Dirichlet Laplacian with the sine diagonal operator with $D (A_{D}) = {u \in H^{2} (0, L) : u^{*} (0) = u^{*} (L) = 0}$ and $A_{D} u = - D^{2} u$ , and let $V \in C ([0, L]; R)$ with $∥ V ∥_{\infty} = \max_{[0, L]} ∣ V ∣$ . Let $A : = A_{D} + M_{V}$ with domain $D (A) : = D (A_{D})$ , where $(M_{V} u) = V u$ is multiplication by $V$ . Then:

(a) $M_{V}$ is bounded self-adjoint with $∥ M_{V} ∥ = ∥ V ∥_{\infty}$ , $A$ is self-adjoint on $D (A_{D})$ , and $A u = - D^{2} u + V u$ for $u \in D (A_{D})$ ;

(b) $A$ has compact resolvent: $(A - z)^{- 1}$ is compact for every $z \in ρ (A)$ ; consequently $A$ has a complete orthonormal family of eigenvectors, all eigenspaces are finite dimensional, the eigenvalues $E_{1} \leq E_{2} \leq \dots$ (listed with multiplicity) satisfy $E_{n} \to \infty$ and $σ_{e s s} (A) = \emptyset$ ;

(c) with $λ_{n} = (n π / L)^{2}$ and $E_{n} (A_{D}) = λ_{n}$ , the min-max eigenvalues satisfy $∣ E_{n} - λ_{n} ∣ \leq ∥ V ∥_{\infty} (n \geq 1) .$

Facts & Assumptions

Given: $L > 0$ ; real continuous $V$ ; the Dirichlet Laplacian $A_{D}$ of Identification of the Dirichlet Laplacian with the sine diagonal operator with spectrum ${λ_{n} : n \geq 1}$ , $λ_{n} = (n π / L)^{2}$ , eigenbasis $(e_{n})$ , and the form identity $⟨ A_{D} u, u ⟩ = ∥ D u ∥_{L^{2}}^{2}$ for $u \in D (A_{D})$ .

[F1]

Multiplication by the bounded measurable function $V$ is a bounded self-adjoint operator on $L^{2} (0, L)$ with norm $∥ V ∥_{\infty}$ , and $V$ is $A_{D}$ -bounded with relative bound $0$ : $∥ M_{V} u ∥ = ∥ V u ∥ \leq ∥ V ∥_{\infty} ∥ u ∥$ , so $D (A_{D}) \subseteq D (M_{V}) = L^{2}$ (Relative boundedness with respect to an operator).

[F2]

Kato-Rellich (Kato-Rellich theorem): if $A_{D}$ is self-adjoint and $B$ is symmetric with $D (A_{D}) \subseteq D (B)$ and $A_{D}$ -bound less than one, then $A_{D} + B$ with domain $D (A_{D})$ is self-adjoint.

[F3]

Finite-rank operators are compact, and a norm limit of compact operators is compact (Bounded finite rank operators are compact, Norm limit of compact operators is compact, Compact linear operator).

[F4]

Compact-resolvent structure (Self-adjoint operators with compact resolvent: eigenbasis and discrete spectrum): if a self-adjoint $A$ on an infinite-dimensional Hilbert space has $c \in ρ (A)$ with $(A - c)^{- 1}$ compact, then $A$ has a complete orthonormal family $(ψ_{n})_{n \geq 1}$ of eigenvectors with eigenvalues $E_{n}$ satisfying $∣ E_{n} ∣ \to \infty$ , all eigenspaces are finite dimensional, $σ (A)$ is exactly the eigenvalue set, $σ_{e s s} (A) = \emptyset$ , and every resolvent is compact.

[F5]

Min-max principle below the essential spectrum (Min-max principle below the essential spectrum, Spectral form domain and core of a semibounded operator): for a self-adjoint operator bounded below with $Λ = \inf σ_{e s s} = + \infty$ , the eigenvalues $E_{1} \leq E_{2} \leq \dots$ , listed with multiplicity, satisfy $E_{n} = \inf_{L^{'} \subseteq Q (A), \dim L^{'} = n} \sup_{x \in L^{'}, ∥ x ∥ = 1} q_{A} [x] = \inf_{L^{'} \subseteq D (A), \dim L^{'} = n} \sup_{x \in L^{'}, ∥ x ∥ = 1} ⟨ A x, x ⟩$ , and $q_{A} [x] = ⟨ A x, x ⟩$ for $x \in D (A)$ .

Proof

1.1F1F2given

[Self-adjointness] By [F1], $M_{V}$ is bounded self-adjoint with norm $∥ V ∥_{\infty}$ and is $A_{D}$ -bounded with relative bound $0$ (take $a = 0$ , $b = ∥ V ∥_{\infty}$ in Relative boundedness with respect to an operator). Since $A_{D}$ is self-adjoint on $D (A_{D})$ , [F2] makes $A = A_{D} + M_{V}$ self-adjoint on $D (A_{D})$ ; and $A u = A_{D} u + V u = - D^{2} u + V u$ by Identification of the Dirichlet Laplacian with the sine diagonal operator.

1.2givenF3algebra

[The free resolvent is a norm limit of finite-rank operators] Let $c < λ_{1}$ be real and let $R_{D} : = (A_{D} - c)^{- 1}$ , the diagonal operator with respect to the orthonormal basis $(e_{n})$ of Identification of the Dirichlet Laplacian with the sine diagonal operator: $R_{D} e_{n} = (λ_{n} - c)^{- 1} e_{n}$ , from $A_{D} e_{n} = λ_{n} e_{n}$ . For $N \geq 1$ let $R_{D}^{(N)}$ be the diagonal operator with the same eigenvectors but coefficients $(λ_{n} - c)^{- 1}$ for $n \leq N$ and $0$ for $n > N$ ; it is finite rank. For $x = \sum_{n} c_{n} e_{n} \in L^{2}$ one has $(R_{D} - R_{D}^{(N)}) x = \sum_{n > N} (λ_{n} - c)^{- 1} c_{n} e_{n}$ , so $∥ (R_{D} - R_{D}^{(N)}) x ∥ \leq (\sup_{n > N} (λ_{n} - c)^{- 1}) ∥ x ∥$ , and the norm identity $∥ R_{D} - R_{D}^{(N)} ∥ = \sup_{n > N} (λ_{n} - c)^{- 1}$ follows by testing on unit eigenvectors; this supremum tends to $0$ because $λ_{n} = (n π / L)^{2} \to \infty$ . Hence $R_{D}$ is a norm limit of finite-rank operators, and by [F3] it is compact.

2.1step 1.1step 1.2givenalgebra

[Compactness of the perturbed resolvent] Choose $c : = λ_{1} - ∥ V ∥_{\infty} - 1 < λ_{1}$ , so that $d : = dist (c, σ (A_{D})) = λ_{1} - c = ∥ V ∥_{\infty} + 1$ and $∥ R_{D} M_{V} ∥ \leq ∥ R_{D} ∥ ∥ V ∥_{\infty} \leq ∥ V ∥_{\infty} / (∥ V ∥_{\infty} + 1) < 1$ . Then $I + R_{D} M_{V}$ is boundedly invertible by the Neumann series, and on $D (A_{D})$ $(A - c) = (A_{D} - c) (I + R_{D} M_{V}),$ because $R_{D} (A_{D} - c) = I$ on $D (A_{D})$ . Hence $(A - c)^{- 1} = (I + R_{D} M_{V})^{- 1} R_{D},$ a product of a bounded operator with the compact operator $R_{D}$ , hence compact (Compositions with a compact operator are compact); the inverse exists because $A$ is self-adjoint and $c < λ_{1} - ∥ V ∥_{\infty} \leq \inf σ (A)$ : indeed $⟨ A u, u ⟩ = ⟨ A_{D} u, u ⟩ + ⟨ V u, u ⟩ \geq λ_{1} ∥ u ∥^{2} - ∥ V ∥_{\infty} ∥ u ∥^{2}$ , so $A \geq (λ_{1} - ∥ V ∥_{\infty}) I$ and $c \in ρ (A)$ .

3.1step 2.1F4

[Eigenbasis and discreteness] The space $L^{2} (0, L; C)$ is infinite dimensional: the vectors $e_{n}$ , $n \geq 1$ , of Identification of the Dirichlet Laplacian with the sine diagonal operator form an infinite orthonormal family (Orthonormal families, complete orthonormal systems and Hilbert bases), while a space with a finite spanning set has no linearly independent subset equinumerous with $N$ , and a finite-dimensional space has a finite basis, hence a finite spanning set (If $V$ has a spanning set with $n$ elements, then every linearly independent subset of $V$ is finite with at most $n$ elements; in particular $V$ has no linearly independent subset equinumerous with $N$ , Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis). Step 2.1 supplies $c \in ρ (A)$ with compact resolvent, so [F4] gives a complete orthonormal eigenfamily $(ψ_{n})_{n \geq 1}$ of eigenvectors with eigenvalues $E_{n}$ satisfying $∣ E_{n} ∣ \to \infty$ , finite dimensional eigenspaces, $σ (A)$ equal to the eigenvalue set and $σ_{e s s} (A) = \emptyset$ ; since $A \geq (λ_{1} - ∥ V ∥_{\infty}) I$ the eigenvalues are bounded below, so $E_{n} \to + \infty$ along any enumeration, and the eigenvalues may be listed with multiplicity in increasing order. This is (b).

4.1step 3.1F5givenalgebra∎

[Min-max comparison] Both $A$ and $A_{D}$ are bounded below and have empty essential spectrum by step 3.1 (for $A_{D}$ : $σ (A_{D}) = {λ_{n}}$ , all isolated eigenvalues of finite multiplicity by Identification of the Dirichlet Laplacian with the sine diagonal operator and the definition of the discrete spectrum, so $σ_{e s s} (A_{D}) = \emptyset$ ). By [F5], for $n \geq 1$ , $E_{n} = \inf_{L^{'} \subseteq D (A), \dim L^{'} = n} \sup_{\begin{array}{c} x \in L^{'} \\ ∥ x ∥ = 1 \end{array}} ⟨ A x, x ⟩, λ_{n} = \inf_{L^{'} \subseteq D (A_{D}), \dim L^{'} = n} \sup_{\begin{array}{c} x \in L^{'} \\ ∥ x ∥ = 1 \end{array}} ⟨ A_{D} x, x ⟩ .$ Since $D (A) = D (A_{D})$ , and $∣ ⟨ M_{V} x, x ⟩ ∣ \leq ∥ V ∥_{\infty} ∥ x ∥^{2}$ , every trial space gives $⟨ A_{D} x, x ⟩ - ∥ V ∥_{\infty} \leq ⟨ A x, x ⟩ \leq ⟨ A_{D} x, x ⟩ + ∥ V ∥_{\infty}$ for $∥ x ∥ = 1$ ; taking suprema over then infima over the same family of $n$ -dimensional trial spaces yields $λ_{n} - ∥ V ∥_{\infty} \leq E_{n} \leq λ_{n} + ∥ V ∥_{\infty},$ which is (c).

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Regular interval IVP: existence, a priori bound and entire spectral-parameter dependence

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $0 < L < \infty$ and let $V \in C ([0, L]; C)$ with $M : = ∥ V ∥_{C^{0}} = \max_{[0, L]} ∣ V ∣$ . For $z \in C$ , $x_{0} \in [0, L]$ and $c_{0}, c_{1} \in C$ consider the initial value problem $(IVP) - u^{''} + V u = z u, u (x_{0}) = c_{0}, u^{'} (x_{0}) = c_{1},$ for $u \in C^{2} ([0, L]; C)$ . Then:

(a) (IVP) has exactly one solution $u (\cdot; z, x_{0}, c_{0}, c_{1})$ ; it is the unique continuous solution of the Volterra equation $u (x) = c_{0} + c_{1} (x - x_{0}) + \int_{x_{0}}^{x} (x - t) (V (t) - z) u (t) d t (x \in [0, L]),$ the integral over $[x_{0}, x]$ for $x < x_{0}$ carrying the negative orientation, and $\sup_{[0, L]} ∣ u ∣ \leq (∣ c_{0} ∣ + ∣ c_{1} ∣ L) e^{(M + ∣ z ∣) L^{2}};$

(b) for fixed $x_{0}, c_{0}, c_{1}$ , put $p (x) : = c_{0} + c_{1} (x - x_{0})$ and, for $f \in C ([0, L]; C)$ , define $K_{z} f (x) : = \int_{x_{0}}^{x} (x - t) (V (t) - z) f (t) d t$ . Then $u (x; z) = \sum_{n = 0}^{\infty} (K_{z}^{n} p) (x)$ and the termwise $x$ -derivative series converge uniformly on $[0, L] \times {z \in C : ∣ z ∣ \leq R}$ for every $R > 0$ . In particular, $(x, z) \mapsto (u (x; z), u^{'} (x; z))$ is continuous, and for each fixed $x$ both $u (x; \cdot)$ and $u^{'} (x; \cdot)$ are entire functions of $z$ . The derivative $w : = \partial_{z} u (\cdot; z)$ is the unique $C^{2}$ solution on $[0, L]$ of $- w^{''} + V w - z w = u (\cdot; z), w (x_{0}) = 0, w^{'} (x_{0}) = 0.$ It satisfies the associated Volterra equation $w (x) = \int_{x_{0}}^{x} (x - t) [(V (t) - z) w (t) - u (t)] d t$ .

Facts & Assumptions

Given: $0 < L < \infty$ ; $V \in C ([0, L]; C)$ ; $z \in C$ ; $x_{0} \in [0, L]$ ; $c_{0}, c_{1} \in C$ ; the Banach space $X = C ([0, L]; C)$ with the supremum norm.

[F1]

Banach fixed point theorem: a strict contraction of a nonempty complete metric space has exactly one fixed point (A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point, Banach space).

[F2]

Gronwall's integral inequality: if $u \geq 0$ is continuous and $u (x) \leq a + \int_{x_{0}}^{x} b (s) u (s) d s$ with $a \geq 0$ and $b \geq 0$ continuous, then $u (x) \leq a \exp (\int_{x_{0}}^{x} b)$ ; the reflected orientation holds for $x < x_{0}$ (Gronwall's integral inequality with variable and constant coefficients).

[F3]

Banach-valued fundamental theorem of calculus: for continuous $f : [a, b] \to X$ the indefinite Bochner integral $G (t) = \int_{a}^{t} f$ is differentiable with $G^{'} = f$ , and if $φ$ is continuous with continuous derivative on $(a, b)$ extending continuously to $[a, b]$ , then $\int_{a}^{b} φ^{'} = φ (b) - φ (a)$ (Fundamental theorem of calculus for Banach-valued continuous curves). For continuous complex scalar integrands, the Bochner integral in this supplier is componentwise Lebesgue integration. Separately, A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion asserts Riemann integrability only for continuous real-valued functions on compact intervals; applying it to the two components gives componentwise Riemann integrability, not a general Banach-valued assertion.

[F4]

Fubini-Tonelli for the completed product measure permits the interchange of iterated integrals of absolutely integrable functions on $[0, L]^{2}$ (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability).

Proof

1.1F3F4given

[Equivalence with the Volterra equation] Let $u \in C^{2} ([0, L])$ . If $u$ solves (IVP), then $(u^{'})^{'} = (V - z) u$ ; [F3] gives $u^{'} (x) = c_{1} + \int_{x_{0}}^{x} (V - z) u$ and, integrating once more and applying [F4] to interchange the order of integration on the triangle ${x_{0} \leq t \leq s \leq x}$ (and its mirror image for $x < x_{0}$ ), $u (x) - c_{0} - c_{1} (x - x_{0}) = \int_{x_{0}}^{x} (x - t) (V (t) - z) u (t) d t .$ Conversely, if $u \in C ([0, L])$ satisfies this Volterra equation, then [F3] applied to the integrand $t \mapsto (V (t) - z) u (t)$ shows that $x \mapsto \int_{x_{0}}^{x} (x - t) (V (t) - z) u (t) d t$ is $C^{1}$ with derivative $\int_{x_{0}}^{x} (V (t) - z) u (t) d t$ , hence $u \in C^{1}$ with $u^{'} (x) = c_{1} + \int_{x_{0}}^{x} (V - z) u$ and, applying [F3] once more, $u \in C^{2}$ and $u^{''} = (V - z) u$ ; the initial conditions hold at $x = x_{0}$ . Thus (IVP) is equivalent to the Volterra equation for $C^{2}$ functions.

1.2F1givenalgebra

[Contraction] The space $X$ is complete: a supremum-norm Cauchy sequence converges pointwise in $C$ , its Cauchy bounds imply uniform convergence, and its uniform limit is continuous by the triangle inequality. Fix $z$ and put $α : = M + ∣ z ∣$ . For $λ > α L$ define on $X = C ([0, L])$ the equivalent norm $∥ u ∥_{λ} : = \sup_{x} ∣ u (x) ∣ e^{- λ ∣ x - x_{0} ∣}$ and the map $(T u) (x) : = c_{0} + c_{1} (x - x_{0}) + \int_{x_{0}}^{x} (x - t) (V (t) - z) u (t) d t .$ For $u, v \in X$ and $x \in [0, L]$ , with $r : = ∣ x - x_{0} ∣$ , substituting $t = x_{0} \pm s$ in the Volterra integral gives $∣ T u (x) - T v (x) ∣ \leq α \int_{0}^{r} (r - s) ∣ u - v ∣ (x_{0} \pm s) d s$ , and $∣ (x_{0} \pm s) - x_{0} ∣ = s$ gives $∣ u - v ∣ (x_{0} \pm s) \leq ∥ u - v ∥_{λ} e^{λ s}$ ; hence $∣ T u (x) - T v (x) ∣ \leq α ∥ u - v ∥_{λ} \int_{0}^{r} (r - s) e^{λ s} d s = α ∥ u - v ∥_{λ} \frac{e^{λ r} - 1 - λ r}{λ^{2}} \leq \frac{α r}{λ} e^{λ r} ∥ u - v ∥_{λ} \leq \frac{α L}{λ} e^{λ r} ∥ u - v ∥_{λ},$ since $e^{λ r} - 1 - λ r \leq λ r e^{λ r}$ for $r \geq 0$ . Hence $∥ T u - T v ∥_{λ} \leq (α L / λ) ∥ u - v ∥_{λ}$ , a strict contraction because $λ > α L$ . By [F1] $T$ has exactly one fixed point $u \in X$ .

2.1step 1.1step 1.2F2algebra

[The fixed point is the solution, and the a priori bound] The fixed point satisfies the Volterra equation, so by step 1.1 it is the unique $C^{2}$ solution of (IVP): uniqueness follows because any $C^{2}$ solution satisfies the Volterra equation by step 1.1 and hence is a fixed point of $T$ , and [F1] gives at most one fixed point. For the bound, put $m (r) : = \sup_{x \in [0, L], ∣ x - x_{0} ∣ \leq r} ∣ u (x) ∣$ for $r \in [0, L]$ ; this is continuous, since uniform continuity of $u$ bounds the change when the interval endpoints move; the Volterra equation gives, for $r = ∣ x - x_{0} ∣$ , $∣ u (x) ∣ \leq ∣ c_{0} ∣ + ∣ c_{1} ∣ L + α \int_{0}^{r} (r - s) m (s) d s \leq ∣ c_{0} ∣ + ∣ c_{1} ∣ L + α L \int_{0}^{r} m (s) d s,$ so $m (r) \leq ∣ c_{0} ∣ + ∣ c_{1} ∣ L + α L \int_{0}^{r} m (s) d s$ by taking suprema, and [F2] gives $m (r) \leq (∣ c_{0} ∣ + ∣ c_{1} ∣ L) e^{α L r} \leq (∣ c_{0} ∣ + ∣ c_{1} ∣ L) e^{α L^{2}}$ . This proves (a).

3.1step 2.1given

[Uniform bounds on a parameter disc] Fix $R > 0$ and let $∣ z ∣ \leq R$ ; put $α_{R} : = M + R$ and $C_{R} : = (∣ c_{0} ∣ + ∣ c_{1} ∣ L) e^{α_{R} L^{2}}$ . By step 2.1, $\sup_{[0, L]} ∣ u (\cdot; z) ∣ \leq C_{R}$ .

4.1step 1.1step 2.1step 3.1F3givenalgebra

[Locally uniform Picard series] Put $p (x) : = c_{0} + c_{1} (x - x_{0})$ and $u_{n} (\cdot; z) : = K_{z}^{n} p$ for $n \geq 0$ . Each $u_{n} (x; \cdot)$ is a polynomial in $z$ with continuous coefficients in $x$ . For $r : = ∣ x - x_{0} ∣$ and $∣ z ∣ \leq R$ , induction in the Volterra integral, using its reversed orientation when $x < x_{0}$ , gives $∣ u_{n} (x; z) ∣ \leq (∣ c_{0} ∣ + ∣ c_{1} ∣ L) \frac{α_{R}^{n} r^{2 n}}{(2 n)!} .$ For $n \geq 1$ , differentiating the Volterra integral gives $∣ u_{n}^{'} (x; z) ∣ \leq (∣ c_{0} ∣ + ∣ c_{1} ∣ L) \frac{α_{R}^{n} r^{2 n - 1}}{(2 n - 1)!} .$ The two majorant series converge uniformly for $r \leq L$ , so $\sum_{n \geq 0} u_{n}$ and $c_{1} + \sum_{n \geq 1} u_{n}^{'}$ converge uniformly on $[0, L] \times {∣ z ∣ \leq R}$ . Their sums satisfy the Volterra equation and its first-derivative equation by uniform convergence under the integrals [F3]; step 2.1 identifies the first sum with $u (\cdot; z)$ , and the derivative equation identifies the second with $u^{'} (\cdot; z)$ . Since this holds for every $R > 0$ , both are locally uniform Picard series in $z$ . Uniform convergence on each product compact, whose summands are continuous in $(x, z)$ , also gives joint continuity.

5.1step 3.1step 4.1step 1.2F2F3algebra∎

[Entire dependence and derivative equation] Fix $z \in C$ and choose $R > ∣ z ∣ + 1$ . For sufficiently small complex $h$ , both $z$ and $z + h$ lie in $∣ ζ ∣ \leq R$ . Put $e : = u (\cdot; z + h) - u (\cdot; z)$ . Subtracting the two Volterra equations gives $e = K_{z} e - h J u (\cdot; z + h)$ , where $J : C ([0, L]; C) \to C ([0, L]; C)$ is defined by $(J f) (x) : = \int_{x_{0}}^{x} (x - t) f (t) d t$ , and $e (x_{0}) = e^{'} (x_{0}) = 0$ . By step 3.1 and the kernel bound, for the continuous running supremum $m (r) : = \sup_{x \in [0, L], ∣ x - x_{0} ∣ \leq r} ∣ e (x) ∣$ we have $m (r) \leq ∣ h ∣ C_{R} L^{2} / 2 + α_{R} L \int_{0}^{r} m (s) d s$ . Gronwall [F2] yields $∥ e ∥_{\infty} \leq ∣ h ∣ C_{R} L^{2} e^{α_{R} L^{2}} / 2$ . The differentiated equation $e^{'} (x) = \int_{x_{0}}^{x} [(V (t) - z) e (t) - h u (t; z + h)] d t$ then gives $∥ e^{'} ∥_{\infty} \leq α_{R} L ∥ e ∥_{\infty} + ∣ h ∣ L C_{R}$ . Let $w$ be the unique continuous solution of $w = K_{z} w - J u (\cdot; z)$ , which exists by the contraction argument of step 1.2; step 1.1 makes it $C^{2}$ and gives $- w^{''} + V w - z w = u (\cdot; z)$ with zero initial data. The quotient $q_{h} : = e / h$ satisfies $q_{h} = K_{z} q_{h} - J u (\cdot; z + h)$ . Thus $d_{h} : = q_{h} - w$ obeys $d_{h} = K_{z} d_{h} - J e$ . Applying the same kernel estimate and [F2] gives $∥ d_{h} ∥_{\infty} \leq (L^{2} / 2) ∥ e ∥_{\infty} e^{α_{R} L^{2}} \leq ∣ h ∣ C_{R} L^{4} e^{2 α_{R} L^{2}} / 4 \to 0$ . Hence the complex derivative $\partial_{z} u (\cdot; z)$ exists in the supremum norm and equals $w$ . Finally, $q_{h}^{'} (x) = \int_{x_{0}}^{x} [(V (t) - z) q_{h} (t) - u (t; z + h)] d t, w^{'} (x) = \int_{x_{0}}^{x} [(V (t) - z) w (t) - u (t; z)] d t,$ so $∥ q_{h}^{'} - w^{'} ∥_{\infty} \leq α_{R} L ∥ q_{h} - w ∥_{\infty} + L ∥ e ∥_{\infty} \to 0$ . Therefore $\partial_{z} u^{'} = w^{'}$ as well; since $z$ was arbitrary, both $u (x; \cdot)$ and $u^{'} (x; \cdot)$ are entire, and the derivative is exactly the stated solution $w$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Prufer angle for the interval equation and strict monotonicity in the energy

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $0 < L < \infty$ and let $V \in C ([0, L]; R)$ . For $E \in R$ let $u (\cdot, E)$ be the unique $C^{2}$ solution of $- u^{''} + V u = E u, u (0, E) = 0, u^{'} (0, E) = 1,$ supplied by Regular interval IVP: existence, a priori bound and entire spectral-parameter dependence (with $x_{0} = 0$ , $c_{0} = 0$ , $c_{1} = 1$ ). Put $r (x, E) : = (u (x, E)^{2} + u^{'} (x, E)^{2})^{1 / 2} > 0, θ (x, E) : = \int_{0}^{x} \frac{u^{'} (t, E)^{2} - u (t, E) u^{''} (t, E)}{u (t, E)^{2} + u^{'} (t, E)^{2}} d t .$ Then:

(a) $u = r \sin θ$ and $u^{'} = r \cos θ$ on $[0, L] \times R$ , $θ$ is $C^{1}$ in $x$ with $θ (0, E) = 0$ and $\partial_{x} θ (x, E) = \cos^{2} θ (x, E) + (E - V (x)) \sin^{2} θ (x, E);$

(b) for every $x \in (0, L]$ the map $E \mapsto θ (x, E)$ is differentiable on $R$ with $\partial_{E} θ (x, E) = \frac{\int_{0}^{x} u (t, E)^{2} d t}{u (x, E)^{2} + u^{'} (x, E)^{2}} > 0;$ in particular $E \mapsto θ (L, E)$ is continuous and strictly increasing on $R$ , and $E \mapsto θ (x, E)$ is continuous on $[0, L] \times R$ ;

(c) for every $E$ , $θ (x, E) > 0$ for $0 < x \leq L$ ; whenever $θ (x_{0}, E) = k π$ for some $k \in Z$ and $x_{0} \in (0, L]$ , one has $\partial_{x} θ (x_{0}, E) = 1$ and $θ (x, E) > k π$ for all $x \in (x_{0}, L]$ ; if additionally $0 < k π < θ (L, E)$ , then the level $k π$ is attained at exactly one point of $(0, L)$ .

Facts & Assumptions

Given: $0 < L < \infty$ ; real continuous $V$ ; the solutions $u (\cdot, E)$ ; $r$ and $θ$ as displayed.

[F1]

For every complex $z$ , the regular IVP supplier gives the unique $C^{2}$ solution, and $z \mapsto (u (\cdot; z), u^{'} (\cdot; z))$ has entire dependence with locally uniform Picard series; its complex derivative satisfies $- w^{''} + V w - z w = u (\cdot; z)$ and $w (0) = w^{'} (0) = 0$ (Regular interval IVP: existence, a priori bound and entire spectral-parameter dependence). For real $E$ , the coefficients and initial data are real, so complex conjugation gives another solution of the same IVP, and uniqueness implies that $u (\cdot; E)$ and $u^{'} (\cdot; E)$ are real-valued.

[F2]

Banach-valued fundamental theorem of calculus: a continuous integrand has a $C^{1}$ indefinite integral with derivative the integrand, and continuous functions are integrable (Fundamental theorem of calculus for Banach-valued continuous curves, A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion).

[F3]

Dominated convergence permits passing to the limit under an integral whenever the integrands converge pointwise and are dominated by an integrable function (Dominated convergence).

[F4]

Gronwall's integral inequality: if $u \geq 0$ is continuous on $[0, L]$ and $u (x) \leq a + \int_{0}^{x} b (s) u (s) d s$ with $a \geq 0$ and $b \geq 0$ continuous, then $u (x) \leq a e^{\int_{0}^{x} b}$ (Gronwall's integral inequality with variable and constant coefficients).

Proof

1.1F1F2F4givenalgebra

[The angle identity] Put $R (x) : = (u (x)^{2} + u^{'} (x)^{2})^{1 / 2}$ . Since $(u, u^{'}) (0) = (0, 1)$ , $R (0) = 1$ ; and $R$ never vanishes: if $u (x_{0}) = u^{'} (x_{0}) = 0$ for some $x_{0}$ , uniqueness in [F1] forces $u \equiv 0$ , contradicting $u^{'} (0) = 1$ . Define $A (x) : = u (x) \cos θ (x) - u^{'} (x) \sin θ (x);$ we show $A \equiv 0$ . Differentiating and using $θ^{'} = (u^{' 2} - u u^{''}) / (u^{2} + u^{' 2})$ , $D : = u^{2} + u^{' 2}$ , $A^{'} = u^{'} \cos θ - u \sin θ θ^{'} - u^{''} \sin θ - u^{'} \cos θ θ^{'} = u^{'} \cos θ (1 - θ^{'}) - (u^{''} + u θ^{'}) \sin θ .$ Now $1 - θ^{'} = (D - u^{' 2} + u u^{''}) / D = u (u + u^{''}) / D$ and $u^{''} + u θ^{'} = u^{''} + \frac{u (u^{' 2} - u u^{''})}{D} = \frac{u^{''} D + u u^{' 2} - u^{2} u^{''}}{D} = \frac{u^{' 2} (u^{''} + u)}{D}$ . Hence $A^{'} = \frac{u^{'} (u + u^{''})}{D} (u \cos θ - u^{'} \sin θ) = \frac{u^{'} (u + u^{''})}{D} A = : c (x) A$ with $c$ continuous, and $A (0) = 0 \cdot 1 - 1 \cdot 0 = 0$ ; by [F2], $A (x) = \int_{0}^{x} c A$ , so $∣ A (x) ∣ \leq \int_{0}^{x} ∣ c ∣ ∣ A ∣$ and [F4] with $a = 0$ gives $A \equiv 0$ . Therefore $u = R \sin θ$ and $u^{'} = R \cos θ$ : the rotation matrix identity $(\begin{array}{c} u \\ u^{'} \end{array}) = (\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array}) (\begin{array}{c} A \\ B \end{array})$ with $B : = u \sin θ + u^{'} \cos θ$ gives $u = B \sin θ$ , $u^{'} = B \cos θ$ and $B^{2} = R^{2} - A^{2} = R^{2}$ ; since $B$ is continuous, $B^{2} = R^{2} > 0$ and $B (0) = u (0) \sin 0 + u^{'} (0) \cos 0 = 1 = R (0)$ , the continuous function $B / R$ has square $1$ and value $1$ at $0$ , so $B / R \equiv 1$ and $B = R$ . This proves the first part of (a).

1.2F1F2F3algebra

[Differentiability in $E$ ] Let $R > 0$ and consider $[0, L] \times [- R, R]$ . By [F1] the maps $(x, E) \mapsto u, u^{'}$ are continuous, so $(x, E) \mapsto q (x, E) : = (u^{' 2} - u u^{''}) / (u^{2} + u^{' 2})$ is continuous on the compact set and, since $u^{2} + u^{' 2}$ is continuous and strictly positive there, its denominator has a positive lower bound. By [F1] $u, u^{'}$ are differentiable in $E$ with derivatives $v, v^{'}$ continuous on $[0, L] \times [- R, R]$ ; hence $\partial_{E} q$ is continuous on the compact set, and the difference quotients of $q$ in $E$ converge uniformly to $\partial_{E} q$ . Since $θ (x, E) = \int_{0}^{x} q (t, E) d t$ , the standard differentiation under the integral sign (difference quotients, [F2] for the inner integration, [F3] for the limit) gives $\partial_{E} θ (x, E) = \int_{0}^{x} \partial_{E} q (t, E) d t .$

2.1step 1.1F1algebra

[The Prufer equation] From the integral definition, $θ$ is $C^{1}$ in $x$ with $θ (0) = 0$ and $θ^{'} = (u^{' 2} - u u^{''}) / (u^{2} + u^{' 2})$ . Substituting $u u^{''} = u (V - E) u = (V - E) u^{2}$ and the identities of step 1.1, $θ^{'} = \frac{u^{' 2} + (E - V) u^{2}}{u^{2} + u^{' 2}} = \cos^{2} θ + (E - V) \sin^{2} θ,$ which completes (a).

2.2step 1.2F1F2algebra

[Wronskian identity] Fix $E$ and let $v : = \partial_{E} u (\cdot, E)$ , so $- v^{''} + V v - E v = u$ and $v (0) = v^{'} (0) = 0$ by [F1]. Then $(u^{'} v - u v^{'})^{'} = u^{''} v - u v^{''} = (V - E) u v - u ((V - E) v - u) = u^{2},$ using $u^{''} = (V - E) u$ and $v^{''} = (V - E) v - u$ . Since $u (0) = v (0) = 0$ , $u^{'} (x) v (x) - u (x) v^{'} (x) = \int_{0}^{x} u (t)^{2} d t (x \in [0, L]) .$

3.1step 1.1step 2.2F1algebra

[Energy derivative] Differentiate the relations $u = r \sin θ$ and $u^{'} = r \cos θ$ of step 1.1 in $E$ : writing $\dot{r} = \partial_{E} r$ and $\dot{θ} = \partial_{E} θ$ , $v = \dot{r} \sin θ + r \cos θ \dot{θ}, v^{'} = \dot{r} \cos θ - r \sin θ \dot{θ} .$ Multiplying the first identity by $u^{'} = r \cos θ$ and the second by $- u = - r \sin θ$ and adding eliminates $\dot{r}$ and gives $u^{'} v - u v^{'} = r^{2} \dot{θ}$ , so by step 2.2 $\partial_{E} θ (x, E) = \frac{u^{'} (x) v (x) - u (x) v^{'} (x)}{u (x)^{2} + u^{'} (x)^{2}} = \frac{\int_{0}^{x} u (t, E)^{2} d t}{u (x, E)^{2} + u^{'} (x, E)^{2}} .$ The denominator is the strictly positive endpoint value $r (x, E)^{2}$ , independent of the integration variable. The numerator has nonnegative integrand and is positive for every $x > 0$ : $u (t, E) = t + O (t^{2})$ near $t = 0$ because $u (0) = 0$ , $u^{'} (0) = 1$ and $u^{'}$ is continuous, so $u (t, E)^{2} > 0$ for small $t > 0$ . Hence $\partial_{E} θ (x, E) > 0$ for every $x \in (0, L]$ : $E \mapsto θ (x, E)$ is strictly increasing, and it is continuous on $[0, L] \times R$ by [F1] and step 1.2. This proves (b).

4.1step 2.1algebra∎

[Crossing structure] By (a), at a point with $θ (x_{0}, E) = k π$ one has $\sin θ = 0$ , $\cos θ = \pm 1$ , hence $\partial_{x} θ (x_{0}, E) = \cos^{2} θ = 1 > 0$ . Also $θ (0, E) = 0$ and $\partial_{x} θ (0, E) = 1 > 0$ , so $θ (x, E) > 0$ for all small $x > 0$ . We claim $θ (x, E) > 0$ for every $x \in (0, L]$ : otherwise let $x_{1}$ be the first point in $(0, L]$ with $θ (x_{1}, E) = 0$ ; then $θ > 0$ on $(0, x_{1})$ and continuity gives $θ (x_{1}) = 0$ , so the left difference quotient of $θ$ at $x_{1}$ is at most $0$ , whence $\partial_{x} θ (x_{1}, E) \leq 0$ , contradicting $\partial_{x} θ (x_{1}, E) = 1$ . The same argument at level $k π$ shows that once $θ$ has crossed the level $k π$ from below it stays strictly above: suppose $θ (x_{0}, E) = k π$ and there is $x_{2} > x_{0}$ with $θ (x_{2}, E) \leq k π$ ; let $x_{1} : = \inf {x > x_{0} : θ (x, E) \leq k π}$ . Then $θ (x_{1}, E) = k π$ and $θ > k π$ on $(x_{0}, x_{1})$ , so the left difference quotient at $x_{1}$ is $\leq 0$ and $\partial_{x} θ (x_{1}, E) \leq 0$ , contradicting $\partial_{x} θ (x_{1}, E) = 1$ . Hence $θ (x, E) > k π$ for every $x \in (x_{0}, L]$ . If $0 < k π < θ (L, E)$ , the intermediate value theorem applied to the continuous function $θ (\cdot, E)$ on $[0, L]$ gives some $x$ with $θ (x, E) = k π$ , and the just-proved strict monotonicity above the level makes this $x$ unique. This proves (c).

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Dirichlet eigenvalues, simplicity and the nodal theorem on an interval

Statement

Assume the Axiom of Choice and Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $0 < L < \infty$ and let $V \in C ([0, L]; R)$ , with $V_{\min} : = \min_{[0, L]} V$ and $V_{\max} : = \max_{[0, L]} V$ . For $E \in R$ let $u (\cdot, E)$ be the solution of $- u^{''} + V u = E u$ , $u (0) = 0$ , $u^{'} (0) = 1$ , and let $θ (\cdot, E)$ be its Prufer angle, as in Prufer angle for the interval equation and strict monotonicity in the energy. Call $E$ a Dirichlet eigenvalue when there is a nonzero $w \in H^{2} (0, L)$ (One-dimensional Sobolev spaces on intervals) with $w^{*} (0) = w^{*} (L) = 0$ satisfying $- w^{''} + V w = E w$ , $w^{*}$ the continuous representative of Absolutely continuous representative of an L2 function with L2 weak derivative. Then:

(a) $E$ is a Dirichlet eigenvalue if and only if $θ (L, E) = k π$ for some integer $k \geq 1$ ;

(b) for every $k \geq 1$ the equation $θ (L, E) = k π$ has exactly one solution $E_{k}$ ; the sequence $(E_{k})_{k \geq 1}$ is strictly increasing, $E_{k} \to \infty$ , and it is exactly the sequence of Dirichlet eigenvalues, each with one-dimensional eigenspace: $\ker (- \frac{d^{2}}{d x^{2}} + V - E_{k}) = C u (\cdot, E_{k})$ ;

(d) nodal count: for every $E$ , the number of zeros of $u (\cdot, E)$ in $(0, L]$ equals $⌊ θ (L, E) / π ⌋$ ; consequently the $k$ -th eigenfunction $u (\cdot, E_{k})$ has exactly $k - 1$ zeros in $(0, L)$ , and the same holds for every nonzero eigenfunction with eigenvalue $E_{k}$ .

Facts & Assumptions

Given: $0 < L < \infty$ ; real continuous $V$ ; the solution family $u (\cdot, E)$ and Prufer angle $θ (\cdot, E)$ of Prufer angle for the interval equation and strict monotonicity in the energy; the spaces One-dimensional Sobolev spaces on intervals.

[F1]

$u (\cdot, E)$ is the unique $C^{2}$ solution of its initial value problem, $r : = (u^{2} + u^{' 2})^{1 / 2} > 0$ , $u = r \sin θ$ , $u^{'} = r \cos θ$ , $θ (0, E) = 0$ , $θ$ is $C^{1}$ in $x$ with $θ^{'} = \cos^{2} θ + (E - V) \sin^{2} θ$ , $E \mapsto θ (L, E)$ is continuous and strictly increasing on $R$ , $θ (x, E) > 0$ for $x > 0$ , and every level $k π$ in $(0, θ (L, E))$ is attained exactly once (Regular interval IVP: existence, a priori bound and entire spectral-parameter dependence, Prufer angle for the interval equation and strict monotonicity in the energy). For the explicit free solution in step 1.4 we use that $\sin^{'} = \cos$ , $\cos^{'} = - \sin$ , $\sin 0 = 0$ , $\cos 0 = 1$ (The derivatives of sine and cosine are cosine and minus sine) and the zero sets $\sin x = 0 \Leftrightarrow x \in π Z$ , $\cos x = 0 \Leftrightarrow x \in (Z + \frac{1}{2}) π$ (The zero sets of sine and cosine and the least positive common period 2 pi).

[F2]

If $w \in H^{2} (0, L)$ and $- w^{''} + V w = E w$ with $w^{*} (0) = w^{*} (L) = 0$ and $w \neq 0$ , then $w^{'} (0) \neq 0$ : otherwise $w$ and the zero function are two $C^{2}$ solutions (by Absolutely continuous representative of an L2 function with L2 weak derivative, the class $w$ has a $C^{2}$ representative because its second weak derivative is the continuous function $(V - E) w^{*}$ , and the representative solves the same initial value problem) with the same initial data, contradicting uniqueness in [F1]. Consequently the $C^{2}$ representative of $w$ equals $w^{'} (0) u (\cdot, E)$ .

[F3]

Comparison lemma for scalar ODEs, a direct consequence of Gronwall's inequality (Gronwall's integral inequality with variable and constant coefficients): let $g : R \to R$ be Lipschitz with constant $K$ , and let $φ, ψ$ be continuous and piecewise $C^{1}$ on $[0, L]$ with $φ (0) = ψ (0)$ , $φ^{'} = g (φ)$ and $ψ^{'} \leq g (ψ)$ wherever differentiable. Then $ψ \leq φ$ . Indeed, for $w : = ψ - φ$ the inequality $w^{'} \leq g (ψ) - g (φ) \leq K w$ holds on every interval where $w > 0$ ; if $w (x_{1}) > 0$ and $x_{0} : = \max {x \leq x_{1} : w (x) = 0}$ , then $w (x_{0}) = 0$ and $w > 0$ on $(x_{0}, x_{1}]$ , so $(w e^{- K (x - x_{0})})^{'} \leq 0$ there with value $0$ at $x_{0}$ , forcing $w (x_{1}) \leq 0$ , a contradiction. The reversed inequality $ψ^{'} \geq g (ψ)$ likewise gives $ψ \geq φ$ .

[F4]

Intermediate value theorem for continuous real functions on $[0, L]$ (Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ ); the integral of a continuous function is continuous (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion).

Proof

1.1F1F2algebra

[Every eigenvalue is a root] Let $E$ be a Dirichlet eigenvalue with nonzero eigenfunction $w$ . By [F2], the $C^{2}$ representative of $w$ equals $w^{'} (0) u (\cdot, E)$ with $w^{'} (0) \neq 0$ . Since $w^{*} (L) = 0$ , $u (L, E) = 0$ , and $u (L, E) = r (L, E) \sin θ (L, E)$ with $r (L, E) > 0$ , so $\sin θ (L, E) = 0$ and $θ (L, E) \in π Z$ ; by [F1] $θ (L, E) > 0$ , hence $θ (L, E) = k π$ for some integer $k \geq 1$ . This also shows that the eigenspace at $E$ is contained in $C u (\cdot, E)$ , hence is one-dimensional.

1.2F1algebra

[Every root is an eigenvalue] Conversely, suppose $θ (L, E) = k π$ , $k \geq 1$ . Then $u (L, E) = r \sin θ (L, E) = 0$ and $u (0, E) = 0$ , while $u (\cdot, E)$ is a nonzero $C^{2}$ solution of $- u^{''} + V u = E u$ . Its class lies in $H^{2} (0, L)$ because $u$ is $C^{2}$ with bounded derivatives on the compact interval, and its continuous representative has the endpoint values $0$ . Hence $E$ is a Dirichlet eigenvalue.

1.3F1F3

[Free comparison] Let $β \in R$ and let $v_{β}$ be the solution of $v^{''} + β v = 0$ , $v (0) = 0$ , $v^{'} (0) = 1$ , with Prufer angle $φ_{β}$ (so $v_{β} = r_{β} \sin φ_{β}$ , $v_{β}^{'} = r_{β} \cos φ_{β}$ , $φ_{β} (0) = 0$ , $φ_{β}^{'} = \cos^{2} φ_{β} + β \sin^{2} φ_{β}$ by [F1] applied with $V \equiv 0$ ). If $a : = E - V (x) \leq β$ on $[0, L]$ , then $θ (x, E) \leq φ_{β} (x)$ for all $x$ ; if $a \geq β$ , then $θ (x, E) \geq φ_{β} (x)$ . Indeed, in the first case $θ^{'} = \cos^{2} θ + a \sin^{2} θ \leq \cos^{2} θ + β \sin^{2} θ = g (θ)$ and $φ_{β}^{'} = g (φ_{β})$ with $g (ϑ) = \cos^{2} ϑ + β \sin^{2} ϑ$ Lipschitz; [F3] with $ψ = θ$ (so $ψ^{'} \leq g (ψ)$ ) and $φ = φ_{β}$ gives $φ_{β} \geq θ$ , and the second case is symmetric.

1.4F1algebra

[The free angle and its zero count] For $β > 0$ the solution is $v_{β} (x) = β^{- 1 / 2} \sin (β^{1 / 2} x)$ , whose zeros are $j π / β^{1 / 2}$ , $j \in N$ , all simple. By [F1] every level $j π$ below $φ_{β} (L)$ is attained exactly once, so with $m : = ⌊ L β^{1 / 2} / π ⌋$ the number of zeros of $v_{β}$ in $(0, L]$ equals $m$ and $m π \leq φ_{β} (L) < (m + 1) π$ ; when $L β^{1 / 2} / π = k \in N$ the $k$ -th zero is exactly at $L$ , so $φ_{β} (L) = k π$ . In particular $φ_{β} (L) \geq k π$ whenever $L β^{1 / 2} \geq k π$ . For $β \leq 0$ the solution satisfies $v_{β}^{''} = - β v_{β} \geq 0$ wherever $v_{β} > 0$ ; since $v_{β} (0) = 0$ , $v_{β}^{'} (0) = 1$ and $v_{β}^{''} \geq 0$ on $[0, L]$ , we get $v_{β}^{'} > 0$ and $v_{β} > 0$ on $(0, L]$ , so $v_{β}$ has no zeros there and $0 \leq φ_{β} (L) < π / 2$ .

2.1step 1.3step 1.4algebra

[Endpoint limits of $θ (L, \cdot)$ ] If $E \leq V_{\min}$ , then $a = E - V (x) \leq 0$ , so by step 1.3 with $β = 0$ , $0 < θ (L, E) \leq φ_{0} (L) < π / 2$ . If $E \geq V_{\max} + (k π / L)^{2}$ , then $a = E - V (x) \geq E - V_{\max} = : β \geq (k π / L)^{2}$ and step 1.3 gives $θ (L, E) \geq φ_{β} (L) \geq k π$ by step 1.4.

3.1step 1.3step 1.4step 2.1F1F4algebra

[Existence, uniqueness and bounds for $E_{k}$ ] Let $k \geq 1$ . Put $E^{-} : = V_{\min} - 1 \leq V_{\min}$ and $E^{+} : = V_{\max} + (k π / L)^{2}$ . By step 2.1, $θ (L, E^{-}) < π / 2 < k π \leq θ (L, E^{+})$ . Since $E \mapsto θ (L, E)$ is continuous and strictly increasing [F1], [F4] gives exactly one $E_{k} \in (E^{-}, E^{+}]$ with $θ (L, E_{k}) = k π$ . Since $k \mapsto k π$ is strictly increasing and $E \mapsto θ (L, E)$ is strictly increasing, the sequence $(E_{k})$ is strictly increasing; and $E_{k} \to \infty$ because $θ (L, E_{k}) = k π \to \infty$ . For the bounds, if $E = V_{\min} + (k π / L)^{2}$ then $a = E - V (x) \leq (k π / L)^{2} = : β$ and step 1.3 with step 1.4 gives $θ (L, E) \leq φ_{β} (L) = k π$ (the free solution has its $k$ -th zero exactly at $L$ ); by strict monotonicity this forces $E \leq E_{k}$ , i.e. $E_{k} \geq V_{\min} + (k π / L)^{2}$ . Similarly, at $E = V_{\max} + (k π / L)^{2}$ one has $θ (L, E) \geq φ_{β} (L) = k π$ with the same $β$ , so $E \geq E_{k}$ , i.e. $E_{k} \leq V_{\max} + (k π / L)^{2}$ .

4.1step 1.1step 1.2step 3.1F1∎

[Zero count and nodal theorem] For fixed $E$ , by [F1] the zeros of $u (\cdot, E) = r \sin θ (\cdot, E)$ in $(0, L]$ are exactly the points where $θ (\cdot, E)$ crosses a level $j π$ , $j \geq 1$ ; each level $j π < θ (L, E)$ is attained exactly once, no level $> θ (L, E)$ is attained, and if $θ (L, E) = m π$ then $L$ itself is a zero. Hence $# {x \in (0, L] : u (x, E) = 0} = ⌊ θ (L, E) / π ⌋ .$ In particular, for $E = E_{k}$ with $θ (L, E_{k}) = k π$ , the solution $u (\cdot, E_{k})$ has exactly $k$ zeros in $(0, L]$ , one of which is $L$ , hence exactly $k - 1$ zeros in $(0, L)$ ; by step 1.1 every nonzero eigenfunction at $E_{k}$ is a nonzero multiple of $u (\cdot, E_{k})$ and has the same zeros. This is (d), and steps 1.1, 1.2 and 3.1 give (a), (b) and (c).

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The one-dimensional delta interface operator

Statement

Assume the Axiom of Choice and Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Fix $α \in R$ . Let $D (A_{α})$ be the set of classes $u \in H^{1} (R)$ whose restrictions to $(0, \infty)$ and $(- \infty, 0)$ have representatives of class $H^{2}$ on each half-line (One-dimensional Sobolev spaces on intervals) and which satisfy the interface condition $u^{' *} (0 +) - u^{' *} (0 -) = α u^{*} (0),$ where $u^{*}$ is the continuous representative of Absolutely continuous representative of an L2 function with L2 weak derivative and $u^{' *} (0 \pm)$ are the one-sided traces of the representative of the weak derivative, which exist by Endpoint trace bound and decay of H1 functions. For $u \in D (A_{α})$ define $A_{α} u$ to be the class equal to $- D^{2} u$ on each half-line. Then:

(a) $D (A_{α})$ is a dense linear subspace of $L^{2} (R; C)$ and $A_{α} : D (A_{α}) \to L^{2} (R)$ is a well-defined linear operator;

(b) $A_{α}$ is symmetric;

(c) $A_{α}$ is self-adjoint; explicitly, for $z \in C ∖ R$ the operator $A_{α} - z$ is a bijection $D (A_{α}) \to L^{2} (R)$ .

Facts & Assumptions

Given: $α \in R$ ; the domain $D (A_{α})$ and operator $A_{α}$ of the statement.

[F1]

Traces: every $H^{1}$ class on an interval has a unique continuous absolutely continuous representative, with $∣ u^{*} (x) - u^{*} (y) ∣ \leq \sqrt{∣ x - y ∣} ∥ u^{'} ∥_{L^{2}}$ and the endpoint bound $∣ u^{*} (x) ∣ \leq δ^{- 1 / 2} ∥ u ∥_{L^{2} (c, d)} + δ^{1 / 2} ∥ u^{'} ∥_{L^{2} (c, d)}$ on each interval $(c, d) ∋ x$ of length $δ$ (Absolutely continuous representative of an L2 function with L2 weak derivative, Endpoint trace bound and decay of H1 functions).

[F2]

Integration by parts on half-lines: for $u, v \in H^{1} (0, \infty)$ , $\int_{0}^{\infty} u \overline{v^{'}} = - u^{*} (0 +) \overline{v^{*} (0 +)} - \int_{0}^{\infty} u^{'} \overline{v}$ ; on $(- \infty, 0)$ the origin contribution instead is $+ u^{*} (0 -) \overline{v^{*} (0 -)}$ ; the boundary products at $\pm \infty$ vanish by the decay of $H^{1}$ representatives (Integration by parts and the endpoint-vanishing space on intervals clause (d), Endpoint trace bound and decay of H1 functions).

[F3]

The free line operator: for $z \notin [0, \infty)$ and $k$ with $k^{2} = - z$ , $Re k > 0$ , the kernel $G_{z} (t) = (2 k)^{- 1} e^{- k ∣ t ∣}$ defines a bounded operator $R_{0} (z) f = G_{z} * f$ on $L^{2} (R)$ with range $H^{2} (R)$ , $R_{0} (z) (A_{0} - z) = I_{H^{2}}$ and $(A_{0} - z) R_{0} (z) = I_{L^{2}}$ , where $A_{0} u = - D^{2} u$ on $H^{2} (R)$ ; also $G_{z}^{'} (0 +) - G_{z}^{'} (0 -) = - 1$ and $r (0) = \int_{R} G_{z} (y) f (y) d y$ for $r = R_{0} (z) f$ (The free-line resolvent kernel and its inverse operator, Real multiplication operators and free plus bounded-potential Schrödinger operators).

[F4]

Smooth compactly supported functions are dense in $L^{2} (R)$ (Compactly supported smooth functions are dense in W^{k,p}(R^n) with $k = 0$ , $p = 2$ ); the range criterion for a densely defined symmetric operator $T$ : $T$ is self-adjoint iff $ran (T - z) = H$ for every $z \in C ∖ R$ (Range criterion for self-adjointness, Symmetric, self-adjoint and essentially self-adjoint operators).

Proof

1.1F1F4algebra

[Density] Let $φ \in C_{c}^{\infty} (R ∖ {0})$ . Then $φ^{*} = φ$ vanishes near $0$ , so $φ^{' *} (0 +) - φ^{' *} (0 -) = 0 = α φ^{*} (0)$ and both half-line restrictions are smooth with compact support: $φ \in D (A_{α})$ . Now let $f \in L^{2} (R)$ and $ε > 0$ . By [F4] there is $ψ \in C_{c}^{\infty} (R)$ with $∥ f - ψ ∥ < ε$ . Choose a function $χ \in C^{\infty} (R)$ with $χ = 0$ near $0$ and $χ = 1$ outside $[- 1, 1]$ , and put $χ_{δ} (x) : = χ (x / δ)$ . Then $χ_{δ} ψ \in C_{c}^{\infty} (R ∖ {0})$ and $∥ χ_{δ} ψ - ψ ∥_{L^{2}} \to 0$ as $δ \to 0$ by dominated convergence, since $∣ χ_{δ} ψ - ψ ∣ \leq 2 ∣ ψ ∣ \in L^{1} \cap L^{2}$ and the left side tends to $0$ pointwise. For small $δ$ , $∥ χ_{δ} ψ - ψ ∥ < ε$ , so $D (A_{α})$ contains a dense subspace.

1.2F1given

[Well-definedness] Let $u \in D (A_{α})$ . By definition the restrictions $u_{\pm}$ lie in $H^{2}$ on the two half-lines, so their second weak derivatives $D^{2} u_{\pm}$ are $L^{2}$ classes on those half-lines; the class $A_{α} u$ obtained by gluing $- D^{2} u_{+}$ and $- D^{2} u_{-}$ is in $L^{2} (R)$ and depends only on the class $u$ . The assignment is linear because weak differentiation and restriction are linear.

1.3F1F2givenalgebra

[Symmetry] Let $u, v \in D (A_{α})$ . Splitting the integrals and integrating by parts twice on each half-line, using [F2] and the vanishing of the boundary products at $\pm \infty$ , $⟨ A_{α} u, v ⟩ = (u^{' *} (0 +) - u^{' *} (0 -)) \overline{v^{*} (0)} + \int_{R} (D u) \overline{D v} = α u^{*} (0) \overline{v^{*} (0)} + \int_{R} (D u) \overline{D v} .$ (On each half-line the two integrations by parts produce the boundary contribution of the derivative jump, and the intermediate terms $\int D u \overline{D v}$ combine.) Hence $⟨ A_{α} u, v ⟩ - ⟨ u, A_{α} v ⟩ = α u^{*} (0) \overline{v^{*} (0)} - \overline{α v^{*} (0) \overline{u^{*} (0)}} = 0,$ so $A_{α}$ is symmetric.

2.1step 1.2F1F3givenalgebra

[Range at nonreal $z$ ] Fix $z \in C ∖ R$ , $f \in L^{2} (R)$ , and put $r : = R_{0} (z) f \in H^{2} (R)$ , so that $(A_{0} - z) r = f$ by [F3]. Let $k$ with $k^{2} = - z$ , $Re k > 0$ , and put $u : = r + c G_{z}, c : = - \frac{α r (0)}{1 + α G_{z} (0)} = - \frac{α r (0)}{1 + α / (2 k)} .$ The denominator does not vanish: if $1 + α / (2 k) = 0$ , then $2 k = - α \in R$ , contradicting $Im k \neq 0$ (which holds because $k^{2} = - z$ is nonreal). Since $r, G_{z} \in H^{1} (R)$ (steps (a) of The free-line resolvent kernel and its inverse operator), $u \in H^{1} (R)$ ; on each half-line $u$ is a linear combination of $H^{2}$ functions, hence satisfies the half-line regularity in the definition of $D (A_{α})$ . The trace of $D u$ at $0$ jumps only through $G_{z}$ : $r \in H^{2} (R)$ has a continuous derivative representative at $0$ , while $G_{z}^{'} (0 +) - G_{z}^{'} (0 -) = - 1$ by [F3]. Therefore $u^{' *} (0 +) - u^{' *} (0 -) = - c = α (r (0) + \frac{c}{2 k}) = α u^{*} (0),$ where the middle equality uses the definition of $c$ and $G_{z} (0) = (2 k)^{- 1}$ . So $u \in D (A_{α})$ , and away from $0$ $(A_{α} - z) u = - D^{2} (r + c G_{z}) - z (r + c G_{z}) = (f + z r) - z r + 0 = f,$ since $- D^{2} r - z r = f$ on $R$ and $- D^{2} G_{z} - z G_{z} = 0$ on each half-line (direct computation from $G_{z}^{''} = k^{2} G_{z}$ and $k^{2} = - z$ ). Hence $ran (A_{α} - z) = L^{2} (R)$ for every nonreal $z$ .

3.1step 1.1step 1.3step 2.1F4∎

[Self-adjointness] Steps 1.1 and 1.3 give a densely defined symmetric operator, and step 2.1 gives $ran (A_{α} - z) = L^{2} (R)$ for $z = \pm i$ ; the range criterion [F4] makes $A_{α}$ self-adjoint.

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Rank-one resolvent formula for the delta interface operator

Statement

Assume the Axiom of Choice and Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $α \in R$ and let $A_{α}$ be the self-adjoint delta interface operator of The one-dimensional delta interface operator; write $A_{0}$ for the operator with $α = 0$ , so that $D (A_{0}) = H^{2} (R)$ and $A_{0} u = - D^{2} u$ . For $z \in C ∖ R$ let $k$ with $k^{2} = - z$ , $Re k > 0$ , let $G_{z} (t) = (2 k)^{- 1} e^{- k ∣ t ∣}$ , and define the bounded linear functional $ℓ_{z} (f) : = \int_{R} G_{z} (y) f (y) d y (f \in L^{2} (R)) .$ Then for every $z \in C ∖ R$ :

(a) $1 + α G_{z} (0) = 1 + α / (2 k) \neq 0$ , and $(A_{α} - z)^{- 1} f = (A_{0} - z)^{- 1} f - \frac{α ℓ_{z} (f)}{1 + α G_{z} (0)} G_{z} (f \in L^{2} (R));$

(b) the difference $K_{z} : = (A_{α} - z)^{- 1} - (A_{0} - z)^{- 1}$ is a bounded operator of rank at most one (rank one if $α \neq 0$ , and zero if $α = 0$ ), $K_{z} f = - \frac{α}{1 + α G_{z} (0)} ℓ_{z} (f) G_{z}$ , its range is contained in $C G_{z}$ , and $∥ K_{z} ∥ \leq \frac{∣ α ∣ ∥ G_{z} ∥_{L^{2}}^{2}}{∣ 1 + α G_{z} (0) ∣} = \frac{∣ α ∣}{4 ∣ k ∣^{2} Re k ∣ 1 + α / (2 k) ∣};$ consequently $∥ (A_{α} - z)^{- 1} ∥ \leq ∥ (A_{0} - z)^{- 1} ∥ + ∥ K_{z} ∥ .$

Facts & Assumptions

Given: $α \in R$ ; the operators $A_{α}$ , $A_{0}$ of The one-dimensional delta interface operator; $z \in C ∖ R$ ; $k^{2} = - z$ , $Re k > 0$ ; the kernel $G_{z}$ .

[F1]

$A_{α}$ is self-adjoint, so $ran (A_{α} - z) = L^{2} (R)$ and $A_{α} - z$ is injective; $(A_{α} - z)^{- 1}$ is bounded (The one-dimensional delta interface operator). The free operator $A_{f r e e} = - D^{2}$ on $H^{2} (R)$ has resolvent $(A_{f r e e} - z)^{- 1} f = G_{z} * f$ with range $H^{2} (R)$ ; $G_{z}^{'} (0 +) - G_{z}^{'} (0 -) = - 1$ and $G_{z} (0) = (2 k)^{- 1}$ (The free-line resolvent kernel and its inverse operator). Step 1.1 proves that $A_{f r e e}$ equals the member $A_{α = 0}$ of the interface family, justifying the notation $A_{0}$ in the Statement.

[F2]

Cauchy-Schwarz: $∣ ℓ_{z} (f) ∣ \leq ∥ G_{z} ∥_{L^{2}} ∥ f ∥_{L^{2}}$ , with $∥ G_{z} ∥_{L^{2}} = (2 ∣ k ∣ \sqrt{Re k})^{- 1}$ (Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors, The free-line resolvent kernel and its inverse operator); a map $f \mapsto ℓ (f) w$ with $ℓ$ bounded linear and fixed $w \neq 0$ is bounded of norm $∥ ℓ ∥ ∥ w ∥$ (A bounded linear operator between normed spaces).

Proof

1.1F1givenalgebra

[Zero-coupling identification and nonvanishing denominator] Let $A_{0}^{i n t}$ be the interface operator with parameter $α = 0$ . If $u \in D (A_{0}^{i n t})$ , its piecewise second derivative $q$ is in $L^{2} (R)$ and the two one-sided derivative traces agree. For every $φ \in C_{c}^{\infty} (R)$ , integration by parts on the two half-lines gives $\int_{R} u^{'} φ^{'} = (u^{' *} (0 -) - u^{' *} (0 +)) φ (0) - \int_{R} q φ = - \int_{R} q φ$ ; hence $D^{2} u = q$ globally and $u \in H^{2} (R)$ . Conversely, if $u \in H^{2} (R)$ , then $u^{'} \in H^{1} (R)$ has a single continuous representative, so its one-sided traces at zero agree; the restrictions of $u$ lie in $H^{2}$ on each half-line and satisfy the zero-jump condition. Thus $D (A_{0}^{i n t}) = H^{2} (R)$ and both $A_{0}^{i n t}$ and $A_{f r e e}$ act as $- D^{2}$ there, proving the identification asserted in the Statement. Now, if $1 + α / (2 k) = 0$ , then $2 k = - α \in R$ ; but $k^{2} = - z$ is nonreal because $z$ is nonreal, so $k$ itself is nonreal. Hence $1 + α G_{z} (0) = 1 + α / (2 k) \neq 0$ .

2.1step 1.1F1givenalgebra

[Construction of the preimage] Fix $f \in L^{2} (R)$ and put $r : = (A_{0} - z)^{- 1} f$ , with $A_{0} = A_{f r e e}$ by step 1.1. Exactly as in the range construction of The one-dimensional delta interface operator, the vector $u : = r + c G_{z}, c : = - \frac{α r (0)}{1 + α G_{z} (0)},$ lies in $D (A_{α})$ and satisfies $(A_{α} - z) u = f$ ; here $r (0) = \int_{R} G_{z} (y) f (y) d y = ℓ_{z} (f)$ because $r (x) = \int G_{z} (x - y) f (y) d y$ evaluated at $x = 0$ .

3.1step 2.1F1

[The resolvent formula] Since $A_{α} - z$ is injective by [F1], the vector $u$ of step 2.1 is the unique preimage of $f$ , hence $(A_{α} - z)^{- 1} f = u = r + c G_{z} = (A_{0} - z)^{- 1} f - \frac{α ℓ_{z} (f)}{1 + α G_{z} (0)} G_{z},$ which is (a).

4.1step 3.1F2algebra∎

[Rank one and bounds] The map $K_{z}$ displayed in (b) equals $f \mapsto ℓ_{z} (f) w$ with $w : = - α G_{z} / (1 + α G_{z} (0))$ ; its range is contained in $C G_{z}$ . If $α \neq 0$ , then $w \neq 0$ , and $f = \overline{G_{z}} \in L^{2}$ satisfies $ℓ_{z} (f) = \int_{R} ∣ G_{z} (y) ∣^{2} d y = ∥ G_{z} ∥_{L^{2}}^{2} > 0$ . Hence the range contains a nonzero multiple of $G_{z}$ , so $K_{z}$ has rank exactly one; if $α = 0$ , then $w = 0$ and $K_{z} = 0$ . By [F2], $∥ ℓ_{z} ∥ \leq ∥ G_{z} ∥_{L^{2}}$ , and therefore [step 3.1, F2, algebra] $∥ K_{z} ∥ \leq ∥ ℓ_{z} ∥ ∥ w ∥ \leq ∥ G_{z} ∥_{L^{2}} \frac{∣ α ∣ ∥ G_{z} ∥_{L^{2}}}{∣ 1 + α G_{z} (0) ∣} = \frac{∣ α ∣ ∥ G_{z} ∥_{L^{2}}^{2}}{∣ 1 + α G_{z} (0) ∣} .$ Substituting $∥ G_{z} ∥_{L^{2}}^{2} = (4 ∣ k ∣^{2} Re k)^{- 1}$ from [F2] gives the displayed value, and the triangle inequality for operator norms gives $∥ (A_{α} - z)^{- 1} ∥ \leq ∥ (A_{0} - z)^{- 1} ∥ + ∥ K_{z} ∥$ . This is (b).

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The unique negative eigenvalue of the attractive delta interface operator

Statement

(a) $A_{α}$ has no negative eigenvalue when $α \geq 0$ ; and $E = 0$ is not an eigenvalue of $A_{α}$ for any $α \in R$ ;

(b) if $α < 0$ , then $E_{α} : = - α^{2} / 4 < 0$ is an eigenvalue of $A_{α}$ with one-dimensional eigenspace $\ker (A_{α} - E_{α}) = C ψ_{α}, ψ_{α} (x) : = \sqrt{- \frac{α}{2}} e^{α ∣ x ∣ / 2};$ the vector $ψ_{α}$ satisfies $∥ ψ_{α} ∥ = 1$ , and $E_{α}$ is the unique negative eigenvalue, that is, if $E < 0$ and $A_{α} u = E u$ for some $0 \neq u \in D (A_{α})$ , then $E = E_{α}$ and $u \in C ψ_{α}$ ;

(c) scaling: for any $ℏ, m > 0$ , the operator $H_{α} : = (ℏ^{2} / 2 m) A_{α}$ on the same domain is self-adjoint, and if $α < 0$ , then $H_{α} ψ_{α} = E_{H} ψ_{α}$ with $E_{H} = - ℏ^{2} α^{2} / (8 m)$ ; writing $g : = ℏ^{2} α / (2 m)$ gives $E_{H} = - m g^{2} / (2 ℏ^{2})$ .

Facts & Assumptions

Given: $α \in R$ ; the operator $A_{α}$ of The one-dimensional delta interface operator; the parametrization $α = - 2 κ$ , $κ > 0$ in part (b).

[F1]

$u \in D (A_{α})$ means $u \in H^{1} (R)$ , both half-line restrictions lie in $H^{2}$ , and $u^{' *} (0 +) - u^{' *} (0 -) = α u^{*} (0)$ ; then $A_{α} u = - D^{2} u$ on each half-line (The one-dimensional delta interface operator). A class in $H^{2}$ on a half-line whose second weak derivative equals the continuous class $κ^{2} u$ is twice classically differentiable with $u^{''} = κ^{2} u$ (Absolutely continuous representative of an L2 function with L2 weak derivative, Endpoint trace bound and decay of H1 functions).

[F2]

The real exponential is differentiable with $(e^{x})^{'} = e^{x}$ (The exponential function is smooth and $(\exp)^{'} = \exp$ ).

Proof

1.1F1F2givenalgebra

[Form of a negative-energy eigenfunction] Let $E < 0$ , put $E = - κ^{2}$ with $κ > 0$ , and let $0 \neq u \in D (A_{α})$ satisfy $A_{α} u = E u$ . On $(0, \infty)$ the function satisfies $u^{''} = κ^{2} u$ by [F1]. Put $g = u^{'} - κ u$ ; then $g^{'} = - κ g$ , so [F2] gives $(e^{κ x} g)^{'} = 0$ and $g = C e^{- κ x}$ . Integrating $(e^{- κ x} u)^{'} = C e^{- 2 κ x}$ shows $u (x) = c_{+} e^{κ x} + d_{+} e^{- κ x}$ . Since $u \in L^{2} (0, \infty)$ , necessarily $c_{+} = 0$ . The same calculation on $(- \infty, 0)$ gives $u (x) = c_{-} e^{κ x} + d_{-} e^{- κ x}$ there, and square integrability forces $d_{-} = 0$ . Continuity of the global representative at $0$ gives $d_{+} = c_{-} = : a$ , so $u (x) = a e^{- κ ∣ x ∣}$ . Since $u \neq 0$ , $a \neq 0$ .

1.2F1algebra

[The eigenvalue at zero] Suppose $A_{α} u = 0$ with $u \neq 0$ . By [F1], $u^{''} = 0$ classically on each half-line, so $u$ is affine on each half-line; an affine function in $L^{2} (0, \infty)$ or $L^{2} (- \infty, 0)$ vanishes, so $u = 0$ , a contradiction. Hence $0$ is not an eigenvalue.

2.1step 1.1F1algebra

[Jump condition forces $α < 0$ ] For $u (x) = a e^{- κ ∣ x ∣}$ the one-sided traces of the derivative are $u^{' *} (0 +) = - κ a$ and $u^{' *} (0 -) = + κ a$ , so the interface condition of [F1] reads $- 2 κ a = α a$ , i.e. $α = - 2 κ < 0$ and $E = - κ^{2} = - α^{2} / 4$ . Hence a negative eigenvalue can exist only for $α < 0$ , and then it equals $- α^{2} / 4$ ; every such eigenfunction is a multiple of $x \mapsto e^{α ∣ x ∣ / 2}$ (because $κ = - α / 2$ ). This proves the nonexistence part of (a) and the uniqueness statement of (b) up to normalisation.

3.1step 2.1step 1.2F1F2algebra

[Existence and normalisation] Let $α < 0$ , $κ : = - α / 2 > 0$ , $E_{α} = - κ^{2} = - α^{2} / 4$ , and $ψ_{α} (x) : = \sqrt{κ} e^{- κ ∣ x ∣}$ . Then $∥ ψ_{α} ∥^{2} = κ \int_{R} e^{- 2 κ ∣ x ∣} d x = κ \cdot \frac{1}{κ} = 1,$ using $\int_{0}^{\infty} e^{- 2 κ x} d x = (2 κ)^{- 1}$ and symmetry. By [F2], $ψ_{α}$ is continuous, $H^{1}$ on each half-line with derivative $- κ sgn (x) ψ_{α}$ a.e., and $H^{2}$ on each half-line with $D^{2} ψ_{α} = κ^{2} ψ_{α}$ ; its derivative jump is $ψ_{α}^{' *} (0 +) - ψ_{α}^{' *} (0 -) = - 2 κ ψ_{α} (0) = α ψ_{α} (0)$ , so $ψ_{α} \in D (A_{α})$ by [F1], and $A_{α} ψ_{α} = - D^{2} ψ_{α} = - κ^{2} ψ_{α} = E_{α} ψ_{α}$ . Hence $E_{α}$ is a negative eigenvalue for every $α < 0$ , and $\ker (A_{α} - E_{α}) = C ψ_{α}$ by step 2.1. This completes (a) and (b).

4.1step 3.1algebra∎

[Scaling] For any $α \in R$ and $ℏ, m > 0$ , let $H_{α} : = (ℏ^{2} / 2 m) A_{α}$ with the same domain. Since the scalar is real and positive, the adjoint identity gives $H_{α}^{*} = (ℏ^{2} / 2 m) A_{α}^{*} = H_{α}$ , so $H_{α}$ is self-adjoint. If $α < 0$ , part (b) gives $H_{α} ψ_{α} = (ℏ^{2} / 2 m) E_{α} ψ_{α} = - ℏ^{2} α^{2} / (8 m) ψ_{α}$ . With $g : = ℏ^{2} α / (2 m)$ , one has $α = 2 m g / ℏ^{2}$ , hence $- ℏ^{2} α^{2} / (8 m) = - m g^{2} / (2 ℏ^{2})$ . No eigenvalue claim for $α \geq 0$ is made in this scaling clause. This proves (c).

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Green resolvent for separated Robin boundary conditions

Statement

Assume the Axiom of Choice and Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $0 < L < \infty$ , let $α, β \in R$ , let $V \in C ([0, L]; R)$ , and put $D (A) : = {u \in H^{2} (0, L) : u^{' *} (0) = α u^{*} (0), u^{' *} (L) = β u^{*} (L)}, A u : = - D^{2} u + V u,$ with $H^{2}$ , weak derivatives and traces as in One-dimensional Sobolev spaces on intervals and Absolutely continuous representative of an L2 function with L2 weak derivative. Then:

(a) $D (A)$ is dense in $L^{2} (0, L; C)$ , $A$ is well defined and symmetric, and $A$ is self-adjoint;

(b) for every $z \in C ∖ R$ there are unique $C^{2}$ solutions $u_{a} (\cdot, z)$ of $- u^{''} + V u = z u$ with $u_{a} (0) = 1$ , $u_{a}^{'} (0) = α$ , and $u_{b} (\cdot, z)$ with $u_{b} (L) = 1$ , $u_{b}^{'} (L) = β$ ; the Wronskian $W (z) : = u_{a} (x, z) u_{b}^{'} (x, z) - u_{a}^{'} (x, z) u_{b} (x, z)$ is independent of $x$ and $W (z) \neq 0$ ;

(c) for $f \in L^{2} (0, L)$ the function $u (x) : = \int_{0}^{L} G_{z} (x, y) f (y) d y, G_{z} (x, y) : = - \frac{u_{a} (\min {x, y}, z) u_{b} (\max {x, y}, z)}{W (z)},$ lies in $H^{2} (0, L)$ , satisfies the two Robin boundary conditions, and $(A - z) u = f$ . Consequently $(A - z)^{- 1} f = u$ , the resolvent is an integral operator with continuous kernel $G_{z}$ on $[0, L]^{2}$ , and $(A - z)^{- 1}$ is a compact operator for every $z$ in the resolvent set of $A$ .

Facts & Assumptions

Given: $L > 0$ ; $α, β \in R$ ; real continuous $V$ ; the operator $A$ and its domain $D (A)$ .

[F1]

The sine system $(e_{n})_{n \geq 1}$ is a complete orthonormal family in $L^{2} (0, L)$ (Completeness of the Dirichlet sine system on an interval). Each $e_{n}$ is in $H^{1} (0, L)$ with zero endpoint traces, so the characterization $H_{0}^{1} (0, L) = {u \in H^{1} : u^{*} (0) = u^{*} (L) = 0}$ in Integration by parts and the endpoint-vanishing space on intervals clause (b) gives $e_{n} \in H_{0}^{1} (0, L)$ . By definition, $H_{0}^{1} (0, L)$ is the $H^{1}$ -norm closure of $C_{c}^{\infty} (0, L)$ (One-dimensional Sobolev spaces on intervals). To see $C_{c}^{\infty} (0, L)$ is dense in $L^{2}$ , take $f \in L^{2}$ and $ε > 0$ , choose a finite sine sum $p$ with $∥ f - p ∥_{2} < ε / 2$ , and then choose $φ \in C_{c}^{\infty} (0, L)$ with $∥ p - φ ∥_{H^{1}} < ε / 2$ ; since the $H^{1}$ norm dominates the $L^{2}$ norm, $∥ f - φ ∥_{2} < ε$ . Thus $C_{c}^{\infty} (0, L) \subseteq D (A)$ is dense.

[F2]

For the Wronskian of two solutions of the same equation, $(u_{a} v_{b}^{'} - u_{a}^{'} v_{b})^{'} = u_{a} v_{b}^{''} - u_{a}^{''} v_{b} = 0$ , so $W$ is constant; this is the standard reduction of order identity for the linear equation.

[F3]

[F4]

A continuous kernel $k$ on $[0, L]^{2}$ defines a bounded integral operator $T_{k} f (x) = \int_{0}^{L} k (x, y) f (y) d y$ with $∥ T_{k} f ∥_{2} \leq L ∥ k ∥_{\infty} ∥ f ∥_{2}$ , and $T_{k}$ is compact: uniform continuity gives piecewise-constant kernels $k_{N}$ with $∥ k - k_{N} ∥_{\infty} \to 0$ , each $T_{k_{N}}$ has finite rank, finite-rank operators are compact, and compact operators are closed in the operator norm (Every continuous map from a nonempty compact Hausdorff space to a uniform space is uniformly continuous, Bounded finite rank operators are compact, Norm limit of compact operators is compact, Compact linear operator); moreover a composite of a compact operator with a bounded operator, in either order, is compact (Compositions with a compact operator are compact).

Proof

1.1F1givenalgebra

[Density, well-definedness, symmetry] $C_{c}^{\infty} (0, L) \subseteq D (A)$ because test functions have vanishing traces at $0$ and $L$ , and $C_{c}^{\infty} (0, L)$ is dense in $L^{2} (0, L)$ by [F1], so $D (A)$ is dense. If $u \in D (A)$ , then $D^{2} u$ and $V u$ are $L^{2}$ classes, so $A u \in L^{2}$ is well defined and depends only on the class $u$ ; linearity is clear. For $u, v \in D (A)$ , the first-order identity of Integration by parts and the endpoint-vanishing space on intervals clause (a), applied to the pairs $(u^{'}, v)$ and $(v^{'}, u)$ , gives $⟨ A u, v ⟩ - ⟨ u, A v ⟩ = [u^{*} \overline{v^{' *}} - u^{' *} \overline{v^{*}}]_{0}^{L} = (u (L) \overline{β v (L)} - β u (L) \overline{v (L)}) - (u (0) \overline{α v (0)} - α u (0) \overline{v (0)}) = 0,$ because both $u$ and $v$ satisfy the two boundary conditions and $α, β$ are real. Hence $A$ is symmetric.

2.1F1F2step 1.1

[The two solutions and the Wronskian] By Regular interval IVP: existence, a priori bound and entire spectral-parameter dependence the solution $u_{a}$ with data $(1, α)$ at $0$ and the solution $u_{b}$ with data $(1, β)$ at $L$ exist uniquely. By [F2] $W$ is constant. If $W = 0$ , then $u_{a}, u_{b}$ are linearly dependent; since $u_{a} (0) = 1 \neq 0$ , $u_{b} = c u_{a}$ for some $c \neq 0$ , and the right boundary condition $u_{b}^{'} (L) = β u_{b} (L)$ gives $u_{a}^{'} (L) = β u_{a} (L)$ , so $u_{a} \in D (A)$ and $A u_{a} = z u_{a}$ with $u_{a} \neq 0$ . A symmetric operator has only real eigenvalues ( $⟨ A u_{a}, u_{a} ⟩ = z ∥ u_{a} ∥^{2}$ equals its own conjugate by symmetry), so $z \in R$ , contradicting $z \notin R$ . Hence $W (z) \neq 0$ .

3.1step 2.1F1algebra

[The Green kernel is well defined and inverts $A - z$ ] For $f \in L^{2} (0, L)$ put $a (x) : = \int_{0}^{x} u_{a} (y) f (y) d y$ and $b (x) : = \int_{x}^{L} u_{b} (y) f (y) d y$ . These functions are absolutely continuous with $a^{'} = u_{a} f$ and $b^{'} = - u_{b} f$ almost everywhere. The kernel formula gives $u (x) = - (u_{b} (x) a (x) + u_{a} (x) b (x)) / W$ . Its derivative is $u^{'} (x) = - (u_{b}^{'} (x) a (x) + u_{a}^{'} (x) b (x)) / W$ , since the two terms containing $f$ cancel. The latter expression is absolutely continuous and, using $u_{a}^{''} = (V - z) u_{a}$ , $u_{b}^{''} = (V - z) u_{b}$ and $u_{b}^{'} u_{a} - u_{a}^{'} u_{b} = W$ , its derivative is $u^{''} = (V - z) u - f$ almost everywhere. Thus $u \in H^{2} (0, L)$ and $- u^{''} + V u - z u = f$ . At zero, $u (0) = - W^{- 1} \int_{0}^{L} u_{b} f$ and $u^{'} (0) = - α W^{- 1} \int_{0}^{L} u_{b} f = α u (0)$ ; at $L$ , $u (L) = - W^{- 1} \int_{0}^{L} u_{a} f$ and $u^{'} (L) = - β W^{- 1} \int_{0}^{L} u_{a} f = β u (L)$ . Therefore $u \in D (A)$ and $(A - z) u = f$ .

4.1step 1.1step 3.1F3

[Self-adjointness] Step 3.1 shows $ran (A - z) = L^{2} (0, L)$ for every nonreal $z$ (in particular for $z = \pm i$ ); with the density and symmetry of step 1.1, the range criterion [F3] makes $A$ self-adjoint.

5.1step 3.1step 4.1F4∎

[Compactness] The kernel $G_{z}$ is continuous on the compact square $[0, L]^{2}$ (it is built from continuous functions, and the apparent singularity at $x = y$ is removable because $\min$ and $\max$ are continuous), hence bounded. By [F4] the resolvent $(A - z)^{- 1}$ is compact for every nonreal $z$ : the uniform-continuity approximation of $G_{z}$ by piecewise-constant kernels exhibits it as a norm limit of finite-rank integral operators. For an arbitrary $z$ in the resolvent set, the second resolvent identity $(A - z)^{- 1} = (A - i)^{- 1} + (z - i) (A - i)^{- 1} (A - z)^{- 1}$ (an identity of bounded operators on $L^{2} (0, L)$ , obtained from $(A - i)^{- 1} - (A - z)^{- 1} = - (z - i) (A - i)^{- 1} (A - z)^{- 1}$ ) writes $(A - z)^{- 1} = (A - i)^{- 1} (I + (z - i) (A - z)^{- 1})$ , a composite of a compact with a bounded operator, hence compact by [F4]. This proves (c).

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Simple complete spectrum of the separated Robin realisation

Statement

Assume the Axiom of Choice and Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $0 < L < \infty$ , $α, β \in R$ , $V \in C ([0, L]; R)$ , and let $A$ be the Robin realisation of Green resolvent for separated Robin boundary conditions: $D (A) = {u \in H^{2} (0, L) : u^{' *} (0) = α u^{*} (0), u^{' *} (L) = β u^{*} (L)}, A u = - D^{2} u + V u .$ Then:

(a) $A$ has compact resolvent; there is a complete orthonormal family $(ψ_{n})_{n \geq 1}$ of eigenvectors of $A$ , every eigenspace of $A$ is one-dimensional, and the spectrum of $A$ consists exactly of the eigenvalues $λ_{n}$ of $A$ , listed with multiplicity, which satisfy $λ_{n} \to \infty$ and $σ_{e s s} (A) = \emptyset$ ;

(b) in particular the eigenvalues can be listed strictly increasingly as $λ_{1} < λ_{2} < \dots$ (each simple), and if $A ψ = λ ψ$ with $ψ \neq 0$ , then $ψ$ is a nonzero multiple of the solution $u_{a} (\cdot, λ)$ of $- u^{''} + V u = λ u$ , $u (0) = 1$ , $u^{'} (0) = α$ , and the eigenvalue condition is $β = \frac{u_{a}^{'} (L, λ)}{u_{a} (L, λ)}$ when $u_{a} (L, λ) \neq 0$ , while $u_{a} (L, λ) = 0$ occurs for no eigenvalue.

Facts & Assumptions

Given: the Robin realisation $A$ of Green resolvent for separated Robin boundary conditions.

[F1]

$A$ is self-adjoint, its resolvent is an integral operator with continuous kernel, and $(A - z)^{- 1}$ is compact for every $z$ in the resolvent set (Green resolvent for separated Robin boundary conditions).

[F2]

Compact-resolvent structure (Self-adjoint operators with compact resolvent: eigenbasis and discrete spectrum): a self-adjoint operator with compact resolvent on an infinite-dimensional Hilbert space has a complete orthonormal family $(ψ_{n})_{n \geq 1}$ of eigenvectors with eigenvalues $μ_{n}$ satisfying $∣ μ_{n} ∣ \to \infty$ , all eigenspaces finite dimensional, spectrum exactly the eigenvalue set, and empty essential spectrum (Orthonormal families, complete orthonormal systems and Hilbert bases).

[F3]

For every $λ \in R$ the unique solution $u_{a} (\cdot, λ)$ of $- u^{''} + V u = λ u$ , $u (0) = 1$ , $u^{'} (0) = α$ , exists and is $C^{2}$ ; every eigenfunction of $A$ with eigenvalue $λ$ satisfies the left boundary condition and hence, together with its value at $0$ , is determined by $u_{a} (\cdot, λ)$ up to the scalar $u (0)$ (Regular interval IVP: existence, a priori bound and entire spectral-parameter dependence).

[F4]

$L^{2} (0, L; C)$ is infinite dimensional: the vectors $e_{n} (x) = \sqrt{2 / L} \sin (n π x / L)$ , $n \geq 1$ , form an infinite orthonormal family (Completeness of the Dirichlet sine system on an interval, Orthonormal families, complete orthonormal systems and Hilbert bases), while a space with a finite spanning set has no linearly independent subset equinumerous with $N$ (If $V$ has a spanning set with $n$ elements, then every linearly independent subset of $V$ is finite with at most $n$ elements; in particular $V$ has no linearly independent subset equinumerous with $N$ ) and a finite-dimensional space has a finite basis, hence a finite spanning set (Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

[F5]

For $u \in D (A)$ , the first-order identity of Integration by parts and the endpoint-vanishing space on intervals clause (a) applied to the pair $(u^{'}, u)$ (both in $H^{1} (0, L)$ because $u \in H^{2} (0, L)$ ), together with the boundary conditions $u^{' *} (0) = α u^{*} (0)$ and $u^{' *} (L) = β u^{*} (L)$ and the reality of $α, β$ , gives $⟨ A u, u ⟩ = \int_{0}^{L} ∣ u^{'} ∣^{2} + \int_{0}^{L} V ∣ u ∣^{2} + α ∣ u (0) ∣^{2} - β ∣ u (L) ∣^{2};$ and the trace bound of Endpoint trace bound and decay of H1 functions clause (a) on the intervals $(0, δ)$ and $(L - δ, L)$ , $0 < δ \leq L$ , gives $∣ u (0) ∣^{2}, ∣ u (L) ∣^{2} \leq 2 δ^{- 1} ∥ u ∥_{L^{2}}^{2} + 2 δ ∥ u^{'} ∥_{L^{2}}^{2}$ .

[F6]

Under Countable Choice, a closed linear subspace $M$ of a Hilbert space satisfies $H = M \oplus M^{⊥}$ (Orthogonal decomposition by a closed subspace). For a densely defined operator, orthogonality of $y$ to $ran (A - c)$ gives $y \in D ((A - c)^{*})$ and $(A - c)^{*} y = 0$ by Adjoint of a densely defined operator; for real $c$ and self-adjoint $A$ , testing the adjoint identity gives $(A - c)^{*} = A - c$ . Cauchy-Schwarz gives $∣ ⟨ (A - c) u, u ⟩ ∣ \leq ∥ (A - c) u ∥ ∥ u ∥$ (Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors).

Proof

1.1F1F4

[Compact resolvent and infinite dimension] By [F1], $(A - i)^{- 1}$ is compact and $i$ lies in the resolvent set of the self-adjoint operator $A$ . By [F4], $L^{2} (0, L; C)$ is infinite dimensional.

1.2F5givenalgebra

[Boundedness below] Put $V_{\infty} : = \max_{[0, L]} ∣ V ∣$ and $δ : = \min {L, (4 (∣ α ∣ + ∣ β ∣ + 1))^{- 1}}$ , so that $2 δ (∣ α ∣ + ∣ β ∣) \leq \frac{1}{2}$ . For $u \in D (A)$ , [F5] gives $⟨ A u, u ⟩ = ∥ u^{'} ∥_{L^{2}}^{2} + \int_{0}^{L} V ∣ u ∣^{2} + α ∣ u (0) ∣^{2} - β ∣ u (L) ∣^{2} \geq (1 - 2 δ (∣ α ∣ + ∣ β ∣)) ∥ u^{'} ∥_{L^{2}}^{2} - (V_{\infty} + 2 δ^{- 1} (∣ α ∣ + ∣ β ∣)) ∥ u ∥_{L^{2}}^{2} \geq - C ∥ u ∥_{L^{2}}^{2},$ where $C : = V_{\infty} + 2 δ^{- 1} (∣ α ∣ + ∣ β ∣)$ : the integral term is at least $- V_{\infty} ∥ u ∥^{2}$ , each of the two boundary terms is at least $- ∣ α ∣$ resp. $- ∣ β ∣$ times the trace bound of [F5], and the coefficient of $∥ u^{'} ∥^{2}$ is at least $\frac{1}{2} \geq 0$ . In particular every eigenvalue $λ$ of $A$ satisfies $λ \geq - C$ , because $λ ∥ u ∥_{L^{2}}^{2} = ⟨ A u, u ⟩ \geq - C ∥ u ∥_{L^{2}}^{2}$ for an eigenvector $u \neq 0$ .

2.1step 1.1F3algebra

[Simplicity] Let $A ψ = λ ψ$ with $ψ \neq 0$ . By [F3] the solution $u_{a} (\cdot, λ)$ is the unique solution of the initial value problem with data $(1, α)$ at $0$ , so the $C^{2}$ representative of $ψ$ equals $ψ^{*} (0) u_{a} (\cdot, λ)$ ; here $ψ^{*} (0) \neq 0$ , because otherwise the representative vanishes at $0$ together with its derivative $α ψ^{*} (0) = 0$ , forcing $ψ = 0$ by uniqueness. Hence $\ker (A - λ) = C u_{a} (\cdot, λ)$ is one-dimensional: every nonzero eigenfunction at $λ$ is a nonzero multiple of $u_{a} (\cdot, λ)$ . The eigenvalue condition is the right boundary condition evaluated on $u_{a} (\cdot, λ)$ ; $u_{a} (L, λ) = 0$ would give $u_{a}^{'} (L, λ) = β u_{a} (L, λ) = 0$ and hence $u_{a} \equiv 0$ by uniqueness, which is impossible since $u_{a} (0) = 1$ .

3.1step 1.1step 1.2step 2.1F1F2F6algebra∎

[A real resolvent point, eigenbasis and increasing list] Put $c = - C - 1$ with $C$ from step 1.2. For $u \in D (A)$ , that lower bound and Cauchy-Schwarz give $∥ (A - c) u ∥ \geq ∥ u ∥$ . The range of $A - c$ is closed: if $(A - c) u_{j} \to f$ , this estimate makes $u_{j}$ Cauchy, and closedness of the self-adjoint operator gives a limit $u \in D (A)$ with $(A - c) u = f$ . By Orthogonal decomposition by a closed subspace, its closed range is all of $L^{2}$ once its orthogonal complement is zero. A vector orthogonal to it belongs to $\ker ((A - c)^{*}) = \ker (A - c) = {0}$ by the adjoint definition and the same estimate. Thus $A - c$ is bijective with inverse norm at most one, so $c \in ρ (A)$ . The resolvent identity gives $(A - c)^{- 1} = (A - i)^{- 1} [I + (c - i) (A - c)^{- 1}]$ ; the compactness for every resolvent in [F1] therefore gives compactness at this real $c$ . Now [F2] applies: a complete orthonormal eigenfamily exists, all eigenspaces are finite dimensional, the spectrum is exactly the eigenvalue set, and the essential spectrum is empty, with $∣ μ_{n} ∣ \to \infty$ . Step 1.2 gives $μ_{n} \geq - C$ , hence $μ_{n} \to + \infty$ . By step 2.1 the eigenvalues are pairwise distinct; their increasing enumeration is consequently $λ_{1} < λ_{2} < \dots$ with $λ_{n} \to \infty$ . This proves (a) and (b).

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Rectangular barrier: transmission coefficients and the thick-barrier error

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $a > 0$ , let $U > E > 0$ and $m, ℏ > 0$ , and put $k : = \frac{\sqrt{2 m E}}{ℏ} > 0, κ : = \frac{\sqrt{2 m (U - E)}}{ℏ} > 0.$ Consider bounded functions $ψ : R \to C$ that are $C^{1}$ on $R$ , twice continuously differentiable on each of $(- \infty, 0)$ , $(0, a)$ , $(a, \infty)$ , and satisfy $- \frac{ℏ^{2}}{2 m} ψ^{''} + V ψ = E ψ, V (x) = {\begin{cases} U, & 0 < x < a, \\ 0 & otherwise, \end{cases}$ on each of the three intervals. Then:

(a) exactly one such $ψ$ has the transmission asymptotics: its restriction to $(- \infty, 0)$ equals $e^{i k x} + r e^{- i k x}$ and its restriction to $(a, \infty)$ equals $t e^{i k x}$ , for uniquely determined $r, t \in C$ ;

(b) with $D : = 2 \cosh (κ a) + i (\frac{κ}{k} - \frac{k}{κ}) \sinh (κ a)$ , $t = \frac{2 e^{- i k a}}{D}, T : = ∣ t ∣^{2} = [1 + \frac{U^{2} \sinh^{2} (κ a)}{4 E (U - E)}]^{- 1}, R : = ∣ r ∣^{2} = 1 - T;$

(c) with $q : = e^{- 2 κ a}$ , $C : = U^{2} / (4 E (U - E))$ and $δ : = (\frac{4}{C} - 2) q + q^{2}$ , $T = \frac{(4 / C) q}{1 + δ}$ . Put $η : = ∣ \frac{4}{C} - 2 ∣ q + q^{2}$ . Whenever $η < 1$ , $∣ T - \frac{16 E (U - E)}{U^{2}} q ∣ \leq \frac{16 E (U - E)}{U^{2}} q \frac{η}{1 - η} .$ In addition, whenever $∣ δ ∣ \leq \frac{1}{2}$ , $∣ T - \frac{16 E (U - E)}{U^{2}} q ∣ \leq \frac{32 E (U - E)}{U^{2}} q ∣ δ ∣ .$

No $L^{2}$ normalisation is asserted: $ψ$ is a bounded generalised solution, not an element of $L^{2} (R)$ .

Facts & Assumptions

Given: $a, U, E, m, ℏ > 0$ with $U > E$ ; $k, κ$ as displayed; the piecewise constant $V$ .

[F1]

For $κ > 0$ , $e^{\pm κ x}$ have derivatives $\pm κ e^{\pm κ x}$ (The exponential function is smooth and $(\exp)^{'} = \exp$ ); for $k > 0$ , $\cos (k x)$ and $\sin (k x)$ have derivatives $- k \sin (k x)$ and $k \cos (k x)$ (Complex sine, cosine, hyperbolic sine, and hyperbolic cosine are entire with their standard derivatives). Step 1.1 proves directly from these identities that every $C^{2}$ solution of $v^{''} = κ^{2} v$ is a linear combination of $e^{κ x}, e^{- κ x}$ , and every solution of $v^{''} = - k^{2} v$ is a linear combination of $\cos (k x), \sin (k x)$ .

[F2]

The exponential definitions of Complex sine, cosine, hyperbolic sine, and hyperbolic cosine from the complex exponential are $\cosh s = (e^{s} + e^{- s}) / 2$ , $\sinh s = (e^{s} - e^{- s}) / 2$ , $\cos s = (e^{i s} + e^{- i s}) / 2$ and $\sin s = (e^{i s} - e^{- i s}) / (2 i)$ . The series in The complex exponential by its power series gives $e^{0} = 1$ , and $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential gives $e^{s} e^{- s} = e^{i s} e^{- i s} = 1$ . Substitution yields $\cosh 0 = 1$ , $\sinh 0 = 0$ , $\cosh^{2} s - \sinh^{2} s = 1$ and $\cos^{2} s + \sin^{2} s = 1$ . For real $s$ , the real extension is real and $e^{s} = (e^{s / 2})^{2} > 0$ , since the reciprocal identity makes $e^{s / 2} \neq 0$ ; thus $\cosh s \geq 1$ by $(e^{s / 2} - e^{- s / 2})^{2} \geq 0$ . Finally, Complex sine, cosine, hyperbolic sine, and hyperbolic cosine are entire with their standard derivatives gives $\frac{d}{d x} \cosh (κ x) = κ \sinh (κ x)$ and $\frac{d}{d x} \sinh (κ x) = κ \cosh (κ x)$ .

[F3]

If $J (x) = Im (\overline{ψ (x)} ψ^{'} (x))$ is piecewise $C^{1}$ and $ψ^{''} = λ (x) ψ$ with $λ$ real-valued on an interval, then $J^{'} = Im (∣ ψ^{'} ∣^{2} + \overline{ψ} ψ^{''}) = Im (λ ∣ ψ ∣^{2}) = 0$ on that interval, so $J$ is constant there (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point); here $λ = 2 m (V - E) / ℏ^{2}$ is real on each of the three intervals.

[F4]

The complex exponential power series differentiates term by term, so $\exp^{'} (z) = \exp z$ (Inside its disc of convergence a complex power series is holomorphic and may be differentiated term by term, The complex exponential by its power series). For real $h$ , the difference quotient along $x \mapsto i k x$ therefore gives $\frac{d}{d x} e^{i k x} = i k e^{i k x}$ and $\frac{d}{d x} e^{- i k x} = - i k e^{- i k x}$ .

[F5]

For real $s$ , Euler's identity gives $e^{i s} = \cos s + i \sin s$ and $∣ e^{i s} ∣ = 1$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ); addition of complex exponentials gives $e^{i s} e^{- i s} = 1$ and hence $\overline{e^{i s}} = e^{- i s}$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential).

Proof

1.1F1F2algebra

[Shape of bounded solutions] On either exterior interval the equation is $ψ^{''} = - k^{2} ψ$ . At any point $x_{0}$ , the matrix sending $(A, B)$ to the value and derivative of $A \cos (k x) + B \sin (k x)$ has determinant $k \neq 0$ , so choose $A, B$ to match any solution's data there. The difference $w$ then solves $w^{''} = - k^{2} w$ with $w (x_{0}) = w^{'} (x_{0}) = 0$ ; differentiating $∣ w^{'} ∣^{2} + k^{2} ∣ w ∣^{2}$ and using [F1] shows it is constant, hence zero, so every solution is $A \cos (k x) + B \sin (k x)$ and is bounded on each half-line. On $(0, a)$ let $v$ solve $v^{''} = κ^{2} v$ and put $g : = v^{'} - κ v$ . Then $g^{'} = - κ g$ , so $(e^{κ x} g)^{'} = 0$ and $g = C e^{- κ x}$ . Solving $(e^{- κ x} v)^{'} = C e^{- 2 κ x}$ gives $v = A e^{κ x} + B e^{- κ x}$ . Thus each interval has a two-dimensional solution space; a function satisfying the regularity in the statement is described by two constants on each interval, with no boundedness restriction on the exterior coefficients.

2.1step 1.1F1F2F4F5algebra

[Matching and the transfer matrix] Write $ψ_{<}, ψ_{0}, ψ_{>}$ for the three restrictions. The conditions that $ψ$ be $C^{1}$ at $0$ and $a$ are the four linear equations $ψ_{<} (0) = ψ_{0} (0), ψ_{<}^{'} (0) = ψ_{0}^{'} (0), ψ_{0} (a) = ψ_{>} (a), ψ_{0}^{'} (a) = ψ_{>}^{'} (a),$ and the transmission asymptotics of (a) impose the incoming coefficient $1$ and the incoming-from-right coefficient $0$ . By [F4], their boundary data are $(1 + r, i k (1 - r))$ at $0$ and $(t e^{i k a}, i k t e^{i k a})$ at $a$ . The map carrying the interior data $(ψ_{0} (0), ψ_{0}^{'} (0))$ to $(ψ_{0} (a), ψ_{0}^{'} (a))$ is the matrix $M : = (\begin{array}{cc} \cosh (κ a) & κ^{- 1} \sinh (κ a) \\ κ \sinh (κ a) & \cosh (κ a) \end{array})$ : its columns are the values at $a$ of the two solutions $x \mapsto \cosh (κ x)$ and $x \mapsto κ^{- 1} \sinh (κ x)$ of $ψ^{''} = κ^{2} ψ$ with data $(1, 0)$ and $(0, 1)$ at $0$ , and by [F1] every solution of the interior equation is a unique linear combination of these two. Hence the matching equations at $0$ and $a$ are equivalent to the single vector equation $(\begin{array}{c} t e^{i k a} \\ i k t e^{i k a} \end{array}) = M (\begin{array}{c} 1 + r \\ i k (1 - r) \end{array}) .$ Multiplying by $M^{- 1} = (\begin{array}{cc} \cosh (κ a) & - κ^{- 1} \sinh (κ a) \\ - κ \sinh (κ a) & \cosh (κ a) \end{array})$ , which is a two-sided inverse because $\det M = \cosh^{2} (κ a) - \sinh^{2} (κ a) = 1 \neq 0$ by [F2], and reading off the two components gives $r = C_{+} t - 1$ and $r = 1 + B t$ , where $C_{+} : = (\cosh (κ a) - i \frac{k}{κ} \sinh (κ a)) e^{i k a}$ and $B : = - (\cosh (κ a) + i \frac{κ}{k} \sinh (κ a)) e^{i k a}$ . Since $C_{+} - B = [2 \cosh (κ a) + i (\frac{κ}{k} - \frac{k}{κ}) \sinh (κ a)] e^{i k a} = D e^{i k a}$ and $∣ D ∣ \geq ∣ Re D ∣ = 2 \cosh (κ a) \geq 2 > 0$ , these two affine relations determine $(r, t)$ uniquely: $t = 2 / (C_{+} - B) = 2 e^{- i k a} / D$ and $r = C_{+} t - 1$ . Consequently a transmission solution has at most one pair $(r, t)$ and hence at most one $ψ$ , since $(r, t)$ determines the interior data $(1 + r, i k (1 - r))$ and the two exterior pieces.

3.1step 2.1F1F2F4F5algebra

[Existence of the transmission solution] Conversely, let $r : = C_{+} t - 1$ and $t : = 2 / (C_{+} - B) = 2 e^{- i k a} / D$ be the unique pair of step 2.1, let $ψ_{0}$ be the solution of $ψ^{''} = κ^{2} ψ$ on $(0, a)$ with $(ψ_{0} (0), ψ_{0}^{'} (0)) = (1 + r, i k (1 - r))$ given explicitly by [F1], and define $ψ (x) : = e^{i k x} + r e^{- i k x}$ for $x \leq 0$ , $ψ : = ψ_{0}$ on $(0, a)$ and $ψ (x) : = t e^{i k x}$ for $x \geq a$ . The relations $r = C_{+} t - 1$ and $r = 1 + B t$ are exactly the two components of the vector equation of step 2.1, so $(ψ_{0} (a), ψ_{0}^{'} (a)) = (t e^{i k a}, i k t e^{i k a})$ ; hence the three pieces match in value and derivative at $0$ and at $a$ , the glued function is $C^{1}$ on $R$ and $C^{\infty}$ on each of the three open intervals with $- \frac{ℏ^{2}}{2 m} ψ^{''} + V ψ = E ψ$ by [F1, F4], and it is bounded, with exactly the prescribed transmission asymptotics. Together with the uniqueness of step 2.1 this proves (a) and gives the first display of (b), $t = 2 e^{- i k a} / D$ .

4.1step 3.1F2F5algebra

[Transmission coefficient] Since [F5] gives $∣ e^{- i k a} ∣ = 1$ , $T = ∣ t ∣^{2} = 4 / ∣ D ∣^{2}$ , and by [F2] $∣ D ∣^{2} = 4 \cosh^{2} (κ a) + (\frac{κ}{k} - \frac{k}{κ})^{2} \sinh^{2} (κ a) = 4 + [4 + (\frac{κ}{k} - \frac{k}{κ})^{2}] \sinh^{2} (κ a) .$ Now $(\frac{κ}{k} - \frac{k}{κ})^{2} = \frac{(κ^{2} - k^{2})^{2}}{k^{2} κ^{2}} = \frac{(2 m (U - 2 E) / ℏ^{2})^{2}}{k^{2} κ^{2}}$ because $κ^{2} - k^{2} = \frac{2 m}{ℏ^{2}} ((U - E) - E)$ , and $k^{2} κ^{2} = (2 m E) (2 m (U - E)) / ℏ^{4}$ , so $4 + (\frac{κ}{k} - \frac{k}{κ})^{2} = \frac{4 E (U - E) + (U - 2 E)^{2}}{E (U - E)} = \frac{U^{2}}{E (U - E)} .$ Therefore $∣ D ∣^{2} = 4 + \frac{U^{2}}{E (U - E)} \sinh^{2} (κ a) = 4 [1 + \frac{U^{2} \sinh^{2} (κ a)}{4 E (U - E)}]$ and $T = [1 + \frac{U^{2} \sinh^{2} (κ a)}{4 E (U - E)}]^{- 1}$ .

5.1step 2.1step 4.1F3F5algebra

[Flux conservation] Put $J (x) = Im (\overline{ψ (x)} ψ^{'} (x))$ . On each of the three intervals $J$ is constant by [F3], and by continuity of $ψ, ψ^{'}$ it is the same constant on all of $R$ . By [F5], $\overline{e^{i k x}} = e^{- i k x}$ and $∣ e^{i k x} ∣ = 1$ . On $(- \infty, 0)$ , $ψ = e^{i k x} + r e^{- i k x}$ gives $J = Im [i k - i k r e^{- 2 i k x} + i k \overline{r} e^{2 i k x} - i k ∣ r ∣^{2}] = k (1 - ∣ r ∣^{2})$ (the two oscillating terms sum to a real number and contribute no imaginary part). On $(a, \infty)$ , $ψ = t e^{i k x}$ gives $J = k ∣ t ∣^{2}$ . Hence $k (1 - ∣ r ∣^{2}) = k ∣ t ∣^{2}$ , i.e. $R + T = 1$ , which is the second display of (b).

6.1step 4.1F2algebra∎

[Thick-barrier form and controlled error] With $q = e^{- 2 κ a}$ one has $\sinh^{2} (κ a) = (e^{κ a} - e^{- κ a})^{2} / 4 = (q^{- 1} - 2 + q) / 4$ , so with $C = U^{2} / (4 E (U - E))$ , $T = [1 + \frac{C}{4} (q^{- 1} - 2 + q)]^{- 1} = \frac{(4 / C) q}{1 + (\frac{4}{C} - 2) q + q^{2}},$ multiplying numerator and denominator by $4 q / C$ . Since $16 E (U - E) / U^{2} = 4 / C$ , the exact difference is $T - \frac{4}{C} q = - \frac{4}{C} q \frac{δ}{1 + δ}, δ = (\frac{4}{C} - 2) q + q^{2} .$ Put $η = ∣ \frac{4}{C} - 2 ∣ q + q^{2}$ , so $∣ δ ∣ \leq η$ . If $η < 1$ , then $∣ 1 + δ ∣ \geq 1 - ∣ δ ∣ \geq 1 - η > 0$ and $\frac{∣ δ ∣}{∣ 1 + δ ∣} \leq \frac{η}{1 - η}$ , which gives the first error bound in (c). If $∣ δ ∣ \leq \frac{1}{2}$ , then $∣ 1 + δ ∣ \geq \frac{1}{2}$ and $∣ T - \frac{4}{C} q ∣ \leq \frac{4}{C} q \cdot 2 ∣ δ ∣$ , which is the second error bound in (c) after substituting $4 / C = 16 E (U - E) / U^{2}$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Finite square well: root conditions, simplicity and the bound-state count

Statement

Assume the Axiom of Choice and Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $a > 0$ , $U > 0$ , $m, ℏ > 0$ and put $K : = \sqrt{2 m U} / ℏ > 0$ . Let $V : R \to R$ be $V (x) = - U$ for $∣ x ∣ \leq a$ and $V (x) = 0$ for $∣ x ∣ > a$ , and let $H ψ : = - \frac{ℏ^{2}}{2 m} ψ^{''} + V ψ, D (H) : = H^{2} (R),$ with $H^{2}$ of One-dimensional Sobolev spaces on intervals. Then:

(a) $H$ is self-adjoint on $H^{2} (R)$ ;

(b) root conditions. $E \in R$ with $- U < E < 0$ is an eigenvalue of $H$ if and only if, writing $E = - ε$ , $κ : = \sqrt{2 m ε} / ℏ$ and $k : = \sqrt{2 m (U - ε)} / ℏ$ (so that $k, κ > 0$ and $k^{2} + κ^{2} = K^{2}$ ), the pair $(k, κ)$ satisfies $k \tan (k a) = κ (even case) or - k \cot (k a) = κ (odd case) .$ Every such eigenvalue is simple, and its eigenfunction is, up to a nonzero scalar, $x \mapsto \cos (k x)$ on $(- a, a)$ continued by $c e^{- κ ∣ x ∣}$ outside in the even case and $x \mapsto \sin (k x)$ on $(- a, a)$ continued by $c sgn (x) e^{- κ ∣ x ∣}$ outside in the odd case;

(c) range of energies. There are no eigenvalues $E \leq - U$ and no eigenvalues $E \geq 0$ ; hence all eigenvalues of $H$ lie in $(- U, 0)$ and are given by (b);

(d) bound count. The number of negative eigenvalues of $H$ equals $⌈ \frac{2 a K}{π} ⌉,$ and more precisely the equations of (b) have exactly one solution $k$ in each interval $(\frac{n π}{2 a}, \frac{(n + 1) π}{2 a})$ , $n \geq 0$ , whose left endpoint satisfies $\frac{n π}{2 a} < K$ , and no solution in $(0, K)$ elsewhere; the solutions are alternatively of even type ( $n$ even) and odd type ( $n$ odd).

Facts & Assumptions

Given: $a, U, m, ℏ > 0$ , $K = \sqrt{2 m U} / ℏ$ , the potential $V$ (bounded measurable real, $∣ V ∣ \leq U$ , support $[- a, a]$ ), and the operator $H$ on $D (H) = H^{2} (R)$ .

[F1]

$V$ is bounded, measurable and real. In clause (b) of Real multiplication operators and free plus bounded-potential Schrödinger operators, take $d = 1$ , $c = 4 π^{2} \frac{ℏ^{2}}{2 m} = \frac{2 π^{2} ℏ^{2}}{m}$ and $b = 0$ ; the resulting free operator is $H_{0} = - \frac{ℏ^{2}}{2 m} D^{2}$ on $D (H_{0}) = H^{2} (R)$ . Clause (c) then gives that $H = H_{0} + M_{V}$ is self-adjoint on this domain and has real spectrum.

[F2]

If $H ψ = E ψ$ with $ψ \in H^{2} (R)$ , then its $H^{2}$ representative is globally $C^{1}$ and is $C^{2}$ on each open interval where $V$ is constant; there $- \frac{ℏ^{2}}{2 m} ψ^{''} + V ψ = E ψ$ holds classically. Indeed $ψ, ψ^{'}$ have locally absolutely continuous representatives and the equation gives the continuous representative of $ψ^{''}$ on each such interval (Absolutely continuous representative of an L2 function with L2 weak derivative). The real exponential and sine/cosine derivative formulas used below are supplied by The exponential function is smooth and $(\exp)^{'} = \exp$ and Complex sine, cosine, hyperbolic sine, and hyperbolic cosine are entire with their standard derivatives; the constant-coefficient solution classifications are proved in steps 1.3 and 2.1. The same regular-IVP supplier gives uniqueness for the central initial-value problem after translating it to a compact interval (Regular interval IVP: existence, a priori bound and entire spectral-parameter dependence).

[F3]

For $ψ \in H^{2} (R)$ , $⟨ H ψ, ψ ⟩ = \frac{ℏ^{2}}{2 m} ∥ ψ^{'} ∥_{2}^{2} - U \int_{[- a, a]} ∣ ψ ∣^{2} \geq \frac{ℏ^{2}}{2 m} ∥ ψ^{'} ∥_{2}^{2} - U ∥ ψ ∥_{2}^{2}$ . In particular, $⟨ H ψ, ψ ⟩ \geq - U ∥ ψ ∥_{2}^{2}$ , with equality exactly when $ψ^{'} = 0$ a.e. and $\int_{[- a, a]} ∣ ψ ∣^{2} = ∥ ψ ∥_{2}^{2}$ , since the difference is the sum of the two nonnegative quantities $\frac{ℏ^{2}}{2 m} ∥ ψ^{'} ∥_{2}^{2}$ and $U \int_{R ∖ [- a, a]} ∣ ψ ∣^{2}$ .

Proof

1.1F1

[Self-adjointness] This is [F1].

1.2F2F3algebra

[Energies $E \leq - U$ are excluded] Let $H ψ = E ψ$ with $∥ ψ ∥_{L^{2}} = 1$ and $E \leq - U$ . By [F3], $E = ⟨ H ψ, ψ ⟩ = \frac{ℏ^{2}}{2 m} ∥ ψ^{'} ∥_{2}^{2} + \int_{[- a, a]} V ∣ ψ ∣^{2}$ and this is at least $- U$ . Thus $E = - U$ and equality holds in the lower bound, so $∥ ψ^{'} ∥_{2} = 0$ . The absolutely continuous representative of an $H^{1}$ function with derivative zero is constant on every bounded interval by [F2], hence is a single constant on $R$ ; the only constant in $L^{2} (R)$ is $0$ . This contradicts $∥ ψ ∥_{2} = 1$ , so $E > - U$ .

1.3F2algebra

[Parity and matching] Let $E = - ε \in (- U, 0)$ , $κ = \sqrt{2 m ε} / ℏ$ , $k = \sqrt{2 m (U - ε)} / ℏ$ , so $k^{2} + κ^{2} = K^{2}$ . On an exterior interval the classical equation is $ψ^{''} = κ^{2} ψ$ . Put $g = ψ^{'} - κ ψ$ ; then $g^{'} = - κ g$ , so $(e^{κ x} g)^{'} = 0$ and $g = C e^{- κ x}$ . Integrating $(e^{- κ x} ψ)^{'} = C e^{- 2 κ x}$ shows that every solution is $A e^{κ x} + B e^{- κ x}$ . The $L^{2}$ condition removes the growing term on each exterior half-line, leaving multiples of $e^{- κ x}$ on $x > a$ and $e^{κ x}$ on $x < - a$ . On $(- a, a)$ the equation is $ψ^{''} = - k^{2} ψ$ . The matrix mapping $(A, B)$ to the value and derivative of $A \cos (k x) + B \sin (k x)$ at any $x_{0}$ has determinant $k \neq 0$ , so choose $A, B$ to match the data of any solution there. The difference $w$ solves $w^{''} = - k^{2} w$ with zero data at $x_{0}$ ; differentiating $∣ w^{'} ∣^{2} + k^{2} ∣ w ∣^{2}$ shows this quantity is constant, hence $w = 0$ . Thus every interior solution is $A \cos (k x) + B \sin (k x)$ . The potential is even, so the reflection operator $(Π ψ) (x) = ψ (- x)$ commutes with $H$ on $H^{2}$ and maps every eigenspace to itself. Eigenspaces are at most one-dimensional: the solutions decaying as $x \to - \infty$ form a one-dimensional space $c e^{κ x}$ ( $x < - a$ ), and the transfer through $(- a, a)$ is invertible by the just-proved initial-data representation, so at most one direction matches the decaying solution at $+ \infty$ . Hence $Π ψ = \pm ψ$ : every eigenfunction is even or odd. For an even eigenfunction, $ψ (x) = A \cos (k x)$ on $(- a, a)$ and $ψ (x) = B e^{- κ ∣ x ∣}$ for $∣ x ∣ > a$ ; matching $ψ$ and $ψ^{'}$ at $a$ gives $A \cos (k a) = B e^{- κ a}, - A k \sin (k a) = - κ B e^{- κ a},$ which eliminates $A, B$ and gives $k \tan (k a) = κ$ . For an odd eigenfunction, $ψ (x) = A \sin (k x)$ on $(- a, a)$ and $ψ (x) = B sgn (x) e^{- κ ∣ x ∣}$ outside; matching gives $A \sin (k a) = B e^{- κ a}$ and $A k \cos (k a) = - κ B e^{- κ a}$ , i.e. $- k \cot (k a) = κ$ . Conversely, if either condition holds with $k, κ > 0$ , $k^{2} + κ^{2} = K^{2}$ , then the indicated piecewise function is $C^{1}$ , piecewise $C^{2}$ , satisfies $H ψ = E ψ$ on each constant-potential interval and decays exponentially. Its restrictions and their first derivatives are square-integrable, and its piecewise classical second derivative is square-integrable because the central interval is bounded and the exterior pieces decay exponentially. To verify the global weak derivatives, integrate against a compactly supported smooth test function separately on $(- \infty, - a)$ , $(- a, a)$ and $(a, \infty)$ : the boundary terms cancel at $\pm a$ for the first derivative because $ψ$ is continuous, and for the second derivative because $ψ^{'}$ matches there. Thus the piecewise first and second derivatives are the global weak derivatives and belong to $L^{2} (R)$ , so $ψ \in H^{2} (R)$ by One-dimensional Sobolev spaces on intervals. It is nonzero and therefore an eigenfunction. This proves (b).

1.4F2algebra

[Monotone branches] For $k \in (0, K)$ put $κ (k) : = \sqrt{K^{2} - k^{2}} > 0$ , a strictly decreasing continuous function. Define on each interval $B_{n} : = (\frac{n π}{2 a}, \frac{(n + 1) π}{2 a})$ the function $L_{n} (k) : = {\begin{cases} k \tan (k a), & n even, \\ - k \cot (k a), & n odd . \end{cases}$ Each $L_{n}$ is strictly increasing on $B_{n}$ : for $n$ even, $L_{n}^{'} (k) = \tan (k a) + k a \sec^{2} (k a) > 0$ because $k a \in (n π / 2, (n + 1) π / 2)$ with $n$ even lies in a range where $\tan (k a) > 0$ ; for $n$ odd, $L_{n} (k) = - k \cot (k a)$ with $\cot (k a) < 0$ and $L_{n}^{'} (k) = - \cot (k a) + k a \csc^{2} (k a) > 0$ . Moreover, $\lim_{k \to (\frac{n π}{2 a})^{+}} L_{n} (k) = 0^{+}$ and $\lim_{k \to (\frac{(n + 1) π}{2 a})^{-}} L_{n} (k) = + \infty$ for every $n \geq 0$ .

2.1step 1.2step 1.3F2algebra

[Energies $E \geq 0$ are excluded] Let $H ψ = E ψ$ with $ψ \neq 0$ and $E \geq 0$ . By [F2], on $(a, \infty)$ the classical equation is $ψ^{''} = - k_{0}^{2} ψ$ with $k_{0} = \sqrt{2 m E} / ℏ \geq 0$ (and $ψ^{''} = 0$ when $E = 0$ ). If $E > 0$ , the classification proved in step 1.3 applies with $k = k_{0}$ and gives $ψ (x) = A \cos (k_{0} x) + B \sin (k_{0} x)$ . The identity $∣ ψ ∣^{2} = ∣ A ∣^{2} \cos^{2} (k_{0} x) + ∣ B ∣^{2} \sin^{2} (k_{0} x) + 2 Re (A \overline{B}) \sin (k_{0} x) \cos (k_{0} x)$ and the antiderivatives of $\cos^{2}$ , $\sin^{2}$ and $\sin \cos$ give $\int_{a}^{X} ∣ ψ ∣^{2} = \frac{∣ A ∣^{2} + ∣ B ∣^{2}}{2} (X - a) + O (1)$ as $X \to \infty$ . Thus $L^{2}$ membership forces $A = B = 0$ , so $ψ = 0$ on $(a, \infty)$ . If $E = 0$ , every solution of $ψ^{''} = 0$ is affine, and an affine $L^{2}$ function vanishes identically, so again $ψ = 0$ there. The same argument on $(- \infty, - a)$ shows $ψ = 0$ outside $[- a, a]$ . The global $C^{1}$ representative then has $ψ (a) = ψ^{'} (a) = 0$ , and on $(- a, a)$ it solves $ψ^{''} = - \frac{2 m (E + U)}{ℏ^{2}} ψ$ with these terminal data. After translating $(- a, a)$ to a compact interval, uniqueness in Regular interval IVP: existence, a priori bound and entire spectral-parameter dependence makes it vanish identically, so $ψ = 0$ , a contradiction. Hence $E < 0$ , and with step 1.2, $E \in (- U, 0)$ .

2.2step 1.4algebra

[Existence and uniqueness of roots] Fix $n \geq 0$ with $\frac{n π}{2 a} < K$ , and consider $F_{n} : = L_{n} - κ$ on $B_{n} \cap (0, K)$ , a strictly increasing continuous function. Near the left endpoint, $L_{n} \to 0^{+}$ while $κ > 0$ (since the left endpoint is $< K$ ), so $F_{n} < 0$ there; put $r_{n} = (n + 1) π / (2 a)$ . If $r_{n} \leq K$ , the one-sided limit $L_{n} (k) \to + \infty$ as $k ↑ r_{n}$ makes $F_{n} > 0$ near that endpoint, including when $r_{n} = K$ ; no value of $L_{n}$ at its pole is taken. If $K < r_{n}$ , then $K$ is interior to the branch, $L_{n} (K) > 0$ is well defined, and $κ (k) \to 0$ as $k ↑ K$ , again giving $F_{n} > 0$ . Hence by the intermediate value theorem $F_{n}$ has a zero, and strict monotonicity makes it unique and interior. If instead $\frac{n π}{2 a} \geq K$ , then $k < K$ on $B_{n}$ never holds, so no root. On each $B_{n}$ the other parity expression is negative and cannot equal $κ > 0$ . At branch endpoints the matching equations of step 1.3 either require a pole denominator to be nonzero or give a zero left side, again excluding a root with $κ > 0$ . Since ${B_{n} : n \geq 0}$ covers $(0, \infty)$ except the endpoints, these are all roots in $(0, K)$ ; the number of $n \geq 0$ with $\frac{n π}{2 a} < K$ is $⌈ 2 a K / π ⌉$ , which is (d).

3.1step 1.1step 2.1step 1.3step 2.2∎

[Assembly] By steps 1.2, 2.1 and 1.3 every negative eigenvalue has one of the two root forms, with simple eigenspace, and by step 2.2 the roots are exactly the described ones with the stated count. Steps 1.1, 1.2, 1.3, 2.1 and 2.2 give (a)-(d).

5 · Examples, experiments and counterexamples

None yet.