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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).

The twisted-periodic momentum operator on an interval

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

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