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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).

The free-line resolvent kernel and its inverse operator

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

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