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Resolvent of a self-adjoint operator: nonreal resolvents and the estimate

Statement

Assume Countable Choice. Let $T$ be a self-adjoint operator on $H$ . Then every nonreal number belongs to $ρ (T)$ : $C ∖ R \subseteq ρ (T)$ . More precisely, for $z = a + i b$ with $a, b \in R$ , $b \neq 0$ , and every $x \in D (T)$ , $∥ (T - z) x ∥^{2} = ∥ (T - a) x ∥^{2} + b^{2} ∥ x ∥^{2},$ and consequently $∥ R_{T} (z) ∥ \leq 1 / ∣ Im z ∣$ . In particular $σ (T) \subseteq R$ .

Facts & Assumptions

[A1]

$T = T^{*}$ ; thus $D (T)$ is dense, $T \subseteq T^{*}$ , and $T$ is closed, $T^{*}$ being closed (Symmetric, self-adjoint and essentially self-adjoint operators, The adjoint is well defined, closed, and reverses inclusions).

[A2]

For $x \in D (T)$ one has $⟨ T x, y ⟩ = ⟨ x, T y ⟩$ whenever $y \in D (T)$ , and $⟨ T x, x ⟩$ is a real number: it equals $⟨ x, T x ⟩ = \overline{⟨ T x, x ⟩}$ (Symmetric, self-adjoint and essentially self-adjoint operators, Real and complex inner-product spaces and their induced length).

[A3]

For $z \in C$ the identity $ran (T - z)^{⊥} = \ker (T^{*} - \overline{z}) = \ker (T - \overline{z})$ holds (The adjoint is well defined, closed, and reverses inclusions, [A1]).

[A4]

If $M$ is a linear subspace of a Hilbert space, then $M^{⊥ ⊥} = \overline{M}$ (The double orthogonal complement of a subspace is its closure, Orthogonality and the orthogonal complement).

[A5]

$z \in ρ (T)$ means that $z - T$ is a bijection of $D (T)$ onto $H$ with bounded inverse, and then $∥ R_{T} (z) ∥$ is the operator norm of that inverse (Resolvent and spectrum of an unbounded operator, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

Proof

technique · direct

Given: Countable Choice, a self-adjoint operator $T$ on $H$ , and a number $z = a + i b$ with $b \neq 0$ .

1.1

For $x \in D (T)$ , expanding $∥ (T - z) x ∥^{2} = ⟨ (T - z) x, (T - z) x ⟩$ gives $∥ T x ∥^{2} - \overline{z} ⟨ T x, x ⟩ - z ⟨ x, T x ⟩ + ∣ z ∣^{2} ∥ x ∥^{2}$ ; by [A2] the two middle terms combine to $- 2 a ⟨ T x, x ⟩$ , so $∥ (T - z) x ∥^{2} = ∥ T x ∥^{2} - 2 a ⟨ T x, x ⟩ + ∣ z ∣^{2} ∥ x ∥^{2}$ .

2.1

Adding and subtracting $a^{2} ∥ x ∥^{2}$ and using $∣ z ∣^{2} - a^{2} = b^{2}$ , step 1.1 becomes $∥ (T - z) x ∥^{2} = ∥ (T - a) x ∥^{2} + b^{2} ∥ x ∥^{2}$ .

step 1.1algebra

3.1

By step 2.1, $∥ (T - z) x ∥ \geq ∣ b ∣ ∥ x ∥$ for every $x \in D (T)$ ; in particular $T - z$ is injective and its range is closed: if $(T - z) x_{n} \to y$ , then $(x_{n})$ is Cauchy, hence $x_{n} \to x$ for some $x \in H$ and $T x_{n} \to z x + y$ , and closedness of $T$ gives $x \in D (T)$ and $T x = z x + y$ , that is $y = (T - z) x$ .

A1step 2.1

4.1

Also by [A3] applied to $z$ , $ran (T - z)^{⊥} = \ker (T - \overline{z})$ , and step 2.1 with $z$ replaced by $\overline{z}$ shows $T - \overline{z}$ is injective, so the kernel is ${0}$ . Hence the closed range of step 3.1 satisfies $ran (T - z) = \overline{ran (T - z)} = (ran (T - z)^{⊥})^{⊥} = H$ .

A3A4step 2.1step 3.1

5.1

By steps 3.1 and 4.1 the map $z - T : D (T) \to H$ is a bijection, and step 2.1 gives $∥ (z - T)^{- 1} y ∥ \leq ∣ b ∣^{- 1} ∥ y ∥$ for every $y \in H$ : applying step 2.1 to $x = (z - T)^{- 1} y$ yields $∥ y ∥^{2} = ∥ (T - a) x ∥^{2} + b^{2} ∥ x ∥^{2} \geq b^{2} ∥ x ∥^{2}$ . Thus $z \in ρ (T)$ and $∥ R_{T} (z) ∥ \leq 1 / ∣ Im z ∣$ by [A5].

A5step 2.1step 3.1step 4.1

6.1

Since $z$ was an arbitrary nonreal number, $C ∖ R \subseteq ρ (T)$ , that is, $σ (T) \subseteq R$ ; the identity and the bound of the statement are steps 2.1 and 5.1.

step 2.1step 5.1∎

Depends on

Used by

Unitary groups converge under strong resolvent convergence Corollary
Cayley transform of a self-adjoint operator Definition
Norm and strong resolvent convergence Definition
Relative compactness with respect to an operator Definition
The generator of a unitary group is closed and skew-adjoint Lemma
The resolvent star algebra is dense in C₀(R) Lemma
Cayley correspondence between self-adjoint operators and unitaries Theorem
Continuous functional calculus under resolvent convergence Theorem
Kato-Rellich theorem Theorem
Range criterion for self-adjointness Theorem

Dependency tree · two levels

34 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Dana P. Williams, Lecture Notes on the Spectral Theorem (standard reference, not scraped)
Gerald Teschl, Mathematical Methods in Quantum Mechanics, second edition (standard reference, not scraped)