Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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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⊆ρ(T). More precisely, for z=a+ib with a,b∈R, b≠0, and every x∈D(T), ∥(T−z)x∥2=∥(T−a)x∥2+b2∥x∥2, and consequently ∥RT(z)∥≤1/∣Im⁡z∣. In particular σ(T)⊆R.

Facts & Assumptions

[A1]

T=T∗; thus D(T) is dense, T⊆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∈D(T) one has ⟨Tx,y⟩=⟨x,Ty⟩ whenever y∈D(T), and ⟨Tx,x⟩ is a real number: it equals ⟨x,Tx⟩=⟨Tx,x⟩‾ (Symmetric, self-adjoint and essentially self-adjoint operators, Real and complex inner-product spaces and their induced length).

[A3]

For z∈C the identity ran⁡(T−z)⊥=ker⁡(T∗−z‾)=ker⁡(T−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⊥⊥=M‾ (The double orthogonal complement of a subspace is its closure, Orthogonality and the orthogonal complement).

[A5]

z∈ρ(T) means that z−T is a bijection of D(T) onto H with bounded inverse, and then ∥RT(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+ib with b≠0.

1.1A2

For x∈D(T), expanding ∥(T−z)x∥2=⟨(T−z)x,(T−z)x⟩ gives ∥Tx∥2−z‾⟨Tx,x⟩−z⟨x,Tx⟩+∣z∣2∥x∥2; by [A2] the two middle terms combine to −2a⟨Tx,x⟩, so ∥(T−z)x∥2=∥Tx∥2−2a⟨Tx,x⟩+∣z∣2∥x∥2.

2.1step 1.1algebra

Adding and subtracting a2∥x∥2 and using ∣z∣2−a2=b2, step 1.1 becomes ∥(T−z)x∥2=∥(T−a)x∥2+b2∥x∥2.

3.1A1step 2.1

By step 2.1, ∥(T−z)x∥≥∣b∣ ∥x∥ for every x∈D(T); in particular T−z is injective and its range is closed: if (T−z)xn→y, then (xn) is Cauchy, hence xn→x for some x∈H and Txn→zx+y, and closedness of T gives x∈D(T) and Tx=zx+y, that is y=(T−z)x.

4.1A3A4step 2.1step 3.1

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

5.1A5step 2.1step 3.1step 4.1

By steps 3.1 and 4.1 the map z−T:D(T)→H is a bijection, and step 2.1 gives ∥(z−T)−1y∥≤∣b∣−1∥y∥ for every y∈H: applying step 2.1 to x=(z−T)−1y yields ∥y∥2=∥(T−a)x∥2+b2∥x∥2≥b2∥x∥2. Thus z∈ρ(T) and ∥RT(z)∥≤1/∣Im⁡z∣ by [A5].

6.1step 2.1step 5.1∎

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

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