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Measure, Integration & Real Analysis, 2021a

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296 Chapter 10 Linear Maps on Hilbert Spaces<br />

10.36 spectrum is closed<br />

The spectrum of a bounded operator on a Banach space is a closed subset of F.<br />

Proof Suppose T is a bounded operator on a Banach space V. Suppose α 1 , α 2 ,...<br />

is a sequence in sp(T) that converges to some α ∈ F. Thus each T − α n I is not<br />

invertible and<br />

lim<br />

n→∞ (T − α nI) =T − αI.<br />

The set of noninvertible elements of B(V) is a closed subset of B(V) (by 10.25).<br />

Hence the equation above implies that T − αI is not invertible. In other words,<br />

α ∈ sp(T), which implies that sp(T) is closed.<br />

Our next result provides the key tool used in proving that the spectrum of a<br />

bounded operator on a nonzero complex Hilbert space is nonempty (see 10.38). The<br />

statement of the next result and the proofs of the next two results use a bit of basic<br />

complex analysis. Because sp(T) is a closed subset of C (by 10.36), C \ sp(T) is<br />

an open subset of C and thus it makes sense to ask whether the function in the result<br />

below is analytic.<br />

To keep things simple, the next two results are stated for complex Hilbert spaces.<br />

See Exercise 6 for the analogous results for complex Banach spaces.<br />

10.37 analyticity of (T − αI) −1<br />

Suppose T is a bounded operator on a complex Hilbert space V. Then the function<br />

α ↦→ 〈 (T − αI) −1 f , g 〉<br />

is analytic on C \ sp(T) for every f , g ∈ V.<br />

Proof Suppose β ∈ C \ sp(T). Then for α ∈ C with |α − β| < 1/‖(T − βI) −1 ‖,<br />

we see from 10.22 that I − (α − β)(T − βI) −1 is invertible and<br />

(<br />

I − (α − β)(T − βI)<br />

−1 ) ∞<br />

−1 = ∑ (α − β) k( (T − βI) −1) k .<br />

k=0<br />

Multiplying both sides of the equation above by (T − βI) −1 and using the equation<br />

A −1 B −1 =(BA) −1 for invertible operators A and B,weget<br />

(T − αI) −1 =<br />

Thus for f , g ∈ V,wehave<br />

〈<br />

(T − αI) −1 f , g 〉 =<br />

∞<br />

∑<br />

k=0<br />

∞<br />

∑<br />

k=0<br />

(α − β) k( (T − βI) −1) k+1 .<br />

〈 ((T − βI)<br />

−1 ) 〉<br />

k+1 f , g (α − β) k .<br />

The equation above shows that the function α ↦→ 〈 (T − αI) −1 f , g 〉 has a power series<br />

expansion as powers of α − β for α near β. Thus this function is analytic near β.<br />

<strong>Measure</strong>, <strong>Integration</strong> & <strong>Real</strong> <strong>Analysis</strong>, by Sheldon Axler

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