Condition for Nonzero Eigenvalue of Compact Operator

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Theorem

Let $H$ be a Hilbert space over $\Bbb F \in \set {\R, \C}$.

Let $T \in \map {B_0} H$ be a compact operator.

Let $\lambda \in \Bbb F, \lambda \ne 0$ be a nonzero scalar.


Suppose that the following holds:

$\inf \set {\norm {\paren {T - \lambda I} h}_H: \norm h_H = 1} = 0$

Then $\lambda \in \map {\sigma_p} T$, that is, $\lambda$ is an eigenvalue for $T$.


Corollary

Suppose $\lambda \notin \map {\sigma_p} T$ and $\bar \lambda \notin \map {\sigma_p} {T^*}$.


Then $T - \lambda I$ is invertible.

Furthermore, $\paren {T - \lambda I}^{-1}$ is a bounded linear operator.


Proof


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By hypothesis, there is a sequence of unit vectors $\sequence{h_n}$ such that $\cmod{(T-\lambda) h_n} \to 0$.

Since $T$ is compact, by definition 2 there is a $f\in H$ and a subsequence $\sequence{h_{n_k}}$ such that $T h_{n_k}\to f$.

But $h_{n_k}=\lambda^{-1}\left[(\lambda-T) h_{n_k}+T h_{n_k}\right] \to\lambda^{-1} f$. So $1=\norm{\lambda^{-1} f}=\cmod\lambda^{-1}\norm{f}$ and $f \neq 0$.

Also, it must be that $T h_{n_k} \to \lambda^{-1} T f$. Since $T h_{n_k} \to f, f=\lambda^{-1} T f$, or $T f=\lambda f$. That is, $f \in \ker(T-\lambda)$ and $f \neq 0$, so $\lambda \in \sigma_p(T)$.

Sources

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