Prove that the approximate point spectrum is a subset of the spectrum

Question

I'm trying to prove that if $\lambda$ an approximate eigenvalue of $T$ then $\lambda \in \sigma(T)$, but I can't work out how to do it. Could someone give me a hint, or point me in the direction of a resource that explains it (fully!).

Many thanks!

Answer 1

Approximate eigenvalues are those for which $ T - \lambda I$ is not bounded from below, i.e there is no $c>0$ such that $ \| T x \| \geq c \| x \| $ for all $x.$

A continuous linear operator is not invertible if it is not bounded from below, and since the spectrum consists of the $\lambda$ such that $ T- \lambda I$ is not injective, approximate eigenvalues lie in the spectrum.

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I assume you are dealing with Banach spaces. The Open mapping theorem gives that if $T$ is continuous, linear and surjective, then it is an open map. Thus, if $T$ is continuous, linear and bijective, then $T^{-1}$ is continuous, so bounded: There exists $C> 0$ such that $$ \| T^{-1} z \| \leq C \| z \| .$$

Replacing $z$ with $Tx$ then gives $$ \| T x\| \geq \frac{1}{C} \|x\|.$$ So every continuous invertible linear operator is bounded from below.

Answer 2

The proof can be made elementary. Assume $\lambda$ is an approximate eigenvalue, i.e. there is a sequence $x_n,$ $\|x_n\|=1$ and $\|(T-\lambda I)x_n\|\to 0.$ Assume by contradiction that $\lambda\notin\sigma(T).$ Then the operator $T-\lambda I$ is invertible. Thus $$1=\|x_n\|=\|(T-\lambda I)^{-1}[(T-\lambda I)x_n]\|\to 0$$ which gives a contradiction.