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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!

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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.


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.

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    That's marvellous - completely understand now! Than$k$ you v$e$ry much!2012-05-12