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Let $X$ be an infinite-dimensional Banach space, and $T$ a compact operator from $X$ to $X$. Why must $0$ then be a spectral value for $T$?

I believe this is equivalent to saying that $T$ is not bijective, but I am not sure how to show that injectivity implies the absence of surjectivity and the other way around (or if this is even the right way to approach the problem).

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If a bounded linear operator is invertible, the image of the closed ball of radius 1 contains an open ball. But if T is compact, the image of the closed ball of radius 1 is relatively compact. –  Qiaochu Yuan Nov 15 '10 at 1:09
a local compact topological vector space is finite dimensional. see rudin's functional analysis –  yaoxiao May 30 '12 at 3:04

1 Answer 1

up vote 10 down vote accepted

You are correct that it is the same as saying that $T$ is not bijective, because it follows from the open mapping theorem that a bounded operator on a Banach space has a bounded inverse if it is bijective. However, more straightforward answers for your question can be given without explicitly thinking in these terms.

You can show that if $T$ is compact and $S$ is bounded, then $ST$ is compact. If $T$ were invertible, this would imply that $I=T^{-1}T$ is compact. This in turn translates to saying that the closed unit ball of $X$ is compact. One way to see that this is impossible in the infinite dimensional case is implicit in this question. There is an infinite sequence of points in the unit ball whose pairwise distances are bounded below, no subsequence of which is Cauchy.

Compact operators on infinite dimensional spaces can be injective, but they can never be surjective. The closed subspaces of the range of a compact operator are finite dimensional.

Also, what Qiaochu said in his comment above.

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