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Brezis shows that a compact operator $T \colon E \to E$ ($E$ a normed vector space) satisfies $0 \in \sigma(T)$ by showing that if $0 \notin \sigma(T)$, then $T$ is bijective and hence $I = T \circ T^{-1}$ is compact, a contradiction to infinite dimensionality. How do I see that 'hence $T \circ T^{-1}$ is compact'?

EDIT: are there any related results when $T$ is only surjecive or injective?

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    Um... do you mean $T: E \to E$ ? – 2017-02-26
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    Also, what do you mean by $\sigma(T)$ ? – 2017-02-26
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    Sorry, $T \colon E \to E$, but $\sigma(T)$ is standard notation for the spectrum... – 2017-02-26

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This follows from the more general result

If $T,S:E\to E$ are bounded linear maps and $T$ is compact, then $TS$ is compact

Indeed, since $S$ is bounded it maps bounded sets in $E$ to bounded sets in $E$. Since $T$ is compact, it maps bounded sets in $E$ to precompact sets in $E$. Thus $TS$ maps bounded sets in $E$ to precompact sets in $E$, and therefore $TS$ is compact.

Now assuming $T$ is compact and $0\notin\sigma(T)$, it has a bounded inverse $T^{-1}$ and applying the above result with $S=T^{-1}$, we see that $I=TT^{-1}$ is compact.

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    Wonderful, thank you. I came up with that proof but was a bit surprised it required the bounded inverse theorem (which is only stated as a corollary and not used again until this chapter) – 2017-02-26
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    Would it be possible to conclude that if $TS = \mathrm{id}$, then neither $T$ nor $S$ is compact? I can see how to use the result in your post to show that $T$ is not compact, but I can't see why $S$ is not compact. – 2017-02-27
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    @user321210 through similar reasoning: Suppose $T$ is bounded, and $S$ is comapct. Thus $S$ maps bounded sets to precompact sets, and $T$ maps precompact sets to precompact sets. So in this case, $TS$ maps bounded sets to precompact sets, and hence is compact. – 2017-02-27