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I am trying to prove this question, which from the book Foundations of Modern Analysis by Avner Friedman pp189 5.1.7. the question is below.

Let T be a compact linear operator from a Banach space X onto itself. If $\mathbf{T^{-1}}$ is a bounded operator, then X is finite-dimensional.

I think this question is not right, because if T is a compact operator then T is a bounded operator and due to the open-mapping theorem, $\mathbf{T^{-1}}$ must be bounded. It seems that this question just tells us that T is one-one and onto.

Cherryblossoms
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Cheburashka
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  • https://math.stackexchange.com/questions/74063/compact-operators-on-an-infinite-dimensional-banach-space-cannot-be-surjective?rq=1 – Mike F Mar 23 '21 at 15:36

1 Answers1

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Two key facts here.

$(1)$ The compact operators form an ideal.

$(2)$ The identity operator on a space $X$ is compact if and only if $X$ has finite-dimension.

By your assumption, $T \circ T^{-1}= 1_X$ and $T$ is compact. Because of $(1)$, we see that $1_X$ is compact. By $(2)$, $X$ has finite-dimension.