I have been reading some posts on here that I think are related such as this and this. I am still having a tough time coming up with a nice proof for my question.

Question: If a compact linear operator $T:X \rightarrow X$ on an infinite dimensional normed space $X$ has an inverse which is defined on all of $X$, show that the inverse cannot be bounded.

I found this in $8.3.8$ of Erwin Kreyszig functional analysis.


Suppose $T^{-1}$ existed and was bounded. Then $T^{-1}(B_X) \subseteq KB_X$ for some $K > 0$. Taking $T$ on both sides yields $$B_X = T(T^{-1}(B_X)) \subseteq T(KB_X) = KT(B_X).$$ Since $\overline{T(B_X)}$ is compact, it follows that $B_X$ is a closed subset of a compact set, and hence is compact, which implies $X$ is finite-dimensional.

  • $\begingroup$ @ Theo thank you for your answer. I like your proof, I am not familiar with the notation of $B_X$. Is that a bounded sequence in X? Thanks. $\endgroup$ – MathIsHard Jul 24 '18 at 2:23
  • $\begingroup$ Do you mind expanding why the closed subset of a compact set is compact implies finite-dimensional? This is a new topic for me... thank you. $\endgroup$ – MathIsHard Jul 24 '18 at 2:25
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    $\begingroup$ Sorry! $B_X$ refers to the closed unit ball (i.e. the set of points with norm less than or equal to $1$). Then $KB_X$ is the set of points with norm less than or equal to $K$. Being bounded means that $B_X$ maps to a bounded a set, i.e. a subset of $KB_X$ for some $K$. Compactness of a map means that $B_X$ maps to a set whose closure is compact. $\endgroup$ – Theo Bendit Jul 24 '18 at 2:30
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    $\begingroup$ @MathIsHard A standard result in topology states that if we have a closed subset of a compact set/space, then the subset is compact (this is true for sequentially compactness as well and is not hard to prove). Less easy to prove, a normed linear space is $X$ is finite-dimensional if and only if $B_X$ is compact. $\endgroup$ – Theo Bendit Jul 24 '18 at 2:32
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    $\begingroup$ @ Theo Bendit thank you very much. I really appreciate the help. $\endgroup$ – MathIsHard Jul 24 '18 at 2:33

It is known that the compact operators $\mathbb{K}(X)$ form a two-sided ideal in the algebra of bounded operators $\mathbb{B}(X)$.

So if $T \in \mathbb{K}(X)$ has an inverse $T^{-1} \in \mathbb{B}(X)$, we have

$$TT^{-1} = I$$

so it would follow that $I \in \mathbb{K}(X)$, and the identity map $I$ is not compact when $X$ is infinite-dimensional.

Therefore $T^{-1}$ cannot be bounded.

Actually, a compact operator $T \in \mathbb{K}(X)$ when $X$ is a Banach space cannot even be surjective, let alone bijective.

Indeed, a surjective bounded map $T \in \mathbb{B}(X)$ is open by the Open mapping theorem, so $T(B(0,1))$ is an open set. Therefore $T(\overline{B}(0,1))$ contains a closed ball which isn't compact in an infinite-dimensional space $X$. Hence $T(\overline{B}(0,1))$ cannot be precompact so $T$ is not compact.

  • $\begingroup$ the identity map I is not compact when X is infinite-dimensional. How to show this? $\endgroup$ – Lucas Jun 15 at 23:23
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    $\begingroup$ @Lucas A compact operator maps the unit ball of $X$ to a relatively compact set in $X$. The closed unit ball is not compact when $X$ is infinite-dimensional. $\endgroup$ – mechanodroid Jun 15 at 23:26

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