Let $X$ and $Y$ be normed spaces and $X$ compact. If $T: X\to Y$ is a bijective closed linear operator, show that $T^{-1}$ is bounded.

I don't know where to start here. Any help would be appreciated.

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    $\begingroup$ A normed space is never compact. What do you mean by $X$ compact? $\endgroup$ – DisintegratingByParts Dec 10 '14 at 6:09
  • $\begingroup$ $X$ is compact w.r.t the metric induced by the norm @TrialAndError $\endgroup$ – Learnmore Mar 3 '17 at 6:38
  • $\begingroup$ Then $X=\{0\}$. $\endgroup$ – DisintegratingByParts Mar 3 '17 at 11:20
  • $\begingroup$ Kreyszig gives this problem in Page 296 where he refers to $X$ as compact $\endgroup$ – Learnmore Mar 3 '17 at 11:23
  • $\begingroup$ @learnmore : If $X$ is compact and $x \in X$ with $x\ne 0$, then $\{ x,2x,3x,4x,\cdots \}$ must have a convergent subsequence, which leads to a contradiction. $\endgroup$ – DisintegratingByParts Mar 3 '17 at 11:24

To show that $T^{-1}$ is bounded it is enough to show that $T^{-1}$ is continuous.

Now $T^{-1}:Y\to X$

Select a closed set $C\subseteq X$ then since $C$ is closed hence it is compact .

Since $T$ is a closed linear operator so image of a compact set under $T$ is closed.

Hence $(T^{-1})^{-1}(C)=T(C)$ is closed .

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    $\begingroup$ 1. The only compact normed space is the null space, so the question makes no sense. 2. By definition, a closed linear operator has closed graph, it does not necessarily send compact sets to closed sets. For example, if $T$ is unbounded, there is a null sequence that is sent to an unbounded sequence. $\endgroup$ – user160629 Mar 3 '17 at 7:09
  • $\begingroup$ Well, yes, I do :) 1. I pointed out that there are no interesting compact normed spaces: if a normed space is non-zero, it contains $\mathbb{R}$ as a closed subspace, hence is non-compact. 2. I argued that closed operators don't necessarily send compact sets to compact sets, hence your penultimate sentence is iffy. I believe that you confuse the notion of a closed operator with that of a closed map. $\endgroup$ – user160629 Mar 3 '17 at 9:52
  • $\begingroup$ I never said closed operators take compact sets to compact sets ;A set is compact w.r.t the metric induced by the norm $\endgroup$ – Learnmore Mar 3 '17 at 11:19
  • $\begingroup$ It's useless to argue here.In case you want to learn refer to Kreyszig Page 296 $\endgroup$ – Learnmore Mar 3 '17 at 11:20
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    $\begingroup$ So your point is that this is a verbatim exercise from a book. Fine, fair enough. Then the solution is simple. Since $X$ is a compact normed space, $X = 0$ and since $T$ is bijective, $Y = 0$. Done, and no need to use that $T$ is closed, so I still think the exercise makes no sense. There's no need to argue (as you claim) that closed operators map compact sets to closed sets (and I still don't think that's true). $\endgroup$ – user160629 Mar 3 '17 at 11:44

Select a compact set $C \subset X$ (which is closed and bounded). Therefore the preimage of $T^{-1}$ is also compact because $(T^{-1})^{-1}(C) = T(C) \subset Y$. Why? Because the preimage of $T$, $T^{-1}T(C) = C$ since $T$ is bijective, so the set $(T^{-1})^{-1}(C)$ is also compact (which is closed and bounded) meaning $T^{-1}$ is continuous, so $T^{-1}$ must be bounded.

I think there is a better argument where you can use the definition of being a closed linear operator, get continuity out of that, and then apply compactness to this.


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