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Let $A : X\to X$ be a bounded operator between Banach spaces and $K : X\to X$ be a compact operator. It is true that if $A(X) \subset K(X)$, then $A$ is also compact. In fact, it was already asked it here: Related.

The point is that some people argue on the comment section that there is a simple proof using open mapping Theorem, but I was not able to complete all the details. The possible proof would follow as:

Consider $A : X\to K(X)$ where $K(X)$ is endowed with the norm $$\|y\| = \inf\{\|x\| : Kx = y\}.$$

The first claim is that $K(X)$ is Banach with such norm. In fact, I made some progress on it, but could not finish: for instance, take $\{y_n\}$ a Cauchy sequence on $K(X)$ with this norm. The, the set $\{x_n\}$ of vectors that realize the norms of each $y_n$ is bounded, since $\{y_n\}$ is Cauchy. But $K$ is compact, which implies that $y_n$ has a convergent subsequence on the norm of $X$. But how can I guarantee that this sequence also converges on the norm I have defined? For instance, if $\{y_{n_k}\}$ is such subsequence, if somehow we could prove that the limit $y^*$ lies on the range of $K$, this would follow easily from the definition of $\|\cdot\|$. I could not finish.

The final step is: use that $K(X)$ is Banach with such norm to prove that $A(B) \subset cK(B)$ for some $c>0$ where $B$ is the unit ball on $X$. This would follow from the Open Mapping Theorem, however, $A$ as we defined is not surjective. How to proceed?

L.F. Cavenaghi
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    To see that $K(X)$ is a Banach space under that norm, note that this norm gives an isometry between $K(X)$ and $X/\ker K$. – Daniel Fischer May 20 '20 at 17:36
  • @DanielFischer, I see, I understand the intuition behind it, but I could not justify by myself. Could you please add some details? – L.F. Cavenaghi May 20 '20 at 17:45
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    Maybe I'm misunderstanding, but there isn't much to add. The coset $\xi + \ker K$ is the set ${ x : K(x) = K(\xi)}$, and the quotient norm is $\lVert \xi + \ker K\rVert = \inf { \lVert x\rVert : x \in \xi + \ker K}$. Then you use the theorem that a quotient of a Banach space modulo a closed subspace is complete with respect to the quotient norm. – Daniel Fischer May 20 '20 at 17:51
  • that is precisely what I was not seeing. Sorry to bother and thanks. Please, could you give add some details to the open mapping part? I would be very happy to understand it. – L.F. Cavenaghi May 20 '20 at 17:53
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    Over there, Jochen spoke of the closed graph theorem. I don't immediately see how that would give a shorter argument (I know how to make a shorter version of Etienne's argument, though). As for the open mapping theorem, nothing jumps out. – Daniel Fischer May 20 '20 at 18:02
  • @DanielFischer yes, he says about the closed map theorem, but I also don't see how to do it, I really thought that maybe he was referring to the open mapping theorem, but I don't know how. – L.F. Cavenaghi May 20 '20 at 18:05
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    The desired conclusion that $A(B) \subset c\cdot K(B)$ for some constant $c$ is equivalent to the continuity of $A$ when we endow $K(X)$ with the quotient norm. By the closed graph theorem, this follows if the graph of $A$ is closed. So consider a sequence $(x_n)$ in $X$ such that i) $x_n \to x$ with respect to the norm on $X$, and ii) $A(x_n) \to y$ with respect to the quotient norm on $K(X)$. We must conclude that $y = A(x)$, then it follows that the graph is closed. Now by the continuity of $A$ with respect to the norm of $Y$ it follows that $A(x_n) \to A(x)$ in $Y$. – Daniel Fischer May 20 '20 at 18:55
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    Since the quotient norm induces a topology on $K(X)$ that is finer than the topology induced by the norm of $Y$ it follows that $A(x_n)$ cannot converge to anything else but $A(x)$ with respect to the quotient norm, and we're done. – Daniel Fischer May 20 '20 at 18:55
  • outstanding. I don't know how to thank you. – L.F. Cavenaghi May 20 '20 at 18:56

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