I have a question. I have Banach spaces $X$ and $Y$, and $Y$ is reflexive. If $T:X\to Y$ is continuous, and $T^{t}:Y^{*}\to X^{*}, T^{t}(\phi)=\phi \circ T$ is compact, is it true that $T$ is compact?
Thanks so much
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Yes, if $T\colon X \to Y$ with Banach spaces $X$ and $Y$, then $T$ is compact if and only if $T^t$ is compact. No reflexivity or anything fancy needed. – Daniel Fischer Aug 12 '13 at 00:43
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an idea to show it – user89940 Aug 12 '13 at 00:47
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Sure, give me a few minutes to write it up. – Daniel Fischer Aug 12 '13 at 00:48
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Or what David posted. Would you then still need an answer here? – Daniel Fischer Aug 12 '13 at 00:51
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@DanielFischer I just noticed that post only proved the other implication. – David Mitra Aug 12 '13 at 00:51
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@DavidMitra But that means we know $T^{tt} \colon X^{\ast\ast} \to Y^{\ast\ast}$ is compact, and that means that $T(B_X)$ is totally bounded (as a subset of a totally bounded set, $T^{tt}(B_{X^{\ast\ast}})$. – Daniel Fischer Aug 12 '13 at 00:54
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@DanielFischer Yes. I was just pointing out my (deleted) link does not give the solution to the OP's problem. – David Mitra Aug 12 '13 at 00:58
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@Daniel Fischer yes, any idea, only, thanks – user89940 Aug 12 '13 at 01:02
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@user89940 Sorry, I don't understand. Does that mean you want an answer here, or the other post together with the comment above is enough? – Daniel Fischer Aug 12 '13 at 01:03
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Here is an alternative proof (I haven't checked it). Although I suggest you first prove that if $T$ is compact,then so is $T^*$ (as in the previous link): then use Daniel's argument. – David Mitra Aug 12 '13 at 01:05
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well, if $B$ is unit closed ball in $X$. i wanna show that $\overline{T(B)}$ is compact in $Y$, and what can i do to use hypothesis of $T^{t}$ is compact. thanks – user89940 Aug 12 '13 at 01:52
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This is known as Schauder's theorem and can be found in most functional analysis textbooks. – Nate Eldredge Aug 12 '13 at 02:07
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@Nate Eldredge thanks now i find it – user89940 Aug 12 '13 at 02:11
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Just to save this answer from being unanswered
1) See this answer for implication $T \mbox{ compact }\implies T^* \mbox{ compact}$
2) For reverse implcation note that from 1) it follows that $T^{**}$ is compact. By $i_E$ we denote isometric embedding of $E$ into its the double dual, then it is easy to check that $i_Y T=T^{**} i_X$. Since $T^{**}$ is compact so does $ T^{**}i_X$ and $i_Y T$. Since $i_Y$ is isometric this implies that $T$ is compact.
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Could you spell out the details as to why $i_Y$ being isometric implies that $T$ is compact, please? – fish_monster Dec 06 '21 at 14:37
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1@maths_student, $T$ is compact if $T(B_X)$ is totally bounded. The notion of total boundedness is invariant under maps that preserve distance. – Norbert Dec 06 '21 at 20:23