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As an exercise I tried to prove the following theorem:

If $X,Y$ are Banach spaces and $u \in B(X,Y)$ is a bounded linear operator then the following are equivalent:

(1) $u$ is compact

(2) for every bounded set $S \subseteq X$ the image $u(S)$ is relatively compact

(3) if $x_n$ is a bounded sequence in $X$ then $u(x_n)$ admits a convergent subsequence in $Y$

Please could someone check my proof?

Proof:

Recall the definition of compact operator: $u$ is compact iff if the image of the unit ball is relatively compact.

$(1) \iff (2)$: Since multiplication by $n$ is a linear homeomorphism, $u$ is compact iff $u(B(0,1))$ is relatively compact iff $u(B(0,n))$ is relatively compact. From this it is obvious that $(1) \iff (2)$.

$(2) \implies (3)$: Let $x_n$ be a bounded sequence. Then $S=\{x_n\}_{n \in \mathbb N}$ is a bounded set hence $u(S)$ is relatively compact hence $u(x_n)$ admits a subsequence that converges in $Y$.

$(2) \Longleftarrow (3)$: Let $S$ be a bounded set. Let $x_n$ be any sequence in $\overline{u(S)}$. Then $x_n=u(s_n)$ for some sequence $s_n$ in $S$. Then $x_n$ admits a convergent subsequence.

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    $\begingroup$ there is a problem with $3)\implies 2)$. If $x\in \overline{u(S)}$ it doesn't mean that there exist $s\in S$ such that $x=u(x)$. $\endgroup$
    – Norbert
    Apr 20, 2014 at 22:37
  • $\begingroup$ @Norbert Thank you, why didn't you post your comment as an answer? If you do I have an answer to which I can award the bounty. $\endgroup$
    – student
    Apr 21, 2014 at 7:09
  • $\begingroup$ ok, soon I'll do it $\endgroup$
    – Norbert
    Apr 21, 2014 at 7:44
  • $\begingroup$ I went over my argument for $(1)$ iff $(2)$ and I did not find any mistakes. I would like to reiterate: an operator is compact if and only if the image of the unit ball is relatively compact. Multiplication by $\lambda\in \mathbb C$ is a linear homeomorphism and if $f$ is a homeomorphism then $f(\overline{A})= \overline{f(A)}$ for all sets $A$. $$\\$$ If $u$ is compact then $\overline{u(B(0,1))}$ is compact and then $n \overline{u(B(0,1))} = \overline{u(B(0,n))}$ is compact. $\endgroup$
    – student
    Apr 21, 2014 at 8:33
  • $\begingroup$ On the other hand, replacing $n$ with ${1 \over n}$ yields that $\overline{u(B(0,1))}$ is compact if $\overline{u(B(0,n))}$ is. Hence $(1) \iff (2)$. $\endgroup$
    – student
    Apr 21, 2014 at 8:33

2 Answers 2

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The only problematic part of your proof is the implication $3)\implies 2)$. You should argue as follows

Assume $S$ is bounded. Consider arbitrary $(y_n)_{n\in\mathbb{N}}\subset\overline{u(S)}$, then $(y_n)_{n\in\mathbb{N}}$ is also bounded. For each $n\in\mathbb{N}$ we can find $x_n\in S$ such that $$ \Vert y_n-u(x_n)\Vert\leq 2^{-n}.\tag{*} $$ Since $(y_n)_{n\in\mathbb{N}}$ is bounded, so does $(u(x_n))_{n\in\mathbb{N}}$. Since $3)$ holds, we have convergent subsequence $u(x_{n_k})_{k\in\mathbb{N}}$. From $(*)$ is follows that $(y_{n_k})_{k\in\mathbb{N}}$ converges to the same limit as $u(x_{n_k})_{k\in\mathbb{N}}$. Thus we constructed convergent subsequence of arbitrary sequence in $\overline{u(S)}$. So $u(S)$ is relatively compact.

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  • $\begingroup$ Yes!! That's what I thought! Sorry I only saw your answer after I posted the last comment above. $\endgroup$
    – student
    Apr 21, 2014 at 9:13
  • $\begingroup$ No problem.${}{}{}$ $\endgroup$
    – Norbert
    Apr 21, 2014 at 9:16
  • $\begingroup$ Is there a reason why you chose $\varepsilon_n = 2^{-1}$? I believe the argument also works if $\varepsilon_n = {1\over n}$. $\endgroup$
    – student
    Apr 21, 2014 at 10:17
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    $\begingroup$ you can choose any $\varepsilon_n\to 0$ $\endgroup$
    – Norbert
    Apr 21, 2014 at 10:26
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    $\begingroup$ @Norbert Hi, I think we don’t need to point out the boundedness of $\{u(x_n)\}$ in the proof, since the condition posted on $(3)$ is merely the boundedness of $\{x_n\}$. $\endgroup$
    – Sam Wong
    Jun 27, 2018 at 7:18
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The definition of a compact operator $u \in B(X,Y)$ is that $u$ maps bounded sets of $X$ into relatively compact sets of Y. Therefore, (1) and (2) are equivalent by definition and no proof is required. This definition is equivalent to saying if $x_n$ is a bounded sequence in $X$, then $u(x_n)$ has a Cauchy subsequence (not necessarily convergent) in $Y$. To prove (3) $\Rightarrow$ (2), let $S$ be a bounded set in $X$. For any bounded sequence $x_n$ in $S$ there exists a Cauchy subsequence $u(x_{n_k}) \in u(S)$. Hence $u(S)$ is relatively compact.

Remember that relatively compactness means closure is compact. There may not exist a convergent subsequence (only definitely Cauchy).

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  • $\begingroup$ No the definition of compact operator is that if $B$ is the unit ball then $u(B)$ is relatively compact. $\endgroup$
    – student
    Apr 19, 2014 at 10:54
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    $\begingroup$ These are two equivalent definitions of a compact operator. $\endgroup$
    – user143921
    Apr 19, 2014 at 11:51
  • $\begingroup$ You are assuming what I want to show (that 1 and 2 are equivalent). $\endgroup$
    – student
    Apr 20, 2014 at 7:06

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