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Definition: A linear operator $T: V \to W$ is compact if and only if the image of the unit ball in $V$ is precompact (= every sequence has a cauchy subsequence $\iff $ totally bounded).

Prove: Let $T: V \to W$ be a compact linear operator. Show that $T$ is bounded.

My attempt:

Suppose $T$ is not bounded. Then,

$$\forall M > 0: \exists v_M \in V: \Vert T v_M \Vert > M \Vert v_M \Vert$$

I then tried to construct a sequence without cauchy sequence in $\{Tv \mid \Vert v \Vert \leq 1\}$

So, let $p > q$. Then, $$\left\Vert T \frac{v_p}{\Vert v_p \Vert} - T \frac{v_q}{\Vert v_q \Vert}\right\Vert \geq \left|\frac{\Vert Tv_p \Vert}{\Vert v_p \Vert} - \frac{\Vert Tv_q \Vert}{\Vert v_q \Vert}\right|$$

but was unable to conlude something because off the minus sign.

Any ideas?

EDIT: This is not a duplicate, as other posts use other definitions of compact operators.

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    $\begingroup$ can you write your definition correctly. You can not have iff in the definition $\endgroup$
    – user537667
    Commented May 31, 2018 at 10:14
  • $\begingroup$ Of course you can have iff in a definition. What is that for nonsense? $\endgroup$
    – user370967
    Commented May 31, 2018 at 10:31
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    $\begingroup$ What do you mean by "What is that for nonsense?" $\endgroup$
    – user537667
    Commented May 31, 2018 at 10:34
  • $\begingroup$ Every definition works in 2 directions. It's nonsense that you can't have an iff in a definition. $\endgroup$
    – user370967
    Commented May 31, 2018 at 10:37
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    $\begingroup$ That is not intended to be a sarcasm. It is just the culture in your place. We do not use iff. We say $A$ iff $B$ if $A$ and $B$ both have a meaning already.. if we are giving meaning for some word, which means there is no meaning for that word already, we use if... We call a map $T:V\rightarrow W$ a linear operator if (** something happens).... We do not write we call a map $T:V\rightarrow W$ linear if and only if (** something happens).. It is just a culture.. $\endgroup$
    – user537667
    Commented May 31, 2018 at 10:45

2 Answers 2

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It follows directly from the definitions. Since $T$ is compact, the image $T(V_1)\subset W$ of the closed unit ball of $V$ is precompact. Consider the cover $W\subset \bigcup_n W_n$, where $W_n$ is the ball of radius $n$. As $\overline{T(V_1)}$ is compact, the cover has a finite subcover, which means that there exists $m$ with $\overline{T(V_1)}\subset W_m$. In particular, $\|Tv\|\leq m$ for all $v\in V_1$, which leads to $$ \|Tv\|\leq m\|v\|,\ \ \ v\in V.$$

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  • $\begingroup$ Why is the image compact? I don't work in a Banach space! $\endgroup$
    – user370967
    Commented Jun 1, 2018 at 7:58
  • $\begingroup$ Take the closure, then. $\endgroup$ Commented Jun 1, 2018 at 16:30
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If $T$ is unbounded, there is a sequence $(x_n)_{n\in\mathbb N}$ of elements of the unit ball in $V$ such that$$(\forall n\in\mathbb N):\bigl\|T(x_n)\bigr\|>n.$$Therefore, the sequence $\bigl(\|T(x_n)\|\bigr)_{n\in\mathbb N}$ is not a Cauchy sequence.

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    $\begingroup$ But maybe it has a cauchy subsequence? $\endgroup$
    – user370967
    Commented May 31, 2018 at 10:00
  • $\begingroup$ I've edited my answer. I hope that everything is clear now. $\endgroup$ Commented May 31, 2018 at 10:03
  • $\begingroup$ Still I don't see how this implies that sequence has no Cauchy subsequence. Sorry. $\endgroup$
    – user370967
    Commented May 31, 2018 at 10:06
  • $\begingroup$ Because the sequence has no bounded subsequence and therefore no Cauchy subsequence. $\endgroup$ Commented May 31, 2018 at 10:20

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