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Note this is a homework problem so I am looking for a hint not a solution:

For normed linear spaces $X$ and $Y$, I'm trying to show that $K(X,Y)$, the set of compact operators $X\to Y$ is a closed subset of $B(X,Y)$ the set of bounded operators $X\to Y$.

At first I thought it might be similar to showing that $c_{0}$ is a closed subspace of $c$. But the standard argument for that (if I am not mistaken) relies on the fact that the scalar field is complete.

Note: It turns out that $Y$ must be complete in order for the result to be true.

I start by assuming $f_{n}\in K(X,Y)$ is compact, and that $f_{n}\to f$ for some $f\in B(X,Y)$.

I want to show $f\in K(X,Y)$ using the criterion that for every sequence $x_{n}\in B_{X}$, $f(x_{n})$ has a convergent subsequence.

For each $m \geq 1$, by the compactness of $f_{m}$, there is a subsequence $x_{n_{k}}$ such that $f_{m}(x_{n_{k}})$ is convergent to some value in $Y$, say $y_{m}$.

If the same subsequence served as an appropriate witness for each $m\geq 1$, I think I may be able to get somewhere by changing orders of limits using an upper bound for the sequence $(f_{n})$. But I doubt this is the case, and thus I am stuck.

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I think it would be easier to show that $f(B_X)$ is totally bounded. To do this, let $\epsilon>0$, then choose $N$ so that $\Vert f_N-f\Vert<\epsilon$. Choose a finite $\epsilon$-net, $\{f_N(x_1),\ldots , f_N(x_n)\}$ for $f_N(B_X)$. Then show this is a $3\epsilon$-net for $f(B_X)$. –  David Mitra Jan 10 '13 at 18:12
Thank you. I will try this. –  Kyle Schlitt Jan 10 '13 at 18:26
I didnt use the concept of totally bounded, but I applied your choice of $N$ and immediately got the convergent subsequence I require. Thanks again! –  Kyle Schlitt Jan 10 '13 at 18:47
If you define a compact operator as one that maps bounded sets to sets with compact closure, then I'm not sure this is true. With $Y$ Banach, this is a standard result. But for arbitrary $Y$ I don't see how to get anything stronger than the result that $f$ maps bounded sets to totally bounded sets. What is your definition of a compact operator? –  David Mitra Jan 10 '13 at 21:25
As you suggested, in my course an operator $T:X\to Y$ is compact if $\overline{T(B_{X})}$ is norm compact in $Y$. The issue you mention is actually why I avoided the totally bounded approach. It seems easy to go from compact to totally bounded, but not the other way around. Some google-browsing reveals that the other direction definitely holds in metric spaces but these facts are not available to me. –  Kyle Schlitt Jan 10 '13 at 21:30
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2 Answers

up vote 1 down vote accepted

A sequential argument that works when $Y$ is a Banach space is given below:

Hint: We take take subsequences of subsequences and diagonalize.

Below are the details:

Let $X_1=(x_n)$ be a sequence in $B_X$. Choose a subsequence $X_2=(x_n^1)_n$ of $X_1$ such that $f_1(x_n^1)$ is convergent. Now choose a subsequence $X_3=(x_n^2)_n$ of $X_2$ such that $f_2(x_n^2)$ converges. Continue in this manner...

We thus have subsequences $$ (x_n^1)\supset (x_n^2)\supset (x_n^3)\supset\cdots $$so that $(f_m x_n^m)_n$ is convergent for each $m$.

Now set $y_n=x_n^n$. Then $(f_m y_n)_n$ is a convergent sequence for each $m$.

We also have $$\eqalign{\Vert f(y_n)-f(y_l)\Vert &\le\Vert (f-f_m)(y_n)\Vert +\Vert f_m(y_n-y_l)\Vert+\Vert (f_m-f)(y_l)\Vert \cr &\le 2\Vert f-f_m\Vert +\Vert f_m(y_n-y_l)\Vert } $$ It follows from the above that $(f(y_n))_n$ is a Cauchy sequence and, thus, convergent.

I'm not sure how to make the above argument go through when $Y$ is only assumed to be a normed space; however, you can show that $f(B_X)$ is totally bounded as in my comment above.

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This looks like the proof of Arzela-Ascoli (or some version of it I saw in an ODE's course). $Y$ is only a Normed space, not a Banach space, so this may not work in general. But I used your first hint in the comment above and the problem worked out fine. –  Kyle Schlitt Jan 10 '13 at 18:52
Unaccepted as you asked, but deleting it doesn't seem right as it is still valuable. –  Kyle Schlitt Jan 10 '13 at 21:25
Reaccepting this as it turns out completeness of $Y$ was a missing hypothesis. –  Kyle Schlitt Jan 12 '13 at 19:37
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The following is incorrect:

Let $x_{n}$ be a sequence in $B_{X}$. Choose $N\geq 1$ so that $\|f_{N} - f\| < \epsilon$.

Choose a subsequence $x_{n_{k}}$ such that $f_{N}(x_{n_{k}})\to y$ for some $y\in Y$, and then $K\geq 1$ such that $\|f_{N}(x_{n_{K}}) - y\| < \epsilon$.

Then $\|f(x_{n_{K}}) - y\| \leq \|f(x_{n_{K}}) - f_{N}(x_{n_{K}})\| + \|f_{N}(x_{n_{K}}) - y\| < 2\epsilon$.

This allows arbitrarily close approximation of $y$ by elements of the sequence $f(x_{n})$, which gives the subsequence which is needed.

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But, for different $\epsilon$, and thus different $N$, how do you know you can use the same limit $y$? Or am I missing something? (Also, would you unaccept my answer? I'd like to delete it as I misread (or rather didn't read carefully) your question.) –  David Mitra Jan 10 '13 at 19:20
You're absolutely right. I was jumping to conclusions. Thanks for pointing out my error. –  Kyle Schlitt Jan 10 '13 at 21:25
I think I can take smaller and smaller values of $\epsilon$, and form a sequence from the resulting $y$ values, and then cross my fingers that it has a convergent subsequence. I think this will fix it. –  Kyle Schlitt Jan 10 '13 at 21:49
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