Compact operator on a Hilbert space

This is a homework problem that I'm having some trouble with, so any hints would be appreciated. Let $H$ be a Hilbert space with an orthonormal basis $\{e_n\}$. Consider the operator $F : H\rightarrow H$ defined by $$Fx = \sum_{n=1}^\infty \beta_n \langle x, e_n\rangle e_n,$$

where $\{\beta_n\}\subset \mathbb C$

Suppose that $F$ is compact. (That is, the image of any bounded sequence contains a convergent subsequence). Show that $\lim_{n\rightarrow\infty} \beta_n = 0$.

The obvious approach is to choose a particular bounded sequence and try and get that to give me the result, and the first sequence I tried was obviously $\{e_n\}$. So the image of this sequence is $\{\beta_ne_n\}$, so this implies that there is a subsequence $\{\beta_{n_j}e_{n_j}\}$ which converges. But this doesn't help me and I can't see how else to approach the question.

I like your idea of building a sequence from the orthonormal basis vectors and proving non-existence of convergent subsequences! But I think we need to be a bit more restrictive in how we select our sequence: rather than selecting all the orthonormal basis vectors to be our sequence, we should select only the "troublesome" ones.

So here is the idea. Suppose, for contradiction, that $\beta_n$ does NOT tend to zero. Then hopefully you can show that there exists a $\epsilon > 0$ and there exists an ascending sequence $n_1, n_2, n_3, \dots$ such that $$|\beta_{n_1}| > \epsilon, \ \ \ | \beta_{n_2} | > \epsilon, \ \ \ | \beta_{n_3} | > \epsilon \dots$$

Now consider the sequence $$e_{n_1}, \ \ e_{n_2}, \ \ e_{n_3}, \ \ \dots$$ Hopefully you can verify that the sequence $$F(e_{n_1}), F(e_{n_2}), F(e_{n_3}), \dots$$ cannot possibly contain a Cauchy subsequence, which would be enough to complete your proof.

To see the problem with using the entire orthonormal basis $e_1, e_2, e_3, \dots$ as the "test" sequence in your argument: Consider the example where $\beta_n$ is $1$ when $n$ is odd and $0$ when $n$ is even. Then $F(e_n)$ clearly contains a convergent subsequence, namely, $F(e_2), F(e_2), F(e_6), \dots$ So we don't get a contradiction. We need to be more restrictive, by picking only the "troublesome" elements $e_1, e_3, e_5, \dots$ to be our "test" sequence, and only then do we find ourselves unable to find a Cauchy subsequence.

• I guess it would also be nice if you take the sequence $b_j = \frac{1}{\beta_{n_j}} e_{n_j}$. It would be bounded by $\frac{1}{\varepsilon}$. And $F(b_j) = e_{n_j}$. – André Caldas May 19 '17 at 11:58

Suppose $lim_n|\beta_n|\neq 0$. There exists $c>0$ such that for every integer $i$, there exists $n_i>i$ such that $|\beta_{n_i}|>c$. Consider the sequence $e_{n_i}$, $F(e_{n_i})$ is not a Cauchy sequence since $\|F(e_{n_i})-F(e_{n_j}\|\geq 2c^2$, so you cannot extract a converging subsequence.

You are more or less on the right track, but you have to be a bit careful. As you have probably noticed, the fact that the sequence $\{\beta_{n} e_{n}\}$ has a convergent subsequence does not imply that $\lim_{n \rightarrow \infty} \beta_{n} = 0$. Consider for example $\beta_{n} = 0$ if $n$ even and $\beta_{n} = 1$ if $n$ odd.

You will have to use that every bounded sequence gets mapped to a sequence with a convergent subsequence.

(1) You can use this to show that every subsequence of $\beta_{n}$ has a convergent subsequence. (How is up to you.)

(2) If you have this fact, the next step is to show that if $\beta_{n_{j}}$ is a convergent subsequence of $\beta_{n}$, then $\beta_{n_{j}} \rightarrow 0$. (Consider what happens when it approaches another number).

Then, finally, you can combine (1) and (2) to prove that $\beta_{n}$ converges to $0$. (Hint: use the fact that if it does not, then there is a subsequence $\beta_{n_{j}}$ that stays away from $0$, and argue that this contradicts (1) and (2)).

Your idea is good. Note that $(\beta_{n_j}e_{n_j})$ converges if and only if $(\beta_{n_j})$ converges to zero, so you are almost done. To get rid of the "almost" use the useful lemma that a sequences $x$ converges to $a$ if and only if every subsequence of $x$ has a subsequence that converges to $a$.

Depending on what information is available to you, you could do the following. Since $F$ is compact, nonzero part of the spectrum consists of eigenvalues, forming a sequence that converge to zero.

So, if you show that the eigenvalues of $F$ are precisely the $\{\beta_n\}$ (easy), you are done.