2

Let $\{e_k\}_{k=1}^\infty$ be an orthonormal set in a Hilbert space $H$. If $\{c_k\}_{k=1}^\infty$ is a sequence of positive real numbers such that $\sum c_k^2<\infty$, then the set: $$A=\left\{\sum_{k=1}^{\infty} a_ke_k :|a_k|\leq c_k\right\}$$ is compact in $H$.

My effort: We need to prove that every sequence $\{f_n\}$ has a subsequence $\{{f_n}_k\}$ so that $\{{f_n}_k\}$ converges to a limit in $A$.I'm trying to build such a sequence:

First, for each $n$ look at $|\langle f_n,e_1\rangle|=|a_{n1}|\leq c_1$ then the sequence $\{a_{n1}\}$ is bounded thus there must be a subsequence that converges and we define the limit as $l_1$. So we continue for all the vectors in the given set and define the limit of these subsets as $f=\sum_{k=1}^{\infty} l_ke_k \in H$.

But something must be wrong with the proof since I havent used $\sum c_k^2<\infty$.

Thanks.

Davide Giraudo
  • 181,608
catch22
  • 3,145

1 Answers1

1

What you did is more incomplete than wrong.

Firstly, we need to justify the sentence "we continue for all the vectors in the given set", writing explicitely the diagonal extraction.

Then we need to prove the limit element we get is indeed in $H$ (otherwise, we would have proved when $c_k=1$ that a set containing the unit ball is compact). That's where the conditions is used.

Also, there are criteria of compactness you can use here.

Community
  • 1
Davide Giraudo
  • 181,608
  • Did you mean that we need to prove that the limit element is in $A$? If so, isn't it obious since each $|l_k|$ is smaller than $c_k$? – catch22 Jun 26 '13 at 21:05
  • @Roy No, the point is that you must prove $\lVert f_n - l\rVert_2 \to 0$. That is where you need the summability of ${c_k}^2$. – Daniel Fischer Jun 26 '13 at 22:46
  • @DanielFischer: I've been trying to prove that but with no success. Can you give me a hint please? – catch22 Jun 28 '13 at 13:18
  • 1
    @Roy Since $\lvert (f_n)k\rvert \leqslant c_k$ for all $n$, the same estimate holds for the pointwise limit, hence $\lvert (f_n - l)_k\rvert \leqslant 2c_k$. Now, given $\varepsilon > 0$, by the summability condition, there exists a $k_0$ such that $\sum{k \geqslant k_0} {c_k}^2 < \frac{\varepsilon^2}{8}$. Split the sum in $\lVert f_n - l\rVert_2^2$ at a convenient place. – Daniel Fischer Jun 28 '13 at 13:25
  • @DanielFischer: I think I understand now, thanks! – catch22 Jun 30 '13 at 11:18