0
$\begingroup$

I have a some fact which seems obvious, but I cannot fill the detail.

Let $\mathcal{H}$ be a Hilbert space which is not necessarily separable. If $A$ is compact set in $\mathcal{H}$, then for every $a$, there is a countable subset of $\{e_i\}$ where $\{e_i\}$ is a orthonormal basis for $\mathcal{H}$ such that $a=\sum_{i=1}^\infty \left<a,e_i\right>e_i$.

The question comes from proof of Lemma 2 of http://www.ams.org/journals/proc/1986-098-02/S0002-9939-1986-0854051-6/S0002-9939-1986-0854051-6.pdf.

My strategy is since $A$ is compact in metric space, it has countable local basis for $A$ (topological sense). Since we can represent $a\in A$ as a Fourier expansion, we can write $$ a =\sum_{i} \left<a,e_i\right> e_i.$$ Since $e_i$ are orthonormal, the sum must be countable sum.

I'm not sure this is right proof. Thanks in advance.

$\endgroup$
1
$\begingroup$

It's easy to show that $A$ is separable. Let $S$ be a countable dense subset. Using the Gram-Schmidt process, construct a countable orthonormal set $E = \{e_i: i \in I\}$ such that each member of $S$ is in the linear span of a finite subset of $E$. Now the function

$$ f(x) = \| x - \sum_i \langle x, e_i \rangle e_i \|$$

is continuous on $\mathcal H$ and $0$ on $S$, therefore $0$ on $A$.

$\endgroup$
  • $\begingroup$ This could be another method. Thanks. $\endgroup$ – Will Kwon Aug 25 '16 at 5:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.