1
$\begingroup$

We can prove that every Banach space with a Schauder basis has the approximation property. I've read that this implies that every Hilbert space $H$ has the approximation property.

It's clear to me that this immediately follows if $H$ is separable, but if $H$ is not separable, it doesn't admit a countable orthonormal basis (and in terms of my definition, a Schauder basis is a countable system).

So, what am I missing?

$\endgroup$
  • 1
    $\begingroup$ I guess that the non-separable situation can be reduced to the separable situation by, e.g., considering the orthogonal complement of the kernel of a compact operator. This orthogonal complement should be separable. $\endgroup$ – gerw Feb 19 '18 at 15:28
0
$\begingroup$

Just look at the singular value decomposition. This is valid for every Hilbert space (also non-seperable) and for every compact operator. If you trucate the sum you have an operator with finite range and it converges to the original operator.

I think the reason is that if you have a orthonormal system $S$, $\langle x,e\rangle\neq 0$ with $e\in S$ and $x\in X$ can be only be the case for finite or countable many $e\in S$. Also if I remember right, you only need the eigenvalues not equal to zero for the decomposition and the only thing that can be uncountable is the basis of $\text{ker}T$ (if $T$ is the operator) that doesn't affect your decomposition.

$\endgroup$
  • $\begingroup$ Yes, it's clear to me that a Hilbert space has the AP and that this can be proved by the SVD. What's confusing me is that I've found multiple books which immediately after the statement "every Banach space with a Schauder basis has the AP" remark that this immediately yields that every Hilbert space has the AP. $\endgroup$ – 0xbadf00d Feb 19 '18 at 16:02
  • $\begingroup$ Ok that's indeed very confusing. Because if the Hilbert space is non-seperable it doesn't have a Schauder basis. Actually a Hilbert space has countable basis iff it is seperable. $\endgroup$ – CostaZach Feb 19 '18 at 17:03

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.