Does every (non-separable) Hilbert space have the approximation property?

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?

• 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. – gerw Feb 19 '18 at 15:28

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.