# An operator is compact only if it maps weakly convergent sequences into strongly convergent sequences.

I have a problem with the proof of the following statement:

A linear operator $A$, defined on a Hilbert space $H$, maps any weakly convergent sequence $(x_n)$ into a strongly convergence sequence $(Ax_n)$ only if it is compact.

I found a proof in "Introduction to Hilbert Spaces with Applications" by Lokenath Debnath and it starts with words "Let $(e_n)$ be a complete orthonormal sequence in $H$,..."

I do not understand why there is such a sequence because $H$ is not necessarily separable. Could you please explain why such a sequence exists or suggest some other book.

• My guess is that the book wants an orthonormal sequence, not necessarily a "complete" such sequence. Check if completeness is ever used later in the proof; I doubt that it is. – Omnomnomnom Oct 19 '16 at 12:10
• @Omnomnomnom He uses the Parseval's formula: $\sum_{k=1}^{\infty} |(x_n, e_k) |^2 = \| x_n\|^2$ which is not true, when $(e_n)$ is not complete. – user374573 Oct 20 '16 at 8:44
• @Omnomnomnom Maybe there is a way to show, that the range of $T$ is separable. Then $H/\ker(T)$ is also separable and we can find such a sequence there. – user374573 Oct 20 '16 at 15:40
• Both $(x_n)$ and $(Tx)$ live in countably infinite-dimensional subspaces of $H$. We can just restrict ourselves to the sum of those two subspaces, no? – anonymous Dec 23 '16 at 14:29