TITLE: Criterion for compactness
QUESTION [7 upvotes]: Let $T:H\to H$ be a continuous operator on a Hilbert space.
Assume there exists an orthonormal base $(e_j)_{j\in\mathbb N}$, such that the sequence $Te_j$ tends to zero.
Must $T$ be compact?

REPLY [4 votes]: We probably don't need another answer, but here it is anyway, especially since Giorgio raised the question about projections.
Lemma: There is an ONB $\{ v_j\}$ of $\mathbb C^n$ such that $|\langle e_1, v_j\rangle |= n^{-1/2}$ for all $j=1,2, \ldots , n$.
Proof: This is trivial: it works if we replace $e_1$ by $w=(n^{-1/2}, \ldots, n^{-1/2})$ and use the standard ONB. Now apply a unitary map that moves $w$ to $e_1$ to everything.
It follows that any projection $P$ on a separable space $H$ with $\dim N(P)=\dim R(P)=\infty$ is a non-compact bounded operator with $Pe_j\to 0$ on a suitable ONB. We can build such an ONB by applying the lemma to reducing subspaces spanned by $n$ orthogonal vectors, with one of them taken from $R(P)$ and the others from $N(P)$ and $n$ becoming larger as we move along.