Does this imply compactness

Question

Suppose H is Hilbert space with a countable orthonormal basis $\{e_n\}$. Let T be a bounded operator on H such that $∥Te_n∥$ tends to 0. Can we conclude that T is compact?

Answer 1

Let $T$ map each of the $2n+1$ unit vectors $e_{n^2},e_{n^2+1},\dots,e_{n^2+2n}$ to $e_n/\sqrt{2n+1}$. So, when restricted to the $(2n+1)$-dimensional subspace spanned by $e_{n^2},e_{n^2+1},\dots,e_{n^2+2n}$, this $T$ maps it into the $1$-dimensional space spanned by $e_n$, and it is easily seen to have norm $1$. Specifically, it maps the unit vector $(e_{n^2}+e_{n^2+1}+\dots+e_{n^2+2n})/\sqrt{2n+1}$ to the unit vector $e_n$. It follows that the whole operator $T$ (defined on the whole $l^2$ space) has norm $1$ (in particular it's bounded) but is not compact (as it sends the bounded sequence of vectors $(e_{n^2}+e_{n^2+1}+\dots+e_{n^2+2n})/\sqrt{2n+1}$ to a sequence with no convergent subsequence), even though the sequence $T(e_k)$ converges to zero.