Show $T$ is compact operator if $\langle Te_n,e_n \rangle$ tend to zero. Suppose $\mathcal{H}$ is a Hilbert space, and $T\in B(\mathcal{H})$. If for each orthonormal (norm 1) basis $\{e_n\}\subseteq \mathcal{H}$, we have $\langle Te_n, e_n \rangle \rightarrow 0$. Can we deduce $T$ is compact?

I guess it may use spectra decomposition. Take adjoin, use the fact that $\langle Ae_n,e_n\rangle \rightarrow 0$ iff $\langle (A+A^*)e_n,e_n \rangle \rightarrow 0$ and $\langle (A-A^*)e_n,e_n\rangle \rightarrow 0$ , we can assume $T$ is self-adjoint. So $T$ can be viewed as a multiple operator(multiple by a $L^{\infty}(X,d\mu)$ function) on $L^2(X,d\mu)$, via unitary equivalence. Here $d\mu$ is an abstract $\sigma$-finite Borel measure on X. But I don’t know how to use the condition.. And, another way, if we use the spectra decomposition via $T=\int z dE$, E is the corresponding spectra measure. To show $T$ is compact, it is sufficient to prove the projection $E(-\infty,-\epsilon)$ and $E(\epsilon, +\infty)$ are all finite rank, for every $\epsilon>0$. And I feel the condition may can be use together with something like dominate converge theorem? But I fail.
Any help or hint? Thanks.
 A: No. Take the real space $H=l^2$ and define $Tx$ for $x=(x_1,x_2,\dots)$ by
$$
Tx=(x_2,-x_1,x_4,-x_3, \dots).
$$
Then $\langle Tx,x\rangle=0$ for all $x$, but $T$ is bijective and cannot be compact.
A: Lemma.  If $T$ is a bounded operator on $H$,  the following are equivalent:

*

*$\langle T(e_n), e_n\rangle \to 0$, for every orthonormal basis $\{e_n\}_n$,


*$\langle T(e_n), e_n\rangle \to 0$, for every orthonormal sequence $\{e_n\}_n$.
Proof.
(1) $\Rightarrow$ (2) Obvious.
(2) $\Rightarrow$ (1)  Given an orhonormal sequence $\{e_n\}_n$, choose another orthonormal set $\{f_i\}_{i\in I}$, such that
$$
  B:= \{e_n:n\in {\mathbb N}\}\cup   \{f_i:i\in I\}
  $$
is an orthonormal basis.  Assuming, as we supposedly are, that $H$ is separable, it is possible to order $B$
into a sequence admiting   $\{e_n\}_n$ as a subsequence.  So the proof follows since a subsequence of a convergent
sequence converges to the same limit.
QED
As already noticed,  in the complex case we may assume without loss of generality that $T$ is self-adjoint.
Let $\mathfrak B(\mathbb R)$ denote the $\sigma $-algebra of all Borel subsets of ${\mathbb R}$, and let
$$
  E:\mathfrak B(\mathbb R)\to B(H)
  $$
be the spectral resolution for $T$.
We then claim that
$$
  P:= E(\varepsilon ,\infty )
  $$
is a finite rank projection for every $\varepsilon >0$.  To see this,
suppose otherwise, so there exists an orthonormal sequence $\{e_n\}_n$ contained in the range of $P$.  Observing that
$TP\geq \varepsilon P$, we then have that
$$
  \langle T(e_n), e_n\rangle  =  \langle TP(e_n), e_n\rangle  \geq  \varepsilon    \langle e_n, e_n\rangle  = \varepsilon \|e_n\|^2 = \varepsilon ,
  $$
contradicting the hypothesis.
In a similar way one proves that  $E(-\infty , -\varepsilon )$ is finite rank.
Finally the compactness of $T$ follows from the fact that
$$
  T = \lim_{\varepsilon \to 0} TE (-\infty , -\varepsilon  )+TE (\varepsilon , \infty  )
  $$
A: Perhaps, the following link provides an affirmative answer for the complex Hilbert space: $T$ is compact if and only if for any sequence $x_n$ weakly converging to $0$, $\langle Tx_n,x_n\rangle\to 0$.
