Show $T$ is compact operator if $\langle Te_n,e_n \rangle$ tend to zero.

Question

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?

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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.

Answer 1

Lemma. If $T$ is a bounded operator on $H$, the following are equivalent:

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

2. $\langle T(e_n), e_n\rangle \to 0$, 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 $$ T = \lim_{\varepsilon \to 0} TE (-\infty , -\varepsilon )+TE (\varepsilon , \infty ) $$

Answer 2

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.

Answer 3

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 $0$, $\langle Tx_n,x_n\rangle\to 0$.