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.
Best Answer
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.