Prove $T$ with bounded basis sum in Hilbert space is compact.

Question

Let $T:\mathcal{H}\rightarrow \mathcal{H}$ be a linear continuous operator between Hilbert spaces and $\{b_i\: |\: i \in I\}$ an orthonormal basis. Prove that if: $$\sum_{i\in I}\lVert T b_i\rVert^2<\infty$$ Then, $T$ is compact.

Hilbert spaces are reflexive, as is easily seen from applying Riesz Representation Theorem twice. This means that if $x_n$ is a sequence whose norm is bounded by $M$, there is a subsequence such that $x_{n_j}\rightharpoonup x$. By continuity of $T$, we obtain that $T(x_{n_j})\rightharpoonup T(x)$. I want to show this convergence is strong. Because we have a basis it is clear that from Parseval we obtain:

$$\lVert T(x_{n_j})-T(x)\rVert^2=\sum_{i\in I}|\langle T(x_{n_j}-x),b_i\rangle |^2=\sum_{i\in I}|\langle x_{n_j}-x,T^*(b_i) \rangle|^2 $$

By our hypothesis there is also $I_o$ with $|I_o|<\infty$ such that $\sum_{i\in I_o^C}\lVert Tb_i\rVert^2<\varepsilon$. Using weak convergence for every $i\in I_o$ we have $|\langle x_{n_j }-x_j, T^*(b_i) \rangle|\rightarrow 0$, and thus one can take $j$ large enough such that:

$$\lVert T(x_{n_j})-T(x)\rVert^2< \varepsilon+4M^2\sum_{i\in I_o^C}\lVert T^*(b_i)\rVert^2$$

If we had a self-adjoint operator, we would be done. But that is not the case... Any thoughts on how to proceed?

Answer 1

I think you can do exactly the same proof, except that you try to show that $T^*$ is compact.

Let $x_n$ be a bounded sequence in $H$. By reflexivity, you find a weakly convergent subsequence $x_{n_j}$. Then $$||T^*(x_{n_j}) - T^*(x)||^2 = \sum_{i\in I} |\langle x_{n_j} - x, T(b_i)\rangle|^2$$ and you can apply your hypothesis to get that $T^*$ is compact, which is equivalent to $T$ itself being compact.

Answer 2

Observe that $$Tx=\sum_{k=1}^\infty \langle x,b_k\rangle Tb_k$$ Let $$T_nx=\sum_{k=1}^n\langle x,b_k\rangle Tb_k$$ The operators $T_n$ are finite dimensional. Moreover $$Tx-T_nx= \sum_{k=n+1}^\infty \langle x,b_k\rangle Tb_k$$ hence by the Cauchy-Schwarz and the Bessel inequalities we get$$\|Tx-T_nx\|^2\le \sum_{k=n+1}^\infty|Tb_k\|^2\,\|x\|^2$$ Thus $$\|T-T_n\|\le \left (\sum_{k=n+1}^\infty\|Tb_k\|^2\right )^{1/2}\underset{n\to \infty}{\longrightarrow} 0$$ Hence $T$ is compact.

Answer 3

A proof by showing $T(B_{\mathcal H})$ is totally bounded: given $\varepsilon >0$, let $F \subset I$ finite such that $$\sum_{i \in I \setminus F}\lVert Tb_i\rVert^ 2 <\varepsilon^2/4.$$ Since $E=\langle b_i : i \in F\rangle$ is finite-dimensional, $T|_E$ is compact so by total boundness there is a finite set $S \subset \mathcal H$ with $\min_{y \in S} \lVert x-y\rVert <\varepsilon/2$ for all $x \in T(B_E)$.

Now, given $x \in B_{\mathcal H}$, write $x=z + w$ where $z \in E$ and $w \in E^\perp$. Notice $|\langle x, b_i\rangle| \le 1$, so the definition of $F$ implies $\lVert T(w)\lVert < \varepsilon/2$. Since there is some $y \in S$ with $\lVert T(z)-y\rVert<\varepsilon/2$, it satisfies $$\lVert T(x) - y\rVert \le \lVert T(z)-y\rVert + \lVert T(w) \rVert <\varepsilon.$$