Let $X= \ell^2$ and $a_j$ be a real sequence in $c_0$. Define $T:X\to X$ as $(Tx)_j =a_jx_j$. Prove that $T$ is a compact operator.

My attempt:

To prove $T$ compact, we need to prove that for any bounded sequence $\{x_n\}\in \ell^2$, $\{Tx_n\}$ has a convergent subsequence.

Let $\{x_n\}$ be a bounded sequence in $\ell^2$. Then there is a subsequence $\{x_{n_j}\}$ weakly converging to $x$. Then we claim there is a convergent subsequence $\{Tx_{n_j}\}$ converging to $Tx$.

Hence we need to show $\lVert Tx_{n_j} - Tx \rVert_{\ell^2}\to 0$.

$\lVert Tx_{n_j} - Tx \rVert_{\ell^2}=\langle Tx_{n_j} - Tx , Tx_{n_j} - Tx \rangle=\langle Tx_{n_j},Tx_{n_j}\rangle -\langle Tx_{n_j} ,Tx\rangle - \langle Tx, Tx_{n_j}\rangle +\langle Tx,Tx\rangle $

Since $T$ is also self-adjoint, and $x_{n_j}$ weakly converges to $x$, we have the above converging to zero.

What confused me is that my approach seems to work for any self-adjoint operator. What is wrong with my approach?

  • $\begingroup$ Do you know that the limit in operator norm of compact operators is compact? $\endgroup$ – Rhys Steele May 17 '17 at 9:32
  • $\begingroup$ @nobody No. Another definition I know is that for compact operators, the image of a bounded set is precompact. $\endgroup$ – userHasNoNumber May 17 '17 at 9:34
  • $\begingroup$ Unfortunately I read your comment only after I posted the answer. @userHasNoNumber $\endgroup$ – user99914 May 17 '17 at 9:40

Note that if $T_n: X\to X$ is a sequence of compact operators on a Banach space $X$ and $\|T_n - T\| \to 0$, then $T$ is also compact. See here for the proof.

In our situation, set

$$T_n(x)_j = \begin{cases} a_j x_j &\text{if } j \le n \\ 0 &\text{if } j>n.\end{cases}$$

Then $T_n$ are comapct for all $n$ since they have finite rank. Note

$$\begin{split} \| T_n - T\| &= \sup_{x\in \ell^2, \| x\|=1} \| T_nx-Tx\| \\ & = \sup_{x\in \ell^2, \| x\|=1} \sqrt{\sum_{j=n}^\infty |a_j x_j|^2} \end{split}$$

for all $\epsilon >0$, since $a\in c_0$, there is $N$ so that $|a_n| <\epsilon$ for all $n\ge N$. Thus for $n\ge N$,

$$\|T_n - T\| \le \epsilon \sup_{x\in \ell^2, \| x\|=1} \sqrt{\sum_{j=n}^\infty | x_j|^2}=\epsilon$$

Thus $T_n$ converges to $T$ in the norm topology.

To remark why your approach does not work, note the weak convergence $T x_{n_j} \to Tx$ does not imply

$$\langle Tx_{n_j}, Tx_{n_j}\rangle \to \langle Tx, Tx\rangle.$$

For example, let $\{e_i\} \in \ell^2$ and $T = I$. Then this sequence converges weakly to $0\in \ell^2$. But

$$\langle e_i, e_i\rangle =1 \neq 0.$$


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