Let $H$ be a Hilbert space and $T\in\mathcal{K}(H)$. Show that if $(x_n)_{n\in\mathbb{N}}$ is a sequence in $H$ that converges weakly to $x_0\in H$ then $\lim_{n\to\infty}||Tx_n-Tx_0||=0$.
My proof:
Since $\overline{T(B_1)}$ is compact and thus separable we conclude by scaling that $\overline{\text{ran}(T)}$ is a separable sub-Hilbert space of $H$ and thus has a countable (Schauder) basis $E:=\{e_i:i\in\mathbb{N}\}$.
Also we have $Tx=\sum_{i\in\mathbb{N}} \langle Tx,e_i\rangle e_i$ for $x\in H$.
Moreover a direct corollary from the Frèchet-Riesz theorem is that $x_n\overset{w}{\to}x_0$ $(n\to\infty)$ if and only if $\langle x_n-x_0,y\rangle\overset{n\to\infty}{\to} 0$ for all $y\in H$.
Also $T$ is bounded and therefore $T^*$ exists on all of $H$ and is bounded.
Now we bring everything together and have
$$\begin{aligned}
||Tx_n-Tx_0||^2 &=||\sum_{i\in\mathbb{N}}\langle T(x_n-x_0),e_i\rangle e_i||^2
&&=\sum_{i\in\mathbb{N}}|\langle T(x_n-x_0),e_i\rangle|^2\\
&= \sum_{i\in\mathbb{N}}|\langle (x_n-x_0),T^*e_i\rangle |^2
&&\overset{n\to\infty}{\to}0
\end{aligned}
$$
My professor told me that I had to give another argument for the last step (taking the limit inside the sum). Can someone tell me, how I would argue?
Best Answer
A few comments:
A Schauder basis is not necessarily an orthonormal one, and so you wouldn't have the equality $Tx=\sum_j\langle Tx,e_j\rangle\,e_j$. You need an orthonormal basis.
You are right that it's a "direct corollary" of Riesz-Fréchet, but over a Hilbert space one just uses weak convergence as $\langle x_n-x,y\rangle\to0$.
As $x_n\to x$ if and only if $x_n-x\to0$, it is simpler (for writing) to assume that $x_n\to0$.
Note that nowhere in your argument have you used that $T$ is compact. The "taking the limit inside the sum" requires you to use that $T$ is compact.
And, also, the result is not true if you consider a weakly convergent net; it has to be a sequence. That's also something that your argument requires. Here, since $\{x_n\}$ is a weakly convergent sequence, it is bounded.
One possibility is to use that $T$, being compact, is a limit of finite-rank operators. If $T_0$ is finite-rank, then $T_0x=\sum_{j=1}^mf_j(x)\,g_j$ for certain bounded linear functionals $f_j$ and linearly independent vectors $g_1,\ldots,g_m$. So $$ \|T_0x_n\|\leq\sum_{j=1}^m|f_j(x_n)|\,\|g_j\|\to0 $$ since $f_j(x_n)\to0$ for all $j$. Then, for any $\varepsilon>0$ there exists $T_0$, finite-rank, with $\|T-T_0\|<\varepsilon$. Then $$ \|Tx_n\|\leq\|T_0x_n\|+\|(T-T_0)x_n\|\leq\|T_0x_n\|+\varepsilon\,\sup_n\|x_n\|. $$ Then $$ \limsup_n\|Tx_n\|\leq \varepsilon\,\sup_n\|x_n\|. $$ As we can do this for every $\varepsilon>0$, we have that the limit exists and is zero.