Compact Operators and Inner Product of Weakly Convergent Sequences

Question

Let $H$ be a Hilbert space and let $x_{n}, y_{n}$ be weakly convergent sequences in $H$, to $x$ and $y$ respectively. In this case I know it's true that if we have a bounded linear operator $T:H \to H$, then T is compact if and only if $$ \lim_{n \to \infty} (T(x_{n}), y_{n}) = (T(x), y) $$ To prove that if $T$ is compact then this holds is not too hard, because we know $T(x_{n})$ is strongly convergent to $T(x)$ if $T$ is compact, and the inner product of a strongly convergent and weakly convergent sequences converges to the inner product of the limits.

How do you prove the converse? My attempt was to show that if $(T(x_{n}), y_{n}) \to (T(x), y)$ for all weakly convergent sequences $x_{n}, y_{n}$, then we can pick an arbitrary $x_{n}$ and plug in a clever $y_{n}$ to conclude that $T(x_{n})$ converges strongly to $T(x)$, which would show $T$ is compact, but I got stuck here. Is this approach correct? If so, what should I do next? If not, what's the right approach?

Answer 1

Given $x_{n}\rightarrow x$ weakly, we are to show $Tx_{n}\rightarrow Tx$, this is an equivalent definition of compact operator.

We have for any $u$ that \begin{align*} (Tx_{n},u)=(Tx_{n},x_{n})+(Tx_{n},u-x_{n}). \end{align*} Keep in mind that $u-x_{n}\rightarrow u-x$ weakly, so by assumption we have \begin{align*} (Tx_{n},x_{n})+(Tx_{n},u-x_{n})\rightarrow(Tx,x)+(Tx,u-x)=(Tx,u), \end{align*} this shows that $Tx_{n}\rightarrow Tx$ weakly, we are not done yet.

We recourse to a fact:

$f_{n}\rightarrow f$ weakly and $\limsup_{n}\|f_{n}\|\leq\|f\|$, then $f_{n}\rightarrow f$.

Assume this fact at a moment, we are to show that $\|Tx\|\geq\limsup_{n}\|Tx_{n}\|$.

We realize the limit supremum to a subsequence, say, $\limsup_{n}\|Tx_{n}\|=\lim_{k}\|Tx_{n_{k}}\|$.

We know that \begin{align*} \|Tx_{n_{k}}\|=(Tx_{n_{k}},y_{n_{k}}) \end{align*} for some $y_{n_{k}}$ such that $\|y_{n_{k}}\|\leq 1$. There is a weakly convergent subsequence $(y_{n_{k_{l}}})$, say, $y_{n_{k_{l}}}\rightarrow y$ weakly, $\|y\|\leq 1$, then \begin{align*} \lim_{k}\|Tx_{n_{k}}\|=\lim_{l}\|Tx_{n_{k_{l}}}\|=\lim_{l}(Tx_{n_{k_{l}}},y_{n_{k_{l}}})=(Tx,y)\leq\|Tx\|\|y\|\leq\|Tx\|, \end{align*} we are done.

Proof of the claim:

We expand $\|f_{n}-f\|^{2}$ to get \begin{align*} \|f_{n}-f\|^{2}&=(f_{n}-f,f_{n}-f)\\ &=\|f_{n}\|^{2}+\|f\|^{2}-(f_{n},f)-(f,f_{n}), \end{align*} so \begin{align*} \limsup_{n}\|f_{n}-f\|^{2}\leq\limsup_{n}\|f_{n}\|^{2}+\|f\|^{2}-\|f\|^{2}-\|f\|^{2}\\ &\leq\|f\|^{2}-\|f\|^{2}\\ &=0, \end{align*} so $\lim_{n}\|f_{n}-f\|=0$.