$H$ is a Hilbert space and $T$ is a linear operator from $H$ to itself. If $x_n$ converges weakly to $0$ and $Tx_n$ converges strongly to $0$, then we have $(Tx_n,x_n)$ converges to $0$.
What about the converse? Namely, if $x_n$ converges weakly to $0$ and $(Tx_n,x_n)$ converges to $0$, do we have $Tx_n$ converges strongly to $0$?
Best Answer
Here is a counterexample for the question you raise in the comment to another answer:
Consider $H = \ell^2$ and define $T$ via $$ Tx = (x_2, -x_1, x_4, -x_3, \ldots). $$ By construction, $(Tx,x) = 0$ for all $x \in \ell^2$.
Hence, for every sequence $(x_n)\subset H$ with $x_n \rightharpoonup 0$ you have $(Tx_n,x_n) = 0 \to 0$.
However, $T$ is an isometry, hence $x_n \not\to 0$ implies $Tx_n \not\to 0$.