If $x_n \rightarrow x$ and $Ax_n \rightharpoonup y$, why does $y = Ax$?

Question

Let $X$ be a Banach space and define a linear operator $A: X \rightarrow X$. Suppose there is a sequence $(x_n) \subseteq X$ which converges to $x$ in norm. Further suppose that the sequence $(Ax_n)$ converges weakly to some $y$. Why must $Ax_n$ converge weakly to $Ax$?

Answer 1

This is false. Let $f$ be a discontinuous linear functional on $X$ and $Ax=f(x)x_0$ where $x_0$ is a fixed non-zero vector. Since $f$ is not continuous its kernel $M$ is not closed. So there is a sequence $\{x_n\}$ in $M$ converging to some point $x$ not belonging to $M$. Now $Ax_n=0$ for all $n$ so $Ax_n\to 0$ weakly, in fact in the norm. But $y \neq Ax$ so $Ax_n$ does not tend to $Ax$.

If $A$ is a bounded operator the conclusion is true and is easy to prove: just show that $x^{*}(y)=x^{*}(Ax)$ for all $x^{*}\in X^{*}$.