Well, the question is in the title. I understand that they are continuous in the weak topology, but can't see that it must hold for the norm topology.
Please help me.
Well, the question is in the title. I understand that they are continuous in the weak topology, but can't see that it must hold for the norm topology.
Please help me.
Let $X$ and $Y$ be normed spaces. Suppose the linear map $T:X\rightarrow Y$ is unbounded. Then there exist norm one elements $x_n\in X$ such that $\Vert Tx_n\Vert\ge n^2$. Consider the sequence $(x_n/n)$. This sequence coverges in norm to $0$ and is thus weakly convergent.
Now, if $T$ is weak-weak continuous, it would follow that $(T(x_n/n))$ is weakly convergent. However, a weakly convergent sequence is norm bounded, and $\Vert T (x_n/n)\Vert \ge n$ for each positive integer $n$.