Claim : a linear function $T$ between Banach spaces is weakly continuous iff norm continuous?
Okay, So I think I have realised weakly continuous implies norm continuous. As weakly continuous implies weakly sequentially continuous. Now suppose that $T$ is unbounded. But we also know that 'weakly convergent implies weakly bounded', which implies norm bounded. But this would then imply that '$T(x_{n})$ converges weakly implies that $\|T(x_{n})\|$ is bounded'.
Hence $T(x_{n})$ is not weakly convergent, so $T$ cannot be weakly continuous. Contradiction! Hence T is bounded.
Any ideas on the converse? I.e How do I show norm continuous is weakly continuous?
Best Answer
For definiteness consider the linear operator $T:X\to Y$. If $T$ is norm continuous, then for each $f \in Y^*$ we have $f \circ T \in X^*$. Thus $f\circ T$ is continuous w.r.t. the strong topology on $X$ and hence also w.r.t. the weak topology on $X$. But if $f \circ T$ is continuous from $X$ with the weak topology to $\mathbb{F}$ for every $f \in Y^*$ then $T$ is weak-weak continuous from $X$ to $Y$ since the weak topology on $Y$ is the induced topology of the continuous linear functionals on $Y$.
The other direction follows since if $T$ is weak-weak continuous then the graph of $T$ is a weakly closed convex set in $X \times X$ and hence a strongly closed set. The Closed Graph Theorem implies that $T$ is norm continuous.