What I want to prove is the following:
Let $X$ and $Y$ be normed vector spaces. If a linear operator $T:(X, \sigma(X,X^*))\to (Y, \sigma(Y,Y^*))$ is continuous, then $T:(X,\|\cdot\|)\to(Y,\|\cdot\|)$ is continuous.
Here, $(X, \sigma(X,X^*))$ is the weak topology and $(X,\|\cdot\|)$ is the topology induced by the norm.
If $X$ and $Y$ were Banach spaces, I can prove it using the Closed Graph Theorem. But, I have no idea how to prove it when $X$ and $Y$ are just normed spaces. Any suggestions?
Notice that a linear functional is weakly continuous iff it's norm continuous. Also, observe that if $\Omega$ is a topological space, $X$ a normed linear space and $f:\Omega \to X $ is a function, then $f$ is weakly continuous iff $x^*f$ is continuous for each $x^*\in X^*$.
Now suppose $T: X\to Y$ is a linear map between normed spaces $X$ and $Y$. Then by the above observations, $T$ is weak-to-weak continuous iff $y^*T$ is weakly continuous for each $y^*\in Y^*$ iff $y^*T$ is norm continuous for each $y^*\in Y^*$. Does $y^*T$ being norm continuous for each $y^*\in Y^*$ imply $T$ is norm-to-norm continuous? (Hint: Any weakly bounded subset of $Y$ is bounded.)
