# Prove that the $T^*$ is Continuous in the Weak${}^*$ Topology

Let $E$ and $F$ be two Banach spaces. Let $T \in \mathscr L(E, F)$ so that $T^* \in \mathscr L(F^*, E^*)$. Prove that $T^*$ is continuous from $F^*$ equipped with $\sigma(F^*, F)$ into $E^*$ equipped with $\sigma(E^*, E)$.

This is my approach: Because of $\sigma(E^*, E)$, we have the following relationship: $$\sigma(F^*, F) \ \ \xrightarrow[]{\text{???}} \ \ \sigma(E^*, E) \ \ \xrightarrow[]{\text{cont.}} \ \ \mathbb R$$

Now, I want to define a mapping from $F^*$ to $\mathbb R$ by: $$f \mapsto \langle T^*f, x \rangle = \langle f, Tx \rangle$$ So, if this is continuous from $\sigma(F^*, F)$ to $\mathbb R$, then $???$ above can be replaced with "cont." also. How do I do that?

Let $\{\phi_i\}$ be a net in $F^\ast$ such that $\phi_i\to \phi$ in $\sigma(F^\ast,F)$. We need to show that $T^\ast\phi_i \to T^\ast\phi$ with respect to $\sigma(E^\ast,E)$. To do this, fix $x\in X$. Then:
$$\hat{x}(T^\ast\phi_i)=(T^\ast\phi_i)x=(\phi_i\circ T)x=\phi_i(Tx)=\widehat{Tx}(\phi_i)\to\widehat{Tx}(\phi)$$
by the $\sigma(F,F^\ast)$ convergence of $\{\phi_i\}$.
• In that case, why don't I just write: $$(T^*\phi_i)x = \phi_i(Tx) \rightarrow \phi(Tx) = (T^*\phi)x$$ The convergence in the middle comes from $\sigma(F^*, F)$. Nov 3, 2016 at 2:35