I'm trying to simplify the proof of Theorem 3.10 in Brezis' book of Functional Analysis. My proof is much simpler than the original. I'm afraid that I made subtle mistakes. Could you have a check on my attempt?
Let $E,F$ be Banach spaces and $E^\star, F^\star$ their continuous dual spaces respectively. Let $\sigma (E, E^\star)$ be the weak topology of $E$. We denote by $E_w$ and $E_s$ the space $E$ with the weak and norm topologies respectively. We do similarly for $F$. Let $T: E \to F$ be linear. If $T:E_w \to F_w$ is continuous, then $T:E_s \to F_s$ is continuous.
Let $\sigma (E, E^\star) \boxtimes \sigma (F, F^\star)$ be the product topology of $\sigma (E, E^\star)$ and $\sigma (F, F^\star)$. Then the author uses the fact that $$\sigma (E, E^\star) \boxtimes \sigma (F, F^\star) = \sigma \big (E \times F, (E \times F)^\star \big).$$
Below I propose my simpler approach. We say "weakly" (resp. "strongly") to refer to topological concepts in weak (resp. norm) topology.
Let $T : E_w \to F_w$ be continuous. Then $\operatorname{graph} T$ is weakly closed in the product of weak topologies. Let $(x_n, y_n)$ be a sequence in $\operatorname{graph} T$ such that $(x_n, y_n) \to (x,y)$ strongly. Then $x_n \to x$ strongly and $y_n \to y$ strongly. Then $x_n \to x$ weakly and $y_n \to y$ weakly. Then $(x_n, y_n) \to (x,y)$ weakly. Then $(x,y) \in \operatorname{graph} T$ and thus $\operatorname{graph} T$ is strongly closed in the product of norm topologies. The claim then follows from closed graph theorem.
Best Answer
It is totally fine to me: indeed note that when you say that "$(x_n, y_n) \to (x,y)$ strongly, then $x_n \to x$ strongly and $y_n \to y$ strongly" you are exactly using properties of the product topology.