# A simpler proof of the continuity equivalence between norm and weak topologies

banach-spacesdual-spacesfunctional-analysissolution-verificationweak-topology

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.

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.