The weak topology of a product

Question

Let $E$ and $F$ be normed vector spaces and let $E^{\sigma}$, resp. $F^{\sigma}$ be $E$, resp. $F$ with the weak topology associated with the elements of the duals $E^*$, resp. $F^*$. Then, for $(E\times F)^{\sigma}$, being $E \times F$ with the weak topology induced by the elements of $(E\times F)^*$, apparently, one should have that $E^{\sigma}\times F^{\sigma}=(E\times F)^{\sigma}$ (I'm not sure about additional conditions on $E$ and $F$, perhaps they should be Banach). I have established that $\Pi : E^*\times F^*\longrightarrow (E\times F)^* $, with $\Pi(f,g)(x,y)= f(x)+g(y), \forall (x,y) \in E \times F $, $\forall (f,g) \in E^* \times F^*$ is a homeomorphism, but after that, I ran out of ideas. Anyone out there with a rather elegant explanation? To avoid further spam, many thanks in advance.

Answer 1

Let $(e_i,f_i)_i$ be a net in $E^\sigma\times F^\sigma$ converging to $(0,0)$ and $T\in (E\times F)^\star$. By what you have established, $T=\prod (R,S)$ for some $R\in E^\star, S\in F^\star$. But as $R(e_i)$ and $S(f_i)$ are both converging to $0$, $R(e_i)+S(f_i) = T(e_i,f_i)$ converges to $0$ too.

Conversely, suppose $(e_i, f_i)_i$ converges to $(0,0)$ in $(E\times F)^\sigma$. For $T\in E^\star$, define $S_T(e,f) = T(e)$, so $S_T \in (E\times F)^\star$. Then $T(e_i) = S_T(e_i,f_i)$ converges to $0$, so the projection of $(e_i,f_i)$ on the first coordinate $E^\sigma$ is continuous. Similarly proceeding for the second coordinate, you conclude $(e_i,f_i)\to 0$ in $E^\sigma\times F^\sigma$.

Essentially, you had already done the main step.