The product topology of weak topologies is the same as the weak topology of the product space?

Question

I am reading Brezis' Functional Analysis. On page 62, in Theorem 3.10, it says

... $E \times F$ equipped with the product topolgy $\sigma(E, E^\star) \times \sigma(F, F^\star)$, which is clearly the same as $\sigma(E\times F, (E \times F)^\star)$.

Here, $E$ and $F$ are Banach spaces and $\sigma(E, E^\star)$ and $\sigma(F, F^\star)$ are their weak topologies respectively.

Actually, I showed that $\sigma(E, E^\star) \times \sigma(F, F^\star) \subset \sigma(E\times F, (E \times F)^\star)$ in the following way.

Let $\pi: E \times F \to E$ with $\pi(x,y) = x$ and $f_1 \in E^\star$. Then, since $f_1 \circ \pi \in (E \times F)^\star$, $(f \circ \pi)^{-1}(O_1) = \pi^{-1}(f_1^{-1}(O_1)) = f_1^{-1}(O_1) \times F \in \sigma(E\times F, (E \times F)^\star)$ for an open set $O_1$ in $\mathbb{R}$.

In the same manner,

if $f_2 \in F^\star$, then $f_2^{-1}(O_2) \times F \in \sigma(E\times F, (E \times F)^\star)$ for an open set $O_2$ in $\mathbb{R}$. Therefore, $f_1^{-1}(O_1) \times f_2^{-1}(O_2) \in \sigma(E\times F, (E \times F)^\star)$. Since $f_1^{-1}(O_1) \times f_2^{-1}(O_2)$ is a basis member of $\sigma(E, E^\star) \times \sigma(F, F^\star)$, $\sigma(E, E^\star) \times \sigma(F, F^\star) \subset \sigma(E\times F, (E \times F)^\star)$.

However, I cannot show $\sigma(E, E^\star) \times \sigma(F, F^\star) \supset \sigma(E\times F, (E \times F)^\star)$. How can we show this? Any help would be really appreciated!

Answer 1

We may prove this by first principles: Let $U \in \sigma(E\times F, (E\times F)^{\ast})$ be an open set in $E\times F$ and $(x_0,y_0) \in U$. Then, there exist finitely many bounded linear functionals $\varphi_i : E\times F \to \mathbb{C}, 1\leq i\leq k$ and finitely many positive real numbers $\epsilon_i > 0$ so that the set $$ W := \bigcap_{i=1}^n \{(x,y) \in E\times F : |\varphi_i(x,y) - \varphi_i(x_0,y_0)| < \epsilon_i\} $$ is contained in $U$. Let $\iota_E : E\to E\times F$ and $\iota_F : F\to E\times F$ denote the inclusions $x\mapsto (x,0)$ and $y\mapsto (0,y)$ respectively. For each $1\leq i\leq k$, define $$ A_i := \{x \in E : |\varphi_i\circ \iota_E(x) - \varphi_i\circ \iota_E(x_0)| < \epsilon_i/2\} $$ and $$ B_i := \{y\in F : |\varphi_i\circ \iota_F(y) - \varphi_i\circ\iota_F(y_0)| < \epsilon_i/2\} $$ Now $A_i \in \sigma(E,E^{\ast})$ and $B_i \in \sigma(F,F^{\ast})$, so if $$ A := \bigcap_{i=1}^n A_i, \text{ and } B := \bigcap_{i=1}^n B_i $$ Then $$ A\times B \in \sigma(E,E^{\ast})\times \sigma(F,F^{\ast}) $$ and if $(x,y) \in A\times B$, then for any $1\leq i\leq k$, $$ |\varphi_i(x,y) - \varphi_i(x_0,y_0)| < \epsilon_i $$ Hence, $A\times B \subset W\subset U$. Thus, $U \in \sigma(E,E^{\ast})\times \sigma(F,F^{\ast})$ as required.

Answer 2

You can use the Theorem 3.2 in the book to show that the map identity map

$$ id: (E \times F, \mathcal P) \rightarrow (E\times F, \mathcal W) , \quad (x,y) \mapsto (x,y) $$

is continuous iff for any bounded linear functionals $L$ on $(E \times F, \|\cdot\|_{E\times F})$ we have the map$ (x,y) \mapsto L \circ id(x,y) = L(x,y) $ is continuous, thus showing that $\mathcal W \subset \mathcal P$. Indeed, there are many ways to do this. The first one is to see

$$ L(x,y) = L(x,0) + L(0,y) $$

and observe $ x \mapsto L(x,0)$, $y \mapsto L(0,y) $ are in $E^*$ and $ F^*$. Now, $(x_n,y_n) \rightarrow (x,y)$ in $\mathcal P$ iff $x\rightharpoonup x$ and $y_n\rightharpoonup y$, which implies $L(x_n,y_n) \rightarrow L(x,y)$, so $L$ is continuous.

Alternatively, you can write directly

$$ L^{-1}(a,b) = U_1 \cap U_2, $$ where

$$ U_1 = \bigcup_{q\in \mathbb R} \pi_E^{-1}\Big(L_x^{-1}( -\infty, b -q)\Big) \cap \pi_F^{-1}\Big(L_y^{-1}( -\infty, q)\Big) $$

and

$$ U_2 = \bigcup_{p\in \mathbb R} \pi_E^{-1}\Big(L_x^{-1}( -a+p, \infty)\Big) \cap \pi_F^{-1}\Big(L_x^{-1}( -p, \infty)\Big). $$

Here I denote the map $L_x (x) = L(x,0)$ and the same for $L_y$. Note that $U_1$ and $U_2$ are open in the product topology of weak topologies $\mathcal P$ (by using the its definition).

Also, a further alternative way is to prove $L$ is continuous at any point $(x,y)$, which reduces to choosing an open set $O$ (containing $(x,y)$) in $\mathcal P$ such that $L(O) \subset (L(x,y)- \epsilon, L(x,y) + \epsilon) $, of the form

$$ \pi_E^{-1}(L_x^{-1}(L(x,0)-c, L(x,0)+ d)) \cap \pi_F^{-1}(L_y^{-1}(L(0,y)-e, L(0,y) +f)) $$

You can check that any $c,d,e,f>0$ satisfying $c+e = \epsilon = d+f$ works and hence $L$ is continuous at $(x,y)$.