$X_1$ and $X_2$ are Banach Spaces.

As stated in the title I want to prove that $$\sigma(X_1 \times X_2, (X_1 \times X_2)^*) = \sigma(X_1,X_1^*) \times \sigma(X_2,X_2^*)$$ where $\sigma(A,A^*)$ denotes the weak topology, i.e. this is the weakest topology on $A$ s.t. every $f \in A^*$ is continuous. And $A^*$ is the dual space of $A$, i.e. in $A^*$ are all bounded linear operators from $A$ to $\mathbb{C}$.

I had to prove a statement for homework and I just used this property but in the correction it was stated that I this property doesn't follow immediately.

I want to give my ideas to this: (As stated in the comments this approach doesn't really work.)

$\subset$ : Let $U$ be in the left hand side, i.e. $U \subset X_1 \times X_2$ s.t. $$U=f^{-1} (W)=(f_1^{-1} (W), f_2^{-1} (W))$$ for $W \subset \mathbb{C}$ open and $f \in (X_1 \times X_2)^*$. Denote $U_1:=f_1^{-1}(W)$ and $U_2:=f_2^{-1}(W)$, i.e. $U_1 \in \sigma(X_1,X_1^*)$ and $U_2 \in \sigma(X_2,X_2^*)$. By the product topology it follows $$U_1+U_2 \in \sigma(X_1,X_1^*) \times \sigma(X_2,X_2^*)$$

I don't know if this really works, this weak topology is new to me and hard to handle.

Would be grateful for any help,

Thanks a lot, Marvin

  • $\begingroup$ The last $+$ should be a $\times $. Also, you should take care of the line $f^{-1}(W)=(f_1^{-1}(W),f_2^{-1}(W))$. That's not clear and probably wrong. $\endgroup$ Commented Feb 19, 2016 at 23:25
  • $\begingroup$ Oh, thank you for your remark, makes sense. Do you or someone else have a correct approach? For the other direction I don't know how to accomplish this. $\endgroup$
    – Cahn
    Commented Feb 20, 2016 at 8:18
  • 1
    $\begingroup$ Other notation, but this: math.stackexchange.com/q/368570/4280 seems pretty similar. $\endgroup$ Commented Feb 20, 2016 at 8:41
  • $\begingroup$ Also sciencedirect.com/science/article/pii/0016660X78900442 is quite a bit more general. $\endgroup$ Commented Feb 20, 2016 at 8:44
  • $\begingroup$ Thank you for your link to the other topic. It is basically the same and very helpful, just transferring the notation right now. $\endgroup$
    – Cahn
    Commented Feb 20, 2016 at 10:40

1 Answer 1


For the purpose of completeness I will post an answer here based on what has been discussed in the comments. This may help someone who stumbles upon this question.

We introduce the homeomorphic operator $$\begin{aligned} &T: X_1^* \times X_2^* \longrightarrow (X_1 \times X_2)^* \\ &T(x_1^*,x_2^*)(x_1,x_2) \equiv x_1^*(x_1)+x_2^*(x_2) \end{aligned}$$

  • We take a net $\{(a_i,b_i)\}_{i\in \mathbb{N}} \subset \sigma(X_1,X_1^*) \times \sigma(X_2,X_2^*)$ such that $(a_i,b_i) \to (0,0)$. Hence $x_1^*(a_i)\to 0$, $x_2^*(b_i) \to 0$ and thus $$T(x_1^*,x_2^*)(a_i,b_i)=x_1^*(a_i)+x_2^*(b_i) \to 0.$$
  • We take a net $\{(c_i,d_i)\}_{i \in \mathbb{N}}$ such that $(c_i,d_i) \to (0,0)$ in $\sigma(X_1 \times X_2, (X_1 \times X_2)^*)$. We define the operator $S_1 \in (X_1 \times X_2)^*$, $S_1(x_1,x_2) \equiv x_1^*(x_1)$ for a fixed $x_1^*$. We have $S_1(c_i,d_i)=x_1^*(c_i) \to 0$. Analogously $S_2(c_i,d_i)=x_2^*(d_i) \to 0$. Hence $$(c_i,d_i) \to 0 \text{ in } \sigma(X_1, X_1^*)\times \sigma(X_2, X_2^*).$$

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