If $X$ is a topological vector space, and $X^*$ is the space of all continuous linear functionals $f:X \rightarrow \mathbb{F}$. Define the operator
$T:X^* \rightarrow \mathbb{F}^X=\prod_{x \in X}\mathbb{F}$
$T(f)=(f(x))_{x\in X}$ We need to show that
- $T$ is bounded linear injective. I was able to prove the linearity and injection but not sure how to show it is bounded (or continuous) operator. I am not even sure how the norm of the range looks like.
Here, we define the weak topology on $X^*$ which has a base of neighborhoods at zeros of the form
$W(x_1,...,x_n;r)=\{f \in X^*: |f(x_i) < r, 1 \leq i \leq n \}$
- Describe the open sets in $\mathbb{F}^X=\prod_{x \in X}\mathbb{F}$
Is it true that the open sets are of the form $\prod_{x\in X}U_x$ where $U$ is an open set in $\mathbb{F}$?
The product topology on $\Bbb F^X$ is the minimal topology such that all maps $\pi_x: \Bbb F^X \to \Bbb F$, where $\pi_x(g)= g(x)$ for all $g$, are continuous, and has the universal property that $F : Y \to \Bbb F^X$ is continuous (for an arbitrary space $Y$ and function $F$) iff for all $x \in X$ we have $\pi_x \circ F$ continuous. See Wikipedia e.g.
Applying it to the present case: $(\pi_x \circ T)^{-1}[U] = \{f \in F^\ast\mid f(x) \in U\}$ where $U \subseteq \Bbb F$ is open (an open disk in $\Bbb C$ or an open interval in $\Bbb R$). The right hand side is just a shifted version of some $W(x; r)$, essentially (if we take $U$ to be $r$-balls, which is sufficient for continuity), and so all $\pi \circ T$ are continuous and so is $T$.
The map $T$ is in fact the "identity" in disguise, and the weak-star topology a relative version of the product topology.