Let be $X$ a normed vectorial space, $X^*_w$ and $X^*_s$ the dual of $X$ weak and strong respectively. I would like to receive a hint in order to show $X^*_w = X^*_s$ is correct.
The following is what I think.
$X^*_w \subset X^*_s:$
Let be $f \in X^*_w$ and $U \subset \mathbb{R}$ an open set, then
$$f^{-1}(U) \in \tau_w \subset \tau_s \Longrightarrow f \in X^*_s$$
because the weak topology $\tau_w$ is strictly contained in the strong topology $\tau_s$, i. e., the topology generated by the norm.
$X^*_s \subset X^*_w:$
Let be $f \in X^*_s$ and $(x_n) \subset X$ such that $x_n \rightarrow x \in \overline{X}$, then $x_n \rightharpoonup x$. I want to show that $f(x_n) \rightharpoonup f(x)$ to conclude that $f \in X^*_w$, but I'm stuck.
I tried prove this inclusion again by other approach:
Assuming the approach that the inverse image of open sets are open in weak topology, is sufficient to show that $f^{-1}(B_{\varepsilon}(f(x))$ is a neighborhood of $x$ since $B_{\varepsilon}(f(x))$ is an open set in $\mathbb{R}$ and the topology of $\mathbb{R}$ is generated by the open balls in $\mathbb{R}$, but it's clear that $f^{-1}(B_{\varepsilon}(f(x))$ is a neighborhood of $x$ because $f^{-1}(B_{\varepsilon}(f(x)) = \{ y \in X \ ; \ |f(y) - f(x)| < \varepsilon \}$ which is a neighborhood of $x$.
I read in a comment here that
weak topology on $W$ is defined as a coarsest topology in which all linear forms are continuous, so $V_w' \supset V'$,
but I really can't see why this fact imply the reciprocal inclusion.
$\{y:|f (y)-f(x)| <\epsilon\}$ is a weak neighborhood of $x$. Hence it contains $x_n$ for all sufficiently large values of $n$. So $|f(x_n)-f(x)| <\epsilon$ for such $n$.
However, this is not a valid proof because sequential continuity is not enough. Weak topology is not given by a metric so it is not enough to show that if a sequence $x_n \to x$ weakly the $f(x_n) \to f(x)$. A correct proof would be to use nets or to prove that inverse image of open sets are open in weak topology. For nets the proof is identical to the proof I have given for sequences.