Proof Request : I am seeking for a proof of the following Lemma defining Weak Convergence. I am aware of a similar statement regarding Hilbert spaces but it seems to differ. I know that for the $(\Rightarrow)$ way, the boundedness of $\{u_n\}_{n \geq 1}$ can be proven via Banach-Steinhaus and Natural Embedding. I am struggling to grasp an intuition on proving the latter part of $(\Rightarrow)$. Also, my search and intuition have both failed to find anything regarding the $(\Leftarrow)$ way.
If there is not a reference for a proof, I'd really appreciate any elaboration or hints so I can work myself at proving it. Thanks in advance.
Lemma – Definition :
Let $X$ be a Banach space and $\{u_n\}_{n \geq 1} \subseteq X$ be a sequence.
Show that $u_n$ converges weakly to $u \in X$, namingly $u_n \xrightarrow{w} u$ if and only if $\{u_n\}_{n \geq 1} \subseteq X$ is bounded and $\langle x^*, u_n \rangle \to \langle x^*, u \rangle$ for all $x^* \in D^* \subseteq X^*$ dense.
Best Answer
The latter half of the direction $\implies$ is straightforward. The weak topology on $X$ is the coarsest topology such that all elements of $X^*$ are continuous. In particular, for $x^* \in D^* \subseteq X^*$, the map $u \mapsto \langle x^*, u \rangle := x^*(u)$ is continuous for the weak topology which implies the desired convergence.
For the converse direction, pick $x^* \in X^*$ and take a sequence $x_n^* \in D^*$ such that $\|x_n^* - x^*\|_{X^*} \to 0$. By assumption, $u_n$ is a bounded sequence and $\langle x_n^*,u_k \rangle \to \langle x_n^*, u \rangle$ for each $n$.
We have \begin{align} | \langle x^*, u_k \rangle - \langle x^*, u \rangle | \leq& |\langle x^* - x_n^*, u_k \rangle| + | \langle x_n^*, u - u_k \rangle | + | \langle x - x_n^*, u \rangle | \\ \leq & \big( \sup_k \|u_k\| + \|u\| \big) \|x^* - x_n^*\| + |\langle x_n^*, u - u_k \rangle| \end{align} You should be able to conclude by fixing $\varepsilon > 0$ and showing that you can pick $n$ and then $K$ such that for $k \geq K$, the right hand side is less than $\varepsilon$.