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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 $u_n xrightarroww u$ 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.

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
beginalign
| 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|
endalign
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 $varepsilon$.

hint:

Wlog $u=0$.
You have to show that $langle x^*,u_nrangleto0$ for all $x^*$, not just the one in $D^*$.
Since $D^*$ is dense, you can approximate $x^*$ by a sequence $x^*_m$ in $D^*$.

Now you know that $langle x^*_m,u_nrangleto0$ for $ntoinfty$.

So it would be natural to somehow combine this term with $langle x^*,u_nrangle$
to show that the latter converges to $0$.