In the book where I'm studying there this exercise:
"(Goldstine's theorem). Recall that $E \subset E''$, or, more precisely, $J_E(E)$ is a subspace of $E''$ where $J_E :E \longrightarrow E''$ is linear isometry. Show that the weak* topology on $E''$ induces on $E$ the weak topology and that the closed unit ball $B_E$ of $E$ is $w*-dense$ in the unit ball $B_{E''}$ of $E''$. Thus $E$ is $w*-dense$ in $E''$."
"Hint: denote by $\overline{B}_E$ the weak* closure in $E''$ of the closed unit ball $B_E$ and suppose that $\mathbb{K}=\mathbb{R}$. To prove that $\overline{B}_E=B_{E''}$, let $w_0 \in E''$ be any point not in $\overline{B}_E$, a weak* closed convex set. Then there is a $u \in E'$ with $ \left \| u \right \|_{E'}=1$, such that $\sup_{w \in \overline{B}_E} \langle u, w \rangle < \langle u, w_0 \rangle$, and it follows that $ \left \| w_0 \right \|_{E''} > 1$. Therefore $B_{E''} \subset \overline{B}_E$."
The first part I have demonstrated as follows:
proof 1) To define the weak* topology $w^*=\sigma(E'',E')$ on $E'':=\mathcal{L}(E',\mathbb{K})$ you can use the same reasoning to build weak* topology on $E':=\mathcal{L}(E,\mathbb{K})$. You may consider $\mathcal{E}$ vector subspace of the algebraic dual of $E''$, i.e. of $(E'')'=\mathcal{L}(E'',\mathbb{K})$, then the weak* topology $w^*$ on $E''$ is the restriction to $\mathcal{E} \subset (E'')'$ of the pointwise convergence on $\mathbb{K}^E$, and it is the weakest topology that makes continuos the linear functional $u:E'' \longrightarrow \mathbb{K}$, and unless the linear isometry $J_E : E \longrightarrow E''$, we can assume that $E \subset E''$, and then weak* topology $w^*$ makes continuous the linear functional $u:E \longrightarrow \mathbb{K}$ (as restriction of $u:E'' \longrightarrow \mathbb{K}$), i.e. it induced on $E$ the weak topology $\sigma(E,E')$, which it is precisely weakest topology with this property (substantially by definition).
It is correct this first part?
I think that the second part is substantially a consequence of the following result:
"A). Assume that $E$ is a real or complex locally convex space. If $K$ is a convex and balanced closed subset of $E$ and $x_0 \in E \setminus K$, then $\exists u \in E'$ such that $|u(K)| \leq 1$ and $u(x_0) >1$."
It is a corollary of the following separation theorem of convex sets:
"Suppose A and B are two nonempty disjoint convex subsets of a real locally convex space $E$.
i) If one of them is open, then $A$ and $B$ are separated, i.e. $f(A) < f(B)$ per some $f \in E'$.
ii) If they are closed and one of them is compact, then $A$ and $B$ are strictly separated, i.e. $\sup f(A) < \inf f(B)$."
B). Another well-known result is that if $C$ is a convex subset of the real or complex LCS $E$, then the weak closure of $C$ is equal to the closure $\overline{C}$ of $C$ for the vectorial topology of $E$.
I continue with the proof of the theorem Goldstine.
proof 2) without considering $J_E$ we must show that $\overline{B}_E=B_{E''}$. Since $B_E \subset B_{E''}$, where $B_{E''}$ is closed unit ball in $E''$, and by Banach-Alaoglu theorem $B_{E''}$ is $w^*$-compact and then $\overline{B}_E \subset B_{E''}$. Reciprocally, suppose by contradiction that $B_{E''} \nsubseteq \overline{B}_E$, then $\exists w_0 \in B_{E''}$ such that $w_0 \notin \overline{B}_E$. Now, $\overline{B}_E$ is $w^*$-closed convex balanced subset of $E''$, i.e. without considering $J_E$ we can consider it as weak-closed convex balanced subset of $E$ by proof 1) (it's correct?), again weak closure of convex subset in LCS is equal to the closure for vector topology: that is, be valid the hypotheses of previous result in A). Then there is a $u \in E'$ with $ \left \| u \right \|_{E'}=1$, such that $\sup_{w \in \overline{B}_E} \langle u, w \rangle < \langle u, w_0 \rangle$ with $\langle u , w_0 \rangle > 1$, but $w_0 \in B_{E''}$ and then $\left \| w_0 \right \|_{E''} \leq 1$, which is a contradiction.
Thanks for any help
This is also exercise 1) page $111$ points a)-b) in Functional Analysis by Rudin. By a discussion in another network the proof it's correct. For second part, it is follows essentially from ii) of convex separation theorem, taking $\lbrace w_0 \rbrace$ as a compact set, and $\overline{B}_E$ as closed set, equivalently considering corollary in A).