Let $X$ be Banach space and $X^*$ be its dual space. Let $S\subset X$ be a convex set containing the origin. Prove that $\bar S = S^{**}.$

Question

I am learning the Dual Space in functional analysis. I have trouble proving this statement.

Let $X$ be Banach space and $X^*$ be its dual space. Let $S\subset X$ be a convex set containing the origin. Define

$$S^*=\{\phi\in X^*: \phi(x)\le 1\ \;\forall x\in S\},$$ $$S^{**}=\{x\in X;\phi(x)\le 1, \text{for all}~\phi\in S^*\}.$$ Prove that $\bar S = S^{**},\bar S~\text{is the closure of} ~S.$

My attempt:

I want to show $\bar S\subset S^{**}$ and $S^{**} \subset \bar S$.

$\forall x,y\in S,(1-a)x+ay\in S$. $\phi((1-a)x+ay)=(1-a)\phi(x)+a\phi(y) \subset\phi(S)$.

I can show that $\phi(S)$ is a convex set. But I don't know what I can do next to prove the statement.

Answer 1

Since $\phi (x) \le 1,\; \forall x \in S,\; \forall \phi \in S^*$ then $S \subset S^{**}$

On the other hand $\forall \phi \in X^*$, $\phi$ is continuous, so $\{x \in X:\; \phi (x) \le 1\}$ is a closed set and $$S^{**} = \bigcap_{\phi \in S^* } \{x \in X :\; \phi(x) \le 1\}$$ so $S^{**}$ is a closed set (intersection of closed sets) then $\bar S \subset \bar {S^{**}} = S^{**}$

Conversly, if $a \not\in \bar S$ and $\epsilon = \frac{d(a, \bar S)}2 > 0$ then $S$ and $B(a,\epsilon)$ are two convex disjoints sets of $X$. By the Hahn–Banach separation theorem (https://en.wikipedia.org/wiki/Hahn%E2%80%93Banach_theorem#Hahn%E2%80%93Banach_Separation_Theorem) since $B(a,\epsilon)$ is open, you can obtain $\psi \in X^*$ and $t \in \mathbb R$ such that $$\forall x \in S, \; \forall y \in B(a,\epsilon), \; \psi (x) \le t < \psi (y)$$ Since $0 \in S$ then $t \ge 0$,

Since $\left(1-\frac{\epsilon}{2\left\|a\right\|}\right)a \in B(a, \epsilon)$, $\psi (a) \left(1-\frac{\epsilon}{2\left\|a\right\|}\right) > t$

let $\phi = \frac1{\psi (a)\left(1-\frac{\epsilon}{2\left\|a\right\|}\right)} \psi$, then $$\forall x \in X,\;\phi(x) = \frac{1}{\psi (a)\left(1-\frac{\epsilon}{2\left\|a\right\|}\right)}\psi (x) \le \frac t{\psi (a) \left(1-\frac{\epsilon}{2\left\|a\right\|}\right)} \le 1 $$ and $\phi (a) = \frac1{\left(1-\frac{\epsilon}{2\left\|a\right\|}\right)} > 1$ then $\phi \in S^*$ and $a \not \in S^{**}$.

Answer 2

First off, notice that if $S$ is convex, then $\overline{S}$ is also convex (this is true on any topological vector space - it simply follows from the continuity of the operations of the vector space structure).

1. $S^{**}\subset \overline{S}$.

   Suppose $x_0\notin \overline{S}$. Then $\left\{x_0\right\}$ and $\overline{S}$ are two disjoint convex subsets of $X$; moreover, $\left\{x_0\right\}$ is compact and $\overline{S}$ is closed. By the strong geometric form of the Hahn-Banach theorem, there is a functional $\phi \in X^*$ such that $$\phi(x_0)>\sup_{x\in \overline{S}}\phi(x) $$ Let $y_0:=\sup_{x\in \overline{S}}\phi(x)$ - we distinguish two cases.

   If $y_0> 0$ then $\phi':=\phi/y_0\in X^*$ is such that $\sup_{x\in \overline{S}}\phi'(x)=1$ , hence $\phi'(x)\leq 1$ for all $x\in \overline{S}$, so $\phi' \in S^{*}$, but $\phi'(x_0)>1$. Thus $x_0\notin S^{**}$.

   If $y_0\leq 0$, then $\phi(x_0)>0$. Choose $\lambda >0$ such that $\lambda \phi(x_0)>1$ and set $\phi':=\lambda \phi\in X^*$. Then we have $\phi'(x)\leq 0 \leq 1$ for all $x\in S$ (so $\phi' \in S^*$), and $\phi'(x_0)>1$. Thus we reach the same conclusion as the previous case, $x_0\notin S^{**}$.

2. $\overline{S}\subset S^{**}$.

   Let $x_0\in \overline{S}$. Then there is a sequence $\left\{x_n\right\}\subset S$ with $x_n\to x$. Now take $\phi \in S^*$. Then $\phi(x_n)\leq 1$ for all $n$, and thus $\phi(x_0)\leq 1$ as well by continuity. Hence $x_0\in {S}^{**}$.