I have the next exercise:
Let $E$ a normal vectorial space and let $C\subseteq E$ a convex set
such that $0\in C$. Lets \begin{eqnarray*}
C^{\star} &=& \{f\in E^{\star}:\langle f,x\rangle\leq1,\forall x\in C\}\\
C^{\star\star} &=& \{x\in E:\langle f,x\rangle\leq1,\forall f\in C^{\star}\}\\ \end{eqnarray*}
- Prove that $C^{\star\star}=\overline{C}$.
- If $C$ is a vector space, what are $C^{\star}$ and $C^{\star\star}$?.
In the first part, I did the following:
Reasoning by double inclusion: Let's see what $\overline{C}\subseteq C^{\star\star}$, let $x\in C$, then for all $f\in C^\star$ verify that $\langle f,x\rangle\leq1$, then $x\in C^\star\star$, i.e., $C\subseteq C^{\star\star}$.
On the other hand, let us note that $C^{\star\star}$ is closed: Let $(x_n)\subseteq C^{\star\star}$, note that for all $f\in C^{\star\star}$
\begin{equation*}
\begin{split}
\langle f,x\rangle
&=\langle f,x-x_n\rangle+\langle f,x_n\rangle\\
&\leq\|f\|\cdot\|x-x_n\|+\langle f,x_n\rangle\\
&\leq\|f\|\cdot\|x-x_n\|+1
\end{split}
\end{equation*}
Then if $n\to\infty$, $x_n\to x$, then
\begin{equation*}
\langle f,x\rangle\leq1
\end{equation*}
Thus $x\in C^{\star\star}$. Then $C^{\star\star}=\overline{C^{\star\star}}$, and how
$\overline{C}\subseteq\overline{C^{\star\star}}=C^{\star\star}$, in particular $\overline{C}\subseteq C^{\star\star}$.
See that $C^{\star\star}\subseteq\overline{C}$ (we reason by contradiction
): Let $x_0\in C^{\star\star}$ such that $x_0\notin\overline{C}$. How $\{x_0\}$ is a compact set and $\overline{C}$ is a closed convex set, by Hahn-Banach-Theorem (separation) exists $f_0\in E^\star$ y $\alpha_0\in\mathbb{R}$ such that
\begin{equation*}
\langle f_0,x\rangle<\alpha_0<\langle f_0,x_0\rangle,\forall x\in\overline{C}
\end{equation*}
Note that $0\in C$, then
\begin{equation*}
\langle f_0,0\rangle<\alpha_0<\langle f_0,x_0\rangle\Rightarrow\alpha_0>0
\end{equation*}
Let $f=f_0/\alpha_0$, then:
\begin{equation*}
\langle f,x\rangle<1<\langle f,x_0\rangle,\forall x\in\overline{C}
\end{equation*}
luego en particular, para todo $x\in C$ tenemos que $\langle f,x_0\rangle<1$, then $f\in C^{\star}$. By other hand, how $x_0\in C^{\star\star}$ there is contradiction
, because $\langle f,x_0\rangle>1$. Then if $x_0\in C^{\star\star}$, then $x_0\in\overline{C}$.
For the next part, I cannot interpret these sets. What would they be? A ball? Anyone have any ideas on this?
Do these sets become relevant in any later content of the functional analysis or convex analysis?
Best Answer
If $C$ is a linear subspace then $C^*$ and $C^{**}$ are as well. The trick is to observe that in this case $$ C^*=\{ f: \ \langle f,x\rangle =0 \ \forall x\in C\}. $$ Assume not: then there is $f\in C^*$ and $x \in C$ such that $\langle f, x\rangle \ne 0$. Now set $y:=\frac 2{\langle f, x\rangle }x \in C$. Since $f\in C^*$, $$ 1\ge\langle f,y\rangle = \frac 2{\langle f, x\rangle }\langle f, x\rangle =2, $$ a contradiction.
What we exploited is that in the condition $\langle f,x\rangle \le 1$ we can replace $x$ by an arbitrary multiple of itself, thus we can make the left-hand side arbitrarily large, while the right-hand side is fixed.