Polars and the Bipolar Theorem

Definition 1.5.1   Let $X$ be a LCS. If $A \subset X$, then the polar of $A$ is

$$A^\circ = \{ f \in X^* : \vert f(x)\vert \leq 1, \forall x \in A \}$$

If $B \leq X^*$, then the prepolar of $B$ is

$${}^\circ B = \{ x \in X : \vert f(x)\vert \leq 1, \forall f \in B$$

Finally, for $A \subset X$, the bipolar of $A$ is ${}^\circ (A^\circ) \subset X$.

Examples:

$$B_1(0)^\circ = \{ f \in X^* : \vert f(x)\vert \leq 1 \forall x \in B_1(0) \}$$

$$=$$    unit ball of $X^*$ $$$$

$$B_\delta (0)^\circ = \{ f \in X^* : \vert f(x)\vert \leq 1, \forall x, \Vert x\Vert < \delta \} = B_{\frac1\delta}(0)$$$$\text( in X^* )$$

$$\{ 0 \}^\circ = X^*$$

$$\{ x \}^\circ = \{ f \in X^* : \vert f(x)\vert \leq 1 \}, x \neq 0$$

Fix $x \neq 0$ and consider

$$(\mathbb{R}_t x)^\circ = \{ f \in X^* : \vert f(\lambda x)\vert \leq 1, \forall \lambda \geq 0 \}$$

$$= \{ f \in X^* : f(x) = 0 \}$$

Thus, if $S \leq X$, then

$$S^\circ = \{ f \in X^* : f\vert _S \equiv 0 \} =: S^\perp.$$

We can think of polar as a generalization of annihilator.

``Obvious" Properties

1. $A^\circ$ is convex and balanced (as are prepolars).
2. If $A_1 \subset A \subset X$, then $A^\circ \subset A_1^\circ$.
   Similarly, if $B_1 \subset B \subset X^*$, then ${}^\circ B \subset {}^\circ B_1$.
3. For $\alpha \neq 0$, $(\alpha A)^\circ = \frac1\alpha (A^\circ)$.
   ${}^\circ (\alpha B) = \frac1\alpha {}^\circ B$
   
4. $A \subset {}^\circ (A^\circ) = \{ x \in X : \vert f(x)\vert \leq 1$    for all $f$    such that $\vert f(a)\vert \leq 1, \forall a \in A \}$
   $$B \subset ({}^\circ B)^\circ$$

5. $A = ({}^\circ A^\circ )^\circ$, $A \subset X$
   ${}^\circ B = {}^\circ ({}^\circ B^\circ)$, $B \subset X^*$.

Wednesday, 1-25-2006:

Recall, we said $A^\circ = ({}^\circ A^\circ)^\circ$ for $A \subset X$.

Proof. As $A \subset {}^\circ A^\circ$, so $A^\circ \supset ({}^\circ A^\circ)^\circ$. Since $B \subset {}^\circ B^\circ$ for $B \subset X^*$, letting $B =A ^\circ$, we have

$$A^\circ \subset {}^\circ (A^\circ)^\circ = ({}^\circ A^\circ)^\circ$$

$\qedsymbol$

Theorem 1.5.2   Bipolar Theorem
Let $X$ be a LCS and $A \subset X$. Then ${}^\circ A^\circ$ is the closed, convex, and balanced hull of $A$ (i.e., the intersection of all closed, convex, and balanced sets containing $A$).

Proof. Let $B$ be the closed, convex, and balanced hull of $A$. Since ${}^\circ A^\circ$ is closed, convex, and balanced, $B \subset {}^\circ A^\circ$. Let $x \notin B$. Then, by the Hahn-Banach separation theorem (on compact sets), there is $f \in X^*$, $\alpha \in \mathbb{R}$, $\epsilon > 0$ such that for all $b \in B$,

    Re $$f(a) \leq \alpha < \alpha + \epsilon <$$    Re $$f(x)$$

Dividing by $\alpha$ (or, if $\alpha = 0$, replacing with $\frac\epsilon 2$)

    Re $$\frac1\alpha f(a) \leq 1 < 1 + \frac\epsilon \alpha <$$    Re $$\frac1\alpha f(x)$$

Replacing $f$ with $\frac1\alpha f$, we may assume $\alpha = 1$.
Given $b \in B$, since $B$ is balanced, there is $b' \in B$ such that

$$\vert f(b)\vert = \Re f(b')$$

So, for all $b \in B$,

$$\vert f(b)\vert < 1 < 1 + \epsilon <$$    Re $$f(x) \leq \vert f(x)\vert$$

So, $f \in B^\circ$ and $x \notin {}^\circ B^\circ$.
Since $A \subset B$,

$$\vert f(a)\vert \leq 1, \forall a \in A$$

so $f \in A^\circ$, and $x \notin {}^\circ A^\circ$. Hence, $B \supset {}^\circ A^\circ$. $\qedsymbol$