The Polar of a Set

Definition: Let $(E, F)$ be a dual pair and let $A \subseteq E$. The Polar of $A$ in $F$ denoted by $A^{\circ}$ is the set of all $y \in F$ such that $\displaystyle{\sup \{ |\langle x, y \rangle | : x \in A \} \leq 1}$.

The following proposition gives us some basic properties of polar sets.

Proposition 1: Let $(E, F)$ be a dual pair and let $A, B, A_{\alpha} \subseteq E$. (1) $A^{\circ}$ is absolutely convex and $\sigma(F, E)$-closed. (2) If $A \subseteq B$ then $B^{\circ} \subseteq A^{\circ}$. (3) For all $\lambda \in \mathbf{F}$ with $\lambda \neq 0$, $\displaystyle{(\lambda A)^{\circ} = \frac{1}{|\lambda|} A^{\circ}}$. (4) $\displaystyle{\left ( \bigcup_{\alpha} A_{\alpha} \right )^{\circ} = \bigcap_{\alpha} A_{\alpha}^{\circ}}$.

• Proof of (1): Let $y, y' \in A^{\circ}$ and let $\lambda, \mu \in \mathbf{F}$ be such that $|\lambda| + |\mu| \leq 1$. Then by the linearity of $\langle \cdot, \cdot \rangle$ in the second component that for all $x \in A$:

(1)

\begin{align} \quad |\langle x, \lambda y + \mu y' | \leq |\lambda| |\langle x, y \rangle | + |\mu| |\langle x, y' \rangle| \leq |\lambda| + |\mu| \leq 1 \end{align}

• Thus $\lambda y + \mu y' \in A^{\circ}$ and so $A^{\circ}$ is absolutely convex. Furthermore, observe that $A^{\circ}$ can be written in the form:

(2)

\begin{align} \quad A^{\circ} = \bigcap_{x \in A} \{ y \in F : |\langle x, y \rangle| \leq 1 \} \end{align}

• But for each $x \in A$, $\{ y \in F : |\langle x, y \rangle | \leq 1 \}$ is $\sigma (F, E)$-closed as the sets of the form $\{ y \in F : |\langle x, y \rangle| \leq 1 \}$ are contained in a base of $\sigma(F, E)$-closed neighbourhoods of the origin for the $\sigma(F, E)$ topology on $F$ (see The Weak Topology on E Determined by F page). Thus $A^{\circ}$ is an arbitrary intersection of $\sigma(F, E)$-closed sets and is consequentially $\sigma(F, E)$-closed itself. $\blacksquare$

• Proof of (2): Suppose that $A \subseteq B$. Then observe that if $y \in F$ is such that $\displaystyle{\sup \{ |\langle b, y \rangle| : b \in B \} \leq 1}$ then trivially, $\displaystyle{\sup \{ |\langle a, y \rangle| : a \in A \} \leq 1}$ too, so that $y \in B^{\circ}$ implies that $y \in A^{\circ}$. Thus $B^{\circ} \subseteq A^{\circ}$. $\blacksquare$

• Proof of (3): Suppose that $\lambda \in \mathbf{F}$ is such that $\lambda \neq 0$. Then observe that $y \in (\lambda A)^{\circ}$ if and only if:

(3)

\begin{align} \quad \sup \{ |\langle x, y \rangle| : x \in \lambda A \} \leq 1 \end{align}

• Or equivalently:

(4)

\begin{align} \quad \sup \left \{ |\langle x, y \rangle| : \frac{1}{\lambda} x \in A \right \} \leq 1 \end{align}

• Substituting $\displaystyle{z := \frac{1}{\lambda} x}$ so that $x = \lambda z$, we get the above is equivalent to:

(5)

\begin{align} |\lambda| \sup \{ |\langle z, y \rangle| : z \in A \} = \sup \{ |\langle \lambda z, y \rangle| : z \in A \} \leq 1 \end{align}

• Which is itself equivalent to saying that $\displaystyle{y \in \frac{1}{|\lambda|} A^{\circ}}$. $\blacksquare$

• Proof of (4): Observe that $\displaystyle{y \in \left ( \bigcup_{\alpha}A_{\alpha} \right )^{\circ}}$ if and only if:

(6)

\begin{align} \quad \sup \left \{ |\langle x, y \rangle| : x \in \bigcup_{\alpha} A_{\alpha} \right \} \leq 1 \end{align}

• And that $\displaystyle{y \in \bigcap_{\alpha} A_{\alpha}^{\circ}}$ if and only if for every $\alpha$:

(7)

\begin{align} \quad \sup \{ |\langle x, y \rangle | : x \in A_{\alpha} \} \leq 1 \end{align}

• So the sets are clearly the same. $\blacksquare$