The Polar of a Subspace

# The Polar of a Subspace

Recall from The Polar of a Set page that if $(E, F)$ is a dual pair and $A \subseteq E$, then the polar of $A$ in $F$ is defined to be the set of all points $y \in F$ such that:

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

The polars of subspaces of $E$ have a particularly nice form.

Proposition 1: Let $(E, F)$ be a dual pair and let $M \subseteq E$ be a subspace of $E$. Then $M^{\circ}$ is the set of all $y \in F$ such that $\langle x, y \rangle = 0$ for all $x \in M$ |

**Proof:**Observe that if $y \in M^{\circ}$ then:

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

- However, for each $\lambda \in \mathbf{F}$, $x \in M$ implies that $\lambda x \in M$ since $M$ is a subspace of $E$. So for all $x \in M$:

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

- Hence, for all $\lambda \in \mathbf{F}$ with $\lambda \neq 0$, $|\langle x, y \rangle| \leq |\lambda|^{-1}$ for all $x \in M$ so that $\langle x, y \rangle = 0$ for all $x \in M$. Thus $M^{\circ}$ is the set of all $y \in F$ such that $\langle x, y \rangle = 0$. $\blacksquare$

Note that in particular, for the dual pair $(E, E^*)$ and for a subspace $M$ of $E$, $M^{\circ} = M^{\perp}$ where $M^{\perp}$ is the set of all linear forms on $E$ such that $f(x) = \langle x, f \rangle = 0$ for all $x \in M$.