The Polar Criterion for Equicontinuity of a Set of Linear Forms

# The Polar Criterion for Equicontinuity of a Set of Linear Forms

Recall that if $E$ is a topological vector space, then a linear form $f$ is continuous on $E$ if and only if for every neighbourhood of real (or complex) numbers $A$ of $0$, there exists a neighbourhood $U \subseteq E$ of the origin $o$ such that:

(1)\begin{align} \quad f(U) \subseteq A \end{align}

We make a similar definition for equicontinuity of a collection of linear forms on $E$.

Definition: Let $E$ be a topological vector space. A collection of linear forms $\mathcal V$ on $E$ is said to be Equicontinuous if for every neighbourhood of real (or complex) numbers $A$ of $0$, there exists a neighbourhood $U \subseteq E$ of the origin $o$ such that $f(U) \subseteq A$ for all $f \in \mathcal V$. |

Equicontinuity of a collection of linear forms in a Hausdorff locally convex topological vector space can be specified in terms of polar sets.

Proposition 1: Let $E$ be a Hausdorff locally convex topological vector space and let $\mathcal V$ be a collection of continuous linear forms on $E$. Then $\mathcal V \subseteq E'$ is equicontinuous if and only if there exists a neighbourhood $U$ of the origin $o$ such that $\mathcal V \subseteq U^{\circ}$ (where the polar $U^{\circ}$ is taken in $E'$). |

**Proof:**Since $E$ is a Hausdorff locally convex topological vector space, $(E, E')$ is a dual pair.

- $\Rightarrow$ Suppose that $\mathcal V$ is equicontinuous. Let $A = \{ \lambda \in \mathbf{F} : |\lambda| \leq 1 \}$, which is a neighbourhood of real (or complex) numbers of $0$. Then by the equicontinuous of $\mathcal V$, there exists a neighbourhood $U$ of the origin $o$ such that $f(U) \subseteq A$ for all $f \in \mathcal V$. So for each $f \in \mathcal V$ we have that:

\begin{align} |\langle u, f \rangle| \leq 1 \end{align}

- so that $\sup \{ |\langle u, f \rangle| : u \in U \} \leq 1$, and hence $f \in U^{\circ}$. Thus $\mathcal V \subseteq U^{\circ}$.

- $\Leftarrow$ Suppose that there exists a neighbourhood $U$ of the origin $o$ such that $\mathcal V \subseteq U^{\circ}$. Given a neighbourhood $A$ of real or complex numbers of $0$, there exists a $\lambda > 0$ such that $\overline{B}(0, \lambda) \subseteq A$.

- So for each $f \in \mathcal V$, $f \in U^{\circ}$ so that $|f(u)| = |\langle u, f \rangle| \leq 1$ for all $u \in U$ and for all $f \in \mathcal V$. Hence by linearity, if $x \in \lambda^{-1} U$, then $|f(x)| \leq \lambda$ for all $f \in \mathcal V$, and thus:

\begin{align} \quad f(\lambda^{-1} U) \subseteq \overline{B}(0, \lambda) \subseteq A \end{align}

- for all $f \in \mathcal V$, so that $\mathcal V$ is equicontinuous. $\blacksquare$