Continuous Linear Forms on a TVS and its Continuous Dual

Table of Contents

Continuous Linear Forms on a Topological Vector Space and its Continuous Dual

Definition: Let

$E$

be a topological vector space. The Continuous Dual of $$E$$ denoted by

$E'$

, is the set of all continuous linear forms on

$E$

Bounded Neighbourhood Criterion for a Linear Form to be Continuous

Proposition 1: Let

$E$

be a topological vector space. Then a linear form

$f$

is continuous if and only if there exists a neighbourhood of the origin

$U$

such that

$f$

is bounded on

$U$

Proof: $\Rightarrow$ Suppose that $$f$$ is continuous. Then $$f$$ is continuous at the origin. Given $\epsilon > 0$ , $(-\epsilon, \epsilon)$ is a neighbourhood of $$f(o) = 0$$ . So by the continuity of $$f$$ at the origin there exists a neighbourhood $$U$$ of the origin such that $f(U) \subseteq (-\epsilon, \epsilon)$ . Thus $$f$$ is bounded on $$U$$ .

$\Leftarrow$ Suppose that there exists a neighbourhood of the origin, call it $$U$$ , such that $$f$$ is bounded by some $$M > 0$$ on $$U$$ . Then $|f(x)| \leq M$ for all $x \in U$ . Then $\left | f \left ( \frac{\epsilon x}{M} \right ) \right | < \epsilon$ for all $x \in U$ . Setting $\displaystyle{y = \frac{\epsilon x}{M}}$ . Then $x = \epsilon^{-1} My$ , and thus $|f(y)| < \epsilon$ for all $\epsilon^{-1}My \in U$ , i.e., for all $y \in \epsilon^{-1}M U$ .

So for each $\epsilon > 0$ there exists a neighbourhood $\epsilon^{-1}M U$ of the origin such that $f(\epsilon^{-1}MU) \subseteq (-\epsilon, \epsilon)$ , and so $$f$$ is continuous at the origin, and thus continuous. $\blacksquare$

Proposition 2: Let

$E$

be a locally convex topological vector space. If

$f$

is a linear form on

$E$

and if

$p$

is a continuous seminorm on

$E$

such that

|f(x)| \leq p(x)

for all

x \in E

then

$f$

is a continuous linear form.

Proof; Let $$f$$ be a linear form and let $$p$$ be a continuous seminorm such that $|f(x)| \leq p(x)$ for all $x \in E$ . Since $$p$$ is a continuous seminorm $p^{-1}((-1, 1)) = \{ x : p(x) < 1 \}$ is an open neighbourhood of the origin, and by setting $U = \{ x : p(x) < 1 \}$ , we have that for all $x \in U$ , $|f(x)| \leq p(x) < 1$ so that $$f$$ is bounded on the neighbourhood $$U$$ . Thus by the previous proposition, $$f$$ is continuous. $\blacksquare$

Corollary 3: Let

$E$

be a locally convex topological vector space. If

$f$

is a continuous linear form on

$E$

then

$|f|$

is a continuous seminorm on

$E$

Proof: Let $$f$$ be a continuous linear form on $$E$$ . Then observe $$|f|(0) = |f(0)| = |0| = 0$$ , $|f|(\lambda x) = |f(\lambda x)| = |\lambda| |f(x)| = |\lambda| |f|(x)$ for all $\lambda \in \mathbf{F}$ and for all $x \in E$ , and furthermore:

(1)

\begin{align} \quad |f|(x + y) = |f(x + y)| = |f(x) + f(y)| \leq |f(x)| + |f(y)| = |f|(x) + |f|(y) \end{align}