Continuous Linear Forms on a TVS and its Continuous Dual

# 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. $\blacksquare$

## Bounded by a Continuous Seminorm Criterion for a Linear Form to be Continuous

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:

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

- So $|f|$ is a continuous seminorm on $E$. $\blacksquare$