Corollaries to the Hahn-Banach Theorem for Vector Spaces 1

Corollary 1: Let $E$ be a locally convex topological vector space and let $M \subseteq E$ be a subspace of $E$. Then every continuous linear form on $M$ can be extended to a continuous linear form on $E$.

• Proof: Let $f$ be a continuous linear form on $M$. Then by the proposition on the Continuous Linear Forms on a TVS and its Continuous Dual page there exists a neighbourhood $W \subseteq M$ of the origin and an $\alpha > 0$ for which $|f(w)| < \alpha$ for all $w \in W$. But since $E$ is a locally convex topological vector space, we know by the proposition on the Every LCTVS Has a Base of Closed Absolutely Convex Absorbent Neighbourhoods of the Origin page that $E$ has a base of absolutely convex absorbent neighbourhoods of the origin. So there exists an absolutely convex absorbent neighbourhood $V \subseteq W$ of the origin for which $|f(v)| < \alpha$ for all $v \in V$.

• Let $U := \alpha^{-1} V \subseteq M$. Then $U$ is an absolutely convex and absorbent neighbourhood of the origin for which $|f(u)| < 1$ for all $u \in U$. Let $p_U : E \to [0, \infty)$ be the gauge of $U$ (which is defined since $U$ is absolutely convex and absorbent). Then recall that:

(1)

\begin{align} \quad \{ x \in E : p_U(x) < 1 \} \subseteq U \subseteq \{ x \in E: p_U(x) \leq 1 \} \end{align}

• So $p_U(x) < 1$ implies that $x \in U$ and thus $|f(x)| \leq 1$. Observing that $f$ being a linear form implies that $|f|$ is a seminorm and by one of the propositions on the Seminorms and Norms on Vector Spaces page, we have that $|f(x)| \leq p_U(x)$ for all $x \in M$.

• By the Hahn-Banach theorem, there exists a linear form $f_1$ on $E$ such that $f_1$ is an extension of $f$, and $|f_1(x)| \leq p_U(x)$ for all $x \in E$. But $p_U$ is a continuous seminorm as it is the gauge of an absolutely convex and absorbent set in a locally convex topological vector space. By a proposition on the Continuous Linear Forms on a TVS and its Continuous Dual page, since $f_1$ is a linear form on $E$ and $p_U$ is a continuous seminorm on $E$ such that $|f_1(x)| \leq p_U(x)$ for all $x \in E$, we have that $f_1$ is continuous. $\blacksquare$

Corollary 2: Let $E$ be a vector space. Then, for seminorm $p$ on $E$ and for each $a \in E$ there exists a linear form $f$ on $E$ with $|f(x)| \leq p(x)$ for all $x \in E$, and $f(a) = p(a)$.

• Proof: Observe that $\mathrm{span} (a) = \mathbf{F} a$ is a vector subspace of $E$. Let $g$ on $\mathrm{span}(a)$ be defined for all $\lambda a \in \mathrm{span}(a)$ by $g(\lambda a) := \lambda p(a)$. Then $g$ is indeed a linear form on $\mathrm{span} (a)$ with the property that for each $x := \lambda a \in \mathrm{span} (a)$ we have that:

(2)

\begin{align} \quad |g(x)| = |g(\lambda a)| = |\lambda p(a)| = |\lambda| p(a) = p(\lambda a) = p(x) \end{align}

• So trivially $|g(x)| \leq p(x)$ for each $x \in \mathrm{span} (a)$. So by the Hahn-Banach theorem, there exists a linear form $f$ on $E$ that extends $g$ and such that $|f(x)| \leq p(x)$ for all $x \in E$. Moreover, we have that $f(a) = g(a) = g(1a) = p(a)$. $\blacksquare$

Corollary 3: Let $E$ be a Hausdorff and locally convex topological vector space and let $a \in E$. If $f(a) = 0$ for all $f \in E'$ then $a = o$.

Here, $E'$ denotes the continuous dual of $E$, i.e., the set of all continuous linear forms on $E$.

It should be remarked that since $E$ is a locally convex topological vector space, there exists a collection $Q$ of seminorms on $E$ for which the locally convex topology on $E$ is the coarsest topology determined by $Q$. This is used in conjunction with the fact that if $E$ is equipped with the coarsest topology determined by a collection $Q$ of seminorms on $E$, then $E$ is Hausdorff if and only if for every nonzero $x \in E$ there exists a $p \in Q$ such that $p(x) > 0$.

• Proof: Let $f(a) = 0$ for all $f \in E'$ and suppose instead that $a \neq o$. By the proposition on the Criterion for the Coarsest Topology Determined by a Set of Seminorms to be Hausdorff page, there exists a continuous seminorm $p$ for which $p(a) > 0$. But by Corollary 2, there must then exists a linear form $g$ on $E$ such that $|g(x)| \leq p(x)$ for all $x \in E$, and $g(a) = p(a) \neq 0$.

• But since $p$ is a continuous seminorm and since $|g(x)| \leq p(x)$ for all $x \in E$, we have that $g$ is continuous by one of the propositions on the Continuous Linear Forms on a TVS and its Continuous Dual page. So $g \in E'$ is such that $g(a) \neq 0$, which is a contradiction.

• So we must have that $a = o$. $\blacksquare$