In Wikipedia, it says that Hahn-Banach Theorem shows there are "enough" continuous linear functionals. But, why is that so in a space that is not necessarily normed? How does the statement of Hahn-Banach show this?
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Because you can easily define a linear functional in a subspace and the theorem ensures the existence of the corresponding extensions to the whole Banach. – Piquito May 12 '18 at 21:24
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Perhaps that what is meant is this: it is a consequence of the theorem that if $V$ is a normed vector space and if $v\in V$, then there exists a continuous linear map $\psi\colon V \longrightarrow {\mathbb K}$ with $ψ(v) = \|v\|$ and $\|\psi\| \leqslant 1$.
José Carlos Santos
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Let $X$ be a normed space and $M$ a finite-dimensional subspace of $X$. Take any linear functional $f : M \to \mathbb{F}$. It will be continuous because $M$ is finite-dimensional.
Hahn-Banach lets you extend $f$ to a continuous linear functional defined on all of $X$, so you obtain a nontrivial element of $X^*$.
mechanodroid
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One can use (the separation versions of) Hahn Banach to prove that in a locally convex topological space $X$, for any $x\in X$ and a subspace $Y\subset X$ with $x\not\in\overline{Y}$, there exists a continuous linear functional $\phi$ with $\phi(x)=1$ and $\phi(y)=0$ for all $y\in \overline{Y}$.
So HB shows that any locally convex topological vector space $X$ has "enough" continuous linear functionals. No norms involved at all (but seminorms instead; as locally convex spaces are precisely those where the topology is given by a family of seminorms).
Martin Argerami
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