 Solution

Use the Hausdorff convention for a locally convex space: a real vector space $X$ is equipped with a separating family $\mathcal{P}$ of seminorms. Thus each $p$ is nonnegative, subadditive and satisfies $p(tx)=|t|p(x)$, and for every $x\ne 0$ some $p\in \mathcal{P}$ has $p(x)>0$. The topology has a neighborhood basis at $a\in X$ consisting ofEquivalently, it is the coarsest vector-space topology making all these seminorms continuous. The separating condition is precisely what makes this topology Hausdorff. If the Hausdorff requirement were omitted, continuous linear functionals could not distinguish points in the common kernel of the seminorms.

The continuous dual space $X^{\ast }$ consists of all continuous linear functionals $X\to \mathbb{R}$. We first prove the needed Hahn-Banach theorem. Let $p$ be a real-valued sublinear function on $X$, meaning $p(a+b)\le p(a)+p(b)$ and $p(ta)=tp(a)$ for $t\ge 0$, and let $g$ be linear on a vector subspace $M$, with $g(m)\le p(m)$. To extend across $v\notin M$, writeThe necessary bounds on $c$ areThey are compatible becauseTaking one variable equal to zero also shows that $L,U$ are finite. Choose $c\in [L,U]$. The upper bound proves domination when $t>0$, after dividing $m+tv$ by $t$; the lower bound proves it when $t<0$, after dividing by $-t$. Thus $\tilde{g}\le p$ on $M+\mathbb{R}v$. This is the one-dimensional dominated extension of a real linear functional.

Order all dominated extensions of $g$ by extension of their domains. A chain has an upper bound obtained by taking the union of the domains and linear functionals. Zorn's lemma gives a maximal extension, and the one-dimensional construction shows that its domain must be all of $X$. We have therefore proved the real dominated-extension theorem. In particular, when $p$ is a seminorm, domination at both $x$ and $-x$ gives $|f(x)|\le p(x)$.

For $x_0\ne 0$, choose a continuous seminorm $p$ with $p(x_0)>0$, and define $g(t{x}_0)=tp(x_0)$ on its one-dimensional span. Then $|g(t{x}_0)|\le p(t{x}_0)$. Extend by the theorem just proved to $f$ with $|f(x)|\le p(x)$. This bound makes $f$ continuous, and $f(x_0)=p(x_0)>0$. Applying this to the difference of two distinct points proves that $X^{\ast }$ separates the points of $X$, the continuous-dual separation theorem for Hausdorff locally convex spaces.

For separation from a closed linear subspace, choose a basic balanced neighborhood $V$ of zero such that $(x_0+V)\cap Y=\varnothing$. WriteThen $q(x_0-y)\ge 1$ for every $y\in Y$. On $Y+\mathbb{R}{x}_0$, define $g(y+t{x}_0)=t$, which is well-defined since $x_0\notin Y$. For $t\ne 0$,for $t=0$ the inequality is immediate. The proved extension theorem gives a continuous $f$ dominated by $q$, withThis proves separation of a point from a closed linear subspace without using any unproved Hahn-Banach extension or separation result.