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In a real normed vector space with a specified countable dense subset, a bounded linear functional on any vector subspace admits a norm-preserving extension without the axiom of choice. Add the dense vectors one at a time using the supremum endpoint in the one-dimensional dominated extension of a real linear functional. This makes the recursion deterministic. Extend from the resulting dense subspace by continuity, using the least index in the dense list which approximates a given vector to within $2^{-n}$.

Let $C=\Vert T\Vert$. We construct the extension by a deterministic recursion over the given dense list, using the one-dimensional dominated extension of a real linear functional. Neither Zorn's lemma nor any selection of a family of extensions is needed.

Suppose $f:W\to \mathbb{R}$ is already a linear functional satisfying $|f(w)|\le C\Vert w\Vert$, and $v\notin W$. An extension to $W+\mathbb{R}v$ must have the formDefineFor any $w,z\in W$, the bound on $f$ and the triangle inequality giveThus every quantity in the supremum defining $L_v$ is at most every quantity in the infimum defining $U_v$. Taking $w=0$ and $z=0$ gives finite bounds, specificallyThe completeness of the real numbers defines these endpoints uniquely. Set $a=L_v$; this is an explicit rule, not an arbitrary choice from a nonempty interval.

For $t>0$, the upper-endpoint inequality with $w/t$ yields $f(w)+ta\le C\Vert w+tv\Vert$. For $t<0$, write $t=-s$ with $s>0$ and use the lower-endpoint inequality with $w/s$ to obtain the same bound. For $t=0$ it is the original bound. Applying the upper bound also to the negative of the vector givesThe representation $w+tv$ is unique because $v\notin W$, so $f_v$ is a well-defined bounded linear functional extending $f$.

Start with $W_0=U$ and $f_0=T$. At step $n$, if $y_n\in W_{n-1}$ leave the functional unchanged; otherwise use exactly the supremum-endpoint rule with $v=y_n$. Put $W_n=U+\mathrm{span}\{y_1,\dots ,y_n\}$. This defines compatible functionals $f_n$ on $W_n$ with the common bound $C$. Their union $f_{\infty }$ on $W_{\infty }={\bigcup }_n{W}_n$ is linear and bounded by $C$. The subspace $W_{\infty }$ is dense in $V$ because it contains every $y_n$.

To make the final passage to the closure equally free of the axiom of choice, for $v\in V$ let $j_n(v)$ be the least positive integer with $\Vert y_{j_n(v)}-v\Vert <2^{-n}$. Density ensures that this integer exists, and leastness specifies it uniquely. DefineThe limit exists becauseAny other approximating sequence in $W_{\infty }$ gives the same limit, by the same norm bound. Applying this independence to sums and scalar multiples of approximating sequences proves linearity of $\tilde{T}$. It also shows $\tilde{T}=f_{\infty }$ on $W_{\infty }$ and $|\tilde{T}(v)|\le C\Vert v\Vert$. Restriction to $U$ therefore givesThe upper bound follows from construction and the lower bound from restriction to $U$. This also covers $C=0$, when all extensions are zero. No completeness of $V$ or closedness of $U$ was assumed. The argument establishes the choice-free Hahn-Banach extension in a separable space by uniquely specified operations at every stage.

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.