= Solution
Let $C=\|T\|$. 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 $|f(w)|\le C\|w\|$, and $v\notin W$. An extension to $W+\mathbb Rv$ must have the form
$$
f_v(w+tv)=f(w)+t a.
$$
Define
$$
L_v=\sup_{w\in W}\bigl(f(w)-C\|w-v\|\bigr),\qquad
U_v=\inf_{w\in W}\bigl(C\|w+v\|-f(w)\bigr).
$$
For any $w,z\in W$, the bound on $f$ and the triangle inequality give
$$
f(w)+f(z)=f(w+z)\le C\|w+z\|\le C\|w-v\|+C\|z+v\|.
$$
Thus 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, specifically
$$
-C\|v\|\le L_v\le U_v\le C\|v\|.
$$
The 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 $f(w)+ta\le C\|w+tv\|$. 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 gives
$$
|f_v(w+tv)|\le C\|w+tv\|.
$$
The 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 $W_n=U+\operatorname{span}\{y_1,\ldots,y_n\}$. This defines compatible functionals $f_n$ on $W_n$ with the common bound $C$. Their union $f_\infty$ on $W_\infty=\bigcup_nW_n$ 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 $\|y_{j_n(v)}-v\|<2^{-n}$. Density ensures that this integer exists, and leastness specifies it uniquely. Define
$$
\widetilde T(v)=\lim_{n\to\infty}f_\infty(y_{j_n(v)}).
$$
The limit exists because
$$
|f_\infty(y_{j_n(v)})-f_\infty(y_{j_m(v)})|
\le C(2^{-n}+2^{-m}).
$$
Any 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 $\widetilde T$. It also shows $\widetilde T=f_\infty$ on $W_\infty$ and $|\widetilde T(v)|\le C\|v\|$. Restriction to $U$ therefore gives
$$
\boxed{\widetilde T|_U=T,\qquad\|\widetilde T\|=\|T\|.}
$$
The 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.
Back to article page
