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For $y=x-Sx\in Y$, the telescoping arithmetic means satisfybecause $x$ is a bounded sequence. Hence $\Vert 1-y{\Vert }_{\infty }\ge |1-N^{-1}{\sum }_{n=1}^N{y}_n|$ for every $N$, and passage to the limit gives $\Vert 1-y{\Vert }_{\infty }\ge 1$. Taking $y=0$ shows $d(1,Y)=1$. The proved Hahn-Banach distance formula therefore yields a bounded linear functional $L$ on the real l-infinity sequence space withWe verify that this single linear functional also has the two remaining properties.

We prove the required Hahn-Banach theorem for a seminorm directly. First suppose the vector space is real. At an intermediate extension stage, let $g:E\to \mathbb{R}$ be a linear functional with $|g(y)|\le p(y)$, and let $z\notin E$. Extending to $E+\mathbb{R}z$ amounts to choosing $c=f(z)$. The upper domination $f\le p$ requiresThe interval is nonempty: for all $y,w\in E$,Its endpoints are finite since inserting zero gives bounds $-p(z)$ and $p(z)$, while every lower candidate is below every upper candidate. Choose $c$ in the interval and put $f(y+tz)=g(y)+tc$. For $t>0$, the upper bound with $w=y/t$ proves $f(y+tz)\le p(y+tz)$. For $t<0$, writing $s=-t>0$ and applying the lower bound with $y/s$ proves the same inequality. The case $t=0$ is already known. Applying this domination to $-x$ and using $p(-x)=p(x)$ gives $|f(x)|\le p(x)$.

Partially order all dominated linear functional extensions of the original $g$ by extension of their domains and values. They form a nonempty set. The union along any chain is a well-defined dominated linear functional on a vector subspace, so every chain has an upper bound. The Zorn lemma gives a maximal extension. If its domain were not $X$, the preceding one-dimensional construction would enlarge it, a contradiction. Thus the required extension exists on $X$.

For a complex vector space, apply the proved real result to $\mathrm{Re}g$ on the underlying real vector subspace. Let $U:X\to \mathbb{R}$ be the resulting real linear functional, with $|U(x)|\le p(x)$, and setIt is additive and real-linear, and $f(ix)=if(x)$, so it is complex-linear. On $Y$, the identity $\mathrm{Re}g(iy)=-\mathrm{Im}g(y)$ proves $f(y)=g(y)$. Choose $\alpha$ with $|\alpha |=1$ and $\alpha f(x)=|f(x)|$ when $f(x)\ne 0$. ThenThe bound is immediate if $f(x)=0$. This proves the real and complex seminorm forms of Hahn-Banach theorem without invoking any version of that theorem.

For the distance to a set assertion, let $d=d(z,Y)>0$. Define on $Y+\mathbb{K}z$ the linear functional $g(y+\alpha z)=\alpha d$. The decomposition is unique because $z\notin Y$, and for $\alpha \ne 0$,For $\alpha =0$ the bound is immediate. Apply the just-proved Hahn-Banach theorem with $p(x)=\Vert x\Vert$ to get $f\in X^{\ast }$, $f{|}_Y=0$, $f(z)=d$, and $\Vert f\Vert \le 1$. Choose $y_n\in Y$ with $\Vert z-y_n\Vert \to d$. Sinceand $d>0$, passage to the limit gives $\Vert f\Vert \ge 1$. Thus $\boxed{f{|}_Y=0,f(z)=d(z,Y),\Vert f\Vert =1}$. This version of the Hahn-Banach distance formula needs neither a closed $Y$ nor an attained distance to a set.

For the Banach limit construction below, take the real l-infinity sequence space, let $S(x_1,x_2,\dots )=(x_2,x_3,\dots )$ be its left shift on bounded sequences, put $Y=(I-S){\ell }^{\infty }$, and let $1=(1,1,\dots )$. This is a vector subspace and its distance to a set from $1$ determines the construction.

Identify $X$ with its canonical image in $X^{\ast \ast }$ and putFor $x\in S_X$ and $\lambda \in \mathbb{R}$, if $|\lambda |\le 1/(d+1)$ thenIf $|\lambda |\ge 1/(d+1)$, thenTherefore $d(x,\mathrm{span}\{\Phi \})\ge c$.

The restriction of $x\in X^{\ast \ast }$ to $\ker \Phi \subseteq X^{\ast }$ has normwhere the first equality is the Hahn-Banach distance formula and $(\ker \Phi {)}^{\perp }=\mathrm{span}\{\Phi \}$. Scaling from $S_X$ givesfor every $x\in X$. Hence $\ker \Phi$ is $c$-norming for $X$.