A closed subspace of a Banach space need not contain a nearest point to every ambient vector

Let $(x^{(m)})$ be a Cauchy sequence in $c_0$ for the supremum norm, with coordinates $x^{(m)}=(x_n^{(m)}{)}_{n\ge 0}$. For each $n$, the scalar sequence $(x_n^{(m)}{)}_m$ is Cauchy because $\mathrm{\mid }{x}_n^{(m)}-x_n^{(\mathrm{\ell })}\mathrm{\mid }\le \mathrm{\parallel }{x}^{(m)}-x^{(\mathrm{\ell })}{\mathrm{\parallel }}_{\mathrm{\infty }}$. Since the scalars are real, let $z_n:={\lim }_m{x}_n^{(m)}$. Choosing $M_0$ with $\mathrm{\parallel }{x}^{(m)}-x^{(\mathrm{\ell })}{\mathrm{\parallel }}_{\mathrm{\infty }}<1$ for $m,\mathrm{\ell }\ge M_0$ and letting $\mathrm{\ell }\to \mathrm{\infty }$ gives $\mathrm{\mid }{z}_n\mathrm{\mid }\le \mathrm{\parallel }{x}^{(M_0)}{\mathrm{\parallel }}_{\mathrm{\infty }}+1$ for every $n$, so $z:=(z_n)$ is bounded. Given $\epsilon >0$, choose $M$ such that $\mathrm{\parallel }{x}^{(m)}-x^{(\mathrm{\ell })}{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon \mathrm{/}2$ whenever $m,\mathrm{\ell }\ge M$. Letting $\mathrm{\ell }\to \mathrm{\infty }$ coordinatewise gives $\mathrm{\mid }{x}_n^{(m)}-z_n\mathrm{\mid }\le \epsilon \mathrm{/}2$ for every $n$ and every $m\ge M$, hence $\mathrm{\parallel }{x}^{(m)}-z{\mathrm{\parallel }}_{\mathrm{\infty }}\le \epsilon \mathrm{/}2$. Fix such an $m$ and choose $N$ with $\mathrm{\mid }{x}_n^{(m)}\mathrm{\mid }<\epsilon \mathrm{/}2$ for all $n\ge N$, since $x^{(m)}\in c_0$. Then $\mathrm{\mid }{z}_n\mathrm{\mid }<\epsilon$ for $n\ge N$, so $z\in c_0$. The displayed uniform estimate also gives $\mathrm{\parallel }{x}^{(m)}-z{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$. Therefore $c_0$ is Banach.

The series defining $\phi$ converges absolutely and $\mathrm{\mid }\phi (z)\mathrm{\mid }\le {\sum }_{n=0}^{\mathrm{\infty }}{2}^{-n-1}\mathrm{\mid }{z}_n\mathrm{\mid }\le \mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}$ for every $z\in c_0$, so $\phi$ is a bounded linear functional with $\mathrm{\parallel }\phi \mathrm{\parallel }\le 1$. If $z^{(m)}\in H$ and $z^{(m)}\to z$ in $c_0$, then $\mathrm{\mid }\phi (z)\mathrm{\mid }=\mathrm{\mid }\phi (z-z^{(m)})\mathrm{\mid }\le \mathrm{\parallel }z-z^{(m)}{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$, so $z\in H$. Thus $H$ is a closed linear subspace. Also $\phi (x)=1$.

For every $y\in H$ one has $\phi (x-y)=1$, hence $1=\mathrm{\mid }\phi (x-y)\mathrm{\mid }\le \mathrm{\parallel }x-y{\mathrm{\parallel }}_{\mathrm{\infty }}$ by step 2.1. Therefore $\mathrm{dist}(x,H)\ge 1$.

For $N\ge 1$, define $z^{(N)}\in c_0$ by $z_n^{(N)}:=(1-2^{-N}{)}^{-1}$ for $0\le n<N$ and $z_n^{(N)}:=0$ for $n\ge N$. Then $\phi (z^{(N)})=(1-2^{-N}{)}^{-1}{\sum }_{n=0}^{N-1}{2}^{-n-1}=1$, so $y^{(N)}:=x-z^{(N)}$ lies in $H$, and $\mathrm{\parallel }x-y^{(N)}{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }{z}^{(N)}{\mathrm{\parallel }}_{\mathrm{\infty }}=(1-2^{-N}{)}^{-1}\to 1$. Hence $\mathrm{dist}(x,H)\le 1$.

Steps 3.1 and 3.2 give $\mathrm{dist}(x,H)=1$. Suppose some $y\in H$ satisfied $\mathrm{\parallel }x-y{\mathrm{\parallel }}_{\mathrm{\infty }}=1$, and put $z:=x-y$. Then $\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}=1$ and $\phi (z)=1$ by step 2.1. Also $1=\phi (z)={\sum }_{n=0}^{\mathrm{\infty }}{2}^{-n-1}{z}_n\le {\sum }_{n=0}^{\mathrm{\infty }}{2}^{-n-1}\mathrm{\mid }{z}_n\mathrm{\mid }\le {\sum }_{n=0}^{\mathrm{\infty }}{2}^{-n-1}\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}=1.$

So equality holds throughout in step 4.1, forcing $\mathrm{\mid }{z}_n\mathrm{\mid }=1$ and $z_n\ge 0$ for every $n$. That contradicts $z\in c_0$, since a sequence converging to $0$ cannot have all coordinates equal to $1$ in modulus.

Thus $H$ is a closed subspace of the Banach space $c_0$, the distance from $x$ to $H$ is $1$, and no $y\in H$ attains it. This refutes the claim.