Nearest-point projection is the tubular retraction after shrinking

If $S$ is empty the assertion is vacuous. Otherwise fix $x=E(p,v)=p+v$ in the tube from [L1]. Since $S$ is closed, the continuous distance function to $x$ attains a minimum on the compact set $S\cap \bar{B}(x,\mathrm{\parallel }v\mathrm{\parallel })$; let $q$ be a minimizer. Because $p\in S$, $\mathrm{\parallel }x-q\mathrm{\parallel }\le \mathrm{\parallel }v\mathrm{\parallel }$. The first derivative of $y\mapsto \mathrm{\parallel }x-y{\mathrm{\parallel }}^2$ along every tangent vector at $q$ vanishes, so $x-q\perp T_{q}S$ and $x=E(q,x-q)$.

The triangle inequality gives $\mathrm{\parallel }q-p\mathrm{\parallel }\le \mathrm{\parallel }q-x\mathrm{\parallel }+\mathrm{\parallel }x-p\mathrm{\parallel }\le 2\mathrm{\parallel }v\mathrm{\parallel }<r(p)\mathrm{/}2$. Also $\mathrm{\parallel }x-q\mathrm{\parallel }\le \mathrm{\parallel }v\mathrm{\parallel }<r(p)\mathrm{/}4$. Choose $a$ with $2\mathrm{\parallel }v\mathrm{\parallel }<a<r(p)$. Both $(p,v)$ and $(q,x-q)$ then lie in $V_a(p)$, on which [L1] makes $E$ injective. Since they have the same image $x$, they coincide, and in particular $q=p$. The same argument applies to any minimizing $q$, so the nearest point is unique.

By [L2], $r(x)=p$. Step 2.1 shows that $p$ is exactly the unique nearest point. The theorem’s constructed tube already has this property, hence so does any smaller tube.