# Retraction by contraction

The Euclidean space ${\mathbb R^n}$ has a nice property: every closed convex subset ${C\subset \mathbb R^n}$ is the image of the whole space under a map ${f}$ that is simultaneously:

• a contraction, meaning ${\|f(a)-f(b)\|\le \|a-b\|}$ for all ${a,b\in\mathbb R^n}$;
• a retraction, meaning ${f(a)=a}$ for all ${a\in C}$.

Indeed, we can simply define ${f(x)}$ to be the (unique) nearest point of ${C}$ to ${x}$; it takes a bit of work to verify that ${f}$ is a contraction, but not much.

In other normed spaces, this nearest point projection does not work that well. For example, take ${X=\ell_1^2}$, the two-dimensional space with the Manhattan metric. Consider the line ${C=\{(2t,t)\colon t\in\mathbb R\}}$ which is a closed and convex set. The nearest point of ${C}$ to ${(2,0)}$ is ${(2,1)}$: moving straight up, since changing the first coordinate doesn’t pay off. Since ${(0,0)}$ remains fixed, the nearest point projection increases some pairwise distances, in this case from ${2}$ to ${3}$.

However, there is a contractive retraction onto this line, given by the formuls ${T(x_1,x_2) = \left(\frac23(x_1+x_2), \frac13(x_1+x_2)\right)}$. Indeed, this is a linear map that fixes the line ${x_1=2x_2}$ pointwise and has norm ${1}$ because $\displaystyle \|T(x)\|_1 = |x_1+x_2|\le \|x\|_1$

More generally, in every normed plane, every closed convex subset admits a contractive retraction. To prove this, it suffices to consider closed halfplanes, since a closed convex set is an intersection of such, and contractions form a semigroup. Furthermore, it suffices to consider lines, because having a contractive retraction onto a line, we can redefine it to be the identity map on one side of the line, and get a contractive retraction onto a halfplane.

Such a retraction onto a line, which is a linear map, is illustrated below.

Given the unit circle (black) and a line (blue), draw supporting lines (red) to the unit circle at the points where it meets the line. Project onto the blue line along the red ones. By construction, the image of the unit disk under this projection is contained in the unit disk. This precisely means that the map has operator norm ${1}$.
In spaces of dimensions ${3}$ or higher, there are closed convex subsets without a contractive retraction. For example, consider the plane in ${\ell_\infty^3}$ passing through the points ${A = (2,2,0)}$, ${B= (2,0,2)}$, and ${C= (0,2,2)}$. This plane has equation ${x_1+x_2+x_3=4}$. The point ${D=(1,1,1)}$ is at distance ${1}$ from each of A,B,C, and it does not lie on the plane. For any point E other than D, at least one of the distances AE, BE, CE exceeds 1. More precisely, the best place to project D is ${(4/3, 4/3, 4/3)}$ which is at distance ${4/3}$ from A, B, and C.
Two natural questions: (a) is there a nice characterization of normed spaces that admit a contractive retraction onto every closed convex subset? (b) what is the smallest constant ${L=L(n)}$ such that every ${n}$-dimensional normed space admits an ${L}$-Lipschitz retraction onto every closed convex subset?