## Projections onto a Convex Body

Given a point ${x} \in {{\bf R}^d}$ and an affine subspace ${H \neq {\bf R}^d}$, the projection of ${x}$ onto ${H}$ is the point ${P_H(x)}$ such that,

$\displaystyle \begin{array}{rcl} \vert \vert x - P_H(x) \vert \vert = \min \{ \vert \vert x y \vert \vert ~ \mid y ~ \in H \} \end{array} .$

One can obviously define other notions of projection but the above is probably the most commonly used in geometry. If ${x \notin H}$ then the projected point ${P_H(x)}$ is the unique point in ${H}$ such that the vector ${P_H(x) - x}$ is orthogonal to ${H}$. Also is well known that projecting any segment to an affine subspace can only shrink its length. The proofs of these facts are easy to see. But in fact these facts are just corollaries of the following two results.

 Given any nonempty closed convex set ${C \subseteq {\bf R}^d}$ and a point ${x \in {\bf R}^d}$, there is a unique point ${P_C(x) \in C}$ such that, $\displaystyle \begin{array}{rcl} \vert \vert x - P_C(x) \vert \vert = \min \{ \vert \vert x - y \vert \vert ~ \mid ~ y \in C\} \end{array} .$

and,

 Let ${C}$ be a nonempty closed convex set. The mapping ${P_C : {\bf R}^d \rightarrow C}$ is a contraction i.e. for any ${x,y \in {\bf R}^d}$ we have, $\displaystyle \begin{array}{rcl} \vert \vert P_C(x) - P_C(y) \vert \vert \leq \vert \vert x - y \vert \vert \end{array}$

Applying these results to affine spaces (which are nonempty closed convex sets) yields the results mentioned earlier. This projection that maps $x$ to $P_C(x)$ is known as the metric projection. The proofs of these facts are in order.

Proof: First we show that the minimum distance to ${C}$ is indeed obtained. This is easy and the details are as follows. Since ${C}$ is nonempty there is some point ${y \in C}$. Let ${l}$ denote ${\vert \vert x - y \vert \vert}$. Clearly a closest point to ${x}$ can only lie in ${C \cap B(x,l)}$ where ${B(x,l)}$ denotes a closed ball of radius ${l}$ with center ${x}$. But ${C \cap B(x,l)}$ is a closed and bounded set and is therefore compact. It is also nonempty. The function ${f(p)}$ defined for ${p \in C \cap B(x,l)}$ as ${f(p) = \vert \vert p - x \vert \vert}$ is a continuous function on a compact set and attains its minimum at some point ${P_C(x)}$.

Next we prove the point ${P_C(x)}$ where ${f}$ attains minimum, is unique. If ${z \in C}$ is another point at which ${\vert \vert x - z \vert \vert = \vert \vert x - P_C(x) \vert \vert}$ then in the triangle ${xP_C(x)z}$, which has two equal sides, the perpendicular from ${x}$ onto segment ${P_C(x)z}$ is clearly shorter than the length ${\vert \vert x - z \vert \vert}$. However the base of this perpendicular lies on the segment itself. By convexity of ${C}$ the entire segment is in ${C}$ and thus we have found a point in ${C}$ closer to ${x}$ than ${P_C(x)}$ which is a contradiction.

Next we prove that the metric projection is a contraction.

Proof: Let the points ${x,y}$ be projected to ${P_C(x)}$ and ${P_C(y)}$ respectively and assume ${P_C(x) \neq P_C(y)}$. Consider the hyperplanes ${H_x, H_y}$ through ${P_C(x),P_C(y)}$ that are orthogonal to the vector ${P_C(x) - P_C(y)}$. See figure below.

The main observation is that $x$ must lie on the other side of $H_x$ as $P_C(y)$. If ${x}$ lies on the same side of ${H_x}$ as ${P_C(y)}$ then one can always find a point on the segment ${(P_C(x) P_C(y)]}$ closer to ${x}$ than ${P_C(x)}$ which would contradict that ${P_C(x)}$ is the closest point to ${x}$ in ${C}$. Similarly ${y}$ cannot lie on same side of ${H_y}$ as ${P_C(x)}$. Thus ${\vert \vert x - y \vert \vert}$ is at least the distance between ${H_x}$ and ${H_y}$ which is ${\vert \vert P_C(x) - P_C(y) \vert \vert}$.