# Nearest point projection, part II: beware of two-dimensional gifts

To avoid the complications from the preceding post, let’s assume that $X$ is a uniformly convex Banach space: such a space is automatically reflexive, and therefore any closed subspace $A$ has a well-defined nearest point projection $P\colon X\to A$.

Recalling that in a Hilbert space $P$ is a linear operator (and a self-adjoint one at that), we might ask if our $P$ is linear. Indeed, the invariance of distance under translations shows that $P(x+a)=P(x)+a$ for all $x\in X, a\in A$. Consequently, all fibers $P^{-1}(a)$ are translates of one another. The map $P$ is also homogeneous: $P(\lambda x)=\lambda P(x)$, which follows from the homogeneity of the norm. In particular, $P^{-1}(0)$ is a two-sided cone: it’s closed under multiplication by scalars.

In the special case $\dim (X/A)=1$ we conclude that $P^{-1}(0)$ is a line, and the direct sum decomposition $X= A+P^{-1}(0)$ identifies $P$ as a (possibly skewed) linear projection.

Well, one of things that the geometry of Banach spaces teaches us is that 2-dimensional examples are often too simple to show what is really going on, while a 3-dimensional example may suffice. For instance, the $\ell_1$ and $\ell_\infty$ norms define isometric spaces in 2 dimensions, but not in 3 or more.

So, let’s take $X$ to be 3-dimensional with the $\ell_p$ norm $\|x\|^p=|x_1|^p+|x_2|^p+|x_3|^p$, where $1. Let $A=\lbrace x_1=x_2=x_3\rbrace$, so that the codimension of $A$ is 2. What is the set $P^{-1}(0)$? We know it is a ruled surface: with each point it contains the line through that point and the origin. More precisely, $P(x)=0$ exactly when the minimum of $d(t):=|x_1-t|^p+|x_2-t|^p+|x_3-t|^p$ is attained at $t=0$. (The minimum point is unique, since the function is strictly convex.) Differentiation reveals that

$\displaystyle P^{-1}(0)=\lbrace x\colon |x_1|^{p-2}x_1+|x_2|^{p-2}x_2+|x_3|^{p-2}x_3 = 0\rbrace$

which is a plane only when $p=2$. Here is this surface for $p=4$, when the equation simplifies to $x_1^3+x_2^3+x_3^3=0$:

The entire 3-dimensional space is foliated by the translates of this surface in the direction of the vector $(1,1,1)$.

The nearest point projection is likely to be the first nonlinear map one encounters in functional analysis. It is not even Lipschitz in general, although in decent spaces such as $\ell_p$ for $1 it is Hölder continuous (I think the optimal exponent is $\frac{p\wedge 2}{p\vee 2}$).

After a little thought, the nonlinearity of NPP is not so surprising: minimization of distance amounts to solving an equation involving the gradient of the norm, and this gradient is nonlinear unless the norm is a quadratic functions, i.e., unless we are in a Hilbert space.