Quasi-projections and isometries

A typical scenario: given a subset {E} of a metric space {X} and a point {x\in X}, we look for a point {y\in E} that is nearest to {x}: that is, {d(x, y) = \mathrm{dist}\,(x, E)}. Such a point is generally not unique: for example, if {E} is the graph of cosine function and {x = (\pi, \pi/2)}, then both {(\pi/2, 0)} and {(3\pi/2, 0)} qualify as nearest to {x}. This makes the nearest-point projection onto {E} discontinuous: moving {x} slightly to the left or to the right will make its projection onto {E} jump from one point to another. Not good.

proj1
Discontinuous nearest-point projection

Even when the nearest point projection is well-defined and continuous, it may not be the kind of projection we want. For example, in a finite-dimensional normed space with strictly convex norm we have a continuous nearest-point projection onto any linear subspace, but it is in general a nonlinear map.

Let’s say that {P\colon X\to E} is a quasi-projection if {d(x, P(x)) \le C \mathrm{dist}\,(x, E)} for some constant {C} independent of {x}. Such maps are much easier to construct: indeed, every Lipschitz continuous map {P\colon X\to E} such that {P(x)=x} for {x \in E} is a quasi-projection. For example, one quasi-projection onto the graph of cosine is the map {(x, y)\mapsto (x, \cos x)} shown below.

proj2
Continuous quasi-projection

If {X} is a Banach space and {E} is its subspace, then any idempotent operator with range {E} is a quasi-projection onto {E}. Not every subspace admits such an operator but many do (these are complemented subspaces; they include all subspaces of finite dimension or finite codimension). By replacing “nearest” with “close enough” we gain linearity. And even some subspaces that are not linearly complemented admit a continuous quasi-projection.

Here is a neat fact: if {M} and {N} are subspaces of a Euclidean space and {\dim M = \dim N}, then there exists an isometric quasi-projection of {M} onto {N} with constant {C=\sqrt{2}}. This constant is best possible: for example, an isometry from the {y}-axis onto the {x}-axis has to send {(0, 1)} to one of {(\pm 1, 0)}, thus moving it by distance {\sqrt{2}}.

proj3
An isometry must incur sqrt(2) distance cost

Proof. Let {k} be the common dimension of {M} and {N}. Fix some orthonormal bases in {M} and {N}. In these bases, the orthogonal (nearest-point) projection from {M} to {N} is represented by some {k\times k} matrix {P} of norm at most {1}. We need an orthogonal {k\times k} matrix {Q} such that the map {M\to N} that it defines is a {\sqrt{2}}-quasi-projection. What exactly does this condition mean for {Q}?

Let’s say {x\in M}, {y\in N} is the orthogonal projection of {x} onto {N}, and {z\in N} is where we want to send {x} by an isometry. Our goal is {\|x-z\|\le \sqrt{2}\|x-y\|}, in addition to {\|z\|=\|x\|}. Squaring and expanding inner products yields {2\langle x, y\rangle - \langle x, z \rangle \le \|y\|^2}. Since both {y} and {z} are in {N}, we can replace {x} on the left by its projection {y}. So, the goal simplifies to {\|y\|^2 \le \langle y, z\rangle}. Geometrically, this means placing {z} so that its projection onto the line through {y} lies on the continuation of this line beyond {y}.

So far so good, but the disappearance of {x} from the inequality is disturbing. Let’s bring it back by observing that {\|y\|^2 \le \langle y, z\rangle} is equivalent to {4(\|y\|^2 - \langle y, z\rangle) + \|z\|^2 \le \|x\|^2}, which is simply {\|2y-z\| \le \|x\|}. So that’s what we want to do: map {x} so that the distance from its image to {2y} does not exceed {\|x\|}. In terms of matrices and their operator norm, this means {\|2P-Q\|\le 1}.

It remains to show that every square matrix of norm at most {2} (such as {2P} here) is within distance {1} of some orthogonal matrix. Let {2P = U\Sigma V^T} be the singular value decomposition, with {U, V} orthogonal and {\Sigma} a diagonal matrix with the singular values of {2P} on the diagonal. Since the singular values of {2P} are between {0} and {2}, it follows that {\|\Sigma-I\|\le 1}. Hence {\|2P - UV^T\|\le 1}, and taking {Q=UV^T} concludes the proof.

(The proof is based on a Stack Exchange post by user hypernova.)

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