A typical scenario: given a subset of a metric space and a point , we look for a point that is nearest to : that is, . Such a point is generally not unique: for example, if is the graph of cosine function and , then both and qualify as nearest to . This makes the nearest-point projection onto discontinuous: moving slightly to the left or to the right will make its projection onto jump from one point to another. Not good.

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 is a **quasi-projection** if for some constant independent of . Such maps are much easier to construct: indeed, every Lipschitz continuous map such that for is a quasi-projection. For example, one quasi-projection onto the graph of cosine is the map shown below.

If is a Banach space and is its subspace, then any idempotent operator with range is a quasi-projection onto . 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 and are subspaces of a Euclidean space and , then there exists anĀ **isometric quasi-projection** of onto with constant . This constant is best possible: for example, an isometry from the -axis onto the -axis has to send to one of , thus moving it by distance .

**Proof**. Let be the common dimension of and . Fix some orthonormal bases in and . In these bases, the orthogonal (nearest-point) projection from to is represented by some matrix of norm at most . We need an orthogonal matrix such that the map that it defines is a -quasi-projection. What exactly does this condition mean for ?

Let’s say , is the orthogonal projection of onto , and is where we want to send by an isometry. Our goal is , in addition to . Squaring and expanding inner products yields . Since both and are in , we can replace on the left by its projection . So, the goal simplifies to . Geometrically, this means placing so that its projection onto the line through lies on the continuation of this line beyond .

So far so good, but the disappearance of from the inequality is disturbing. Let’s bring it back by observing that is equivalent to , which is simply . So that’s what we want to do: map so that the distance from its image to does not exceed . In terms of matrices and their operator norm, this means .

It remains to show that every square matrix of norm at most (such as here) is within distance of some orthogonal matrix. Let be the singular value decomposition, with orthogonal and a diagonal matrix with the singular values of on the diagonal. Since the singular values of are between and , it follows that . Hence , and taking concludes the proof.

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