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.)