凸集上投影

the following content is from https://cstheory.wordpress.com/2010/08/23/projections-onto-a-convex-body/#more-203.

 

Projections onto a Convex Body

Given a point  and an affine subspace , the projection of  onto  is the point  such that,

One can obviously define other notions of projection but the above is probably the most commonly used in geometry. If  then the projected point  is the unique point in  such that the vector  is orthogonal to . 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  and a point , there is a unique point  such that,

and,

Let  be a nonempty closed convex set. The mapping  is a contraction i.e. for any  we have,

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

Proof: First we show that the minimum distance to  is indeed obtained. This is easy and the details are as follows. Since  is nonempty there is some point . Let  denote . Clearly a closest point to  can only lie in  where denotes a closed ball of radius  with center . But  is a closed and bounded set and is therefore compact. It is also nonempty. The function  defined for  as is a continuous function on a compact set and attains its minimum at some point .

Next we prove the point  where  attains minimum, is unique. If  is another point at which  then in the triangle , which has two equal sides, the perpendicular from  onto segment  is clearly shorter than the length . However the base of this perpendicular lies on the segment itself. By convexity of  the entire segment is in  and thus we have found a point in  closer to  than  which is a contradiction.

Next we prove that the metric projection is a contraction.

Proof: Let the points  be projected to  and respectively and assume . Consider the hyperplanes  through  that are orthogonal to the vector . See figure below.

The main observation is that  must lie on the other side of  as . If  lies on the same side of  as  then one can always find a point on the segment  closer to  than  which would contradict that  is the closest point to  in . Similarly  cannot lie on same side of  as . Thus  is at least the distance between  and  which is .