# Existence and uniqueness of projections on closed convex sets

Let $$X$$ be a normed vector space. I'm interested in determining what are the minimal assumptions on $$X$$ that guarantee the existence and uniqueness of projections on closed convex sets and in counterexamples showing that those assumptions are indeed necessary.

In particular let P1 and P2 be the following statements

P1: (existence) for every convex closed set $$C\subseteq X$$ and every $$x\in X$$ there exist $$y\in C$$ with $$\|x-y\|=d(x,C)$$.

P2: (uniqueness) for every convex closed set $$C\subseteq X$$ and every $$x\in X$$ there exist a unique $$y\in C$$ with $$\|x-y\|=d(x,C)$$.

What I know so far is that P2 holds in all uniformly convex Banach spaces, while to get P1 is enough to assume that $$X$$ is a reflexive Banach space, but I don't have an example of a reflexive Banach space not satisfying P2 and I don't know if those assumptions can be further weakened.

• A theorem of Joram Lindenstrauss and Lior Tzafriri states that a normed space is isomorphic to a Hilbert Space iff it possess bounded projections on every closed subspace. Shouldn't this be enough? – pitariver Jun 14 '19 at 10:45
• I'm not seeing why is this enough, can you elaborate? Also when you say "isomorphic" which kind of isomorphism are you talking about? @pitariver – Alessandro Codenotti Jun 14 '19 at 11:47

Take $$R^2$$ with norm $$\|(x,y)\| = \max \{|x| , |y| \}$$ it is clearly reflexive space.

now consider the line $$x = 1$$ in $$R^2$$ as our convex set $$C$$, and the point $$(0 , 0)$$. Then all points in the form $$(1 , y)$$ with $$-1 \leq y \leq 1$$ on the line serve as nearest point to $$(0 , 0)$$.

So being reflexive in not enough. You have to seek conditions explicitly on the norm of the space, topological behavior is not enough. I think norm has to be strictly convex.

• I don't see how your reply contradicts any statement made in the question. OP mentioned existence for reflexive spaces and you provided an example where the projection exists (i.e. there exists a point realizing the minimal distance) but is not unique. – xel Jun 18 '19 at 15:21
• @xel what do you mean? He said " I don't have an example of a reflexive Banach space not satisfying P2". And I provided that example. And he asked for some minimal conditions , I said strictly convex norm. – Red shoes Jun 18 '19 at 16:21
• Sorry misread the question. You are correct. – xel Jun 20 '19 at 18:01

Answering my own question since I found a good reference.

Let $$X$$ be a normed space and $$C\subseteq X$$. Then $$C$$ is called

• A set of uniqueness iff for every $$x\in X$$ there is at most one $$c\in C$$ with $$d(x,C)=d(x,c)$$
• A set of existence iff for every $$x\in X$$ there is $$c\in C$$ with $$d(x,C)=d(x,c)$$
• A Chebyshev set iff it is both a set of uniqueness and a set of existence

We have the following results:

Theorem: Let $$X$$ be a normed space, then the following are equivalent

• $$X$$ is strictly convex
• Every nonempty convex set in $$X$$ is a set of uniqueness
• Every nonempty closed convex set in $$X$$ is a set of uniqueness

Theorem: Let $$X$$ be a normed space, then the following are equivalent

• $$X$$ reflexive
• Every nonempty closed convex set in $$X$$ is a set of existence

Theorem (Day-James): Let $$X$$ be a normed space, then the following are equivalent

• $$X$$ reflexive and strictly convex
• Every nonempty closed convex set in $$X$$ is a Chebyshev set

All of these results can be found in Megginson's "An Introduction to Banach Space Theory" as theorem 5.1.17, corollary 5.1.19 and subsequent remarks.

• Don't you think this is exactly what mentioned in the answer I wrote down 3 days ago !! " Reflexive is not enough, you also need to assume them norm of the space is strictly convex" – Red shoes Jun 18 '19 at 18:06
• You wrote you thought strictly convex would be enough, but provided no proof nor reference. I found one in the meantime and wrote it down – Alessandro Codenotti Jun 18 '19 at 18:24