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Let $E$ be a Banach space, which need not be a Hilbert space, and let $F$ be a finite-dimensional subspace of $E$. Suppose that for all $x \in E$, there exists a $y \in F$ realizing the minimal distance of $x$ to $F$. Does this imply that there is a continuous projection $p : E \rightarrow F$ minimizing the distance?

Branimir Ćaćić
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gin
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  • I can't seem to delete my own question, sorry – gin Aug 08 '13 at 18:09
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    What's the other relevant question? If you can't flag it as a duplicate yourself, I can flag it for you. – Branimir Ćaćić Aug 08 '13 at 18:14
  • @BranimirĆaćić this question : http://math.stackexchange.com/questions/156609/continuous-projections-to-finite-dimensional-subspaces-of-normed-spaces?rq=1 is related. but the question is now : is the projection defined on this space minimizing distance ? – gin Aug 08 '13 at 21:43
  • It's related, but I definitely agree that your (modified) question is interesting in its own right. – Branimir Ćaćić Aug 08 '13 at 21:47

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$\newcommand{\P}{\mathscr{P}}$Let $E$ be a normed vector space, and let $F$ be a subset. For $x \in E$, let $$ \P_F(x) = \left\{ x_0 \in E \mid \|x-x_0\| = \inf_{y \in E} \|x-y\| \right\}. $$ Now, suppose that $F$ is proximinal, i.e., that $\P_F(x) \neq \emptyset$ for all $x \in E$. Then a metric selection of $E$ is a function $\pi : E \to F$ such that $\pi(x) \in \P_F(x)$ for all $x \in E$. If I have now correctly understood your question, what it amounts to is the following:

Let $E$ be a Banach space, and let $F$ be a finite-dimensional proximinal subspace of $E$. Does there exist a continuous metric selection $\pi : E \to F$ of $F$?

Poking about the literature yields the following characterisation of the existence of a continuous metric selection, for what it's worth.

[Deutsch–Li–Park, 1989] Let $E$ be a Banach space, and let $F$ be a proximinal subspace of $E$. Let $\ker \P_F := \{x \in E \mid \P_F(x) = \{0\}\}$. Then the following are equivalent:

  1. $F$ admits a continuous metric selection.
  2. $F$ admits a continuous metric selection which is homogeneous and additive modulo $F$.
  3. $\ker \P_F$ contains a closed homogeneous subset $N$ such that the canonical map $\omega : E \to E/F$ restricts to a homeomorphism $\left.\omega\right|_N : N \to E/F$.

In general, however, continuous metric selections need not exist:

[Lazar–Wulbert–Morris, 1969] Let $E = L^1(X,\mu)$ for $(X,\mu)$ a non-atomic positive measure space. Then for any finite-dimensional (and hence necessarily proximinal) subspace $F$ of $E$, there does not exist a continuous metric selection.

In general, there's a fair bit of literature on the existence of continuous metric selections, particularly for finite-dimensional proximinal subspaces, but a cursory look suggests that there might not be all that much in the way of general theory. The moment, however, that $F$ is Chebyshev (i.e., $\P_F(x)$ is a singleton for all $x \in E$), for instance when $E$ is reflexive and strictly convex, then the unique metric selection, the metric projection, is necessarily continuous.

Branimir Ćaćić
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  • Thank you, but my question was actually a little bit stronger, namely I only assume the existence (and not the unicity) of the best approximation element. I guess if the norm is strictly convex then unicity is automatic, but I am not making the assumption of strict convexity. however it seems (pending verifications) that there is a theorem stating that any separable banach can be equipped with an equivalent, strictly convex norm. – gin Aug 08 '13 at 23:38
  • I guess that would imply that existence implies unicity for a convex set (in particular for a vector space) in the context of a separable banach ? interesting... – gin Aug 08 '13 at 23:39
  • If you care about linearity, Cheney–Price's example of $E = C(X,\mathbb{R})$ for $X$ perfect compact should serve, a fortiori, as a counterexample to linearity at your desired level of generality (i.e., proximinality instead of Chebyshev). If it's continuity that you care about, I'm getting the impression, superficially poking about this literature, that the results about continuity of minimal projections really do assume uniqueness of best approximation elements (i.e., Chebyshev and not just proximinality)... – Branimir Ćaćić Aug 08 '13 at 23:59
  • ...That said, given that for a Banach space, reflexivity and strict convexity is apparently equivalent to every closed subspace being Chebyshev, perhaps this isn't too restrictive, depending on the Banach spaces you have in mind? – Branimir Ćaćić Aug 09 '13 at 00:01
  • unfortunately the banach space I have in mind is not reflexive (it's the bergman space A^1). but I am actually more interested in continuity than linearity (which was more curiosity). so my main question was really : is the metric projection to a finite-dimensional subspace continuous ? and the answer seems to be that in general a finite dimensional subspace is proximal but not chebyshev, so there is in general no single-valued projection, but if there is one, it must be continuous but needs not be linear. – gin Aug 09 '13 at 00:35
  • I've updated my answer to try to answer your actual question. As it turns out, in $L^1$ of a non-atomic positive measure space, finite-dimensional subspaces never admit continuous metric selections. – Branimir Ćaćić Aug 09 '13 at 01:36