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If $H$ is a Hilbert space, we have the Hilbert Projection Theorem, which tells us that given a nonempty, closed, convex subset $K \subset H$, and a point $x \in H$, there is a unique point $y \in K$ which minimizes $\lVert x-y \rVert$.

In the $L^{p}(X,\mathcal{X},\mu)$ spaces, for $1 < p < \infty$, we get the same result, even though these are not Hilbert spaces for $p\neq 2$ (assuming that $(X,\mathcal{X})$ is sufficiently non-trivial). This can be proved using the Hanner inequalities.

I am interested in the case of $L^1$ or $L^\infty$. It is easy to construct examples where distance minimizers (in some closed, convex, nonempty subset) exist, but they are not unique. However, I am wondering whether or not existence can fail as well. I have thought about this a fair bit, and tried searching online, but I could not resolve this question.

Can anyone share any insight? Thanks.

Elchanan Solomon
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  • What do you think is a good measure for distance in $L^{p}(X)$? Do you want to use $|f-g|=|f-g|_{p}$? – Bombyx mori Jan 24 '13 at 22:48
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    The $L^p$ norm. – Elchanan Solomon Jan 24 '13 at 22:49
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    @Tomás If the set is compact you always have existence. Not closed. for example in $\ell^2(\mathbb{N})$ take the set $K = \left{ \frac{n+1}{n} e_n \right}_{n\in\mathbb{N}}$ it has no minimum norm. – Deven Ware Jan 24 '13 at 22:54
  • Yes you are right. Let me delete my comments. Just for the sake of clarity, what I have said does work in space with finite dimensional? – Tomás Jan 24 '13 at 23:13
  • @Tomás Yes because in a finite dimensional case, you can bound to be within some distance of $x$, then the set is compact – Deven Ware Jan 24 '13 at 23:17
  • Now I understood, so the problem is the lack of compactness of the ball in spaces with infinite dimension. Thank you @DevenWare – Tomás Jan 24 '13 at 23:19
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    Some terms that might help you searching: sets of existence for minimizers are called proximinal (or proximal) and sets for which the minimizer is unique are called Chebyshev sets (you'll find some references there). By a standard exercise to James's theorem every non-reflexive Banach space contains a closed convex set which is not proximinal. See also this introduction by Borwein. – Martin Jan 25 '13 at 04:58
  • @Martin: Thanks for the references! – Elchanan Solomon Jan 25 '13 at 15:58

1 Answers1

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Here is an example in $L^1[0,1]$. Let $K$ be the set of functions $f\in L^1[0,1]$ such that $\int_0^1 xf(x)\,dx=0$. This is a closed subspace of $L^1$. Consider the distance from $g(x)=1$ to $K$. For any $f\in K$ we have $$ \left|\int_0^1 x(f(x)-1)\,dx \right| = \int_0^1 x\,dx =\frac12 \tag{1} $$ Since $x<1$ a.e., it follows that $$ \int_0^1 |f(x)-1|\,dx > \int_0^1 x|f(x)-1|\,dx \ge \frac12 \tag{2} $$ On the other hand, the sequence $$ f_n(x) = 1 - \frac{n^2}{2n-1}\chi_{[1-1/n,1]} \tag{3} $$ belongs to $K$ and satisfies $\|f_n-g\|_{L^1}\to \frac12$. Therefore, $\operatorname{dist}(g,K)=1/2$ and this distance is not attained.

I don't have an explicit example in $L^\infty$, but since $L^\infty$ contains an isometric copy of every separable Banach space, one can use an isometric embedding $\phi : L^1\to L^\infty$ to produce an implicit example: $\phi(g)$ and $\phi(K)$. Since $K$ is a complete metric space, its image $\phi(K)$ is a closed subspace of $L^\infty$.

Jonas Meyer
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    This is a nice example. I think the closed convex hull of ${\frac{n+1}{n} e_n \mid n \in \mathbb{N}}$ should be an example in $\ell_1$. Since $\ell_1$ embeds quite explicitly into $\ell_\infty$ this would give another example. – Martin Jan 25 '13 at 05:09
  • Will the image $\phi(K)$ be convex? – Elchanan Solomon Jan 25 '13 at 05:26
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    @Martin Also, the subspace example works in $\ell^1$ as follows: let $K = {x\in\ell_1:\sum \frac{n}{n+1}x_n=0}$; its distance to $e_1$ is $1/2$, and is not realized for a similar reason. However, I do not know an explicit embedding into $\ell_\infty$... –  Jan 25 '13 at 05:32
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    @IsaacSolomon I meant a linear isometric embedding $X\to L^\infty$. Such a thing can be produced by considering elements of $X$ as continuous functions on the closed unit ball of $X^$ equipped with the weak topology. This gives a linear isometric embedding into $C(B_{X^})$, and one can then move $C(B_{X^})$ into $C[0,1]$ (hence, into $L^\infty[0,1]$) using a space-filling curve. –  Jan 25 '13 at 05:38
  • Thanks for your explanation! I hereby declare this counterexample 'lovely' and accept your answer. – Elchanan Solomon Jan 25 '13 at 05:43
  • Hence the qualifier "quite". The closest to explicit I know is to take a sequence $(A_n){n \in \mathbb{N}}$ of independent subsets of $\mathbb{N}$, set $\varphi_n(k) = 1$ if $k \in A_n$ and $\varphi_n(k) = -1$ if $k \notin A_n$. Then $e_n \mapsto \varphi_n$ yields an isometric embedding of $\ell_1$ into $\ell\infty$. – Martin Jan 25 '13 at 05:49