# Distance minimizers in $L^1$ and $L^{\infty}$

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.

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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
The $L^p$ norm. – Isaac Solomon Jan 24 '13 at 22:49
@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
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 4:58

## 1 Answer

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 $L^\infty$.

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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 5:09
Will the image $\phi(K)$ be convex? – Isaac Solomon Jan 25 '13 at 5:26
@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 $\ell_\infty$... – user53153 Jan 25 '13 at 5:32
@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. – user53153 Jan 25 '13 at 5:38
Thanks for your explanation! I hereby declare this counterexample 'lovely' and accept your answer. – Isaac Solomon Jan 25 '13 at 5:43