Does this functional admit an absolute minimizer?

Question

This is a close relative of the following problem.

Let $\Omega$ be an open, bounded subdomain of $\mathbb R^n$ with smooth boundary, and $f_i \in W^{1, \infty} (\Omega)$ a sequence of functions equibounded in $W^{1, \infty}$ norm - that is, there exists some $M > 0$ such that $\|f_i\|_{W^{1, \infty} (\Omega)} < M$ for each $i$.

Define the functional $L: W^{1, \infty} (\Omega) \to \mathbb R$ by

$$L(h) = \limsup_{i \to \infty} \|f_i - h\|_{L^\infty (\Omega)},$$

and write

$$L_{|U} (h) := \limsup_{i \to \infty} \|f_i - h\|_{L^\infty (U)}$$

for the restriction of the functional $L$ to an open set $U \subset \mathbb R^n$.

We say $f \in W^{1, \infty} (\Omega)$ is an absolute minimizer of $L$ if for every open ball $B$, and every $g \in W^{1, \infty}(\Omega)$ such that $f = g$ on $\partial B$, we have

$$L_{|B} (f) \leq L_{|B} (g)$$

Question: Does the functional $L$ admit an absolute minimizer?

Answer 1

I think everything is clear to you by now, anyway here are some elementary fact that seem relevant.

• For an open set $\Omega\subset \mathbb R^n$, $f\in W^{1,\infty}(\Omega)$ iff the restrictions $f_{|B}$ to balls $B\subset \Omega$ are uniformly bounded and uniformly Lipschitz, the best Lipschitz constant being $\|Df\|_{\infty,\Omega}$ .

• As a consequence, for a bounded sequence $(f_i)_{i\ge0}\subset W^{1,\infty}(\Omega)$, one has that $\displaystyle\sup_{i\ge0}f_i$, $\displaystyle\inf_{i\ge0}f_i$ and $\displaystyle\limsup_{i\to\infty}f_i=\inf_{k\ge0}\sup_{i\ge k}f_i$ are in $\in W^{1,\infty}(\Omega)$. Moreover, the decreasing sequence $\sup_{i\ge k} f_i$ converges uniformly on every bounded set $B\subset \Omega$ by the Ascoli-Arzelà theorem or by the Dini's monotone convergence theorem, so $$\sup_{x\in B}\limsup_{i\to\infty}f_i(x)= \sup_{x\in B}\lim_{k\to\infty} \sup_{i\ge k} f_i(x)= \lim_{k\to\infty} \sup_{x\in B} \sup_{i\ge k} f_i(x)= $$$$=\lim_{k\to\infty} \sup_{i\ge k} \sup_{x\in B}f_i(x)=\limsup_{i\to\infty} \sup_{x\in B}f_i(x).$$

• For a sequence of real numbers $(a_i)_{i\ge0}$ and $b\in\mathbb R$, $$\limsup_{i\to\infty}|a_i-b|= |\limsup_{i\to\infty}a_i-b|\vee |\liminf_{i\to\infty}a_i-b| $$

• In particular, your functional just depends from $f^*:=\limsup_{i\to\infty}f_i$ and $f_*:=\liminf_{i\to\infty}f_i$ rather than on the whole sequence $(f_i)$, namely, for every $g$ $$L(g,B):=\limsup_{i\to\infty}\|f_i-g\|_{\infty,B}= \|f^*-g\|_{\infty,B}\vee \|f_*-g\|_{\infty,B} $$

• If $\tilde f:=\frac12(f^*+f_*)$, $$ L(\tilde f,B)=\|\frac12(f^*-f_*)\|_{\infty,B}\le \frac12\|f^*-g\|_{\infty,B}+ \frac12\|f_*-g\|_{\infty,B} \le$$

$$\|f^*-g\|_{\infty,B}\vee \|f_*-g\|_{\infty,B}=L(g,B).$$

• An absolute minimiser according to the above definition is not unique in general. For instance, for $n:=1$, $\Omega:=]0,1[$, suppose $f^*(t)=t$ and $f_*(t)=0$. Then $f^*$ is also an absolute minimiser: for every $B:=[a,b]\subset\Omega$ and for every $g$ such that $g(a)=a$ and $g(b)=b$,one has $L(f^*,B)=\|f^*-f_*\|_{\infty,B}=b\le\|f^*-g\|_{\infty,B}\vee\|f_*-g\|_{\infty,B}=F(g,B)$.