# The approximation of supremum of a set.

Given $\Omega \subset \mathbb R^N$ be open and let $g$: $\Omega\to \mathbb R^+$ be a $l.s.c$ function such that $g\geq 1$, but not necessarily bounded above. Also assume that there exists a sequence of Lipschitz continuous function $(g_k)$ such that $1\leq g_k(x)\leq g(x)$ and $g_k\nearrow g$ for every $x\in\Omega$.

Also assume that there exists a continuous linear functional $L$: $C_c(\Omega)\to \mathbb R$ we have $$a:=\sup\{ L(f),\,\,f\in C_c(\Omega),\,\, |f(x)|\leq g(x),\,\,\operatorname{spt}(f)\subset U\}<\infty$$ for each $U\subset \Omega$ compact.

Also we define $$a_k:=\sup\{ L(f),\,\,f\in C_c(\Omega),\,\, |f(x)|\leq g_k(x),\,\,\operatorname{spt}(f)\subset U\}$$ (Yea, this is the Riesz assumption. Section 1.8 in Evans & Gariepy)

My question: do we have $a_k\nearrow a$ as $k\to \infty$?

A quick observation shows that $a_k\leq a_{k+1}\leq \ldots\leq a$ for all $k$ and hence $a_k\to a'$ for some $a'\leq a$. But I am not sure if there will be a gap between $a'$ and $a$.

What I was trying is to assume $a'<a$ and try to build some $g_K$ such that $a_K=\frac{a'+a}{2}$, but I failed.

Any help is really welcome!

Pick any $0 < q < 1$. Then pick an $f\in C_c(\Omega)$ such that $\lvert f\rvert \leqslant g$, $\operatorname{spt} f \subset U$, and

$$L(f) > q\cdot a.$$

Then consider $h = q\cdot f$. Let $F_k = \{ x : g_k(x) \leqslant \lvert h(x)\rvert \}$. $F_k$ is a closed - since $g_k$ and $\lvert h\rvert$ are continuous - subset of the compact $U$, hence compact. Further, since $g_k \leqslant g_{k+1}$, we have $F_k \supset F_{k+1}$. Since $\lim\limits_{k\to\infty} g_k(x) = g(x) > q\cdot g(x) \geqslant \lvert h(x)\rvert$ for all $x$, we have $\bigcap\limits_{k = 1}^\infty F_k = \varnothing$, and by the nestedness it follows that there is an $m$ with $F_m = \varnothing$, i.e. $\lvert h(x)\rvert < g_m(x)$ for all $x\in \Omega$. Thus we have

$$a' \geqslant a_m \geqslant L(h) = L(q\cdot f) = q\cdot L(f) > q^2\cdot a.$$

Hence

$$a' \geqslant \sup \{ q^2\cdot a : 0 < q < 1\} = a.$$

• So nice!!! How you come up with this idea at the first place? – spatially Aug 17 '15 at 13:25
• It's a variation of the proof Rudin gives for the monotone convergence theorem [I have no idea who first gave that proof]. The situation here is somewhat similar, so it seemed natural to look whether the same idea works here too. – Daniel Fischer Aug 17 '15 at 13:38
• Ah right the monotone convergence! I thought it looks similar somewhat but I can not recall where I saw such similar proof... Thx again! Really nice proof. – spatially Aug 17 '15 at 13:39