# Weil's Haar measure construction from below

Weil's construction of a Haar measure on a locally compact group rests on approximating a function from above by sums of translates of another function. I would need to know something similar for an approximation from below. Actually, I need something stronger:

Let $G$ be a locally compact group and let $C_c^+(G)$ be the set of all compactly supported functions $f$ on $G$ with $f\ne 0$ and $f(x)\ge 0$ for all $x\in G$. For $\phi,f\in C_c^+(G)$ let $(\phi,f)$ be the infimum over all sums $\sum_{j=1}^nc_j$, where $c_j>0$ such that there exist $s_1,\dots, s_n\in G$ with $$\phi(x)\le\sum_{j=1}^nc_j\ f(s_jx),\quad \forall\ x\in G.$$ Likewise, let $[\phi,f]$ be the supremum over all $\sum_{j=1}^nc_j$ such that there exist $s_1,\dots, s_n\in G$ with $$\phi(x)\ge\sum_{j=1}^nc_j\ f(s_jx),\quad \forall\ x\in G.$$ Then Weil's proof of the existence and uniqueness of the Haar measure implies that for every net $(f_\alpha)_{\alpha\in A}$ with the properties $$\alpha\le\beta\quad\Rightarrow\quad \mathrm{supp} (f_\alpha)\supset \mathrm{supp} (f_\beta)$$ and $$\bigcap_{\alpha\in A}\mathrm{supp}(f_\alpha)=\{1\},$$ and any two $\phi,\psi\in C_c^+(G)$, the quotient $\frac{(\phi,f_\alpha)}{(\psi,f_\alpha)}$ converges to $\frac{\int_G\phi\,d\mu}{\int_G\psi\,d\mu}$. My question is this: For given $\phi\in C_c^+(G)$, does the quotient $$\frac{[\phi,f_\alpha]}{(\phi,f_\alpha)}$$ converge to 1?

In case this is not true for any net as above, does there exist one net with this property?

It would be enough to assume that $G$ is first countable, so instead of nets you may as well use sequences.

The answer is Yes, provided that for every $\alpha$ and every $x\in G$, $f_\alpha(x^{-1})=f_\alpha(x)$.

Since we already completed the construction of the Haar measure we can actually use it in our proof. I suppose that one can show this also elementarily, but why bother?

Choose a left Haar measure on $G$. Let me assume, as the expression in consideration is homogeneous, that the $L^1(G)$ norm of all $f_\alpha,\phi$ is 1. The answer will follow from the claims $\lim[\phi,f_\alpha]=1$ and $\lim(\phi,f_\alpha)= 1$. I will prove the first claim, the second being similar.

Let me first note that for every $\alpha$, $\phi(x)\leq \sum c_jf_\alpha(s_jx)$ implies by taking integrals that $1\leq \sum c_j$. Therefore $[\phi,f_\alpha]\geq 1$. We are left to show that for every $m>1$ there exists $\alpha_0$ such that for every $\alpha\geq \alpha_0$, $[\phi,f_\alpha] \leq m$.

Fix $m>1$. Choose a compactly supported $[0,1]$-valued continuous function $\psi$ which is 1 on the support of $\phi$. Denote $\epsilon=(m-1)/\|\psi\|_1$. Set $\phi'=\phi+\epsilon \psi$ and note that $\|\phi'\|_1=m$.

Note that the net $(f_\alpha)$ forms an approximate identity and therefore $\phi'*f_\alpha \to \phi'$ in the sup norm. See e.g: Folland, "A course in Abstract Harmonic Analysis" (2.42) - this is why I needed the assumption $f_\alpha(x^{-1})=f_\alpha(x)$. Find $\alpha_0$ such that for every $\alpha\geq \alpha_0$, $\|\phi'*f_\alpha- \phi'\|_\infty <\epsilon/2$.

We now fix $\alpha\geq\alpha_0$ and denote for convenience $f=f_\alpha$. We are left to show that $[\phi,f]\leq m$.

Let $K$ be the support of $\phi'$ and $K'$ be the support of $\phi'*f$. Let $Q$ be the convex set of positive measures on $G$ with total mass $m$ and support contained in $K$ and let $Q'$ be the set of finitely supported measures in $Q$. Identify $Q$ with a subset of $C(K)^*$ and note that $Q'$ is weak* dense in $Q$. Observe that $Q*f$ is a subset of $C(K')$.

I claim that $Q'*f$ is weakly dense in $Q'*f$ within $C(K')$. Assume not. Then there exists $\nu\in Q$ such that $\nu*f$ is not in the weak closure of $Q'*f$. Find a (signed) measure $\lambda$ which separates $\nu* f$ from $Q'*f$ in $C(K')$ and observe that the continuous function $\lambda*\check{f}|_K$ separates $\nu$ from $Q'$ in $C(K)^*$. But this is absurd, as $Q'$ is weak*-dense in $Q$. This proves the claim.

Recall that for convex sets the weak closure is the same as the norm closure. We deduce that $Q'*f$ is dense in $Q*f$ wrt the sup norm of $C(K')$. Observe that $\phi'\in Q$. Thus $\phi'*f\in Q*f$ and there exists a finitely supported measure $\mu\in Q'$ such that $\|\mu*f-\phi'*f\|_\infty<\epsilon/2$.

It follows that $\|\mu*f-\phi'\|_\infty<\epsilon$. Since $\phi'-\phi=\epsilon\psi$ equals $\epsilon$ on the support of $\phi'$ and $\mu*f\geq 0$ everywhere we get that $\mu*f\geq \phi$ everywhere.

Writing $\mu=\sum_{j=1}^n c_j\delta_{s_j}$ we get that for every $x\in G$, $\sum c_jf_\alpha(s_jx) \geq \phi(x)$. Thus $[\phi,f]\leq \sum c_j=m$.

• @Anton, I noticed two things I must comment about. 1) I wrote things using right shifts instead of left for some unclear reason... 2) More seriously, in order for $f_\alpha$ to be an approximate identity you need $f_\alpha(x^{-1})=f_\alpha(x)$ (according to Folland, "A course in Abstract Harmonic Analysis" (2.42)) so I guess I have to assume that. – Uri Bader Apr 29 '16 at 11:03
• I will edit accordingly my answer when I find the time... – Uri Bader Apr 29 '16 at 11:04
• @Anton - edited. I filled in many details. – Uri Bader Apr 29 '16 at 12:36
• @Anton, I realize that if you consider expressions as $Q'*\check{f}*f$ you can get away without the assumption $f_\alpha(x^{-1})=f_\alpha(x)$. I will edit my answer if you care about the details. – Uri Bader Apr 29 '16 at 16:31