Haar integral comparison inequality

For each cover $f\le \sum c_j{L}_{x_j}g$, positivity and left invariance give $I(f)\le \sum c_{j}I(g)$. Taking the infimum gives $I(f)\le (f:g)I(g)$, including $f=0$.

Write $K=\mathrm{supp}f$, $S=\mathrm{supp}u$. The continuous kernel $f(x)u(x^{-1}y)$ has support in the compact set $K\times KS$; compactness here follows from compact products and continuous multiplication as in [F3]. Integrating $y$ by left invariance yields $I(f)J(u)$. Interchanging integrals and substituting $x=yz$ by left invariance of $I$ yields $J_y{I}_{z}f(yz)u(z^{-1})=J_y{I}_{z}f(yz)u(z)$. The transformed kernel is supported in $K{S}^{-1}\times S$, so a second application of [F4] gives $I(f)J(u)=I_{z}u(z)J_{y}f(yz)$.

Subtract $I(u)J(f)$. For every $z$, $\mathrm{\mid }J(R_{z}f-f)\mathrm{\mid }\le J(\mathrm{\mid }{R}_{z}f-f\mathrm{\mid })$ by positivity. The latter is continuous in $z$: [F3] and $\mathrm{\mid }\mathrm{\mid }h\mathrm{\mid }-\mathrm{\mid }k\mathrm{\mid }\mathrm{\mid }\le \mathrm{\mid }h-k\mathrm{\mid }$, dominated on a common compact support by a cutoff, give the same continuity bound as in [F4]. Hence its supremum on compact $S$ is finite. Positivity applied to $u$ times this bound proves the stated comparison inequality.

On a compact identity neighbourhood, all $R_{z}f-f$ have support in one compact set $C$. Choose a cutoff $c=1$ on $C$. Then $J(\mathrm{\mid }{R}_{z}f-f\mathrm{\mid })\le \mathrm{\parallel }{R}_{z}f-f{\mathrm{\parallel }}_{\mathrm{\infty }}J(c)\to 0$. If $J(c)=0$ the bound is already zero. Given $ϵ$, choose a neighbourhood with this bound $<ϵ\mathrm{/}2$, ensuring the supremum is at most $ϵ\mathrm{/}2<ϵ$. Inside a smaller symmetric open neighbourhood $W$, take $0\le v\le 1_W$ with $v(e)=1$, and set $w=(2v-1{)}^{+}$ and $u(z)=w(z)w(z^{-1})$. Its support is contained in $W$, it is symmetric, and $u(e)=1$. Strict positivity gives $I(u),J(u)>0$.