Haar covering ratios are finite and positive

If $f=0$, the empty cover gives ratio zero. Otherwise put $M=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}>0$. Choose $y$ with $\varphi (y)>0$ and $0<t<\varphi (y)$. The open set $W=\{\varphi >t\}$ contains $y$. For every $x\in \mathrm{supp}f$, the translate $x{y}^{-1}W$ contains $x$. Finitely many such translates cover the compact support, and coefficients $M\mathrm{/}t$ on them dominate $f$, so the infimum is finite.

Every cover satisfies $f(x)\le \mathrm{\parallel }\varphi {\mathrm{\parallel }}_{\mathrm{\infty }}\sum c_j$ for every $x$. Taking the supremum and then the infimum yields $(f:\varphi )\ge \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}\mathrm{\parallel }\varphi {\mathrm{\parallel }}_{\mathrm{\infty }}>0$ when $f\mathrm{\ne }0$. In particular the normalizing denominator is finite and positive.

A cover of a larger function covers a smaller one. Multiplying coefficients by $t>0$ and dividing them back by $t$ proves $(tf:\varphi )=t(f:\varphi )$; scalar zero follows from the empty cover. Combining covers of $f$ and $g$ and taking independently arbitrarily close upper approximations to the two infima gives $(f+g:\varphi )\le (f:\varphi )+(g:\varphi )$. Translating a cover by $a$ replaces each centre $x_j$ by $a{x}_j$; translation by $a^{-1}$ reverses this operation. Thus $(L_{a}f:\varphi )=(f:\varphi )$.

If $f\le {\sum }_j{c}_j{L}_{x_j}\varphi$ and $\varphi \le {\sum }_k{d}_k{L}_{y_k}\psi$, substitution gives $f\le {\sum }_{j,k}{c}_j{d}_k{L}_{x_j{y}_k}\psi$. Choose the two coefficient sums below their finite infima plus $\delta >0$ and let $\delta \downarrow 0$. Their product proves the claimed composition inequality, also for $f=0$. Applying it to $(f,f_0,\varphi )$ gives the upper coordinate bound. Applying it to $(f_0,f,\varphi )$ and dividing the positive factors gives the lower bound.