Radially decreasing kernels are dominated by the maximal function

Fix $x$ and put $g(y):=\mathrm{\mid }f(x-y)\mathrm{\mid }$ for $y\in {\mathbb{R}}^n$; then $g\ge 0$ is measurable and locally integrable, $\int \mathrm{\mid }f(x-y)\mathrm{\mid }\omega (y)dy=\int g\omega d\lambda$, and substitution $z=x-y$ (with $B(x,r)=x-B(0,r)$ and translation invariance of $\lambda$) gives ${\sup }_{r>0}\lambda (B(0,r){)}^{-1}{\int }_{B(0,r)}gd\lambda ={\sup }_{r>0}\lambda (B(x,r){)}^{-1}{\int }_{B(x,r)}\mathrm{\mid }f\mathrm{\mid }d\lambda =Mf(x)$, that is, $Mg(0)=Mf(x)$; if $\mathrm{\parallel }\omega {\mathrm{\parallel }}_1=0$, the nonnegative integrand vanishes almost everywhere and both sides are zero (with the usual zero-times-infinity convention). Otherwise the case $Mf(x)=+\mathrm{\infty }$ makes the desired inequality trivial, so assume $Mf(x)<\mathrm{\infty }$ and fix a height $t>0$.

For $t>0$ put $S_t:=\{\omega >t\}$ and $r_t:=\sup \{\mathrm{\mid }y\mathrm{\mid }:y\in S_t\}\in [0,\mathrm{\infty }]$, using $r_t=0$ when $S_t=\mathrm{\varnothing }$. Then $r_t<\mathrm{\infty }$: if $r_t=\mathrm{\infty }$, then for every $\rho >0$ radial monotonicity and the definition of the supremum give $B(0,\rho )\subseteq S_t$, so $\mathrm{\parallel }\omega {\mathrm{\parallel }}_1\ge t\lambda (B(0,\rho ))$ for all $\rho$, which is impossible because $\lambda (B(0,\rho ))\to \mathrm{\infty }$ as $\rho \to \mathrm{\infty }$. Moreover $B(0,r_t)\subseteq S_t\subseteq B(0,r_t+\epsilon )$ for every $\epsilon >0$: the first inclusion uses that $\mathrm{\mid }z\mathrm{\mid }<r_t$ provides $y\in S_t$ with $\mathrm{\mid }y\mathrm{\mid }>\mathrm{\mid }z\mathrm{\mid }$ and then $\omega (z)\ge \omega (y)>t$, and the second uses $\mathrm{\mid }y\mathrm{\mid }\le r_t$ for $y\in S_t$. Consequently, by monotonicity [F2] and the scaling identity [F5], $\lambda (S_t)\le \lambda (B(0,r_t+\epsilon ))=(r_t+\epsilon {)}^n\lambda (B(0,1))(\epsilon >0),\lambda (B(0,r_t))=r_t^n\lambda (B(0,1)),$ so letting $\epsilon \downarrow 0$ along $\epsilon =1\mathrm{/}k$ and using continuity of $t\mapsto t^n$ yields $\lambda (S_t)=\lambda (B(0,r_t))$.

For every $\rho >0$ one has ${\int }_{B(0,\rho )}gd\lambda \le \lambda (B(0,\rho ))Mg(0)=\lambda (B(0,\rho ))Mf(x)$ by [F1] and step 1.1, hence for every $\epsilon >0$ the inclusions of step 1.2 give ${\int }_{S_t}gd\lambda \le {\int }_{B(0,r_t+\epsilon )}gd\lambda \le (r_t+\epsilon {)}^n\lambda (B(0,1))Mf(x);$ letting $\epsilon \downarrow 0$ as in step 1.2 gives ${\int }_{S_t}gd\lambda \le \lambda (B(0,r_t))Mf(x)$.

The layer-cake identity [F3] applied to $\omega$ with $q=1$, together with $\lambda (S_t)=\lambda (B(0,r_t))$ from step 1.2, gives $\mathrm{\parallel }\omega {\mathrm{\parallel }}_1={\int }_0^{\mathrm{\infty }}\lambda (S_t)dt={\int }_0^{\mathrm{\infty }}\lambda (B(0,r_t))dt$.

The pointwise identity $\omega (y)={\int }_0^{\mathrm{\infty }}{\mathbf{1}}_{S_t}(y)dt$ for $y\in {\mathbb{R}}^n$, Tonelli's theorem [F4] applied to the nonnegative product-measurable integrand $(y,t)\mapsto g(y){\mathbf{1}}_{S_t}(y)$, and steps 2.1 and 2.2 give $\int g\omega d\lambda ={\int }_0^{\mathrm{\infty }}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\int }_{S_t}gd\lambda dt\le {\int }_0^{\mathrm{\infty }}\lambda (B(0,r_t))Mf(x)dt=Mf(x){\int }_0^{\mathrm{\infty }}\lambda (B(0,r_t))dt=Mf(x)\mathrm{\parallel }\omega {\mathrm{\parallel }}_1,$ which is the asserted inequality; the case $Mf(x)=+\mathrm{\infty }$ was already trivial in step 1.1.