The truncated Riesz kernel is bounded on $L^p$ of a bounded set

The truncated kernel has finite mass. Put $k(z):=\mathrm{\mid }z{\mathrm{\mid }}^{1-n}{\mathbf{1}}_{B(0,2\rho )}(z)$ for $z\mathrm{\ne }0$ and $k(0):=0$. By [F2] applied to the nonnegative Borel function $k$ and by [F1], [F3], $\mathrm{\parallel }k{\mathrm{\parallel }}_{L^1({\mathbb{R}}^n)}={\int }_{S^{n-1}}{\int }_0^{2\rho }{t}^{1-n}{t}^{n-1}dtd\sigma (\omega )=2\rho \sigma (S^{n-1})=2\rho n{\lambda }_n(B(0,1))=:C_0(n)\rho$, so $\mathrm{\parallel }k{\mathrm{\parallel }}_1$ is finite and equals a dimension-only multiple of $\rho$.

John-domain volume control. Put $a:=\mathrm{dist}(x_0,\mathrm{\partial }\mathrm{\Omega })>0$. The ball $B(x_0,a)$ lies in $\mathrm{\Omega }$: a segment from $x_0$ to any point outside $\mathrm{\Omega }$ first meets the boundary, so an outside point cannot be closer than $a$. Polar coordinates [F2] therefore give $\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }\ge \sigma (S^{n-1})a^n\mathrm{/}n$. Evaluating [F7] at the endpoint of the curve gives $\mathrm{\mid }x-x_0\mathrm{\mid }\le c_{J}a$ for every $x\in \mathrm{\Omega }$, hence $\mathrm{diam}\mathrm{\Omega }\le 2c_{J}a\le 2c_J(n\mathrm{/}\sigma (S^{n-1}){)}^{1\mathrm{/}n}\mathrm{\mid }\mathrm{\Omega }{\mathrm{\mid }}^{1\mathrm{/}n}$.

The convolution bound. For $g\in L^p({\mathbb{R}}^n;\mathbb{K})$, choose a finite Borel representative $g_0$ by [F11] (replace any infinite values on a Borel null set by zero). The function $F_0(x,z)=k(z)g_0(x-z)$ is Borel, hence measurable for the product of the Lebesgue sigma-algebras. The two Lebesgue spaces are sigma-finite, being exhausted by bounded balls of finite measure [F2, F3]. Translation invariance [F5] and step 1.1 give $\int \mathrm{\parallel }{F}_0(\cdot ,z){\mathrm{\parallel }}_{p}dz=\int k(z)\mathrm{\parallel }{g}_0(\cdot -z){\mathrm{\parallel }}_{p}dz=\mathrm{\parallel }k{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_p<\mathrm{\infty }$. Minkowski [F4] therefore shows that the absolute integral is finite almost everywhere and $\mathrm{\parallel }T{g}_0{\mathrm{\parallel }}_p\le \mathrm{\parallel }k{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_p$, where $T{g}_0(x)=\int k(z)g_0(x-z)dz$. This also defines $Tg$ for any Lebesgue representative $g$: for each fixed $x$, the exceptional set of $z$ is the translate and reflection $x-N$ of the null set $N$ where $g\mathrm{\ne }{g}_0$, and has measure zero by reflection invariance [F12] and translation invariance [F5]. Thus the section integrals agree wherever finite, and the measurable almost-everywhere representative $T{g}_0$ supplies $\mathrm{\parallel }Tg{\mathrm{\parallel }}_p\le \mathrm{\parallel }k{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_p$.

In particular step 1.2 gives $\mathrm{diam}\mathrm{\Omega }\le 2c(n,c_J)\mathrm{\mid }\mathrm{\Omega }{\mathrm{\mid }}^{1\mathrm{/}n}$ with $c(n,c_J):=c_J(n\mathrm{/}\sigma (S^{n-1}){)}^{1\mathrm{/}n}$; taking a supremum does not require that a farthest point exist.

The bound on $\mathrm{\Omega }$ for a set inside a ball. Extend $f$ by zero to ${\mathbb{R}}^n$ and put $g:=\mathrm{\mid }f\mathrm{\mid }{\mathbf{1}}_{\mathrm{\Omega }}$, a measurable function with $\mathrm{\parallel }g{\mathrm{\parallel }}_{L^p({\mathbb{R}}^n)}=\mathrm{\parallel }f{\mathrm{\parallel }}_{L^p(\mathrm{\Omega })}$ by [F9]. By translation and reflection invariance [F5, F12], the substitution $y=x-z$ gives $Tg(x)=\int k(x-y)g(y)dy$ wherever finite. Since $x,y\in \mathrm{\Omega }\subseteq B(z_0,\rho )$ implies $\mathrm{\mid }x-y\mathrm{\mid }\le 2\rho$, for almost every $x\in \mathrm{\Omega }$ one has $Tg(x)={\int }_{\mathrm{\Omega }}\mathrm{\mid }x-y{\mathrm{\mid }}^{1-n}\mathrm{\mid }f(y)\mathrm{\mid }dy$: the kernel truncation in the convolution is inactive exactly on the pairs with $\mathrm{\mid }x-y\mathrm{\mid }<2\rho$. Hence, by step 2.1, $\parallel {\int }_{\mathrm{\Omega }}\mathrm{\mid }x-y{\mathrm{\mid }}^{1-n}\mathrm{\mid }f(y)\mathrm{\mid }dy{\parallel }_{L^p(\mathrm{\Omega })}\le \mathrm{\parallel }Tg{\mathrm{\parallel }}_{L^p({\mathbb{R}}^n)}\le \mathrm{\parallel }k{\mathrm{\parallel }}_{L^1}\mathrm{\parallel }f{\mathrm{\parallel }}_{L^p(\mathrm{\Omega })}\le C(n)\rho \mathrm{\parallel }f{\mathrm{\parallel }}_{L^p(\mathrm{\Omega })}$.

The John-domain form of the bound. Apply step 3.1 with $z_0:=x_0$ and $\rho :=2c(n,c_J)\mathrm{\mid }\mathrm{\Omega }{\mathrm{\mid }}^{1\mathrm{/}n}$, which is admissible by step 2.2 because then $\mathrm{\Omega }\subseteq B(x_0,\rho )$. The resulting coefficient is $C(n)\rho =C(n,c_J)\mathrm{\mid }\mathrm{\Omega }{\mathrm{\mid }}^{1\mathrm{/}n}$.