Pointwise potential bound for compactly supported smooth functions

The radial primitive. Fix $\omega \in S^{n-1}$ and put $g(t):=u(x+t\omega )$ for $t\ge 0$. Since $u$ is smooth and compactly supported, $g$ is differentiable with $g^{\mathrm{\prime }}(t)=Du(x+t\omega )\cdot \omega$, and $g(t)=0$ for all $t\ge T$ once $T$ is so large that $x+[0,\mathrm{\infty })\omega$ leaves the support of $u$. Applying the fundamental theorem [F4] on $[0,T]$ and letting $T\to \mathrm{\infty }$ gives $g(0)=-{\int }_0^{\mathrm{\infty }}{g}^{\mathrm{\prime }}(t)dt$, hence $\mathrm{\mid }u(x)\mathrm{\mid }\le {\int }_0^{\mathrm{\infty }}\mathrm{\mid }Du(x+t\omega )\mathrm{\mid }dt$.

Surface normalisation. By [F2] applied to ${\mathbf{1}}_{B(0,1)}$, ${\lambda }_n(B(0,1))=\sigma (S^{n-1}){\int }_0^1{t}^{n-1}dt=\sigma (S^{n-1})\mathrm{/}n$. Thus [F6] gives $0<\sigma (S^{n-1})=n{\lambda }_n(B(0,1))<\mathrm{\infty }$.

Translation to polar coordinates at $x$. Define $G(y):=\mathrm{\mid }Du(y)\mathrm{\mid }\mathrm{\mid }x-y{\mathrm{\mid }}^{1-n}$ for $y\mathrm{\ne }x$ and $G(x):=0$; this is Borel measurable because $\mathrm{\mid }Du\mathrm{\mid }$ is continuous and $y\mapsto \mathrm{\mid }x-y{\mathrm{\mid }}^{1-n}$ is Borel. Applying [F2] to the nonnegative Borel function $z\mapsto G(x+z)$ and then [F3] gives ${\int }_{S^{n-1}}{\int }_0^{\mathrm{\infty }}\mathrm{\mid }Du(x+t\omega )\mathrm{\mid }dtd\sigma (\omega )={\int }_{S^{n-1}}{\int }_0^{\mathrm{\infty }}G(x+t\omega )t^{n-1}dtd\sigma (\omega )={\int }_{{\mathbb{R}}^n}G(y)dy,$ where the first equality uses $\mathrm{\mid }x-(x+t\omega ){\mathrm{\mid }}^{1-n}=t^{1-n}$ and the two integrations are the iterated polar integral of the nonnegative function $z\mapsto G(x+z)$; the singularity at $y=x$ is a single point and does not affect the value of the integral.

Integrating the pointwise bound over the sphere. The function $(t,\omega )\mapsto \mathrm{\mid }Du(x+t\omega )\mathrm{\mid }$ is continuous on $[0,\mathrm{\infty })\times S^{n-1}$, hence product measurable, and it is nonnegative; by Tonelli [F5] its iterated integral over the sigma-finite product $[0,\mathrm{\infty })\times S^{n-1}$ is well defined. Integrating the inequality of step 1.1 over $S^{n-1}$ against $\sigma$ therefore gives $\mathrm{\mid }u(x)\mathrm{\mid }\sigma (S^{n-1})\le {\int }_{S^{n-1}}{\int }_0^{\mathrm{\infty }}\mathrm{\mid }Du(x+t\omega )\mathrm{\mid }dtd\sigma (\omega )$.

Conclusion. Combining steps 2.1 and 1.3 with the positivity of $\sigma (S^{n-1})$ from step 1.2 gives $\mathrm{\mid }u(x)\mathrm{\mid }\le \frac{1}{\sigma (S^{n-1})}{\int }_{{\mathbb{R}}^n}\mathrm{\mid }Du(y)\mathrm{\mid }\mathrm{\mid }x-y{\mathrm{\mid }}^{1-n}dy$, and $C(n):=1\mathrm{/}\sigma (S^{n-1})=1\mathrm{/}(n{\lambda }_n(B(0,1)))$ is the asserted dimension-only constant.