I'm reading a book that contains the following statements:
Observe that by integrating over the unit sphere the expression: $$u(x) = -\int_0^{\infty}D_ru(x+r\omega)\,dr$$ is valid for all unit vectors $\omega$ in $\mathbb{R}^n$, and by changing variables, we arrive at the pointwise representation $$u(x) = C(n) \int_{\mathbb{R}^n}\frac{\nabla u(y)\cdot(x-y)}{|x-y|^n}\,dy$$ for any smooth, compactly support function $u$ in $\mathbb{r}^n$. The actual equality in the expression above is not as important to us as the following inequality: $$|u(x) |\leq C(n) \int_{\mathbb{R}^n}\frac{|\nabla u(y)|}{|x-y|^{n-1}}\,dy$$ The above inequality is called a potential estimate.
I'm having trouble understanding how these three expression are obtained -- can anyone provide some direction? Thanks!
Here $u$ is assumed to be a smooth function with compact support. Therefore, by the fundamental theorem of calculus we have $$-u(x)= u(\infty)-u(x) = \int_{\gamma} \nabla u \,ds$$ where $\gamma(r)=x+r\omega$, $0\leq r<\infty$, and $\omega$ is a unit vector in $S^{n-1}$. Hence, $$-u(x)=\int_0^\infty \nabla u(x+r\omega)\cdot \omega \,dr.$$ We integrate over $\omega\in S^{n-1}$. We let $\sigma$ denote the spherical measure of $S^{n-1}$. Then $$-\sigma(S^{n-1})u(x)=\int_{S^{n-1}}\int_0^\infty \nabla u(x+r\omega)\cdot \omega \,drd\sigma(\omega)=\int_{S^{n-1}}\int_0^\infty \nabla u(x+r\omega)\cdot \omega r \cdot r^{-n} \cdot r^{n-1} \,drd\sigma(\omega).$$ If we set $y=x+r\omega$, then the right hand side is $$\int_{\mathbb R^n} \frac{\nabla u(y)\cdot (y-x)}{|y-x|^n}dx.$$ See Folland's Real Analysis book, p.78 for polar coordinates integration in $\mathbb R^n$. Finally, we take absolute values and we have $$\sigma(S^{n-1}) |u(x)|\leq \int_{\mathbb R^n} \frac{|\nabla u(y)|}{|y-x|^{n-1}}dx.$$
Let us call the right hand side $I_1(|\nabla u|)$, following Heinonen's book Lectures on Analysis on Metric Spaces. Here, $$I_{1}(f)(x)= (|y|^{1-n} \ast f)(x).$$ In Proposition 3.19 it is proved that the sublinear operator $f\mapsto I_1(f)$ is bounded from $L^p$ to $L^{np/(n-p)}$ if $1<p<n$. Equivalently, there is a constant $C>0$ such that $$\|I_1(f)\|_{np/(n-p)}\leq C \|f\|_{p}$$ for all $f\in L^p(\mathbb R^n)$. In our case, since $|u|\leq CI_1(|\nabla u|)$ as we proved above, we have $$\|u\|_{np/(n-p)} \leq C \|\nabla u\|_p$$ for all smooth functions $u$ with compact support. In particular, the above inequality also holds for all functions in $W^{1,p}(\mathbb R^n)$, $1<p<n$.