## Regularity of Newtonian Potential

Suppose ${\Gamma(\cdot)}$ is the fundamental solution of laplace equation.

$\displaystyle \Gamma(x-y)=\begin{cases}\frac{1}{n(2-n)\omega_n}|x-y|^{2-n}, n>2\\\frac{1}{2\pi}\log|x-y|, n=2.\end{cases}$

We know it has property

$\displaystyle |D_{y_i}\Gamma(x-y)|\leq C\frac{1}{|x-y|^{n-1}}$

We can define the Newtonianial potential of ${f}$

$\displaystyle Nf(x)=\int_{\Omega}\Gamma(x-y)f(y)dy\quad x\in\mathbb{R}^n$

${Nf}$ is well defined. We have the following propery of ${Nf}$

Thm: Suppose ${\Omega\subset\mathbb{R}^n}$ is a bounded domain, ${n\geq 2}$. Assume ${f}$ is bounded and local integrable in ${\Omega}$, then ${Nf}$ is ${C^1(\mathbb{R}^n)}$ and

$\displaystyle D_{x_i}Nf=\int_{\Omega} D_{x_i}\Gamma(x-y)f(y)dy$

Proof: Fix ${0<\epsilon<1}$, let ${B_\epsilon(x)=\{y\in\Omega||x-y|<\epsilon\}}$ and ${B_\epsilon^c(x)=\{y\in\Omega||x-y|\geq\epsilon\}}$.

$\displaystyle Nf=\int_{B_\epsilon(x)}\Gamma(x-y)f(y)dy+\int_{B^c_\epsilon(x)}\Gamma(x-y)f(y)dy=I+II$

Since ${D_{x_i}\Gamma(x-y)}$ is uniformly bounded in ${B^c_\epsilon(x)}$, by the Lebesgue dominating theorem, ${II}$ is differentiable and

$\displaystyle D_{x_i}II=\int_{B^c_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy$

Considering ${I}$, we will prove if ${|x-z|<\epsilon/2}$

$\displaystyle |I(x)-I(z)|\leq \begin{cases}C\epsilon|x-z|\quad\quad\text{ if } n>2\\ C\epsilon|\log \frac{\epsilon}{2}||x-z|\text{ if }n=2\end{cases}$

where ${C=C(||f||_{L^\infty},n,\Omega)}$.

$\displaystyle \int_{B_\epsilon(x)}\Gamma(x-y)f(y)dy-\int_{B_\epsilon(z)}\Gamma(z-y)f(y)dy=\int_{B_\epsilon(x)\cap B^c_\epsilon(z)}\Gamma(x-y)f(y)dy$

$\displaystyle +\int_{B_\epsilon(x)\cap B_\epsilon(z)}[\Gamma(x-y)-\Gamma(z-y)]f(y)dy-\int_{B^c_\epsilon(x)\cap B_\epsilon(z)}\Gamma(z-y)f(y)dy$

${=\mathcal{A}+\mathcal{B}+\mathcal{C}}$

For any ${y\in B_\epsilon(x)\cap B^c_\epsilon(z)}$, ${|y-x|\geq |y-z|-|z-x|\geq \epsilon/2}$. So

$\displaystyle |\mathcal{A}|\leq C\frac{1}{\epsilon^{n-2}} |B_\epsilon(x)\cap B^c_\epsilon(z)|\leq C\frac{\epsilon^n-(\epsilon-|x-z|/2)^n}{\epsilon^{n-2}}\leq C\epsilon |x-z|\text{ if } n>2$

$\displaystyle |\mathcal{A}|\leq C|\log\frac{\epsilon}{2}| |B_\epsilon(x)\cap B^c_\epsilon(z)|\leq C\epsilon|\log\frac{\epsilon}{2}||x-z|\text{ if }n=2$

Similarly for ${C}$, we also have

$\displaystyle |\mathcal{C}|\leq\begin{cases}C\epsilon|x-z|\quad \quad \text{ if } n>2\\ C\epsilon|\log\frac{\epsilon}{2}||x-z|\text{ if }n=2\end{cases}$

Considering ${\mathcal{B}}$, we have

$\displaystyle |\mathcal{B}|\leq C\int_{B_\epsilon(x)\cap B_\epsilon(z)}\int_0^1|D_{x_i}\Gamma(tx+(1-t)z-y)||x-z|dtdy$

$\displaystyle =C|x-z|\int_0^1\int_{B_\epsilon(x)\cap B_\epsilon(z)}|D_{x_i}\Gamma(tx+(1-t)z-y)|dydt$

$\displaystyle \leq C|x-z|\int_0^1\int_{B_\epsilon(x)\cap B_\epsilon(z)} \frac{1}{|tx+(1-t)z-y|^{n-1}}dydt$

$\displaystyle \leq C|x-z|\int^1_0 C\epsilon dt\leq C\epsilon |x-z|$

Combing the fact about ${\mathcal{A},\mathcal{B},\mathcal{C}}$, we get

$\displaystyle |I(x)-I(z)|\leq \begin{cases}C\epsilon|x-z|\quad\quad\text{ if } n>2\\ C\epsilon|\log \frac{\epsilon}{2}||x-z|\text{ if }n=2\end{cases}$

So

$\displaystyle \left|\frac{Nf(x)-Nf(z)}{x-z}-\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy\right|$

$\displaystyle \leq \left|\frac{II(x)-II(z)}{x-z}-\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy\right|+\left|\frac{I(x)-I(z)}{x-z}\right|$

Applying the fact that ${II}$ is differentiable and the fact about ${I}$

$\displaystyle \overline{\lim\limits_{z\rightarrow x}}\left|\frac{Nf(x)-Nf(z)}{x-z}-\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy\right|$

$\displaystyle \leq \left|\int_{B_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy\right|+\begin{cases}C\epsilon\quad\quad\text{ if } n>2\\ C\epsilon|\log \frac{\epsilon}{2}|\text{ if }n=2\end{cases}$

$\displaystyle \leq C\epsilon+\begin{cases}C\epsilon\quad\quad\text{ if } n>2\\ C\epsilon|\log \frac{\epsilon}{2}|\text{ if }n=2\end{cases}$

Let ${\epsilon\rightarrow 0}$, we get ${D_{x_i}Nf}$ exists and

$\displaystyle D_{x_i}Nf=\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy$

Next we shall prove ${D_{x_i}Nf}$ is continuous for ${i=1,\cdots,n}$, then ${Nf\in C^1(\mathbb{R}^n)}$

$\displaystyle \left|\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy-\int_{\Omega}D_{x_i}\Gamma(z-y)f(y)dy\right|$

$\displaystyle \leq\left|\int_{B^c_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy-\int_{B^c_{\epsilon}(z)}D_{x_i}\Gamma(z-y)f(y)dy\right|$

$\displaystyle +\left|\int_{B_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy\right|+\left|\int_{B_\epsilon(z)}D_{x_i}\Gamma(z-y)f(y)dy\right|$

By the dominating theorem

$\displaystyle \lim\limits_{x\rightarrow z}\left|\int_{B^c_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy-\int_{B^c_{\epsilon}(z)}D_{x_i}\Gamma(z-y)f(y)dy\right|=0$

$\displaystyle \left|\int_{B_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy\right|\leq C\epsilon$

$\displaystyle \left|\int_{B_\epsilon(z)}D_{x_i}\Gamma(z-y)f(y)dy\right|\leq C\epsilon$

where ${C=C(||f||_{L^\infty}, n)}$. So ${\displaystyle \int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy=D_{x_i}Nf}$ is continuous everywhere.

Remark: This revised version also have errors. It is a shame. What do you mean ${II}$ is differentiable? Don’t you notice that the integral domain of ${II}$ also has relation with ${x}$?

To prove this theorem rigorously and neatly, we need the following lemma from calculus

Lemma: Suppose ${u_n{x}}$ is a sequence of differntiable function on ${[a,b]\rightarrow \mathbb{R}}$. If ${u_n}$ converges to another function ${u}$ uniformly(${u}$ is finite somewhere) and ${u'_n}$ converges uniformly to ${v}$ in ${\Omega}$ then ${u}$ is differentiable and ${u'=v}$.

Proof: Let ${\xi(r):C^1(\mathbb{R}_+^1)\rightarrow \mathbb{R}^1}$ and ${0\leq\eta'\leq 2}$, $\xi(r)=0\text{ if } r\leq 1$, $\xi(r)=1\text{ if }r>2$.

Define

${\displaystyle w_\epsilon=\int_{\Omega}\xi\left(\frac{|x-y|}{\epsilon}\right)\Gamma(x-y)f(y)dy}$

Let ${\xi_\epsilon(r)=\xi_{\epsilon}(r/\epsilon)}$ Since

$\displaystyle \left|D_{x_i}\left(\xi_\epsilon\left(|x-y|\right)\Gamma(x-y)f(y)\right)\right|\leq \frac{C}{\epsilon^{n-1}}f(y)$

which is in ${L^1(\Omega)}$. By the Lebesgue Differentiable theorem, ${w_\epsilon}$ is differntiable and

$\displaystyle D_{x_i}w_\epsilon=\int_{\Omega}D_{x_i}\left(\xi_\epsilon\left(|x-y|\right)\Gamma(x-y)\right)f(y)dy$

$\displaystyle \left|D_{x_i}w_\epsilon-\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy\right|\leq \sup|f|\int_{\Omega}\left|D_{x_i}((1-\xi_\epsilon)\Gamma(x-y))\right|dy$

$\displaystyle \leq \sup|f|\int_{|x-y|\leq 2\epsilon}|D_{x_i}\Gamma|+\frac{2}{\epsilon}|\Gamma|dy$

$\displaystyle \leq \sup|f|\begin{cases}C\epsilon\quad \quad \text{ if }n>2\\ C\epsilon (1+|\log\epsilon|)\text{ if }n=2\end{cases}$

So ${D_{x_i}w_\epsilon}$ converges to ${\displaystyle \int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy}$ uniformly on any compact subset of ${\mathbb{R}^n}$. And it is easy to prove ${w_\epsilon}$ converges to ${Nf}$ uniformly on any compact subset of ${\mathbb{R}^n}$. So by the lemma, we have ${Nf}$ is differentiable and has

$\displaystyle D_{x_i}Nf(x)=\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy,\quad i=1,2\cdots,n$

• Noah  On February 27, 2017 at 7:44 pm

Why is it that we can only conclude uniform convergence on compact sets at the end, and not uniform convergence on all of R^n?

• Sun's World  On February 27, 2017 at 9:36 pm

I think we do not define Nf on all R^n. we need to make some assumption of f in order to define the convolution.

• Noah  On February 27, 2017 at 10:26 pm

Hmm..but doesn’t the statement of the theorem say that Nf is in C^1(R^n)? Also, in the third line of your proof, you define the domains of integration to be subsets of Omega, so I don’t think there’s a problem in defining the convolution for all of R^n.

btw is your source from Gilbarg and Trudinger? I ask because there’s a very similar derivation in that book and they also conclude uniform convergence in compact subsets of R^n. It’s a rather minor issue in the context of this proof imo and doesn’t change the result but still just curious…

• Sun's World  On February 28, 2017 at 2:09 pm

Thank you for pointing that out. I was sloppy. Here the second proof is mostly copied from GT’s book.

Firstly, Nf is define for whole R^n, but f is defined in a bounded domain(my previous comment just means this).

Secondly, if x is strictly away from \Omega, then Nf is differential obviously. The convolution has no singularity near x.

Last, I think the convergence can be made uniformly on R^n. Take a larger domain than \Omega and D_{x}w_\epsilon will converge to DNf. Outside that domain, you have control obviously. I think that works, but I did not really write it down.