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

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Comments

  • 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.

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