Suppose $u \in C^2(\overline{U})$ is an arbitrary function. Fix $x \in U$, choose $\epsilon >0$ such that $B(x, \epsilon) \subset U$, and apply Green's formula to the region $V_{\epsilon} := U \setminus B(x,\epsilon)$ to $u(y)$ and $\Phi(y-x)$, where $\Phi(y-x)$ is the fundamental solution to Laplace's equation. We then have, $$ \int_{V_{\epsilon}} u(y) \Delta \Phi(y-x) - \Phi(y-x) \Delta u(y) dy $$ $$ = \int_{\partial V_{\epsilon}} u(y) \frac{\partial \Phi}{\partial \nu} (y-x) - \Phi(y-x) \frac{\partial u}{\partial \nu}(y) dS(y)$$ where $\nu$ is the outward facing normal vector on $\partial V_{\epsilon}$. Recall that $\Delta \Phi(y-x)=0$ for $x \neq y$. We observe also, $$ \bigg| \int_{\partial B(x, \epsilon)} \Phi(y-x) \frac{\partial u}{\partial v} (y-x) dS(y) \bigg| \le C \epsilon^{n-1} \max_{\partial B(0,\epsilon)} |\Phi| = o(1) \hspace{10mm} (*)$$

The above is part of the derivation of Green's function from Evan's PDE. Where is the inequality denoted by $(*)$ coming from?

  • $\begingroup$ I had the exact same question! $\endgroup$ – yoshi Feb 13 '17 at 16:32

You are really just seeing the basic inequality:

$$\left | \int_A f(x) dm(x) \right | \leq m(A) \max_{x \in A} |f(x)|.$$

Here $A$ is the sphere of radius $\epsilon$ in $n$ dimensions. This has surface area $C_n \epsilon^{n-1}$, where $C_n$ is the surface area of the unit sphere in $n$ dimensions. Now the rest is just the max of $\left |\frac{\partial u}{d \nu} \right | |\Phi|$. The former is bounded by the assumption that $u \in C^2(\bar{U})$, so that gets built into the $C$.

  • $\begingroup$ Very simple. Thank you! $\endgroup$ – user7090 Apr 19 '15 at 20:49

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