# Harmonic analysis, compute that this integral tends to $0$

We have the following setting.

• $U$ is a bounded Lipschitz domain in the complex plane.
• Consider the following classical Dirichlet problem for the Laplace operator: \begin{align} \Delta{}u&=0 \text{ } \text{in}\text{ } U, \quad u =f \,\partial U.\end{align}
• $F(x)=-\frac{1}{2\pi}\log|x|$ denotes the fundamental solution for the Laplacian in the complex plane.
• For $x\in{}U$, $B(x,\epsilon)\subseteq U$ denotes the usual ball centered at $x$ with raidus $\epsilon>0$. $U_\epsilon{}=U\backslash B(x,\epsilon)$.

So what I want to get is the following \begin{align} \int\limits_{\partial{}B(x,\epsilon)}F(x-y)\frac{\partial{u}}{\partial{\nu}}(y)d\sigma(y)=\int\limits_{\partial{}B(x,\epsilon)}-\frac{1}{2\pi}\log|x|\frac{\partial{u}}{\partial{\nu}}(y)d\sigma(y) \end{align} tends to zero as $\epsilon\rightarrow0$. Perhaps we can derive this by using the explicit formula for $F$, but I don't know how to.

My thoughts so far are as follows. Note that $y \mapsto F(x - y)$ is constant on $\partial B(x, \epsilon)$, and is equal to $-1/2\pi\,\log(\epsilon)$. Hence, we can factor it, and we need a final inequality of the form$$\left|\int_{\partial B(x, \epsilon)} {{\partial \mu}\over{\partial \nu}}\right| \le c\epsilon,$$but I can not tell much about that.

If $u$ is smooth on $B(x,\epsilon)$, then $$\int_{\partial B(x,\epsilon)} \frac{\partial u}{\partial \nu} = \int_{B(x,\epsilon)} \Delta u = 0$$ by the divergence theorem.