Is this function bounded? Next question about integral $\int_{\partial M} \frac{1}{||y-x||} n_y \cdot \nabla_y \frac1{||y-x||} dS_y$. 
Let $\partial M$ be $C^2$ closed surface in $\mathbb{R}^3$, $M$ is open. Show that 
   $$
f(x) = \frac{\int_{\partial M} \left| \frac{1}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} \right| dS_y}{\left| \int_{\partial M}  \frac{1}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||}  dS_y\right|}
$$
  is bounded in $\overline{M}$


I already know few things about this kind of integral, I already had two questions about it here and here. But I still can't show that $f$ is bounded.
$f$ is continuous in $M$. So if I show that 
$$
\lim_{x\rightarrow x_0} f(x) = 1
$$
for every $x_0 \in \partial M$. Than $f$ is continuous in $\overline{M}$ hence it is bounded.

My idea of proving it is:

Find $U(x)\subset \partial M$ that $\left( n_y \cdot \nabla_y \frac{1}{||y-x||} \right)  > 0$ for all $y\in U(x)$ and that
  $$
\frac{\int_{\partial M\setminus U(x)} \left| \frac{1}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} \right| dS_y}{ \int_{U(x)}  \frac{1}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||}  dS_y} \rightarrow 0\qquad \text{as dist}(x,\partial M) \rightarrow 0
$$

If you show that map $U(x)$ exists than you can easily show the limit  $\lim_{x\rightarrow x_0 \in \partial M} f(x) = 1 $
 A: Why not thinking it the other way around? Integrals like this are typical of electromagnetic theory.
Let 
\begin{multline}
\iint_M \mbox{div}\left(\frac{1}{\|y-x\|}\nabla_y \frac{1}{\|y - x\|}\right) dy = \\
\iint_M \left\{\frac{1}{\|y-x\|}\Delta_y \frac{1}{\|y-x\|} + \left\|\nabla_y \frac{1}{\|y-x\|}\right\|^2\right\}d y \\
= \iint_M \left\{\frac{1}{\|y-x\|}\Delta_y \frac{1}{\|y-x\|} + \frac{1}{\|y-x\|^4}\right\}d y
\end{multline}
This integral presents problems when $x = y$, hence we take $M = M_\epsilon \cup B_\epsilon$, where $B_\epsilon$ is the ball with center in $x$ and radius $\epsilon$, and $M_\epsilon = M \setminus B_\epsilon$.
Given that for $x \neq y$, 
$$
\Delta_y \frac{1}{\|y - x\|} = 0
$$
we have, for the first integral,
$$
\iint_M \frac{1}{\|y-x\|}\Delta_y \frac{1}{\|y-x\|} dy = \iint_{B_\epsilon} \frac{1}{\|y-x\|}\Delta_y \frac{1}{\|y-x\|} dy.
$$
Using Green's first identity,
\begin{multline}
\iint_M \frac{1}{\|y-x\|}\Delta_y \frac{1}{\|y-x\|} dy = \int_{\partial B_\epsilon} \frac{1}{\|y-x\|} \frac{\partial}{\partial n_y} \left(\frac{1}{\|y-x\|}\right) d S_y \\
- \iint_{B_\epsilon} \left\|\nabla_y \frac{1}{\|y-x\|}\right\|^2 d y,
\end{multline}
and then
\begin{multline}
\iint_M \mbox{div}\left(\frac{1}{\|y-x\|}\nabla_y \frac{1}{\|y - x\|}\right) dy = \\
\int_{\partial B_\epsilon} \frac{1}{\|y-x\|} \frac{\partial}{\partial n_y} \left(\frac{1}{\|y-x\|}\right) d S_y + \iint_{M_\epsilon} \frac{1}{\|y-x\|^4} d y.
\end{multline}
Now,
$$
\frac{\partial}{\partial n_y} \left(\frac{1}{\|y-x\|}\right)_{\|y - x\| = \epsilon} = \frac{\partial}{\partial r} \left(\frac{1}{r}\right)_{r = \epsilon} = - \frac{1}{\epsilon^2}
$$
then
$$
\iint_{B_\epsilon} \frac{1}{\|y-x\|}\Delta_y \frac{1}{\|y-x\|} dy = -\frac{1}{\epsilon^2} \int_{\|y - x\| = \epsilon} \frac{1}{\|y-x\|} d S_y
$$
Taking the change of variables $y = x + \epsilon \eta$, where $\|\eta\| = 1$,
$$
\iint_{B_\epsilon} \frac{1}{\|y-x\|}\Delta_y \frac{1}{\|y-x\|} dy = - \int_{\|\eta\| = 1} \frac{1}{\epsilon} d S_\eta = -\frac{4 \pi}{\epsilon},
$$
while
$$
\iint_{B_{2\epsilon}\setminus B_\epsilon} \frac{1}{\|y-x\|^4} d y \le \iint_{M_\epsilon} \frac{1}{\|y-x\|^4} d y \le \iint_{B_R \setminus B_\epsilon} \frac{1}{\|y-x\|^4} d y,
$$
where $B_\epsilon \subseteq B_{2\epsilon} \subseteq M_\epsilon \subseteq B_R$.
Taking the change of variable $y = x + \xi$ on the first integral, and $y = x + \eta$ on the last, where $\epsilon \le \|\xi\| \le 2\epsilon$, and $\epsilon \le \|\eta\| \le R$
$$
\iint_{\epsilon \le \|\xi\| \le 2\epsilon} \frac{1}{\|\xi\|^4} d\xi \le \iint_{M_\epsilon} \frac{1}{\|y-x\|^4} d y \le \iint_{\epsilon \le \|\xi\| \le R} \frac{1}{\|\eta\|^4} d\eta
$$
and then
$$
\frac{2 \pi}{\epsilon} \le \iint_{M_\epsilon} \frac{1}{\|y-x\|^4} d y \le 4 \pi \left(\frac{1}{\epsilon} - \frac{1}{R}\right).
$$
Finally,
$$
\frac{2 \pi}{\epsilon} \le \left| \iint_M \mbox{div}\left(\frac{1}{\|y-x\|}\nabla_y \frac{1}{\|y - x\|}\right) dy \right|
$$
and
$$
\iint_M \left| \mbox{div}\left(\frac{1}{\|y-x\|}\nabla_y \frac{1}{\|y - x\|}\right)  \right| dy \le 4 \pi \left(\frac{2}{\epsilon} - \frac{1}{R}\right).
$$
Using the Divergence theorem, we conclude that
$$
1 \le f(x) = \frac{\int_{\partial M} \left|\left(\frac{1}{\|y-x\|}\nabla_y \frac{1}{\|y - x\|}\right)\cdot n_y \right| dS_y}{\left|\int_{\partial M} \left(\frac{1}{\|y-x\|}\nabla_y \frac{1}{\|y - x\|}\right)\cdot n_y dS_y \right|} \le 4
$$
Of course, the argument can be refined to improve the upper bound.
