Limit with integral or is this function continuous? Hello I need to show one identity and one limit. I am having problems with it.
notation:
$x_i$ is i-th coordinate of $x$
$B(x,r)$ ball with center $x$ and radius $r$
$S(x,r)$ sphere with center $x$ and radius $r$
$n_y$ in integral it means unit outer normal at point $y$
$dS_y$ standard surface measure with $y$ as integration variable

Let $\partial M$ be closed surface in $\mathbb{R}^3$. Than show that this identity hold
  $$ x_i = \frac{\int_{\partial M} \frac{y_i}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y}{\int_{\partial M} \frac{1}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y} $$
  And let $f$ be continuous function defined only on $\partial M$. Than show that $g$ is continuous in $\overline{M}$. Where $g$ is:
   $$ g(x) = \frac{\int_{\partial M} \frac{f(y)}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y}{\int_{\partial M} \frac{1}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y} $$

Reference where I got this problem. I'm reading this paper where they define function $g$ and they just comment that is continuous because $\frac{1}{\| y-x \|}$ goes to infinity as $y$ approaches $x$. 
Plus they wrote down those integrals in very funny way which I do not completely understand. I am having problems when the surface $\partial M$(they denote it $P$) is not strictly convex. But that is not that important.  

Ok so the first identity. It is more convinient to write it in this form:
$$  \int_{\partial M} \frac{y_i-x_i}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y = 0$$
Intuitively this part: $n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y$ is $dS_y$ projected onto sphere of unit radius and center $x$
and $\frac{y_i-x_i}{||y-x||}$ is just outer normal of that sphere.
So the integral is almost this: $\int_{S(x,1)}n_y dS_y$ which is zero. But problem is that in original integral you run over some places multiple times and even in reverse orientation.

The only problem with $g$ is to show that it is continuous on the boundary $\partial M$. So we deal with limit:
$$ \lim_{x \rightarrow x_0, x\in M} \frac{\int_{\partial M} \frac{f(y)}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y}{\int_{\partial M} \frac{1}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y} \overset{?}{=} f(x_0) $$
Which we can rewrite as:
$$ \lim_{x \rightarrow x_0, x\in M} \frac{\int_{\partial M} \frac{f(y)-f(x_0)}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y}{\int_{\partial M} \frac{1}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y} \overset{?}{=} 0 $$
What I try: For give $\epsilon$ I take $\delta$ that $|x_0-y|<\delta \Rightarrow |f(x_0)-f(y)|<\epsilon$
Then I split the integral:
$$\left| \frac{\int_{\partial M\setminus B(x_0,\delta)} \frac{f(y)-f(x_0)}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y + \int_{\partial M \cap B(x_0,\delta)} \frac{f(y)-f(x_0)}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y }{\int_{\partial M} \frac{1}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y} \right| \leq $$
$$  \frac{ \left| \int_{\partial M\setminus B(x_0,\delta)} \frac{f(y)-f(x_0)}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y \right| +  \left|\int_{\partial M \cap B(x_0,\delta)} \frac{f(y)-f(x_0)}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y \right| }{ \left|\int_{\partial M} \frac{1}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y \right|} \leq $$
$$ \frac{ \left| \int_{\partial M\setminus B(x_0,\delta)} \frac{f(y)-f(x_0)}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y \right| }{ \left|\int_{\partial M} \frac{1}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y \right|}  + \epsilon $$
But I can't find any bound for the second part. I'm probably doing it completely wrong. Maybe even this inequality I used is wrong.
$$ \frac{  \int_{\partial M \cap B(x_0,\delta)}  \left| \frac{1}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} \right| dS_y  }{ \left|\int_{\partial M} \frac{1}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y \right|} \leq 1  $$
The can maybe go something terribly wrong with the surface $\partial M$ so this inequality fails. I don't know. Any help would be very much appreciated.
 A: Ok I think I can answer my self :D After few days of trials and errors.
If anyone checks my answer and posts answer with notes on my answer I will give him the bounty.

Let's assume that $M$ is convex. Than for every $x\in M^0$ there is bijection $p_x$  between unit sphere and $\partial M$, $p_x : S(x,1) \rightarrow  \partial M$ that $y \in S(x,1)$ lies on ray given by $x,p_x(y)$.
Than integral:
$$g(x) = \frac{\int_{\partial M} \frac{f(y)}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y}{\int_{\partial M} \frac{1}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y}$$
can be rewriten as
$$g(x) = \frac{\int_{S(x,1)} \frac{f(p_x(y))}{||p_x(y)-x||}  dS_y}{\int_{S(x,1)} \frac{1}{||p_x(y)-x||} dS_y}$$
This is because infinitesimal surface area $dS_y$, of $\partial M$ at point $y$, when projected on to sphere $S(x,1)$ has area $n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y$.
For any $\epsilon > 0$ there is $\delta >0$ such that:
$$ ||x-y||<\delta \Rightarrow ||f(x)-f(y)||<\epsilon $$
Because $f$ is continuous and $\partial M$ is compact, than there is $K$ that 
$$\forall x\in \partial M:  |f(x)|\leq K$$
Denote:
$$ U_{x,\delta} = \{ y \in S(x,1): ||p_x(y)-x_0|| < \delta \}$$
Now we want to show that this limit holds:
$$\lim_{x\rightarrow x_0,x\in M^0} \frac{\int_{S(x,1)} \frac{f(p_x(y))-f(x_0)}{||p_x(y)-x||}  dS_y}{\int_{S(x,1)} \frac{1}{||p_x(y)-x||} dS_y} = 0$$
Let's start:
 $$\left| \frac{\int_{S(x,1)} \frac{f(p_x(y))-f(x_0)}{||p_x(y)-x||}  dS_y}{\int_{S(x,1)} \frac{1}{||p_x(y)-x||} dS_y} \right| \leq$$
$$\frac{ \int_{S(x,1) \setminus U_{x,\delta}} \left| \frac{f(p_x(y))-f(x_0)}{||p_x(y)-x||}\right|  dS_y  +  \int_{S(x,1) \cap U_{x,\delta}} \left| \frac{f(p_x(y))-f(x_0)}{||p_x(y)-x||} \right|  dS_y}{\int_{S(x,1)} \frac{1}{||p_x(y)-x||} dS_y} \leq$$
$$\frac{4\pi \frac{1}{\delta} 2K}{\int_{S(x,1)} \frac{1}{||p_x(y)-x||} dS_y} + \epsilon \frac{  \int_{S(x,1) \cap U_{x,\delta}}  \frac{1}{||p_x(y)-x||}   dS_y}{\int_{S(x,1)} \frac{1}{||p_x(y)-x||} dS_y} \leq$$
$$\frac{4\pi \frac{1}{\delta} 2K}{\int_{S(x,1)} \frac{1}{||p_x(y)-x||} dS_y} + \epsilon \overset{x\rightarrow x_0}{\rightarrow} \epsilon$$
Because 
$$\int_{S(x,1)} \frac{1}{||p_x(y)-x||} dS_y \overset{x\rightarrow x_0}{\rightarrow} \infty $$

Ok so I proved the second question the first question is really simple for convex $M$
$$\int_{\partial M} \frac{y-x}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y = 0$$
can be rewriten as
$$ \int_{S(x,1)} \frac{p_x(y)-x}{||p_x(y)-x||} dS_y = 0$$
But $\frac{p_x(y)-x}{||p_x(y)-x||} = \frac{y-x}{||y-x||}$ and that is normal to the sphere $S(x,1)$ at point $y$.

So if $M$ can be written as union of finitely many convex sets 
$$ M = \bigcup_i M_i$$
that $M^0_i \cap M^0_j = \emptyset$ for $i\neq j$, we can use preceding proof. 
Let $f$ is vector valued function defined on $\partial M$. Integrals of type :
$$\int_{\partial M} f(y)\cdot n_y dS_y$$
can be rewriten to 
$$ \sum_i \int_{\partial M_i} f(y) \cdot n_y dS_y $$
Note that you have to use some extension theorem, because not all of points on $\partial M_i$ has to lie on $\partial M$ where $f$ is defined. But the integral does not depend on the extension because you integrate twice(with different orientations) over all point where $f$ needs to be extended.

