Limit with integral or is this function continuous?

Solution 1:

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.