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Here is my question:

Does the following limit exist? $ \lim_{y\to\xi}\frac{(\xi_i-y_i)(\xi_j-y_j)({{\xi}-y})\cdot n(y)}{|\xi-y|^5},\quad 1\leq i,j\leq 3,\tag{*} $ where $S\subset{\mathbb R}^3$ is a surface which has a continuously varying normal vector, $\xi=(\xi_1,\xi_2,\xi_3)\in S$, $y=(y_1,y_2,y_3)\in S$, $n(y)$ is the [EDITED: unit] normal vector at point $y$. Here $(\xi-y)\cdot n(y)$ is the dot product.

In the spirit of Polya, I find a simpler case where $S$ is a unit sphere. Then we have $n(y)=y$. But I don't have a strategy to go on.

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    @Blah, I think the plane is a trivial case since we always have $(\xi-y)\cdot n(y)=0$ for a plane.2012-03-02

1 Answers 1

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"In general" this will be unbounded. You may assume $S$ in the form $(x,y)\mapsto \bigl(u,v,f(u,v)\bigr), \qquad f(0,0)=f_u(0,0)=f_v(0,0)=0\ ,$ and $(0,0,0)$ is your $\xi\in S$. Let $y=\bigl(u,v,f(u,v)\bigr)$ be an arbitrary point of $S$. One has $n(y)=(-f_u(u,v),-f_v(u,v),1)\ ,$ and as $|f_u|$, $|f_v|$ are $\ll 1$ near $(0,0)$ we may forget about the normalization. Now we may assume $f(u,v)=\lambda u^2 +\mu v^2 +O(r^3),\ f_u(u,v)=2\lambda u + O(r^2),\quad f_v(u,v)=2\mu v +O(r^2)\qquad\bigl( r:=\sqrt{u^2+v^2}\to 0\bigr)\ ,$ where "generically" $(\lambda, \mu)\ne(0,0)$. Therefore $(y-\xi)\cdot n(y)=\bigl(u,v,f(u,v)\bigr)\cdot\bigl(-f_u(u,v),-f_v(u,v),1)=-\lambda u^2-\mu v^2 +O(r^3)\ .$ It follows that one of your quotients will look like ${u\ v\ (\lambda u^2 +\mu v^2+O(r^3))\over \bigl( r^2+O(r^4)\bigr)^{5/2}}\ , $ which in the case $u=v:={r\over\sqrt{2}}$ is of order ${1\over r}$ for $r\to 0$.

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    I think one can always do this because changing the ordinates only in the $xOy$ plane can kill the "mixed term" and keep the normal at the origin to be $(0,0,1)$. More precisely, one can write $f(x,y)=ax^2+2bxy+cy^2=a(x+\frac{b}{a}y)^2+(c-\frac{b^2}{a})y^2$ and let $u=x+\frac{b}{a}y, v=y$.2012-03-14