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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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    You should write 9 limits (one for each i,j). Have you tried the even simpler case of $S=\left\{x_3=0\right\}$?2012-03-01
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    @Blah, Hmm, I should have noticed the even simpler case. Yes, there are 9 limits indeed.2012-03-02
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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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    Thanks for your answer. The limit is supposed to be $\lim_{y\to\xi}$ instead of $\lim_{\xi\to y}$. Sorry for the mistake. But this does not change much, does it?2012-03-02
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    @Jack: See my edit.2012-03-02
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    I think you mean $\lvert f_u \rvert$, $\lvert f_v \rvert$ are *zero* near $(0,0)$.2012-03-02
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    @ChristianBlatter You've added three terms each $O(r^2)$ as $r$ goes to $0$ and said that generically this will be $O(r^2)$, but isn't there a distinct chance for cancellation here that would leave the sum $O(r^3)$ as $r\rightarrow 0$? It seems like you'd really want to write out the height function explicitly as a quadratic plus cubic-or-better terms to get an explicit bound on the top term, as well as showing when the limit _can_ exist...2012-03-02
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    @Steven Stadnicki: I said "in general", i.e., for a "nonspecial" surface. As Jack has remarked in the case of a plane the considered quotient is trivially $\equiv0$.2012-03-02
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    @ChristianBlatter But it could easily still be the case that the terms cancel out 'generically' - for a non-trivial example, consider the central derivative $(f(x+h)-f(x-h))/2h$. 'Generically' this difference should be $f'(x)+O(h)$ because _e.g._ the $f(x+h)$ term is $f(x)+hf'(x)+O(h^2)$, but explicit cancellation on the second-order term of the Taylor expansion means that this is in fact $f'(x)+O(h^2)$ for any function $f$. To show that your limit can't converge you can't just say '$O(r^2)$', you have to exhibit an explicit $k\cdot r^2$ term.2012-03-02
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    @Steven Stadnicki: You are right, I was a little sloppy. I hope it is watertight now.2012-03-02
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    @ChristianBlatter: what do you mean by "'generically'$(\lambda,\mu)\neq(0,0)$"?2012-03-08
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    @Jack: When $\lambda$ and $\mu$ are both zero then we have a flat umbilic point at $(0,0,0)$, and in this case the limit in question may exist. Wikipedia has to say the following about such points: "At flat umbilic points both principal curvatures are zero. A generic surface will not contain flat umbilic points". See here: http://en.wikipedia.org/wiki/Principal_curvature2012-03-09
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    @ChristianBlatter: I understand that by "translation", one can assume that $f(0,0)=0$. But how do you come up with $f_v(0,0)=f_u(0,0)=0$, which is a key assumption I think in your answer?2012-03-11
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    @Jack: I choose the $(x,y,z)$-coordinate system such that the point $\xi$ has coordinates $(0,0,0)$, that the tangent plane to $S$ at $\xi$ is the plane $z=0$, and that the second derivative of $f$ has no "mixed term".2012-03-11
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    @ChristianBlatter: I see how to choose the coordinate system which satisfies the first two conditions. But how can I guarantee that there is no "mixed term" for the second derivative of $f$ *at the same time*?2012-03-14
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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