This is a updated question closely related to the one I posted several days ago in math.SE. (I've put on math.SE, but there is no answer so far.)

Thanks to Christian Blatter's answer to that question, the limit (there are 9 limits here indeed.) $$ \lim_{y\to\xi}\frac{(\xi_i-y_i)(\xi_j-y_j)({{\xi}-y})\cdot n(y)}{|\xi-y|^5},\quad 1\leq i,j\leq 3,\tag{1} $$ does not exist in general. Here $S\subset{\mathbb R}^3$ 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 unit normal vector at point $y$. Here $(\xi-y)\cdot n(y)$ is the dot product.

The key point in the counterexample is that the quotient is of order $\frac{1}{|\xi-y|}$. I am interested in the following "updated" limit: $$ \lim_{y\to\xi}\,[\psi_j(\xi)-\psi_j(y)]\frac{(\xi_i-y_i)(\xi_j-y_j)({{\xi}-y})\cdot n(y)}{|\xi-y|^5},\quad 1\leq i\leq 3,\tag{2} $$ where the Einstein summation convention is applied for $j$ here and $\psi_j:S\to{\mathbb R}$ is assumed to be $C^{\infty}$. Or without normalization, consider the limit $$ \lim_{y\to\xi}\,[\psi_j(\xi)-\psi_j(y)]\frac{(\xi_i-y_i)(\xi_j-y_j)({{\xi}-y})\cdot \frac{\partial y}{\partial\alpha}\times\frac{\partial y}{\partial\beta}}{|\xi-y|^5},\quad 1\leq i\leq 3,\tag{3} $$ where $y(\alpha,\beta)=(y_i(\alpha,\beta))_{1\leq i\leq3}$ is a parameterization of $S$.

Here is the new question:

Does the updated limit (2) or (3) exist?

Intuitively, the $[\psi_j(\xi)-\psi_j(y)]$ term may compensate the order of the numerator. In the trivial case where $S$ is a plane, the quotient is $0$. But I don't have a strategy for the general case (e.g., when $S$ is a unit ball).


In general, these limits won't exist, though they might exist for special choices of the $\psi_j$. You can see this as follows: By translation, you might as well assume that $\xi=0$. Moreover, by rotation, you might as well assume that $n(0)=(0,0,1)$, and then the surface can be parametrized near $0\in\mathbb{R}^3$ in the form $$ y = \bigl(u,v,h(u,v)\bigr) $$ where $h(u,v) = a\ u^2 + b\ v^2 + O(3)$ for some real constants $a$ and $b$. Then $y\cdot n(y) = - a\ u^2 - b\ v^2 + O(3)$ and $|y| = \sqrt{(u^2 + v^2)} + O(2)$.

Without loss of generality, you can assume that $\psi$ vanishes at $0\in S$, so $y\cdot\psi(y) = r\ u^2 + 2s\ uv + t\ v^2 + O(3)$ for some real numbers $r, s, t$, which could be any real numbers, depending on your choice of $\psi$. So now, you are basically asking whether the limits $$ \lim_{(u,v)\to(0,0)} \frac{u\ (r\ u^2 + 2s\ uv + t\ v^2)(a\ u^2 + b\ v^2)}{(u^2 + v^2)^{5/2}} $$ and $$ \lim_{(u,v)\to(0,0)} \frac{v\ (r\ u^2 + 2s\ uv + t\ v^2)(a\ u^2 + b\ v^2)}{(u^2 + v^2)^{5/2}} $$ exist, whatever the values of $a,b,r, s,t$, and it is clear that, for most such values, these limits will not exist.

  • $\begingroup$ @Robert, thank you for your answer. I guess $O(2)$ in your answer means $O((\sqrt{u^2+v^2})^2)$, correct? $\endgroup$ – Jack Mar 10 '12 at 21:49
  • 1
    $\begingroup$ @Jack: Yes, that's correct. $O(n)$ means 'terms that are bounded by a constant times $(u^2+v^2)^{(n/2)}$. $\endgroup$ – Robert Bryant Mar 10 '12 at 22:54

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