Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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).

share|cite|improve this question

1 Answer 1

up vote 3 down vote accepted

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.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.