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This is a follow up question (which is unanswered in math.SE) for a previous one. Consider the following functions $F_{ij}:S\subset{\mathbb R}^3\to{\mathbb R}$, $$ F_{ij}(y) = \begin{cases} \frac{(y_i-x_i)(y_j-x_j)(y-x)\cdot n(y)}{|y-x|^3},&y\neq x; \\ 0,& y=x.\end{cases} \quad i,j = 1,2,3 $$ where $S$ is a surface which has a continuously varying normal vector, $x=(x_1,x_2,x_3)\in S$ is given, $y=(y_1,y_2,y_3)\in S$, $n(y)$ is the normal vector at point $y$ which is assumed to be smooth. Here $(y-x)\cdot n(y)$ is the dot product. Using the method in the answer to the previous question, I conclude that given $i,j$ $$ \lim_{y\to x}F_{ij}(y)=0 $$ which implies that $F_{ij}(y)$ is continuous at $y=x$.

Here is my question:

  • Is $F_{ij}(y)$ smooth at $x$? If it is not, what would be the key properties to fail the smoothness?

An immediate idea is that I should test the smoothness of $F_{ij}$ by definition. The difficulty is that with a parameterization $y=y(\alpha,\beta)$, it is not trivial to find the high order partial derivatives for $F_{ij}(y(\alpha,\beta))$. I am not even able to determine whether $F_{ij}$ is $C^1(S)$ or not.

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Have you tried the following: Place the surface so that $x = (0,0,0)$ and the tangent plane at $x$ is spanned by the $xy$ plane. Then the surface can be parameterized locally as a graph $z = h(x,y)$. You can then find a formula for $F$ in terms of $h$, ad $x$, and $y$ (my notation, not yours) that you should be able to analyze. – Deane Yang Mar 20 '12 at 18:25
@Deane Yang: Thanks for your comment. I am able to get something like $$F_{12}=\frac{-xy(xh_x+yh_y)}{(x^2+y^2+h^2)^{3/2}}.$$ Due to the root term in the denominator, I guess the the function can not be smooth. It seems that there is no choice but calculate $\frac{\partial F}{\partial x}$ and see what's $\lim_{(x,y)\to(0,0)}\frac{\partial F}{\partial x}$. On the other hand, however, if the function is smooth, I can't imagine what $\frac{\partial^n F}{\partial x^n}$ would look like. – Jack Mar 20 '12 at 20:24
Oops, $(xh_x+yh_y)$ should be $(xh_x+yh_y-h)$ above. – Jack Mar 20 '12 at 20:26
Yes, you should be able to show that if $h$ is smooth and vanishes at the origin, then $F_{12}$ cannot be smooth due to the $3/2$ power. – Deane Yang Mar 20 '12 at 21:57

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