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This is a question closely related to the one I posted two days ago.

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 my question:

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

Intuitively, the $[\psi_j(\xi)-\psi_j(y)]$ term may compensate the order of the numerator. But I don't have an idea for the general case even for unit ball.

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The following does not solve the problem. Anyway, some effort for attacking it:

First of all, we parameterize the surface $\partial S$, and denote $y\in\partial S$ as $$ y(\alpha,\beta)=(y_1(\alpha,\beta),y_2(\alpha,\beta),y_3(\alpha,\beta)). $$ and $x=y(\alpha_0,\beta_0)$.

Without loss of generality, assume that $x=y(0,0)$. It follows that $$ \psi_j(y)-\psi_j(x)=\psi_{j}(y(\alpha,\beta))-\psi_j(y(0,0)) =\nabla\widetilde{\psi_{j}}(0,0)\cdot[\alpha,\beta]^T+O(\alpha^2+\beta^2) $$

where $\widetilde{\psi_{j}}=\psi_j\circ y^{-1}:{\mathbb R^2}\to{\mathbb R}$ What's more, $$ y_i(\alpha,\beta)-y_i(0,0)=\nabla y_i(0,0)\cdot[\alpha,\beta]^T+O(\alpha^2+\beta^2), $$ $$ |y(\alpha,\beta)-y(0,0)|=\sqrt{\sum_{i=1}^3(\nabla y_i(0,0)\cdot[\alpha,\beta]^T+O(\alpha^2+\beta^2))^2} $$ Denote $\nabla\widetilde{\psi_{j}}(0,0)=(a_j,b_j)$, $\nabla y_i(0,0)=(\lambda_i,\mu_i)$. Then we have $$ \psi_j(y)-\psi_j(x)=\psi_{j}(y(\alpha,\beta))-\psi_j(y(0,0)) =a_j\alpha+b_j\beta+O(\alpha^2+\beta^2) $$ $$ y_i(\alpha,\beta)-y_i(0,0)=\lambda_i\alpha+\mu_i\beta+O(\alpha^2+\beta^2), $$ $$ |y(\alpha,\beta)-y(0,0)|=\sqrt{\sum_{i=1}^3(\lambda_i\alpha+\mu_i\beta+O(\alpha^2+\beta^2))^2} $$

It seems that this kind of expansion does not help.

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up vote 0 down vote accepted

Thanks to @Robert's answer on MO, the limit generally do not exit.

The key point is that one should use a "special" parameterization of the surface in order to simplify the quotient.

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