I have done a lot in this problem, but unfortunately it is not enough to solve it, answers or hints are very welcome.

Let $B$ be a rectangle in $\mathbb R^2$ and consider $\varphi\colon B\to\mathbb{R}^3$ of class $C^1$.

For any $y\in\mathbb{R}^3\setminus\varphi(B)$, define $$\Omega(y) = \int_\varphi\frac{1}{|y-x|^3}(y-x)\,dS_x.$$ Show that $\Omega$ is of class $C^1$ as a function in variable $y$.

At this point I could find that $$\frac{\partial\Omega}{\partial y_1}(y)=\int_{\partial\varphi}\frac{1}{|y-x|^3}(0,x_3-y_3,x_2-y_2)$$ $$\frac{\partial\Omega}{\partial y_2}(y)=\int_{\partial\varphi}\frac{1}{|y-x|^3}(x_3-y_3,0,x_1-y_1)$$ and $$\frac{\partial\Omega}{\partial y_3}(y)=\int_{\partial\varphi}\frac{1}{|y-x|^3}(x_2-y_2,x_1-y_1,0).$$ Now, all I have to do is show that any one of these is continuous as a function of $y$, and that is where things are getting hard.


  • $\begingroup$ Interesting question. I'm wondering, though, do you mean $y \notin R^3 \\ \phi(B)$? The reason I ask is: $y \in R^3 \\ \phi(B)$ means $y \in R^3$ and $y \notin \phi(B)$; the logical negation of this assertion is $y \notin R^3$ or $y \in \phi(B); $\endgroup$ Jul 18, 2013 at 5:59
  • $\begingroup$ wrong typing....should be $y\in R^3\backslash \varphi(B)$, going to fix this. Thanks. $\endgroup$
    – Integral
    Jul 18, 2013 at 12:45
  • $\begingroup$ Is the rectangle $B$ closed? (If so, then it's compact and the complement of $\varphi(B)$ is open, which helps.) $\endgroup$
    – 40 votes
    Jul 18, 2013 at 14:08
  • $\begingroup$ If I understand the problem correctly, you have $$\Omega(y)=\int_B \frac{y-\varphi(u,v)}{|y-\varphi(u,v)|^2}\,|\varphi_x\times \varphi_y|\,du\,dv$$ As a function of $y$, the integrand is real-analytic on the complement of $\varphi(B)$. The power series can be integrated term by term (standard theorem), showing that $\Omega$ is real analytic as well. $\endgroup$
    – 40 votes
    Jul 18, 2013 at 14:15
  • 1
    $\begingroup$ @Integral: the question of the rectangle $R$ being closed or not actually makes a substantial difference in the difficulty of the promlem, IMO. Could you please clarify this issue for us? $\endgroup$ Jul 18, 2013 at 22:15

1 Answer 1


I'm going to sketch what I believe to be a valid proof of this assertion. Unfortunately, I haven't the time at the moment to flesh out all the details; I'm hoping I'll provide enough detail that the fleshing can be done by my readers without too much work. Please feel free to leave comments if you feel further discussion is necessary or in any way called for.

I think there are two critical facts needed in proving this. The first, mentioned in the comments, is the compactness of $\varphi(B)$ in $R^3$. This follows directly from the well-known and basic fact that the continuous image of a compact set is compact. Now fix for the moment $y \in R^3 \setminus \varphi(B)$, and consider the Euclidean distance function $\vert x - y \vert$ where $x \in \varphi(B)$. This is a continuous function of $x \in \varphi(B)$, an assertion which follows easily from the triangle inequality $\vert y - z \vert \le \vert y - x \vert + \vert x - z \vert$, which readily implies $\vert \vert y - z \vert - \vert y - x \vert \vert \le \vert x - z \vert$, so that if $\vert x - z \vert$ is sufficienly small, $\vert y - z \vert$ is as close as we please to $\vert y - x \vert$. As a continuous function on the compact set $\varphi(B)$, $\vert x - y \vert$ attains its minimum value $d_1$ at some point $x_0 \in \varphi(B)$, and we must have $d_1 > 0$ since $y \notin \varphi(B)$. Now pick some real $d_2$, with $0 < d_2 < d_1$, and consider the closed ball $\bar B(y, d_2)$ of radius $d_2$ around $y \in R^3 \setminus \varphi(B)$. Clearly $\bar B(y, d_2) \cap \varphi(B) = \phi$, otherwise there would be a point $w \in \varphi(B)$ with $\vert y - w \vert \le d_2$, contradicting the fact that the minimum distance between $y$ and any point $x \in \varphi(B)$ is $d_1$. In point of fact, we can go a bit further: for $w \in \bar B(y, d_2)$, and $x \in \varphi(B)$, writing $w - x = (w - y) - (x - y)$, we have $\vert w - x \vert = \vert (w - y) - (x - y) \vert \ge \vert \vert w - y \vert \ - \vert x - y \vert \vert$; but $\vert x - y \vert \ge d_1$ by the choice of $x_0$, and $\vert w - y \vert \le d_2$ since $w \in \bar B(y, d_2)$, we have $\vert w - x \vert \ge \vert d_1 - d_2 \vert > 0$, for any $x \in \varphi(B)$ and any $w \in \bar B(y, d_2)$.

The preceding argument in fact shows that the integrand in the expression for $\Omega(y)$, that is,

$\frac{1}{\vert y - x \vert^3}(y - x)$,

is uniformly bounded for $x \in \varphi(B)$ and $w \in \bar B(y, d_2)$; indeed,

$\vert \frac{1}{\vert w - x \vert^3}(w - x) \vert = \frac{1}{\vert w - x \vert^3} \vert (w - x) \vert = \frac {1}{\vert w - x \vert^2} \le \frac{1}{\vert d_1 - d_2 \vert^2} < \infty$.

We're getting there.

The second critical fact concerns the measure $dS_x$ on $\varphi(B)$. To make things fly, we need it to be reasonably well-behaved. By this I mean mostly that a set of finite measure in $B$ cannot map, under the action of $\varphi$, to something of infinite measure. But I think it is pretty clear, given the differentiability of $\varphi$, and again calling on the compactness of $B$, that this can't happen. Indeed, the expression for $dS_x$ given by 40 votes in his comment, viz. $\vert \varphi_u \times \varphi_v \vert du dv$, rules out this kind of pathology since $\varphi_x$, $\varphi_y$ are continuous ($\varphi$ is of class $C^1$ by hypothesis) and $B$ is compact. Indeed, in "performing" the requisite integrations we can work in $u-v$ coordinates in $B$.

I think we've got enough stuff together to kick this one in the head.

By what we have seen, the entire integrand, which thanks to 40 votes and his astute comment is, in terms of coordinates in $B$,

$\frac{(w - \varphi(u, v))}{\vert w - \varphi(u, v) \vert^3} \vert \varphi_u \times \varphi_v \vert$,

is uniformly bounded for $x \in \varphi(B)$, that is, for $(u, v) \in B$ with $x = \varphi(u, v)$ and $w \in \bar B(y, d_2)$. So we can thus invoke the standard theorems, which sort of go by such names as the Leibniz integral rule, the dominated convergence theorem, the bounded convergence theorem, see this link, Leibniz integral rule, for the details.

Finally, I have in the above made repeated use of the inequality $\vert \vert a \vert - \vert b \vert \vert \le \vert a - b \vert$, which holds in any normed space and follows from the triangle inquality: if $\vert a + b \vert \le \vert a \vert + \vert b \vert$ via $\vert a \vert = \vert (a - b) + b \vert \le \vert a - b \vert + \vert b \vert$, whence $\vert a \vert - \vert b \vert \le \vert a - b \vert$, and by switching the roles of $a$ and $b$ the assertion $\vert \vert a \vert - \vert b \vert \vert \le \vert a - b \vert$ follows.

Hope this is clear; hope I haven't made too many typos. Gotta run, my night job beckons. If any errors aren't so large as to destroy my answer, I'll try and correct them later.


  • $\begingroup$ @Integral: didn't this question once have a bounty, back in ages past? I remember busting my ass to finish it in time, then missing the deadline or some such. Ha! Better late than never! Thanks for the "acceptance"! ;-) $\endgroup$ Apr 15, 2014 at 22:26

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