Let $\Omega$ be a bounded open set in $\mathbb{R}^n$ with $C^\infty$ boundary, $n\ge 3$. Define $$V(z)=\int_\Omega \frac{1}{|z-y|^{n-2}}dy$$ Is it true that $V(z) \in C^{\infty}(\partial \Omega)$?

Motivation: Classical Holder estimate says that if $f \in C_c^\alpha(\mathbb{R}^n)$, then the Newtonian potential given by $$V(z)=\int_{\mathbb{R}^n} \frac{f(y)}{|z-y|^{n-2}}dy$$ is $C^2$ with respect to $z$ variable. If $f$ is merely bounded, then $V$ is $C^1$ and we cannot expect better regulartiy. This can be seen from the classical example that $f=\chi_{B_1}$, where $\chi$ is the characteristic function and $B_1$ is the unit ball.

However, for this example, one can see that $V\in C^{\infty}(\partial B_1)$. From this example, it is natural to study the smoothness of $V$ along smooth boundaries. This motivates the original question. Of course, one can consider similar problems for more general Riesz potentials.

Further Update:

The question actually was motivated from a talk yesterday given by Jian Lu from South China Normal University. The talk is on chord log Minkowski problem. The speaker actually have already proved the smoothness of $V(z)$ along boundaries of any smooth convex bodies, and the details can be seen in http://arxiv.org/abs/2304.14220 (2nd version). Hence I was led to ask about the regularity on generic smooth sets.


3 Answers 3


Sure. By a smooth dyadic decomposition it suffices to show that convolutions of the form $$ \varepsilon^{-n} \int_\Omega \varphi\left(\frac{y-z}{\varepsilon}\right)\ dy$$ for $0 < \varepsilon \lesssim 1$ and $\varphi$ a fixed bump function are smooth on $\partial \Omega$ uniformly in $\varepsilon$ (multiply by $\varepsilon^2$ and sum over dyadic $\varepsilon>0$ for a suitably chosen $\varphi$ to recover the Newton potential). This is trivial for large $\varepsilon$, so we may assume $\varepsilon$ small.

The strategy here is to transform this expression to eliminate all negative powers of $\varepsilon$, as this is the only obstruction to non-uniformity.

Locally we may parameterize $\Omega$ as a half-space $\{ (y', y_n): y_n \geq f(y') \}$ for some smooth function $f$, and then for $z = (z',f(z'))$ and $\varepsilon$ small enough the above expression becomes $$ \varepsilon^{-n} \int_{{\bf R}^{n-1}} \int_0^\infty \varphi\left(\frac{(y'-z', f(y')-f(z')+t)}{\varepsilon}\right)\ dy' dt$$ which after a change of variables $y' = z'+\varepsilon w$, $t = \varepsilon s$ becomes $$ \int_{{\bf R}^{n-1}} \int_0^\infty \varphi\left(\left(w, \frac{f(z'+\varepsilon w)-f(z')}{\varepsilon}+s\right)\right)\ dw ds. \quad (1) $$ By the fundamental theorem of calculus we have $$ \frac{f(z'+\varepsilon w)-f(z')}{\varepsilon} = \int_0^1 w \cdot \nabla f(z' + \varepsilon \theta w)\ d\theta$$ which can then be seen to (locally) be a smooth function of $z'$ and $w$ uniformly in $\varepsilon$. From this it follows from repeated differentiation under the integral sign and the chain rule that the expression in (1) is a smooth function of $z'$ uniformly in $\varepsilon$, giving the claim.

  • $\begingroup$ Professor Tao: This makes perfect sense. Thank you so much! $\endgroup$
    – student
    May 9, 2023 at 0:45
  • 2
    $\begingroup$ Could you please say more about the smooth dyadic decomposition in the first sentence: what it is and why "it suffices to show that [...]"? $\endgroup$ May 9, 2023 at 4:09
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    $\begingroup$ Starting with a standard Littlewood-Paley decomposition $1 = \sum_j \phi(2^j \xi)$ (with $\phi$ smooth and supported on an annulus $\{ \xi: |\xi| \sim 1\}$) we have $\frac{1}{|z-y|^{n-2}} = \sum_\varepsilon \varepsilon^2 \varepsilon^{-n} \varphi(\frac{y-z}{\varepsilon})$, where $\varepsilon$ ranges over powers of two and $\varphi(x) := \phi(x)/|x|^{n-2}$. Because of the boundedness of $\Omega$ we can discard the contribution of all sufficiently large $\varepsilon$. $\endgroup$
    – Terry Tao
    May 9, 2023 at 7:07
  • $\begingroup$ To check if I have understood: you need $\epsilon$ small to sum $\sum_\epsilon \epsilon^2$ (over small diadic numbers) and for $\epsilon$ large you sum $\sum_\epsilon \epsilon^{2-n-k}$ where $k$ denotes a number of derivatives. I do not see a restricion on $\epsilon$ when localizing. Is this correct? Thank you $\endgroup$ May 9, 2023 at 9:11
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    $\begingroup$ $\Omega$ is bounded by hypothesis, so $y-z$ is bounded, hence any summand with $\varepsilon$ sufficiently large vanishes. $\endgroup$
    – Terry Tao
    May 9, 2023 at 14:28

This is a direct proof which gives $V \in C^\infty (\bar \Omega)$ whenever $g \in C^\infty (\bar \Omega)$, $V$ being the Newtonian potential of $g$. As in the proof by @Terry Tao assume that locally $\Omega=\{(y',y_n): y_n \geq f(y')\}$ and for $x=\{(x', x_n), x_n \geq f(x') \}$, $$ V(x)=\int_{\mathbb R^{n-1}}\int_{\{y_n \geq f(y')\}}|x-y|^{2-n}g(y',y_n) dy' dy_n=\\\int_{\mathbb R^{n-1}} dz' \int_0^\infty (|z'|^2+(f(x')-f(x'-z')+s-t)^2)^{(2-n)/2}g(x'-z', f(x'-z')+t )dt $$ with $y'=x'-z'$, $y_n=f(y')+t$, $x_n=f(x')+s$ and $t,s \geq 0$.

Since $f,g$ are smooth (we may assume also that $g$ has a compact support and that $x'$ runs in a compact set) this formula allows to differentiate with respect to $x'$ infinitely many times under the integral sign. Note that the singularity of the kernel has a power $2-n$ and does not increase by differentiating with respect to $x'$, because of the term $f(x')-f(x'-z')$. This shows that all derivatives of any order with respect to the $x'$ variables are continuous on $\bar \Omega$.

At this point the continuity of the derivatives with respect to $x_n$ follows from the equation $\Delta V=g$. This is immediate for $V_{x_n x_n}$. Having $D^\alpha_{x'}V$ and $\Delta D^\alpha_{x'}V=D^\alpha_{x'}\Delta V=D^\alpha_{x'}g$ continuous, it follows that $D^\alpha_{x'}D_{x_n x_n}V$ is continuous and so on for even numbers of derivatives with respect to $x_n$. The continuity of those having odd numbers of $x_n$ derivatives follows from elementary interpolation inequalities.


One more perspective: after performing a diffeomorphism that locally flattens the boundary, we get an equation of the form $$a_{ij}(x)w_{ij} + b_i(x)w_i = f(x_n) \text{ in } B_1,$$ where the coefficients are smooth (with bounds on derivatives depending only on $\Omega$) and $f$ is the Heaviside step function. Assume first for simplicity that $f$ is smooth (so that $w$ is smooth) and bounded between $0$ and $1$. Calderon-Zygmund theory gives $w \in W^{2,\,p}$ for any $p$ with an estimate. Differentiating the equation in the $e_k$ direction for $k < n$ we see that $a_{ij}(w_k)_{ij}$ is a linear combination of $D^2w, \nabla w$. Calderon-Zygmund theory thus gives $W^{2,p}$ estimates for the horizontal derivatives of $w$ for any $p$. One can continue to differentiate horizontally and apply the Calderon-Zygmund estimates to get bounds on the derivatives of $w$ of all orders in the horizontal directions, independent of the regularity of $f$. By approximation, the same holds when $f$ is the heaviside function. Reversing the diffeomorphism we see that $V$ is smooth along the boundary.


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