# Regularity of Newtonian potential along smooth boundary

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.

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.

• Professor Tao: This makes perfect sense. Thank you so much! May 9, 2023 at 0:45
• Could you please say more about the smooth dyadic decomposition in the first sentence: what it is and why "it suffices to show that [...]"? May 9, 2023 at 4:09
• 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 $\varepsilon$. May 9, 2023 at 7:07
• 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 May 9, 2023 at 9:11
• $\Omega$ is bounded by hypothesis, so $y-z$ is bounded, hence any summand with $\varepsilon$ sufficiently large vanishes. 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.