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 $V(z) \in C^{\infty}(\partial \Omega)$?




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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.



Regularity of Newtonian potential along smooth boundary
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.


