Higher dimensional Cauchy's integral formula looking thing for a harmonic function

The question in full: Show that if $f$ is a harmonic function in a closed sphere of radius $r$ with center $(x_0,y_0,z_0)$ then $$f(x_0,y_0,z_0)=\frac{1}{4\pi r^2}\int \int_Sfd\sigma=\frac{3}{4\pi r^2}\int \int\int_TfdV$$ First, I think the coefficient on the second integral should be $\frac{3}{4\pi r^3}$, is that correct?

I think the divergence theorem may be involved, given the second equality, but I do not have much facility with the theorem. If this were a problem about discs, I could relate back to complex analysis, but I am not sure that is germane here.

Any hints would be great. Also any relationship with harmonic functions being the real part of holomorphic functions would be really helpful. Thanks!

edit: The double integral over $S$ denotes the integral over the sphere, the triple integral over $T$ denotes integrating over the ball.

• What are $S$ and $T$? Commented Mar 12, 2017 at 3:27
• $S$ is the sphere, $T$ is the ball, will edit Commented Mar 12, 2017 at 3:28

This is just the $\mathbb{R}^3$ case of the mean value property for harmonic functions. If $\Omega$ is a domain in $\mathbb{R}^n$ and $x = (x_1,\dots,x_n) \in \Omega$, then for any $r \in (0,\mathrm{dist}(x, \Omega))$, one has $$f(x) = \frac{1}{n\alpha(n)r^{n-1}} \int_{\partial B} f \;\mathrm{d}\sigma = \frac{1}{\alpha(n)r^n} \int_B f \; \mathrm{d}V,$$ where $B = B(x, r)$ and $\alpha(n)$ is the volume of the unit ball in $\mathbb{R}^n$. This property also characterizes harmonic functions, namely that if the mean value property holds for every $x \in \Omega$, then the function is harmonic. The proof in any dimension is the same: use a change of variables to move the radius into the argument of the function in the volume integral, take derivatives to show that the integral is independent of $r$ (pass the derivative inside the integral to do this; that's the point of the change of variables), apply the divergence theorem, and use continuity to show that the integral is equal to the value of the function at the center of the ball.
Let $u \colon \Omega \to \mathbb{R}$ be harmonic where $\Omega \subset \mathbb{R}^d$ is an open set, $d \geq 2$. Without loss of generality, let $x_0 = 0$ (otherwise, change variables via translation). For $r > 0$, let $f(r)$ be the mean value of $u$ over the sphere $rS^{d-1}$, i.e. $$f(r) = \frac{1}{Cr^{d-1}} \int_{\partial B(0,r)} u(y) \;d\sigma.$$ By the change of variables $y \mapsto rx$, $$f(r) = \frac{1}{Cr^{d-1}} \int_{\partial B(0,1)} u(rx) r^{d-1} \;d\sigma = C' \int_{\partial B(0,1)} u(rx) \;d \sigma.$$ Since the integrand is smooth, we can differentiate under the integral sign and apply the divergence theorem \begin{align*} f'(r) &= C' \int_{\partial B(0,1)} \nabla u(rx) \cdot x \;d\sigma \\ &= C' \int_{\partial B(0,r)} \nabla u(y) \cdot \frac{y}{r} \;d\sigma \\ &= C' \int_{\partial B(0,r)} \nabla u(y) \cdot \vec{n} \;d\sigma \\ &= C' \int_{B(0,r)} \operatorname{div}(\nabla u(y)) \;dV \\ &= C' \int_{B(0,r)} \Delta u(y) \;dV \\ &= 0. \end{align*} Note that by continuity, $f(0) = u(0)$, so $f(r) = u(0)$ for all $r \geq 0$. This is the desired result. The proof of the reverse direction is much shorter and can be done easily by contradiction: if $u$ is not harmonic, then there is some point $x_0$ such that $\Delta u(x_0) > 0$ (without loss of generality). By continuity, there is a ball where the Laplacian is strictly positive. Now apply the divergence theorem.