show that $u$ is harmonic with a condition

Question

Let $D$ be a domain in $\mathbb{C}$ and let $u : D \to \mathbb{R}$ be a continuous function. I suppose that for each $a \in D$ there exists $r_a > 0$ with $\overline{D}(a , r_a) \subset D$ and such that $$ u(a) = \frac1{\pi r^2} \int \int_{\overline{D}(a , r)} u(z) dx dy $$ for all $r \in (0 , r_a)$. With this condition, we have that $u$ is harmonic, but how can I show it? I tried some things but no result.

Answer 1

You'll want to proceed in a couple steps. First, you'll want to show that $u$ having the mean value property implies that $u$ is smooth. You need to know that you can even compute $\Delta u$ since $u$ is, a priori, only continuous. For this, you can show that $u=\eta_\varepsilon*u$, where $\eta_\varepsilon$ is a standard mollifier.

To show that the Laplacian of $u$ vanishes, start by considering $$\int_{D_r(a)} \Delta u \ dA.$$ Then, use the divergence theorem, change variables so you're integrating over the unit sphere and try to follow your nose from there. I can add more details if you have trouble.

Edit (to address the comment): I don't want to say that an easier approach doesn't exist, but nothing comes to mind immediately. Any approach will need to accomplish the two steps I outlined above. Namely, you have to (1) prove that $\Delta u$ makes sense (a priori $u$ is only continuous) and then (2) prove that $\Delta u = 0$.

The approach I described is fairly easy. Each step only takes a few lines of calculations. Let me maybe say a few words about why I think that this approach is natural.

For the first step, you want to show that $u \in C^2(D)$. You are given a condition comparing $u$ to its integral, the mean value property (MVP): $$u(x) = \int_{D_r(x)} u \ dA.$$ So, given this it seems natural to compare $u$ to a mollified version $u_\varepsilon = \eta_\varepsilon*u$. You could perhaps try to work from difference quotients or something to prove the desired regularity, but this seems likely to be more messy and technical rather than easier.

For the second step, again what you have to go off of is the MVP. So, it makes sense to begin by considering the integral of $\Delta u$ as opposed the just $\Delta u$. Again, if you make the right moves it doesn't take that many calculations to show that the Laplacian vanishes.

Edit 2: So, let $\eta_\varepsilon$ be a standard mollifier (in particular, radially symmetric). I'm going to mildly abuse notation and write $\eta_\varepsilon(x) = \eta_\varepsilon(|x|)$. For $x \in D$, pick $R>0$ sufficiently small so that $D_R(x) \Subset D$. We begin with $$u_\varepsilon(x) = \int_{D_R(0)} \eta_\varepsilon(y)u(x-y) \ dm(y) = \int_0^R\int_{C_1(0)} \eta_\varepsilon(rP)u(x-rP) \ d\sigma(P) \ r \ dr,$$ where I introduced the notation $C_r(Q) = \partial D_r(Q)$ and we just switched to polar coordinates in the last integral. We can now exploit the radial symmetry of the mollifier and change coordinates so that we're integrating $u$ over the circle of radius $r$ centered at $x$: $$\int_0^R\int_{C_1(0)} \eta_\varepsilon(rP)u(x-rP) \ d\sigma(P) \ r \ dr = 2\pi\int_0^R \left( \frac{1}{2\pi r}\int_{C_r(x)} u \ d\sigma \right) \eta_\varepsilon(r)r \ dr.$$ Now, we exploit the MVP by noticing that the quantity in parentheses is just $u(x)$: $$2\pi\int_0^R \left( \frac{1}{2\pi r}\int_{C_r(x)} u \ d\sigma \right) \eta_\varepsilon(r)r \ dr = 2\pi u(x)\int_0^R \eta_\varepsilon(r) r \ dr.$$ To finish it off, we'll change variables one last time and use the fact that $\int \eta_\varepsilon = 1$: $$2\pi u(x)\int_0^R \eta_\varepsilon(r) r \ dr = u(x)\int \eta_\varepsilon \ dm = u(x).$$

Note that in the above, I use $m$ to denote (two-dimensional) Lebesgue measure.

Edit 3: To see that the above implies that $u \in C^2$, notice that $$\partial_x^k(\eta_\varepsilon*u)(x) = ((\partial_x^k \eta_\varepsilon)*u)(x).$$ Since $\eta_\varepsilon$ is $C^\infty$, this implies $\eta_\varepsilon*u$ is actually $C^\infty$.

Edit 4: We know that $u \in C^\infty$, so for any $D_r(x) \Subset D$, it makes sense to consider $\int_{D_r(x)} \Delta u$. The divergence theorem implies that $$\int_{D_r(x)} \Delta u \ dm = \int_{C_r(x)} \partial_n u \ d\sigma,$$ where $n$ is the outward pointing unit normal. Now, we can change coordinates so that we are integrating over $C_1 = C_1(0)$: $$\int_{C_r(x)} \partial_n u \ d\sigma = r\int_{C_1}\partial_r u(x + rP) \ d\sigma(P) = r\partial_r\left(\int_{C_1} u(x+rP) \ d\sigma(P)\right).$$ Now, we will change coordinates one more time: $$r\partial_r\left(\int_{C_1} u(x+rP) \ d\sigma(P)\right) = 2\pi r\partial_r\left( \frac{1}{2\pi r} \int_{C_r(x)} u \ d\sigma \right).$$ Finally, we can apply the MVP: $$2\pi r\partial_r\left( \frac{1}{2\pi r} \int_{C_r(x)} u \ d\sigma \right) = 2\pi r \partial_r u(x).$$ But, recall that $x$ is fixed. We therefore have $$\int_{D_r(x)} \Delta u \ dm = 2\pi r\partial_r u(x) = 0.$$ Then, since the above holds for any $D_r(x) \Subset D$, we deduce that $$\Delta u = 0 \text{ in } D.$$