# Equivalence of local mean value property and mean value property

Let $$\Omega \subseteq \mathbb{R}^n$$ open and connected. Let $$u$$ be continuous.

We say $$u$$ satifies the Mean-Value Property (MVP) if $$\forall x \in \Omega, B_{\epsilon}(x) \subseteq \Omega$$, we have that

$$$$u(x) = \frac{1}{n \omega_n R^{n-1}} \int_{\partial B_{\epsilon}(x)}u(y) dS(y).$$$$

We say $$u$$ satisfies the Local Mean-Value Property (LMVP) if $$\forall x \in \Omega, \exists \epsilon(x) > 0$$ such that $$B_{\epsilon}(x) \subseteq \Omega$$ and $$\forall 0 < \delta < \epsilon(x)$$

$$$$u(x) = \frac{1}{n\omega_n R^{n-1}} \int_{\partial B_{\delta}(x)} u(y) dS(y)$$$$

Clearly, $$MVP \Rightarrow LMVP$$, but $$LMVP \Rightarrow MVP$$?

Say I fix $$x \in \Omega$$. I wish to show that $$u$$ is harmonic in $$B_{\epsilon(x)}(x)$$. I have continuity of $$u$$ and LMVP, which gives me MVP on the smaller domain and so the function is harmonic on $$B_{\epsilon(x)}(x)$$ (1). Since $$\epsilon(x) > 0$$ for each $$x$$, $$u$$ is harmonic on $$\Omega$$. So MVP is satisfied. But returning to (1), this logic is invalid.I don't know that MVP is satisfied. If I fix $$y \in B_{\epsilon(x)}(x)$$, I then have to pass to $$\epsilon(y)$$, and these could get arbitrarily small. Any ideas?

I was wondering if that is true as well. I think you can take the ideas from the answer of this question (Does a weaker form of the mean value property already imply harmonicity for continuous functions?) to solve it.

Edit: I was told to instead provide a detailed answer in this thread. Still the main ideas come from the aforementioned post.

I think either I am missing something in your original argument, or it has a mistake in it: Fixing $$\varepsilon(x)$$ so that the MVP with respect to $$x$$ holds in $$B_{\varepsilon(x)}(x)$$ does not immediately give you the MVP on the ball $$B_{\varepsilon(x)}(x)$$. At least not in the sense that for any point $$y$$ in this ball and any $$R>0$$ so that $$B_{R}(y) \subseteq B_{\varepsilon(x)}(x)$$ it holds $$u(y) = \frac{1}{n\omega_n R^{n-1}} \int_{\partial B_{R}(y)} u(y) dS(y).$$ In fact, if that were the case, you would directly get that $$u$$ is harmonic in $$\Omega$$, which directly implies that the MVP holds. The idea is that harmonicity, i.e. $$\Delta u = 0$$ (and $$u$$ twice differentiable) is a property that you can check locally. The reason why you don't get the statement for all $$y$$, $$R$$ as above is that the LMVP is only applicable if $$R \leq \varepsilon(y)$$, which (as you said) could get arbitrarily small.

There is a non-continuous 'counterexample' that highlights this problem. Take $$\Omega = \mathbb{R}^2$$ and define $$u$$ by letting $$u(x) := 1$$ if $$x_2>0$$, $$u(x) := 0$$ if $$x_2 = 0$$ and $$u(x) := -1$$ if $$x_2 < 0$$. This function satisfies the LMVP, but it is clearly not harmonic. The problem is that as you approach the $$x_1$$-axis the $$\varepsilon(x)$$ get arbitrarily small. A continuous counterexample does not exist, since (I believe) LMVP really does imply MVP, which I will try to sketch a proof for.

As we have seen, all we really need to show is that $$u$$ is harmonic. Take any $$x\in \Omega$$ and $$r>0$$ so that $$\overline{B_r(x)} \subseteq \Omega$$. Let $$\tilde{u}$$ be the solution to the Dirichlet problem $$\Delta \tilde{u} = 0 \quad \text{in B_r(x)},$$ $$\tilde{u} = u \quad \text{on \partial B_r(x)}.$$ Since $$u$$ is continuous and $$B_r(x)$$ has a sufficiently smooth boundary, a solution exists. The function $$\tilde{u}$$ is harmonic and thus satisfies the MVP, so trivially also the LMVP. Thus the function $$v = u - \tilde{u}$$ (as a function on $$\overline{B_r(x)}$$) is continuous and satisfies the LMVP. Note also that because of the boundary conditions, $$v$$ vanishes on $$\partial B_r(x)$$.

Next we show that $$v$$ takes its maximum on $$\partial B_r(x)$$ (this is the maximum principle, I will take a proof from one of my lectures, I believe it is similar in Evans' book). Since $$v$$ is a continuous function on a compact domain, it clearly obtains its maximum $$M$$ somewhere. Let $$B_r(x)$$ denote the open ball and define $$\Sigma := \{ y \in B_r(x) \colon v(y) = M \}.$$ Since $$v$$ is continuous, $$\Sigma$$ is relatively closed in $$B_r(x)$$. For any $$y$$ in $$\Sigma$$, choose $$\varepsilon(y)$$ as in the statement of the LMVP and let $$\varepsilon' \leq \varepsilon(y)$$ be so that $$B_{\varepsilon'}(y) \subseteq B_r(x)$$. Then for any $$\delta \in (0,\varepsilon')$$:

$$M = v(y) = \frac{1}{n\omega_n \delta^{n-1}} \int_{\partial B_{\delta}(y)} v(y) dS(y) \leq \frac{1}{n\omega_n \delta^{n-1}} \int_{\partial B_{\delta}(y)} M dS(y) = M,$$ So the two integrals in the middle are equal. Since $$v$$ is contnuous and $$v \leq M$$, this gives $$v \equiv M$$ on $$\partial B_\delta(y)$$. By the arbitrariness of $$\delta$$ we get $$v \equiv M$$ on $$B_{\varepsilon'}(y)$$, i.e. $$B_{\varepsilon'}(y) \subseteq \Sigma$$. This shows that $$\Sigma$$ is relatively open in $$B_r(x)$$. Since $$B_r(x)$$ is connected, $$\Sigma = \emptyset$$ or $$\Sigma = B_r(x)$$, in both cases it attains its maximum on the boundary. We know that $$v$$ vanishes on the boundary, so we get $$v \leq 0$$ on $$B_r(x)$$.

Repeating the same argument for the minimum yields that $$v \geq 0$$ on $$B_r(x)$$, so in summary we get that $$v \equiv 0$$. This means that $$u$$ and $$\tilde{u}$$ are identical on $$B_r(x)$$, so $$u$$ is harmonic on $$B_r(x)$$. This is precisely what we wanted to show.

• While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review Nov 4, 2021 at 19:40