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

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

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 $\forall 0 < \delta < \epsilon(x)$

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

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?


1 Answer 1


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.

  • $\begingroup$ 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 $\endgroup$
    – MH.Lee
    Commented Nov 4, 2021 at 19:40

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