Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $u$ is only continuous and satisfies Mean value property , is it true that $u$ is harmonic in $\Omega \subset \mathbb{R}^n$ . $\Omega$ is bounded and open. What basically here should I know to prove it . Hints are appreciated . Thanks

share|cite|improve this question
What is $\Omega$? What exactly do you intend by mean value property? – Davide Giraudo Jul 11 '12 at 12:24
@DavideGiraudo : i've edited. – Theorem Jul 11 '12 at 12:28
Roughly speaking, you need to approximate $u$ by mollifiers, and then use the Mean Value Property to show that $u$ is harmonic. I learned this many years ago, and I have no reference right now. – Siminore Jul 11 '12 at 12:31

You can find a (sketch) of the proof on Wikipedia, based on approximation by convolutions. A different approach, based on the solvability of the Dirichlet problem for the laplacian, can be read in Axler's book here.

share|cite|improve this answer

One approach is to convolve with a radially symmetric mollifier and show that the convolution actually agrees with $u$ (it's better than an approximation). Intuitively, radial symmetry means that integrating over spheres gives you $u(x)$, and the for the radial integration use that the mollifier has weight one. This shows that $u$ is in fact smooth.

To show it is harmonic, an interesting approach is to use second-order Taylor approximation to show that for $C^2$ functions $u$, we have $$\Delta u(x) = \lim_{r \rightarrow 0} \frac{2}{r^2}\left(\frac{1}{|\partial B_r|}\int_{\partial B_r(x)} u - u(x)\right).$$ Applying the mean value property gives harmonicity.

share|cite|improve this answer

There are authors who define Harmonic functions as continuous functions satisfying the mean value equality, like Doob in his 'Classical Potential Theory and it's Probabilistic Counter Part'. This admits interesting generalizations, if, e.g, the Lebesgue Measure is replaced by other measures.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.