MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

This is probably a very basic matter, but I am looking for a proof of the Poisson inequality for subharmonic functions, which reads

$$\varphi(r \mathrm{e}^{\mathrm{i} \theta})\leq\frac{1}{2\pi} \int_0^{2\pi}\mathrm{d} t\,P_r\left(\theta-t\right)\varphi\left(\mathrm{e}^{\mathrm{i} t}\right)$$

where the Poisson kernel is


Extensive search on the internet has not been very successful. Perhaps because this is trivial, but I don't see it. Of course, a fairly self-contained proof would be dynamite, but any help will be very much appreciated.



share|cite|improve this question
This is just a minor point and more for the sake of people reading your question, but could you be explicit and precise about the domain of $\varphi$ as well as what you are assuming about $\varphi$? – Deane Yang Jul 10 '13 at 18:18
No you're right, that's the way math should be done. According to Wikipedia, for this inequality to hold, $\varphi$ should be subharmonic, continuous and nonnegative in an open subset containing the closed unit disk. And also, I should make clear that $r<1$ here. – Vincent D Jul 11 '13 at 16:22
If you want your accounts to be merged, you should go here: – S. Carnahan Aug 20 '13 at 0:34

I don't have Ransford's book in front of me, but perhaps I can give some hints about how the proof goes. It is quite standard if you know the basic results on subharmonic functions and Poisson integrals.

The right hand-side of the inequality is the Poisson integral $P[\phi]$ of $\phi$ at the point $re^{i\theta}$. This gives a harmonic function in the unit disk with boundary values equal to $\phi$ (See e.g. Rudin's real and complex analysis, chapter on harmonic functions).

Hence the function $\phi - P[\phi]$ is subharmonic on the unit disk and $0$ on the unit circle, so that $\phi - P[\phi] \leq 0$ on the unit disk by the maximum principle for subharmonic functions. (This is where the term "subharmonic" comes from : a subharmonic function which is less than equal to a harmonic function on the boundary of some compact set must be less than or equal to the same harmonic function in the interior).

Hope this helps, Malik

share|cite|improve this answer
I would add that the RHS is harmonic, because $P_r(\theta)$ as a function of $z = r(\cos\theta + i\sin\theta)$ is harmonic. The limit of the RHS is equal to $\phi(e^{i\theta})$, because $P_r(\theta)$, viewed as family of functions of $\theta$ only and parameterized by $r$, converges to the delta function centered at $\theta = 0$. – Deane Yang Jul 13 '13 at 23:18
Thanks for all the answers. I don't really have time to look this up now, but I'll keep you guys informed of my progress. Cheers. – Vincent D Jul 15 '13 at 8:36
Okay Malik I finally took the time to look at what you had written. I followed your method and it was fairly simple, requiring much less intermediate results than the proof in Ransford's book. So thank you very much and thanks to everybody else. – user37377 Jul 19 '13 at 14:55
You're welcome! You might want to accept my answer then by clicking on the checkmark next to it. – Malik Younsi Jul 19 '13 at 16:48

The book Potential Theory in the Complex Plane by Thomas Ransford has a proof of this fact on page 35. I learned potential theory in two dimensions from this book.

share|cite|improve this answer
"I learned potential theory in two dimensions from this book." You're not the only one ;) It is a very nice book. – Malik Younsi Jul 10 '13 at 22:47
Thank you very much for this quick answer. I will try and understand what is done in the book. The proof does not look very self-contained, but this is not an urgent matter, so I can try and tackle it. Thanks again. I am still open to other suggestions though. – Vincent D Jul 11 '13 at 16:28
@MalikYounsi Another vote for the book here. Though I really must finish my abandoned project of working through the exercises... – Yemon Choi Jul 13 '13 at 18:02

protected by Community Aug 20 '13 at 0:34

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

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