0
$\begingroup$

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

$$P_r\left(\theta-t\right)\equiv\frac{1-r^2}{1-2r\cos\left(t-\theta\right)+r^2}.$$

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.

Thanks,

V.D.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ 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 $\varphi$? $\endgroup$
    – Deane Yang
    Jul 10, 2013 at 18:18
  • $\begingroup$ 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. $\endgroup$
    – Vincent D
    Jul 11, 2013 at 16:22
  • 1
    $\begingroup$ If you want your accounts to be merged, you should go here: mathoverflow.net/help/merging-accounts $\endgroup$
    – S. Carnahan ♦
    Aug 20, 2013 at 0:34

2 Answers 2

Reset to default
4
$\begingroup$

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

Improve this answer
$\endgroup$
4
  • $\begingroup$ 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 $\theta = 0$. $\endgroup$
    – Deane Yang
    Jul 13, 2013 at 23:18
  • $\begingroup$ 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. $\endgroup$
    – Vincent D
    Jul 15, 2013 at 8:36
  • $\begingroup$ 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. $\endgroup$
    – user37377
    Jul 19, 2013 at 14:55
  • $\begingroup$ You're welcome! You might want to accept my answer then by clicking on the checkmark next to it. $\endgroup$
    – Malik Younsi
    Jul 19, 2013 at 16:48
3
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • 3
    $\begingroup$ "I learned potential theory in two dimensions from this book." You're not the only one ;) It is a very nice book. $\endgroup$
    – Malik Younsi
    Jul 10, 2013 at 22:47
  • $\begingroup$ 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. $\endgroup$
    – Vincent D
    Jul 11, 2013 at 16:28
  • 1
    $\begingroup$ @MalikYounsi Another vote for the book here. Though I really must finish my abandoned project of working through the exercises... $\endgroup$
    – Yemon Choi
    Jul 13, 2013 at 18:02

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