In Axler's book "Harmonic function theory" on this link http://www.latp.univ-mrs.fr/~chaabi/ARTICLES%20IMPORTANTS/Autres%20articles%20interessant/Livres/Harmonic%20Function%20theory.pdf on page 112 the author say: "We know that if $f\in C\left(S\right)$, then $P\left[f\right]$ has a continuous extension to $\overline{B}$. What can be said of the more general Poisson integrals defined above? We begin to answer this question in the next two theorems." I tried to understand the idea of this, but I couldn't understand. So, my first question is: "What is the idea and answer about this problem?" Then, I tried to think in other way. If we suppose that $P\left[\mu\right]$ can be extended continuously to $\overline{B}$, and if we denote by $u$ such extension, then from theorem 1.21 on page 15 we will have $u=P\left[f\right]$, where $f$ is a restriction of $u$ to $S$. Now we have that $P\left[\mu\right]=P\left[f\right]$ on $B$, and this implies that $$\int_{S}P\left(x,\xi\right)d\mu\left(\xi\right)=\int_{S}P\left(x,\xi\right)d\mu_{f}\left(\xi\right),$$ where $d\mu_{f}\left(\xi\right)=f\left(\xi\right)d\sigma\left(\xi\right)$. My intuition tells me that I need to prove that $\mu=\mu_{f}$ but I don't know how to prove that. Are there any theorems in Measure theory related to this kind of problem? So we have the same integrals of a "nice" function in two measures. What are the conditions for this "nice" function and on the set to imply the equality of these measures?


The book you are reading deals with harmonic functions from the point of view of "classical analysis", whereas the question you are asking is best answered in terms of the more abstract "potential theory". The point is that for any fixed boundary point $\xi$ the Poisson kernel $P(x,\xi)$ as a function of $x$ is positive harmonic, moreover, all functions $P(\cdot,\xi)$ are minimal in the cone of positive harmonic functions. Now, it is a pretty general fact (by which I mean that it applies to other definitions of harmonicity as well) that the cone of positive harmonic functions is a lattice, which by Choquet's theorem implies that any such function has a unique decomposition into an integral of minimal ones.

  • 2
    $\begingroup$ I really don't understand your answer. $\endgroup$ – Alem Jan 19 '14 at 17:47
  • $\begingroup$ What concretely? $\endgroup$ – R W Jan 20 '14 at 8:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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