If we have a function $\tilde{f}\in L^p(T)$ on the unit circle $T$ with $p \geq 1$ we can regain a harmonic function $f$ on the unit disk using the Poisson kernel $P_r$:

$$ f\left(re^{i\theta}\right)=\frac{1}{2\pi} \int_0^{2\pi} P_r(\theta-\phi) \tilde f\left(e^{i\phi}\right) \,\mathrm{d}\phi, \quad r < 1 $$

It looks somewhat similar to the Cauchy's Integral formula. The latter states that a holomorphic function defined on a disk is completely determined by its values on the boundary $\gamma$ of the disk.

$$ f(a)={\frac {1}{2\pi i}}\oint _{\gamma }{\frac {f(z)}{z-a}}\,dz $$


1) Could you please explain the logic behind the integral transformation with the Poisson kernel? Why does the kernel have the form $P_{r}(\theta )=\operatorname {Re} \left({\frac {1+re^{i\theta }}{1-re^{i\theta }}}\right),\ 0\leq r<1$ ?

2) What is the difference between the integral transformation with the Poisson kernel and the Cauchy's integral formula?

Thank you for your help in advance.


1 Answer 1


The usual Cauchy kernel is a member of a wider family: we can add an arbitrary holomorphic function. If $h$ is holomorphic on the closed disk then $$ f(a) = \frac1{2\pi i}\oint_{|z|=1} f(z) \left(\frac1{z-a}+h(z)\right) \mathrm{d}z = \frac1{2\pi}\oint_{|z|=1} f(z) \left(\frac1{1-a\bar{z}}+zh(z)\right) \frac{\mathrm{d}z}{iz}. $$ $$ = \frac1{2\pi}\oint_{|z|=1} f(z) \left(1+\frac{a\bar{z}}{1-a\bar{z}}+zh(z)\right) \frac{\mathrm{d}z}{iz}. $$

From this point we can look for a function $zh(z)$ (which must have a root at $0$) such that the modified kernel $1+\frac{a\bar{z}}{1-a\bar{z}}+zh(z)$ is real along the unit circle. (Notice that $\frac{\mathrm{d}z}{iz} = \mathrm{d}(\arg z)$ is real.) This can be done by choosing $$ zh(z) = \overline{\left(\frac{a\bar{z}}{1-a\bar{z}}\right)} = \frac{\bar{a}z}{1-\bar{a}z}, $$ so $$ h(z) = \frac{\bar{a}}{1-\bar{a}z}, $$ that is holomorphic in the disk $|z|<\frac1{|a|}$.

Hence, the modified kernel is $$ 1+\frac{a\bar{z}}{1-a\bar{z}} + \overline{\left(\frac{a\bar{z}}{1-a\bar{z}}\right)} = \mathrm{Re}\left(1+2\frac{a\bar{z}}{1-a\bar{z}}\right) = \mathrm{Re}\left(\frac{1+\bar{a}z}{1-\bar{a}z}\right). $$

Therefore, the difference between the Cauchy and the Poission kernels is a holomorphic function. :-)

  • $\begingroup$ Thank you for the reply! Could you please explain: 1) why do we need the modified kernel to be real along the unit circle? 2) how did you do this step $\frac1{2\pi}\oint_{|z|=1} f(z) \left(\frac1{1-a\bar{z}}+zh(z)\right) \frac{\mathrm{d}z}{iz}$? $\endgroup$
    – Konstantin
    Feb 24, 2017 at 16:44
  • 1
    $\begingroup$ 1) If the kernel is real then the same formula holds for the real part, so we achieve Poisson's formula. 2) First we want something purely real instead of $dz$. Then we apply $\bar{z}z=1$ along the circle. $\endgroup$
    – G.Kós
    Feb 24, 2017 at 21:57
  • $\begingroup$ @user141614 very nice answer. I am quite rusty on complex analysis, however, so I can't remember how or why the first sentence of your answer is true. Could you suggest a reference I can look up? Many thanks $\endgroup$
    – pdini
    Apr 22, 2020 at 18:40
  • $\begingroup$ @pdini this follows from Cauchy's theorem namely the fact that $\oint_{|z|=1} g(z) dz = 0$ for any $g$ holomorphic on $|z|\le 1$. If this is not clear, then please ask a new question $\endgroup$ Sep 20, 2020 at 7:41
  • 1
    $\begingroup$ Thank you @CalvinKhor. After asking this question I went back to Tristan Needham's great book Visual Complex Analysis and reminded myself why one can add an arbitrary holomorphic function -- precisely as you say. I meant to come back here and add this reference but forgot. Thanks again for your reply. $\endgroup$
    – pdini
    Sep 21, 2020 at 8:15

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