I'm trying to prove that for any harmonic function $u$, we have :

let $ \Omega \subset \mathbb{R}^2$ and $ \overline B(0,R) \subset \Omega $

$$ u \colon \Omega \to \mathbb R $$

$$\forall z \in B(0,R) : u(z) = \Re \ \ \frac 1 {2 \pi i}\int_{ |\zeta| = R } \frac{ \zeta+z}{ \zeta ( \zeta -z)} u( \zeta ) d\zeta $$

I've tried a few things:

I've shown that $$ \int_{|\zeta| = R } \frac{ \zeta+z}{ \zeta ( \zeta -z)} u( \zeta ) d\zeta = 2 \int_{|\zeta| = R } \frac{ u( \zeta ) }{ ( \zeta -z)} d\zeta - \int_{|\zeta| = R } \frac{ u( \zeta ) }{ \zeta} d\zeta $$

Which ( I'm not sure about this point... but in complex analysis I think it would have made sense ) is proportional to

$$ 2 \operatorname{Res}_z(u) - \operatorname{Res}_0(u) $$

But I don't think I'm going anywhere ...

So I started again and I studied the function :

$$L_f(z) = \frac 1 {2 \pi i}\int_{ \mathbb{T} } \frac{ \zeta+z}{ \zeta ( \zeta -z)} f( \zeta ) d\zeta$$

I would still need to prove that this is holomorphic, but the derivative is given by :

$$L'_f(z) = \frac 1 { \pi i}\int_{ \mathbb{T} } \frac{ f( \zeta ) }{ ( \zeta -z)^2 }d\zeta$$

I was very surprised because the RHS is exactly the expression of $ 2 f'(z) $ according to Cauchy Integral Formula. So I was believing that $L_f \equiv f$. But in order to prove the equality, I would still need to prove that :

$$- \int_{ \mathbb{T} } \frac{ f( \zeta ) }{ \zeta} d\zeta = \frac 1 { \pi i}\int_{ \mathbb{T} } \frac{ f( \zeta ) }{ \zeta -z }d\zeta $$

So could you please tell me if my reasoning is true/ going somewhere, if I'm allowed to talk about residues for harmonic functions... Or if you have another solution for the main problem, I would also be very grateful to read it :)

  • 1
    $\begingroup$ Where did you encounter this claim? I could very well be wrong but it doesn't seem true to me. The reason I am skeptical is that on a simply connected domain, for $u$ harmonic you may always find $v$ harmonic such that $u+iv$ is analytic. Using this fact and the Cauchy integral formula, this seems to imply that $u(z)=2u(z)-u(0)$ which is definitely not true in general. $\endgroup$
    – user293794
    Dec 16, 2018 at 23:48
  • $\begingroup$ You are saying u(z) =2u(z)-u(0) because of my formula with the Residues? Is this what I wanted to say? $\endgroup$ Dec 16, 2018 at 23:50
  • 1
    $\begingroup$ I'm saying because if $f=u+iv$ with $v$ as in my previous comment then $\frac{1}{2\pi i}\int_{|\zeta=1|}\frac{\zeta+z}{\zeta(\zeta-z)}f(\zeta)\,d\zeta=2f(z)-f(0)$ by the decomposition you have above and the Cauchy integral formula. $\endgroup$
    – user293794
    Dec 16, 2018 at 23:53

1 Answer 1


This is basically Poisson's formula for harmonic functions in the unit disk, namely $$ u(z) = \frac{1}{2\pi} \int_0^{2\pi} u(e^{i\theta}) \frac{1-|z|^2}{|e^{i\theta}-z|^2} \, d\theta = \Re \left[\frac{1}{2\pi} \int_0^{2\pi} u(e^{i\theta}) \, \frac{e^{i\theta} +z}{e^{i\theta}-z} \, d\theta \right]. $$ Rewriting the integral in complex form, with $\zeta = e^{i\theta}$, we get that $d\theta = \frac{dz}{iz}$, so $$ u(z) = \Re \left[\frac{1}{2\pi} \int_0^{2\pi} u(\zeta) \, \frac{\zeta +z}{\zeta-z} \, \frac{d\zeta}{i\zeta} \right] = \Re \left[\frac{1}{2\pi i} \int_0^{2\pi} u(\zeta) \, \frac{\zeta +z}{\zeta (\zeta-z)} \, d\zeta \right]. $$


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