Krdytkyu

Let $f(z) = log|F(z)|$, where $F: mathbb{C} rightarrow mathbb{C}$ is entire. Then $f$ defines a distribution on $mathbb{R}^2$, and we want to show that its distributional Laplacian is

$$Delta f = sum_{n in I} 2 pi , d_n , delta_{z_n}$$

where ${z_n mid n in I}$ are the zeros of $F$, which have respective degree (multiplicity) $d_n$.

This is what I have got so far (though I suspect that there could be a completely different approach which is easier, so solutions that don't use this work are more than welcome!):

Let $U_varepsilon = mathbb{R}^2 backslash left( bigcuplimits_{n in I} B(z_n , varepsilon) right)$.

Then, for a test function $varphi in D(mathbb{R}^2)$,

$$leftlangle f,Deltavarphirightrangle=lim_{varepsilonrightarrow0^+}int_{U_varepsilon} fDeltavarphi dx,$$
and by Green's second identity
$$int_{U_varepsilon} fDeltavarphi dx=int_{U_varepsilon} varphiDelta f dx+int_{partial U_varepsilon} ffrac{partialvarphi}{partial n}ds-int_{partial U_varepsilon}varphifrac{partial f}{partial n} ds.$$

The first integral on the RHS vanishes since $f = Re(log(F))$, so is harmonic on $U_varepsilon$ where $log(F)$ is holomorphic.

We can obviously split $partial U_varepsilon$ into circles around each singularity of $f$.

This is where I get stuck. How can we deal with the terms $ffrac{partialvarphi}{partial n}$ and $frac{partial f}{partial n}$.

Any suggestions would be appreciated!

About the term $int_{partial U_varepsilon} ffrac{partialvarphi}{partial n}ds$, note that $partial U_epsilon = bigcup_n partial B(z_n,epsilon)$. Fix $n$ and observe that
$$
F(z) = (z-z_n)^d G(z)
$$ where $G(z)$ does not vanish near $z=z_n$. This implies
$$
f(z) = dlog |z-z_n| + log|G(z)|=dlog epsilon + log|G(z)|
$$ on $|z-z_n|=epsilon$. Note that $log |G(z)|$ is a continuous harmonic function and hence $|log |G(z)||leq C$ for some $C>0$. Now, we see that
$$
int_{|z-z_n|=epsilon} ffrac{partialvarphi}{partial n}ds =dlogepsilon int_{|z-z_n|=epsilon} langle nabla varphi,nrangle ds +int_{|z-z_n|=epsilon} log|G(z)|langle nabla varphi,nrangle ds.
$$ From
$$
dBig|logepsilon int_{|z-z_n|=epsilon} langle nabla varphi,nrangle dsBig| = d logepsilon^{-1} int_{|z-z_n|=epsilon} |nabla varphi| dsleq d logepsilon^{-1} cdot 2piepsilon cdotsup|nabla varphi|to 0,
$$ and
$$
Big|int_{|z-z_n|=epsilon} log|G(z)|langle nabla varphi,nrangle dsBig|leq 2pi Cepsilon sup|nablavarphi|to 0,
$$ we get $$
lim_{epsilonto 0}int_{|z-z_n|=epsilon} ffrac{partialvarphi}{partial n}ds =0.
$$ This holds for all $n$, and we conclude $lim_{epsilonto 0}int_{partial U_epsilon} ffrac{partialvarphi}{partial n}ds=0.$

Next, note that $$frac{partial f(z_n + re^{itheta})}{partial n} = frac{partial}{partial r} f(z_n + re^{itheta})= frac{partial}{partial r}[dlog r + log |G(z_n+re^{itheta})|] = frac{d}{r}+partial_rlog |G(z_n+re^{itheta})|. $$ This is because the unit outward normal vector $n$ of the circle $|z-z_n|=r$ at $z = z_n +re^{itheta}$ is just $e^{itheta}$. Now observe that
$$
int_{|z-z_n|=epsilon} varphifrac{partial f}{partial n}ds = frac{d}{epsilon}int_{|z-z_n|=epsilon} varphi ds + int_{|z-z_n|=epsilon} varphi left(partial_r|_{r=epsilon}log |G(z_n +re^{itheta})|right)ds .
$$ The first term is
$$
frac{d}{epsilon}int_{|z-z_n|=epsilon} varphi ds= frac{d}{epsilon}int_0^{2pi} varphi(z_n + epsilon e^{itheta} )epsilon dtheta =dint_0^{2pi} varphi(z_n + epsilon e^{itheta} )dtheta to 2pi d varphi(z_n).
$$ The integrand in the second term is a continuous function and thus
$$
Big|int_{|z-z_n|=epsilon} varphi left(partial_r|_{r=epsilon}log |G(z_n +re^{itheta})|right)dsBig| leq sup |varphi|cdot sup|partial_r log |G||cdot 2piepsilon to 0.
$$ Summing up over all $n$, we see that
$$
lim_{epsilonto 0}int_{U_varepsilon} fDeltavarphi dx = sum_n 2pi d_n varphi(z_n),
$$ as desired.