Extending a harmonic function satisfying a growth condition at an isolated singularity Consider a harmonic function u on the punctured disc $\Delta(\rho)^*:= \{ z\in \mathbb C: 0 <|z|<\rho\}$ with $\lim\limits_{z \rightarrow 0} z*u(z) = 0$. 
Prove that $u$ can be written in the form 
$$u(z) = \alpha \ log|z| + u_0(z)$$
with a constant $\alpha \in \mathbb C$ and a function $u_0$, which extends to a harmonic function on the disk $\Delta(\rho)$.
Note. The question has already been posted some days ago by user https://math.stackexchange.com/users/187162/cora as a problem from Ahlfors’ textbook. The question had been set on hold and later deleted by the moderator. Because I could not find a duplicate I repeat the question.
 A: Note: I answer my own question because the question results from a different user, see note in question.
According to the mean value formula for harmonic functions (Ahlfors’ textbook IV,6 Theor. 22): Setting
$$\alpha := \frac {1}{2 \pi} \int r\frac {\partial u(r,\theta)}{\partial r}d\theta$$
gives 
$$\frac {1}{2 \pi} \int u(r, \theta) d\theta=\alpha \ log|z|+ \beta$$
with a suitable constant $\beta$. The function $h := u - \alpha \ log|z|$ is harmonic for $z \in \Delta(\rho)^*$. It also satisfies $\lim\limits_{z \rightarrow 0} z*h(z) = 0$. Furthermore, 
$$0 = \frac {1}{2 \pi} \int r\frac {\partial h(r,\theta)}{\partial r}d\theta.$$
Due to the latter condition $h$ has a harmonic conjugate, i.e. $h = \Re(f)$ for a holomorphic function $f$ defined in $\Delta(\rho)^*$. Consider the Laurent decomposition $f=f_1 + f_2$ with
$$f_1(z)= \sum_{n=0}^{\infty}a_n z^n$$
holomorphic in $\Delta(\rho)$ and 
$$f_2(z)=\sum_{n=1}^{\infty}b_n \frac{1}{z^n}$$
holomorphic in $\mathbb C^*$. The substitution $w=\frac {1}{z}$ defines a a function
$$F(w):=f_2(\frac {1}{w}) = \sum_{n=1}^{\infty}b_n w^n$$
holomorphic on $\mathbb C$. For the real part derives the Laurent series
$$h= h_1 + h_2$$
with 
$$h_1(z):= \Re(f_1(z))= \sum_{n=0}^{\infty}\Re(a_n z^n)$$ 
harmonic in $\Delta(\rho)$ and
$$h_2(z):= \Re(f_2(z))= \sum_{n=1}^{\infty}\Re(b_n \frac{1}{z^n}).$$
harmonic in $\mathbb C^*$. Define 
$$H(w):= h_2(\frac {1}{w}) = \sum_{n=1}^{\infty}\Re(b_n w^n), w\in \mathbb C.$$
We have 
$$\Re(b_n w^n)= \Re(b_n)\Re(w^n) - \Im(b_n)\Im(w^n), n\geq 1.$$
For $w=x+iy$ and $n \geq 1$ the two functions $\Re((x+iy)^n)$ and $\Im((x+iy)^n)$ are homogeneous polynomials of degree n in the two variables $x,y$. Because $h$ is harmonic, the homogenous polynomials are harmonic too. Summing up,
$$H(w)= \sum_{n=1}^{\infty}p_n(x,y), w = x + iy,$$
with real-valued homogenous harmonic polynomials $p_n$ of degree $n$. From $\lim\limits_{z \rightarrow 0} z*h(z) = 0$ we conclude
$$\lim\limits_{w \rightarrow \infty} \frac {H(w)}{w} = 0.$$
We show $H=0$: For any $\epsilon > 0$ a number $R$ exists with $|\frac {H(w)}{w}| < \epsilon$ for $|w| \geq R$, i.e. $|H(w)| < \epsilon |w|$ for $|w| \geq R$. Because both functions $H(w)$ and $|w|$ are harmonic and assume minimum and maximum on the boundary of $\Delta(R)$ we conclude 
$$|H(w)| < \epsilon |w|$$
for all $w\in \mathbb C$. This holds for arbitrary $\epsilon > 0$, hence $H = 0$. 
As a consequence $h_2 = 0$, i.e. $h_1$ is a harmonic entension of h to the whole disc $\Delta(\rho)$, q.e.d. 
