Can any harmonic function on $\{z:0<|z|<1\}$ be extended to $z=0$?

Question

After looking at this question for quite some time, I've asked a couple of other students, and they also couldn't seem to come up with an answer. This is from an old qualifying exam at our university.

Let $u$ be a harmonic function bounded on the set $\{z:0 < |z| < 1\}$. Can it always be defined at the point $ z= 0$ to become harmonic on the whole unit disk?

The standard argument with the logarithm doesn't work here, as it must be bounded on the punctured disk, and the logarithm blows up at $0$. Also, we can't use any kind of harmonic conjugate argument because our domain is not simply connected. Thus, I think it's probably true, but I haven't been able to come up with a proof.

Thanks in advance for any help!

Answer 1

Yes, you can extend $u$ to a harmonic function $U$ defined on the whole disk $D=\{z\in \mathbb C:0 < |z| < 1\}$.

Take a circle of radius $0\lt r\lt 1$. The key to the problem is the remark that if the harmonic extension $U$ exists, it will be given for $|z|\lt r$ by the Poisson formula $$U(z) =\frac{1}{2\pi}\int_0^{2\pi} u(re^{i\theta})P(r,\theta,z) d\theta$$ where $$ P(r,\theta,z)=\frac{r^2-|z|^2}{|re^{i\theta}-z|}$$ is the Poisson kernel.
Conversely, $U$ defined by the formula above is harmonic on $\lbrace z\in \mathbb C:0 < |z| < r\rbrace$ : this is known as the solution to Dirichlet's problem by Poisson's integral formula.
That $U$ coincides with $u$ on $\lbrace z\in \mathbb C:0 < |z| < r\rbrace$ and thus harmonically extends $u$ through $0$ necessitates a little approximation argument that you will find on page 33 of this freely available book on harmonic functions.

Answer 2

Let $M = \sup_{0<|z|<1} |u(z)|$. By the hypothesis, $M$ is finite.

Now note that if $z \neq 0$, then for all $0 < r < |z|$ and $0 < \epsilon < \min\left(r, \frac{1}{2}|z|\right)$, we have

$$\left|\oint_{|\zeta|=r}\frac{u(\zeta)}{\zeta-z}\;d\zeta\right| = \left|\oint_{|\zeta|=\epsilon}\frac{u(\zeta)}{\zeta-z}\;d\zeta\right| \leq \oint_{|\zeta|=\epsilon}\left|\frac{u(\zeta)}{\zeta-z}\right|\;\left|d\zeta\right| \leq \oint_{|\zeta|=\epsilon}\frac{M}{|z|/2}\;\left|d\zeta\right| = \frac{4\pi M}{|z|}\epsilon,$$

which implies, by taking $\epsilon \to 0$, that

$$ \oint_{|\zeta|=r}\frac{u(\zeta)}{\zeta-z}\;d\zeta = 0. $$

Thus if we fix $0 < R < 1$ and define

$$\tilde{u}(z) = \frac{1}{2\pi i}\oint_{|\zeta|=R}\frac{u(\zeta)}{\zeta-z}\;d\zeta$$

for $|z| < R$, then by a simple application of Cauchy integration formula, for all $0 < |z| < R$ we have

$$ \begin{align*} \tilde{u}(z) = \frac{1}{2\pi i}\oint_{|\zeta|=R}\frac{u(\zeta)}{\zeta-z}\;d\zeta &= \frac{1}{2\pi i}\oint_{\partial (B_R \setminus B_{|z|/3})}\frac{u(\zeta)}{\zeta-z}\;d\zeta + \frac{1}{2\pi i}\oint_{|\zeta|=|z|/3}\frac{u(\zeta)}{\zeta-z}\;d\zeta \\ &= u(z) + 0 = u(z). \end{align*}$$

Now since $\tilde{u}(z)$ is analytic on $|z| < R$, by gluing $u$ and $\tilde{u}$ we obtain an analytic function on the unit disk which coincides with $u$ on the punctuated unit disk.