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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!

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marked as duplicate by ᴡᴏʀᴅs, martini, 5xum, William, LeGrandDODOM Aug 5 '14 at 8:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

What do you mean by "bounded on"? Do you mean that its domain is the punctured open disc and that it is bounded? Anyway, if $u$ is harmonic then its value at $0$ is uniquely specified by convolving with the Poisson kernel... – Qiaochu Yuan Aug 20 '12 at 6:14
Yes that's what I mean. And probably no one could solve it because we didn't learn about Poisson kernel in our class... – Euler....IS_ALIVE Aug 20 '12 at 6:24
I've edited the title to be more descriptive. Please check that it accurately reflects the question. – Rahul Aug 20 '12 at 6:41
@QiaochuYuan An additional argument is needed to show in this situation that the original function is the same as what you get by convolving with the Poisson kernel – Zarrax Aug 20 '12 at 13:29
@900, undoubtedly you noticed that the question you linked is newer. I can support the idea of closing one with the other (or merging if the answers are sufficiently different and worth keeping), but I am curious about your reasons for suggesting that this should be closed as a dup. Normally I vote to keep the older one (and possibly merge the newer answers). – Jyrki Lahtonen Aug 5 '14 at 7:26
up vote 6 down vote accepted

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 $\mid z\mid \lt r$ 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 $$ P(r,\theta,z)=\frac{r^2-\mid z\mid }{\mid re^{i\theta}-z\mid}^2 $$ 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.

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@GE; thanks; where the solution uses the boundedness assumption? Is the hypothesis $u$ bounded is necessary? (would you kindly explain me, I am trying to understand the solution): thanks – Inquisitive Jul 21 '15 at 16:33

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.

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dist $\mapsto$ disk $\:$ – Ricky Demer Aug 20 '12 at 7:04
I'm probably missing something, but the question was about harmonic functions, not analytic functions. – t.b. Aug 20 '12 at 7:46
@t.b. Sorry for an inadequate answer... I must be overdruknen when I wrote down this! Recently I feel my reading comprehension ability is going into a nosedive. :( – Sangchul Lee Aug 20 '12 at 8:17
No need to feel bad, I don't think it is inadequate, it deals nicely with a special case of the problem at hand. – t.b. Aug 20 '12 at 8:30
From Wikipedia: "Harmonic functions are infinitely differentiable. In fact, harmonic functions are real analytic." – celtschk Aug 20 '12 at 10:21

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