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"If $u$ is harmonic and bounded in the punctured disk $0<|z|< \rho$, then $u$ can be extended harmonically to the disk $|z|<\rho$ harmonically." This fact has been shown here.

My Question is: Is the hypothesis bounded necessary in the above fact? Or can find counter example?

[In the answer given by Georges Elencwajg, I am unable see where he is using bounded assumption]

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Inquisitive
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    The function need not be bounded on the whole punctured disk, but it must be bounded on a smaller punctured disk $0 < \lvert z\rvert \leqslant r < \rho$ by continuity. – Daniel Fischer Jul 24 '15 at 16:14
  • @DF; thanks; Is this known as theorem of Scwartz? (Every continuous function on $|z|=r$ can be extended harmonically into $|z|<r$) – Inquisitive Jul 24 '15 at 16:25

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If a harmonic extension to the disk exists, then that extension is in particular continuous, and hence bounded on all disks $\{ z : \lvert z\rvert \leqslant r\}$ for $0 < r < \rho$, so it is a necessary condition that $u$ be bounded on some punctured disk $\{ z : 0 < \lvert z\rvert \leqslant r\}$, but it need not be bounded on the whole punctured disk $\{ z : 0 < \lvert z\rvert < \rho\}$, $u$ can be unbounded as $z$ approaches the boundary circle $\lvert z\rvert = \rho$.

In the linked answer, the boundedness is used in the "little approximation argument that you will find on page 33" that shows that $U$ coincides with $u$, I suppose. Without the boundedness, you cannot show that. I haven't looked at the linked book, but the usual approximation argument is as follows:

$v := U - u$ is a bounded harmonic function on $\{ z : 0 < \lvert z\rvert < r\}$, and on the circle $\lvert z\rvert = r$, it has boundary values $0$. Now let $\varepsilon > 0$ and consider

$$v_\varepsilon(z) = v(z) + \varepsilon \cdot \log \frac{\lvert z\rvert}{r}.$$

Then $v_\varepsilon$ is a harmonic function on $0 < \lvert z\rvert < r$, has boundary values $0$ on the circle $\lvert z\rvert = r$, and, since $v$ is bounded,

$$\lim_{z\to 0} v_\varepsilon(z) = -\infty.$$

For all sufficiently small $s > 0$, $v_\varepsilon$ is therefore strictly negative on the circle $\lvert z\rvert = s$, and by the maximum principle for harmonic functions, it follows that $v_\varepsilon(z) < 0$ for all $z$ in the annulus $s < \lvert z\rvert < r$. Letting $s \to 0$, we see that $v_\varepsilon < 0$ on the punctured disk $0 < \lvert z\rvert < r$.

This holds for all $\varepsilon > 0$, and now letting $\varepsilon \to 0$ it follows that $v(z) \leqslant 0$ for $0 < \lvert z\rvert < r$.

The same argument for $w_\varepsilon(z) = v(z) - \varepsilon\cdot \log \frac{\lvert z\rvert}{r}$ shows $v \geqslant 0$ on $0 < \lvert z\rvert < r$, so together we have $v \equiv 0$, or $U \equiv u$.

Daniel Fischer
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    So a priori, you don't need to know $v$ is bounded. As long as $v(z) \leq o(\log |z|)$, it will be still true that $v_\epsilon(z)$ will go to $-\infty$. It is the analysis that forced $v$ to be bounded. That means, the statement can be weakened as: "if $u$ is harmonic in the punctured disk and $u(z) \leq o(\log |z|)$, then $u$ can be extended harmonically at the origin." – henryforever14 Jul 24 '15 at 17:01
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    True, just like an isolated singularity of a holomorphic function where $\lvert f(z)\rvert \in o(\lvert z-z_0\rvert^{-1})$ is in fact a removable singularity. – Daniel Fischer Jul 24 '15 at 17:03
  • Why are the conditions different between harmonic functions and holomorphic function? – henryforever14 Jul 24 '15 at 17:11
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    @henryforever14 Not sure how to answer "why". But $\log \lvert z\rvert$ is harmonic on $\mathbb{C}\setminus {0}$, so harmonic functions can have logarithmic isolated singularities, while holomorphic functions can't, since $0$ is a branch-point of $\log z$, not an isolated singularity. – Daniel Fischer Jul 24 '15 at 17:16
  • @DF; wow, thanks a lot; would you please tell me: If $v_{\epsilon}$ is negative on circle, then why does it follows that it is negative on annulus. (How you have used the maximum principle); thanks – Inquisitive Jul 24 '15 at 17:27
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    @Inquisitive $v_\varepsilon$ is negative on the inner boundary circle of the annulus, and it is $0$ on the outer boundary circle. If there were a point $z_0$ with $s < \lvert z_0\rvert < r$ and $v_\varepsilon(z_0) \geqslant 0$, then $v_\varepsilon$ would have a maximum (global, and hence also local) in the interior of the annulus, hence be constant. But $v_\varepsilon$ is not constant, since the boundary values on the two boundary circles are different. – Daniel Fischer Jul 24 '15 at 17:33
  • @DF; thanks for everything; – Inquisitive Jul 24 '15 at 18:02
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If it goes to $\infty$ at the origin, then you can not even extend it by continuity. So it has to be bounded in a punctured neighborhood of the origin.

Svetoslav
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If you assume $u(z)/\ln|z|\to 0$ as $z\to 0,$ then $0$ is a removable singularity for $u.$

zhw.
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  • Interesting downvote. The OP asked "Is the hypothesis bounded necessary in the above fact?" My answer shows that you do not have to assume $u$ is bounded to see that $u$ extends harmonically to the whole disc, and this answer contains the "bounded" version. – zhw. Jul 24 '15 at 19:12
  • Maybe we disagree about the meaning of "Is the hypothesis bounded necessary". Suppose $u$ is analytic in a punctured disk about the origin, and $u$ has a removable singularity at $0$. Then $u$ is bounded in a neighborhood of the origin. The first Comment on the Question and the other Answers are stating this. – hardmath Jul 25 '15 at 01:18