Is the boundedness necessary to extend harmonically? "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]
 A: 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 $U \equiv u$.
A: 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.
A: If you assume $u(z)/\ln|z|\to 0$ as $z\to 0,$ then $0$ is a removable singularity for $u.$
