Does such a function $f$ exist?
I think the answer is no.
My attempt:
Take a harmonic conjugate of $u$, say $v'$ so that $g = u + iv'$ is holomorphic on the closed unit disk (not sure what this means). Then $v$ and $v'$ differ by a constant since they are harmonic conjugates of $u$, so $f$ and $g$ differ by a constant and thus $f$ is in fact holomorphic at 0. Therefore, no such $f$ can exist.
Questions:
1) What does it mean for a function to be holomorphic on a closed set? Does it just mean it is holomorphic on an open set containing this closed set?
2) Is the last step correct where I conclude that $v$ and $v'$ differ by a constant? In particular, I don't see how this can hold at 0 since $v(0)$ is not defined (as $f(0)$ is a singularity). I have a feeling this is true, but I don't know how to justify it.
Best Answer
Unfortunately the hypothesis confuses you by by stating something that is not necessary for the proof. (Closed disk is unnecessary). There is no analytic function on the punctured open disk such that $u=\Re f$ is harmonic at $0$ and $f$ has a non-removable singularity at $0$. The reason is $u$ is harmonic on the unit disk which is simply connected, so there is a harmonic function $w$ such that $g=u+iw$ is analytic on the unit disk. Now the real part of $f-g$ is $0$ on the punctured disk which is connected so (using C-R equations) we can show that $f-g$ is a constant $c$. Thus $f=c+g$ on the punctured disk which implies that $f$ has a removable singularity at $0$.