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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.

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  • $\begingroup$ 1 means the function is holomorphic on an open neighborhood of the closed set, so on a slightly larger disc here by compacity; 2 is correct, the apparent contradiction coming precisely from the fact that such $f$ doesn't exist. Note that the problem holds if we work on a small open disc around $0$ only as it contains a smaller closed disc, so the original hypothesis holds by a homothety $\endgroup$ – Conrad Aug 10 at 23:10
  • $\begingroup$ Sorry, I don't understand why 2 is correct. How do I prove that $v$ is a harmonic conjugate of $u$? $v$ is harmonic everywhere but $0$, so how do I get around $0$? $\endgroup$ – Saad Aug 10 at 23:16
  • $\begingroup$ We know $g=u+iv'$ is holomorphic on the disc and $g-f=c$ except possibly at zero. It follows $f$ is bounded near zero so it has a removable singularity (same thing is true for harmonic functions though they can have logarithmic singularities too, not only singularities coming from poles but if they are bounded singularity is removable, but proof is harder and here we can use the easy proof for holomorphic functions) $\endgroup$ – Conrad Aug 10 at 23:24
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    $\begingroup$ In the above ending paragraph all singularities are assumed isolated of course, the point being that harmonic functions have extra kinds of isolated singularities as opposed to holomorphic ones but they are still unbounded there $\endgroup$ – Conrad Aug 10 at 23:30
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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 $0$.

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  • $\begingroup$ Why does $f = c + g$ imply that $f$ has a removable singularity at $0$? Also, did you mean to say the real part of $f - g$? $\endgroup$ – Saad Aug 10 at 23:38
  • $\begingroup$ Is it because $g$ is holomorphic at $0$, so it is bounded near $0$ and thus $f$ is also bounded near $0$? $\endgroup$ – Saad Aug 10 at 23:39
  • $\begingroup$ $g$ is analytic, so it is bounded in some neighborhood of $0$. So $f$ is bounded in some punctured disk around $0$. This implies that it has a removable singularity. $\endgroup$ – Kabo Murphy Aug 10 at 23:41
  • $\begingroup$ Sorry about the typo. I did mean real part of $f-g$. $\endgroup$ – Kabo Murphy Aug 10 at 23:42

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