Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 24F (Riemann Surfaces)

Suppose first that $u$ is harmonic. Sincethe Cauchy-Riemann equations show thatis holomorphic on $D$. An open disc is a simply connected domain, so $g$ has a holomorphic primitive $f$. If $f=p+iq$, thenThus $p-u$ has zero gradient and is constant on the connected disc. Subtracting that real constant from $f$ gives

Conversely, if $u=\mathrm{Re}f$ for a holomorphic $f=u+iv$, differentiating the Cauchy-Riemann equations givesHence $u$ is harmonic. This proves the harmonic function as the real part of a holomorphic function criterion on $D$.

Take the punctured complex planeandIn polar coordinates, $u(r,\theta )=\log r$, soThus $u$ is harmonic on $D$.

If $u=\mathrm{Re}f$ for a holomorphic function $f$, the calculation from part (a) would giveThe integral of a derivative around a closed curve is zero, whereas the residue theorem givesThis contradiction proves that no such $f$ exists. It is precisely the log modulus has no global harmonic conjugate on the punctured plane obstruction.

Lift $u$ through the complex exponential by settingThe conformal invariance of harmonicity makes $v$ harmonic. Since $e^{w+2\pi i}=e^w$ and $u(2z)=u(z)$,Thus $v$ is bounded on the compact rectangleand its two periods make it bounded on the whole plane. The harmonic Liouville theorem makes $v$ constant. Since the exponential map is surjective onto ${\mathbb{C}}^{\ast }$, $u$ is constant.

For the requested counterexample after deleting a countable set, takeand let ${\wp }_{\Lambda }$ be the Weierstrass elliptic function for the latticeDefineChanging a branch of the logarithm adds $2\pi i$, a period of ${\wp }_{\Lambda }$, so $u$ is well defined. Its poles project precisely to $S$, and the conformal invariance of harmonicity makes $u$ harmonic away from them. The other period givesMoreover, $z\notin S$ implies $2z\notin S$. Finally, $u$ is nonconstant because as positive real $z\to 1$,which is real and unbounded. This is a scale-periodic harmonic function from an elliptic function.

Since $\mathbb{C}$ is simply connected, writefor an entire function $f$. If $u$ were bounded above, thenwould be bounded, so the Liouville theorem would make $e^f$ constant. Its derivative $f^{\prime }{e}^f$ would then vanish, whence $f$ and $u$ would be constant. Applying the same argument to $e^{-f}$ rules out a lower bound. Thus a nonconstant $u$ is unbounded both above and below.

Given any $c\in \mathbb{R}$, choose $z_{-},z_{+}\in \mathbb{C}$ withThe restriction of $u$ to the line segment from $z_{-}$ to $z_{+}$ is continuous, so the intermediate value theorem supplies a point on that segment where $u=c$. ThereforeThis proves the surjectivity of a nonconstant entire harmonic function.