Prove that $0$ is not an essential singularity (UW Madison Qualifying exam)

Since $f$ is holomorphic on the punctured unit disk $\mathbb{D}^*$, it admits the following Laurent expansion

$$ f(z) = \sum_{n\in\mathbb{Z}} a_n z^n = g(z) + a_0 + h(z), $$

where $a_n \in \mathbb{R}$ for all $n \in \mathbb{Z}$ and

$$ g(z) = \sum_{n > 0} a_n z^n \qquad \text{and} \qquad h(z) = \sum_{n < 0} a_n z^n. $$

Note that $g(z)$ converges on all of the unit disk $\mathbb{D}$, whereas $h(1/z)$ defines an entire function. Now assume that $0 < r < 1$ and $r < |z| < 1$. By using the relation $\operatorname{Im}\{g(z)\} = -\operatorname{Im}\{g(\overline{z})\}$,

\begin{align*} &\int_{|\xi|=r} i \operatorname{Im}\{f(\xi)\} \biggl( \frac{z+\xi}{z-\xi} \biggr) \, \frac{|\mathrm{d}\xi|}{2\pi r} \\ &= \int_{|\xi|=r} i \operatorname{Im}\{h(\xi) - g(r^2/\xi)\} \biggl( \frac{z+\xi}{z-\xi} \biggr) \, \frac{|\mathrm{d}\xi|}{2\pi r} \\ &= \int_{|\zeta|=r} i \operatorname{Im}\{h(r^2/\zeta) - g(\zeta)\} \biggl( \frac{\zeta+r^2/z}{\zeta-r^2/z} \biggr) \, \frac{|\mathrm{d}\zeta|}{2\pi r} \tag{$\zeta=r^2/\xi$} \\ &= h(z) - g(r^2/z). \end{align*}

Here, the last step is a consequence of the Schwarz integral formula (of OP's version) applied to the holomorphic function $h(r^2/z) - g(z)$ on $\mathbb{D}$ and $|r^2/z| < r$. Then by substituting $\xi = re^{i\theta}$,

\begin{align*} h(z) - g(r^2/z) &= \int_{-\pi}^{\pi} i \operatorname{Im}\{f(re^{i\theta})\} \biggl( \frac{z+re^{i\theta}}{z-re^{i\theta}} \biggr) \, \frac{\mathrm{d}\theta}{2\pi} \\ &= \int_{0}^{\pi} i \operatorname{Im}\{f(re^{i\theta})\} \biggl( \frac{z+re^{i\theta}}{z-re^{i\theta}} - \frac{z+re^{-i\theta}}{z-re^{-i\theta}} \biggr) \, \frac{\mathrm{d}\theta}{2\pi} \\ &= -\frac{2}{\pi} \int_{0}^{\pi} \frac{z}{r^2 + z^2 - 2rz \cos\theta} \, \varphi_r(\theta) \, \mathrm{d}\theta, \tag{*} \end{align*}

where $\varphi_r(\theta)$ is defined by

$$ \varphi_r(\theta) = r \operatorname{Im}\{f(re^{i\theta})\} \sin \theta $$

and the relation $\operatorname{Im}\{f(\overline{z})\} = -\operatorname{Im}\{f(z)\}$ is utilized in the second step. Now define

$$ C_r = \frac{2}{\pi} \int_{0}^{\pi} \varphi_r(\theta) \, \mathrm{d}\theta.$$

The assumption tells that $\varphi_r$ is non-negative, and so, $C_r \geq 0$. Moreover, $\text{(*)}$ applied to $z = R$ with a fixed $R > 0$ and $0 < r < R$ shows that

\begin{align*} C_r &= \frac{2(R + r)^2}{\pi R} \int_{0}^{\pi} \frac{R}{(R+r)^2} \, \varphi_r(\theta) \, \mathrm{d}\theta \\ &\leq \frac{2(R + r)^2}{\pi R} \int_{0}^{\pi} \frac{R}{r^2 + R^2 - 2rR \cos\theta} \, \varphi_r(\theta) \, \mathrm{d}\theta \\ &= \frac{(R + r)^2}{R} \left| h(R) - g(r^2/R) \right| \end{align*}

and so, $C_r$ is bounded as $r \to 0^+$. Finally,

\begin{align*} \left| z h(z) \right| &\leq \left| z g(r^2/z) \right| + \left| z ( h(z) - g(r^2/z)) \right| \\ &\leq \left| z g(r^2/z) \right| + \frac{2}{\pi} \int_{0}^{\pi} \frac{\left| z \right|^2}{\left|z\right|^2 - r^2 - 2r\left|z\right|} \, \varphi_r(\theta) \, \mathrm{d}\theta \\ &= \left| z g(r^2/z) \right| + \frac{\left| z \right|^2}{\left|z\right|^2 - r^2 - 2r\left|z\right|} C_r, \end{align*}

and so, taking limit superior as $r \to 0^+$ gives

$$ \left| z h(z) \right| \leq \limsup_{r\to 0^+} C_r < \infty. $$

Now this inequality holds for any $0 < |z| < 1$, and so, $zh(z)$ has a removable singularity at $0$ and therefore $f(z)$ cannot have an essential singularity at $0$.

This problem (the part with simple pole or analytic) has an elementary solution since first $f'(x) \ne 0$ if $x \in (-1,0) \cup (0,1)$ by the local form of an analytic function (if $f'(r)=0, a_n, n \ge 2$ is the first non-zero coefficient at $r, f(z)=f(r)+a_n(z-r)^n+O((z-r)^{n+1}), a_n \in \mathbb R$ and $(z-r)^n$ maps each half circle centered at $r$ on the full circle, which means that $f$ does so near $r$ and that contradicts the hypothesis on $\Im f$)

But then $f$ is monotonic on $(-1,0)$ and on $(0,1)$ which means that $f$ has limits at zero from both left and right (possibly $\pm \infty$ of course) so for $r>0$ small, $\mathbb R -(f(-r,0) \cup f(0,r))$ contains a non-degenerate interval $I$. But this means $\mathbb C - f(D(0,r)^*)$ contains $I$ too and that shows that $0$ cannot be an essential singularity (directly without Picard by mapping $\mathbb C-I$ into the unit disc with a meromorphic $g$ etc)

But then if $0$ is removable, $f'(0) \ne 0$ by the same argument as for $f'(r)$ above, while if $0$ is a pole, the local form of it shows again it must be simple (same reason that $1/z^n, n \ge 2$ scrambles imaginary parts) with negative residue.

Note that $f$ may have zeroes on the real axis (at most two of course by monotonicity) so one cannot apply directly the removable singularity result to $-1/f$ though indeed one can restrict to a neighborhood of zero where $f$ is non-zero and do it there since of course, the same arguments (showing that it has a simple pole or a removable singularity) apply if $f$ is typically real on a small punctured disc only