Suppose that on the punctured disk $0 < |z| < 1, \ f(z)$ is analytic with its first derivative bounded by $$ |\,f'(z)| \leq \frac{1}{|z|}. $$

Show that $f$ has a removable singularity at $z=0$.

I suspect this question may have been answered elsewhere—perhaps someone could point me to an existing answer.

I suspect there may be a clever way to use the Schwarz lemma or perhaps Cauchy's integral formula.


The hypotheses imply $f'(z)$ has at worst a pole of order $1$ at $0.$ So we can write

$$f'(z) = \frac{c}{z} +g(z)$$

for $0<|z|<1,$ where $g$ is analytic in $D(0,1).$ Now $g$ has an antiderivative $G$ in $D(0,1).$ It follows that

$$(f-G)'(z) = \frac{c}{z}, \,\, 0<|z|<1.$$

If $c\ne 0,$ we have discovered an antiderivative of $1/z$ in $\{0<|z|<1\}.$ Is that possible? Insert "no" here: ___. Thus $c=0,$ leading to the desired conclusion.

  • $\begingroup$ Or $$c = \frac{1}{2\pi i} \int_{\lvert z\rvert = 1/2} f'(z)\,dz = 0\,.$$ $\endgroup$ – Daniel Fischer Jun 19 '18 at 20:30
  • $\begingroup$ @MarkViola Fundamental theorem of calculus, essentially. The integral of a derivative over a closed curve is always $0$. $\endgroup$ – Daniel Fischer Jun 19 '18 at 21:50
  • $\begingroup$ @MarkViola But $\frac{1}{z}$ isn't a derivative in any domain that has a loop winding about $0$. $\endgroup$ – Daniel Fischer Jun 20 '18 at 11:08
  • $\begingroup$ @DanielFischer Yes, I understand that $\log(z)$ is not analytic on $\mathbb{C}$. I was wondering where you begin . The problem statement is that $f$ is analytic on the punctured disk and that $|f'(z)|\le \frac{1}{|z|}$. So, did you begin with $f(z)=\frac cz+g(z)$ for some analytic function $g$? $\endgroup$ – Mark Viola Jun 20 '18 at 18:04
  • $\begingroup$ @MarkViola We begin with $f$, which by assumption is holomorphic on the punctured unit disk. Hence $\int_{\lvert z\rvert = 1/2} f'(z)\,dz = 0$. Next we use the assumption about $f'$, which gives the decomposition in the answer. And then we use $\int_{\lvert z\rvert = 1/2} \frac{c}{z} + g(z)\,dz = 2\pi i c$. $\endgroup$ – Daniel Fischer Jun 20 '18 at 18:48

Let us consider the Laurent series of $f(z)$ and $f'(z)$ in the punctured disk:

$$ f(z) = \sum_{n\in\mathbb{Z}} a_n z^n,\qquad f'(z)=\sum_{n\in\mathbb{Z}}na_n z^{n-1}. $$ If we assume that some $a_n$ with $n<0$ is $\neq 0$, we have a violation of the inequality $\left|f'(z)\right|\leq \frac{1}{|z|}$ for some $z$ sufficiently close to the origin, since in such a case $z f'(z)$ has a singularity at the origin, which is either a pole or an essential singularity. In both cases $zf'(z)$ cannot stay bounded as $z\to 0$. It follows that the only non-zero coefficients are the ones with $n\geq 0$ and $f(z)$ is holomorphic over $|z|<1$.


Set $\,g(z)=z\,f'(z)$. Clearly, $$ |g(z)|=|z\,f'(z)|\le 1, $$ and hence $g$ is bounded in the punctured unit disc $\mathbb D\setminus\{0\}$, and hence, $g$ has removable singularity at $z=0$, and consequently, $g$ extends analytically in the unit disc.

Expand $g$ in the unit disc as $g(z)=\sum_{n=0}^\infty a_nz^n$. Then $$ f'(z)=\frac{a_0}{z}+\sum_{n=0}^\infty a_{n+1}z^n. $$ Let $\gamma\subset\mathbb D\setminus\{0\}$ be a closed curve. Then $$ 0=\int_{\gamma}\big(z\,f(z)\big)'\,dz=\int_{\partial B_r}\big(\,f(z)+z\,f'(z)\big)\,dz=\int_{\partial B_r}\big(\,f(z)+g(z)\big)\,dz=\int_{\partial B_r}f(z)\,dz. $$ Hence, there exist an analytic function $F$ in $D\setminus\{0\}$, such that $F'=f$. Such an $F$ has a Laurent expansion of the form $$ F(z)=\sum_{k\in \mathbb Z}b_kz^k $$ and hence $$ f'(z)=F''(z)=\sum_{k\in \mathbb Z}k(k-1)b_kz^{k-2}. $$ This implies that the coefficient of $z^{-1}$ is the expansion of $f'$ is $0$ and hence $a_0=0$, and thus $f$ is analytic in the disc.


Note, applying L'Hopital's rule:

$$\lim_{z\to 0} zf(z)=\lim_{z\to0} \frac{f(z)}{1/z}=\lim_{z\to 0} \frac{f'(z)}{-1/z^2}=\lim_{z\to 0} -z^2f'(z)$$

Note that

$$\lim_{z\to 0} |-z^2f'(z)|\le \lim_{z\to0}|z|=0$$

  • $\begingroup$ You need not assume that $f\to \infty$ to apply LHR. $\endgroup$ – Mark Viola Jun 19 '18 at 19:14
  • $\begingroup$ @MarkViola Wait, really? What do you mean? $\endgroup$ – user223391 Jun 19 '18 at 19:15
  • 2
    $\begingroup$ LHR does not require that the numerator approach $\infty$ (in fact, the limit of the numerator need not even exist) when all of the smoothness conditions are met and the limit of the denominator is $\infty$. $\endgroup$ – Mark Viola Jun 19 '18 at 20:14
  • $\begingroup$ @MarkViola Wow very cool, thanks for that! It makes my answer better. $\endgroup$ – user223391 Jun 19 '18 at 20:53

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