Let $\mathcal{F}$ be a family of holomorphic functions on $\mathbb{D}$ so that for any $f\in\mathcal{F}$, $$|f'(z)|\left(1-|z|^2\right)+|f(0)|\leq 1,$$ for all $z\in\mathbb{D}$. Prove that $\mathcal{F}$ is a normal family on $\mathbb{D}$.

Attempt: Let $0<R<1$ and consider the open disk centered at $0$ of radius $1-R$, $D_{1-R}(0)$. For $z\in \overline{D_{1-R}(0)}$ we have by assumption $$|f'(z)|\left(1-|z|^2\right)+|f(0)|\leq 1$$ $$\implies |f'(z)|\leq\frac{1-|f(0)|}{1-|z|^2}.$$

Thus $$|f(z)|=\left|\int f'(z)\ dz\right|\leq \int|f'(z)|\ dz\leq \int \frac{1-|f(0)|}{1-|z|^2}\ dz.$$

My question is how to bound the last integral to conclude that $\mathcal{F}$ is uniformly bounded on compact subsets of $\mathbb{D}$

  • $\begingroup$ Where are you integrating . $f(z)=\int f'(z)dz$ is not correct. $\endgroup$ – Kavi Rama Murthy Jan 12 at 5:12
  • $\begingroup$ You're right. We are integrating over the entire closed disk, not just the boundary. $\endgroup$ – Sham Jan 12 at 5:13

Let $K=\{z:|z| \leq 1-r\}$, $0<r<1$. Then $|f'(z)|$ is bounded on $K$ by $\frac 1{1-r^{2}}$. We can write $f(z)=f(0)+\int_{\gamma} f'(\zeta) d\zeta$ where $\gamma$ is the line segment from $0$ to $z$. It follows from this that $|f(z)|$ is bounded by $\frac {2-r^{2}} {1-r^{2}}$ on $K$. Since any compact subset of $D$ is contained in $K=\{z:|z| \leq 1-r\}$ for some $r <1$ we have proved that the given family is uniformly bounded on compact sets. Hence it is a normal family by Montel's Theorem.

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  • $\begingroup$ I estimated that $|f(z)|\leq 1+\frac{1}{2}\ln\left|\frac{1+|z|}{1-|z|}\right|$. This should achieve a maximum on $K$. How did you find that particular bound? @KaviRamaMurthy? $\endgroup$ – Sham Jan 12 at 5:42
  • $\begingroup$ @Sham $|\int_{\gamma} f'(\zeta) d\zeta| \leq \frac 1 {1-r^{r}} L$ where $L$ is the length of the path $\gamma$. The length of the line segment from $0$ to $z$ is $|z|$ which is bounded by $1$. Hence we get the bound $1+\frac 1 {1-r^{r}}=\frac {2-r^{2}} {1-r^{r}}$. $\endgroup$ – Kavi Rama Murthy Jan 12 at 5:52

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