# Show that $\mathcal{F}$ is a normal family on $\mathbb {D}$.

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 $$0 and consider the open disk centered at $$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 $$\mathbb{D}$$

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

Let $$K=\{z:|z| \leq 1-r\}$$, $$0. Then $$|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.
• 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? – Sham Jan 12 at 5:42
• @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 $1+\frac 1 {1-r^{r}}=\frac {2-r^{2}} {1-r^{r}}$. – Kavi Rama Murthy Jan 12 at 5:52