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I'm trying to determine whether or not the following families of functions in $H(\mathbb{D})$ are normal or not.

The first family is

$$\mathcal{F}=\left\{f\in H(\mathbb{D}); f(0)=0,\ \sup_{z\in\mathbb{D}}\ |f''(z)|\cdot(1-|z|^2)<+\infty\right\}$$

The second family is

$$\mathcal{G}=\left\{ g\in H(\mathbb{D}); \sup_{z\in\mathbb{D}}\left|g(z)-\dfrac{z^3}{(1-z^2)} \right|<+\infty\right\}$$

I imagine the idea is trying to prove these are locally uniformly bounded in $\mathbb{D}$ and then apply Montel's theorem, but I can't make heads or tails of how to bound these. Also, I'm supposed to provide additional conditions for normality in case these families aren't normal. Any hints?

Thank you.

Alain Remillard
  • 3,625
Surek
  • 13

1 Answers1

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  • First family:

Let $\mathcal{F}_1\subset \mathcal{F}$ be the family of linear functions such that $f(0)=0$, i.e. $\mathcal{F}_1=\{az; a\in \mathbb{C}\}$. This family is not normal, as it is not locally uniformly bounded ($\underset{ n\in \mathbb{N}}{\forall}\underset{z\in \mathbb{D}-\{0\}}{\forall} \exists f\in \mathcal{F}:|f(z)|=n$, namely $f(w):=\frac{n}{z}w$), and so neither is $\mathcal{F}$. However, if one imposes the condition $|f'(0)|<k$ and a stricter condition like $\sup_{\mathbb{D}}|f''(z)|(1-|z|^2)<h$ the result follows. In fact, given a compact set $\Gamma \subset \mathbb{D}$ of diameter $\rho<1$, we have $$|f'(z)|\le |f'(0)|+\sup_{\Gamma}|f''(z)||z|\le |f'(0)|+|z|\frac{h}{1-\rho^2}\le k+\frac{\rho h}{1-\rho^2}\\ |f(z)|\le |f(0)|+\sup_{\Gamma}|f'(z)||z|\le \left(k+\frac{\rho h}{1-\rho^2}\right)|z|\le \left(k+\frac{\rho h}{1-\rho^2}\right)\rho$$

  • Second family

Note that $\mathcal{G}$ contains the family $\mathcal{G}_1:=\left\{f(z)=a+\frac{z^3}{1-z^2}; a\in \mathbb{C}\right\}$, which is not locally bounded (as it is not bounded in $0$). However, one condition which is sufficient to ensure normality is to modify the definition of $\mathcal{G}$ as follows:

$$\mathcal{G}=\left\{f\in H(\mathbb{D});\sup_{z\in \mathbb{D}}\left|f(z)-\frac{z^3}{1-z^2}\right|<k\right\}$$

The normality of these redefined $\mathcal{G}$ follows from the fact that, given a compact set $\Gamma\subset \mathbb{D}$ of diameter $\rho<1$, we have

$$|f(z)|\le \left|\frac{z^3}{1-z^2}\right|+k\le \frac{\rho^3}{1-\rho^2}+k$$

  • Damn, that was a quick response, thanks a lot @Caffeine I often have a hard time finding counterexamples and such, but I'm definitely taking notes on your approach and will study it carefully so I can come up with similar strategies. Appreciate it :) – Surek Apr 27 '20 at 12:46
  • Hi again @Caffeine, I'm trying to work out the inequality for $\left| f'(z)\right|$ but I don't quite follow. I'm under the assumption the Taylor series is involved, but I don't understand where did the rest of the terms of the series go, and why are we taking sup for the second derivative instead of using the usual $f''(0)$ that would come up. I tried using Cauchy''s inequality for the n-th derivative but I get a divergent series :/ Thanks – Surek Apr 27 '20 at 16:32
  • @Surek there is no need to do this. The result I used is called the mean value inequality, and can be derived as follows: let $\gamma_z$ be the straight line joining $0$ and $z$. We then have $f(z)=f(0)+\int_{\gamma_z}f'(t)dt$, and thus (by triangle inequality and the fact that $|\int f|\le \int |f|$) we get $|f(z)|\le |f(0)|+\int |f'|$, and if the derivative is bounded by $M$ $|f(z)|\le |f(0)|+M|z|$. I used this inequality twice: first to derive a bound for $f'$ (since its derivative, $f''$, is bounded we can apply the inequality) and then, once obtained the bound for $f'$, for $f$ –  Apr 27 '20 at 16:53
  • This result (which can be easily extended to vector valued functions of a real variable) is true provided the domain is convex (as otherwise we cannot link two points by a straight line), but there is a slight generalization that still holds, see https://math.stackexchange.com/questions/208609/mean-value-theorem-for-complex-functions . –  Apr 27 '20 at 16:57
  • Ok I will look into that, thank you very much :) – Surek Apr 27 '20 at 17:22