Is the set of derivatives of analytic functions of the unit disk equicontinuous?

Let $F$ be the set of analytic functions from the (open) unit disk $\mathbb{D}$ to itself. Let $G$ be the set of derivative functions of $F$. I'm trying to show that $G$ is not an equicontinuous family of functions by showing that $G$ contains a sequence of functions that doesn't contain a subsequence that converges to a continuous function, thereby demonstrating that the conclusion of the Arzelà–Ascoli theorem fails for $G$. Since $G$ is a uniformly pointwise bounded family of functions, we would be done.

My suggestion for such a candidate sequence of functions is as follows: Let $f_n(z) = e^{inp} z$ where $p$ is a positive irrational number and $n \geq 1$. I conjecture that the sequence of derivatives $f'_n(z) = e^{inp}$ would do the job, but not sure how to prove it.

• Are you asking for equicontinuity on the entire disk or for local equicontinuity, i.e. on compact subsets? – Martin R Oct 30 '17 at 9:01
• Equicontunity on the disk $\mathbb{D}$ – Dennis Chew Oct 30 '17 at 10:06
• Then $f_n(z) = z^n$ should be a counter-example. – Note that even $F$ need not be equicontinuous on the entire disk. – Martin R Oct 30 '17 at 10:08
• Wouldn't the sequence of functions $z^n$ converge point wise to the 0 function? – Dennis Chew Oct 30 '17 at 10:55
• Are you asking for equicontinuity in each point of the disk, or uniform equicontinuity in the disk? – Martin R Oct 30 '17 at 11:18

Cauchy's integral formula for the second derivative $$f''(a) = \frac{2!}{2 \pi i} \int_{|z|=r} \frac{f(z)}{(z-a)^3} dz \, .$$ for $|a| < r < 1$ implies that the family $\{ f'' \mid f \in F \}$ of second derivatives is uniformly bounded on each closed disk $K = \{ z: |z-a| \le \varepsilon \} \subset \Bbb D$.

It follows that the family $G$ of first derivatives is uniformly equicontinuous on $K$, and in particular equicontinuous at each point $a \in \Bbb D$.

In your example $f'_n(z) = e^{inp}$ there is a subsequence converging to the constant $1$, because $np$ comes arbitrarily close to multiples of $2 \pi$.

• By the same reasoning then won't $F$ also be uniformly equicontinuous on $\mathbb{D}$? – Dennis Chew Oct 30 '17 at 12:18
• @DennisChew: No, it works only on compact subsets of the unit disk (so that you have a lower bound for the denominator in the integral). – Martin R Oct 30 '17 at 12:21

By the compactness of the unit circle, there is a subsequence $n_k$ such that $e^{in_k p}$ converges to some $e^{i\theta},$ so your trial balloon is fast losing altitude. I would suggest looking at Montel's theorem.

• Thank you. But doesn't Montel's theorem require uniform boundedness and not the weaker uniform pointwise boundedness? – Dennis Chew Oct 31 '17 at 8:38
• @DennisChew I'm thinking of this version of Montel: If $f_n$ is a sequence of holomorphic functions on $\mathbb D$ that is uniformly bounded on each compact subset of $\mathbb D,$ then there exists a subsequence $n_k$ such that $D^m(f_{n_k})$ converges uniformly on compact sets for each $m=0,1,\dots.$ So every sequence in $G$ contains a subsequence that converges uniformly on compact sets, hence converges to a holomorphic function in $\mathbb D.$ – zhw. Oct 31 '17 at 17:59