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.

  • $\begingroup$ Are you asking for equicontinuity on the entire disk or for local equicontinuity, i.e. on compact subsets? $\endgroup$ – Martin R Oct 30 '17 at 9:01
  • $\begingroup$ Equicontunity on the disk $\mathbb{D}$ $\endgroup$ – Dennis Chew Oct 30 '17 at 10:06
  • $\begingroup$ Then $f_n(z) = z^n$ should be a counter-example. – Note that even $F$ need not be equicontinuous on the entire disk. $\endgroup$ – Martin R Oct 30 '17 at 10:08
  • $\begingroup$ Wouldn't the sequence of functions $z^n$ converge point wise to the 0 function? $\endgroup$ – Dennis Chew Oct 30 '17 at 10:55
  • $\begingroup$ Are you asking for equicontinuity in each point of the disk, or uniform equicontinuity in the disk? $\endgroup$ – 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 $2 \pi$.

  • $\begingroup$ By the same reasoning then won't $F$ also be uniformly equicontinuous on $\mathbb{D} $? $\endgroup$ – Dennis Chew Oct 30 '17 at 12:18
  • $\begingroup$ @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). $\endgroup$ – 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.

  • $\begingroup$ Thank you. But doesn't Montel's theorem require uniform boundedness and not the weaker uniform pointwise boundedness? $\endgroup$ – Dennis Chew Oct 31 '17 at 8:38
  • $\begingroup$ @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 $\mathbb D.$ $\endgroup$ – zhw. Oct 31 '17 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.