MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

a) Is true the following statement. Let $h$ be analytic in the unit disk such that $$|h(z)|\le \frac{|z|^2}{1-|z|^2},$$ then $$|h'(z)|\le \frac{2}{(1-|z|^2)^2}.$$ a') Is true the following statement. Let $h$ be analytic in the unit disk such that $$|h(z)|\le \frac{|z|^2}{1-|z|^2},$$ then the inequality $$|h'(z)|\le \frac{8}{\pi(1-|z|^2)^2}$$ is sharp. The inequality can be proved by using Schur test, and Riesz-Thorin convexity type theorem (Dunford & Schwartz 1958, §VI.10.11).

b) If $$|h(z)|\le \frac{|z|^2}{|1-z^2|}$$ then we have better conclusion $$|h'|\le \frac{2|z|}{(1-|z|^2)|1-z^2|}$$ and this follows by using Schwarz lemma. Namely in this case $$|H(z)|=|(1-z^2) h(z)/z^2|\le 1.$$ Then $$|H'(z)|\le \frac{1-|H(z)|^2}{1-|z|^2}.$$

As $$H'(z)=(1-z^2) h'(z)/z^2-2/z^3 h(z),$$ it follows that $$|(1-z^2) h'(z)/z^2|\le \frac{2(1-|z|^2)/|z|^3 h(z)+1-|H(z)|^2}{1-|z|^2}$$ $$\le \frac{2|H(z)|/|z| +1-|H(z)|^2}{1-|z|^2}\le \frac{2|z|^{-1}}{1-|z|^2}.$$

The question a) is related to precise estimation of norm of a Bergman projection into Bloch space and is far for being a homework.

share|cite|improve this question
Why do you ask? Without background this looks like homework. – Igor Rivin Nov 14 '12 at 23:44
It is far less obvious than it looks, Igor (what is the answer in your opinion?). I'm not so sure it needs to be closed. Anyway, if it gets closed here, it will make a cute problem for AoPS->College Playground->Complex Analysis :). @djoke If you think that $\frac{z^2}{1-z^2}$ is "obviously" the worst case scenario, you may want to think again! ;) – fedja Nov 15 '12 at 0:14
I agree with Igor. If the author conjectures an exact inequality (not just asks for an estimate), he should have reasons for this, and I expect him to tell us these reasons. Perhaps he has a conjectured extremal function, or whatever. I agree that this is not trivial, but this is not sufficient justification. – Alexandre Eremenko Nov 15 '12 at 3:26
Or, perhaps, he is just curious because he fails to see it himself (and this is one of those cases when even a person with brains can get perplexed for quite a while; if you don't agree, let's just see how quickly the answer comes). What's wrong with being curious for no apparent reason? – fedja Nov 15 '12 at 4:36
This is not the first unmotivated question that we've seen from this poster. Do we want to encourage this? – Anthony Quas Nov 15 '12 at 8:49
up vote 1 down vote accepted

Let $|z|=r$, apply the Cauchy estimate to the disc $|\zeta-z|<(1-r)/2$. We obtain $$|f'(z)|\leq \frac{2}{1-r}\frac{(1+r)^2}{(1-r)(3+r)}.$$ Maximizing the factor $(1+r)^2/(3+r)$ by Calculus, we obtain that is it at most $1$.

This gives $$|f'(z)|\leq\frac{2}{(1-|z|)^2}$$ which is worse than conjectured only by a factor of $(1+|z|)^2$, which is at most $4$. Perhaps one can improve the constant by applying Cauchy to a disc of radius $t\in(0,1-r)$, and then optimizing in $t$, which leads to solving a cubic equation.

It is not likely that a simple extremal function exists, and probably for each $z$ there will be a different extremal function.

share|cite|improve this answer
I have tried by Cauchy inequality, Cauchy integral, Schwarz lemma, and all lead to the constant 4. It is interesting that, up to now the "best example" is $f(z)=z^2/(1-z^2)$, and some similar combinations but all reach only the constant 2. So it is very interesting that no better constant than 4 and no better example than 2. – djoke Nov 16 '12 at 19:56

Looks like we are closing the question anyway, so I'll just provide a counterexample quickly before the final vote is cast.

If you think a bit of what is asked and what the natural freedoms and scalings are present here, you'll see that it is enough to get an analytic $f$ in the right half-plane $x>0$ ($z=x+iy$ as usual) such that $|f|<1/x$ and $|f'(1)|>1$. Now just take something like $f(z)=\frac 1z-aze^{-\sqrt{z}}$ with sufficiently small positive $a$. I leave it to somebody else to beat $4$ in the upper bound.

As to "motivation" in general, look up in the evening. You'll see the stars in the sky. What other motivation do you need?

share|cite|improve this answer
@fedja If guess that you are taking a composition of $h$ with a Mobius transformation of the half-plane onto the unit disk. If $f$ is the composition, then you request that $|f(z)<1/x$. But $|z^2|/(1-|z|^2)$ is not Poincare metric. I cannot see how this gives counterexample. Could you provide much details. – djoke Nov 17 '12 at 9:46
No, I just put the disk of radius $R$ centered at $R$ inside the plane and rescale. Near the boundary the estimate is trivial (up to a $1+\varepsilon$ factor) because the circumference is closer than the line. To get everything right deep inside the disk, just multiply by $z^10$. Since the jump can be made arbitrarily close to the boundary, that won't hurt. The whole story unfolds near the boundary, so you can blow-up fully at the boundary point from the very beginning. As I said, you just need to "think a bit" here :). – fedja Nov 17 '12 at 12:34
@fedja. Probably you have an idea which is not written in this post. When you construct a function $h$, the domain should be the unit disk. The disk $|z-R|<R$ is not unit disk, it is a part of right-half plane. The mapping $(z-R)/R$ maps $|z-R|<R$ into $|z|<1$. The condition $|h(z)|\le |z|^2/(1-|z|^2)$, implies that $h'(0)=0$. It can be rewritten assuming that $h$ is analytic in the half plane What is your condition $|f'(1)|>1$? – djoke Nov 17 '12 at 19:17
OK, I'll spell the details out but later. The condition $|f'(1)|>1$ is the condition that $|h'(z)|\ge \frac{A}{(1-|z\^2)^2}$ with $A>2$, of course. – fedja Nov 18 '12 at 1:30
@djoke: I have already told you: it is not the best! I don't really know what the best one is at the moment (essentially you need to find the extremal function in my "blown-up" formulation to figure it out), but to improve upon $4$ is not hard at all. Just think a bit! – fedja Nov 18 '12 at 23:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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