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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.

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    $\begingroup$ Why do you ask? Without background this looks like homework. $\endgroup$
    – Igor Rivin
    Commented Nov 14, 2012 at 23:44
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    $\begingroup$ 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! ;) $\endgroup$
    – fedja
    Commented Nov 15, 2012 at 0:14
  • $\begingroup$ 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. $\endgroup$
    – Alexandre Eremenko
    Commented Nov 15, 2012 at 3:26
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    $\begingroup$ 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? $\endgroup$
    – fedja
    Commented Nov 15, 2012 at 4:36
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    $\begingroup$ This is not the first unmotivated question that we've seen from this poster. Do we want to encourage this? $\endgroup$
    – Anthony Quas
    Commented Nov 15, 2012 at 8:49

2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$
    – djoke
    Commented Nov 16, 2012 at 19:56
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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?

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  • $\begingroup$ @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. $\endgroup$
    – djoke
    Commented Nov 17, 2012 at 9:46
  • $\begingroup$ 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 :). $\endgroup$
    – fedja
    Commented Nov 17, 2012 at 12:34
  • $\begingroup$ @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 $|f'(1)|>1$? $\endgroup$
    – djoke
    Commented Nov 17, 2012 at 19:17
  • $\begingroup$ 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. $\endgroup$
    – fedja
    Commented Nov 18, 2012 at 1:30
  • $\begingroup$ @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! $\endgroup$
    – fedja
    Commented Nov 18, 2012 at 23:36

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