Schwarz type inequality - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T10:20:20Z http://mathoverflow.net/feeds/question/112432 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/112432/schwarz-type-inequality Schwarz type inequality djoke 2012-11-14T23:02:15Z 2012-11-19T19:48:32Z <p>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 &amp; Schwartz 1958, §VI.10.11).</p> <p>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 $$|H'(z)|\le \frac{1-|H(z)|^2}{1-|z|^2}.$$</p> <p>As $$H'(z)=(1-z^2) h'(z)/z^2-2/z^3 h(z),$$ it follows that $$|(1-z^2) h'(z)/z^2|\le \frac{2(1-|z|^2)/|z|^3 h(z)+1-|H(z)|^2}{1-|z|^2}$$ $$\le \frac{2|H(z)|/|z| +1-|H(z)|^2}{1-|z|^2}\le \frac{2|z|^{-1}}{1-|z|^2}.$$</p> <p>The question a) is related to precise estimation of norm of a Bergman projection into Bloch space and is far for being a homework.</p> http://mathoverflow.net/questions/112432/schwarz-type-inequality/112605#112605 Answer by Alexandre Eremenko for Schwarz type inequality Alexandre Eremenko 2012-11-16T18:36:59Z 2012-11-16T20:29:52Z <p>Let $|z|=r$, apply the Cauchy estimate to the disc $|\zeta-z|&lt;(1-r)/2$. 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 $1$.</p> <p>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.</p> <p>It is not likely that a simple extremal function exists, and probably for each $z$ there will be a different extremal function.</p> http://mathoverflow.net/questions/112432/schwarz-type-inequality/112647#112647 Answer by fedja for Schwarz type inequality fedja 2012-11-17T05:07:34Z 2012-11-17T05:07:34Z <p>Looks like we are closing the question anyway, so I'll just provide a counterexample quickly before the final vote is cast. </p> <p>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 $|f|&lt;1/x$ 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. </p> <p>As to "motivation" in general, look up in the evening. You'll see the stars in the sky. What other motivation do you need?</p>