Let $f(z)$ be a holomorphic function defined on the disk $|z|\le 2$. Suppose $|f(z)|<1$ for $|z|\le 2$. It looks like there is a constant $c>0$ such that $|f(z)'|<c$ on the disk $|z|\le 1$ (for example, $c=1$?). I wonder if this is true. If yes, is an optimal such constant can be found?
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$\begingroup$ Cauchy integral formula gives $\lvert f'(z)\rvert\leq 2$ (integrate over a circle centered at 0, of radius $2-\varepsilon $). This is certainly not optimal. $\endgroup$– abxCommented Mar 26, 2018 at 16:42
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3$\begingroup$ The optimal inequality should be the one you get from the invariant form of the Schwarz lemma: en.wikipedia.org/wiki/Schwarz_lemma $\endgroup$– Christian RemlingCommented Mar 26, 2018 at 17:16
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$\begingroup$ If you rescale so that your big disk is $|z|<1$ and the small one is $|z|<1/2$, then this gives $c=4/3$. $\endgroup$– Christian RemlingCommented Mar 26, 2018 at 17:17
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Yes, this is true. The simple reason is that bounded functions form a normal family. Therefore their derivatives are uniformly bounded on every compact.
To obtain the estimate $|f'(z)|<1$, apply Cauchy theorem: $$|f'(z)|=\left|\frac{1}{2\pi}\int_{|\zeta-z|=1}\frac{f(\zeta) d\zeta}{(\zeta-z)^2}\right|\leq 1.$$ Equality in the right inequality can happen only if $f$ is constant, so we always have a strict inequality.
So this estimate is not exact. To obtain the exact one, follow @Christian Remling's suggestion and use the Schwarz lemma. For convenience, let $g(z)=f(2z)$, then $g$ maps the unit disk into itself, so $$|g'(z)|\leq \frac{1-|g(z)|^2}{1-|z|^2}\leq\frac{1}{1-|z|^2}\leq \frac{4}{3}.$$ Thus the exact estimate us $|f'(z)|=|g'(z)|/2\leq 2/3.$ This is attained on a Mobius map.
answered Mar 26, 2018 at 18:22
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$\begingroup$ I don't understand your final statement. Isn't it obvious from my comment above that $|f'(z)|$ is maximized by the Mobius transformations that send $z\mapsto 0$ ? $\endgroup$ Commented Mar 26, 2018 at 18:45
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1$\begingroup$ @Christian Remling: Of course, but when you rescale the derivative changes by a factor of $2$. $\endgroup$ Commented Mar 27, 2018 at 2:08
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$\begingroup$ The "final statement" I was referring to was the final statement when I wrote my comment, before your edit. $\endgroup$ Commented Apr 4, 2018 at 23:41