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Question

Let $\mathcal{F}$ be the family of holomorphic functions $f$ on the open unit disk such that $\Re f>0$ for all $z$ and $f(0)=1$. Compute $$\alpha=\sup\{|f'(0)|: f \in \mathcal{F}\}.$$ Determine whether or not the supremum $\alpha$ is attained.

Attemp

I am unsure where to begin. I feel that since the function has image in the right half plane, I can map it back conformally to the disk and then apply Schwarz lemma. In particular, let $g(z)=\frac{1-z}{1+z}$. Then $g(1)=0$. Consequently $g \circ f(0)=0$. By Schwarz lemma, if $g \circ f=a_1z+ \mathrm{higher \ order\ terms}$, then $|(g\circ f)'(0)|=|a_1|\leq 1$. I tried taking the derivative and I also tried expanding the power series but I didn't get anything meaningful so I think this might be the wrong approach.

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    $\begingroup$ $\Re f>0$ describes the right halfplane, not the first quadrant – so which one do you actually mean? $\endgroup$ – Martin R Nov 19 '17 at 20:03
  • $\begingroup$ @MartinR you are right! I have made the appropriate edits. I meant the right half plane. $\endgroup$ – user210552 Nov 19 '17 at 20:04
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    $\begingroup$ See math.stackexchange.com/questions/950682/…. $\endgroup$ – Martin R Nov 19 '17 at 20:07
  • $\begingroup$ I am able to derive this bound but I do not know how to find the supremum and to show that it is attained. $\endgroup$ – user210552 Nov 19 '17 at 20:10
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    $\begingroup$ What about $f = g^{-1}$? $\endgroup$ – Daniel Fischer Nov 19 '17 at 20:23
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$T(z) = \frac{1-z}{1+z}$ maps the right half plane conformally to the unit disk with $T(1) = 0$, therefore $g = T \circ f$ satisfies the conditions of the Schwarz Lemma, so that $$ 1 \ge |g'(0)| = |T'(f(0))f'(0) | = |T'(1)f'(0)| = \frac 12 |f'(0)| \\ \Longrightarrow |f'(0)| \le 2 \, . $$ Equality in the Schwarz Lemma holds exactly for the functions $$ g(z) = \lambda z $$ with $|\lambda| = 1$, and therefore $|f'(0)|=2$ holds exactly for the functions $$ f(z) = T^{-1}(\lambda z) = \frac{1 - \lambda z}{1 + \lambda z} \, . $$

Therefore $$ \sup\{|f'(0)|: f \in \mathcal{F}\} = 2 $$ and the supremum is attained.

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Let $f(z)=1+a_1z+a_2z^2+\cdots$ then Schwarz formula (see Ahlfors p.167 - print 1966) says $$f(z)=\dfrac{1}{2\pi}\int_0^{2\pi}\dfrac{re^{i\theta}+z}{re^{i\theta}-z}{\bf Re}\,f(re^{i\theta})\,d\theta$$ for $|z|<r<1$ and thus $\displaystyle f(0)=\dfrac{1}{2\pi}\int_0^{2\pi}{\bf Re}\,f(re^{i\theta})\,d\theta$. Then $$f'(z)=\dfrac{1}{2\pi}\int_0^{2\pi}\dfrac{2re^{i\theta}}{(re^{i\theta}-z)^2}{\bf Re}\,f(re^{i\theta})\,d\theta$$ and $$|f'(0)|\leq\dfrac{1}{2\pi}\int_0^{2\pi}\dfrac{2}{r}{\bf Re}\,f(re^{i\theta})\,d\theta=\dfrac{2}{r}|f(0)|\leq2$$ as $r\to1^-$.

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