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This question is a old question but in that question one condition was not explained well.

Let $f$ be analytic on the unit disk $D$. Assume that $f(r)=\max\limits_{|z|=r} |f(z)|$. (Note that here we are not defining a new function. It just means that $f(z)$ attains its maximum at a point $z=r$.)

Why $f′(r)>0$, if $f$ is not a constant?

And why if $f(0)=0$, then $rf'(r)\geq f(r)$ and the equality holds if and only if $f(z)=cz$ for some nonnegative constant $c$ ?

Why $f'(r)$ is real number and even positive? Why not negative or some complex number? It is pretty strange for me!

Tom-Tom
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user98619
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  • Try looking at Schwarz Lemma: http://en.wikipedia.org/wiki/Schwarz_lemma – Paul Hurst May 14 '14 at 04:59
  • @PaulHurst Thanks!I know this lemma but how can this help? – user98619 May 14 '14 at 05:01
  • I just thought it might be applicable because the second part- not sure if it helps, though. Is this the complete problem. I'm sure that you need $0<r<1$. Are there any other conditions? – Paul Hurst May 14 '14 at 06:51
  • By the maximum modulus principle, $f(r) \geq |f(z)| \forall |z| \leq r$. – Paul Hurst May 14 '14 at 07:08
  • Here is the logic for the first part. $f'(z) = u_x - iv_y$. Since $f(r) \geq |f(z)|$ for all $|z| \leq r$, then if $f$ is not a constant, it is increasing along the $x$- axis at $x = r$. Thus $u_x > 0$. If $u_y \neq 0$ then as you go either up or down the modulus will increase one of those directions. So $u_y = 0$ and $f'(r)>0$. This is sort of the logic behind it. I don't have time to put it into a proof, though. – Paul Hurst May 14 '14 at 07:58

3 Answers3

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Define $g(z) = f(z)/z$ if $z \neq 0$ and $g(z) = f'(0)$ if $z = 0$.

$g$ is analytic in the unit disk. Also, $|g(z)| \leq g(r)/r$ on the circle $|z| = r$.

By the maximum modulus principle, $|g(z)| \leq g(r)/r$ for all $|z| \leq r$.

Converting back to $f$, we have $|f(z)/z| \leq f(r)/r^2$, or $|f(z)| \leq |z|f(r)/r^2$ for all $|z| \leq r$.

I'm kind of stuck at this point. I'm guessing you've probably gotten at least to here.

Paul Hurst
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  • Thanks.I got the first part,f(r) is real along the real line.I missed this.But about the part two,why the equality in Schwarz lemma holds.1 is not in the disk,it is in the boundary.So I guess we can not use this to get g(z)=cz. – user98619 May 14 '14 at 14:26
  • Sorry I got confused. It's been several years since I've used Schwarz's Lemma. I'm trying to re-learn this. I'll try to modify my answer to get it to work. – Paul Hurst May 15 '14 at 19:10
  • @PaulHurstThanks.I guess we can use mean value theory and use maximum principle again.Then every thing is fine. – user98619 May 17 '14 at 03:19
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If the derivative were ever negative, we would contradict the maximum principle: increase r slightly and you get a boundary with smaller modulus than in the interior. If it were ever zero, then by Rolle's theorem (the one from calculus) there are distinct $r_1,r_2$ with $f(r_1)=f(r_2)$. This implies $f$ is constant by the maximum modulus principle.

Ask yourself exactly which of the assumptions are actually required here. The other part seems to be something you are making progress on. It's definitely Schwarz.

zibadawa timmy
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We assume that $|f(r)|>0$ otherwise $f=0$ and the rest is trivial. Define $g(z)=f(z)/f(r)$. Therefore $g:D\to D$ is analytic, so we can apply Schwartz's lemma. This tells us that $|g(z)|\leq |z|$ on $D$ and we have $|g(r)|=1$. This can only hold if $|r|=1$. Let us write $r=\mathrm e^{\mathrm i\rho}$. As we have the equality $|g(z)|=|z|$ for $z=r$, the lemma says that $g(z)=az$ with $|a|=1$. Let us write $a=\mathrm e^{\mathrm i\alpha}$.
We have obtained $f(z)=f(\mathrm e^{\mathrm i\rho})\mathrm e^{\mathrm i\alpha}z$, which solves the problem. (Note in particular that $f'(r)=f(r)$)

Tom-Tom
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