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Suppose $f$ is an analytic function on $|z|\leq 1$ with $f(0)=0$, and let $|f(z)|$ have a maximum for $|z|\leq 1$ at 1, show that $f'(1)\neq 0$ unless $f$ is a constant.


Remarks:

1, At first attempt, I tried to construct some function related to $f$, and then try to use Schwarz's lemma, but I got stuck; now I guess there exists an direction from $1$ such that the modulus locally increase.

2, To be precise, when applying Schwarz's lemma to $\frac{f}{|f(1)|}$, I got stuck because 1 is not in the open disk.

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    $\begingroup$ The exchange of knowledge is more effective when both sides participate. What do you know about the behavior of an analytic function near a point where its derivative is zero? $\endgroup$
    – user53153
    Jan 2, 2013 at 23:02
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    $\begingroup$ I think the condition $f(0)=0$ is redundant. $\endgroup$
    – 23rd
    Jan 2, 2013 at 23:04
  • $\begingroup$ at least as a continuous function, it is local maximum or local minimum at the 1 if $f'(1)=0$. $\endgroup$
    – ougao
    Jan 2, 2013 at 23:17

2 Answers 2

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Suppose that $f$ is not a constant function and $f'(1)=0$. Let $f(1)=Me^{i\theta}$, where $M=|f(1)|$. It suffice to show that there exists $z$ with $|z|<1$, such that $|f(z)|>M$.

Since $f$ is analytic at $z=1$, $f'(1)=0$ and $f$ is not a constant function, there exists $n\ge 2$ and $a\ne 0$, such that when $z$ is close to $1$, $$f(z)=f(1)+a(z-1)^n+R_n(z),$$ where $\lim_{z\to 1}\frac{|R_n(z)|}{|z-1|^n}=0$. In particular, when $|z-1|$ is small, $|R_n(z)|\le\frac{|a|}{4}|z-1|^n$, and hence

$$|f(z)|\ge |f(1)+a(z-1)^n|-|R_n(z)|\ge |M+ae^{-i\theta}(z-1)^n|-\frac{|a|}{4}|z-1|^n.\tag{1}$$

Since $n\ge 2$, there exists $\frac{\pi}{2}<\phi<\frac{3\pi}{2}$, such that

$$\mathrm{Re}(ae^{-i\theta}e^{in\phi})>\frac{|a|}{2}.\tag{2}$$ Since $\cos\phi<0$, when $r>0$ is sufficiently small, for $z=1+re^{i\phi}$, $|z|<1$. Therefore, according to $(1)$ and $(2)$, $$|f(z)|\ge \mathrm{Re}(M+ae^{-i\theta}(z-1)^n)-|R_n(z)|\ge M+\frac{|a|}{4}r^n>M,$$ which completes the proof.

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  • $\begingroup$ thanks! It is a little strange to me that you have proved that for a holomorphic function $f$, if $f'(z_0)=0$ for some point $z_0$ in its domain, then for any neighborhood of $z_0$, it contains a point $z_n$ in this neighborhood such that $|f(z_n)|>|f(z_0)|$. $\endgroup$
    – ougao
    Jan 9, 2013 at 12:15
  • $\begingroup$ @ougao: You are welcome. In fact, due to maximum modulus principle, your statement is still valid without assuming that $f'(z_0)=0$, so what I proved is a little stronger than your statement. $\endgroup$
    – 23rd
    Jan 9, 2013 at 16:33
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Take $ g = \dfrac{f}{M} $, where $ M $ is the maximum of $ f $ on the unit disk. Observe that $ |g(z)| = |z| $ holds iff $ g'(z) = az $, so $ |a| = 1 $ by Schwarz's Lemma. But this implies that $ |g'(0)| > 0 $, so $ |f'(0)| > 0 $.

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  • $\begingroup$ nice edit thanks $\endgroup$
    – Koushik
    Jan 3, 2013 at 0:43
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    $\begingroup$ The question is about $f'(1)$, not $f'(0)$. Moreover, your argument is incorrect, because you cannot conclude $g'(0)\ne 0$. For example, consider $f(z)=z^2$. $\endgroup$ Apr 12, 2013 at 18:40

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