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I know the following is a well-known result.

Let $D = B(0,1) \subset \mathbb{C} $ a disc, $f$ holomorphic on $D$. Show that $$ 2|f^{'}(0)| \le \sup_{z, w \in D} |f(z)-f(w)|$$ Furthermore, there is equality if and only if $f$ is linear.

I need some reference about the second part, i.e. there is equality if and only if $f$ is linear.

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Out of curiosity, would not it be simpler to give a direct proof to the fact rather than looking for the reference? –  Alvin Feb 14 at 20:39
    
@Alvin: if you know how to prove that equality if possible only for linear $f$, then tell us the proof. –  Alexandre Eremenko Feb 14 at 21:38

2 Answers 2

up vote 3 down vote accepted

This was first proved by Landau and Toeplitz in 1907. A reference for the proof (and for generalizations) is the paper Area, capacity and diameter versions of Schwarz's lemma by Burckel, Marshall, Minda, Poggi-Corradini and Ransford.

See Theorem 1.3 here

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From the reference you give, I conclude that the statement about extremal functions was not in Landau and Toepliz paper; it is due to Gehring and Hag. Inequality itself has one-line proof (see my answer). –  Alexandre Eremenko Feb 16 at 21:17
    
@AlexandreEremenko : After Theorem 1.3 of the paper mentioned in my answer, it is written The main contribution of the Landau-Toeplitz paper is perhaps its elucidation of the extremal case. –  Malik Younsi Feb 17 at 19:40
    
Unfortunately, I hardly read German:-( But anyway, we have several proofs of the extremal case now. –  Alexandre Eremenko Feb 17 at 20:56

Edit.

Let us first prove the inequality: $$\sup_{z,w}|f(z)-f(w)|\geq\sup_z|f(z)-f(-z)|=\sup|g(z)|\geq|g'(0)|=2|f'(0)|,$$ where the Schwarz Lemma was applied to $g(z)=f(z)-f(-z)$. Equality in Schwarz lemma can happen only if $g(z)=kz$, thus $f(z)=kz+\phi(z)$, where $\phi$ is even.

Now let us see when equality is possible in the first inequality. We must have $$\sup_{|z|<1,|w|<1}|k(z-w)+\phi(z)-\phi(w)|=2|k|,$$ which an even function $\phi$ analytic in the unit disc. We have to derive from here that $\phi$ is constant.

In the reference http://www.math.wustl.edu/~geknese/schwarzpoly.pdf, where extremal functions for the Schwarz lemma in the polydisc are described. But those functions are extremal at every point. And our function is extremal at only one point, the origin.

I can prove that $\phi$ is constant only under the additonal restriction that $\phi$ is differentiable in the closed disc. WLOG $k=1$. Let $|z|=1$ and put $w=-ze^{it}$, where $t$ is small. Then, neglecting the high powers of $t$, we must have $$|2+it-\phi'(z)it|\leq 2$$ This implies that $\phi'(z)$ must be real for all $z$ on the unit circle. But such function must be constant, and as $\phi$ is even, we conclude that $\phi$ is constant.

To get rid of the additional assumption of differentiability, one may combine this and and this.

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Why can you apply the Schwarz Lemma to $g(z) $ ? –  user33122 Feb 15 at 7:38
    
And I don't understand why $\sup_{|z|<1,|w|<1}|k(z-w)+\phi(z)-\phi(w)|\leq 2|k|$ –  user33122 Feb 15 at 7:41
    
1. Because $g$ is analytic and $g(0)=0$. 2. This is the statement that the first inequality in the first line becomes equality. –  Alexandre Eremenko Feb 15 at 14:44

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