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An improved Schwarz Lemma at the boundary

Peter R. Mercer
Published Online: 2018-10-19 | DOI: https://doi.org/10.1515/math-2018-0096

Abstract

We obtain an new boundary Schwarz inequality, for analytic functions mapping the unit disk to itself. The result contains and improves a number of known estimates.

Keywords: Schwarz Lemma; Julia’s Lemma

MSC 2010: 30C80

1 Introduction

Denote by ⊂ ℂ the open unit disk, and let f: be analytic. We assume that there is x > ∂∆ and β > ℝ such that

$liminfz→x1−|f(z)|1−|z|=β.$(1)

By pre-composing with a rotation we may suppose that x = 1, and by post-composing with a rotation we may suppose that f(1) = 1. Then Julia’s Lemma (e.g. [1, 2]) gives

$|1−f(z)|21−|f(z)|2≤β|1−z|21−|z|2 ∀z∈Δ.$

This inequality has an appealing geometric interpretation, which we do not use here. But two immediate consequences which we do use, are that β > 0 and that the radial derivative of f exists at 1 ∈ > ∂∆:

$limr↗1f(r)−f(1)r−1=f′(1) with |f′(1)|=β.$(2)

(There are many other consequences of Julia’s Lemma, the most important being contained in the Julia-Carathéodory Theorems.)

Assuming the normalization f(0) = 0, we evidently have β ≥ 1. But even better, Osserman  showed that in this case

$β≥1+1−|f′(0)|1+|f′(0)|.$(3)

(A proof of (3) can also be found in , which is motivated by the influential paper .) Now Osserman’s inequality was in fact anticipated by Ünkelbach , who had already obtained the better estimate

$β≥2(1−Ref′(0))1−|f′(0)|2=1+|1−f′(0)|21−|f′(0)|2.$(4)

However,  also contains a non-normalized version, which reduces to (3) if f(0) = 0, viz.

$β≥2(1−|f(0)|)21−|f(0)|2+|f′(0)|.$(5)

Since the appearance of Osserman’s paper, a good number of authors have refined and generalized these estimates – as discussed in the next section. The aim here is to provide a different and very elementary approach, which contains and improves many of these modification. But first we recall some results which are of use in the sequel.

The well-known Schwarz’s Lemma, which is a consequence of the Maximum Principle, says that if f: is analytic with f(0) = 0, then

$|f(z)|≤|z| ∀z∈Δ, and consequently |f′(0)|≤1.$

To remove the normalization f(0) = 0, one applies Schwarz’s Lemma to ϕf(a)fϕa where ϕa is the automorphism of which interchanges a and 0:

$ϕa(z)=a−z1−a¯z.$

This gives the Schwarz-Pick Lemma which says that for f: analytic,

$|f(w)−f(z)1−f(w)¯f(z)|≤|w−z1−w¯z| ∀z,w∈Δ.$

Consequently, the hyperbolic derivative satisfies

$|f∗(z)|≤1 ∀z∈Δ, wheref∗(z)=1-|z|21-|f(z)|2f′(z).$

It is the Schwarz-Pick Lemma that does most of the work in proving Julia’s Lemma. But another consequence of the Schwarz-Pick Lemma is the following (e.g. [7–9]), which we shall also rely upon.

Lemma 1.1

(Dieudonné’s Lemma). Let f: ∆ be analytic, with f(z) = w and f(z1) = w1. Then

$|f′(z)−c|≤r,$

where

$c=ϕw(w1)ϕz(z1)1−|ϕz(z1)|21−|ϕw(w1)|21−|w|21−|z|2,r=|ϕz(z1)|2−|ϕw(w1)|2|ϕz(z1)|2(1−|ϕw(w1)|2)1−|w|21−|z|2.$

2 Main result

We remove the dependence on f(0), while improving many estimates which do contain f(0). We shall rely on Dieudonné’s Lemma, the Schwarz-Pick Lemma, and Julia’s Lemma.

Theorem 2.1

Let f: ∆ be analytic with f(z) = w and f(1) = 1 as in (1). Then

$β≥2|1−w|21−|w|21−|z|2|1−z|21−Re(f∗(z)1−w¯1−w1−z1−z¯)1−|f∗(z)|2.$(6)

Proof

Using the easily verified identity

$1−|ϕa(λ)|2=(1−|a|2)(1−|λ|2)|1−a¯λ|2,$(7)

we get, in Dieudonné’s Lemma,

$c=w1−w1−w¯w11−z¯z1z1−z1−|z1|2|1−z¯z1|2|1−w¯w1|21−|w1|2=w1−wz1−z1−ww1¯1−zz1¯1−|z1|21−|w1|2,$

and

$r=(1−|ϕw(w1)|2)−(1−|ϕz(z1)|2)|ϕz(z1)|2(1−|ϕw(w1)|2)1−|w|21−|z|2=1|ϕz(z1)|2(1−1−|z|21−|w|21−|z1|21−|w1|2|1−w¯w1|2|1−z¯z1|2)1−|w|21−|z|2.$

then having z1 → 1 along a sequence for which β in (1) is attained, we get

$c→ c˜=(1−w1−z)21β and r→ r˜=1-|w|21-|z|2-1β|1-w|2|1-z|2.$

That is,

$|f′(z)− c˜|≤r˜.$(8)

Now, upon squaring both sides of this inequality, there is some cancellation:

$|f′(z)|2−2Re(f′(z)¯(1−w1−z)21β)≤(1−|w|21−|z|2)2−2β1−|w|21−|z|2|1−w¯|2|1−z¯|2.$

That is,

$(1−|w|21−|z|2)2(|f∗(z)|2−1)≤2β|1−w¯|2|1−z¯|21−|w|21−|z|2[Re(f∗(z)1−w¯1−w1−z1−z¯)−1].$

By the Schwarz-Pick Lemma each side of this last inequality is nonpositive, so isolating β we get (6).

Remark 2.2

Having z → 1 radially in line (8), and using (2), we obtain

$limr↗1f′(r)=f′(1).$

From this, and using $|\tau |=1⇒\frac{1-\mathrm{Re}\left(\sigma \tau \right)}{1-|\sigma {|}^{2}}\ge \frac{1}{1+|\sigma |}$, follows the rather comforting fact that the right-hand side of (6) tends to β as z → 1 radially.

Remark 2.3

In Lemma 6.1 of  is the estimate

$β≥21+|f∗(z)|1−|f(z)|1+|f(z)|1−|z|1+|z|,$(9)

which contains (5), but is quite mild if |z| or |f(z)| is near 1. Anyway, $|\tau |=1⇒\frac{1-\mathrm{Re}\left(\sigma \tau \right)}{1-|\sigma |}\ge 1$ shows that (6) improves (9).

Remark 2.4

Now take z = 0, so that (6) reads

$β≥2|1−f(0)|21−|f(0)|21−Re(f∗(0)1−f(0)¯1−f(0))1−|f∗(0)|2.$(10)

This may be regarded as an non-normalized version of (4). Indeed, taking also f(0) = 0 recovers (4). This is the same estimate which results from having z = 0 in Theorem 5 of . However, that result (which is arrived at by very nonelementary means) contains f(0) even for z ≠ 0, a deficiency from which Theorem 2.1 does not suffer.

Remark 2.5

Using again $|\tau |=1⇒\frac{1-\mathrm{Re}\left(\sigma \tau \right)}{1-|\sigma {|}^{2}}\ge \frac{1}{1+|\sigma |}$ in (10), we get

$β≥2|1−f(0)|21−|f(0)|2+|f′(0)|,$

which improves (5), analogously to how (4) improves (3).

Remark 2.6

But using just $|\tau |=1⇒\frac{1-\mathrm{Re}\left(\sigma \tau \right)}{1-|\sigma {|}^{2}}\ge \frac{1}{1+\mathrm{Re}\left(\sigma \tau \right)}$ in (10), then $\frac{1-|f\left(0\right){|}^{2}}{|1-f\left(0\right){|}^{2}}=\mathrm{Re}\frac{1+f\left(0\right)}{1-f\left(0\right)}$, we get

$β≥2Re1−f(0)2+f′(0)(1−f(0))2,$(11)

which improves (5) more effectively. Estimate (11) was obtained differently in each of  and .

3 Consequences

Cases for which z = w = 0 (i.e. f(0) = 0) are obviously contained in the remarks above, but when this holds we can do a little better, as follows.

Corollary 3.1

Let f: ∆ be analytic with f(0) = 0 and f(1) = 1 as in (1). Then

$β≥1+2|1−f′(0)|21−|f′(0)|2+|f′′(0)|/21+Re(f′′(0)2(1−|f′(0)|2))1−|f′′(0)|2(1−|f′(0)|2).$(12)

Proof

We introduce f″(0), in standard fashion: Set

$g(λ)=f(λ)λ (with g(0):=f′(0)), and h(λ)=ϕg(0)(g(λ)).$

Then h is analytic on with h(0) = 0, and by Schwarz’s Lemma h: . Here we have

$h′(0)=−f′′(0)2(1−|f′(0)|2).$(13)

A calculation using the identity (7) and the assumption (1) gives

$liminfz→11−|h(z)|1−|z|=(β−1)1−|f′(0)|2|1−f′(0)|2=β^, say.$(14)

Then in (6), i.e. (4), replacing f with h and β with $\stackrel{^}{\beta }$, we obtain

$β≥1+|1−f′(0)|21−|f′(0)|22(1−Reh′(0))1−|h′(0)|2.$

Inserting (13) and a little tidying yields (12), as desired.

Remark 3.2

Corollary 3.1 improves

$β≥1+2(1−|f′(0)|2)1−|f′(0)|2+|f′′(0)|/2,$(15)

which was obtained by Dubinin  using a proof which relies directly on (3). (Incidentally, Schwarz’s Lemma applied to h gives |f″(0)|/2 ≤ 1 − |f′(0)|2, from which it is readily seen that (15) improves (3).)

Remark 3.3

We add finally using that (4) in the form

$β≥1+|1−f′(0)|21−|f′(0)|2,$

then replacing f with h and β with $\stackrel{^}{\beta }$ here, and using (13) and (14), we get another way of expressing (12):

$β ≥1+|1−f′(0)|21−|f′(0)|2(1+|1+f′′(0)2(1−|f′(0)|2)|21−|f′′(0)2(1−|f′(0)|2)|2)=1+|1−f′(0)|21−|f′(0)|2+|1+f′′(0)2(1−|f′(0)|2)|21−|f′′(0)2(1−|f′(0)|2)||1−f′(0)|21−|f′(0)|2+|f′′(0)|/2.$

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Accepted: 2018-09-05

Published Online: 2018-10-19

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 1140–1144, ISSN (Online) 2391-5455,

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