Schwarz Lemma - Complex Analysis, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


The Schwarz Lemma and
HyperbolicGeometry
This short chapter is devoted to the Schwarz lemma, which is a simple
consequence of the power series expansion and the maximum principle.
The Schwarz lemma is proved in Section 1, and it is used in Section 2
to determine the conformal self-maps of the unit disk. In Section 2 we
formulatetheSchwarzlemmatobeinvariantundertheconformalself-maps
oftheunitdisk, therebyobtainingPick’slemma. ThisleadsinSection3
tothehyperbolicmetricandhyperbolicgeometryoftheunitdisk.
1.TheSchwarzLemma
TheSchwarzlemmaiseasytoprove,yetithasfar-reachingconsequences.
Theorem (Schwarz Lemma). Letf(z)beanalyticfor|z|<1. Suppose
|f(z)|=1 for all |z|<1,and f(0)=0. Then
(1.1) |f(z)|=|z|, |z|<1.
Further, if equality holds in(1.1) at some pointz
0
=0, thenf(z)=?z for
some constant ? of unit modulus.
For the proof, we factor f(z)= zg(z), where g(z) is analytic, and we
applythemaximumprincipleto g(z). Letr<1. If |z|=r,then |g(z)|=
|f(z)|/r =1/r. Bythemaximumprinciple,|g(z)|=1/rforallzsatisfying
|z|= r.Ifwelet r ?1, we obtain |g(z)|=1 for all |z| <1. This yields
(1.1). If |f(z
0
)|= |z
0
|forsome z
0
=0,then |g(z
0
)|=1,andbythestrict
maximumprinciple, g(z)isconstant,say g(z)=?. Then f(z)=?z.
Ananalogousestimateholdsinanydisk. Iff(z)isanalyticfor|z-z
0
|<
R, |f(z)|=M,and f(z
0
)=0,then
(1.2) |f(z)|=
M
R
|z-z
0
|, |z-z
0
|<R,
Page 2


The Schwarz Lemma and
HyperbolicGeometry
This short chapter is devoted to the Schwarz lemma, which is a simple
consequence of the power series expansion and the maximum principle.
The Schwarz lemma is proved in Section 1, and it is used in Section 2
to determine the conformal self-maps of the unit disk. In Section 2 we
formulatetheSchwarzlemmatobeinvariantundertheconformalself-maps
oftheunitdisk, therebyobtainingPick’slemma. ThisleadsinSection3
tothehyperbolicmetricandhyperbolicgeometryoftheunitdisk.
1.TheSchwarzLemma
TheSchwarzlemmaiseasytoprove,yetithasfar-reachingconsequences.
Theorem (Schwarz Lemma). Letf(z)beanalyticfor|z|<1. Suppose
|f(z)|=1 for all |z|<1,and f(0)=0. Then
(1.1) |f(z)|=|z|, |z|<1.
Further, if equality holds in(1.1) at some pointz
0
=0, thenf(z)=?z for
some constant ? of unit modulus.
For the proof, we factor f(z)= zg(z), where g(z) is analytic, and we
applythemaximumprincipleto g(z). Letr<1. If |z|=r,then |g(z)|=
|f(z)|/r =1/r. Bythemaximumprinciple,|g(z)|=1/rforallzsatisfying
|z|= r.Ifwelet r ?1, we obtain |g(z)|=1 for all |z| <1. This yields
(1.1). If |f(z
0
)|= |z
0
|forsome z
0
=0,then |g(z
0
)|=1,andbythestrict
maximumprinciple, g(z)isconstant,say g(z)=?. Then f(z)=?z.
Ananalogousestimateholdsinanydisk. Iff(z)isanalyticfor|z-z
0
|<
R, |f(z)|=M,and f(z
0
)=0,then
(1.2) |f(z)|=
M
R
|z-z
0
|, |z-z
0
|<R,
withequalityonlywhen f(z)isamultipleof z-z
0
. Thiscanbeproved
directly, based on the factorization f(z)=(z -z
0
)g(z). It can also be
obtained from (1.1) by scaling in both the z-variable and the w-variable,
w=f(z),andbytranslatingthecenterofthediskto z
0
,asfollows. The
changeofvariable ? ?R?+z
0
mapstheunitdisk {|?|<1}ontothedisk
{|z-z
0
|<R}. If we de?ne h(?)= f(R?+z
0
)/M, then h(?) is analytic
ontheopenunitdiskandsatis?es |h(?)|=1and h(0)=0. Theestimate
|h(?)|=|?|becomes(1.2).
TheSchwarzlemmagivesanexplicitestimateforthe“modulusofcon-
tinuity”ofananalyticfunction. Itshowsthatauniformlyboundedfamily
of analytic functions is “equicontinuous” at each point. We will return
in Chapter XI to treat the ideas of equicontinuity and compactness for
familiesofanalyticfunctions.
Thereisanin?nitesimalversionoftheSchwarzlemma.
Theorem. Let f(z) be analytic for |z|<1.If |f(z)|=1 for |z|<1,and
f(0)=0, then
(1.3) |f

(0)|= 1,
with equality if and only if f(z)=?z for some constant ? with |?|=1.
Theestimate(1.3)followsbytaking z ?0intheSchwarzlemma. For
the case of equality, we consider the factorization f(z)= zg(z)usedin
the proof of the Schwarz lemma, and we observe that g(0) = f

(0). If
|f

(0)|=1, we then have |g(0)|=1, and we conclude as before from the
strictmaximumprinciplethat g(z)isconstant. Hence f(z)=?z.
Note that the estimate (1.3) is the same as the Cauchy estimate for
f

(0) derived in Section IV.4, without the hypothesis that f(0)=0. See
alsoExercise7.

Page 3


The Schwarz Lemma and
HyperbolicGeometry
This short chapter is devoted to the Schwarz lemma, which is a simple
consequence of the power series expansion and the maximum principle.
The Schwarz lemma is proved in Section 1, and it is used in Section 2
to determine the conformal self-maps of the unit disk. In Section 2 we
formulatetheSchwarzlemmatobeinvariantundertheconformalself-maps
oftheunitdisk, therebyobtainingPick’slemma. ThisleadsinSection3
tothehyperbolicmetricandhyperbolicgeometryoftheunitdisk.
1.TheSchwarzLemma
TheSchwarzlemmaiseasytoprove,yetithasfar-reachingconsequences.
Theorem (Schwarz Lemma). Letf(z)beanalyticfor|z|<1. Suppose
|f(z)|=1 for all |z|<1,and f(0)=0. Then
(1.1) |f(z)|=|z|, |z|<1.
Further, if equality holds in(1.1) at some pointz
0
=0, thenf(z)=?z for
some constant ? of unit modulus.
For the proof, we factor f(z)= zg(z), where g(z) is analytic, and we
applythemaximumprincipleto g(z). Letr<1. If |z|=r,then |g(z)|=
|f(z)|/r =1/r. Bythemaximumprinciple,|g(z)|=1/rforallzsatisfying
|z|= r.Ifwelet r ?1, we obtain |g(z)|=1 for all |z| <1. This yields
(1.1). If |f(z
0
)|= |z
0
|forsome z
0
=0,then |g(z
0
)|=1,andbythestrict
maximumprinciple, g(z)isconstant,say g(z)=?. Then f(z)=?z.
Ananalogousestimateholdsinanydisk. Iff(z)isanalyticfor|z-z
0
|<
R, |f(z)|=M,and f(z
0
)=0,then
(1.2) |f(z)|=
M
R
|z-z
0
|, |z-z
0
|<R,
withequalityonlywhen f(z)isamultipleof z-z
0
. Thiscanbeproved
directly, based on the factorization f(z)=(z -z
0
)g(z). It can also be
obtained from (1.1) by scaling in both the z-variable and the w-variable,
w=f(z),andbytranslatingthecenterofthediskto z
0
,asfollows. The
changeofvariable ? ?R?+z
0
mapstheunitdisk {|?|<1}ontothedisk
{|z-z
0
|<R}. If we de?ne h(?)= f(R?+z
0
)/M, then h(?) is analytic
ontheopenunitdiskandsatis?es |h(?)|=1and h(0)=0. Theestimate
|h(?)|=|?|becomes(1.2).
TheSchwarzlemmagivesanexplicitestimateforthe“modulusofcon-
tinuity”ofananalyticfunction. Itshowsthatauniformlyboundedfamily
of analytic functions is “equicontinuous” at each point. We will return
in Chapter XI to treat the ideas of equicontinuity and compactness for
familiesofanalyticfunctions.
Thereisanin?nitesimalversionoftheSchwarzlemma.
Theorem. Let f(z) be analytic for |z|<1.If |f(z)|=1 for |z|<1,and
f(0)=0, then
(1.3) |f

(0)|= 1,
with equality if and only if f(z)=?z for some constant ? with |?|=1.
Theestimate(1.3)followsbytaking z ?0intheSchwarzlemma. For
the case of equality, we consider the factorization f(z)= zg(z)usedin
the proof of the Schwarz lemma, and we observe that g(0) = f

(0). If
|f

(0)|=1, we then have |g(0)|=1, and we conclude as before from the
strictmaximumprinciplethat g(z)isconstant. Hence f(z)=?z.
Note that the estimate (1.3) is the same as the Cauchy estimate for
f

(0) derived in Section IV.4, without the hypothesis that f(0)=0. See
alsoExercise7.
 2.ConformalSelf-MapsoftheUnitDisk
WedenotebyDtheopenunitdiskinthecomplexplane,D= {|z| <1}.
A conformal self-map of the unit diskisananalyticfunctionfromD
to itself that is one-to-one and onto. The composition of two conformal
self-maps is again a conformal self-map, and the inverse of a conformal
self-map is a conformal self-map. The conformal self-maps form what is
called a “group,” with composition as the group operation. The group
identityistheidentitymap g(z)=z.
For ?xed angle ?, the rotation z ? e
i?
z is a conformal self-map ofD
that?xestheorigin,andthesearetheonlyconformalself-mapsthatleave0
?xed.
Lemma. Ifg(z)isaconformalself-mapoftheunitdiskDsuchthatg(0)=
0,theng(z)isarotation,thatis,g(z)=e
i?
z forsome?xed?,0=?=2p.
Toseethis,weapplytheSchwarzlemmatog(z)andtoitsinverse. Since
g(0)=0 and |g(z)| <1, the Schwarz lemma applies, and |g(z)|=|z|.If
we apply the Schwarz lemma also to g
-1
(w), we obtain |g
-1
(w)|=|w|,
which for w= g(z)becomes |z|=|g(z)|.Thus |g(z)|= |z|. Since g(z)/z
has constant modulus, it is constant. Hence g(z)= ?z for a unimodular
constant ?.
Theorem. The conformal self-maps of the open unit diskD are precisely
the fractional linear transformations of the form
(2.1) f(z)= e
i?
z-a
1-az
, |z|<1,
where a is complex, |a|<1,and0=?=2p.
De?neg(z)=(z-a)/(1-¯ az). Sinceg(z)isafractionallineartransfor-
mation, it is a conformal self-map of the extended complex plane, and it
mapscirclestocircles. From
|e
i?
-a| = |e
-i?
-¯ a| = |1-¯ ae
i?
|, 0=? =2p,
weseethat|g(z)|=1forz=e
i?
,sothatg(z)mapstheunitcircletoitself.
Since g(a)=0, g(z)mustmaptheopenunitdisktoitself. Consequently,
g(z)isaconformalself-mapoftheunitdisk,andsoisf(z)de?nedby(2.1).
Leth(z)beanarbitraryconformalself-mapofD,andseta=h
-1
(0). Then
h?g
-1
isaconformalself-mapofD,and(h?g
-1
)(0)=h(a)=0. Bythe
lemma, (h ?g
-1
)(w)= e
i?
w for some ?xed ?,0 = ? = 2p. Writing
w=g(z),weobtain h(z)=e
i?
g(z),and h(z)hastheform(2.1).
Page 4


The Schwarz Lemma and
HyperbolicGeometry
This short chapter is devoted to the Schwarz lemma, which is a simple
consequence of the power series expansion and the maximum principle.
The Schwarz lemma is proved in Section 1, and it is used in Section 2
to determine the conformal self-maps of the unit disk. In Section 2 we
formulatetheSchwarzlemmatobeinvariantundertheconformalself-maps
oftheunitdisk, therebyobtainingPick’slemma. ThisleadsinSection3
tothehyperbolicmetricandhyperbolicgeometryoftheunitdisk.
1.TheSchwarzLemma
TheSchwarzlemmaiseasytoprove,yetithasfar-reachingconsequences.
Theorem (Schwarz Lemma). Letf(z)beanalyticfor|z|<1. Suppose
|f(z)|=1 for all |z|<1,and f(0)=0. Then
(1.1) |f(z)|=|z|, |z|<1.
Further, if equality holds in(1.1) at some pointz
0
=0, thenf(z)=?z for
some constant ? of unit modulus.
For the proof, we factor f(z)= zg(z), where g(z) is analytic, and we
applythemaximumprincipleto g(z). Letr<1. If |z|=r,then |g(z)|=
|f(z)|/r =1/r. Bythemaximumprinciple,|g(z)|=1/rforallzsatisfying
|z|= r.Ifwelet r ?1, we obtain |g(z)|=1 for all |z| <1. This yields
(1.1). If |f(z
0
)|= |z
0
|forsome z
0
=0,then |g(z
0
)|=1,andbythestrict
maximumprinciple, g(z)isconstant,say g(z)=?. Then f(z)=?z.
Ananalogousestimateholdsinanydisk. Iff(z)isanalyticfor|z-z
0
|<
R, |f(z)|=M,and f(z
0
)=0,then
(1.2) |f(z)|=
M
R
|z-z
0
|, |z-z
0
|<R,
withequalityonlywhen f(z)isamultipleof z-z
0
. Thiscanbeproved
directly, based on the factorization f(z)=(z -z
0
)g(z). It can also be
obtained from (1.1) by scaling in both the z-variable and the w-variable,
w=f(z),andbytranslatingthecenterofthediskto z
0
,asfollows. The
changeofvariable ? ?R?+z
0
mapstheunitdisk {|?|<1}ontothedisk
{|z-z
0
|<R}. If we de?ne h(?)= f(R?+z
0
)/M, then h(?) is analytic
ontheopenunitdiskandsatis?es |h(?)|=1and h(0)=0. Theestimate
|h(?)|=|?|becomes(1.2).
TheSchwarzlemmagivesanexplicitestimateforthe“modulusofcon-
tinuity”ofananalyticfunction. Itshowsthatauniformlyboundedfamily
of analytic functions is “equicontinuous” at each point. We will return
in Chapter XI to treat the ideas of equicontinuity and compactness for
familiesofanalyticfunctions.
Thereisanin?nitesimalversionoftheSchwarzlemma.
Theorem. Let f(z) be analytic for |z|<1.If |f(z)|=1 for |z|<1,and
f(0)=0, then
(1.3) |f

(0)|= 1,
with equality if and only if f(z)=?z for some constant ? with |?|=1.
Theestimate(1.3)followsbytaking z ?0intheSchwarzlemma. For
the case of equality, we consider the factorization f(z)= zg(z)usedin
the proof of the Schwarz lemma, and we observe that g(0) = f

(0). If
|f

(0)|=1, we then have |g(0)|=1, and we conclude as before from the
strictmaximumprinciplethat g(z)isconstant. Hence f(z)=?z.
Note that the estimate (1.3) is the same as the Cauchy estimate for
f

(0) derived in Section IV.4, without the hypothesis that f(0)=0. See
alsoExercise7.
 2.ConformalSelf-MapsoftheUnitDisk
WedenotebyDtheopenunitdiskinthecomplexplane,D= {|z| <1}.
A conformal self-map of the unit diskisananalyticfunctionfromD
to itself that is one-to-one and onto. The composition of two conformal
self-maps is again a conformal self-map, and the inverse of a conformal
self-map is a conformal self-map. The conformal self-maps form what is
called a “group,” with composition as the group operation. The group
identityistheidentitymap g(z)=z.
For ?xed angle ?, the rotation z ? e
i?
z is a conformal self-map ofD
that?xestheorigin,andthesearetheonlyconformalself-mapsthatleave0
?xed.
Lemma. Ifg(z)isaconformalself-mapoftheunitdiskDsuchthatg(0)=
0,theng(z)isarotation,thatis,g(z)=e
i?
z forsome?xed?,0=?=2p.
Toseethis,weapplytheSchwarzlemmatog(z)andtoitsinverse. Since
g(0)=0 and |g(z)| <1, the Schwarz lemma applies, and |g(z)|=|z|.If
we apply the Schwarz lemma also to g
-1
(w), we obtain |g
-1
(w)|=|w|,
which for w= g(z)becomes |z|=|g(z)|.Thus |g(z)|= |z|. Since g(z)/z
has constant modulus, it is constant. Hence g(z)= ?z for a unimodular
constant ?.
Theorem. The conformal self-maps of the open unit diskD are precisely
the fractional linear transformations of the form
(2.1) f(z)= e
i?
z-a
1-az
, |z|<1,
where a is complex, |a|<1,and0=?=2p.
De?neg(z)=(z-a)/(1-¯ az). Sinceg(z)isafractionallineartransfor-
mation, it is a conformal self-map of the extended complex plane, and it
mapscirclestocircles. From
|e
i?
-a| = |e
-i?
-¯ a| = |1-¯ ae
i?
|, 0=? =2p,
weseethat|g(z)|=1forz=e
i?
,sothatg(z)mapstheunitcircletoitself.
Since g(a)=0, g(z)mustmaptheopenunitdisktoitself. Consequently,
g(z)isaconformalself-mapoftheunitdisk,andsoisf(z)de?nedby(2.1).
Leth(z)beanarbitraryconformalself-mapofD,andseta=h
-1
(0). Then
h?g
-1
isaconformalself-mapofD,and(h?g
-1
)(0)=h(a)=0. Bythe
lemma, (h ?g
-1
)(w)= e
i?
w for some ?xed ?,0 = ? = 2p. Writing
w=g(z),weobtain h(z)=e
i?
g(z),and h(z)hastheform(2.1).
The parameters a and e
i?
are uniquely determined by the conformal
self-map f(z)ofD. Theparameter ais f
-1
(0),andsince
f

(z)= e
i?
1-|a|
2
(1-¯ az)
2
, |z|<1,
theparameter?isuniquelyspeci?ed(modulo2p)astheargumentoff

(0).
Thusthereisaone-to-onecorrespondencebetweenpointsoftheparameter
spaceD×?Dandconformalself-mapsoftheopenunitdisk.
NextwetakeagiantstepbyprovingaformoftheSchwarzlemmathat
isinvariantunderconformalself-mapsoftheopenunitdisk.
Theorem (Pick’s Lemma). If f(z) is analytic and satis?es |f(z)| < 1
for |z|<1, then
(2.2) |f

(z)|=
1-|f(z)|
2
1-|z|
2
, |z|<1.
Iff(z)isaconformalself-mapofD,thenequalityholdsin(2.2);otherwise,
there is strict inequality for all |z|<1.
To prove (2.2), our strategy is to transport z and f(z)to0usingcon-
formal self-maps, and to apply the Schwarz lemma to the resulting com-
position. Fix z
0
?Dandset w
0
=f(z
0
). Let g(z)and h(z)beconformal
self-mapsofDmapping0to z
0
and w
0
to0,respectively,say
g(z)=
z+z
0
1+z
0
z
,h(w)=
w-w
0
1-w
0
w
.
Then h?f ?gmaps0to0. Theestimate(1.3)andthechainruleyield
(2.3) |(h?f ?g)

(0)| = |h

(w
0
)f

(z
0
)g

(0)|= 1,
hence |f

(z
0
)|= 1/|g

(0)||h

(w
0
)|. Substituting g

(0) = 1-|z
0
|
2
and
h

(w
0
)=1/(1-|w
0
|
2
),weobtain(2.2).
O
O
z
0
w
0
gf h
Iff(z)isaconformalself-mapofD,thensoish?f?g,sowehaveequality
in (2.3), which yields equality in (2.2). Conversely, suppose that f(z)is
ananalyticfunctionfromDtoDsuchthatequalityholdsin(2.2)atone
point z
0
. Thenthecalculationsabovegive |(h?f ?g)

(0)|=1. According
toSection1, h?f ?gismultiplicationbyaunimodularconstant,hencea
conformalself-mapofD. Composingbyh
-1
ontheleftandbyg
-1
onthe
right,weconcludethat f isaconformalself-mapofD.