C OMPOSITIO M ATHEMATICA
G. A. C AMERA J. G IMENEZ
The nonlinear superposition operator acting on Bergman spaces
Compositio Mathematica, tome 93, n
o1 (1994), p. 23-35
<http://www.numdam.org/item?id=CM_1994__93_1_23_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
http://www.numdam.org/
The nonlinear superposition operator acting
onBergman spaces
G. A.
CAMERA
& J.GIMÉNEZ
Department of Mathematics, I. v.I. C., Caracas, Venezuela Received 30 August 1992; accepted in revised form 20 June 1993
1. Introduction
Let A denote the unit disc
{z: Izi 1}
in thecomplex plane
and letH(0)
denote the space of
analytic
functions in A with thetopology
of the uniform convergence incompact
subsets of A. Given a functionf :
C - C weassociate to it the
operator F f
definedby
This
operator
is known as the autonomous nonlinearsuperposition (or composition) operator [1].
If A and B are linearsubspaces
ofH(I1)
andFf(u)
E B whenever u E A we shall say thatFf
acts from A to B. It is easyto see that if
F f
acts fromH (A)
toH(0),
thenf
must be an entire functionand
conversely.
In this case mere actionimplies
thecontinuity
and theboundedness of the
operator [2].
That mere actionimplies continuity
hasalready
beenproved
for various spaces of realfunctions,
for instance Lp spaces[6]
and Sobolev spaces[7]. Necessary
and sufficient conditions have beengiven
in[2]
in order thatFf
acts from HP toHq,
0p, q
+ oo, where HPdenotes the classical
Hardy
space in the unit disc. It is also true in this casethat mere action
implies continuity [2].
If N denotes the Nevanlinna space of functions inH (0)
of bounded characteristic then the actions fromupq H"
to N and from N to N have been studied in[3].
In this note we shall consider the
problem
of action andcontinuity
between the
Bergman space Bp
definedby
The space
B 00
is the usual one of boundedanalytic
functions in A. Thetopology
in these spaces isgiven by
the metric induced(when p > 1) by
If p 1
thetopology
inducedby
the metricII u I I p
is used. We also consider the action betweenBp
and theHardy
space Hq and vice versa. Forfunctions in Hp we use the standard notation
The
symbol
BN(which
stands forBergman-Nevanlinna)
shall denote theset of functions u in
H(0)
such thatClearly
HP ciBp
andBp
c BN for all p.Finally,
westudy
the actionbetween
Hardy
functions andBergman-Nevanlinna
functions.We would like to thank the referee for his
helpful
comments.2. The action in
B,
We shall need the
following
lemma.Proof.
This is an easy consequence of thesubharmonicity
ofjulp.
Next we are
ready
to prove thefollowing
result. In what follows thesymbol "[s]"
denotes theinteger part
of s.THEOREM 1. Let
f :
C - C be an entirefunction.
Thenh’f
actsfrom Bp
toBq,
0 p,q x
ooif
andonly if f is
apolynomial of degree
less than orequal
Proof.
Iff
is apolynomial
ofdegree n p
thenfOUEBq, VUEBp.
Inq
fact,
it isenough
to see that ifk p ,k c N, then uk E Bq.
This is true sinceq
Next,
we assume thatFf
acts fromBp
toBq,
0 p, q oo. Let s > 0 and define(z) - 1 - - 1 2/P+E
.Clearly uÉl belongs
toBp .
Thereforef 0 Ul) E Bq
andby
Lemma 1 one can writeSet wi =
u. (1)(z).
Assume first that p 1 and take e 1 - p. Then the range set ofu(l)
isC B0.
Given wic- C B0
let z G A such that wi =U(l)(z)
andThus,
from(2.1)
weget
Ifwi
is such that!wJ
>el 12(p + E)
then z1 =W)+t:/2( -’TC Arg w1 n)
satis-fies
Re z 1
>e 1 4 cos( - n12(p
+8))
> 0. Infact,
sinceArg wi a -n
then((p
+e)/2) Arg Wi à - 7t/2W
+8).
On the other handHence,
iflw,l
>el / 2 (p +,,)
then Re zi >e’/’ cos(7r/2(p
+8))
> 0.Therefore,
one can find a
positive
constant c(depending
on 8 andp)
such thatIf we use this
inequality
in(2.2)
we obtainfor all wi such
that IWll
>el 12(p +").
Thusf
is apolynomial
ofdegree
lessthan or
equal to p
+Elq. By letting
e --> 0 we obtain the desired result whenp 1.
Now,
we assume thatp >,
1. LetIf w 1
ESI
we choose z 1 such that w 1 =zi/p+£ with 20137T/4 Arg
zn14.
Thus
(lw)+£/2
+11
-wi +E2 - 21)2/q >
c >0,
for some constant c. Combin-ing
thisinequality
with(2.2)
we obtainLet
and
Given
W2 E 52
we choose Z2 such thatIZ21
> 1 and w2 =z/p+¿: ein/p+¿:
with-
n/4 x Arg z2 n14.
From(2.4)
we obtainSince
Hence,
from(2.5)
we obtainBy repeating
the sameargument n times,
where n is such thatA
=CBtJ?= Si,
we obtainwhere ci
depends on f,
E and q. This proves thatf
is an entire function of order atmost p
+elq. Letting
e tend to zerogives
the desired result forp > 1.
3.
Continuity of Ff
We shall prove in this section that if
F f
acts fromBp
toBq
then it isnecessarily
continuous. We also prove localLipschitzness.
First of all let us prove the
following
lemma.Proof.
Theproof is
similar to theanalogue
lemmagiven
in[2].
Wegive
it here somewhat
simplified.
The case n = 1 is obvious. Let us assume thatn > 1. The functions
uk
and Unbelong
toB9.
Infact,
sinceiiq x p
then forevery
u E B p
we haveOn the other
hand,
Now we use Hôlder
inequality
and obtain
where c is a constant.
Again by
a refined version of Hôlderinequality
weget
thatUk-l-IUI E Bp/n- b
=0, 1,...,
n - 1 andThis
inequality implies
that all summands on theright-hand
side of(3.1 )
arebounded
(for
allk).
Henceuk
-+ u" inBq
asrequired.
THEOREM 2.
If Ff acts from BP to Bq
then it isnecessarily continuous,
bounded andlocally Lipschitz.
Proof. Since-Ff
actsfrom Bp3
toBq then, by
Theorem1, f
is apolynomial
ofdegree n p .
Setf (z)
=an Z"
+ - - - + a,, LetMjz)
--+u (z),
as k --+ oo, inB...
q
Then
Thus
by
Lemma 2. The boundedness ofFf
comes from theinequality
which can be deduced from
Il u, 1 Il B, Il u Il ",, B
for allu c- B p
andall n q LJ
In order to prove that
F f
islocally lipschitz
we must see that if u,v E B(o, R) c Bp
then there exists a constantC = C(p, q, R, f)
such thatIlFf (U) - FF(V)II B, 1 CIIU - VIIB,-
On the one
hand,
On the other
hand,
in the same way that we deduced(3.1)
and(3.2)
we obtain4. The action
from Bp
to theBergman-Nevanlinna
space TheBergman-Nevanlinna
space is definedby
It is easy to see that
Bp
cBN, Vp
> 0. Thefollowing
result is an easy consequence of Theorem 1.COROLLARY 1.
Let f be
an entirefunction
such thatFf
actsfrom
BN toBq,
0q
oo.Then f is
constant.Proof.
Inparticular Ff
acts fromBp
toBq,
for all p > 0. Then from Theorem 1 we conclude thatf
is apolynomial
ofdegree
atmost q -q-
Taking
p lessthan q
one obtains the desired conclusion.LEMMA 3.
If u c- BN
thenThis lemma is a consequence of the
subharmonicity
oflog’lul.
THEOREM 3.
Let f be
anentire function.
ThenFf acts from U,,B q
to BN(0
poo ) if
andonly if f
has order at most p.Proof.
Assume thatF f
acts fromUpqBq
to BN. Let B > 0. The functionsbelong
toUpq Bq Hence, fOUt; E BN
and so,by
Lemma 3If p 1 we take e so
that p
+ e 1. Then the rangeof uE
isCBO. Thus, given
wleCB0
we take z c- A such thatFrom
(4.1)
weget
Therefore,
there is apositive
constant C such thatCombining (4.2)
and(4.3)
we obtainfor a suitable
positive
constantCi
and all w 1 suchthat lwll
>el / 2 (p +,E)
Thisconclude that
f
is of order at most p.Next,
we assume thatp >
1. In this case we argue as in Theorem 1 and obtain from(4.2)
thatfor all w outside a ball. Hence
f
is an entire function of order at most p.Let us suppose now that
f
is an entire function of order less than orequal
to p andUE UpqBq.
Take E > 0 such thatuEBp+E.
There is aconstant C so that
To prove that
fou
E BN we writesince u E
Bp+t:.
T H EO R E M 4. Let
f
be an entirefunction of
order less than p orof
order pand finite
type(0
poo).
ThenF f acts from Bp
to BN.Proof.
We may assume thatf
is of order p and finitetype u - ô
> 0.Hence,
there is a constant C such thatIf u c- Bp then
as
required.
5.
Transforming Hardy
functions intoBergman
functions and vice versaWe shall
begin by stating
thefollowing
classical resultsby Hardy
andLittlewood
[5].
LEMMA 4. Let u be
analytic in 0
andThen there is a constant K =
K (p, f3)
such thatA
proof
of this result can be found in[4,
p.84].
LEMMA 5.
If
0 pq 1
oo,u c- Hl’, Â >,
p, and a =1/p - 1/ql
1 thenThe reader can find a
proof
of this result in[4,
p.87].
The next result shows that one cannot transform
Bergman
functions intoHardy
functionsby
means of nonlinearsuperposition.
In case p = oo it is trivial thatFf
acts forany f
THEOREM 5.
Let f be
anentire function. Ifp #
oo thenF f acts from Bp
toHq
if and only if f is
constant.Proof.
IfFf
acts fromBp
to Hqthen, by
Theorem1, f
is apolynomial.
Now we
get
the desired conclusionby noting
that for a non-constantpolynomial f
it is not true thatfou
EHq,
b’u EBp.
If this were true then thezeros of all
Bergman
functions would have tosatisfy
the Blaschke condi-tion,
and this is false.THEOREM 6.
Let f be an entire function.
ThenF f acts from
HP toBq if and only if f is a polynomial of degree
at most2p
.If p
= oo thenF f
acts from HP toBq
forany f
In theproof
of thistheorem we shall rule out this case.
COROLLARY 2. The
operator F f
actsfrom
HP toBp if
andonly if f is
apolynomial of degree
one or two.less than or
equal
to2p
can be done as in Theorem 1. Let us assume now thatf
is apolynomial
ofdegree n 2p
We shall prove that ifq
U G HP then
un E Bq.
Let us suppose, first ofall,
that n’q.Then 2p
qNow, using
Lemma 4 withfi
=0,
ql =2p
we obtainfor some constant C.
Combining
thisinequality
with(5.1)
onegets
Thus
u" E Bq,
asrequired.
Next we assume that n =
2p/q.
In this case we use Lemma 5 with q 1 =2p,
oc =1/2p,
and =2p
to conclude thatas
required.
6. The action from IF to BN
If p = oo and
f
is any entire function thenFf
acts from H°° to H°° andconsequently
it acts from HOO to BN.When p
oo we have thefollowing
result.
THEOREM 7. Let f : C - C. Then Ff acts from UpqHq to BN if
andonly
if f is
an entirefunction of
order at most2p.
Proof.
Let us assume thatF f
acts fromUpqHq
to BN.Clearly f
mustbe an entire function. On the other
hand, given
e >0,
the functionbelongs
toup q Hq. Using
these functions and Lemma 3 andarguing
as inTheorem 1 we deduce
that f
has order at most2p. Conversely,
let ussuppose that
f
has order at most2p.
Let U Eu P qHq
and e > 0 such thatu c- HP".
Next we take a constant C such thatand use this
inequality
toget
since
u2 E Bp+E
in view ofCorollary
2.As a
corollary
we haveCOROLLARY 3.
If u c- N, u(z) * 0 in A,
andlog u(z) * 0 in A,
thene(log u)2 - ec-BN, Ve, 0 e 2.
Proof.
For e = 0 the result breaks down as it is shownby
theexample
To prove the
corollary
weproceed
as follows. Since u does not vanish inA then
log u
isanalytic
there. Moreoverlog lui E hl,
wherehl
is the space of harmonic functions in A whichsatisfy
theRiesz-Herglotz representation.
This can be seen from the
following
relationsand the fact that u E N. Then
log u E Hp, b’p
1. Thus(log U)2-t:
EU112qHq,
7, e(log E BN,
asrequired.
Finally,
we haveTHEOREM 8. Let
f
be an entirefunction of
order less than2p
orof
order2p and finite type.
ThenFf acts from
HP to BN.Proof.
We may assume thatf
is of order2p
and finitetype
u - l5 > 0. There is a constant C such thatIf u E HP we
get
from the lastinequality
since
U2
eBp
in view ofCorollary
2.7. Some open
questions
We finish this article
by posing
somequestions. (1)
IfF f
acts fromB p
toBN
then,
inparticular,
it acts fromU pq Bq to
BNand, by
Theorem3,
ithas order at most p. In case that
f
has order p is it true thatf
has finitetype? (2)
One may ask thecorresponding question
for the action between HP and BN.References
1. Appel, J., Zabrejko, P. P.: Nonlinear Superposition Operators. Cambridge University Press, 1990.
2. Cámera, G. A.: The autonomous nonlinear superposition operator in Hp (to appear).
3. Cámera, G. A.: On entire functions acting via nonlinear superposition on Nevanlinna
functions (to appear).
4. Duren, P.: Theory of Hp Spaces. Academic Press, New York, 1970.
5. Hardy, G. H. and Littlewood, J. E.: Some properties of fractional integrals. II. Math. Z.
34 (1932) 403-439.
6. Krasnosel’skij, M. A.: On the continuity of the operator Fu(x) = f(x, u(x)) (Russian).
Doklady Akad. Nauk SSSR 77(2) (1951) 185-188.
7. Marcus, M. and Mizel, V. J.: Every superposition operator mapping one Sobolev space into another is continuous. J. Funct. Anal. 33 (1979) 217-229.