Internat. J. Math. & Math. Scl.
VOL. 15 NO. 3 (1992) 441-448
441
A MAJORANT PROBLEM
RONEN PERETZ
Department of Mathematics
University of Michigan
Ann Arbor, MI 48109
(Received September 25, 1990)
Let
ABSTRACT.
f(z)
k=O
e
Inflnlte complex vector
k
induces a
0,1,2
akzk
a
o
be analytlc in the unlt disc.
0
(eo,et,e2 )
function re(Z)
such that
k=O
akekZk
lekl
Any
1,
whlch is still analytic
In the unit disc.
In this paper we study the problem of maximizing the p-means:
e
over all possible vectors
and for values of
close to
r
0
p<2.
It is proved that a maxlmlzlng function Is
and that
r
-laoi
f,{z}
+
and for all
.
laklZ
k=l
k
could be taken to be any positive number which Is smaller than
the radius of the largest disc centered at the orlgln which can be Inscrlbed
in the zero sets of
f
Thls problem is originated by a well known
maJorant
problem for Fourier coefficients that was studied by Hardy and Llttlewood.
One consequence of our paper is that for
for the Hardy-Llttlewood problem should be
p < 2
-[ao[
the extremal function
r. laklz
+
k=1
We also give some appllcatlons to derive some sharp Inequalltles for the
classes of Schllcht functions and of functions of positive real part.
Key Words and Phrases:
Fourier coefficients, analytic continuation,
Schllcht functions, functions of positive real part
1980 AMS SUBJECT CLASSIFICATION CODES:
1.
42A05, 30BlO, 30D55
DEFINITIONS AND NOTATIONS.
{z e C
We will denote D{a,r)
Iz-al < r} and
H{U} is the set of all functions analytic In U.
If
f(z}
kffiO
akzk
complex vector such that
e H(U)
lekl
and If
I, k
e
(eo,el,e2
0,1,2
U
)
D(0,1).
Is an infinite
then we will define
442
R. PERETZ
fO (z)
For fixed
p
n e Z
If
(Usually
f
R
by the formuIa:
J
(0)
r
then we wilI denote:
p
and
z e U.
p
and
{fe}
functional on
laklOk zk,
E
k:O
such that
(f)
min(1,min((lzl
Akzk
H(U)
F(z)
If(retO)lP e -i nodO.
will be fixed so we will shorten
f
z,
and satisfies
k=O
then we will say that
we will define a
II.
fs(reiO),PdO.
cf, p,n (r)
The radius of the zero-free region of
If
0 s r < 1,
and
c (r)
c_
(r)).
I ,p0n
n
will be defined by
o,
fc-)
lak[
-<
Ak.
0,1,2
k
F majorizes f and write f
F.
Some familiar families of analytic functions in U are:
S
f(O)
O, f’ (0)
(f H(U)
1, f is univalent in
H(U)
f(O)
P
Re{f
(z)}
> 0 V z e U}.
1,
{f
2.
U}
THE MAIN RESULTS.
f(z)
If
k=m
Let
k
s
f=(z)
k’
akzk
e H(U)
-I +
k=l
k
a
m
Denote
0
k zk, f(z)
-lam Jzm+k=l lam+kl m+k
f (z}
-I +
k=l
k zk
H(U)
satisfy
1,2,3
The following theorems will be proved in the paper.
For every p < 2
and every
0 s r < 6(f ), we have
o.f(rei’)IPd# If1(rei’)lPd,.
s
The inequality is usually not valid for
For every
p < 2
For every p < 0
n =I,2,3
THEOREM D.
n
1,2,3
we
and every
0 s r <
and
0 s r <
6(fl),
(fB),
we have
and for all
haveloJf(rei#)JPe-in#d#l }f,(rei,)jPe-in#d#.
For every p < 0 and every 0 s r <
we have
o
3.
r >
f,(re i#)’ Pe-in#d#
s
o. f,(rei#).
8(fB)
and for all
Pe- in#d#.
A MAJORANT PROBLEM
TO IIARI}Y AND L ITTLEWOOD.
In this paper we solve the following extremal
problem:
max J
(8)
for every p < 2 and every 0
p,r
We also find the extremal directions e’ for
which J
(8’)
p,r
(The compactness of
{re}
in the topology of uniform
max J
O
p.r
(e].
convergence on compact
MAJORANT PROBLEM
U
subsets of
443
A majorant
guarantees the existence of maximizing functions).
problem for Fourier coefficients, that was studied by Hardy and Littlewood,
goes as follows:
p
If
P
P
Are there any other values of
ffl
F e L
and
p,
f e L
then
p
and,
FII
f
in fact,
F
f
is an even integer and if
for which an inequality of the type
p
B F
F ?
holds whenever f
P
P
P
In his paper [2] from 1975, Harold Shapiro characterizes the
class of all smooth functions
2
o
l(elldo
n
p
function unless
@
R
on
u
+
Jo @..f(eit}..
I)
dtCI
for which
x
he ddu h
x
p
o
uoh
is an even integer.
One consequence of our paper in connection with this problem is that for
p < 2
new
the majorant function of Hardy and Littlewood should be replaced by a
maJorant
function:
-1o{
4.
PRELIMINARY
(a}
If
and every
/
k=l
instead of
k=O
{aklzk
0).
(a
REM/ugKS.
H(u) and f
f,g
r <
0
{akizk
g,
then for every
p
2k,
0,1,2
k
we have
o
lf(rei)IPd
o
lg(rei)lPd.
Even more is true:
.f(reiO}.Pe-inbd@
l.g(rei}.Pe-inbdO,
(In view of the fact that for every
z
k(z}
f
If{n)(rei@}12kd
Ih(n)(rei@)12kd
n,k
(b) If
0, I,2
and
mln
and in particular
(c)
J
{e’}
6(f)
p v O,
If
max J
e
p,r
f1(rei)l
{e)
0
fS
hP
r < I.)
0
H(U), a
0
k=O
then (by the triangle inequality} we have
o-2
we have
l+z
1-z
t(z}
oIk{n)(rei@}12kd@,
olt(n)(rei@)12kd@,
akzk
f(z)
h e P
and every
h
(l_z)2
we can deduce that
for all
f e S
O, 1,2
n
-< rain
o--2
and if
f(rei)l
-[ao[
fl(z}
0
r <
+
.
k=l
(f)
-< Sff).
r <
and e’=
satisfies
2
and if we have the Fourier expansions:
$(fx)
(8o,8,e
444
R. PERETZ
k=O
p-2
Ife,(Zll
then for all
0,1,2
k
bkzk
fe,(z)
arg(bk)
we have
{arg(k), arg(Ok)
8J
see this we use the fact that for all
Al AltCLYTIC EXTIOI FOR
S.
Let
PIOPOSlTIO 1.
(a)
If
Cf,p,n
(b}
(r)
If
(a)
If
gn
. atzt
.
It is enough to consider the case
f(z
then
)k
t=O
homogeneous polynomlal of degree
l
i--[P6[ 1/6= I-
Co{r)
extension of
(b)
Let
8Ch)
H{U)
h
8(f)
[a611/6.
k
r < 8(f).
D(0,8(f)).
0
F.
6=0
R
Hence
(h(zei@))
But
II---J JdtJ2J
point on
Izl
-
t2
1-
]d61
1/6
8(f) (since
at)z6
where
is an
P6
coefficients such
E
6=0
a6)]2z26.
lP6(a0
that
zmh(z),
.
6--0
IdtlZz26,
h(zei@)h(ei@)) is an
(h(ze I@)h(e I@)) P/2d@
d6(zei@}t,
Izl
Izl
Co{r)
< (f)
for
then by
< ,scf:.
,h(zei@)] p/2
--(- because
Since
p/2
and clearly
p/2
U.
z
o
=-rmpgo{r)
go{Z}
Parseval’s equality we have
go (z)
0.
f(z)
and
D(0,8{f))
Finally, if
n
with positive integer
we have for each fixed
is an analytic function in
for every
gn H(D(O,@}).
p6(ao
go(Z)
h{0)
satisfy
O,
z
By Parseval’s equality the desired analytic
is given by
analytic function in
0
gn(r)
such that
rmpgn{r} Cf,p,n(r}
such that
t=O
that
H(U)
} and if f has a zero of order m at
gn H(D(0,8{f)))
f(z)
=0.)
k
Z.
n
and
0 <- r < 8(f}. }4oreover for each } > 8(f} we have
Proof.
80
n
) then there Is a
{2klk=0,1,2
then there is a
p,r
0 -< r < 1.
for every
p
we have
c (r).
H(U)
f
(2klk=O,1,2
p
0,1,2
k
(To
}.
/
has a singular
Is not a nonnegatlve integer).
We remark that in (b) we have proved the representatlon
6.
o
)[
.f(rel@-.Pd@
r
mp
E
6=0
A SOLUTION FOR THE PROBLEN max J
Let
f(z}
e
1 +
F. k zk
k=l
H(U)
Idtl z2t,
p,r
(O)
satisfy
0
r < 8(f).
FOR p < 2, 0
0
f. Let
O
r < 8(f).
(10 O2
’
..)
MAJORANT PROBLEM
lekl
be an infinite complex vector such that
<
(fe(z))q
and
LEIIA 2.
dek zk, Izl
ldekl
Denote
Proof.
<
e
then for every
ldkl,
we have
then
(fe(z))q
(l+x)
q,
and
this implies
Hence we have a representation of the form:
det
where
’j
e R
=
n +... +n
and depends onIy on
j(lel )nl
in
(iz
monomlals of the form
(-!
)l-n et
ntJ
j
e).
(not on
q
zz
z
the coefficient of
3z
.)n
3
Yn,t
If
then
+ i
+
et n I1
(2)
n
(1
is
is a sum of
Y
t.
t}
n
But
(qn)
sgn
and by (1) we have
(-1)n
q<O
nl Cqn) Yn’
d,
Hence using the representatlon (2) we have
(3)
+nj
n lj+...
and the conclusion therefore follows.
PROOF OF THEOREN A.
(fez})p/2=
We can assume
E dkzk, I1
< Cf)
By the representation In Proposition
2
y
Le..a
-
<
th
2
O,
a
o
> O, a
(ritZ))p/2=
k=O
Let
O.
dzk, I1
<
(f}.
we have
Z
k=O
lf1(reillPd
m
and
k=O
lf(rei)lPd
E
Idk 12 r2k,
0
IdOl r2k,
Idkl Idl,
r < 6(f}
0 s r <
k=O
e hve
k
0,,-
nd e
obtain the desired inequality.
For the final part of the theorem we consider the following example:
Fix
0 <
We have
<
and define
gO(z]
(g)
(V- I).
Denote
Gl(r
Then
and
o
z
d@
+(R).
+ 2z
e
iezz, g1(z]
o
Ge{r
g (reiq)
G(6(g)) Ge(a(ge)
lim 6(ge)
6(gI), we
e_)o
q
1,2,3
k
kek zk, Izl < 8(fe ),
x
Suppose
1,2,3
1, k
(fe ).
(1,1,-1,1,-1,1,-1
@
If
k--O
445
Since
6(g)
<
can do the following:
6(ge)
2
+ 2z
z
d@
ge(rei)
for all
0
e
2
446
R. PERETZ
For every r
So
Gl(r o)
<
>
(g)
G
while
(r
eo
eo
tlcular we can choose
dz k, Izl
<
Let
{f (z) )p/2
<
{ge-
GOo (to)
k=O
to.
o
dk zk’
By by formula
0,1,2
idkl EIjI.11j...
we have
<
G1(ro
kj
done.
it suffices to prove
(2) In the proof of Lemma 2
EIjIIj...
idol
f ,(f (z) )p/2
<
z
By representation in Proposition
(f).
Idl, k
Idkl
o)
(gl)
0 such that
o
could be made as large as we want and in par-
so that
PROOF OF IIIEORE B.
k=O
that
eo >
we can choose a
kj
and we are
o
The combination of Theorems A and B gives us, for example, the
following:
COROLLARY 3.
modulus
(a)
Let
p < 2
and let
be an Inflnlte complex vector of
8
components.
f e P
For every
and every
r <
0
we have
3
d.
l-{re i@)
Moreover for every
n
there is a
0
n
> 0
such that for all 0
r <
n
we have
}/{n)(rei@)
{n)(rei@)IPd@
(b}
For every
f e S
and every
r <
0
o"
2.n!
lPd@ (t(z)
vr)
(2
we have
(rel’)
If
is the minimal modulus of the zeros of
every
0 s r <
i
P
2(1-z}
+ z
3
then for
we have
o
Flnally for every
there Is a
n
on
l+(re i@)
> 0
P
2
such that for all
0
r <
we have
Je (reilJPd
Let
COROLLARY 4.
Jk (n)(reiqb)
2n.n!
ak zk
H(U)
f(z)
.
k=O
fCz)
-IoI
+
(0)
lakl zk,
f (z)’)
then for every
p
0
IPdq
a
o
(k(z)
0
f(rel’) jpd’ s
and
and for all
we have
flcO)flcrei’)
o" 2f-(rel,)_fl
(0)
p
z
(l-z)
0 s r < mln
n
MAJORANT PROBLEM
PROOF.
Since
f{z)
0
"- r <
rain
(h(z))
usual let
.
p/2
:Imk
z
I=kl -Ikl,
and
zmh(z)
f(z)
k=O
dkzk,
Izl
where
< (f).
lklZk)_
+
f (Z)
fl{rei)
H(U)
h
and
we obtain
1,2,3
k
fl(O
f(ret
2
(5(fl)’ 5(f (0) fl (-’"
Let
PROOF OF THEOREM C.
k
D(O,8(fl)
both are analytic in
for every
+
a
447
-Pdo
O.
h(O)
and
As
Then, just as in Proposition 1,
we have
y [f(rei)[Pe-ind
For
dk+nkr2k)rmP,
2(
0 -< r < (f).
k=O
small enough we have, as in Lemma 2:
]z]
(1- z- z
-
3
z
3
)q
+
2]z
[-q2 + q (q-l)
2!
z
(22) q(q-1)q(q-2)=
3!
(-ql)z
+
f-q=3
+ q(q-1)
2!
+
[_q=4
+
q(q-1)
2
+
q(q-1)q(q-2)
(21+)
2a
3
(3
+ q(q-1)q(q-2)(q-3)
4!
and if
for all
d
k
z
is the coefficient of
k
0,1,2
en consideri
if we choose
Ire(re
max
e
k
then
maxldkl
arg(=j)
)I
a,4]z 4
+
is solved simultaneously
J
O,
12.)
1,2,3
when
q < I.
we
solve simultaneously
aXldkl
let us check
0
q <
then
then
sgn(dk)
choice
arg(j)
follows,
and
arg(dk+nk "
arg(dk+nk}
sgn(do)
+I,
O,
k
+1
for the choice
while
0, I,2
J
const.
1,2,3
sgn(dk)
O, 1,2
k
J
arg(j)=
O,
-1,
1,2,3,’
k
1,2,3
If
If
q < 0
q < O, the
This proves that when
solves (I) and the conclusion
u
The proof of Theorem D goes along these lines in analogy to the proof
of Theorem B.
[11
[2]
G.H. Hardy and J.E. Llttlewood, "Notes on the theory of series (XIX):
a problem concernlng majorants of Fourler serles", Quart. J. Hath.
(Oxford) 6 (1935), 304-315.
H.S. Shapiro, "Majorant problems for Fourier coefficients", Quart. J.
Math. (Oxford) {2}, 26 (1975), 9-18.