FRITZ CARLSON'S INEQUALITY AND ITS APPLICATION
AMIR KAMALY
Abstract
A Carlson-type inequality is proved and it is applied to show a Babenko-Beckner type of the Hausdorff-Young inequality onn-dimensional torus.
Introduction
Fritz Carlson's inequality (1934) states, [4], that X1
n1
an<
p X1
n1
an2
!14 X1
n1
n2an2
!14
holds for any positive sequence an1n1 and not allan are 0. Let an:bf n, for a periodic functionf. Then, there can be equality only iff is a multiple off0, and therefor an exponential functionC0ebx. This is plainly impossible, [7].
Note that the sumsP1
n1an2 andP1
n1n2an2 are supposed to be finite.
The corresponding integral inequality, [4], [7], is Z 1
0 f xdx
p Z 1
0 f2 xdx
14 Z 1
0 x2f2 xdx
14
: Here there is equality whenf x:abx1 2, for any positivea;b.
Forf 2A Tandbf 0 0, the other expression of Carlson's inequality is kfkA TC
kfk2kf0k2 12
: 1
HerekfkA T:P
m2Zjbf mjandA Tis the space of continuous functions on T having an absolutely convergent Fourier series. The variety of the constantCin 1depends on the definitions ofTand the Fourier series off.
MATH. SCAND. 86 (2000), 100^108
Received May 20, 1997.
B. Kjellberg, [11], and D. Mu«ller, [14] (Lemma 3.1) proved a multi- dimensional extension of Carlson's inequality of the integral type. By using the idea1 of Theorem 2.7.6. in [15], Carlson's inequality can be carried over fromRn to Tn. Our proof of the multi-dimensional case of 1(for the case Rnsee [10]) is new and more direct.
The well-known classical Hausdorff-Young inequality (1912^1923) states that, for any complex-valued functiongin the Banach spaceLp ,T
b g
k kp0k kg p 2
holds for 1p2. Here and throughout the paper,p0is the dual exponent of p. Also, k kbg p0: P
n2Zjbg njp0
1
p0
and k kg p: R
Tjg xjpdx ÿ 1p
are supposed to be finite.
Titchmarsh, [18], proved 2 for the space Lp R in 1924. In fact, 2 is true for locally compact unimodular groups , [13]. The result is due to R.A.
Kunze (1957). Hardy and Littlewood, [8], showed that 2is sharp and there is equality if and only ifgC0e2mix form2Z.
For the space Lp Rn and for the even integer p0, [2], the improvement is due to K.I. Babenko (1961) and for allp, [3], it is due to W. Beckner (1975).
That is
kbfkp0 Bpnk kf p 3
holds for p2 1;2. Bp:
p1p p0p10
s
is called the Babenko-Beckner constant.
bf :R
Rnf xeÿ2i<;x>dx is the Fourier transform of f and h;xi:
Pn
1x.
B. Russo (1974), [16], and J.J.F. Fournier (1977), [6], proved 3 for cer- tain classes of locally compact unimodular groups.
The extension of 3is due to J. Inoue (1992), [9]. For certain classes of nilpotent Lie groups he improved 3and obtained the constant
Bpdim Gÿm2:
HereG:exp gandgis Lie algebras with the dual spacebg. dim Gis the dimension of nilpotent Lie groups G and m is the dimension of generic coadjoint orbits ofGinbg.
For the even integerp0, [1], M.E. Andersson (1994) and for all p, [17], P.
1The referee made kindly this idea clear to me. He also informed me of the references [11] and [14] and gave me valuable comments on this paper (see the remark).
Sjo«lin (1995) proved a Babenko-Beckner type inequality 3for functions in the spaceLp , with small supports.T
The purpose of this paper is to prove Carlson's inequality of type 1onn- dimensional torus and applying it to prove a Babenko-Beckner type of the Hausdorff-Young inequality for periodic functions with small supports.
Theorems and Proofs
Let the multi-indices andbe vectors inRn with componentskandkin N0 such that is equivalent to kk for all 1kn. Define m:Qn
k1mkk form2Znand 00:1.
Throughout this paper,jj:Pn
k1kand :Qn
k1kk. The operator D:Qn
k1 @k
@xkk. Let also
Hp;a:sup k kbg p0
g
k kp : g2Lp ;Tn suppgB 0;a; k kg p60
( )
and defineHp :lima!0Hp;a. Here and everywhere in the paperaobeys the restriction 0<a<12 and B 0;ais a closed ball of radius a, centered at the origin. Also,Tn:x2Rn:jxj 12; 1n
. Assume ' x: 1 jxj 12
0 jxj 1
such that '2C01 Rn; 0'1 and 'a:'ÿ xa
. Define x:ÿ
eÿ2i<b;x>ÿ1
'a x. Here b: b1;b2; ;bn andj j bk 12; 1kn.
With the previous notation, we prove the following:
Theorem 1 (Generalisation of Carlson's inequality). Let f 2A Tn and bf 0 0. Let the absolute value of the multi-index be equal to the positive integersuch that1and >nqwhere1<q2. Then we get
kfkA TnKn;q kfkq1ÿqn X
jj
kDfkq 0
@
1 A
qn
:
In the casebf 0 60, we obtain
kfkA Tn kfk1Kn;q kfkq1ÿqn X
jj
kDfkq 0
@
1 A
qn
:
The positive constantKn;q depends only onn; andq.
Proof of Theorem1. The technique is analogous to the casen1, due to Hardy, [7]. Letbf 0 0 andq0 be the dual exponent ofq. Define
102 amir kamaly
S: kbfkqq00
T : X
jj
kDdfkqq00: Fort>0 we also define
P: X
jj
t j 2mjq0
ThenT P
jjkDdfkq0
q0
. By Ho«lder's inequality we get
kfkA Tn X
jmj>0
jbf mjPq10Pÿq10 4:1
X
jmj>0
jbf mjq0P 0
@
1 A
q10
X
jmj>0
Pÿqq0 0
@
1 A
1q
tÿq10
tcn;ST 1
q0 X
jmj>0
1Cn;
t jmjq0
ÿq
q0
2 4
3 5
1q
: Because
X
jj
t j 2mjq0
cn;tX
jj
j 2mjq0 ttCn;jmjq0: Here cn;:P
jj1 and Ddf m 2imbf m. The positive constant Cn; does depend onnand.
It is not hard to see that the sum P
jmj>0 1 1jmjq0
ÿ q
q0
" #1q
is finite for >nqand
Z 1
0
dx
1xq0nqq0ÿn qÿ1q
ÿ qÿ1 qÿnq
n qÿ1q ÿ qÿ1 : 4:2
Now, by 4:1and 4:2we obtain
kfkA Tnc0tÿq10
tcn;ST 1
q0 Z
Rn
dx 1Cn;t jxjq0
q
q0
2 64
3 75
1q
c0tÿq10 t Cn;
n
qq0
tcn;ST 1
q0 Z
Rn
dx 1 jxjq0qq0
!1q
c0tÿq10 t Cn;
n
qq0
tcn;ST 1
q0 Z 1
0
Z
fx2Rnÿ1:jxj1g
rnÿ1drdx 1rq0 qq0
!1q
c0 wnÿ1
n
1
qtÿq10 t Cn;
n
qq0
tcn;ST
ÿ 1
q0
Z 1
0
dx 1xq0n
q
q0
0 B@
1 CA
1q
c0A n;qtnqq0 cn;ST t
1
q0
; for a positive constantc0. Here
A n;q :
qÿ1wnÿ1ÿn qÿ1q
ÿ qÿ1 qÿnq
qÿ qÿ1 ÿCn;n 1ÿqq
q
vu ut
; andwnÿ1 is the surface area of the unit sphere inRnÿ1.
Choose tST, then by using (two times) the classical Hausdorff-Young inequality 2we get
kfkA Tnc0A n;q
cn;1 ÿ qÿ1 qq
kbfkq0
1ÿqn X
jj
kDdfkq0
0
@
1 A
qn
c0A n;q
cn;1 ÿ qÿ1 qq
kfkq1ÿqn X
jj
kDfkq 0
@
1 A
qn
Kn;q kfkq1ÿqn X
jj
kDfkq 0
@
1 A
qn
:
For the casebf 0 60 the proof is similar and we know thatjbf 0j kfk1:
104 amir kamaly
Application of Theorem 1 for estimating of theA Tn-norm of andHp;a Lemma (An upper bound for k kA Tn). There exists a positive constant C0, does not depend on a, such that
k kA TnC0a:
Proof of Lemma. It is obvious that 2C01 Rnand form2Zn we get j mj jb
Z
jxja xeÿ2i<m;x>dxj an Z
jyj1jeÿ2ia<b;y>ÿ1jdy
pn an1
Z
jyj1dxnan1; because
jeÿ2ia<b;y>ÿ1j 2aj<b;y>
pn a:
Heren:
pn
wn andwnÿ2nn2
2 is the surface area of the unit sphere in Rn.
Furthermore, by Leibniz's formula, together with Minkowski's inequality we obtain
X
jj
kD kq
pn aX
jj
kD'akqX
jj
X
jj60
jjkDÿ'akq 5
pn
a1ÿnq X
jj
kD'kqX
jj
X
jj60
jjajjÿjjnqkDÿ'kq
a1ÿnq
pn X
jj
kD'kqX
jj
X
jj60
jjkDÿ'kq 8>
<
>:
9>
=
>;
An;q;a1ÿnq;5 because
X
jj
D'a
qanqÿX
jj
D' q: Now, by Theorem 1 and invoking 5, we get
kbk1 X
m2Zn
jb mj nan1 X
jmj>0
jb mj
nan1K n;qk kq1ÿqn X
jj
kD kq 0
@
1 A
qn
nan1 Kn;q
pn k'kq 1ÿn
qAqn;q;n
a nKn;q An;q;qn
pn k'kq 1ÿqn
a
C0a;
because
k kq
pn
k'kqa1np:
Note that is the positive integer defined in Theorem 1 and kbk1: k kA Tn.
Theorem 2 (An upper bound for Hp;a). For a fixed n2N, there exists a positive constant C0which does not depend on a, such that
Hp;a
1C0a
Bpn; 1p2:
Proof of Theorem2. The technique is analogous to the casen1, due to Y. Domar, [5]. Choosef 2Lp ;Rn g2Lp Tn, such that f g on the ball B 0;aand zero outside of the ball. Definegb x:eÿ2i<x;b>g x. Then
gbb m bg mb
kfkp kgkp fbb m gbb m:
Also, we get
gbb m ÿbg m Z
eÿ2ihm;xiÿ
eÿ2ihb;xiÿ1 g xdx
Z
B 0;a xg xeÿ2ihm;xidx
Z
B 0;ag x X
m02Zn
mb 0e2ihm0;xi
eÿ2ihm;xidx
X
m02Zn
mb 0bg mÿm0:
106 amir kamaly
Thus, we obtain b gbÿbg
k kp0 X
m02Zn
jb m0j X
m2Zn
bg m
j jp0 1
p0
k kbg p0k kb 1: By triangle inequality we have
b g
k kp0ÿk kgbb p0kgbbÿbgkp0k kbg p0k kb 1: Similarly, fort2Rn, we obtain
bg k kp0
1ÿ k kb 1
k kgbb p0 X
m2Zn
gbb m
j jp0 1
p0
X
m
fbb m
p0 1
p0
:
That is
bg k kp0
ÿ1ÿ kbk1
X
m
Z
fb:jbkj12g
fbb m
p
0
db 1
p0
X
m
Z
ftÿm:jtkÿmkj12g
bf t
p
0
dt 1
p0
kbfkp0: Now, by Lemma we get
bg
k kp0 kbfkp0
1ÿ kbk1 kbfkp0
1ÿC0a: 6
By 6, thus, we obtain
Hp;a sup
g
bg k kp0
g
k kp Bpn 1ÿC0a: Because
supf
kbfkp0
f
k kpBpn;
(see [3], p. 160). Chooseasuch thatC0a<12, then we get 1
1ÿC0a1C0a;
because1ÿC10a1C0aO Cÿ 02a2 . Hence Hp;aÿ
1C0a Bpn:
Remark. The arguments in this proof can be used to prove that the quo- tient of the normsbgandbf inlp0 andLp0, respectively, is 1O a, asa!0. The Babenko-Beckner type of the Hausdorff-Young inequality for periodic functions with small supports
Theorem3 HpBpn.
The proof is a consequence of Theorem 2.
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ROSA VILLAN REDUTTV. 13 SE-187 68 TØBY SWEDEN
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