on the occasion of his 70th birthday
INEQUALITIES IN MATRIX SPACES
IRINA POPA and NICOLAE POPA
We extend an inequality of A. Shields from 1983, proved for the Schatten classC1, toCpfor 1≤p≤2. We discuss also some more inequalities in Bergman-Schatten classes, extending results from [6].
AMS 2000 Subject Classification: 47B10.
Key words: Hardy-Littlewood inequalities, Schatten class, Bergman-Schatten classes.
1. INTRODUCTION
It is well-known that the Hardy-Littlewood inequalities
∞
X
n=0
(n+ 1)p−2|an|p
!1/p
≤C(p)kfkHp, 0< p≤2 hold for the functions f(t) =P
n≥0aneint belonging to Hardy spaces Hp(T).
See [5].
The similarity between functions and infinite matrices was remarked for the first time by Arazy in [1] and exploited further by Shields in [12].
More precisely, for an infinite matrix A = (aij)i,j≥1, we consider the diagonal matrixAk def= (a0ij)i,j≥1,wherea0ij =
aij ifj−i=k
0 otherwise ,k∈Z,and we identify the later to the kth Fourier coefficientof the matrixA.
On the other hand, we have enough reasons to consider the Schatten class of orderp, denoted byCp,to be the similar notion of the Lebesgue space Lp(T),0< p <∞.See [12]. (We note that the above analogy is quite imperfect since for 0 < p1 < p2 < ∞,it follows that Lp2(T) ⊂Lp1(T) contrarily to the known inclusion Cp1 ⊂Cp2.)
Then the analogue of Riesz projection P0(f) def= P
n≥0aneint, where f is a trigonometric polynomial, is the triangular projection PT(A) = (a0ij)i,j≥1,
MATH. REPORTS12(62),2 (2010), 169–180
where
a0ij =
aij if 0≤j−i≤k
0 otherwise ,
A= (aij)i,j≥1 and aij = 0 if |i−j|> kfor some fixed k∈N.
Therefore, the analogue of the Hardy space Hp(T), 0 < p < ∞, is the space Tp def= {A|A upper triangular matrix;A∈Cp},with the normkAkTp= kAkCp.
On the other hand, if f(t) = aneint, it follows that kfkHp = |an|, thus the corresponding object to|an|would bekAnkCp =k(ai,i+n)ik`p,whereAn= (ai,i+n)i≥1, n∈N.
Then we may expect that the inequalities (1)
∞
X
n=0
(n+ 1)p−2kAnkpC
p ≤K(p)kAkpT
p
hold for all A∈Tp,1≤p≤2,or, equivalently, (2)
∞
X
n=0
(n+ 1)p−2
∞
X
i=1
|ai,i+n|p
!
≤K(p)kAkpT
p, 1≤p≤2.
Forp= 1 the inequality (2) with K(1) =π, that is,
∞
X
n=0
(n+ 1)−1
∞
X
i=1
|ai,i+n|
!
≤πkAkT1, A∈T1
was proved by Shields in 1983. See [12].
For another, more general, proof see [4, Theorem 2.2 a)].
Now we describe the content of the paper. In Section 2 we consider first the case 1< p ≤2 and we prove (only for 1< p <2) a weaker result as (1).
We conjecture that (1) holds also in the case 1 < p <2.Of course, (1) is easy to prove in the casep= 2 withK(2) = 1.Also we get a result dual to the first inequality. In Section 3 we introduce the Bergman-Schatten classes and prove the similar inequalities to those proved by Nakamura, Ohya and Watanabe for Bergman function spaces. See [6, pp. 83–84].
2. INEQUALITIES IN SCHATTEN CLASSES
In order to state and prove the result we have to recall some notation from the paper [10].
Let A be an upper triangular matrix and let (Ak)k≥0 be the sequence of its diagonal matrices. We denote by Tp(`2R) the completion of the space of all finite sequences (Ak)nk=0 with respect to the norm k(Ak)nk=0kT
p(`2R) def=
k(Pn
k=0A∗kAk)1/2kCp, where 1≤p ≤2.Here A∗k is the adjoint matrix of Ak. It is clear that Tp(`2R) is a space of upper triangular matrices.
Similarly, Tp(`2C) is the completion of the space of all finite sequences (Ak)nk=0 with respect to the norm k(Ak)nk=0kT
p(`2C)
def= k(Pn
k=0AkA∗k)1/2kCp. Tp(`2C) is a space of upper triangular matrices too.
Now, let us denote by Tp(`2R) +Tp(`2C) the space of all upper triangular matrices A such that there exist A0 ∈ Tp(`2R) and A00 ∈ Tp(`2C) with A = A0+A00.We introduce on this space the norm
kAkT
p(`2R)+Tp(`2C)
def= inf
Ak=A0k+A00k
(
X
k≥0
A0∗kA0k
!1/2 Cp
+
X
k≥0
A00kA00∗k
!1/2 Cp
) ,
1≤p≤2.Let us remark that
∞ X
k=0
A∗kAk 1/2
Cp
=
" ∞ X
j=1
∞ X
i=1
|aij|2
p/2#1/p
, 1≤p≤2, and similarly
∞
X
k=0
AkA∗k 1/2
Cp
=
" ∞ X
i=1
∞
X
j=i
|aij|2
!p/2#1/p
,
so that kAkT
p(`2R)+Tp(`2C) = inf
Ak=A0k+A00k, k≥0
(" ∞ X
j=1
∞
X
i=1
|aij|2
!p/2#1/p
+
+
" ∞ X
i=1
∞
X
j=i
|aij|2
!p/2#1/p) .
Moreover, the relation (I.13) in [10] implies that
(3) X
k≥0
kAkk2C
p
!1/2
≤ kAkT
p(`2R)+Tp(`2C). It is useful to remark that
(4) kAnkTp(`2
R)+Tp(`2C) =kAnkCp = X
i≥1
|ai,i+n|p
!1/p
, 1≤p≤2.
Now we state and prove the following result.
Theorem 1. Let 1< p <2 andA∈Tp(`2R) +Tp(`2C). Then there exists a positive constant K(p) such that
(5)
∞
X
n=0
(n+ 1)p−2kAnkpC
p ≤K(p)kAkT
p(`2R)+Tp(`2C).
Proof. We follow the idea of the proof in [5, pp. 95–97].
LetA be an upper triangular matrix such thatA=Pn
k=0Ak,and letµ be the measure onN defined by
µ(n) = 1
(n+ 1)2, n= 0,1,2, . . . . Let us consider A(t) = P∞
n=0Aneint = A∗Et, where t ∈ T, ∗ means the Schur product of matrices and Et = (ekj)k,j≥1 is the Toeplitz matrix, where ekj = ei(j−k)t,for all j, k≥1.Therefore, kA(t)kCp ≤ kAkTp.
Now, denote (n+ 1)Anby Aen and let us fixs >0.ThenA(t) = Φs(t) + Ψs(t), t∈T,where
Φs(t) =
( A(t) ifkA(t)kTp > s, 0 ifkA(t)kTp ≤s, and Ψs(t) =A(t)−Φs(t), t∈T.
It is clear that we have [A(t)]n = [Φs(t)]n+ [Ψs(t)]n, n = 0,1,2, . . .; therefore, An∗[Et]n = [Φs]n ∗[Et]n+ [Ψs]n∗[Et]n, where [Φs]n = An and [Ψs]n= 0 if kAkTp = sup
t∈T
kA(t)kTp > s. The same holds for Ψs. Then
X
n≥0
(n+ 1)p−2kAnkpC
p=X
n≥0
kA˜nkpC
pµ(n)≤ (6)
≤2p
X
n≥0
k[ ˜Φs]nkpC
pµ(n) +X
n≥0
k[ ˜Ψs]nkpC
pµ(n)
.
Put nowα(s) =µ{n;k[ ˜Φs]nkCp > s} and β(s) =µ{n;k[ ˜Ψs]nkCp > s}= µ(Es).Then
(7) s2β(s)≤X
n≥0
[Ψs]n
2 Cp.
Let us denote by Fs the set {n≥ 0;k[ ˜Φs]nkCp > s}.Since kAnkCp ≤ kAkCp, n∈N,(see [7]) we have
µ(Fs) =µ{n; (n+ 1)k[Φs]nkCp > s} ≤µ{n; (n+ 1)kΦskCp > s}=
= X
{n;kΦskCp>n+1s }
1
(n+ 1)2 ≤4 X
{n;kΦskCp>n+1s }
Z n+1 n
1
x2dx≤ 4 n0+ 1, where n0= min{n;kΦskCp > n+1s }.
Thus
(8) α(s) =µ(Fs)≤ 4kΦskCp
s , s >0.
By (9) we have
∞
X
n=0
k[ ˜Φs]nkpC
pµ(n) =− Z ∞
0
spdα(s) =p Z ∞
0
sp−1α(s)ds≤
≤4p Z ∞
0
sp−1kΦskCp
s ds≤4p
Z kAkCp 0
kAkCpsp−2ds≤ 4p p−1kAkpT
p, for all t∈T.
By (8) it follows
∞
X
n=0
k[ ˜Ψs]nkpC
pµ(n) =p Z ∞
0
sp−1β(s)ds≤p Z ∞
0
sp−3
∞
X
n=0
k[Ψs]nk2C
pds=
=p Z ∞
kAkCp
∞
X
n=0
kAnk2Cpsp−3ds≤ 2p
2−pkAkp−2C
p
∞
X
n=0
kAnk2Cp ≤
(by (4))
≤ p
2−pkAkp−2T
p kAk2T
p(`2R)+Tp(`2C). Using (7) we get
(9)
∞
X
n=0
(n+ 1)p−2kAnkpC
p ≤K(p) h
kAkpT
p+kAkp−2T
p kAk2T
p(`2R)+Tp(`2C)
i .
But in [8] it is proved that kAkCp ≤ kAkTp(`2
R)+Tp(`2C), consequently we get the inequality of Hardy-Littlewood type
(10) X
n≥0
(n+ 1)p−2kAnkpC
p ≤K(p)kAkp
Tp(`2R)+Tp(`2C), 1< p <2.
SinceTp(`2R)+Tp(`2C) is the completion of the space of all finite sequences, Theorem 1 is proved.
Remark. Theorem 1 is weaker than thestrong version of Hardy-Littlewood inequality
(11) X
n≥0
(n+ 1)p−2kAnkpC
p ≤K(p)kAkpT
p,
for all upper triangular matrices A. Indeed, in [8] it is proved that there is an upper triangular matrix A ∈ Cp such that A 6∈ Tp(`2R) +Tp(`2C), 1 < p <2.
Then it is natural to ask the following:
Question. Does Theorem 1 hold for 1< p <2 with (12) instead of (6)?
Using the duality betweenTp(`2R) +Tp(`2C) and Tp(`2R)∩Tp(`2C) (see [9]) given by
hA, Bi=
∞
X
k=0
tr(AkBk∗), where A=P
k≥0Ak, B =P
k≥0Bk,we have Theorem 2. Let 2 ≤ q < ∞ and A = P
k≥0Ak such that P
n≥0(n+ 1)q−2kAnkqC
q <∞. ThenA∈Tq(`2R)∩Tq(`2C) and kAkT
q(`2R)∩Tq(`2C)=
def= max ∞
X
i=1
∞ X
j=i
|ai,j|2
q/21/q
, ∞
X
j=1
j X
i=1
|ai,j|2
q/21/q!
≤
≤C(q) X
n≥0
(n+ 1)q−2 ∞
X
i=1
|ai,i+n|q !1/q
=C(q)
X
n≥0
(n+ 1)q−2kAnkqC
q
1/q
.
Proof. Let p = q/(q −1) and G = Pn
k=0Gk be a finite type band matrix with
kGkT
p(`2R)∩Tp(`2C)≤1.
Let nowSn(A) =Pn
k=0Ak.Then
|hG, Sn(A)i|=
n
X
k=0
tr(GkA∗k) ≤
n
X
k=0
∞
X
i=1
gi,k+iai,k+i ≤
(by H¨older inequality)
≤
n
X
k=0
X
i
|gi,k+i|p 1/p
X
i
|ai,k+i|q 1/q
=
=
n
X
k=0
kGkkCpkAkkCq (again by H¨older inequality)
≤
n
X
k=0
kGkkpC
p k+ 1p−21/p
n
X
k=0
kAkkqC
q(k+ 1)q−2 1/q
≤
(by the previous theorem)
≤ C(p)kGkTp(`2
R)+Tp(`2C) n
X
k=0
kAkkqC
q(k+ 1)q−2
!1/q
≤
≤C(p)
n
X
k=0
kAkkqC
q(k+ 1)q−2
!1/q
.
Hence
kSn(A)kT
q(`2R)∩Tq(`2C) = sup
kGkTp(`2 R)+Tp(`2
C)≤1
|hG, Sn(A)i| ≤
≤C(p)
n
X
k=0
kAkkqC
q(k+ 1)q−2
!1/q
,
for all n and the theorem is proved.
Now we discuss an inequality of Hausdorff-Young type, which is probably known. We include here for the convenience of the reader.
Theorem 3(Hausdorff-Young Theorem). For 1 ≤p ≤ ∞, let q be the conjugate index, with 1p +1q = 1.
(i)If 1≤p≤2 thenA∈Tp implies P∞
n=0kAnkqT
p
1/q
≤ kAkTp. (ii)If2≤p≤ ∞then{kAnkTp} ∈`qimplieskAkTp≤
P∞
n=0kAnkqT
p
1/q
. Proof. In case (ii), ifp= 2 we havep=q= 2 and if P∞
n=0
P∞
l=1|al,n+l|21/2
< ∞ then, clearly, A= (aij)∈ T2 and kAkT2 ≤ P∞ n=0
P∞
l=1|al,n+l|21/2
, in other words the map T({An}n≥0) =P
n≥0An has the norm less than 1 from the space `2(`2) into T2. Here `p(`q) means the space of all matrices (aij)i,j
such that P∞ n=1
P∞
i=1|ai,i+n|2 <∞.
Ifp=∞, q = 1 and the mapT has the norm less than 1 from`1(`∞) into T∞, since, clearly, kAkT∞ ≤ P∞
n=0kAnkT∞. Using the complex interpolation we get the conclusion.
Case (i) results by duality, since`p(`q)∗ =`q(`p),and (Tp)∗ =Tq.
3. INEQUALITIES IN BERGMAN-SCHATTEN CLASSES
Let 0≤p <∞.We denote Lp(D, `2) :=
n
r→A(r) a strong measurableCp-valued function defined on [0,1); such that kA(r)kLp(D,`2):=
2
Z 1 0
kA(r)kpC
prdr 1/p
<∞o , where Cp is the Schatten class of orderp,and we denote
L˜pa(D, `2) ={A(r) :=A∗P(r); A upper triangular matrices | kAkLp(D,`2)<∞}, which is a subspace of Lp(D, `2).
By Lpa(D, `2) we mean the space of all upper triangular matrices such thatkA(r)kLp
a(D,`2) <∞, whereA(r) =P(r)∗A,r∈[0,1).We call the spaces Lpa(D, `2),the Bergman-Schatten classes.
We can state and prove the matrix version of Theorem 3, [6, p. 83]:
Theorem 4.1.Ifp= 1andA∈L1a(D, `2), thenP∞
n=1n−2kAnkC1 <∞.
2. If 1< p≤2 and A∈Lp,unca (D, `2) =n
A| kAkp,uncdef= Z 1
0
Z 1 0
X
k≥0
k(t)Akrk
p Cp
rdrdt 1/p
<∞o ,
where k means thekth Rademacher function, then we have
∞
X
n=1
np−3kAnkpC
p <∞.
3. If 2≤p <∞ and P∞
n=1np−3kAnkpC
p <∞, thenA∈Lp,unca (D, `2).
4. If 2≤p <∞ and A∈Lpa(D, `2),then P∞
n=1n−1kAnkCp <∞.
5. If 1< p≤2 and P
n≥0 1
n+1kAnkpC
p <∞,thenA∈Lpa(D, `2).
Proof. 1. Let A ∈ L1a(D, `2). We use the results of Shields [11] applied to the function z→A(rz),with a fixed 0< r <1.Then we have
∞
X
n=0
(n+ 1)−1kAnrnkC1 ≤C{M1,1(r, A)}, where M1,1(r, A) = 2π1 R2π
0 kP
n≥0Anrneinθk1dθ. Multiplying the above ine- quality by r and integrating on (0,1) we have
∞
X
n=0
(n+ 1)−1kAnkC1 Z 1
0
rn+1dr≤ kAkL1 a,
or ∞
X
n=0
(n+ 1)−2kAnkC1 ≤CkAkL1 a.
2. Let nowA∈Ap,unc and 1< p≤2.By Theorem 1 applied forAr(z) = A(rz), 0< r <1,with Ar∈Tp(`2R) +Tp(`2C),we have
∞
X
n=0
(n+ 1)p−2kAnrnkpC
p ≤C(p) Z 1
0
∞
X
k=0
k(t)Akrk
p Cp
dt.
Proceeding as in 1. we get
∞
X
n=0
(n+ 1)p−2kAnkpC
p
Z 1 0
rnp+1dr ≤C(p)kAkpp,unc,
or ∞
X
n=0
(n+ 1)p−3kAnkpC
p≤C(p)kAkpp,unc.
3. By Theorem 3 applied toAr(z) =A(rz),for 0< r <1,we have kArkpp,unc≤C(p)
∞
X
n=0
(n+ 1)p−2kAnrnkpC
p
and, consequently, Z 1
0
∞
X
k=0
k(t)Akrk
p Cp
dt≤C(p)
∞
X
n=0
(n+ 1)p−2kAnrnkpC
p. Hence
Z 1 0
Z 1 0
∞
X
k=0
k(t)Akrk
p Cp
rdr
dt≤C(p)
∞
X
n=0
(n+ 1)p−2kAnkpC
p
Z 1 0
rnp+1dr and finally
kAkpp,unc≤C(p)
∞
X
n=0
(n+ 1)p−3kAnkpC
p.
4. We have the following interpolation theorem (see [11]): Let1< p <∞ and 1−θ= 1/p.Then Lpa(D, `2) = [L1a(D, `2),B(`2)]θ with equivalent norms.
In the same manner we get Lpa(D, `2) = [L2a(D, `2),B(`2)]θ with 2/p = 1 −θ. Now, let T : B(`2) → `∞(`∞); given by T(A) = (An)n≥0. T maps continuouslyL2a(D, `2) into`2(`2, w),wherew(n) = 1/(n+1),for alln≥0,and (`2, w) is the space of all sequences x= (xn)n≥0 such thatP
n≥0|xn|2w(n)<
∞,with natural norm. Indeed, similarly to Theorem at page 84 in [6], we get kT(A)k2 =X
n≥0
1
n+ 1kAnk2∞≤C(2)kAk2L2 a(D,`2).
Hence T :Lpa(D, `2) →[`2(`2, w), `∞(`∞)]θ =`p(`p, w), whereθ = 1−2/p, is a bounded operator, see [13] for the last equation, and
X
n≥0
1
n+ 1kAnkpC
p
1/p
≤C(p)kAkLp
a(D,`2).
5. We have [Lpa(D, `2)]∗ = Lqa(D, `2), 1/p+ 1/q = 1. See [11]. On the other hand [`p(`p, w)]∗ = `q(`q, w) for 1/p+ 1/q = 1. Then the conclusion follows from 4.
Now, we give the Hausdorff-Young Theorem for Bergman-Schatten classes extending Theorem 2 [6, pages 81–82].
Theorem 5. Let 1≤p≤ ∞ and letq = p−1p be its conjugate exponent.
(i)If 1≤p≤2 thenA∈Lpa(D, `2) implies that
X
n≥0
(n+ 1)1−qkAnkqT
p
1/q
≤ kAkTp.
(ii)If 2≤p≤ ∞then P
n≥0(n+ 1)1−qkAnkqT
p <∞ implies that kAkTp ≤
∞
X
n=0
(n+ 1)1−qkAnkqT
p
!1/q
.
Proof. In case (ii), let µ be the discrete measure on the set N which assigns the massµ(n) =n+1 to the integern= 0,1,2, . . .. Consider the linear operatorT that maps the sequence{n+11 An}to the formal seriesP
n≥0Anzn. We want to show that T is bounded as an operator from Lq(N,dµ;Tp) to Lp(D,dσ;Tp), with norm kTk ≤ 1, where Lp(D,dσ;Tp) is the space of all p-Lebesgue Bochner integrable Tp-valued functions defined on the unit disk D with respect to the area measure dσ on D. For p = 2 this follows from the relation
kT({An/(n+ 1)})k2L2(D,dσ;T2)= 2X
n≥0
kAnk2C
2(n+ 1)−1=
=k{An/(n+ 1)}k2L2(N,dµ;T2). For p=∞ it is trivial that sup|z|≤1kP
n≥0 AnznkT∞ ≤P
n≥0kAnkT∞. We use now the complex interpolation of vector valuedLp spaces.
Case (i). Let A(z) ∈ Lp(D,dσ;Tp). Define the linear operator T(A) = {bn}, wherebn =R
DA(z)zndσ. IfA∈Lpa(D, `2) thenbn =An/(n+ 1).With the measureµdefined as before, we want to show thatT is a bounded operator from Lp(D,dσ;Tp) toLq(N,dµ;Tp),with norm kTk ≤1.Forp= 1 this is the
trivial inequality kbnkT1 ≤ kAkL1(D,dσ;T1) for all n ∈ N. For p = 2 it follows from the relation
∞
X
n=0
(n+ 1)kbnk2T
2 =
∞
X
n=0
(n+ 1) Z
D
A(z)zndσ
2
=
= Z
D
Z
D
A(z)A(ζ)
∞
X
n=0
(n+ 1)(zζ)ndσ(z)dσ(ζ) =
= Z
D
Z
D
A(z)A(ζ)
(1−zζ)2dσ(z)dσ(ζ) =hP A(·), A(·)i, where P denotes the Bergman projection. Since
|hP A(·), A(·)i| ≤ kP A(·)k2kA(·)k2≤ kA(·)k22, this gives the desired inequality for p= 2.
One can conclude by interpolation as before that
∞
X
n=0
(n+ 1)kbnkqT
p
!1/q
≤ Z
D
kA(z)kpT
pdσ 1/p
, A(·)∈Lp(D,dσ;Tp).
Specializing toA∈Lpa(D, `2) and recalling thatbn=An/(n+ 1),we arrive at the desired result.
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Received 25 September 2009 Technical University of Civil Engineering Department of Mathematics and Informatics
020396, Bucharest, Romania [email protected]
and Romanian Academy
“Simion Stoilow” Institute of Mathematics P.O. Box 1-764
014700 Bucharest, Romania [email protected]