On Besov spaces and absolute convergence of the Fourier transform on Heisenberg groups
Leszek Skrzypczak
Abstract. In this paper the absolute convergence of the group Fourier transform for the Heisenberg group is investigated. It is proved that the Fourier transform of functions belonging to certain Besov spaces is absolutely convergent. The function spaces are defined in terms of the heat semigroup of the full Laplacian of the Heisenberg group.
Keywords: Besov spaces, Heisenberg groups, group Fourier transform Classification: 46E35, 43A80
1. Introduction
Different problems in harmonic analysis on Heisenberg groups attracted attention in the last two decades. As far as the group Fourier transform is regarded a basic reference is Geller’s fundamental work [5], confer also Folland’s book [3].
D. Geller obtained in his paper, among other things, a characterization for the Fourier transform of rapidly decreasing Schwartz functions. On the other hand a natural analogue of the Paley-Wiener theorem for Heisenberg groups was proved by S. Thangavelu [13]. In the classical setting the third type of results that describe the connection between the smoothness of functions and the behaviour of their Fourier transform are the Bernstein type theorems. The last ones assert that the Fourier transform of a function is absolutely convergent if the function belongs to a suitable Besov space, cf. [1], [7]. In this paper we want to prove the similar results for the group Fourier transform of the Heisenberg group.
The Heisenberg groupHn=Rn×Rn×Ris a nilpotent Lie group whose group law is defined by
(1) (x1, y1, t1)·(x2, y2, t2) = (x1+x2, y1+y2, t1+t2+1
2(y1·x2−x1·y2)).
The Lie algebrahnofHnis spanned by the left invariant vector fields
(2) Xi = ∂
∂xi +1 2yi ∂
∂t, Yi= ∂
∂yi −1 2xi∂
∂t, T = ∂
∂t, i= 1, . . . n.
We have the commutation relations
(3) [Yi, Xi] =T, i= 1, . . . , n,
and all other commutators vanish. The Haar measure ofHn coincides with the Lebesgue measure onR2n+1.
For each nonzero real numberλthere is a irreducible unitary representation of HnonL2(Rn) given by
(4) πλ(x, y, t)φ(ξ) =eπ√−1(2tλ+2y·ξ+λx·y)φ(ξ+λx)
whereφ∈L2(Hn). There are also one dimensional unitary representation ofHn, π(a,b),a, b∈Rn, given by
(5) π(a,b)(x, y, t)ξ=e2π√−1(a·x+b·y)ξ , ξ∈C.
Any irreducible unitary representation ofHn is unitary equivalent to one of just described representations. The representations π(a,b) are of less importance for us since they form a set of Plancherel measure zero.
For a functionf on the Heisenberg group, sayf ∈L1(Hn), the Fourier trans- form is defined to be the operator valued function
(6) fˆ(λ) =
Z
Hn
f(x, y, t)πλ(−x,−y−t)dx dy dt.
Ifφ∈L2(Hn), then ˆf(λ)φis given by (7) ˆf(λ)φ(x) =|λ|n
Z
f(λ−1(x−y), w, t)e−π√−1(y+x)·w−2π√−1tλφ(y)dy dw dt.
So it is an integral operators with kernel
(8) Kfλ(x, y) =|λ|−nF2,3f(λ−1(x−y),1
2(x+y), λ),
whereF2,3 denotes Fourier transformation in the second and third variables.
The Plancherel formula forHnlooks as follows (9)
Z
Hn|f(x, y, t)|2dx dy dt=cn
Z ∞
−∞kfˆ(λ)k2HS|λ|ndλ.
Herek · kHS stands for the Hilbert-Schmidt norm.
2. Besov spaces on Heisenberg groups
To describe the smoothness of functions we use the Besov spaces related to the full Laplacian onHn. Let ∆ =Pn
i=1Xi2+Yi2+T2 be the sum of squares of left invariant vector fields. If we equipHn with the left invariant Riemannian metric gsuch that the vector fields are the orthonormal basis in any tangent space, then
the operator ∆ coincides with the Laplace-Beltrami operator corresponding to this metric, and the Riemannian measure coincides with the Haar measure onHn.
The Bessel-potentials (I−∆)−s/2withs >0 can be defined inL2(Hn) via the spectral theory. They can be extended afterwards fromL2(Hn) to Lp(Hn) with 1< p <∞, cf. [12]. For 1< p <∞,s∈Rwe define the Sobolev spacesWps(Hn) in the following way:
- ifs >0, then Wps(Hn) is the collection of all functionsf ∈ Lp(Hn) such that f = (I−∆)−s/2hfor someh∈Lp(Hn), with the normkf|Wps(Hn)k=khkp, - ifs < 0, thenWps(Hn) is the collection of all distributionsf ∈D′(Hn) of the form f = (I−∆)mhwithh∈Wp2m+s(Hn), where m is a natural number such that 2m+s >0, andkf|Wps(Hn)k=kh|Wp2m+s(Hn)k,
- ifs= 0, thenWp0(Hn) =Lp(Hn).
The spacesWps(Hn) withs <0 are independent ofm(equivalent norms). Ifs is a positive integer, then one can use left-invariant vector fields onHn to define equivalent norms inWps(Hn), cf. [14].
Fors∈R, 1< p <∞and 1≤q≤ ∞we define the Besov spacesBsp,q(Hn) via the real interpolation
(10) Bsp,q(Hn) = (Wps0(Hn), Wps1(Hn))θ,q, s= (1−θ)s0+θs1, 0< θ <1.
The norm in the Besov spaces can be described by the heat semigroupHt=et∆
([11]). The heat semigroupHt is given by a right convolution:
(11) Htf(x) =
Z
G
f(y)ht(y−1x)dy
where (t, x) −→ ht(x) is a C∞ function on R+×G and a positive solution of (∂t∂ + ∆)u = 0. The semigroup is symmetric submarkovian, hence analytic in Lp(G) if 1< p <∞. If s∈ R, 1< p <∞, 1≤q≤ ∞ and m > |s2|, then the expression
(12) kf|Bsp,q(Hn)kH =kf∗h0,mkp+ Z 1
0
t(m−s/2)qkdm
dtmf∗htkqp
dt t
1/q
is an equivalent norm inBp,qs (Hn). Hereh0,m is aC∞ function onHngiven by
(13) h0,m=
m−1
X
l=0
cl ∂l
∂tlht|t=1.
Using the above norm one can define the Besov spaces forp= 1 andp=∞. The definition if independent ofm(equivalent norms). Ifs >0, then one can usekfkp
instead ofkf∗h0,mkp in (12), 1≤p≤ ∞.
We have the following elementary topological embeddings Bsp,q11(Hn)⊂Bp,qs22(Hn) if s1> s2, (14)
Bsp,q1(Hn)⊂Bp,qs 2(Hn) if q1< q2, (15)
Bp,p0 (Hn)⊂Lp(Hn) if 1≤p≤2.
(16)
More information about function spaces of Hardy-Sobolev-Besov type on Lie groups and more general on Riemannian manifolds with bounded geometry may be found in [15] and [10], [11].
3. Absolute convergence of the Fourier transform We start with the following standard lemma.
Lemma 1. Let0 < r < c. Then there is a positive constant C > 0 such that a geodesic ballΩ(0, r) is contained in the euclidean box centered at0 with sides parallel to the coordinate axes and sizesCr.
Proof: We give a short proof of the lemma for completeness. Let (x1, . . . , x2n+1)
∈ Ω(0, r) be a point inside the geodesic ball. Then the is a geodesic γ(t) = (γ1(t), . . . , γ2n+1(t)) with γi(0) = 0, γi(1) = xi and kγ(t)˙ k ≤ r. There are functionsβi(t) such that
˙
γ(t) = ( ˙γ1(t), . . . ,γ˙2n+1(t)) =
n
X
i=1
βi(t)Xi+βn+1(t)Yi+β2n+1(t)T
and 2n+1
X
i=1
β2i(t)≤r2.
Thus |βi| ≤ r for each i. But (2) implies ˙γi(t) = βi(t) for i = 1, . . .2n and
˙
γ2n+1(t) = β2n+1(t) + 2Pn
i=1γn+i(t)βi(t) − γi(t)βn+i(t). Now |γi(1)| ≤ R1
0 |γ˙i(t)|dt≤Cr sinceγi(0) = 0 andr≤c.
The main result reads us follows.
Theorem 1. The following inequalities Z
Rkfˆ(λ)kHS|λ|ndλ≤Ckf|Bn+[
n 2]+2 1,1 (Hn)k (17)
Z
Rkfˆ(λ)kHS|λ|n2 dλ≤Ckf|B1,1n+2(Hn)k (18)
hold.
Remark 1. The measure|λ|ndλseems to be more natural since it is a Plancherel measure, but the measure|λ|n2 dλ better describes the behaviour of the Fourier transform near zero, cf. the proof of the theorem. So we decide to state the both inequalities.
The next corollary follows by real interpolation from the above theorem and Plancherel theorem sinceB2,20 (Hn) =L2(Hn).
Corollary 1. Let sp = 2−pp(n+ [n2] + 2) and σp = 2−pp(n+ 2) for 1 < p < 2.
Then the following inequalities Z
Rkfˆ(λ)kpHS|λ|ndλ 1/p
≤Ckf|Bp,psp(Hn)k (19)
Z
Rkfˆ(λ)kpHS|λ|np2 dλ 1/p
≤Ckf|Bp,pσp(Hn)k (20)
hold.
Proof of Theorem 1: Step1. To prove the theorem we use the atomic decom- position for the spacesBsp,q, cf. [10]. For anyj,j= 0,1, . . ., let{Ω(zj,i,2−j)}∞i=0 be the uniformly locally finite covering of Hn with the multiplicity independent ofj. The atomic decomposition theorem asserts thatf ∈B1,1s (Hn),s >0, if and only iff can be decomposed into the sum
(21) f =
X∞
i,j=0
sj,iaj,i, convergence in S′(R2n+1),
where functions aj,i ∈C0∞(R2n+1) are smooth atoms and the real numberssj,i satisfy the condition
(22)
X∞
i,j=0
|sj,i|<∞.
The function aj,i is called a smooth atom if the following two conditions are fulfilled
suppaj,i⊂Ω(xj,i,2−j+1), (23)
|Zm1. . . Zmkaj,i| ≤C2−j(s−k−N), k≤L (24)
where Zm denotes any left invariant vector field Xi, Yi or T and N = 2n+ 1.
The constantLis a fixed real numberL >[s] + 1, andC is an absolute constant.
Moreover, the infimum of (22) taken over all possible decompositions (21) is an
equivalent norm in Bs1,1(Hn). The details about the atomic decomposition of Besov spaces on Riemannian manifolds can be found in [10].
Step2. Letaj,i be the smooth atom supported in Ω(zj,i,2−j+1) Thena(z) = aj,i(zj,i·z) is a smooth atom supported in Ω(0,2−j+1). LetF3 denotes Fourier transformation in the third variable. Letzj,= (x0, y0, t0). Then
kaˆj,i(λ)k2HS=|λ|−n Z
|F3aj,i(x, y, λ)|2dx dy
=|λ|−n Z
|F3aj,i(x+x0, y+y0, λ)|2dx dy, cf. [3, p. 39]. But
F3aj,i(x+x0, y+y0, λ) =C Z
R
aj,i(x+x0, y+y0, t)e−√−1tλdt
=C Z
R
a(x, y, t−t0−1
2(y·x0−x·y0))e−√−1tλdt
=Ce−√−1λ(t0−12(y·x0−x·y0)) Z
R
a(x, y, t)e−√−1tλdt,
and
e−√−1tλ= (−√
−1λ)−mdm
dtme−√−1tλ. Thus Lemma 1 implies
kˆaj,i(λ)k2HS ≤C|λ|−n−m Z
Z
R
e−√−1tλTma(x, y, t)dt
2
dx dy
≤ |λ|−n−2m Z
( Z C2−j
−C2−j|Tma(x, y, t)|dt)2dx dy
≤C|λ|−n−2m2−j Z Z Z
|Tma(x, y, t)|2dx dy dt.
Now from the definition of an atom we get
(25) kaˆj,i(λ)kHS≤C|λ|−m−n22−j(s−m−N−21)=C|λ|−m−n22−j(s−m−n). Using these estimates withm= 0 for small values of|λ|andm= 2 for|λ|big we get
(26)
Z
R\{0}kˆaj,i(λ)kHS|λ|n2 dλ≤C2−j(s−2−n).
Similarly, takingm= [n+42 ] for big values of|λ|we get (27)
Z
R\{0}kˆaj,i(λ)kHS|λ|ndλ≤C2−j(s−[n2]−n−2).
Step 3. If f ∈ L1(Hn), then the operator ˆf(λ) is an integral operator with kernel
Kfλ(x, y) =|λ|−nF2,3f(λ−1(x−y),1
2(x+y), λ).
Ifaj,iis an atom, then the corresponding kernelsKj,iλ are smooth functions.
Iff ∈B1,1n+2(Hn) andf =P∞
j,i=0sj,iaj,i, thenf ∈L1(Hn) and Z
f(x, y, t)ψ(x, y, t)dx dy dt= X∞
j,i=0
sj,i Z Z Z
aj,i(x, y, t)ψ(x, y, t)dx dy dt,
ψ∈S(RN). Thus
|Kfλ(x, y)| ≤ X∞
j,i=0
|sj,i| |λ|−n| |F2,3aj,i(λ−1(x−y),1
2(x+y), λ)|= X∞
j,i=0
|sj,i|Kj,iλ
whereKj,iλ is a kernel of ˆaj,i(λ). Thus kfˆ(λ)kHS = (
Z
|Kfλ(x, y)|2dx dy)1/2
≤ X∞
j,i=0
|sj,i|( Z
|Kj,iλ (x, y)|2dx dy)1/2= X∞
j,i=0
|sj,i|kˆaj,ikHS. Now the theorem follows from inequalities (26)–(27).
Theorem 2. Let1≤q≤2andsq= 2−qq(n+12). Then the following inequality (28)
Z
Rkfˆ(λ)kqHS|λ|ndλ 1/q
≤Ckf|B2,qsq (Hn)k holds.
Proof: Forq= 2 the theorem is obvious sinceB02,2(Hn) =L2(Hn), cf. [14]. We prove the theorem for q = 1. The rest follows by interpolation. Any function f ∈L2(Hn) can be decomposed in the following way
(29) f(x) =C
f∗h0,m+ Z 2
0
tmdm
dtmf ∗htdt t
whereh0,m∈S(R2n+1) is given by (13), cf. [11]. Thus kfˆ(λ)kHS ≤ kfˆ(λ)ˆh0,m(λ)kHS+
Z 1
0
tmkfˆ(λ)ˆhmt/2(λ)ˆht/2(λ)kHS
dt t
≤ kfˆ(λ)kHSkˆh0,m(λ)kHS+ Z 1
0
tmkf[∗hmt/2kHSkˆht/2(λ)kHS
dt t Z
Rkfˆ(λ)kHS|λ|ndλ≤ Z
Rkfˆ(λ)k2HS|λ|ndλ
1/2Z
Rkˆh0,m(λ)k2HS|λ|ndλ 1/2
+ Z 2
0
tm Z
Rkf[∗hmt/2k2HS|λ|ndλ
1/2Z
Rkhˆt/2(λ)k2HS|λ|ndλ 1/2 dt
t
=kfk2kh0,mk2+ Z 1
0
tmkf∗hmt/2k2kht/2k2
dt t
≤ kfk2khm,0k2+ Z 2
0
t(m−N4)kf∗hmt/2k2
dt
t ≤Ckf|Bn+
1 2
2,1 (Hn)k.
Corollary 2. Let1≤q≤p≤2andsp,q=n+2−pp[n+22 ] +p1+1q−1. Then the following inequality
(30)
Z
Rkfˆ(λ)kqHS|λ|ndλ 1/q
≤Ckf|Bsp,qpq(Hn)k holds.
Remark 2. There is also a heat semigroup version of Bernstein theorem on a unimodular Lie group, cf. [8] and [4].
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Faculty of Mathematics and Computer Science, A. Mickiewicz University, Matejki 48-49, 60-769 Pozna´n, Poland
E-mail: [email protected]
(Received February 23, 1998)