l1 - h1 bounds for a generalized strichartz potential
TRANSCRIPT
ISSN 1066-369X, Russian Mathematics (Iz. VUZ), 2011, Vol. 55, No. 9, pp. 7–14. c© Allerton Press, Inc., 2011.Original Russian Text c© A.V. Gil’ and V.A. Nogin, 2011, published in Izvestiya Vysshikh Uchebnykh Zavedenii. Matematika, 2011, No. 9, pp. 10–18.
L1 − H1 Bounds for a Generalized Strichartz Potential
A. V. Gil’1* and V. A. Nogin2**
1Southern Federal University, ul. Mil’chakova 8a, Rostov-on-Don, 344090 Russia2Southern Federal University, ul. Mil’chakova 8a, Rostov-on-Don, 344090 Russia; Southern
Mathematical Institute of the Vladikavkaz Scientific Center of the Russian Academy of Sciences,ul. Markusa 22, Vladikavkaz, 362027 Russia
Received July 5, 2010
Abstract—We obtain the necessary and sufficient conditions for the boundedness of a generalizedStrichartz potential.
DOI: 10.3103/S1066369X11090027
Keywords and phrases: asymptotics, multiplier, Fourier transform, convolution, symbol.
1. INTRODUCTIONIn this paper we obtain L1 − H1 and BMO − L∞ bounds for the generalized Strichartz potential
Kβθ ϕ = kβ
θ ∗ ϕ (1.1)
with the kernel
kβθ (y) =
1Γ(β)
θ(|y|)(1 − |y|2)β−1+ , β ≥ 0,
where θ(1) �= 0; see (3.18) for the treatment of convolution (1.1) with β = 0. Here θ(r) is a smoothfunction which is said to be the characteristic of the operator Kβ
θ .At the present time there are several papers devoted to bounds for potentials (1.1) and close to them
operators from Hp to Hq, from Hp to BMO, and others [1–5]. Evidently, the operator Kβθ is also
bounded in L1. However, the question about the boundedness of this operator from L1 to H1 (H1 is aclosed subspace in L1) is rather nontrivial. Theorem 3.1 gives an answer to this question. We prove it bya rather fine technique using oscillating multipliers.
Note that the operator symbol in (1.1) contains multipliers that oscillate at infinity (see equalities(3.9) and (3.11)). This fact is essentially used in the proof of the corresponding theorems. The methoddeveloped in this paper for estimating potential (1.1) is based on obtaining special presentations for theoperator symbol in (1.1). Namely, we represent it as the sum of some integrals that contain oscillatingexponents and then apply to these integrals the stationary phase method and results obtained in [2] for“model” multipliers
m±b (|ξ|) = v(|ξ|2)|ξ|−be±i|ξ|, b > 0, (1.2)
where v(r) ∈ C∞(0,∞), 0 ≤ v(r) ≤ 1; v(r) = 0, if r ≤ 1, and v(r) = 1, if r ≥ 2.Note that the indicated bounds were obtained earlier in [2] only for multiplier operators with symbols
(1.2) (see Theorem 2.1).
Below we use the following denotations: Sn−1 is the unit sphere in R
n; f(ξ) =∫
Rn
f(x)eiξxdx is
the Fourier transform of the function f ; (F−1f)(ξ) = ˜f(ξ) = (2π)−n(Ff)(−ξ) is the inverse Fouriertransform; S is the Schwartz class of quickly decreasing smooth functions.
*E-mail: [email protected].**E-mail: [email protected].
7
8 GIL’, NOGIN
2. AUXILIARY INFORMATION
2.1. Some spaces of functions and distributions. We denote by H1 = H1(Rn) the set of summablefunctions f ∈ L1 such that for any function a(x) ∈ S it holds
∫
a(x)dx �= 0, the maximal function isdefined as follows:
Maf(x) = supt>0
|(at ∗ f)(x)| ∈ L1,
where at(x) = t−na(x/t). If f ∈ H1, then f ∈ L1 and∫
f(x)dx = 0. Evidently, H1 is a closed subspacein L1.
We denote by BMO = BMO(Rn) the set of all locally integrable functions such that
‖f‖BMO = supB
{
1|B|
∫
B|f(x) − fB|dx
}
< ∞,
where fB = 1|B|
∫
B
f(x)dx and the supremum is taken over all balls B in Rn. Note that the space BMO
is conjugate to H1 ([6], P. 142).In the proof of Lemma 2.1 we essentially use the Fefferman inequality ([6], P. 142): If f ∈ H1,
g ∈ BMO, and fg ∈ L1, then∣
∣
∣
∣
∫
Rn
f(x)g(x)dx
∣
∣
∣
∣
≤ C‖f‖H1‖g‖BMO. (2.1)
Let X and Y coincide with H1 or L1. Following [2], we denote by K(X,Y ) the space of all k ∈ S ′
such that
‖k‖K(X,Y ) = sup{
‖k ∗ f‖Y
‖f‖X: f ∈ S ∩ X, ‖f‖X �= 0
}
< ∞.
We denote by M(X,Y ) the set of generalized functions m ∈ S ′ such that
‖m‖M(X,Y ) = sup
{
‖F−1(m f)‖Y
‖f‖X: f ∈ S ∩ X, ‖f‖X �= 0
}
< ∞.
Therefore,
‖m‖M(X,Y ) = ‖F−1m‖K(X,Y ). (2.2)
Below we use the following equality from theorem 3.3 in [2] (P. 278):
K(L1,H1) = H1. (2.3)
2.2. Some Hp − Hq multipliers. For multiplier (1.2) the following theorem is valid.
Theorem 2.1 ([2], P. 284). The following equivalences take place:
a) m±b (|ξ|) ∈ M(L1,H1) = M(BMO,L∞) if and only if b > (n − 1)/2;
b) m±b (|ξ|) ∈ M(H1,H1) = M(BMO,BMO) if and only if b ≥ (n − 1)/2;
c) m±b (|ξ|) ∈ M(Hp,Hq), 0 < p ≤ q < ∞, if and only if 1/p + 1/q ≤ 1, 1/p − n/q ≤ b − (n −
1)/2 or 1/p + 1/q ≥ 1, n/p − 1/q ≤ b + (n − 1)/2.
Below we also use the following assertions.
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L1 − H1 BOUNDS FOR A GENERALIZED STRICHARTZ POTENTIAL 9
Theorem 2.2 ([7], P. 167). Let 0 < p ≤ 1 and k = 1 + max{[
n(1p − 1
2)]
,[
n2
]}
. If m(ξ) is a bounded
function from the class Ck(Rn \ {0}) and∣
∣
∣
∣
(
∂
∂ξ
)α
m(ξ)∣
∣
∣
∣
≤ B|ξ|−|α|
with |α| ≤ k, then
m(ξ) ∈ M(Hp,Hp) and ‖m‖M(Hp,Hp) ≤ Ap,
where the constant Ap depends only on B, p, and n.
Lemma 2.1. If k(y) ∈ S and∫
Rn
k(y)dy = 0, then k(y) ∈ K(BMO,L∞) = K(L1,H1).
Proof. In accordance with inequality (2.1) we have k(y) ∈ K(BMO,L∞), because k(y) ∈ H1 and thenorm of the space BMO(Rn) is invariant with respect to the shift. Then due to the duality we getk(y) ∈ K(L1,H1).
2.3. The uniform asymptotic expansion of the Bessel function Jν(z). Let −π/2 < α < π/2and Ω = {z ∈ C : |z| > η, | arg z| < θ}, where η > 0 and θ ∈ (0, π/2). We represent Jν(z) as a linear
combination of the Hankel functions H(1)±ν (z) and H
(2)±ν (z) (here we use +ν, if ν > −1/2, and −ν
otherwise), apply the results of [8] (P. 220), and for z ∈ Ω obtain the equality
Jν(z) =(
πz
2
)−1/2[
e−iz
( N∑
l=0
C(ν)l,−z−l + R
(ν)N,−(z)
)
+ eiz
( N∑
l=0
C(ν)l,+z−l + R
(ν)N,+(z)
)]
, (2.4)
where C(ν)0,± = 1
2e∓(iπ/4)(2ν+1) ,
R(ν)N,±(z) =
B±N
zN+1Q
(ν)N,±(z), (2.5)
Q(ν)N,±(z) =
∫ 1
0(1 − t)Ndt
∫ ∞ exp(iα)
0e−uuν+N+1/2
(
1 − ut
−2iz
)ν−N− 32
du. (2.6)
2.4. The asymptotic expansion of some integrals containing an oscillating exponent. Theanalysis of the proof of the Erdelyi lemma in [9] (pp. 97–100) shows that the following assertion is true.
Lemma 2.2. Let β > 0, f(x) ∈ C∞([0, a]), and f (j)(a) = 0 (j = 0, 1, . . . ). Then∫ a
0xβ−1f(x)e±iλx dx = a±β λ−β + W±,β
1 (λ), λ ≥ 1, (2.7)
where a±β = f(0)Γ(β)(±i)β ;
|(W±,β1 (λ))(j)| ≤ C±,j
λ1+β+j, λ > 1, j = 0, 1, 2, . . . , (2.8)
and constants C±,j are independent of λ.
3. MAIN RESULTS
For operator (1.1) the following theorem is true.
Theorem 3.1. Let β > 0 and∫ 1
0ρn−1(1 − ρ2)β−1θ(ρ) dρ = 0. (3.1)
Then the operator Kβθ is bounded from L1 to H1. The operator K0
θ is not bounded from L1 to H1.
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10 GIL’, NOGIN
Proof. Let β > 0. Let us show that
kβθ (ξ) ∈ M(L1,H1). (3.2)
Then, taking into account formula (2.2) and remark 2.3 in [2], we obtain
‖Kβθ ϕ‖H1 ≤ ‖kβ
θ ‖K(L1,H1)‖ϕ‖L1 , ϕ ∈ L1.
Represent the operator Kβθ in the form
(Kβθ ϕ)(x) = (Kβ,1
θ ϕ)(x) + (Kβ,2θ ϕ)(x), (3.3)
where
(Kβ,1θ )(x) =
1Γ(β)
∫
1−δ≤|y|≤1
ω(|y|)kβθ (y)ϕ(x − y)dy, (3.4)
(Kβ,2θ ϕ)(x) =
1Γ(β)
∫
|y|≤1−δ/2
(1 − ω(|y|))kβθ (y)ϕ(x − y)dy,
the function ω(r) ∈ C∞(0,∞) is such that 0 ≤ ω(r) ≤ 1; ω(r) = 0, if r /∈ (1 − δ, 1); and ω(r) = 1, ifr ∈ [1 − δ/2, 1], 0 < δ < 1.
Consider the operator Kβ,1θ . We write the Fourier transform of its kernel m1
β,θ(ξ) = (ω · kβθ )(ξ) in the
form
m1β,θ(ξ) =
1Γ(β)
∫ 1
1−δρn−1(1 − ρ2)β−1θ(ρ)ω(ρ) dρ
∫
Sn−1
eiρ(ξ·σ) dσ.
We have
m1β,θ(ξ) = (1 − v(|ξ|2))m1
β,θ(ξ) + v(|ξ|2)m1β,θ(ξ) ≡ m0
β,θ(ξ) + m∞β,θ(ξ).
Consider m∞β,θ(ξ). Applying the equality
∫
Sn−1
ei(x·σ) dσ =(2π)
n2
|x|n2−1
Jn2−1(|x|) (3.5)
and formula (2.4) with N =[
n+12
]
+ 1, we obtain
m∞β,θ(ξ) =
N∑
k=0
(hβ,k,−1 (|ξ|) + hβ,k,+
1 (|ξ|)) + Rβ,N,−1 (|ξ|) + Rβ,N,+
1 (|ξ|),
where
hβ,k,±1 (|ξ|) =
v(|ξ|2)|ξ|n−1
2+k
∫ 1
1−δ(1 − ρ)β−1gk(ρ)e±iρ|ξ| dρ, 0 ≤ k ≤ N, (3.6)
gk(ρ) = γk,±ρn−1
2−k(1 + ρ)β−1θ(ρ)ω(ρ), γ0,± =
1Γ(β)
(2π)n−1
2 e∓iπ4
(n−1),
Rβ,N,±1 (|ξ|) =
γN+1,±γ0,±
v(|ξ|2)|ξ|n−1
2
∫ 1
1−δ(1 − ρ)β−1g0(ρ)e±iρ|ξ|R
(n−22 )
N,± (ρ|ξ|) dρ. (3.7)
Consider multiplier (3.6). The replacement 1 − ρ = τ gives
hβ,k,±1 (|ξ|) =
e±i|ξ|v(|ξ|2)|ξ|n−1
2+k
∫ δ
0τβ−1gk(1 − τ)e∓i|ξ|τ dτ. (3.8)
RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 55 No. 9 2011
L1 − H1 BOUNDS FOR A GENERALIZED STRICHARTZ POTENTIAL 11
Let a function v(|ξ|2) be such that v(r) ∈ C∞(0,∞); 0 ≤ v(r) ≤ 1; v(r) = 0, if r ≤ 1; and v(r) = 1, ifr ≥ 2. Then v(r2)v(r2) = v(r2). We transform multiplier (3.8), using Lemma 2.2. Taking into accountformula (2.7), we get
hβ,k,±1 (|ξ|) = v(|ξ|2)e±i|ξ| · |ξ| 1−n
2−k−β · v(|ξ|2)(ak,∓
β + |ξ|βW∓,β1 (|ξ|)), (3.9)
where ak,∓β = γk,±θ(1)(∓i)βΓ(β) and W∓,β
1 (|ξ|) satisfies inequality (2.8).
Note that by Theorem 2.1 a) we have
v(|ξ|2)e±i|ξ||ξ|1−n
2−β ∈ M(L1,H1). (3.10)
In addition, according to Theorem 2.2,
v(|ξ|2)(ak,∓β + |ξ|βW∓,β
1 (|ξ|)) ∈ M(H1,H1).
Then
hβ,0,±1 (|ξ|) ∈ M(L1,H1).
Analogously we prove that hβ,k,±1 (|ξ|) ∈ M(L1,H1), 1 ≤ k ≤ N .
Consider formula (3.7). After the replacement 1 − ρ = τ , taking into account equality (2.5), we get
Rβ,N,±1 (|ξ|) =
B−Ne±i|ξ|v(|ξ|2)|ξ|n−1
2+N+1
∫ δ
0τβ−1gN+1(1 − τ)e∓i|ξ|τQ
(n−22 )
N,± ((1 − τ)|ξ|) dτ.
Applying Lemma 2.2, we obtain
Rβ,N,±1 (|ξ|) = B−
Nv(|ξ|2)e±i|ξ||ξ| 1−n2
−N−1−β v(|ξ|2)(aN+1,∓β (|ξ|) + |ξ|βW∓,β
1 (|ξ|)), (3.11)
where aN+1,∓β (|ξ|) = γN+1,±θ(1)(∓i)βΓ(β)Q
(n−22
)
N,± (|ξ|) and Q(n−2
2)
N,± (z) takes form (2.6). Then by Theo-rem 2.1 a) we have
B−Nv(|ξ|2)e±i|ξ||ξ|−
n+12
−N−β ∈ M(L1,H1).
In addition, in view of Theorem 2.2,
v(|ξ|2)(
aN+1,∓β (|ξ|) + |ξ|βW∓,β
1 (|ξ|))
∈ M(H1,H1).
Hence Rβ,N,±1 (|ξ|) ∈ M(L1,H1).
Summing up the above reasoning, we obtain
m∞β,θ(ξ) ∈ M(L1,H1). (3.12)
Consider the operator Kβ,2θ . We write the Fourier transform of its kernel m2
β,θ(ξ) = ((1 − ω)kβθ )(ξ)
in the form
m2β,θ(ξ) =
1Γ(β)
∫ 1−δ/2
0ρn−1(1 − ρ2)β−1θ(ρ)(1 − ω(ρ)) dρ
∫
Sn−1
eiρ(ξ·σ) dσ.
We have
m2β,θ(ξ) = (1 − v(|ξ|2))m2
β,θ(ξ) + v(|ξ|2)m2β,θ(ξ).
Consider the multiplier v(|ξ|2)m2β,θ(ξ). Applying formula (3.5), we obtain
v(|ξ|2)m2β,θ(ξ) =
v(|ξ|2)|ξ|n
2−1
∫ 1−δ/2
0u(ρ)ρ
n2 Jn
2−1(ρ|ξ|) dρ, (3.13)
where u(ρ) = 1Γ(β)(2π)n/2(1 − ρ2)β−1θ(ρ)(1 − ω(ρ)).
RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 55 No. 9 2011
12 GIL’, NOGIN
Integrating integral (3.13) by parts l times, taking into account the formula∫
zn+1Jn(z)dz = zn+1Jn+1(z)
([10], P. 648, formula 5.52(1)), we get
v(|ξ|2)m2β,θ(ξ) =
(−1)lv(|ξ|2)|ξ|n−2
2+l
∫ 1−δ/2
0
(
d
dρ
1ρ
)l
(ρu(ρ)) ρn−2
2+lJn−2
2+l(ρ|ξ|) dρ
=v(|ξ|2)ei|ξ|
|ξ|n−12
+β
(−1)lv(|ξ|2)e−i|ξ|
|ξ|l−β− 12
∫ 1−δ/2
0
(
d
dρ
1ρ
)l
(ρu(ρ)) ρn−2
2+lJn−2
2+l(ρ|ξ|) dρ.
Let l = [β] + 3 + max{[
n(1p − 1
2)]
,[
n2
]}
. Note that by Theorem 2.1 a) it holds v(|ξ|2)ei|ξ||ξ| 1−n2
−β ∈M(L1,H1). In addition, according to Theorem 2.2,
(−1)lv(|ξ|2)e−i|ξ|
|ξ|l−β− 12
∫ 1−δ/2
0
(
d
dρ
1ρ
)l
(ρu(ρ)) ρn−2
2+lJn−2
2+l(ρ|ξ|) dρ ∈ M(H1,H1)
and is a multiplier in S. Then
v(|ξ|2)m2β,θ(ξ) ∈ M(L1,H1). (3.14)
In view of Lemma 2.1 we obtain
F−1(m0β,θ(ξ))(x) + F−1
(
(1 − v(|ξ|2))m2β,θ(ξ)
)
(x) ∈ K(L1,H1). (3.15)
Then formulas (3.12), (3.14), and (3.15) give (3.2).Let us now consider the case β = 0. Let us construct an analytic continuation of convolution (1.1)
from the half-plane Re β > 0 to that Re β > −1, taking into account correlation (3.3). Assuming thatRe β > 0 and ϕ ∈ L1, passing to polar coordinates in integral (3.4), and integrating by parts, we obtain
(Kβ,1θ ϕ)(x) =
12Γ(β + 1)
∫ 1
1−δ(1 − ρ2)β
∂
∂ρ
(
ρn−2θ(ρ)ω(ρ)∫
Sn−1
ϕ(x − ρσ) dσ
)
dρ. (3.16)
Represent the operator Kβ,2θ in the form
(Kβ,2θ ϕ)(x) =
β
Γ(β + 1)
∫
|y|<1−δ/2(1 − ω(|y|))kβ
θ (y)ϕ(x − y)dy. (3.17)
From correlations (3.16) and (3.17) we see that the right-hand side of (3.3) is an analytic function onthe half-plane Re β > −1. In formulas (3.16) and (3.17) we set β = 0 and obtain
(K0θ ϕ)(x) =
θ(1)2
∫
Sn−1
ϕ(x − σ) dσ. (3.18)
In what follows, K0θ is the operator defined by equality (3.18). Passing to Fourier images in (3.18),
we get
where
k0θ(ξ) =
θ(1)2
∫
Sn−1
eiσξ dσ.
We write the symbol of the operator K0θ in the form
k0θ(ξ) = (1 − v(|ξ|2)) k0
θ(ξ) + v(|ξ|2) k0θ(ξ) ≡ k0
θ,0(ξ) + k0θ,∞(ξ). (3.19)
RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 55 No. 9 2011
L1 − H1 BOUNDS FOR A GENERALIZED STRICHARTZ POTENTIAL 13
Consider k0θ,∞(ξ). Applying formulas (3.5) and (2.4) with N =
[
n+12
]
+ 2, we obtain
k0θ,∞(ξ) =
θ(1)v(|ξ|2)2
(2π)n/2
|ξ|n−22
Jn−22
(|ξ|)
=N
∑
k=0
(hk,−0 (|ξ|) + hk,+
0 (|ξ|)) + RN,−0 (|ξ|) + RN,+
0 (|ξ|), (3.20)
where
hk,±0 (|ξ|) =
θ(1)γk,±v(|ξ|2)e±i|ξ|
2|ξ|n−12
+k, 0 ≤ k ≤ N, γ0,± = (2π)
n−12 e∓
iπ4
(n−1),
RN,±0 (|ξ|) =
θ(1)γ0,±v(|ξ|2)e±i|ξ|
2|ξ|n−12
R(n−2
2)
N,± (|ξ|).
Denote
μ0(|ξ|) ≡ h0,+0,2 (|ξ|) + h0,−
0,2 (|ξ|) =b−0 v(|ξ|2)ei|ξ|
|ξ|n−12
+b+0 v(|ξ|2)e−i|ξ|
|ξ|n−12
,
where b∓0 = (2π)n−1
2 e∓iπ4
(n−1) θ(1)2 . Consider
μ0(|x|) =1
(2π)n
∫ ∞
0μ0(|ρ|)ρn−1dρ
∫
Sn−1
e−iρσxdσ.
Applying theorem 5.1 (iii) from [2], we conclude that the function μ0(|x|) is smooth in Rn \ {|x| = 1} and
μ0(|x|) = O(|x|−l) as |x| → ∞ for any l > 0. Let us calculate the asymptote of the function μ0(|x|) as|x| → 1. Applying formula (3.5) and the asymptotic expansion of the Bessel function (2.4) with N = 1,after simple transformations we obtain
μ0(|x|) = limε→0
[
1
|x|n−12
I1 +1
|x|n+12
I2
]
+ (a bounded function),
where
I1 =∫ ∞
0(a−0 C0,−eiρ(1−|x|)−ερ + a+
0 C0,+e−iρ(1−|x|)−ερ)dρ,
I2 =∫ ∞
1(a−0 C1,−eiρ(1−|x|)−ερ + a+
0 C1,+e−iρ(1−|x|)−ερ)dρ
ρ,
a∓0 = e∓iπ4
(n−1) θ(1)2π , C0,± = 1
2e∓(iπ/4)(n−1), and C1,± = ∓n2−4n+34i e∓(iπ/4)(n−1).
Calculating I1 and applying formulas 3.351(5) and 8.214(2) from [10] (pp. 324, 941) to I2, we get
μ0(|x|) =1
|x|n−12
(
a−0 C0,−eiπ2 + a+
0 C0,+e−iπ2
)
1 − |x| +C
|x|n+12
(
a−0 C1,− + a+0 C1,+
)
− 1
|x|n+12
(
a−0 C1,− + a+0 C1,+
)
ln(i(1 − |x|)) − a−0 C1,−iπ
|x|n+12
+ (a bounded function),
where C is the Euler constant. Due to equalities a−0 C1,−+a+0 C1,+ = 0 and a−0 C0,−e
iπ2 +a+
0 C0,+e−iπ2 = 0
we conclude that the function μ0(|x|) is bounded in the neighborhood Sn−1.
Therefore, μ0(|x|) ∈ L1.Taking into account correlations (3.19) and (3.20) and theorem 5.1 (iii) from [2], we conclude that
the function k0θ(x) − μ0(|x|) has the same properties as μ0(|x|). Therefore, k0
θ(x) ∈ L1. However,
RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 55 No. 9 2011
14 GIL’, NOGIN
k0θ(0) = θ(1)
2
∫
Sn−1
dσ �= 0, whence k0θ(y) /∈ H1. Due to formula (2.3) we conclude that the operator K0
θ
is not bounded from L1 to H1.
Due to the duality Theorem 3.1 implies the next assertion.
Theorem 3.2. Let β > 0 and let condition (3.1) be fulfilled. Then the operator Kβθ is bounded from
BMO to L∞.
Remark 3.1. As is mentioned in the proof of Theorem 3.1, the operator K0θ is not bounded from L1 to
H1. Therefore, it is not bounded from BMO to L∞.
Remark 3.2. Condition (3.1) is necessary for the boundedness of the operator Kβθ from L1 to H1
and from BMO to L∞. Really, if Kβθ ϕ ∈ H1, then Kβ
θ ϕ ∈ L1 and Kβ
θ ϕ(0) = 0. We have Kβ
θ ϕ(ξ) =
kβθ (ξ)ϕ(ξ). Due to the arbitrariness of the function ϕ(∈ L1) we obtain
Kβ
θ ϕ(0) = 0 ⇐⇒
kβθ (0) = 0 ⇐⇒ condition (3.1) is fulfilled.
Based on the above representations for
kβθ (ξ), by applying Theorem 2.1 b), c) we obtain the following
assertion.
Theorem 3.3. Let β ≥ 0. The operator Kβθ is bounded from H1 to Hp, 0 < p ≤ 1, if and only
if p = 1. In addition, the operator Kβθ is bounded from H1 to Lp, 1 ≤ p ≤ ∞, if and only if
1 − β ≤ 1/p.
Note that the multiplier that equals (ei|ξ| + e−i|ξ|)v(|ξ|2)|ξ| 1−n2
−β (see (3.10)) belongs to M(H1, Lp),
if 1 − β ≤ 1/p ≤ 1. And the multiplier that equals
kβθ (ξ) − (ei|ξ| + e−i|ξ|)v(|ξ|2)|ξ| 1−n
2−β belongs to
M(H1, Lp), if 1 ≤ p ≤ ∞.
ACKNOWLEDGMENTSThis work was supported by the South Mathematical Institute of the Vladikavkaz Scientific Center
of the Russian Academy of Sciences.
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Translated by O. V. Pinyagina
RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 55 No. 9 2011