arXiv:math/9210207v1 [math.FA] 19 Oct 1992
SCHOENBERG’S PROBLEM ON POSITIVE DEFINITE
FUNCTIONS
ALEXANDER KOLDOBSKY
University of Missouri–Columbia
Department of Mathematics
Columbia, MO 65211
1. Introduction
In 1938, I. J. Schoenberg [28] posed the following problem: For which
numbers β > 0 is the function exp(−kxkβq ) positive definite on R⋉?
Here q > 2 and kxkq = (|x1 |q + · · · + |xn |q )1/q . Denote by Bn (q) the set
of such numbers β.
We prove in this article that the functions exp(−kxkβq ) are not positive definite for all n ≥ 3, q > 2 and β > 0, i.e. Bn (q) = ∅ for every
q > 2 and n ≥ 3.
Besides, B2 (q) = (0, 1] for every q > 2. In the case n = 2 we have
to prove only that B2 (q) ∩ (1, ∞) = ∅. In fact, it is well–known that
exp(−kxkβ ) is a positive definite function for every two–dimensional
norm and every β ∈ (0, 1], see Ferguson [9], Hertz [12], Lindenstrauss
and Tzafriri [20], Dor [7], Misiewicz and Ryll–Nardziewski [23], Yost
[31], Koldobsky [14] for different proofs.
Thus we give a complete answer to Schoenberg’s question.
1
The case q ∈ (0, 2] was settled by Schoenberg [28]. Here one has
Bn (q) = (0, q] for every n ≥ 2.
Let us mention some facts which are valid for all q > 0, see Schoenberg [28]:
(i) B1 (q) = (0, 2];
(ii) Bn (q) ⊃ Bn+1 (q) for every n ∈ N;
(iii) if β ∈ Bn (q) then (0, β] ⊂ Bn (q).
Thanks to (ii) and (iii) it is enough to prove that B3 (q) ∩ (0, 2) = ∅
to verify that Bn (q) = ∅ for every n ≥ 3.
However, we treat here a more general situation. Denote by φn (q)
the class of all even functions f : R → R which are such that f (kxkq )
is a characteristic function, i.e. there exists a probability measure µ on
R⋉ with µ̂(x) = f (kxkq ) for all x ∈ R⋉.
It is clear that φn (q) ⊃ φn+1 (q). If f ∈ φn (q) then by Bochner’s
theorem the function f (kxkq ) is positive definite, as well as f itself.
Hence there exists a probability measure ν on R with ν̂ = f .
We prove the following
Theorem . Let q > 2, n ∈ N, f ∈ φn (q), f 6≡ 1 and ν̂ = f . Then in
every one of the cases:
a) n ≥ 3, β ∈ (0, 2);
b) n = 2, β ∈ (1, 2);
c) n ≥ 4, β ∈ (−1, 0)
the β–th moment of the measure ν is infinite, i.e.
R
R
|t|β dν(t) = ∞.
Using a) one can immediately get an answer to Schoenberg’s question
for n ≥ 3. Indeed, for every β ∈ (0, 2), exp(−|t|β ) is the characteristic
function of the β–stable measure on R. This measure has finite moments of all positive orders less than β, see Zolotarev [30] or the text
2
preceding Lemma 4 below. By a) the function exp(−|t|β ) 6∈ φ3 (q) for
every q > 2. By Bochner’s theorem the function exp(−kxkβq ) is not
positive definite on R3. Thus B3 (q) ∩ (0, 2) = ∅ for every q > 2 and we
are done.
For the same reason, it follows from b) that B2 (q) ∩ (1, 2) = ∅ for
every q > 2.
The statements a) and c) show that the measure ν corresponding to
a function f ∈ φn (q), n ≥ 4, q > 2 must have a very special behavior
at infinity and at zero. Nevertheless, the following question is open:
For n ≥ 3 and q > 2, are there any functions in the class φn (q) besides
the function f ≡ 1?
The classes φn (q) have been investigated by several authors. Schoenberg [29] described the classes φn (2) and φ∞ (2) =
T
n
φn (2) completely.
A similar result was obtained for the classes φn (1) by Cambanis et al
[5]. Bretagnolle et al [4] described the classes φ∞ (q) for all q > 0. In
particular, for very q > 2 the class φ∞ (q) contains no functions besides f ≡ 1. For some partial results on the classes φn (q), 0 < q < 2,
see Richards [26, 27] and Misiewicz [21]. Misiewicz [22] proved that
for n ≥ 3 a function f (max(|x1 |, . . . , |xn |)) is positive definite only if
f ≡ 1. This result gives an answer to Problem 2 from Schoenberg [28].
References related to the topic include also Askey [2], Berg and Ressel
[3], Kuelbs [16], Eaton [8], Christensen and Ressel [6], Kuritsyn [17],
Kuritsyn and Shestakov [18], Misiewicz and Scheffer [24], Aharoni et
al [1].
The well known theorem of Bretagnolle et al [4] states that a Banach
space (E, k · k) is isometric to a subspace of Lβ ([0, 1]), β ∈ [1, 2] iff
the function exp(−kxkβ ) is positive definite. Dor [7] pointed out all
the pairs of numbers β, q ∈ [1, ∞) for which the space lqn , n ≥ 2,
3
is isometric to a subspace of Lβ ([0, 1]). Combining these results one
can prove that Bn (q) ∩ (1, ∞) = ∅ for every n ≥ 2 and q > 2. So
Schoenberg’s problem was open only for β ∈ (0, 1).
In spite of the connection between Schoenberg’s problem and isometries, in this paper we don’t deal with isometries directly. Our main
tool is the Fourier transform of distributions.
2. Some Applications of the Fourier Transform
Let (E, k · k) be an n–dimensional Banach space and f : R → R
be an even function such that there exists a probability measure µ on
R⋉ with µ̂(x) = f (kxk) for all x ∈ R⋉. Measures with characteristic
functions of the form f (kxk) are called E–stable or pseudo–isotropic,
since all one–dimensional projectional of such measures are equal up
to a scale parameter. We give an easy proof of this fact.
Lemma 1 (see Levy [19], Eaton [8]). If µ is a probability measure on
R⋉ with the characteristic function f (kxk) then:
a) there exists a probability measure ν on R with ν̂ = f ;
b) for every x ∈ R⋉, x 6= 0, the image of the measure µ under the
mapping ξ → hx, ξi/kxk from R⋉ to R coincides with the measure
ν (here hx, ξi stands for the scalar product).
Proof. Fix an element x ∈ R⋉, x 6= 0. Let ν be the image of the
measure µ under the mapping ξ → hx, ξi/kxk. Put y = hx, ξi/kxk.
Then for every k ∈ R
f (kkxk) = µ̂(kx) =
=
Z
R⋉
Z
R⋉
exp(−ihkx, ξi) dµ(ξ)
exp(−ikkxk(hx, ξi/kxk)) dµ(ξ) =
Z
exp(−ikkxky) dν(y) = ν̂(kkxk).
R
Hence ν̂ = f and ν doesn’t depend on the choice of x ∈ R⋉, x 6= 0.
4
Let S(R⋉) be, as usual, the space of rapidly decreasing infinitely
differentiable functions. Denote by S ′ (R⋉) the space of distribution
over S(R⋉).
Let Ω be an open subset of R⋉. A distribution g ∈ S ′ (R⋉) is called
positive (negative) on Ω if hg, ψi ≥ 0 (hg, ψi ≤ 0) for every non–
negative function ψ ∈ S(R⋉) with supp ψ ⊂ Ω.
Lemma 2 (cf. Koldobsky [13]). Let β ∈ (−1, ∞), β 6= 0, 2, 4, . . . . Let
ψ ∈ S(R⋉) be a function with 0 6∈ supp ψ. Then for every ξ ∈ R⋉,
ξ 6= 0,
Z
β
R⋉
|hx, ξi| ψ̂(x) dx = cβ
Z
R
|t|−1−β ψ(tξ) dt, cβ =
2β+1 π 1/2 Γ((β + 1)/2)
.
Γ(−β/2)
Proof. It is well–known that (|x|β )∧ (t) = cβ |t|−1−β , t 6= 0, for all β ∈
(−1, ∞), β 6= 0, 2, 4, . . . , see Gelfand and Shilov [10]. By the Fubini
theorem
Z
R⋉
β
|hx, ξi| ψ̂(x) dx =
Z
R
|z|
β
Z
hx,ξi=z
β
ψ̂(x) dx dz = |z| ,
Z
hx,ξi=z
ψ̂(x) dx .
(2.1)
The function t → 2πψ(−tξ) is the Fourier transform of the function
z →
R
hx,ξi=z
ψ̂(x) dx (it is a simple property of the Radon transform,
see Gelfand et al [11]). Therefore, we can continue the equality (2.1):
= cβ h|t|−1−β , ψ(tξ)i = cβ
Z
R
|t|−1−β ψ(tξ) dt.
Lemma 3. Let f , µ and ν be as in Lemma 1. Additionally, assume
that f 6≡ 1 and
R
R
|t|β dν(t) < ∞ for a number β ∈ (−1, 2), β 6= 0.
Then, if β ∈ (−1, 0) the distribution (kxkβ )∧ is positive on R⋉\{0}. If
β ∈ (0, 2) the distribution (kxkβ )∧ is negative on R⋉\{0}.
5
Proof. Since f 6≡ 1, the measure ν is not supported at zero and
0. By Lemma 1 we have
Z
R⋉
β
|hx, ξi| dµ(ξ) = kxk
β
Z
R⋉
hx, ξi β
kxk dµ(ξ) = kxk
β
Z
R
R
R
|t|β dν(t) >
|t|β dν(t)
for every x ∈ R⋉. It follows from Lemma 2 and the Fubini theorem
that
1
β
R |t| dν(t)
h(kxkβ )∧ , ψi = hkxkβ , ψ̂i = R
cβ
=R
β
R |t| dν(t)
Z
R
dµ(ξ)
Z
R
Z
R⋉
dµ(ξ)
Z
R⋉
|hx, ξi|β ψ̂(x) dx
|t|−1−β ψ(tξ) dt
for every function ψ ∈ S(R⋉) with 0 6∈ supp ψ. If the function ψ is
non–negative, the quantity in the right–hand side has the same sign as
the number cβ . It suffices to remark that cβ > 0 if β ∈ (−1, 0) and
cβ < 0 if β ∈ (0, 2).
For arbitrary q > 0, define the function γq on R by γq (t) = (exp(−|x|q ))∧ (t),
t ∈ R. Then
lim t1+q γq (t) = 2Γ(q + 1) sin(πq/2),
t→∞
see Polya and Szego [25], Part 3, Problem 154. Therefore, the integral
Sq (α) =
Z
R
|t|α γq (t) dt
converges absolutely for every α ∈ (−1, q).
Lemma 4 (for the case q ∈ (0, 2), see Zolotarev [30] or Koldobsky [13]).
Let
q > 2. Then for every α ∈ (−1, q), α 6= 0, 2, . . . , 2[q/2],
Sq (α) = 2α+2 π 1/2 Γ(−α/q)Γ((α + 1)/2)/(qΓ(−α/2)).
In particular, Sq (α) > 0 if α ∈ (−1, 2), α 6= 0, and Sq (α) < 0 if
α ∈ (2, min(4, q)).
6
Proof. Assume −1 < α < 0. By the Parseval theorem
Sq (α) =
=
2
Z
R
|t|α γq (t) dt = cα
Z
|z|−1−α exp(−|z|q ) dz
R
(2.2)
α+1 1/2
π
Γ((α + 1)/2) 2Γ(−α/q)
·
.
Γ(−α/2)
q
If we allow α to assume complex values then all functions of α in (2.2)
are analytic in the domain {−1 < Reα < q, α 6= 0, 2, . . . , 2[q/2]}. Since
analytic continuation from the interval (−1, 0) is unique, the equality
(2.2) remains valid for all α ∈ (−1, q), α 6= 0, 2, . . . , 2[q/2]. To complete
the proof, note that Γ(z) > 0 if z > 0 or z ∈ (−2, −1), and Γ(z) < 0 if
z ∈ (−1, 0).
Let us compute the Fourier transform of the function kxkβq .
Lemma 5. Let q > 0, n ∈ N, −n < β < qn, β/q 6∈ N ∪ {0}, ξ =
(ξ1 , . . . , ξn ) ∈ R⋉, ξk 6= 0, 1 ≤ k ≤ n. Then
Z ∞
n
Y
q
tn+β−1
γq (tξk ) dt
Γ(−β/q) 0
k=1
(2.3)
((|x1 |q + · · · + |xn |q )β/q )∧ (ξ) =
Proof. Assume −1 < β < 0. By definition of the Γ–function
q
q β/q
(|x1 | +· · ·+|xn | )
q
=
Γ(−β/q)
Z
∞
0
y −1−β exp(−y q (|x1 |q +· · ·+|xn |q )) dy.
For every fixed y > 0,
(exp(−y q (|x1 |q + · · · + |xn |q ))∧ (ξ) = y −n
n
Y
γq (tξk ).
k=1
Put t = 1/y. We have
((|x1 |q + · · · + |xn |q )β/q )∧ (ξ) =
=
q
Γ(−β/q)
Z
∞
0
tt+β−1
n
Y
q
Γ(−β/q)
γq (tξk ) dt.
k=1
7
Z
0
∞
y −n−β−1
n
Y
k=1
γq (ξk /y) dy
The latter integral converges if −n < β < qn because the function
t→
Qn
k=1
γq (tξk ) decreases at infinity like t−n−nq (remind that ξk 6= 0,
1 ≤ k ≤ n; see the text preceding Lemma 4).
If β is allowed to assume complex values then the both sides of (2.3)
are analytic functions of β in the domain {−n < Reβ < nq, β/q 6∈
N ∪ {0}}. These two functions admit unique analytic continuation
from the interval (−1, 0). Thus the equality (2.3) remains valid for all
β ∈ (−n, qn), β/q 6∈ N ∪ {0} (see Gelfand and Shilov [10] for details of
analytic continuation in such situations).
For q ∈ [1, 2], Lemma 5 was proved in Koldobsky [15].
3. Proof of Theorem
Let n ∈ N, n ≥ 2, −n < β < qn, q > 2. Consider the integral
Jn (α1 , . . . , αn−1 )
=
=
Z
R⋉−1
Z
∞
Z
∞
0
=
0
|ξ1 |
α1
· · · · · |ξn−1|
tn+β−1 γq (t)
n−1
YZ
k=1 R
αn−1
·
Z
∞
n+β−1
t
0
γq (t)
n−1
Y
γq (tξk ) dt dξ1 · · · · · dξn−1
k=1
|ξk |αk γq (tξk ) dξk dt
t−α1 −···−αn−1 +β γq (t) dt ·
n−1
YZ
k=1 R
(3.1)
|z|αk γq (z) dz
= Sq (α1 ) · Sq (α2 ) · · · · · Sq (−α1 − α2 − · · · − αn−1 + β).
If all the numbers α1 , . . . , αn−1 , −α1 − · · · − αn−1 + β belong to the
interval (−1, q) then the integrals in (3.1) converge absolutely. Therefore, the Fibini theorem is applicable and the integral Jn (α1 , . . . , αn−1)
converges.
Case a): n ≥ 3, β ∈ (0, 2).
8
Note that φn (q) ⊃ φn+1 (q), so it is enough to prove the theorem for
n = 3.
Let n = 3 and assume that there exists a function f ∈ φ3 (q), f 6≡ 1,
with
R
R
|t|β dν(t) < ∞, where ν̂ = f and β ∈ (0, 2).
Put α1 = α2 = −1 + δ with δ ∈ (0, 1) and (2 + β − min(4, q)) <
δ < β/2. Then α1 , α2 ∈ (−1, 0) and −α1 − α2 + β ∈ (2, min(4, q)). By
Lemma 4, J3 (α1 , α2 ) = Sq (α1 ) · Sq (α2 ) · Sq (−α1 − α2 + β) < 0.
On the other hand, by Lemma 3 the function (kxkβq )∧ is negative on
R⋉\{0}. Since Γ(−β/q) < 0, it follows from Lemma 5 that
h(ξ1 , ξ2 ) =
Z
∞
0
tβ+2 γq (tξ1 )γq (tξ2 )γq (t) dt ≥ 0
for every ξ1 , ξ2 6= 0. Hence
J3 (α1 , α2 ) =
Z
R2
|ξ1|α1 · |ξ2 |α2 h(ξ1 , ξ2) dξ1 dξ2 > 0
and we get a contradiction.
In the cases b) and c) proofs are similar. In the case b) we put α1 =
−1 + δ with δ ∈ (0, 1) and 1 + β − min(4, q) < δ < β − 1. In the case c)
we may restrict ourselves to n = 4 and put α1 = α2 = α3 = −1+δ with
δ ∈ (0, 1) and (3 + β − min(4, q))/3 < δ < (β + 1)/3. Then the numbers
J2 (α1 ) and J4 (α1 , α2 , α3 ) are negative. On the other hand, if there
exists a function f for which the measure ν has finite β–th moment,
then Lemma 3 and Lemma 5 imply J2 (α1 ) ≥ 0 and J4 (α1 , α2 , α3 ) ≥ 0
and we get a contradiction (note that Γ(−β/q) > 0 if β ∈ (−1, 0) and
Γ(−β/q) < 0 if β ∈ (1, 2).)
Acknowledgment . I am grateful to Jolanta Misiewicz for helpful discussions during my stay in Technical University of Wroclaw in 1989.
9
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