A tempered Radon measure is a &#963;-finite Radon measure in Rn which generates a tempered distribution. We prove the following assertions. A Radon measure &#956; is tempered if, and only if, there is a real number &#946;such that ……. finite. A Radon measure is finite if, and only if, it belongs to the positive cone…….. (equivalent norms).A tempered Radon measure is a &#963;-finite Radon measure in Rn which generates a tempered distribution. We prove the following assertions. A Radon measure &#956; is tempered if, and only if, there is a real number &#946;such that ……. finite. A Radon measure is finite if, and only if, it belongs to the positive cone…….. (equivalent norms)

Kabanava, Maryia

Portal de Revistas Científicas Complutenses

Tempered Radon Measures
Maryia KABANAVA
Mathematical Institute
Friedrich Schiller University Jena
D-07737 Jena, Germany
kabanova@minet.uni-jena.de
Received: December 11, 2007
Accepted: January 1, 2008
ABSTRACT
A tempered Radon measure is a σ-ﬁnite Radon measure in Rn which generates
a tempered distribution. We prove the following assertions. A Radon measure
μ is tempered if, and only if, there is a real number β such that (1 + |x|2) β2 μ is
ﬁnite. A Radon measure is ﬁnite if, and only if, it belongs to the positive cone
+
B01∞(R
n) of B01∞(R
n). Then μ(Rn) ∼ ‖μ | B01∞(Rn)‖ (equivalent norms).
Key words: Radon measure, tempered distributions, Besov spaces
2000 Mathematics Subject Classiﬁcation: 42B35, 28C05.
Introduction
A substantial part of fractal geometry and fractal analysis deals with Radon measures
in Rn (also called fractal measures) with compact support. One may consult [5]
and the references given there. In the present paper we clarify the relation between
arbitrary σ-ﬁnite Radon measure in Rn, tempered distributions and weighted Besov
spaces. It comes out that a σ-ﬁnite Radon measure μ in Rn can be identiﬁed with a
tempered distribution μ ∈ S′(Rn) if and only if there is a real number β such that
μβ(Rn) < ∞, where μβ = (1 + |x|2)
β
2 μ.
Radon measures μ with μ(Rn) < ∞ are called ﬁnite. These ﬁnite Radon mea-
sures can be identiﬁed with the positive cone
+
B01∞(R
n) of the distinguished Besov
space B01∞(R
n) and
‖μ | B01∞(Rn)‖ ∼ μ(Rn)
Rev. Mat. Complut.
20 (2008), no. 2, 553–564 553 ISSN: 1139-1138
http://dx.doi.org/10.5209/rev_REMA.2008.v21.n2.16418
Maryia Kabanava Tempered Radon measures
(equivalent norms).
This paper is organised as follows. In section 1 we collect the deﬁnitions and
preliminaries. We introduce the well-known weighted Besov spaces Bspq(R
n, 〈x〉α)
and prove that for ﬁxed p, q with 0 < p, q ≤ ∞
S(Rn) =
⋂
α,s∈R
Bspq(R
n, 〈x〉α)
and
S′(Rn) =
⋃
α,s∈R
Bspq(R
n, 〈x〉α).
Although known to specialists we could not ﬁnd an explicit reference. In section 2 we
prove in the Theorems 2.1 and 2.2 the above indicated main results.
1. Deﬁnitions and preliminaries
Let N be the collection of all natural numbers and N0 = N∪{0}. Let Rn be Euclidean
n-space, where n ∈ N. Put R = R1, whereas C is the complex plane. Let S(Rn) be
the Schwartz space of all complex-valued, rapidly decreasing, inﬁnitely diﬀerentiable
functions on Rn. By S′(Rn) we denote its topological dual, the space of all tempered
distributions on Rn. Lp(Rn) with 0 < p ≤ ∞, is the standard quasi-Banach space
with respect to Lebesgue measure, quasi-normed by
‖f | Lp(Rn)‖ =
( ∫
Rn
|f(x)|p dx
) 1
p
, 0 < p < ∞
with the standard modiﬁcation if p = ∞.
If ϕ ∈ S(Rn) then
ϕˆ(ξ) = Fϕ(ξ) = (2π)−
n
2
∫
Rn
ϕ(x)e−ixξ dx, ξ ∈ Rn,
denotes the Fourier transform of ϕ. The inverse Fourier transform is given by
ϕˇ(x) = F−1ϕ(x) = (2π)−
n
2
∫
Rn
ϕ(ξ)eixξ dξ, x ∈ Rn.
One extends F and F−1 in the usual way from S to S′. For f ∈ S′(Rn),
Ff(ϕ) = f(Fϕ), ϕ ∈ S(Rn).
Let ϕ0 ∈ S(Rn) with
ϕ0(x) = 1, |x| ≤ 1 and ϕ0(x) = 0, |x| ≥ 32 , (1)
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Maryia Kabanava Tempered Radon measures
and let
ϕk(x) = ϕ0(2−kx)− ϕ0(2−k+1x), x ∈ Rn, k ∈ N. (2)
Then, since
1 =
∞∑
j=0
ϕj(x) for all x ∈ Rn, (3)
the ϕj form a dyadic resolution of unity in Rn. (ϕkfˆ )ˇ is an entire analytic function
on Rn for any f ∈ S′(Rn). In particular, (ϕkfˆ )ˇ (x) makes sense pointwise.
Deﬁnition 1.1. Let ϕ = {ϕj}∞j=0 be the dyadic resolution of unity according to
(1)–(3), s ∈ R, 0 < p ≤ ∞, 0 < q ≤ ∞, and
‖f | Bspq(Rn)‖ϕ =
( ∞∑
j=0
2jsq‖(ϕkfˆ )ˇ | Lp(Rn)‖q
) 1
q
(with the usual modiﬁcation if q = ∞). Then the Besov space Bspq(Rn) consists of all
f ∈ S′(Rn) such that ‖f | Bspq(Rn)‖ϕ < ∞.
We denote by Lp(Rn, 〈x〉α), where
〈x〉α = (1 + |x|2)α2 ,
the weighted Lp-space quasi-normed by
‖f | Lp(Rn, 〈x〉α)‖ = ‖〈·〉αf | Lp(Rn)‖.
Deﬁnition 1.2. Let ϕ = {ϕj}∞j=0 be the dyadic resolution of unity according to
(1)–(3), s ∈ R, 0 < p ≤ ∞, 0 < q ≤ ∞. Then the weighted Besov space Bspq(Rn, 〈x〉α)
is a collection of all f ∈ S′(Rn) such that
‖f | Bspq(Rn, 〈x〉α)‖ϕ =
( ∞∑
j=0
2jsq‖(ϕkfˆ )ˇ | Lp(Rn, 〈x〉α)‖q
) 1
q
(with the usual modiﬁcation if q = ∞) is ﬁnite.
Remark 1.3. If α = 0 then we have the space Bspq(R
n) as introduced in Deﬁnition 1.1.
It is also known from [1, ch. 4.2.2] that the operator f → 〈x〉αf is an isomorphic
mapping from Bspq(R
n, 〈x〉α) onto Bspq(Rn). In particular,
‖〈·〉αf | Bspq(Rn)‖ ∼ ‖f | Bspq(Rn, 〈x〉α)‖.
Next we review some special properties of weighted Besov spaces.
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Maryia Kabanava Tempered Radon measures
Proposition 1.4. For ﬁxed 0 < p, q ≤ ∞
S(Rn) =
⋂
α,s∈R
Bspq(R
n, 〈x〉α) (4)
and
S′(Rn) =
⋃
α,s∈R
Bspq(R
n, 〈x〉α).
Proof. Step 1. The inclusion
S(Rn) ⊂
⋂
α,s∈R
Bspq(R
n, 〈x〉α)
is clear.
To prove that any f ∈ ⋂α,s∈R Bspq(Rn, 〈x〉α) belongs to S(Rn), it is suﬃcient to
show that for any ﬁxed N ∈ N there are α(N) ∈ R and s(N) ∈ R such that
sup
|β|≤N
sup
x∈Rn
〈x〉2N |Dβf(x)| ≤ c‖f | Bspq(Rn, 〈x〉α)‖.
For any multiindex β there are polynomials P βγ , degP
β
γ ≤ 2N such that
〈x〉2NDβf(x) =
∑
γ≤β
Dγ [(P βγ f)(x)].
Hence
sup
|β|≤N
sup
x∈Rn
〈x〉2N |Dβf(x)| = sup
|β|≤N
sup
x∈Rn
∣∣∣∑
γ≤β
Dγ [(P βγ f)(x)]
∣∣∣
≤ sup
|β|≤N
∑
|γ|≤N
sup
x∈Rn
|Dγ [(P βγ f)(x)]|
≤ sup
|β|≤N
∑
|γ|≤N
‖P βγ f | CN (Rn)‖. (5)
Due to the embedding theorems [3, ch. 2.7.1],
‖P βγ f | CN (Rn)‖ ≤ c
∥∥∥P βγ f
∣∣∣ BN+ np +εpq (Rn)
∥∥∥
= c
∥∥∥∥ P
β
γ
〈x〉2N 〈x〉
2Nf
∣∣∣∣ BN+
n
p +ε
pq (Rn)
∥∥∥∥ (6)
for any ε > 0. P
β
γ
〈x〉2N is a pointwise multiplier for B
N+ np +ε
pq (Rn) [3, ch. 2.8.2]. Therefore
∥∥∥∥ P
β
γ
〈x〉2N 〈x〉
2Nf
∣∣∣∣BN+
n
p +ε
pq (Rn)
∥∥∥∥
≤ c
∥∥∥∥ P
β
γ
〈x〉2N
∣∣∣∣ CN+ np +ε(Rn)
∥∥∥∥ ·
∥∥∥〈x〉2Nf
∣∣∣ BN+ np +εpq (Rn)
∥∥∥. (7)
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Maryia Kabanava Tempered Radon measures
According to Remark 1.3
∥∥∥〈x〉2Nf
∣∣∣ BN+ np +εpq (Rn)
∥∥∥ ∼
∥∥∥f
∣∣∣ BN+ np +εpq (Rn, 〈x〉2N )
∥∥∥. (8)
Combining (5)–(8), one gets
sup
|β|≤N
sup
x∈Rn
〈x〉2N |Dβf(x)| ≤ c
∑
|γ|≤N
∥∥∥〈x〉2Nf
∣∣∣ BN+ np +εpq (Rn)∥∥
≤ c
∥∥∥f ∣∣ BN+ np +εpq (Rn, 〈x〉2N )
∥∥∥ (9)
and it follows (4).
Step 2. Let 1 < p ≤ ∞, 1 < q ≤ ∞ and let p′ and q′ be deﬁned in the standard
way by
1
p
+
1
p′
= 1,
1
q
+
1
q′
= 1.
The inclusion ⋃
α,s∈R
Bspq(R
n, 〈x〉α) ⊂ S′(Rn)
is evident.
As far as the opposite inclusion is concerned, we recall that f ∈ S′(Rn) if and only
if there are l ∈ N and m ∈ N such that
|f(ϕ)| ≤ c sup
|α|≤m
sup
x∈Rn
〈x〉l|Dαϕ(x)|,
for all ϕ ∈ S(Rn). By (9),
sup
|α|≤m
sup
x∈Rn
〈x〉l|Dαϕ(x)| ≤ c
∥∥∥ϕ
∣∣∣ Bm+ np +εp′q′ (Rn, 〈x〉l)
∥∥∥.
According to our choice of p and q, it follows that 1 ≤ p′ < ∞ and 1 ≤ q′ < ∞. Thus,
by [3, ch. 2.11.2],
f ∈
(
B
m+ np +ε
p′q′ (R
n, 〈x〉l)
)′
= B
−(m+ np +ε)
pq (Rn, 〈x〉−l).
This means
S′(Rn) ⊂
⋃
α,s∈R
Bspq(R
n, 〈x〉α).
Step 3. Let 0 < p ≤ 1, 1 < q ≤ ∞. By the arguments above, for f ∈ S′(Rn)
there are α ∈ R and s ∈ R such that
f ∈ Bs∞q(Rn, 〈x〉α).
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We want to show that
f ∈ Bspq(Rn, 〈x〉α−γ), γ >
n
p
.
Indeed,
‖f | Bspq(Rn, 〈x〉α−γ)‖ =
( ∞∑
j=0
2jsq‖〈x〉α−γ(ϕj fˆ )ˇ | Lp(Rn)‖q
) 1
q
≤
( ∞∑
j=0
2jsq sup
x∈Rn
[〈x〉α|(ϕj fˆ )ˇ (x)|]q
(∫
Rn
〈x〉−γp dx
) q
p
) 1
q
≤ c‖f | Bs∞q(Rn, 〈x〉α)‖.
Step 4. When 0 < q ≤ 1, ﬁrst we may ﬁnd α ∈ R and s ∈ R such that
f ∈ Bspq∗(Rn, 〈x〉α),
q∗ > 1, and then use the fact that
Bspq∗(R
n, 〈x〉α) ⊂ Bs−εpq (Rn, 〈x〉α), ε > 0.
Next we recall some notation. A measure μ is called σ-ﬁnite in Rn if for any
R > 0,
μ({x : |x| < R}) < ∞.
A measure μ is a Radon measure if all Borel sets are μ measurable and
(i) μ(K) < ∞ for compact sets K ⊂ Rn,
(ii) μ(V ) = sup{μ(K) : K ⊂ V is compact} for open sets V ⊂ Rn,
(iii) μ(A) = inf{μ(V ) : A ⊂ V, V is open} for A ⊂ Rn.
Let μ be a positive Radon measure in Rn. Let Tμ,
Tμ : ϕ −→
∫
Rn
ϕ(x)μ(dx), ϕ ∈ S(Rn),
be the linear functional generated by μ.
Deﬁnition 1.5. A positive Radon measure μ is said to be tempered if Tμ ∈ S′(Rn).
Proposition 1.6. Let μ1 and μ2 be two tempered Radon measures. Then
Tμ1 = Tμ2 in S′(Rn) if, and only if, μ1 = μ2.
Proof. The Proposition is valid by the arguments in [5, p. 80].
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Maryia Kabanava Tempered Radon measures
This justiﬁes the identiﬁcation of μ and correspondent tempered distribution Tμ
and we may write μ ∈ S′(Rn).
Deﬁnition 1.7. f ∈ S′(Rn) is called a positive distribution if
f(ϕ) ≥ 0 for any ϕ ∈ S(Rn) with ϕ ≥ 0.
If f ∈ Lloc1 (Rn) then f ≥ 0 means f(x) ≥ 0 almost everywhere.
Remark 1.8. If f is a positive distribution, then f ∈ C0(Rn)′ and it follows from the
Radon-Riesz theorem that there is a tempered Radon measure μ such that
f(ϕ) =
∫
Rn
ϕ(x)μ(dx)
[2, pp. 61, 62, 71, 75].
2. Main assertions
Our next result refers to tempered measures.
Theorem 2.1.
(i) A Radon measure μ in Rn is tempered if, and only if, there is a real number β
such that 〈x〉βμ is ﬁnite.
(ii) Let μ be a tempered Radon measure in Rn. Let j ∈ N,
Aj = {x : 2j−1 ≤ |x| ≤ 2j+1 }, A0 = {x : |x| ≤ 2 }.
Then for some c > 0, α ≥ 0,
μ(Ak) ≤ c2kα for all k ∈ N0.
Proof. Step 1. First we prove part (ii). Suppose that the assertion does not hold.
Then for c = 1 and l ∈ N there is kl ∈ N0 such that
μ(Akl) > 2
kll. (10)
As soon as it is found one kl with (10), it follows that there are inﬁnitely many kml ,
m ∈ N, that satisfy (10).
With j ∈ N,
A∗j = {x : 2j−2 ≤ |x| ≤ 2j+2 }, A∗0 = {x : |x| ≤ 4 }.
For l = 1 take any of km1 , let it be k1. For l = 2 choose k2  k1 in such a way
that A∗k1 and A
∗
k2
have an empty intersection. For arbitrary l ∈ N take
kl  kl−1 and A∗kl−1 ∩A∗kl = ∅.
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Maryia Kabanava Tempered Radon measures
Let ϕ0 be a C∞ function on Rn with
ϕ0(x) = 1, |x| ≤ 2 and ϕ0(x) = 0, |x| ≥ 4.
Let k ∈ N and
ϕk(x) = ϕ0(2−kx)− ϕ0(2−k+3x), x ∈ Rn.
Then we have
suppϕk ⊂ A∗k
and
ϕk(x) = 1, x ∈ Ak.
Let
ϕ(x) =
∞∑
l=1
2−lklϕkl(x).
For any ﬁxed N ∈ N0
sup
|α|≤N
sup
x∈Rn
(1 + |x|2)N |Dαϕ(x)|
= sup
|α|≤N
sup
x∈Rn
(1 + |x|2)N
∣∣∣∣Dα
( ∞∑
l=1
2−lklϕkl(x)
)∣∣∣∣
≤ sup
l∈N
sup
|α|≤N
sup
x∈Rn
2−lkl2−|α|kl2|α|(1 + |x|2)N |(Dαϕ1)(2−kl+1x)|.
The last inequality holds, since the functions ϕkl have disjoint supports. With the
change of variables
x′ = 2−kl+1x
one gets
sup
|α|≤N
sup
x∈Rn
(1 + |x|2)N |Dαϕ(x)|
≤ sup
l∈N
sup
|α|≤N
2−lkl2−|α|kl2|α|22(kl−1)N sup
x∈Rn
(1 + |x|2)N |Dαϕ1(x)|
≤ c sup
l∈N
sup
|α|≤N
2−kl(l+|α|−2N) ≤ c sup
l∈N
2−kl(l−2N).
Since N is ﬁxed and l is tending to inﬁnity, 2−kl(l−2N) is bounded. Thus ϕ ∈ S(Rn).
According to the deﬁnition of tempered Radon measures
∫
Rn
ψ(x)μ(dx) < +∞
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for any ψ ∈ S(Rn), but
∫
Rn
ϕ(x)μ(dx) ≥
∞∑
l=1
∫
Akl
ϕ(x)μ(dx) ≥
∞∑
l=1
2−lkl2lkl = +∞.
This means that our assertion (10) is false.
Step 2. We prove part (i). Since 〈x〉βμ is ﬁnite, it is tempered. Then μ is also
tempered. To prove the other direction we take β = −(α + 1). Then we get
〈·〉βμ(Rn) =
∫
Rn
〈x〉−(α+1) μ(dx) ≤
∞∑
k=0
∫
Ak
〈x〉−(α+1) μ(dx)
≤ c
∞∑
k=0
2−k(α+1)
∫
Ak
μ(dx) ≤ c
∞∑
k=0
2−k(α+1)2kα < ∞.
In order to characterize ﬁnite Radon measures we deﬁne the positive cone
+
Bspq(R
n)
as the collection of all positive f ∈ Bspq(Rn).
Theorem 2.2. Let M(Rn) be the collection of all ﬁnite Radon measures. Then
M(Rn) =
+
B01∞(R
n)
and
μ(Rn) ∼ ‖μ | B01∞(Rn)‖, μ ∈ M(Rn). (11)
Proof. By the proof in [5, pp. 82, 83, Proposition 1.127],
‖μ | B01∞(Rn)‖ ≤ μ(Rn) if μ ∈ M(Rn).
In order to prove the converse inequality, one use the characterisation of Besov
spaces via local means. Let k0 be a C∞ non-negative function with
supp k0 ⊂ {x : |x| ≤ 1 } and
 
k0(0) = 0.
If f ∈
+
B01∞(R
n), then f = μ is a tempered measure. By [5, p. 10, Theorem 1.10],
‖μ | B01∞(Rn)‖ ≥ c‖k0(1, μ)|L1(Rn)‖ = c
∫
Rn
∫
Rn
k0(x− y) dμ(y) dx.
Applying Fubini’s theorem, one gets
‖μ | B01∞(Rn)‖ ≥ cμ(Rn).
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Maryia Kabanava Tempered Radon measures
Corollary 2.3. Let f ∈ L1(Rn) and f(x) ≥ 0 almost everywhere. Then
‖f | L1(Rn)‖ ∼ ‖f | B01∞(Rn)‖.
Proof. Let μ = fμL, where μL is the Lebesgue measure. Then
μ(Rn) =
∫
Rn
f(x)μL(dx) = ‖f | L1(Rn)‖
and
‖μ|B01∞(Rn)‖ = ‖f | B01∞(Rn)‖.
From (11) follows the statement in the Corollary.
The question arises whether Corollary 2.3 can be extended to all f ∈ L1(Rn). We
have
L1(Rn) ↪−→ B01∞(Rn), hence ‖f | B01∞(Rn)‖ ≤ c‖f | L1(Rn)‖
for all f ∈ L1(Rn). But the converse is not true even for functions f ∈ L1(Rn) with
compact support in the unit ball.
Proposition 2.4. There are functions fj ∈ L1(Rn) with
supp fj ⊂ { y : |y| ≤ 1 }, j ∈ N,
such that {fj} is a bounded set in B01∞(Rn), but
‖fj | L1(Rn)‖ → ∞ if j →∞.
Proof. We may assume n = 1.
Let a ∈ C1(R) be an odd function with
supp a ⊂ {x : |x| ≤ 2 }, a(x) ≥ 0, x ≥ 0
and
max
−2≤x≤2
|a(x)| = |a(−1)| = a(1) = 1.
If c = max−2≤x≤2|a′(x)|, then c ≥ 1. Deﬁne a0 ∈ C1(R) by
a0(x) = c−1a(x).
Then one has for any x ∈ R,
|a0(x)| ≤ c−1 ≤ 1, |a′0(x)| ≤ 1, and
∫
R
a0(x) dx = 0.
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Deﬁne a function aν , ν ∈ N, by
aν(x) = 2νa0(2νx).
Then
supp aν ⊂ [−2−ν+1, 2−ν+1]
and
|aν(x)| ≤ c−12ν , |a′ν(x)| ≤ 22ν ,
∫
R
aν(x) dx = 0.
According to [5, p. 12, Deﬁnition 1.15], a0 is an 11-atom and aν are (0, 1)1,1-atoms. It
follows from [4, Theorem 13.8] that
∑∞
ν=1 aν(x) converges in S
′(Rn) and represents
an element of B01∞(R
n). Let f S
′
=
∑∞
ν=1 aν .
Let
fj(x) =
j∑
ν=1
aν(x).
Then supp fj ⊂ [−1, 1],
‖fj | L1(Rn)‖ ≥
+∞∫
0
fj(x) dx =
+∞∫
0
j∑
ν=1
aν(x) dx
= j
+∞∫
0
a0(x) dx →∞, j →∞.
On the other hand one has by the above atomic argument
‖fj | B01∞(R)‖ ≤ 1 for j ∈ N.
Corollary 2.5. Not any characteristic function of a measurable subset of Rn is a
pointwise multiplier in B01∞(R
n).
Proof. Let f ∈ L1(Rn) real. Let M+ be a set of points x such that f(x) ≥ 0 and
M− = {x : f(x) < 0 }. Then
‖f | L1(Rn)‖ = ‖χM+f | L1(Rn)‖+ ‖χM−f | L1(Rn)‖,
where χM+ , χM− are characteristic functions of sets M+ and M− respectively. One
may apply Corollary 2.3 to the functions χM+f and χM−f and get
‖f | L1(Rn)‖ ≤ c‖χM+f | B01∞(Rn)‖+ c‖χM−f | B01∞(Rn)‖.
563
Revista Matema´tica Complutense
2008: vol. 20, num. 2, pags. 553–564
Maryia Kabanava Tempered Radon measures
If any characteristic function of a set in Rn would be a pointwise multiplier in B01∞(R
n),
then
‖χM+f | B01∞(Rn)‖ ≤ c‖f | B01∞(Rn)‖, ‖χM−f | B01∞(Rn)‖ ≤ c‖f | B01∞(Rn)‖,
hence
‖f | L1(Rn)‖ ≤ c‖f | B01∞(Rn)‖.
Since for any function f ∈ L1(Rn) holds
‖f | B01∞(Rn)‖ ≤ c‖f | L1(Rn)‖,
one gets
‖f | L1(Rn)‖ ∼ ‖f | B01∞(Rn)‖, for real f ∈ L1(Rn).
This can be also extended to complex functions f ∈ L1(Rn). But acoording to the
Proposition 2.4 this is not true.
References
[1] D. E. Edmunds and H. Triebel, Function spaces, entropy numbers, diﬀerential operators, Cam-
bridge Tracts in Mathematics, vol. 120, Cambridge University Press, Cambridge, 1996.
[2] P. Malliavin, Integration and probability, Graduate Texts in Mathematics, vol. 157, Springer-
Verlag, New York, 1995.
[3] H. Triebel, Theory of function spaces, Monographs in Mathematics, vol. 78, Birkha¨user Verlag,
Basel, 1983.
[4] , Fractals and spectra, Monographs in Mathematics, vol. 91, Birkha¨user Verlag, Basel,
1997.
[5] , Theory of function spaces, III, Monographs in Mathematics, vol. 100, Birkha¨user Verlag,
Basel, 2006.
Revista Matema´tica Complutense
2008: vol. 20, num. 2, pags. 553–564 564


Tempered Radon Measures

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Tempered Radon Measures

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