Georgian Mathematical Journal
1(94), No. 4, 395-403
ON THE BOUNDEDNESS OF CAUCHY SINGULAR
OPERATOR FROM
THE SPACE Lp TO Lq , p > q ≥ 1
G. KHUSKIVADZE AND V. PAATASHVILI
Abstract. It is proved that for a Cauchy type singular operator,
given by equality (1), to be bounded from the Lebesgue space Lp (Γ)
to Lq (Γ), as Γ = ∪∞
n=1 Γn , Γn = {z : |z| = rn }, it is necessary and
sufficient that either condition (4) or (5) be fulfilled.
1. Let Γ be a plane rectifiable Jordan curve, Lp (Γ), p ≥ 1, a class of
functions summable to the p-th degree on Γ, and SΓ a Cauchy singular
operator
Z
f (τ ) dτ
1
, f ∈ Lp (Γ), t ∈ Γ.
(1)
SΓ (f )(t) =
πi
τ −t
Γ
Numerous studies have been devoted to problems of the existence of
SΓ (f )(t) and boundedness of the operator SΓ : f → SΓ (f ) in the space
Lp (Γ) (see, e.g., [1–3]). The final solution of these problems is given in [4,5].
It was proved by G.David that for the operator SΓ to be bounded in Lp (Γ),
it is necessary and sufficient that the condition
l(t, r) ≤ Cr
(2)
be fulfilled, where l(t, r) is a length of the part of Γ contained in the circle
with center at t ∈ Γ and radius r, and C is a constant.1
The present paper is devoted to the problem of boundedness of the operator SΓ from Lp (Γ) to Lq (Γ), p > q ≥ 1 (see also [6–9]).
a strictly
2. Throughout the rest of this paper by {rn }∞
n=1 is meantP
n
decreasing sequence of positive numbers satisfying the condition k=1 rk <
∞, and by Γ, the family of concentric circumferences on a complex plane
Γn = {z : |z| = rn }, n = 1, 2, . . . .
1991 Mathematics Subject Classification. 47B38.
1 Following [7,8], the necessity of condition (2) is shown also in [6]. In the same work
its sufficiency is proved for special classes of curves.
395
c 1994 Plenum Publishing Corporation
1072-947X/94/0700-0395$12.50/0 
396
G. KHUSKIVADZE AND V. PAATASHVILI
It has been shown in [10,11] that for the operator SΓ to be bounded in
Lp (Γ), p > 1, it is necessary and sufficient that the conditions
∞
X
k=n
rk ≤ Crn ,
n = 1, 2, . . . ,
(3)
be fulfilled, where C is an absolute constant.
We shall prove
Theorem. Let p > q ≥ 1 and σ = pq/(p − q). Then the following
statements are equivalent:
(A) operator SΓ is bounded from Lp (Γ) to Lq (Γ);
“σ
∞ ’ P∞
X
k=n rk
rn < ∞;
(4)
(B)
rn
n=1
∞
X
(C)
nσ rn < ∞.
(5)
n=1
Remark. A family of concentric circumferences the sum of whose lengths
is finite, as a set of integration, principally, “simulates” rectifiable curves
with isolated singularities. Analogy of conditions (2) and (3) also indicates
this fact. Taking into account the above, we assume that the following
statement (an analogue of the theorem from Subsection 2) is valid: for the
operator SΓ to be bounded from Lp (Γ) to Lq (Γ), where Γ is an arbitrary
rectifiable curve, p > q ≥ 1, it is necessary and sufficient that the condition
Z
[χ(t)]pq/(p−q) |dt| < ∞
Γ
be fulfilled, where
χ(t) = sup
r
l(t, τ )
,
r
t ∈ Γ.
3. In proving this theorem, use will often be made of the well-known
Abel equality (see, e.g., [12], p.307)
∞ X
n
X
n=1
k=1
∞
∞
X
‘
‘
X
uk vn =
un
vk ,
n=1
where {un } and {vn } are sequences of positive numbers and
as well as of its particular case
∞
X
n=1
We shall also need
nvn =
∞ X
∞
X
n=1 k=n
(6)
k=n
vk .
P∞
k=1
vk < ∞,
(7)
BOUNDEDNESS OF THE CAUCHY SINGULAR OPERATOR
397
Lemma. Let p > 0. If f is a function analytic in the circle |z| < 1, then
for r < R < 1,
Z
Z
r
p
|f (z)| |dz| ≤
|f (z)|p |dz|.
(8)
R
|z|=R
|z|=r
If f is a function analytic in the domain |z| > 1 and f (∞) = 0 then for
1<R<r
Z
 R ‘p−1 Z
p
|f (z)|p |dz|.
(9)
|f (z)| |dz| ≤
r
|z|=R
|z|=r
If, in addition, f belongs to the Hardy class Hp in the domains |z| < 1 or
R 2π
|z| > 1, i.e., supρ 0 |f (ρeiϑ )|p dϑ < ∞ (in particular, if f is represented
by a Cauchy type integral), then we can take R = 1 in inequalities (8) and
(9).
Proof. Since |dz| = |dρeiϑ | = ρdϑ, inequality (8) follows from the fact that
R 2π
1
|f (ρeiϑ )|p dϑ of |f (ρeiϑ )|p is a nondecreasing function
the mean value 2π
0
of ρ (see, e.g., [13], p.9).
Under the conditions of the lemma, if |z| > 1, then the function g(ζ) =
1
1
f
ζ ( ζ ) is analytic in the circle |ζ| < 1. Using inequality (8) for g, we get
Z
 ‘p+1 Z Œ € 1 Œ
Œ € 1 Œp
Œf
Œf
Œ |dζ| ≤ R
Œp |dζ|.
ζ
r
ζ
|ζ|= r1
1
|ζ|= R
Applying the transformation of ζ = z1 , the latter inequality reduces to (9).
If f ∈ Hp , then by the Riesz theorem
lim
ρ→1
Z2π
0
iϑ
p
|f (ρe )| dϑ =
Z2π
0
|f (eiϑ )|p dϑ
(see, e.g., [13], p.21), which enables us to suppose that R = 1.
„
4. Let us prove the equivalence of conditions (B) and (C). This follows
from equality (7) for σ = 1 and therefore we shall assume that σ > 1.
(C) follows from (B). We usePAbel–Dini’s theorem (see, e.g., [12], p. 292):
∞
parif a series with positive terms
P∞ ann=1 an diverges and Sn means its
Pn-th
∞
n
tial sum, then the series n=1 Sn also diverges, while the series n=1 Sa1+ε
n
P∞
σ
(ε > 0) converges. Assume Žthat
n=1 n rn diverges. Then, setPnthe series
σ
,
we
shall
see by this theorem that
k
r
ting an = nσ rn and ωn = 1
k
k=1
P∞
P∞
0
the series n=1 ωn nσ rn diverges while the series n=1 ωnσ rn nσ converges,
σ
where σ 0 = σ−1 > 1.
398
G. KHUSKIVADZE AND V. PAATASHVILI
Using equality (6) and the Hölder inequality, we obtain
∞
X
n=1
ωn nσ rn ≤ 2
=2
∞
X
ωn
n=1
∞
X
n
X
k=1
ωn nσ−1
n=1
"
n
∞ X
‘
‘
X
k σ−1 rn ≤ 2
ωk k σ−1 rn =
’ P∞
k=n rk
rn
“σ
∞ ’ P∞
X
k=n rk
rn
≤2
rn
n=1
“
n=1
k=1
rn ≤
#1/σ
∞
X
0
ωnσ nσ rn
n=1
‘1/σ0
< ∞.
The obtained contradiction shows that (C) follows from (B).
Let us now show that (B) follows from (C). If m ≤ n, then
Am =
+
P∞
k=m rk
r
P∞ m
k=n rk
=
rm + rm+1 + · · · + rn−1
+
rm
≤ (n − m) + An .
rm
(10)
Let 1 ≤ s ≤ σ. Using equality (6) and inequality (10), we get
∞
X
n
s−1
Anσ−s+1 rn
=
∞ X
n
X
n=1
≤ 2σ
≤ 2σ
≤ 2σ
ns−1 Aσ−s
n
n=1
n=1
=
∞
X
k=1
n=1
∞
X
rk =
k=n
∞ 
‘
X
‚
ƒσ−s ‘
s−1
k
+
(n
k)
rn ≤
r
≤
A
−
k s−1 Aσ−s
n
n
k
∞ X
n
X
n=1
∞
X
∞
X
k=1
Aσ−s
n
n=1
k
‚
s−1
n
X
k=1
Aσ−s
+ (n − k)σ−s
n
ƒ‘
rn ≤
n
∞ X
‘
‘
X
k s−1 (n − k)σ−s rn ≤
k s−1 rn + 2σ
s
σ
Aσ−s
n n rn + 2
n=1
∞
X
k=1
n σ rn .
(11)
n=1
n=1
Let [σ] be the integer part of σ and α = σ − [σ]. Using inequality (11)
successively [σ] times for s = 1, 2, . . . , [σ], we arrive at the inequality
∞
X
n=1
Aσn rn ≤ C1
∞
X
[σ]
Aα
n n r n + C2 ,
n=1
where the constants C1 and C2 depend on σ only.
(12)
BOUNDEDNESS OF THE CAUCHY SINGULAR OPERATOR
399
If σ is an integer, then α = 0, and consequently the proof is completed.
Let α > 0. Then making use of the Hölder inequality and equality (7), we
obtain
∞
X
[σ]
Aα
n n rn =
n=1
n=1
≤
=
≤
=
∞
X
∞
X
n=1
∞
X
n=1
∞
X
n=1
∞
X
An nσ−1 rn
α(σ−1) σ(1−α)
Aα
n
rn ≤
nn
∞
‘α  X
n σ rn
n=1
nσ rn
nσ rn
nσ r n
n=1
∞
‘1−α  X
nσ−1
n=1
∞
∞ X
‘1−α  X
∞
X
rk
k=n
k σ−1 rk
n=1 k=n
∞
‘1−α  X
n=1
‘1−α
=
‘α
‘α
≤
=
∞
‘α X
nσ rn < ∞,
n(nσ−1 rn ) =
n=1
which completes the proof.
5. Let us show that (A) follows from (B) or (C). Consider first the case
when q = 1 and show that if p > 1 and σ = p0 = p/(p − 1), then SΓ is
bounded from Sp (Γ) to L1 (Γ).
Let φni and φln be the functions determined respectively in Int Γn and
Ext Γn by the Cauchy type integral
Z
1
ϕn (t) dt
, ϕn ∈ Lp (Γn ), p ≥ 1, z 6∈ Γn .
(13)
2πi
t−z
Γn
Using the Sokhotsky–Plemelj formula
φin (t) − φen (t) = ϕn (t),
φin (t) + φen (t) = SΓ (ϕn )(t)
and the Cauchy formula
1
2πi
Z
Γn
ϕn (t) dt
1
=
t−z
2πi
Z
Γn
φni (t) − φne (t)
dt =
t−z
(
φni (z), z ∈ Int Γn ,
φen (z), z ∈ Ext Γn ,
we obtain by direct calculations
SΓ (ϕ)(t) = 2
n−1
X
k=1
for t ∈ Γn .
φki (t) + [φin (t) + φen (t)] + 2
∞
X
k=n+1
φek (t)
(14)
400
G. KHUSKIVADZE AND V. PAATASHVILI
Let us evaluate the integrals of the sums
S1 (t) = 2
n−1
X
φik (t)
+
φni (t),
S2 (t) =
φen (t)
+2
k=1
∞
X
φek (t).
k=n+1
Using the lemma from Subsection 3 and the Hölder inequality, we can
write
Z
Z
n Z
n
X
X
rn
|φik (t)| ds ≤
|φki (t)| ds ≤ 2
|S1 (t)| ds ≤ 2
rk
k=1Γ
n
Γn
≤ 2(2π)1/p
k=1
0
n
X
k=1
rk
1/p
rk
Z
Γk
|φki (t)|p ds
Γk
‘1/p
,
where φki is a limiting function of the Cauchy type integral (13) on Γk ,
k = 1, 2, . . . , n.
Next, changing the order of summation and using the Riesz’s inequality
for the Cauchy singular operator in the case of the circle as well as the
Hölder inequality, we get
Z
∞ Z
X
|S1 (t)| ds ≤
|S1 (t)| ds =
n=1Γ
n
Γ
≤ 2(2π)1/p
0
n
∞ X
X
n=1 k=1
Z
rn 
1/p
rk
|φik (t)|p ds
Γk
‘1/p
=
∞ P∞
Z
‘1/p
X
i
p
1/p0
k=n rk
|φ
(t)|
ds
≤
= 2(2π)
n
1/p
rn
n=1
Γn
∞ P∞
Z
‘1/p
X
i
p
1/p0
k=n rk
|ϕ
≤
(t)|
ds
Cp
≤ 2(2π)
n
1/p
rn
n=1
Γn
1/p0
≤ 2(2π)
Cp
"
“p0
∞ ’ P∞
X
k=n rk
n=1
rn
rn
#1/p0
Z
Γ
|ϕ(t)|p ds
‘1/p
,
(15)
where Cp is the constant from the Riesz inequality (which depends on p
only).
The integral of S2 (t) can be evaluated analogously. Using inequality (9),
as well as the Hölder and Riesz inequalities, we obtain
Z
∞ Z
∞ Z
‘1/p
X
0 X
0
|S2 (t)|dt ≤ 2
|φke (t)|ds ≤ 2(2π)1/p
|φek (t)|p ds
rn1/p ≤
Γn
k=nΓ
n
k=n
Γn
BOUNDEDNESS OF THE CAUCHY SINGULAR OPERATOR
≤ 2(2π)1/p
0
Z
∞ 
p−1
X
rk ‘ p 
rn
k=n
0
≤ 2(2π)1/p Cp
∞
X
1/p
rk
k=n
0
Z
|φek (t)|p ds
Γk
|ϕk (t)|p ds
Γk
‘1/p
‘1/p
401
0
rn1/p ≤
.
Next, changing the order of summation and using the Hölder inequality,
we can write
Z
|S2 (t)| dt =
Γ
∞ Z
X
0
n=1Γ
n
|S2 (t)| ds ≤ 2(2π)1/p Cp
∞ X
∞
X
1/p0
rk
n=1 k=n
×
∞
‘1/p
Z
Z
‘1/p
X
0
0
|ϕn (t)|p ds
≤
|ϕk (t)|p ds
nrn1/p
= 2(2π)1/p Cp
×
n=1
Γk
0
≤ 2(2π)1/p Cp
∞
X
0
np rn
n=1
0
= 2(2π)1/p Cp
∞ Z
‘1/p h X
0
n=1
∞
X
n=1
0
n p rn
Γn
|ϕn (t)|p ds
Γn
‘1/p0  Z
Γ
|ϕ(t)|p ds
‘ p1 ·p i p1
‘1/p
.
=
(16)
It follows from (14),(15) and (16) that if conditions (B) and (C) are
fulfilled for σ = p0 , then the operator SΓ is bounded from Lp (Γ) to L1 (Γ).
Let us now consider the general case. Let conditions (B) and (C) be
fulfilled for p > q ≥ 1 and σ = pq/(p − q). Then, by virtue of the above
σ
arguments, SΓ is continuous from Lσ0 (Γ), σ 0 = σ−1
, to L1 (Γ). But then
SΓ is also continuous from L∞ (Γ) to Lσ (Γ) (L∞ (Γ) is a class of functions
essentially bounded on Γ). This statement can be proved by the well-known
method using the Riesz equality
Z
Z
ϕSΓ ψ dt = − ψSΓ ϕ dt, ϕ ∈ Lσ0 (Γ), ψ ∈ L∞ (Γ),
Γ
Γ
whose validity in our case can be immediately verified.
Further, since SΓ is bounded from Lσ0 (Γ) and L∞ (Γ) to L1 (Γ) and Lσ (Γ),
respectively, according to Riesz–Torin’s theorem on interpolation of linear
operators (see, e.g., [14], p.144), it follows that SΓ is bounded from Lα (Γ),
σ 0 ≤ α ≤ ∞, to Lασ/(α+σ) (Γ). Letting α = p, we get that SΓ is bounded
from Lp (Γ) to Lq (Γ).
6. Let us now show that (C) and consequently (B) followPfrom (A).
∞
Let for a pair p and q, p > q ≥ 1, σ = pq/(p − q), the series n=1 nσ rn
402
G. KHUSKIVADZE AND V. PAATASHVILI
diverge.
PnThen, according to the above-mentioned Abel–Dini’s theorem, if
ωn = ( k=1 k σ rk )−1/q , then
∞
X
ωnp nσ rn
=
n=1
∞
X
nσ rn
p/q
n=1
∞
X
Sn
< ∞, Sn =
n
X
k σ rk ,
k=1
∞
X
n σ rn
ωnq nσ rn =
= ∞.
Sn
n=1
n=1
Consider, on Γ, the function ϕ(t) = ωn nσ/p for t ∈ Γn , n = 1, 2, . . . .
Then
Z
∞
∞ Z
X
X
p
ωnp nσ rn < ∞.
|ϕ(t)| |dt| =
|ϕ(t)|p ds = 2π
(17)
n=1Γ
n
Γ
n=1
Next, by equality (14) we have
SΓ (ϕ)(t) = 2
n−1
X
ωk k σ/p + ωn nσ/p >
k=1
n
X
ωk k σ/p
k=1
for t ∈ Γn . Consequently,
Z
∞ X
n
∞ Z
‘q
X
X
q
q
ωk k σ/p rn >
|SΓ (ϕ)(t)| |dt| =
|SΓ (ϕ)(t)| |dt| > 2π
Γ
>
n=1 k=1
n=1Γ
n
∞
∞
n
‘q
X
X
X
σ
ωnq
ωnq n( p +1)q rn
2π
k σ/p rn ≥ 2π
n=1
n=1
k=1
∞
X
= 2π
ωnq nσ rn = ∞.
n=1
=
(18)
It follows from (17) and (18) that if condition (C) is not fulfilled for
p > q ≥ 1, then there exists a function ϕ ∈ Lp (Γ) for which SΓ (ϕ) 6∈ Lq (Γ).
Consequently, for condition (A) to be fulfilled, it is necessary that condition
(C) (and hence (B)) be fulfilled.
References
1. B.V. Khvedelidze, Method of Cauchy type integrals in discontinuous
boundary value problems of the theory of holomorphic functions of one complex variable. (Russian) Current problems of mathematics, v. 7 (Russian),
5-162. Itogi nauki i tekhniki. Akad. Nauk SSSR, Vsesoyuzn. Inst. Nauchn.
i Tekhn. Inform. Moscow, 1975.
2. E.M. Dynkin, Methods of the theory of singular integrals (Hilbert
transform and Calderon–Zygmund theory). (Russian) Current problems of
BOUNDEDNESS OF THE CAUCHY SINGULAR OPERATOR
403
mathematics. Fundamental directions, v.15 (Russian) 197-292, Itogi nauki
i tekhniki. Akad. Nauk SSSR, Vsesoyuzn. Inst. Nauchn. i Tekhn. Inform.
Moscow, 1987.
3. ”—-”, Methods of the theory of singular integrals, II. Littlewood–
Paley theory and its applications. (Russian) Current problems of mathematics. Fundamental directions, v.42 (Russian), 105-198, Itogi nauki i
tekhniki. Akad. Nauk SSSR, Vsesoyuzn. Inst. Nauchn. i Tekhn. Inform.
Moscow, 1989.
4. A.P. Calderon, Cauchy integrals on Lipschitz curves and related operators. Proc. Nat. Acad. Sci. USA, 1977, 4, 1324-1327.
5. G. David, L’integrale de Cauchy sur le courbes rectifiables. Prepublications Univ. Paris–Sud; Dept. Math. 82 T 05(1982).
6. V.A. Paatashvili and G.A. Khuskivadze, On boundedness of Cauchy
operator in Lebesgue spaces in the case of non-smooth contours. Trudy
Tbiliss. Mat. Inst. Razmadze, 69(1982), 93-107.
7. V.P. Havin, Boundary properties of Cauchy type integrals and harmonically conjugate functions in domains with rectifiable boundary. (Russian) Mat. Sbornik 68(110)(1965), No. 4, 499-517.
8. G.A. Khuskivadze, On singular Cauchy integral and on Cauchy type
integral. (Russian) Trudy Tbiliss. Mat. Inst. Razmadze, 53(1976), 75-86.
9. T.S. Salimov, Singular Cauchy integral in spaces Lp , p ≥ 1. (Russian)
Dokl. Akad. Nauk Azerbaijan SSR, XI(1985), No. 3, 3-5.
10. A.V. Aizens̆tat, On singular integral equations on a countable set of
closed curves imbedded in one another. (Russian) Dokl. Akad. Nauk SSSR,
232(1977), No. 5, 985-988.
11. V.A. Paatashvili and G.A. Khuskivadze, On singular Cauchy operator on a countable set of concentric circles. (Russian) Trudy Tbiliss. Mat.
Inst. Razmadze, 69(1982), 80-92.
12. G.M. Fichtenholz, A course of differential and integral calculus, v.II.
(Russian) Fizmatgiz, Moscow, 1959.
13. P.L. Duren, Theory of Hp spaces. Academic Press, New York and
London, 1970.
14. A. Zygmund, Trigonometric series, v.2. (Translated into Russian)
Mir, Moscow, 1965; English original: Cambridge, 1989.
(Received 28.04.1993)
Authors’ address:
A.Razmadze Mathematical Institute
Georgian Academy of Sciences
1, Z.Rukhadze St., 380093 Tbilisi
Republic of Georgia