Georgian Mathematical Journal 1(94), No. 4, 395-403

ON THE BOUNDEDNESS OF CAUCHY SINGULAR OPERATOR FROM THE SPACE L_{p} TO L_{q}, 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 L_{p}(Γ) to L_{q}(Γ), as Γ = ∪_{n=1}^{∞}Γ_{n}, Γ_{n} = {z : |z| = r_{n}}, it is necessary and sufficient that either condition (4) or (5) be fulfilled.

1. Let Γ be a plane rectifiable Jordan curve, L_{p}(Γ), p ≥ 1, a class of functions summable to the p-th degree on Γ, and S_{Γ} a Cauchy singular operator

S_{Γ}(f)(t) = \frac{1}{πi} ∫_{Γ} \frac{f(τ) dτ}{τ − t} , f ∈ L_{p}(Γ), t ∈ Γ.  (1)

Numerous studies have been devoted to problems of the existence of S_{Γ}(f)(t) and boundedness of the operator S_{Γ} : f → S_{Γ}(f) in the space L_{p}(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 S_{Γ} to be bounded in L_{p}(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 S_{Γ} from L_{p}(Γ) to L_{q}(Γ), p > q ≥ 1 (see also [6–9]).

2. Throughout the rest of this paper by {r_{n}}_{n=1}^{∞} is meant a strictly decreasing sequence of positive numbers satisfying the condition ∑_{k=1}^{n} r_{k} < ∞, and by Γ, the family of concentric circumferences on a complex plane Γ_{n} = {z : |z| = r_{n}}, n = 1, 2, . . ..

395

1991 Mathematics Subject Classification. 47B38.

1Following [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.

1072-947X/94/0700-0395$12.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1994 Plenum Publishing Corporation

It has been shown in [10,11] that for the operator S_{Γ} to be bounded in L_{p}(Γ), p > 1, it is necessary and sufficient that the conditions

∑_{k=n}^{∞} r_{k} ≤ Cr_{n}, 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 S_{Γ} is bounded from L_{p}(Γ) to L_{q}(Γ);

(B) ∑_{n=1}^{∞} (\frac{∑_{k=n}^{∞} r_{k}}{r_{n}} )^{σ} r_{n} < ∞; (4)

(C) ∑_{n=1}^{∞} n^{σ}r_{n} < ∞. (5)

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 S_{Γ} to be bounded from L_{p}(Γ) to L_{q}(Γ), where Γ is an arbitrary rectifiable curve, p > q ≥ 1, it is necessary and sufficient that the condition

∫_{Γ} [χ(t)]^{pq/(p−q)}|dt| < ∞

be fulfilled, where

χ(t) = sup_{r} \frac{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)

∑_{n=1}^{∞} ( ∑_{k=1}^{n} u_{k})v_{n} = ∑_{n=1}^{∞} u_{n}( ∑_{k=n}^{∞} v_{k}),  (6)

where {u_{n}} and {v_{n}} are sequences of positive numbers and ∑_{k=1}^{∞} v_{k} < ∞, as well as of its particular case

∑_{n=1}^{∞} nv_{n} = ∑_{n=1}^{∞} ∑_{k=n}^{∞} v_{k}.  (7)

We shall also need

Lemma. Let p > 0. If f is a function analytic in the circle |z| < 1, then for r < R < 1,

∫_{|z|=r} |f(z)|^{p}|dz| ≤ \frac{r}{R} ∫_{|z|=R} |f(z)|^{p}|dz|.  (8)

If f is a function analytic in the domain |z| > 1 and f(∞) = 0 then for 1 < R < r

∫_{|z|=r} |f(z)|^{p}|dz| ≤ (\frac{R}{r})^{p−1} ∫_{|z|=R} |f(z)|^{p}|dz|.  (9)

If, in addition, f belongs to the Hardy class H_{p} in the domains |z| < 1 or |z| > 1, i.e., sup_{ρ} ∫_{0}^{2π} |f(ρe^{iϑ})|^{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| = |dρe^{iϑ}| = ρdϑ, inequality (8) follows from the fact that the mean value \frac{1}{2π} ∫_{0}^{2π} |f(ρe^{iϑ})|^{p}dϑ of |f(ρe^{iϑ})|^{p} is a nondecreasing function of ρ (see, e.g., [13], p.9).

Under the conditions of the lemma, if |z| > 1, then the function g(ζ) = \frac{1}{ζ}f(\frac{1}{ζ}) is analytic in the circle |ζ| < 1. Using inequality (8) for g, we get

∫_{|ζ|= \frac{1}{r}} |f(\frac{1}{ζ})|^{p} |dζ| ≤ (\frac{R}{r})^{p+1} ∫_{|ζ|= \frac{1}{R}} |f(\frac{1}{ζ})|^{p} |dζ|.

Applying the transformation of ζ = \frac{1}{z}, the latter inequality reduces to (9). If f ∈ H_{p}, then by the Riesz theorem

lim_{ρ→1} ∫_{0}^{2π}|f(ρe^{iϑ})|^{p}dϑ = ∫_{0}^{2π}|f(e^{iϑ})|^{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 use Abel–Dini’s theorem (see, e.g., [12], p. 292): if a series with positive terms ∑_{n=1}^{∞} a_{n} diverges and S_{n} means its n-th partial sum, then the series ∑_{n=1}^{∞} \frac{a_{n}}{S_{n}} also diverges, while the series ∑_{n=1}^{∞} \frac{a_{n}}{S_{n}^{1+ε}} (ε > 0) converges. Assume that the series ∑_{n=1}^{∞} n^{σ}r_{n} diverges. Then, setting a_{n} = n^{σ}r_{n} and ω_{n} = 1/ ∑_{k=1}^{n} k^{σ}r_{k}, we shall see by this theorem that the series ∑_{n=1}^{∞} ω_{n}n^{σ}r_{n} diverges while the series ∑_{n=1}^{∞} ω_{n}^{σ′}r_{n}n^{σ} converges, where σ′ = \frac{σ}{σ−1} > 1.

398 G. KHUSKIVADZE AND V. PAATASHVILI

Using equality (6) and the Hölder inequality, we obtain

∑_{n=1}^{∞} ω_{n}n^{σ}r_{n} ≤ 2 ∑_{n=1}^{∞} ω_{n}( ∑_{k=1}^{n} k^{σ−1})r_{n} ≤ 2 ∑_{n=1}^{∞} ( ∑_{k=1}^{n} ω_{k}k^{σ−1})r_{n} =

= 2 ∑_{n=1}^{∞} ω_{n}n^{σ−1}(\frac{∑_{k=n}^{∞} r_{k}}{r_{n}} )r_{n} ≤

≤ 2 [∑_{n=1}^{∞} (\frac{∑_{k=n}^{∞} r_{k}}{r_{n}} )^{σ}r_{n}]^{1/σ} ( ∑_{n=1}^{∞} ω_{n}^{σ′}n^{σ}r_{n})^{1/σ′} < ∞.

The obtained contradiction shows that (C) follows from (B).

Let us now show that (B) follows from (C). If m ≤ n, then

A_{m} = \frac{∑_{k=m}^{∞} r_{k}}{r_{m}} = \frac{r_{m} + r_{m+1} + ⋯ + r_{n−1}}{r_{m}} +
+ \frac{∑_{k=n}^{∞} r_{k}}{r_{m}} ≤ (n − m) + A_{n}.  (10)

Let 1 ≤ s ≤ σ. Using equality (6) and inequality (10), we get

∑_{n=1}^{∞} n^{s−1}A_{n}^{σ−s+1}r_{n} = ∑_{n=1}^{∞} n^{s−1}A_{n}^{σ−s} ∑_{k=n}^{∞} r_{k} =

= ∑_{n=1}^{∞} ( ∑_{k=1}^{n} k^{s−1}A_{k}^{σ−s})r_{n} ≤ ∑_{n=1}^{∞} (k^{s−1}[A_{n} + (n − k)]^{σ−s}r_{n}) ≤

≤ 2^{σ} ∑_{n=1}^{∞} ( ∑_{k=1}^{n} k^{s−1}[A_{n}^{σ−s} + (n − k)^{σ−s}])r_{n} ≤

≤ 2^{σ} ∑_{n=1}^{∞} A_{n}^{σ−s}( ∑_{k=1}^{n} k^{s−1})r_{n} + 2^{σ} ∑_{n=1}^{∞} ( ∑_{k=1}^{n} k^{s−1}(n − k)^{σ−s})r_{n} ≤

≤ 2^{σ} ∑_{n=1}^{∞} A_{n}^{σ−s}n^{s}r_{n} + 2^{σ} ∑_{n=1}^{∞} n^{σ}r_{n}.  (11)

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

∑_{n=1}^{∞} A_{n}^{σ}r_{n} ≤ C_{1} ∑_{n=1}^{∞} A_{n}^{α}n^{[σ]}r_{n} + C_{2},  (12)

where the constants C_{1} and C_{2} depend on σ only.

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

∑_{n=1}^{∞} A_{n}^{α}n^{[σ]}r_{n} = ∑_{n=1}^{∞} A_{n}^{α}n^{α(σ−1)}n^{σ(1−α)}r_{n} ≤

≤ ( ∑_{n=1}^{∞} A_{n}n^{σ−1}r_{n})^{α}( ∑_{n=1}^{∞} n^{σ}r_{n})^{1−α} =

= ( ∑_{n=1}^{∞} n^{σ}r_{n})^{1−α}( ∑_{n=1}^{∞} n^{σ−1} ∑_{k=n}^{∞} r_{k})^{α} ≤

≤ ( ∑_{n=1}^{∞} n^{σ}r_{n})^{1−α}( ∑_{n=1}^{∞} ∑_{k=n}^{∞} k^{σ−1}r_{k})^{α} =

= ( ∑_{n=1}^{∞} n^{σ}r_{n})^{1−α}( ∑_{n=1}^{∞} n(n^{σ−1}r_{n}))^{α} = ∑_{n=1}^{∞} n^{σ}r_{n} < ∞,

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 = p′ = p/(p − 1), then S_{Γ} is bounded from S_{p}(Γ) to L_{1}(Γ).

Let ϕ_{n}^{i} and ϕ_{n}^{l} be the functions determined respectively in Int Γ_{n} and Ext Γ_{n} by the Cauchy type integral

\frac{1}{2πi} ∫_{Γ_{n}} \frac{φ_{n}(t) dt}{t − z}, φ_{n} ∈ L_{p}(Γ_{n}), p ≥ 1, z ∉ Γ_{n}.  (13)

Using the Sokhotsky–Plemelj formula

ϕ_{n}^{i}(t) − ϕ_{n}^{e}(t) = φ_{n}(t), ϕ_{n}^{i}(t) + ϕ_{n}^{e}(t) = S_{Γ}(φ_{n})(t)

and the Cauchy formula

\frac{1}{2πi} ∫_{Γ_{n}} \frac{φ_{n}(t) dt}{t − z} = \frac{1}{2πi} ∫_{Γ_{n}} \frac{ϕ_{n}^{i}(t) − ϕ_{n}^{e}(t)}{t − z} dt = \begin{cases}ϕ_{n}^{i}(z), & z ∈ Int Γ_{n}, \\ ϕ_{n}^{e}(z), & z ∈ Ext Γ_{n}, \end{cases}

we obtain by direct calculations

S_{Γ}(φ)(t) = 2 ∑_{k=1}^{n−1} ϕ_{k}^{i}(t) + [ϕ_{n}^{i}(t) + ϕ_{n}^{e}(t)] + 2 ∑_{k=n+1}^{∞} ϕ_{k}^{e}(t)  (14)

for t ∈ Γ_{n}.

Let us evaluate the integrals of the sums

S_{1}(t) = 2 ∑_{k=1}^{n−1} ϕ_{k}^{i}(t) + ϕ_{n}^{i}(t), S_{2}(t) = ϕ_{n}^{e}(t) + 2 ∑_{k=n+1}^{∞} ϕ_{k}^{e}(t).

Using the lemma from Subsection 3 and the Hölder inequality, we can write

∫_{Γ_{n}} |S_{1}(t)| ds ≤ 2 ∑_{k=1}^{n} ∫_{Γ_{n}} |ϕ_{k}^{i}(t)| ds ≤ 2 ∑_{k=1}^{n} \frac{r_{n}}{r_{k}} ∫_{Γ_{k}} |ϕ_{k}^{i}(t)| ds ≤

≤ 2(2π)^{1/p′} ∑_{k=1}^{n} \frac{r_{k}}{r_{k}^{1/p}} ( ∫_{Γ_{k}} |ϕ_{k}^{i}(t)|^{p} ds)^{1/p},

where ϕ_{k}^{i} is a limiting function of the Cauchy type integral (13) on Γ_{k}, 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

∫_{Γ} |S_{1}(t)| ds = ∑_{n=1}^{∞} ∫_{Γ_{n}} |S_{1}(t)| ds ≤

≤ 2(2π)^{1/p′} ∑_{n=1}^{∞} ∑_{k=1}^{n} \frac{r_{n}}{r_{k}^{1/p}} ( ∫_{Γ_{k}} |ϕ_{k}^{i}(t)|^{p}ds)^{1/p} =

= 2(2π)^{1/p′} ∑_{n=1}^{∞} \frac{∑_{k=n}^{∞} r_{k}}{r_{n}^{1/p}} ( ∫_{Γ_{n}} |ϕ_{n}^{i}(t)|^{p}ds)^{1/p} ≤

≤ 2(2π)^{1/p′}C_{p} ∑_{n=1}^{∞} \frac{∑_{k=n}^{∞} r_{k}}{r_{n}^{1/p}} ( ∫_{Γ_{n}} |φ_{n}^{i}(t)|^{p}ds)^{1/p} ≤

≤ 2(2π)^{1/p′}C_{p} [∑_{n=1}^{∞} (\frac{∑_{k=n}^{∞} r_{k}}{r_{n}} )^{p′} r_{n}]^{1/p′} ( ∫_{Γ} |φ(t)|^{p} ds)^{1/p}, (15)

where C_{p} is the constant from the Riesz inequality (which depends on p only).

The integral of S_{2}(t) can be evaluated analogously. Using inequality (9), as well as the Hölder and Riesz inequalities, we obtain

∫_{Γ_{n}} |S_{2}(t)|dt≤2 ∑_{k=n}^{∞} ∫_{Γ_{n}} |ϕ_{k}^{e}(t)|ds≤2(2π)^{1/p′} ∑_{k=n}^{∞} ( ∫_{Γ_{n}} |ϕ_{k}^{e}(t)|^{p}ds)^{1/p}r_{n}^{1/p′} ≤

≤ 2(2π)^{1/p′} ∑_{k=n}^{∞} (\frac{r_{k}}{r_{n}} )^{\frac{p−1}{p}} ( ∫_{Γ_{k}} |ϕ_{k}^{e}(t)|^{p} ds)^{1/p}r_{n}^{1/p′} ≤

≤ 2(2π)^{1/p′}C_{p} ∑_{k=n}^{∞} r_{k}^{1/p′}( ∫_{Γ_{k}} |φ_{k}(t)|^{p} ds)^{1/p}.

Next, changing the order of summation and using the Hölder inequality, we can write

∫_{Γ} |S_{2}(t)| dt = ∑_{n=1}^{∞} ∫_{Γ_{n}} |S_{2}(t)| ds ≤ 2(2π)^{1/p′}C_{p} ∑_{n=1}^{∞} ∑_{k=n}^{∞} r_{k}^{1/p′} ×

×( ∫_{Γ_{k}} |φ_{k}(t)|^{p} ds)^{1/p} = 2(2π)^{1/p′}C_{p} ∑_{n=1}^{∞} nr_{n}^{1/p′}( ∫_{Γ_{n}} |φ_{n}(t)|^{p} ds)^{1/p} ≤

≤ 2(2π)^{1/p′}C_{p}( ∑_{n=1}^{∞} n^{p′}r_{n})^{1/p′}[ ∑_{n=1}^{∞} ( ∫_{Γ_{n}} |φ_{n}(t)|^{p} ds)^{\frac{1}{p} ·p}]^{\frac{1}{p}} =

= 2(2π)^{1/p′}C_{p}( ∑_{n=1}^{∞} n^{p′}r_{n})^{1/p′}( ∫_{Γ} |φ(t)|^{p} ds)^{1/p}.  (16)

It follows from (14),(15) and (16) that if conditions (B) and (C) are fulfilled for σ = p′, then the operator S_{Γ} is bounded from L_{p}(Γ) to L_{1}(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 arguments, S_{Γ} is continuous from L_{σ′}(Γ), σ′ = \frac{σ}{σ−1}, to L_{1}(Γ). But then S_{Γ} is also continuous from L_{∞}(Γ) to L_{σ}(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

∫_{Γ} φS_{Γ}ψ dt = − ∫_{Γ} ψS_{Γ}φ dt, φ ∈ L_{σ′}(Γ), ψ ∈ L_{∞}(Γ),

whose validity in our case can be immediately verified.

Further, since S_{Γ} is bounded from L_{σ′}(Γ) and L_{∞}(Γ) to L_{1}(Γ) and L_{σ}(Γ), respectively, according to Riesz–Torin’s theorem on interpolation of linear operators (see, e.g., [14], p.144), it follows that S_{Γ} is bounded from L_{α}(Γ), σ′ ≤ α ≤ ∞, to L_{ασ/(α+σ)}(Γ). Letting α = p, we get that S_{Γ} is bounded from L_{p}(Γ) to L_{q}(Γ).

6. Let us now show that (C) and consequently (B) follow from (A). Let for a pair p and q, p > q ≥ 1, σ = pq/(p − q), the series ∑_{n=1}^{∞} n^{σ}r_{n} diverge. Then, according to the above-mentioned Abel–Dini’s theorem, if ω_{n} = (∑_{k=1}^{n} k^{σ}r_{k})^{−1/q}, then

∑_{n=1}^{∞} ω_{n}^{p}n^{σ}r_{n} = ∑_{n=1}^{∞} \frac{n^{σ}r_{n}}{S_{n}^{p/q}} < ∞, S_{n} = ∑_{k=1}^{n} k^{σ}r_{k},

∑_{n=1}^{∞} ω_{n}^{q}n^{σ}r_{n} = ∑_{n=1}^{∞} \frac{n^{σ}r_{n}}{S_{n}} = ∞.

Consider, on Γ, the function φ(t) = ω_{n}n^{σ/p} for t ∈ Γ_{n}, n = 1, 2, . . .. Then

∫_{Γ} |φ(t)|^{p} |dt| = ∑_{n=1}^{∞} ∫_{Γ_{n}} |φ(t)|^{p} ds = 2π ∑_{n=1}^{∞} ω_{n}^{p}n^{σ}r_{n} < ∞.  (17)

Next, by equality (14) we have

S_{Γ}(φ)(t) = 2 ∑_{k=1}^{n−1} ω_{k}k^{σ/p} + ω_{n}n^{σ/p} > ∑_{k=1}^{n} ω_{k}k^{σ/p}

for t ∈ Γ_{n}. Consequently,

∫_{Γ} |S_{Γ}(φ)(t)|^{q} |dt| = ∑_{n=1}^{∞} ∫_{Γ_{n}} |S_{Γ}(φ)(t)|^{q} |dt| > 2π ∑_{n=1}^{∞} ( ∑_{k=1}^{n} ω_{k}k^{σ/p})^{q}r_{n} >

> 2π ∑_{n=1}^{∞} ω_{n}^{q}( ∑_{k=1}^{n} k^{σ/p})^{q}r_{n} ≥ 2π ∑_{n=1}^{∞} ω_{n}^{q}n^{( \frac{σ}{p} +1)q}r_{n} =

= 2π ∑_{n=1}^{∞} ω_{n}^{q}n^{σ}r_{n} = ∞.  (18)

It follows from (17) and (18) that if condition (C) is not fulfilled for p > q ≥ 1, then there exists a function φ ∈ L_{p}(Γ) for which S_{Γ}(φ) ∉ L_{q}(Γ). Consequently, for condition (A) to be fulfilled, it is necessary that condition (C) (and hence (B)) be fulfilled.

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(Received 28.04.1993)

Authors’ address:

A.Razmadze Mathematical Institute

Georgian Academy of Sciences

1, Z.Rukhadze St., 380093 Tbilisi

Republic of Georgia