Georgian Mathematical Journal 1(1994), No. 5, 537-559

FRACTIONAL TYPE OPERATORS IN WEIGHTED GENERALIZED HÖLDER SPACES

S.G. SAMKO AND Z.U. MUSSALAEVA

Abstract. Weighted Zygmund type estimates are obtained for the continuity modulus of some convolution type integrals. In the case of fractional integrals this is strengthened to a result on isomorphism between certain weighted generalized Hölder type spaces.

1. Introduction

A great number of results is known concerning boundedness of convolution type operators in spaces of summable functions, including the weighted case. In the spaces of continuous functions such as H_{0}^{ω}(ρ) the convolution type operators are less investigated. The goal of this paper is to fill a gap to a certain extent in investigations of such a kind.

We consider here the Volterra convolution type operators

Kφ = ∫_{a}^{x} k(x − t)φ(t) dt, a < x < b,  (1.1)

in the weighted generalized Hölder spaces H_{0}^{ω}(ρ) (see definitions in Sec.2), −∞ < a < b < ∞. The kernel k(x) is assumed to be close in a sense to a power function.

The result of the type

K : H_{0}^{ω}(ρ) → H_{0}^{ω_{1}}(ρ)  (1.2)

for certain characteristic functions ω(h) and ω_{1}(h) was earlier known in the case of the power kernel k(x) = x^{α−1}, 0 < α < 1, and a power weight function ρ(t). We deal here with arbitrary kernels and weights, i.e. not necessarily power ones.

We introduce a certain class V_{λ} of kernels and the class w_{μ} of weight functions for which we manage to give the weighted Zygmund type estimate, that is, to estimate the modulus of continuity ω(ρK_{φ}, h) by the modulus of

537

1072-947X/1994/0900-0537$07.00/0 \begin{matrix}c \\ ◯ \end{matrix} 1994 Plenum Publishing Corporation

continuity ω(ρφ, h). This estimate provides the general result of the type (1.2).

In the case of purely power kernel, i.e. in the case of the fractional integration operator

I_{a+}^{α}φ = \frac{1}{Γ(α)} ∫_{a}^{x}(x − t)^{α−1}φ(t)dt, 0 < α < 1,  (1.3)

the result (1.2) is extended to isomorphism:

I_{a^{+}}^{α}[H_{0}^{ω}(ρ)] = H_{0}^{ω_{α}}(ρ)  (1.4)

with ω_{α}(h)=h^{α}ω(h). This is achieved by the preliminary derivation of Zygmund type estimate for fractional differentiation. The latter is treated in a difference form due to A.Marchaud [9] and G.H.Hardy and J.E.Littlewood [2]: (See [17], Sec.13, in this connection.)

\begin{matrix}D_{a+}^{α}f(x) = \frac{f(x)}{Γ(1 − α)(x − a)^{α}} + \frac{α}{Γ(1 − α)} ∫_{a}^{x} \frac{f(x) − f(t)}{(x − t)^{1+α}}dt, \\ 0 < α < 1. \end{matrix} (1.5)

The paper is organized as follows. In Sec.2 we give necessary preliminaries. Sec.3 contains Zygmund type estimates for the operator (1.2) in the case of kernels in V_{λ} in the non-weighted case first (Theorem 1) and afterwards in the weighted case (Theorem 2). In Theorem 3 we give conditions of Zygmund–Bari–Stechkin type on a characteristic function ω(h) guaranteeing the result (1.2) for k(x) ∈ V_{λ} and weighted functions in w_{μ}. The characteristic function ω_{1}(h) in (1.2) proves to be equal to hk(h)ω(h). We note corollary of Theorem 3 for k(x) = x^{α−1}(ln \frac{γ}{x})^{β}, γ > b − a.

In Sec.4 we establish the weighted Zygmund type estimate for D_{a+}^{α}f with a weight function in w_{μ} (Theorem 4). We prove the assertion D_{a+}^{α} : H_{0}^{ω}(ρ) → H_{0}^{ω_{−α}}(ρ) with ω_{−α}(h) = h^{−α}ω(h) under appropriate assumptions on ω(h) and ρ(x) (Theorem 5).

As a corollary of Theorems 3 and 5 we give conditions for validity of the second index law of E.R.Love within the framework of the spaces H_{0}^{λ}(x^{μ})

Finally, in Sec.5 we prove the isomorphism (1.4) (Theorem 6).

Presented theorems generalize the results of the papers [10]–[12],[18], where the power case for both k(x) = x^{α−1} and ρ(x) = (x − a)^{α} was considered. The presentation of the results of [10] in the non-weighted case can be also found in [18], Sec.13. Note that in [12] the case ρ(x) = (x − a)^{μ}(b − x)^{v} was also considered, not contained in the results of the present paper. The origin of the statement (1.4) is the classical result by G.H.Hardy and J.E.Littlewood [2] for the fractional integration concerning the case ω(h) = h^{λ}, ρ(x) ≡ 1, α + λ < 1. (As for the case ω(h) = ∏_{k=1}^{n} |x − x_{k}|^{μ_{k}}, see [13] and [17]. Sec.13.)

We also note the papers [4], [5] where Zygmund type estimates are given for the fractional integrodifferentiation in the case of L_{p}-moduli of continuity.

The question we finally note as open is whether I_{a+}^{α}[H_{0}^{ω}(ρ)] = H_{0}^{ω}(ρ) in the case of purely imaginary α, under the appropriate assumptions on ω(h) and ρ(x). We refer to the paper [7] by E.R.Love concerning such fractional integrals (see also [17], Sec.2, n^{0}4).

2. Preliminaries

We follow the papers [14], [15] in the definitions below.

Definition 1. We say that ψ(x) ∈ W_{μ} = W_{μ}([0, l]) if ψ(x) ∈ C([0, l]), ψ(0) = 0, ψ(x) > 0 for x > 0, ψ(x) is almost increasing, while ψ(x)/x^{μ} is almost decreasing and there exists a constant c > 0 such that

|\frac{ψ(x) − ψ(y)}{x − y}| ≤ c\frac{ψ(x^{∗})}{x^{∗}}, x^{∗} = max(x, y).  (2.1)

We remind that a non-negative function ψ(x), 0 ≤ x ≤ l, 0 < l ≤ ∞, is called almost increasing (decreasing) if ψ(x) ≤ cψ(y) for all x ≤ y (x ≥ y, resp.), this notion being due to S.Bernstein.

Definition 2. We say that ψ(x) ∈ W_{μ}^{∗} if ψ(x) ∈ W_{μ} and ψ(x)/x^{μ−ε} is almost increasing for all ε > 0.

We shall also need the following modification of the Definition 2.

Definition 3. We say that a non-negative function k(x) on [0, l] belongs to the class V_{λ}, λ > 0 if

i) k(x) ≢ 0, x^{λ}k(x) is almost increasing and x^{λ}k(x)|_{x=0} = 0;

ii) there exists ε, 0 < ε < λ, such that x^{λ−ε}k(x) is almost decreasing; iii) there exists c > 0 such that

|\frac{k(x) − k(y)}{x − y}| ≤ c\frac{k(x^{∗})}{x^{∗}}, x^{∗} = max(x, y).  (2.2)

Remark 1. x^{λ}k(x) ∈ W_{λ}^{∗} ⇒ k(x) ∈ V_{λ} and k(x) ∈ V_{λ} ⇒ k(x) ∈ W_{λ}. Remark 2. If the almost monotonicity in Definitions 1 and 3 is replaced

by the usual monotonicity, then conditions (2.1) and (2.2) are satisfied automatically.

Indeed, let us prove e.g. (2.1), following [14]. If φ(x)/x^{μ} is decreasing, so that 1 − φ(x)/φ(y) ≤ 1 − x^{μ}/y^{μ} for y ≥ x, then φ(y) − φ(x) ≤ \frac{y^{μ}−x^{μ}}{y^{μ}} φ(y). Since y^{μ} − x^{μ} ≤ c(y − x)y^{μ−1}, we obtain (2.1).

Definition 4 ([1]). A non-negative function φ(t) on [0, l] belongs to Zygmund class Z = Z([0, l]) if ∫_{0}^{h} \frac{φ(t)}{t} dt ≤ cφ(h), 0 ≤ h ≤ l.

Definition 5 ([1]). A non-negative function φ(t) belongs to Zygmund class Z_{1} = Z_{1}([0, l]) if ∫_{h}^{l} \frac{φ(t)}{t^{2}} dt ≤ c\frac{φ(h)}{h}.

Definition 6. A function φ(x) belongs to the generalized Hölder space H^{ω} = H^{ω}([a, b]) if

ω(φ, h) =^{def} sup_{0<t<h} sup_{x,x+t∈[a,b]} |φ(x + t) − φ(x)| ≤ cω(h),  (2.3)

where ω(h) is a given positive function on [0, l], ω(0) = 0; we set ∥φ∥_{H^{ω}} = ∥φ∥_{C} + sup_{h>0}[ω(φ, h)/ω(h)].

By H_{0}^{ω} we denote the subspace of functions in H^{ω} which vanish at x = a. The function ω(h) is called a characteristic of the space H^{ω}. Definition 7. By H_{0}^{ω}(ρ) we denote the space of functions f(x) such

that ρ(x)f(x) ∈ H_{0}^{ω}, ∥f∥_{H_{0}^{ω}(ρ)} = ∥ρf∥_{H_{0}^{ω}}, where ρ(x) is a non-negative weight function.

In the sequel we shall use the following inequalities:

1) if ω(φ, h) is the continuity modulus (2.3), then

\frac{ω(φ, x)}{x} ≤ c\frac{ω(φ, y)}{y}, x ≥ y;  (2.4)

2) if 0 < α < 1, then

\frac{ω(φ, h)}{h^{α}} ≤ c ∫_{0}^{h} \frac{ω(φ, t)}{t^{1+α}} dt;  (2.5)

3) if ψ(x) ∈ W_{μ}, then

ψ(x) ≤ c(\frac{x}{y})^{μ}ψ(y), x ≥ y;  (2.6)

4) if ψ(x) ∈ W_{μ} with 0 < μ < 1, then (2.1) holds with x^{∗} replaced both by x and y:

\begin{matrix}|\frac{ψ(x) − ψ(y)}{x − y}| ≤ c\frac{ψ(x)}{x}, \\ |\frac{ψ(x) − ψ(y)}{x − y}| ≤ c\frac{ψ(y)}{y}; \end{matrix}  (2.7)

5) if k(x) ∈ V_{λ}, then

k(x) ≤ c(\frac{y}{x})^{λ}k(y), x ≤ y,  (2.8)

and there exists ε > 0 such that

k(x) ≤ c(\frac{y}{x})^{λ−ε}k(y), x ≥ y;  (2.9)

6) if λ ≤ 1, then

|x^{λ} − y^{λ}| ≤ c(x − y)y^{λ−1}, x ≥ y > 0,  (2.10)

and if λ ≥ 0, then

|x^{λ} − y^{λ}| ≤ c(x − y)x^{λ−1}, x ≥ y > 0.  (2.11)

Lemma 1. Let k(x) ∈ V_{λ}, λ > 0 and let ω(x) ≥ 0 be an almost increasing function. Then ω(x)k(x) ≤ c ∫_{x}^{l} \frac{ω(t)k(t)}{t}dt for 0 < x < l/2.

Proof. By (2.8) we have

∫_{x}^{l} \frac{ω(t)k(t)}{t}dt≥cω(x)x^{λ}k(x)∫_{x}^{l} \frac{dt}{t^{1+λ}} ≥cω(x)x^{λ}k(x)∫_{x}^{2x}\frac{dt}{t^{1+λ}} =cω(x)k(x). ■

3. Mapping Properties of Convolution Operators in the Space H_{0}^{ω}(ρ)

The following theorem provides a Zygmund type estimate for the integral (1.1).

Theorem 1. Let k(x) ∈ V_{λ}, 0 < λ < 1 and φ(x) ∈ C([a, b]), φ(a) = 0. Then

ω(Kφ, h) ≤ chk(h)ω(φ, h) + ch ∫_{h}^{b−a}\frac{k(t)ω(φ, t)}{t} dt.  (3.1)

Proof. Let a = 0 for simplicity. We denote g(x) = φ(x) − φ(0) and f(x) = ∫_{0}^{x} k(x − t)g(t)dt. For all x, x + h ∈ [0, b] we have

f(x + h) − f(x) = ∫_{−h}^{x}[g(x − t) − g(x)]k(t + h)dt −

− ∫_{0}^{x}[g(x − t) − g(x)]k(t)dt + g(x)[ ∫_{−h}^{x} k(t + h)dt − ∫_{0}^{x} k(t)dt].

So,

|f(x + h) − f(x)| ≤ | ∫_{−h}^{0}[g(x − t) − g(x)]k(t + h)dt| +

+|∫_{0}^{x}[g(x−t)−g(x)][k(t)−k(t+h)]dt| + |g(x) ∫_{x}^{x+h}k(t)dt| = A_{1}+A_{2}+A_{3}.

Taking (2.8) and increasing of ω(φ, t) into account, we have for A_{1}:

A_{1} ≤∫_{0}^{h}ω(φ, t)k(h−t)dt≤cω(φ, h)k(h)∫_{0}^{h}(\frac{h}{h − t})^{λ}dt≤chk(h)ω(h). (3.2)

For A_{2}, applying (2.2) and (2.9), we obtain in the case h ≥ x:

A_{2} ≤ ch ∫_{0}^{x} \frac{ω(φ, t)k(t + h)}{t + h} dt ≤ chk(h)h^{λ−ε} ∫_{0}^{x} \frac{ω(φ, t)dt}{(t + h)^{1+λ−ε}} =

= chk(h) ∫_{0}^{x/h}\frac{ω(φ, ht)dt}{(t + 1)^{1+λ−ε}} ≤ chk(h)ω(φ, h) ∫_{0}^{1} \frac{dt}{(t + 1)^{1+λ−ε}} ≤

≤ chk(h)ω(φ, h).  (3.3)

In the case h < x we write A_{2} ≤ ∫_{0}^{h} + ∫_{h}^{x} = B_{1} + B_{2}. For B_{1} the estimate (3.3) is valid, while for B_{2} we have

B_{2} ≤ ch ∫_{h}^{b} \frac{ω(φ, t)k(t)}{t} dt.  (3.4)

As regards A_{3}, we have in the case h ≥ x:

A_{3} ≤ cω(φ, h)k(x + h)(x + h)^{λ} ∫_{x}^{x+h}\frac{dt}{t^{λ}} ≤

≤ cω(φ, h)k(h)h^{λ} ∫_{0}^{2h} \frac{dt}{t^{λ}} ≤ cω(φ, h)hk(h).  (3.5)

If h < x, we use Lemma 1 to obtain

A_{3} ≤ cω(φ, x)hk(x) ≤ ch ∫_{x}^{b} \frac{ω(φ, t)k(t)dt}{t} ≤ ch ∫_{h}^{b} \frac{ω(φ, t)k(t)}{t} dt. (3.6)

Gathering all the estimates for A_{i}, i = 1, 2, 3, we arrive at (3.1).

Theorem 2. Let k(x) ∈ V_{λ}, 0 < λ < 1, ρ(x) = ψ(x − a), ψ(x) ∈ W_{μ}, 0 < μ < 2. Assume that

i) ρ(x)φ(x) ∈ C([a, b]) and ρ(x)φ(x)|_{x=a} = 0;

ii) ∫_{0}^{b−a} t^{−γ}ω(ρφ, t)dt < ∞, γ = max(1, μ). Then the following Zygmund type estimate holds:

ω(ρKφ, h) ≤ ch^{γ}k(h) ∫_{0}^{h} \frac{ω(ρφ, t)}{t^{γ}} dt + ch ∫_{h}^{b−a}\frac{ω(ρφ, t)k(t)}{t} dt, (3.7)

if 0 < μ < 1 + λ and

ω(ρKφ, h) ≤ ch ∫_{0}^{h} \frac{ω(ρφ, t)}{t^{μ}} dt + h ∫_{h}^{b−a}\frac{ω(ρφ, t)}{t} dt,  (3.8)

if 1 + λ ≤ μ < 2.

Proof. Let φ_{0}(x) = ρ(x)φ(x) and a = 0 for simplicity. We have

ρ(x)(Kφ)(x) = ∫_{0}^{x} k(x − t)φ_{0}(t)dt + ∫_{0}^{x} \frac{ψ(x) − ψ(t)}{ψ(t)} k(x − t)φ_{0}(t)dt =
= f_{1}(x) + f_{2}(x).

Since φ_{0} ∈C([0, b]) and φ_{0}(0) = 0, the first term f_{1}(t) is covered by Theorem 1. To estimate ω(f_{2}, h) we represent the difference f_{2}(x+h)−f_{2}(x) as

f_{2}(x + h) − f_{2}(x) = ∫_{x}^{x+h}\frac{ψ(x + h) − ψ(t)}{ψ(t)} φ_{0}(t)k(x + h − t)dt +

+ ∫_{0}^{x} \frac{ψ(x + h) − ψ(x)}{ψ(t)} φ_{0}(t)k(x + h − t)dt +

+∫_{0}^{x} \frac{ψ(x)−ψ(t)}{ψ(t)} [k(x−t+h)−k(x−t)]φ_{0}(t)dt = I_{1} + I_{2} + I_{3}.

Estimate for I_{1}. A) Let 0 < μ < 1 at first. Taking (2.4), (2.7) and (2.8) into account, we have

|I_{1}| ≤ ∫_{x}^{x+h}\frac{(x + h − t)k(x + h − t)ω(φ_{0}, t)dt}{t} ≤

≤ chk(h) ∫_{x}^{x+h}\frac{ω(φ_{0}, t − x)dt}{t − x} ≤ chk(h) ∫_{0}^{h} \frac{ω(φ_{0}, t)dt}{t}.  (3.9)

544 S.G. SAMKO AND Z.U. MUSSALAEVA

B) If 1 ≤ μ < 2, then by (2.4),(2.6) and (2.8) we obtain

|I_{1}| ≤ c ∫_{0}^{h} \frac{(h − t)k(h − t)(x + h)^{μ−1}}{(x + t)^{μ−1}} \frac{ω(φ_{0}, t)}{x + t} dt ≤

≤ chk(h) ∫_{0}^{h} \frac{(x + h)^{μ−1}}{t(x + h)^{μ−1}} ω(φ_{0}, t)dt.  (3.10)

In the case h < x we derive from (3.10)

|I_{1}| ≤ chk(h) ∫_{0}^{h} \frac{x^{μ−1}ω(φ_{0}, t)dt}{t(x + t)^{μ−1}} ≤ chk(h) ∫_{0}^{h} \frac{ω(φ_{0}, t)dt}{t}. (3.11)

In the case h > x the inequality (3.10) yields

|I_{1}| ≤ chk(h)h^{μ−1} ∫_{0}^{h} \frac{ω(φ_{0}, t)dt}{t(x + t)^{μ−1}} ≤ ch^{μ}k(h) ∫_{0}^{h} \frac{ω(φ_{0}, t)dt}{t^{μ}}. (3.12)

So from (3.9), (3.11) and (3.12) there follows the estimate

|I_{1}| ≤ ch^{γ}k(h) ∫_{0}^{h} \frac{ω(φ_{0}, t)dt}{t^{γ}}, γ = max(1, μ).  (3.13)

Estimate for I_{2}. A) Let 0 < μ < 1. By (2.6) and (2.7) we have

|I_{2}| ≤ ch ∫_{0}^{x} \frac{k(x + h − t)ω(φ_{0}, t)dt}{t}.  (3.14)

In the case h < x we represent (3.14) as |I_{2}| ≤ ∫_{0}^{h}+∫_{h}^{(x+h)/2}+∫_{(x+h)/2}^{x} = I′_{2}+I′′_{2}+I′′′_{2}. It is clear that

I′_{2} ≤ chk(h) ∫_{0}^{h} \frac{ω(φ_{0}, t)dt}{t}.  (3.15)

Since x + h − t ≥ t in I′′_{2}, we obtain

I′′_{2} ≤ ch ∫_{h}^{b} \frac{k(t)ω(φ_{0}, t)dt}{t}.  (3.16)

FRACTIONAL INTEGRALS IN HÖLDER-TYPE SPACES 545

Further, x + h − t ≤ t in I′′′_{2}, so by (2.4)

I′′′ ≤ ch ∫_{(x+h−t)/2}^{x} \frac{k(x + h − t)ω(φ_{0}, x + h − t)dt}{x + h − t} ≤
≤ ch ∫_{h}^{b} \frac{ω(φ_{0}, t)k(t)dt}{t}.  (3.17)

If h ≥ x,then (3.14) immediately yields

|I_{2}| ≤ chk(h) ∫_{0}^{h} \frac{ω(φ_{0}, t)dt}{t}.  (3.18)

B) Let now 1 ≤ μ < 2. Taking (2.1) into account, we have

|I_{2}| ≤ ch ∫_{0}^{x} \frac{(x + h)^{μ−1}k(x + h − t)ω(φ_{0}, t)dt}{t^{μ}}.  (3.19)

Hence

|I_{2}| ≤ ch^{μ}k(h) ∫_{0}^{h} \frac{ω(φ_{0}, t)dt}{t^{μ}}

in the case h ≥ x. If h < x, we represent (3.19) as

|I_{2}| ≤ ∫_{0}^{h} + ∫_{h}^{(x+h)/2}+ ∫_{(x+h)/2}^{x} = B_{1} + B_{2} + B_{3}.

Taking into account that x+h ≤ 2(x+h−t) in B_{1}, in the case 0 < μ−1 < λ we obtain

|B_{1}| ≤ ch ∫_{0}^{h} \frac{(x + h − t)^{μ−1}k(x + h − t)ω(φ_{0}, t)dt}{t^{μ}} ≤
≤ ch^{μ}k(h) ∫_{0}^{h} \frac{ω(φ_{0}, t)dt}{t^{μ}}  (3.20)

by (2.9). If μ − 1 > λ, the function t^{μ−1}k(t) is bounded. So

B_{1} ≤ ch ∫_{0}^{h} \frac{ω(φ_{0}, t)dt}{t^{μ}}.  (3.21)

Since x + h ≤ 2(x + h − t) again, we have

B_{2} ≤ ch ∫_{h}^{(x+h)/2}\frac{(x + h − t)^{μ−1}k(x + h − t)}{t^{μ}} dt.

Here x + h − t ≥ t, so that

B_{2} ≤ ch ∫_{h}^{b} \frac{k(t)ω(φ_{0}, t)dt}{t}  (3.22)

by (2.9), if μ − 1 < λ. If μ − 1 ≥ λ, by boundedness of t^{μ−1}k(t) we have

B_{2} ≤ ch ∫_{h}^{b} \frac{ω(φ_{0}, t)dt}{t^{μ}}.  (3.23)

To estimate B_{3} we notice that t ≥ x + h − t in B_{3}. So by (2.4) we have

B_{3} ≤ ch ∫_{x+h−t/2}^{x}\frac{k(x + h − t)ω(φ_{0}, x + h − t)dt}{x + h − t} ≤ ch∫_{h}^{b} \frac{k(t)ω(φ_{0}, t)dt}{t}. (3.24)

Thus, I_{2} admits the estimate

|I_{2}|≤ch^{γ}k(h)=∫_{0}^{h} \frac{ω(φ_{0}, t)dt}{t^{μ}} +ch∫_{h}^{b} \frac{k(t)ω(φ_{0}, t)dt}{t}, γ =max(1, μ),(3.25)

if μ < 1 + λ and

|I_{2}| ≤ ch ∫_{0}^{h} \frac{ω(φ_{0}, t)dt}{t^{μ}} + ch ∫_{h}^{b} \frac{ω(φ_{0}, t)dt}{t^{μ}}  (3.26)

if μ ≥ 1 + λ.

Estimate for I_{3}. Let 0 < μ < 1. By (2.2) and (2.4) we have

|I_{3}| ≤ ch ∫_{0}^{x} \frac{k(x + h − t)ω(φ_{0}, t)dt}{t}

which coincides with the estimate in (3.14). If 1 ≤ μ < 2, we derive from (2.1),(2.2) and (2.6):

|I_{3}|≤ch∫_{0}^{x} \frac{x^{μ−1}k(x+h−t)ω(φ_{0}, t)dt}{t^{μ}} ≤ch∫_{0}^{x} \frac{(x+h)^{μ−1}k(x+h−t)ω(φ_{0}, t)dt}{t^{μ}}.

The latter coincides with (3.19). Gathering estimates for I_{1}, I_{2} and I_{3}, we obtain (3.7)–(3.8).

Theorem 3. Let ρ(x) = ψ(x − a), ψ(x) ∈ W_{μ}, 0 < μ < 2, k(t) ∈ V_{λ}, 0 < λ < 1. Assume that

i) μ < λ + 1;

ii) t^{− max(0,μ−1)}ω(t) ∈ Z , tk(t)ω(t) ∈ Z_{1}. Then the operator K is bounded from H_{0}^{ω}(ρ) into H_{0}^{ω_{k}}(ρ) with ω_{k}(h) = hk(h)ω(h).

Proof. Let f = Kφ with φ ∈ H_{0}^{ω}(ρ) and let a = 0. To prove that f ∈ H_{0}^{ω_{k}}(ρ) we remark at first that

∫_{0}^{b} \frac{ω(ρφ, t)}{t^{γ}} dt < ∞, γ = max(1, μ).  (3.27)

Therefore Zygmund type estimate (3.27) concerning the case 0 < μ < 1 + λ holds which gives

\frac{ω(ρf, h)}{hk(h)ω(h)} ≤∥ρφ∥_{H_{0}^{ω}}\{\frac{h^{γ−1}}{ω(h)} ∫_{0}^{h} \frac{ω(t)}{t^{γ}} dt+ \frac{1}{k(h)ω(h)}∫_{h}^{b} \frac{ω(t)k(t)dt}{t}\}. (3.28)

Hence by the condition ii) we have

\frac{ω(ρf, h)}{hk(h)ω(h)} ≤ c∥φ∥_{H_{0}^{ω}(ρ)}.  (3.29)

It remains to prove that ρ(x)f(x)|_{x=0} = 0. After the change of variable t = x − ξx we have

|ρ(x)f(x)| ≤ xψ(x) ∫_{0}^{1} \frac{|φ_{0}(x − xξ)|k(xξ)dξ}{ψ(x − xξ)} .

Since φ_{0}(0) = 0, this yields

|ρ(x)f(x)| ≤ cxψ(x) ∫_{0}^{1} \frac{ω(φ_{0}(, 1 − ξ)dξ}{ψ(x − xξ)} .  (3.30)

According to (2.6) and (2.8) we see that

|ρ(x)f(x)| ≤ cxk(x) ∫_{0}^{1} \frac{ω(φ_{0}, 1 − ξ)dξ}{ξ^{λ}(1 − ξ)^{γ}} = c_{1}xk(x) → 0  (3.31)

as x → 0 in view of (3.27). So ρ(x)f(x)|_{x=0} = 0.

Corollary 1. The operator (1.1) with the kernel k(t) = t^{α−1}(ln \frac{γ}{t})^{β}, γ > b−a, 0 < α < 1, β ≥ 0 is bounded from H_{0}^{ω}(ρ) into H_{0}^{ω_{α,β}}(ρ), where ρ(x) = ψ(x − a), ψ(x) ∈ W_{μ} and ωα, β(h) = ω(h)h^{α}(ln \frac{γ}{h})^{β} under the assumption that 0 < μ < 2 − α and h^{− max(0,μ−1)}ω(h) ∈ Z, h^{α}(ln \frac{γ}{h})^{β}ω(h) ∈ Z_{1}.

In the case ψ(x) = x^{μ} and ω(h) = h^{λ} the assertion of Corollary 1 was proved in [6] (see [17], Theorem 21.2).

Corollary 2. The operator of the form ∫_{x}^{b} k(t−x)φ(t)dt is bounded from H_{0}^{ω}(ρ) into H_{0}^{ω_{k}}(ρ) under the assumptions of Theorem 3 if the requirement ρf|_{x=a} = 0 in the definition of the space H_{0}^{ω}(ρ) is replaced by ρf|_{x=b} = 0.

4. Mapping Properties of Fractional Differentiation in the Spaces H_{0}^{ω}(ρ)

Now we give Zygmund type estimate for the fractional derivative (1.5).

Theorem 4. Let ρ(x) = ψ(x − a), ψ(x) ∈ W_{μ}, 0 < μ < 2, and

∫_{0}^{b−a}\frac{ω(ρf, t)}{t^{α+μ}} dt < ∞, γ = max(1, μ).

Then

ω(ρD_{a+}^{α}f, h) ≤ ch^{γ−1} ∫_{0}^{h} \frac{ω(ρf, t)}{t^{α+γ}} dt.  (4.1)

Proof. According to (1.5) we have

ρ(x)(D_{a+}^{α}f)(x) = \frac{ρ(x)f(x)}{Γ(1 − α)(x − a)^{α}} +

+\frac{α}{Γ(1 − α)} ρ(x) ∫_{0}^{x−a}\frac{f(x) − f(x − t)}{t^{1+α}} dt.  (4.2)

We set a = 0 and denote θ(x) = ψ(x) ∫_{0}^{x} \frac{f(x)−f(x−t)}{t^{1+α}}dt. To estimate the difference θ(x + h) − θ(x) we represent it in the form θ(x + h) − θ(x) = ∑_{k=1}^{8} A_{k}(x) (as in [13] in the purely power case), where

A_{1}(x) = [1 − \frac{ψ(x)}{ψ(x + h)}] ∫_{0}^{x} \frac{g(x + h) − g(y)}{(x + h − y)^{1+α}} dy,

A_{2}(x) = [ψ(x + h) − ψ(x)] ∫_{0}^{x+h}\frac{g(y)}{(x + h − y)^{1+α}} [\frac{1}{ψ(x + h)} − \frac{1}{ψ(y)}] dy,

A_{3}(x) = \frac{ψ(x)}{ψ(x + h)} ∫_{x}^{x+h}\frac{g(x + h) − g(y)}{(x + h − y)^{1+α}} dy,

A_{4}(x) = ψ(x) ∫_{x}^{x+h}\frac{g(y)}{(x + h − y)^{1+α}} [\frac{1}{ψ(x + h)} − \frac{1}{ψ(y)}] dy,

A_{5}(x) = ∫_{0}^{x}[g(x) − g(y)][(x + h − y)^{−1−α} − (x − y)^{−1−α}] dy,

A_{6}(x)=ψ(x) ∫_{0}^{x}g(y)[\frac{1}{ψ(x)} − \frac{1}{ψ(y)}][(x+h−y)^{−1−α}−(x−y)^{−1−α}]dy,

A_{7}(x) = \frac{1}{α} \frac{ψ(x)}{ψ(x + h)}[g(x + h) − g(x)][h^{−α} − (x + h)^{−α}],
A_{8}(x) = \frac{1}{α} ψ(x)g(x)[\frac{1}{ψ(x + h)} − \frac{1}{ψ(x)}][h^{−α} − (x + h)^{−α}].

Estimate for A_{1}. By (2.1) we have

|A_{1}|≤c\frac{h}{x+h} ∫_{0}^{x+h}\frac{ω(g, x+h−y)}{(x+h−y)^{1+α}}dy =c\frac{h}{x+h} ∫_{0}^{x+h}\frac{ω(g, t)}{t^{1+α}}dt.  (4.3)

If h ≥ x, it is obvious that

|A_{1}| ≤ c ∫_{0}^{h} \frac{ω(g, t)}{t^{1+α}} dt.  (4.4)

If h < x represent (4.3) as |A_{1}| ≤ ∫_{0}^{h} + ∫_{h}^{x+h} = A′_{1} + A′′_{1}. For A′_{1} the estimate (4.4) holds. As regards A′′_{1}, applying (2.4) and (2.5), we have

A′′_{1} ≤ ch ∫_{x}^{x+h}\frac{ω(g, t)}{t} \frac{dt}{t^{1+α}} ≤ cω(g, h) ∫_{h}^{∞} \frac{dt}{t^{1+α}} ≤

≤ c\frac{ω(g, h)}{h^{α}} ≤ c ∫_{0}^{h} \frac{ω(g, t)}{t^{1+α}} dt.  (4.5)

550 S.G. SAMKO AND Z.U. MUSSALAEVA

Estimate for A_{2}. A) In the case 0 < μ < 1, using (2.7), we obtain

|A_{2}| ≤ c\frac{h}{x + h} ∫_{0}^{x+h}\frac{ω(g, y)}{y(x + h − y)^{α}} dy = ∫_{0}^{h/2}+ ∫_{h/2}^{x+h}= A′_{2} + A′′_{2}.

Obviously, A′_{2} ≤ c ∫_{0}^{h} \frac{ω(g,t)}{t^{1+α}}dt. Using (2.4) and (2.5), we derive the following estimate for A′′_{2}:

A′′_{2} ≤ c\frac{ω(g, h)}{x + h} ∫_{0}^{x+h}\frac{dy}{(x + h − y)^{α}} = c_{1} \frac{ω(g, h)}{(x + h)^{α}} ≤ c_{2} ∫_{0}^{h} \frac{ω(g, t)}{t^{1+α}} dt.

B) If 1≤μ<2−α, taking (2.1) and (2.6) into account, we obtain for A_{2}:

|A_{2}| ≤ c\frac{h}{(x + h)^{2−μ}} ∫_{0}^{h} \frac{ω(g, y)dy}{y^{μ}(x + h − y)^{α}} = ∫_{0}^{h/2}+ ∫_{h/2}^{x+h}= B_{1} + B_{2}.

It is obvious that

B_{1} ≤ ch^{μ−1} ∫_{0}^{h} \frac{ω(g, t)}{t^{μ+α}} dt.  (4.6)

As regards B_{2}, we apply (2.4) and obtain

B_{2} ≤ c\frac{ω(g, h)}{(x + h)^{2−μ}} ∫_{h/2}^{x+h}\frac{dy}{y^{μ−1}(x + h − y)^{α}} ≤

≤ c\frac{ω(g, h)}{(x + h)^{α}} ∫_{0}^{1} \frac{dt}{t^{μ−1}(1 − t)^{α}} = c_{1} \frac{ω(g, h)}{(x + h)^{α}} .

Using then (2.5), we notice that the estimate for B_{2} is the same as in (4.6). Estimate for A_{3}. Since ψ(x) is almost increasing, we have

|A_{3}| ≤ c ∫_{x}^{x+h}\frac{ω(g, x + h − y)}{(x + h − y)^{1+α}} dy = c ∫_{0}^{h} \frac{ω(g, t)}{t^{1+α}} dt.

Estimate for A_{4}. Let 0 < μ < 1 at first. In view of (2.7) we have

|A_{4}| ≤ c ∫_{0}^{h} \frac{ω(g, x + h − t)}{t^{α}(x + h − t)} dt.  (4.7)

FRACTIONAL INTEGRALS IN HÖLDER-TYPE SPACES 551

In the case h < x we have t ≤ x+h−t in (4.7). So by (2.4) we obtain |A_{4}| ≤ c ∫_{0}^{h} \frac{ω(g,t)}{t^{1+α}} dt. If h ≥ x, we represent (4.7) as |A_{4}| ≤ ∫_{0}^{(x+h)/2} + ∫_{(x+h)/2}^{h}. Since t ≤ x + h − t and t ≥ x + h − t in the first and second terms, respectively, by (2.4) we derive that

|A_{4}| ≤ c ∫_{0}^{(x+h)/2}\frac{ω(g, t)}{t^{1+α}} dt + c ∫_{(x+h)/2}^{h} \frac{ω(g, x + h − t)}{(x + h − t)^{1+α}} dt ≤ c ∫_{0}^{h} \frac{ω(g, t)}{t^{1+α}} dt.

Let now 1 ≤ μ < 2 − α. Using (2.1) and (2.6), we get

|A_{4}| ≤ \frac{c}{(x + h)^{1−μ}} ∫_{0}^{h} \frac{ω(g, x + h − t)}{(x + h − t)^{μ}t^{α}} dt.

If h < x, by (2.4) we have

|A_{4}| ≤ c(x + h)^{μ−1} ∫_{0}^{h} \frac{ω(g, t)dt}{(x + h − t)^{μ−1}t^{1+α}} ≤ c ∫_{0}^{h} \frac{ω(g, t)}{t^{1+α}} dt.

If h ≥ x, then |A_{4}| ≤ ∫_{0}^{(x+h)/2} + ∫_{(x+h)/2}^{h}. We use (2.4) in the first term and the inequality t^{α} ≥ (x + h − t)^{α} in the second. This yields

|A_{4}| ≤ ∫_{0}^{(x+h)/2}\frac{ω(g, t)}{t^{1+α}} dt c(x + h)^{μ−1} ∫_{(x+h)/2}^{h} \frac{ω(g, x + h − t)}{(x + h − t)^{α+μ}} dt ≤

≤ c ∫_{0}^{h} \frac{ω(g, t)}{t^{1+α}} dt + ch^{μ−1} ∫_{x}^{(x+h)/2}\frac{ω(g, t)}{t^{α+μ}} dt ≤ ch^{μ−1} ∫_{0}^{h} \frac{ω(g, t)}{t^{α+μ}} dt.

Estimate for A_{5}. Applying (2.11), we have

|A_{5}| ≤ ch ∫_{0}^{h} ω(g, x − y) \frac{dy}{(x + h − y)(x − y)^{1+α}} = ch ∫_{0}^{x} \frac{ω(g, t)dt}{t^{1+α}(t + h)}.

In the case h ≥ x it is clear that

|A_{5}| ≤ c ∫_{0}^{h} \frac{ω(g, t)}{t^{1+α}} dt.  (4.8)

If h < x, then |A_{5}| ≤ ∫_{0}^{h} + ∫_{h}^{x} = A′_{5} + A′′_{5} with the same estimate as in (4.8) for A′_{5}. As regards A′′_{5}, we have

A′′_{5} ≤ cω(g, h) ∫_{h}^{∞} \frac{dt}{t^{1+α}} ≤ c\frac{ω(g, h)}{h^{α}} ≤ c ∫_{0}^{h} \frac{ω(g, t)}{t^{1+α}}dt  (4.9)

by (2.4) and (2.5).

Estimate for A_{6}. A) Let 0 < μ < 1 at first. Applying (2.7) and (2.11) we arrive at

|A_{6}| ≤ ch ∫_{0}^{x} \frac{ω(g, y)dy}{y(x + h − y)(x − y)^{α}} = ∫_{0}^{x/2}+ ∫_{x/2}^{x} = A′_{6} + A′′_{6}.

For A′_{6} we have

A′_{6} ≤ ch ∫_{0}^{x/2}\frac{ω(g, y)dy}{(y + h)y^{1+α}} .  (4.10)

If h < x, A′_{6} ≤ ∫_{0}^{h/2} + ∫_{h/2}^{x/2} = K_{1} + K_{2}. It is evident that K_{1} admits the same estimate as in (4.9). For K_{2} the application of (2.5) provides the same result: K_{2} ≤ cω(g, h) ∫_{h/2}^{∞} \frac{dt}{t^{1+α}} ≤ c ∫_{0}^{h} \frac{ω(g,t)}{t^{1+α}}dt. If h ≥ x, then immediately

A′_{6} ≤ c ∫_{0}^{h} \frac{ω(g, t)}{t^{1+α}} dt.  (4.11)

To estimate A′_{6} we remark that y ≥ x − y so that

A′′_{6} ≤ ch ∫_{x/2}^{x} \frac{ω(g, x − y)dy}{(x − y)^{1+α}(x + h − y)} = ch ∫_{0}^{x/2}\frac{ω(g, t)dt}{t^{1+α}(t + h)},

which is the same as in (4.10) and so A′′_{6} admits the same estimates as in (4.5).

B) Let 1 ≤ μ < 2 − α. Using (2.1), (2.6) and (2.11), we have

A_{6} ≤ chx^{μ−1} ∫_{0}^{x} \frac{ω(g, y)dy}{y^{μ}(x + h − y)(x − y)^{α}} = ∫_{0}^{x/2}+ ∫_{x/2}^{x} = U_{1} + U_{2}.

If h < x, we set U_{1} = ∫_{0}^{h/2} + ∫_{h/2}^{x/2} = U′_{1} + U′′_{1}, whence easy calculations yield the inequality

U_{1} ≤ ch ∫_{0}^{h} \frac{ω(g, t)dt}{t^{1+α}} + ch^{μ−1} ∫_{0}^{h} \frac{ω(g, t)dt}{t^{μ+α}} ≤ ch^{μ−1} ∫_{0}^{h} \frac{ω(g, t)dt}{t^{μ+α}}. (4.12)

For U_{2} by (2.5) we have

U_{2} ≤ ch ∫_{x/2}^{x} \frac{ω(g, y)dy}{y(x + h − y)(x − y)^{α}} ≤ cω(g, h) ∫_{0}^{x/2}\frac{dt}{t^{α}(t + h)} ≤

≤ c\frac{ω(g, h)}{h^{α}} ∫_{0}^{∞} \frac{dξ}{ξ^{α}(1 + ξ)} ≤ c ∫_{0}^{h} \frac{ω(g, t)dt}{t^{1+α}}.

If h ≥ x, the estimation of U_{1} and U_{2} is easy and provides the same as in (4.2). Gathering all the estimates, we obtain |A_{6}| ≤ ch^{γ−1} ∫_{0}^{h} \frac{ω(g,t)}{t^{α+γ}} dt, γ = max(1, μ).

Estimate for A_{7}. Applying (2.5) and (2.10) and almost increasing of ψ(x), we easily obtain

|A_{7}| ≤ c ∫_{0}^{h} \frac{ω(g, t)}{t^{1+α}} dt.  (4.13)

Estimate for A_{8}. Using the inequalities (2.1) and (2.10) for 0 < μ < 2−α, we make sure of validity of the estimate (4.13) for A_{8} as well.

It remains to consider the first term

r(x) = \frac{ψ(x − a)f(x)}{Γ(1 − α)(x − a)^{α}} = \frac{g(x)}{Γ(1 − α)(x − a)^{α}}

in (4.2). Since g(x) ∈ H_{0}^{ω}, we have the estimate

|r(x + h) − r(x)| ≤ c ∫_{0}^{h} \frac{ω(g, t)dt}{t^{1+α}},  (4.14)

which is derived by direct calculations under the assumptions of the theorem.

Collecting all the estimates for A_{i}, i = 1, …, 8, and (4.14), we obtain the required inequality (4.1).

Theorem 5. Let ρ(x) = ψ(x − a), ψ(x) ∈ W_{μ}, 0 < μ < 2 − α and let 1) ω(t) ≠ 0, t > 0,

2) ω(t)t^{−α}|_{t=0} = 0, (4.15)

3) ω(t)t^{1−α−γ} ∈ Z, γ = max(1, μ). Then the operator D_{a+}^{α} continuously maps H_{0}^{ω}(ρ) into H_{0}^{ω_{−α}}(ρ) with ω_{−α}(h) = h^{−α}ω(h).

Proof. Let f(x) ∈ H_{0}^{ω}(ρ) and φ(x) = D_{a^{+}}^{α}f(x). To show that

sup_{0<h≤b−a} \frac{h^{α}ω(ρφ, h)}{ω(h)} = c < ∞

we observe that the inclusion ω(t)t^{1−α−γ} ∈ Z implies convergence of the integral ∫_{0}^{b−a} ω(t)t^{−α−γ}dt, so Theorem 4 is applicable. Using the estimate (4.1) of Theorem 4, we obtain

\frac{h^{α}ω(ρφ, h)}{ω(h)} ≤ \frac{h^{α+γ−1} ∫_{0}^{1} ω(ρf, t)t^{−α−γ}dt}{ω(h)} ≤ c∥f∥_{H_{0}^{ω}(ρ)}. (4.16)

It remains to show that ρ(x)φ(x)|_{x=a} = 0. By (4.2) we have

|ρ(x)φ(x)| ≤ \frac{ω(ρf, x − a)}{Γ(1 − α)(x − a)^{α}} + \frac{α}{Γ(1 − α)} ∫_{0}^{x−a}\frac{ω(ρf, t)dt}{t^{1+α}} +

+\frac{α}{Γ(1 − α)} ∫_{0}^{x−a}\frac{|ψ(x − a) − ψ(x − a − t)|ω(ρf, x − a − t)dt}{t^{1+α}ψ(x − a − t)} =

= D_{1} + D_{2} + D_{3}.  (4.17)

Here D_{1} ≤c∥f∥H_{0}^{ω}(ρ)\frac{ω(x−a)}{(x−a)^{α}}, the condition (4.16) implies ω(x)x^{−α}|_{x=0} = 0. So lim_{x→a} D_{1} = 0. The equality lim_{x→a} D_{2} = 0 is obvious by the existence of the integral in D_{2}.

For the term D_{3} in the case 0 < μ < 1 we have by (2.7)

D_{3} ≤ c ∫_{0}^{x−a}\frac{ω(ρf, x − a − t)dt}{t^{α}(x − a − t)}.

We evaluate this separately for x−a−t≥t and x−a−t≤t by means of (2.4):

D_{3} ≤ c ∫_{0}^{(x−a)/2}\frac{ω(ρf, t)dt}{t^{1+α}} + c ∫_{(x−a)/2}^{x−a} \frac{ω(ρf, x − a − t)}{(x − a − t)^{1+α}}dt,

whence lim_{x→a} D_{3} = 0.

If 1 ≤ μ ≤ 2 − α, we use (2.1) and (2.5) and obtain

D_{3} ≤ c ∫_{0}^{x−a}\frac{ω(ρf, x − a − t)}{(x − a)^{1−μ}(x − a − t)^{μ}t^{α}} = ∫_{0}^{(x−a)/2}+ ∫_{(x−a)/2}^{x−a}

and similar to what we did above we have

D_{3} ≤ c ∫_{0}^{(x−a)/2}\frac{(x − a)^{μ−1}ω(ρf, t)}{t^{α+μ}}dt + c ∫_{(x−a)/2}^{x−a} \frac{(x − a)^{μ−1}ω(ρf, x − a − t)}{(x − a − t)^{α+μ}}dt ≤

≤ (x − a)^{μ−1} ∫_{0}^{(x−a)/2}\frac{ω(ρf, t)dt}{t^{α+μ}},

so that lim_{x→a} D_{3} = 0. Therefore, lim_{x→a} ρ(x)φ(x) = 0.

Corollary 1. Let φ(x) ∈ W_{μ} 0 < μ < 2 − α, and let ω(t) be an almost increasing function on [0, b − a] such that

1) ω(0) = 0, ω(t) ≠ 0 for t ∈ (0, b − a];

2) ω(t)t^{−1−γ} ∈ Z, γ = max(1, μ).

Then the operator D_{a^{+}}^{α} of fractional differentiation continuously maps H_{0}^{ω_{α}}(ρ) into H_{0}^{ω} with ρ(x) = φ(x − a), ω_{α}(h) = h^{α}ω(h).

Another corollary (of Theorems 5 and 3) will be related to the following Love’s index law [8]:

I_{0+}^{γ}x^{α}I_{0+}^{β}x^{γ}I_{0+}^{α}x^{β}f(x) = f(x), α + β + γ = 0  (4.18)

well known in fractional calculus. This corollary will provide conditions guaranteeing validity of (4.18) for functions f ∈ H_{0}^{ω}(ρ). For simplicity we restrict ourselves with the cases ω(x) = x^{λ} and ρ(x) = x^{μ}. The notation

α^{∗} = \begin{cases}α, & α ≤ 1 \\ 1, & α ≥ 1 \end{cases}

is used below.

Corollary 2. Relation (4.18) is valid for all functions f(x) ∈ H_{0}^{λ}(x^{μ}) and all α, β, γ ∈ R^{1} such that α+β +γ = 0, if the number λ ∈ (0, 1] satisfies the conditions λ > −α, (λ + α)^{∗} + β > 0 [(λ + α)^{∗} + β]^{∗} + γ > 0 while the weight exponent μ satisfies the conditions

1) μ < (λ + α)^{∗} + 1
2) μ < [(λ + α)^{∗} + β]^{∗} + 1
3) μ < {[(λ + α)^{∗} + γ}^{∗} + 1.

5. A Theorem on Isomorphism

In Theorem 3 and Corollary 1 of Theorem 5 it was proved that

I_{a+}^{a} : H_{0}^{ω}(ρ) ⟶ H_{0}^{ω_{α}}(ρ),  (5.1)
D_{a+}^{a} : H_{0}^{ω_{α}}(ρ) ⟶ H_{0}^{ω}(ρ)  (5.2)

under the appropriate assumptions on ω(h) and ρ(x). To derive the assertion I_{a+}^{α}[H_{0}^{ω}(ρ)] = H_{0}^{ω_{α}}(ρ) it remains to show that any function in H_{0}^{ω_{α}}(ρ) is representable by the fractional integral of a function in H_{0}^{ω}(ρ). This will be the goal of Theorem 6 below. Preliminarily we state two auxiliary assertions we need.

Lemma 2 ([17], p.185; p.231 in English ed.). In order a function f(x) to be representable as f = I_{a+}^{α}φ, φ ∈ L_{p}(a, b), −∞ < a < b < ∞, it is necessary and sufficient that

i) f(x)(x − a)^{−α} ∈ L_{p}(a, b);

ii) ∥ψ_{ε}∥_{L_{p}} ≤ c < ∞ with c not depending on ε, where

ψ_{ε}(x) = ∫_{a}^{x−ε}\frac{f(x) − f(t)}{(x − t)^{1+α}}dt,

for a + ε < x < b and ψ_{ε}(x) = 0 for a < x < a + ε.

A close version of Lemma 2 can be found in [16]. (See also [3] for another version under additional assumptions that f ∈ L_{p} and 1 < p < 1/2α).

Lemma 3. Let ω(t)t^{−δ} ∈ Z. There exists p > 1 such that ω(t)t^{−1−δ} ∈ L_{p}(0, 1).

Proof. It is known that the inclusion φ(t) ∈ Z implies existence of ε ∈ (0, 1) such that t^{−ε}φ(t) is almost increasing (see, e.g. [1]). Therefore, there exists ε ∈ (0, 1) such that t^{−ε−δ}ω(t) is bounded. So ω(t) ≤ ct^{δ+ε} and to have a finite L_{p}-norm for ω(t)/t^{1+δ} we must choose p < \frac{1}{1−ε}.

Theorem 6 (On isomorphism). Let ψ(x) ∈ W_{μ}, 0 < μ < 2 − α and let ω(t) be a continuous function such that ω(t)t^{1−γ} ∈ Z, ω(t)t^{α} ∈ Z_{1}, γ = max(1, μ), 0 < α < 1. Then the fractional integration operator I_{a+}^{α} isomorphically maps the weighted space H_{0}^{ω}(ρ) with ρ(x) = ψ(x−a) onto the space H_{0}^{ω_{α}}(ρ) with the same weight and the characteristic ω_{α}(h) = h^{α}ω(h).

Proof. In view of (5.1)–(5.2) it is sufficient to prove the representability of a function f ∈ H_{0}^{ω_{α}}(ρ) by a fractional integral. Aiming to apply Lemma 2 we shall prove that there exists p > 1 such that conditions i)–ii) of Lemma 2 are satisfied.

The estimate

\frac{|f(x)|}{(x − a)^{α}} ≤ c∥f∥_{H_{0}^{ωα}} \frac{ω(x − a)}{(x − a)^{μ}}  (5.3)

is valid for any f(x) ∈ H_{0}^{ω_{α}}(ρ). Really, by (2.6) we have

\frac{|f(x)|}{(x − a)^{α}} ≤ c\frac{|ρ(x)f(x)|}{(x − a)^{μ+α}} ,  (5.4)

which immediately provides (5.3).

Since ω(x)/x^{δ−1} ∈ Z, from (5.3) and Lemma 3 we conclude that there exists p > 1 such that the condition i) of Lemma 2 is satisfied.

For this p we shall show that a constant c > 0 exists such that

∥ψ_{ε}∥_{L_{p}} ≤ c ≤ ∞.  (5.5)

We set g(x) = f(x)ψ(x − a) and have

|ψ_{ε}(x)| ≤ \frac{1}{ψ(x − a)} ∫_{a}^{x−ε}\frac{|g(x) − g(t)|}{(x − t)^{1+α}} dt + ∫_{a}^{x−ε}\frac{|g(t)[\frac{1}{ψ(x−a)} − \frac{1}{ψ(t−a)}]|}{(x − t)^{1+α}} dt.

To estimate F_{1} we use (2.6) and obtain

F_{1} ≤ \frac{c}{(x − a)^{μ}} ∫_{a}^{x} \frac{ω(g, x − t)}{(x − t)^{1+α}} dt ≤ \frac{c}{(x − a)^{γ}} ∫_{a}^{x} \frac{ω(g, x − t)}{(x − t)^{1+α}} dt. (5.6)

Since g(x) ∈ H_{0}^{ω_{α}}(ρ), it is easily proved that (5.6) yields

F_{1} ≤ \frac{c}{(x − a)^{γ}} .  (5.7)

For F_{2} in the case 0 < μ < 1 we use (2.6) and (2.7) and obtain

F_{2} ≤ \frac{c}{(x − a)} ∫_{0}^{x−a}\frac{ω(g, x − t − a)}{t^{α}(x − t − a)} dt = ∫_{0}^{(x−a)/2}+ ∫_{(x−a)/2}^{x−a} .

Hence after simple calculations

F_{2} ≤ \frac{c}{x − a} ∫_{0}^{(x−a)/2}\frac{ω(t)dt}{t} + \frac{c}{x − a} ∫_{(x−a)/2}^{x−a} \frac{ω(x − t − a)dt}{x − t − a} ≤

≤ c\frac{ω(x − a)}{x − a} + \frac{c}{x − a} ∫_{0}^{x−a}\frac{ω(t)}{t} dt.  (5.8)

Since the condition ω(t)t^{1−μ} ∈ Z with 0 < μ < 1 implies ω(t) ∈ Z, we derive from (5.8) the estimate F_{2} ≤ cω(x − a)/(x − a).

Let now 1 ≤ μ < 2 − α. We use inequalities (2.1) and (2.6) to obtain

F_{3} ≤ \frac{c}{x − a} ∫_{0}^{x−a}\frac{ω(g, x − t − a)}{t^{α}(x − t − a)^{μ}} dt ≤ \frac{c}{x − a} ∫_{0}^{x−a}\frac{ω(x − t − a)dt}{t^{α}(x − t − a)^{μ−a}} =

= ∫_{0}^{(x−a/2}+ ∫_{(x−a)/2}^{x−a} .

Calculations and arguments similar to those in the case 0 < μ < 1 give the estimate F_{2} ≤ c\frac{ω(x−a)}{(x−a)^{μ}}. Therefore,

F_{2} ≤ c\frac{ω(x − a)}{(x − a)^{γ}}, γ = max(1, μ).  (5.9)

So, from (5.8), (5.9) we obtain |ψ_{ε}(x)|≤c\frac{ω(x−a)}{(x−a)^{γ}} ∈L_{p}. Hence ∥ψ∥_{L_{p}} ≤ c.

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(Received 15.07.1993)

Author’s address:

Rostov University

Department of Mathematics and Mechanics

105, ul. Bol’shaya Sadovaya,

Rostov-on-Don, 344711

Russia