GEORGIAN MATHEMATICAL JOURNAL: Vol. 5, No. 6, 1998, 565-574

SOLUTION OF SOME WEIGHT PROBLEMS FOR THE RIEMANN–LIOUVILLE AND WEYL OPERATORS

A. MESKHI

Abstract. The necessary and sufficient conditions are found for the weight function v, which provide the boundedness and compactness of the Riemann–Liouville operator R_{α} from L^{p} to L_{v}^{q}. The criteria are also established for the weight function w, which guarantee the boundedness and compactness of the Weyl operator W_{α} from L_{w}^{p} to L^{q}.

In this paper, the necessary and sufficient conditions are found for the weight function v (w), which provide the boundedness and compactness of the Riemann-Liouville transform R_{α}f(x) = ∫_{0}^{x} \frac{f(t)}{(x−t)^{1−α}} dt (of the Weyl

transform W_{α}f(x) = ∫_{x}^{∞} \frac{f(t)}{(t−x)^{1−α}} dt) from L^{p} to L_{v}^{q} (from L_{w}^{p} to L^{q}) when 1 < p, q < ∞, \frac{1}{p} < α < 1 or α > 1 (\frac{q−1}{q} < α < 1 or α > 1).

A complete description of the weight pairs (v, w) providing the boundedness of the operators R_{α} and W_{α} from L_{w}^{p} to L_{v}^{q} when 1 < p < q < ∞ and 0 < α < 1 is given in [1]. For 1 < p ≤ q < ∞ and α > 1 a similar problem has been solved by many authors (see, e.g., [2, 3]).

The necessary and sufficient conditions for pairs of weights, which provide the boundedness of the above-mentioned operators when 1 < q < p < ∞ and α > 1, are obtained in [4].

For 1 < q ≤ p < ∞ and 0 < α < 1, the two-weight problem for the operators R_{α} and W_{α} remains unsolved and in this context the results presented here are interesting.

Let v and w be positive almost everywhere, locally integrable functions defined on ℝ_{+}.

565

1991 Mathematics Subject Classification. 42B20, 46E40, 47G60.

Key words and phrases. Weight, Riemann–Liouville and Weyl operators, boundedness and compactness.

1072-947X/98/1100-0565$15.00/0 \begin{matrix}c \\ ◯ \end{matrix} 1998 Plenum Publishing Corporation

Denote by L_{v}^{p} (1 < p < ∞) a class of all Lebesgue-measurable functions defined on ℝ_{+} for which

∥f∥_{L_{v}^{p}} = ( ∫_{0}^{∞}|f(x)|^{p}v(x) dx)^{\frac{1}{p}} < ∞.

First, let us recall some familiar results.

Theorem A ([5–10]). Let 1 ≤ p ≤ q < ∞. The inequality

( ∫_{0}^{∞}| ∫_{0}^{x} f(t)dt|^{q}v(x)dx)^{\frac{1}{q}} ≤ c( ∫_{0}^{∞}|f(x)|^{p}w(x)dx)^{\frac{1}{p}} ,  (1)

where the positive constant c does not depend on f, is fulfilled iff

D = sup_{t>0} ( ∫_{t}^{∞}v(x)dx)^{\frac{1}{q}} ( ∫_{0}^{t} w^{1−p′}(x)dx)^{\frac{1}{p′}} < ∞ (p′ = \frac{p}{p − 1}).

Moreover, if c is the best constant in (1), then c ≈ D (the symbol ≈ here denotes a two-sided inequality).

Theorem B ([10]). Let 1 ≤ q < p < ∞. Then inequality (1) holds iff

D_{1} = ( ∫_{0}^{∞}[( ∫_{t}^{∞}v(x)dx)( ∫_{0}^{t} w^{1−p′}(x)dx)^{q−1}]^{\frac{p}{p−q}} w^{1−p′}(t)dt)^{\frac{p−q}{pq}} < ∞.

Moreover, if c is the best constant in (1), then c ≈ D_{1}.

We also need Kolmogorov’s theorem formulated as follows (see, e.g., [11]): Theorem C. Let 1 < p, q < ∞ and K : L^{p} → L_{v}^{q} be an integral operator

of the form Kf(x) = ∫_{0}^{∞} k(x, y)f(y)dy. If

∥ ∥k(x, ·)∥_{L^{p′}} ∥_{L_{v}^{q}} < ∞,

then the operator K is compact.

Theorem 1. Let 1 < p ≤ q < ∞, \frac{1}{p} < α < 1 or α > 1. The inequality

∥R_{α}f∥_{L_{v}^{q}} ≤ A∥f∥_{L^{p}},  (2)

where the positive constant A does not depend on f, is fulfilled iff

B = sup_{t>0} B(t) = sup_{t>0} ( ∫_{t}^{∞} \frac{v(x)}{x^{(1−α)q}} dx)^{\frac{1}{q}} t^{\frac{1}{p′}} < ∞.  (3)

Moreover, if A is the best constant in (2), then A ≈ B.

Proof. Sufficiency. Denoting I_{1α}f(x) = ∫0^{\frac{x}{2}} \frac{f(t)}{(x−t)^{1−α}} dt and I_{2α}f(x) =

∫ x_{\frac{x}{2}} \frac{f(t)}{(x−t)^{1−α}} dt for f ∈ L^{p} we write R_{α}f as R_{α}f(x) = I_{1α}f(x) + I_{2α}f(x).

We obtain

∥R_{α}f∥_{L_{v}^{q}}^{q} ≤c_{1} ∫_{0}^{∞}|I_{1α}f(x)|^{q}v(x)dx+c_{1} ∫_{0}^{∞}|I_{2α}f(x)|^{q}v(x)dx=S_{1}+S_{2}.

If 0 < t < \frac{x}{2}, then (x−t)^{α−1} ≤ bx^{α−1}, where the positive constant b depends only on α. Consequently, using Theorem A with w ≡ 1, we have

S_{1} ≤ c_{2} ∫_{0}^{∞} \frac{v(x)}{x^{(1−α)q}} ( ∫_{0}^{x} |f(t)|dt)^{q}dx ≤ c_{3}B^{q}∥f∥_{L^{p}}^{q}.

Now we shall estimate S_{2}. Using the Hölder inequality and the condition \frac{1}{p} < α, we obtain

S_{2} = c_{1} ∫_{0}^{∞}v(x)| ∫_{\frac{x}{2}}^{x} \frac{f(t)}{(x − t)^{1−α}} dt|^{q}dx ≤

≤ c_{1} ∫_{0}^{∞}v(x)( ∫_{\frac{x}{2}}^{x} |f(t)|^{p}dt)^{\frac{q}{p}} ( ∫_{\frac{x}{2}}^{x} \frac{dt}{(x − t)^{(1−α)p′}} )^{\frac{q}{p′}} dx =

= c_{4} ∑_{k∈ℤ} ∫_{2^{k}}^{2^{k+1}}v(x) · x^{(α−1)q+ \frac{q}{p′}} ( ∫_{\frac{x}{2}}^{x} |f(t)|^{p}dt)^{\frac{q}{p}} dx ≤

≤ c_{4} ∑_{k∈ℤ} ( ∫_{2^{k−1}}^{2^{k+1}}|f(t)|^{p}dt)^{\frac{q}{p}} ( ∫_{2^{k}}^{2^{k+1}}v(x) · x^{(α−1)q+ \frac{q}{p′}} dx) ≤

≤ c_{5} ∑_{k∈ℤ} ( ∫_{2^{k−1}}^{2^{k+1}}|f(t)|^{p}dt)^{\frac{q}{p}} ( ∫_{2^{k}}^{2^{k+1}}v(x) · x^{(α−1)q}dx) · 2^{\frac{kq}{p′}} ≤

≤ c_{5}B^{q} ∑_{k∈ℤ} ( ∫_{2^{k−1}}^{2^{k+1}}|f(t)|^{p}dt)^{\frac{q}{p}} ≤ c_{6}B^{q}∥f∥_{L^{p}}^{q}

which proves the sufficiency.

Necessity. Let f(x) = χ_{(0, \frac{t}{2} )}(x). Note that if 0 < y < \frac{t}{2} and x > t, then (x − y)^{α−1} ≥ b_{1}x^{α−1}, where the positive constant b_{1} depends only on α.

We have

∥R_{α}f∥_{L_{v}^{q}} ≥ ( ∫_{t}^{∞}v(x)( ∫_{0}^{\frac{t}{2}} \frac{dy}{(x − y)^{1−α}} )^{q}dx)^{\frac{1}{q}} ≥ c_{7}( ∫_{t}^{∞} \frac{v(x)}{x^{(1−α)q}} dx)^{\frac{1}{q}} · t.

On the other hand, ∥f∥_{L^{p}} = c_{8}t^{\frac{1}{p}} and by virtue of inequality (2) we find that B(t) ≤ c_{9}A for all t > 0.

A most complicated proof of a similar theorem is given in [12] for the case p = q = 2.

Remark 1. Condition (3) is equivalent to the condition

B̃ = sup_{k∈ℤ} ( ∫_{2^{k}}^{2^{k+1}} \frac{v(x)}{x^{(1−α)q− \frac{q}{p′}}} dx)^{\frac{1}{q}} < ∞.  (4)

Moreover, B ≈ B̃.

Indeed, the fact that (3) implies (4) follows from the proof of Theorem 1. Now let condition (4) be satisfied and t ∈ (0, ∞). Then t ∈ (2^{m}, 2^{m+1}] for some m ∈ ℤ. We have

B(t)^{q} = ( ∫_{t}^{∞} \frac{v(x)}{x^{(1−α)q}} dx)t^{\frac{q}{p′}} ≤ ( ∫_{2^{m}}^{∞} \frac{v(x)}{x^{(1−α)q}} dx)2^{\frac{(m+1)q}{p′}} =

= c_{1}2^{\frac{mq}{p′}} ∑_{k=m}^{∞} ( ∫_{2^{k}}^{2^{k+1}} \frac{v(x)}{x^{(1−α)q}} dx) ≤ c_{2}2^{\frac{mq}{p′}} ∑_{k=m}^{∞} 2^{− \frac{kq}{p′}} ∫_{2^{k}}^{2^{k+1}} \frac{v(x)x^{\frac{q}{p′}}}{x^{(1−α)q}} dx ≤

≤ c_{2}B̃^{q}2^{\frac{mq}{p′}} ∑_{k=m}^{∞} 2^{− \frac{kq}{p′}} ≤ c_{3}B̃^{q}

and therefore B ≤ c_{4}B̃ < ∞.

By the duality argument and Theorem 1 we obtain

Theorem 2. Let 1 < p ≤ q < ∞, \frac{1}{q′} < α < 1 or α > 1. For the inequality

∥W_{α}f∥_{L^{q}} ≤ \overline{A}∥f∥_{L_{w}^{p}},  (5)

where the positive constant \overline{A} does not depend on f, to be valid it is necessary and sufficient that

\overline{B} = sup_{t>0} \overline{B}(t) = sup_{t>0} ( ∫_{t}^{∞} \frac{w^{1−p′}(x)}{x^{(1−α)p′}}dx)^{\frac{1}{p′}} t^{\frac{1}{q}} < ∞.  (6)

SOLUTION OF SOME WEIGHT PROBLEMS 569

Moreover, if \overline{A} is the best constant in inequality (5), then \overline{A} ≈ \overline{B}.

We shall now consider the case 1 < q < p < ∞. Applying the integration by parts, we obtain

Lemma 1. Let 1 < q < p < ∞ and u be a locally integrable function on ℝ_{+}. Then the equality

( ∫_{a}^{b} u(x)dx)^{\frac{p}{p−q}} = \frac{p}{p − q} ∫_{a}^{b} ( ∫_{x}^{b} u(t)dt)^{\frac{q}{p−q}} u(x)dx

holds, where 0 ≤ a < b < ∞.

Theorem 3. Let 1 < q < p < ∞, \frac{1}{p} < α < 1 or α > 1. The inequality

∥R_{α}f∥_{L_{v}^{q}} ≤ A_{1}∥f∥_{L^{p}}  (7)

is fulfilled iff

B_{1} = ( ∫_{0}^{∞}( ∫_{x}^{∞} \frac{v(t)}{t^{(1−α)q}} dt)^{\frac{p}{p−q}} x^{\frac{(q−1)p}{p−q}} dx)^{\frac{p−q}{pq}} < ∞.  (8)

Moreover, if A_{1} is the best constant in inequality (7), then A_{1} ≈ B_{1}. Proof. Sufficiency. In the notation introduced in the proof of Theorem 1 we have

∥R_{α}f∥_{L_{v}^{q}}^{q} ≤ S_{1} + S_{2}.

Using Theorem B with w ≡ 1 and the argument from the proof of Theorem 1, we obtain

S_{1} ≤ c_{2}B_{1}^{q}∥f∥_{L^{p}}^{q}.

Applying the Hölder inequality twice and the fact that \frac{1}{p} < α, we have

S_{2} ≤ c_{1} ∫_{0}^{∞}( ∫_{\frac{x}{2}}^{x} |f(t)|^{p}dt)^{\frac{q}{p}} ( ∫_{\frac{x}{2}}^{x} \frac{dt}{(x − t)^{(1−α)p′}} )^{\frac{q}{p′}} v(x)dx =

= c_{3} ∑_{k∈ℤ} ∫_{2^{k}}^{2^{k+1}}( ∫_{\frac{x}{2}}^{x} |f(t)|^{p}dt)^{\frac{q}{p}} v(x)x^{(α−1)q+ \frac{q}{p′}} dx ≤

≤ c_{3} ∑_{k∈ℤ} ( ∫_{2^{k−1}}^{2^{k+1}}|f(t)|^{p}dt)^{\frac{q}{p}} ( ∫_{2^{k}}^{2^{k+1}}v(x)x^{(α−1)q+ \frac{q}{p′}} dx) ≤

570 A. MESKHI

≤ c_{3}( ∑_{k∈ℤ} ∫_{2^{k}}^{2^{k+1}}|f(t)|^{p}dt)^{\frac{q}{p}} ( ∑_{k∈ℤ} ( ∫_{2^{k}}^{2^{k+1}}v(x)x^{(α−1)q+ \frac{q}{p′}} dx)^{\frac{p}{p−q}} )^{\frac{p−q}{p}} ≤

≤c_{4}∥f∥_{L^{p}}^{q}( ∑_{k∈ℤ} ( ∫_{2^{k}}^{2^{k+1}}v(x)x^{(α−1)q+\frac{q}{p′}} dx)^{\frac{p}{p−q}} )^{\frac{p−q}{p}} =c_{4}∥f∥_{L^{p}}^{q}( ∑_{k∈ℤ} S_{2k})^{\frac{p−q}{p}} .

By Lemma 1, we find for S_{2k} that

S_{2k} ≤ 2^{\frac{(k+1)qp}{p′(p−q)}} ( ∫_{2^{k}}^{2^{k+1}}v(x)x^{(α−1)q}dx)^{\frac{p}{p−q}} ≤

≤ c_{5}2^{\frac{kqp}{p′(p−q)}} ∫_{2^{k}}^{2^{k+1}}( ∫_{x}^{2^{k+1}} \frac{v(t)}{t^{(1−α)q}} dt)^{\frac{q}{p−q}} \frac{v(x)}{x^{(1−α)q}} dx ≤

≤ c_{5} ∫_{2^{k}}^{2^{k+1}}( ∫_{x}^{∞} \frac{v(t)}{t^{(1−α)q}} dt)^{\frac{q}{p−q}} \frac{v(x)}{x^{(1−α)q}} · x^{\frac{q(p−1)}{p−q}} dx.

Using integration by parts we get

S_{2} ≤ c_{6}∥f∥_{L^{p}}^{q}( ∫_{0}^{∞}( ∫_{x}^{∞} \frac{v(t)}{t^{(1−α)q}} dt)^{\frac{q}{p−q}} \frac{v(x)}{x^{(1−α)q}} · x^{\frac{q(p−1)}{p−q}} dx)^{\frac{p−q}{p}} =

= c_{7}∥f∥_{L^{p}}^{q}( ∫_{0}^{∞}( ∫_{x}^{∞} \frac{v(t)}{t^{(1−α)q}} dt)^{\frac{p}{p−q}} x^{\frac{p(q−1)}{p−q}} dx)^{\frac{p−q}{p}} = c_{7}∥f∥_{L^{p}}^{q}B_{1}^{q}

and finally we obtain inequality (7).

Necessity. Let \frac{1}{p} < α < 1 and v_{0}(t) = v(t) · χ_{(a,b)}(t), w_{0}(t) = χ_{(a,b)}(t), where 0 < a < b < ∞, and let

f(x) = ( ∫_{x}^{∞} \frac{v_{0}(t)}{t^{(1−α)q}} dt)^{\frac{1}{p−q}} ( ∫_{0}^{x} w_{0}(t)dt)^{\frac{q−1}{p−q}} w_{0}(x).

Then we have

∥f∥_{L^{p}} = ( ∫_{a}^{b} ( ∫_{x}^{∞} \frac{v_{0}(t)}{t^{(1−α)q}} dt)^{\frac{p}{p−q}} ( ∫_{0}^{x} w_{0}(t)dt)^{\frac{(q−1)p}{p−q}} dx)^{\frac{1}{p}} < ∞.

On the other hand,

∥R_{α}f∥_{L_{v}^{q}} = ( ∫_{0}^{∞}v(x)( ∫_{0}^{x} \frac{f(t)}{(x − t)^{1−α}} dt)^{q}dx)^{\frac{1}{q}} ≥

≥ c_{8}( ∫_{0}^{∞} \frac{v(x)}{x^{(1−α)q}} ( ∫_{0}^{x} f(t)dt)^{q}dx)^{\frac{1}{q}} ≥ c_{9}( ∫_{0}^{∞} \frac{v(x)}{x^{(1−α)q}} ×

×( ∫_{x}^{∞} \frac{v_{0}(y)}{y^{(1−α)q}} dy)^{\frac{q}{p−q}} ( ∫_{0}^{x} ( ∫_{0}^{t} w_{0}(y)dy)^{\frac{q−1}{p−q}} w_{0}(t)dt)^{q}dx)^{\frac{1}{q}} ≥

≥ c_{10}(∫_{0}^{∞} \frac{v_{0}(x)}{x^{(1−α)q}}(∫_{x}^{∞} \frac{v_{0}(y)}{y^{(1−α)q}} dy)^{\frac{q}{p−q}} ( ∫_{0}^{x} w_{0}(y)dy)^{\frac{(p−1)q}{p−q}} dx)^{\frac{1}{q}} =

= c_{11}( ∫_{0}^{∞}( ∫_{x}^{∞} \frac{v_{0}(t)}{t^{(1−α)q}} dt)^{\frac{p}{p−q}} ( ∫_{0}^{x} w_{0}(t)dt)^{\frac{(q−1)p}{p−q}} w_{0}(x)dx)^{\frac{1}{q}} =

= c_{11}( ∫_{a}^{b} ( ∫_{x}^{∞} \frac{v_{0}(t)}{t^{(1−α)q}} dt)^{\frac{p}{p−q}} ( ∫_{0}^{x} w_{0}(t)dt)^{\frac{(q−1)p}{p−q}} dx)^{\frac{1}{q}} .

From inequality (7) we have

( ∫_{a}^{b} ( ∫_{x}^{∞} \frac{v_{0}(t)}{t^{(1−α)q}} dt)^{\frac{p}{p−q}} ( ∫_{0}^{x} w_{0}(t)dt)^{\frac{(q−1)p}{p−q}} dx)^{\frac{q−p}{pq}} ≤ c_{12}A_{1},

where c_{12} does not depend on a and b. By Fatou’s lemma we finally obtain condition (8). The case α > 1 is proved similarly.

By the duality argument and Theorem 3 we have

Theorem 4. Let 1 < q < p < ∞, \frac{1}{q′} < α < 1 or α > 1. The inequality

∥W_{α}f∥_{L^{q}} ≤ \overline{A}_{1}∥f∥_{L_{w}^{p}},  (9)

where the positive constanr \overline{A}_{1} does not depend f, holds iff

\overline{B}_{1} = ( ∫_{0}^{∞}( ∫_{x}^{∞} \frac{w^{1−p′}(t)}{t^{(1−α)p′}}dt)^{\frac{q(p−1)}{p−q}} x^{\frac{q}{p−q}} dx)^{\frac{p−q}{pq}} < ∞.  (10)

Moreover, if \overline{A}_{1} is the best constant in inequality (9), then \overline{A}_{1} ≈ \overline{B}_{1}. Let us now investigate the compactness of the operators R_{α} and W_{α}.

Theorem 5. Let 1 < p ≤ q < ∞, \frac{1}{p} < α < 1 or α > 1. The operator R_{α} is compact from L^{p} to L_{v}^{q} iff condition (3) and the condition

lim_{t→0} B(t) = lim_{t→∞} B(t) = 0

is satisfied.

Proof. Sufficiency. Let 0 < a < b < ∞. We write R_{α}f as

R_{α}f = χ_{[0,a]}R_{α}(f · χ_{(0,a)}) + χ_{(a,b)}R_{α}(f · χ_{(0,b)}) + χ_{[b,∞)}R_{α}(f · χ_{(0, \frac{b}{2} )}) +

+χ_{[b,∞)}R_{α}(f · χ_{( \frac{b}{2} ,∞)}) = P_{1α}f + P_{2α}f + P_{2α}f + P_{4α}f.

For P_{2α}f we have P_{2α}f(x) = χ_{(a,b)}(x) ∫_{0}^{∞} k_{1}(x, y)f(y)dy, with k_{1}(x, y) = (x − y)^{α−1} for y < x and k_{1}(x, y) = 0 for y ≥ x. Consequently

∫_{a}^{b} v(x)( ∫_{0}^{∞}(k_{1}(x, y))^{p′}dy)^{\frac{q}{p′}} dx ≤ ( ∫_{a}^{b} \frac{v(x)}{x^{(1−α)q}} dx)b^{\frac{q}{p′}} < ∞

and by Theorem C we conclude that P_{2α} is compact from L^{p} to L_{v}^{q}. In a similar manner we show that P_{3α} is compact too. Using Theorem 1 for the operators P_{1α} and P_{4α}, we obtain

∥P_{1α}∥ ≤ c_{1} sup_{0<t<a} B(t) and ∥P_{4α}∥ ≤ c_{2} sup_{t> \frac{b}{2}} B(t).

Consequently

∥R_{α} − P_{2α} − P_{3α}∥ ≤ ∥P_{1α}∥ + ∥P_{4α}∥ ≤ c_{1} sup_{0<t<a} B(t) + c_{2} sup_{t> \frac{b}{2}} B(t) → 0

as a → 0 and b → ∞.

Thus the operator R_{α} is compact, since it is a limit of compact operators. The sufficiency is proved.

Necessity. Note that the fact B < ∞ follows from Theorem 1. Thus we need to prove the remaining part. Let f_{t}(x) = χ_{(0,t)}(x)t^{−1/p}. Then the sequence f_{t} is weakly convergent to 0. Indeed, assuming that φ ∈ L^{p′}, we obtain

| ∫_{0}^{∞}f_{t}(x)φ(x)dx| ≤ ( ∫_{0}^{t} |φ(x)|^{p′}dx)^{\frac{1}{p′}} → 0 as t → 0.

On the other hand, we have

∥R_{α}f_{t}∥_{L_{v}^{q}} ≥ ( ∫_{t}^{∞}v(x)( ∫_{0}^{t} \frac{f_{t}(y)}{(x − y)^{1−α}} dy)^{q}dx)^{\frac{1}{q}} ≥

≥ c_{3}( ∫_{t}^{∞} \frac{v(x)}{x^{(1−α)q}} dx)^{\frac{1}{q}} t^{\frac{1}{p′}} = c_{3}B(t).

Using the fact that a compact operator maps a weakly convergent sequence into a strongly convergent form, we find that B(t) → 0 as t → 0.

Keeping in mind that the operator W_{α} is compact from L_{v^{1−q′}}^{q′} to L^{p′} and arguing as above, we prove the remaining part of the theorem.

In [12] a similar theorem is proved for the case p = q = 2.

Since the operator W_{α} is compact from L_{w}^{p} to L^{q} iff the operator R_{α} is compact from L^{q′} to L_{w^{1−p′}}^{p′}, by Theorem 5 we obtain

Theorem 6. Let 1 < q < p < ∞, \frac{1}{q′} < α < 1 or α > 1. The operator W_{α} is compact from L_{w}^{p} to L^{q} iff condition (6) and the condition

lim_{t→0} \overline{B}(t) = lim_{t→∞} \overline{B}(t) = 0

are fulfilled.

Theorem 7. Let 1 < q < p < ∞, \frac{1}{p} < α < 1 or α > 1. The operator R_{α} is compact from L^{p} to L_{v}^{q} iff condition (8) is satisfied.

Proof. The sufficiency is proved as in proving Theorem 5 while the necessity follows from Theorem 3.

By the duality argument we have

Theorem 8. Let 1 < q < p < ∞, \frac{1}{q′} < α < 1 or α > 1. The operator W_{α} is compact from L_{w}^{p} to L^{q} iff condition (10) is fulfilled.

In [13, 14] the necessary and sufficient conditions are found for the operators R_{α} and W_{α} to be compact when 1 < p ≤ q < ∞ and α = 1.

An analogous problem for α > 1 was investigated in [15].

Remark 2. In Theorems 1 and 5 it suffices to consider v as a measurable almost everywhere, positive function. The same assumption can be made for w in Theorems 2 and 6.

Acknowledgement

The author expresses his gratitude to Professor V. Kokilashvili and also to A. Gogatishvili for helpful remarks.

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(Received 25.12.1997)

Author’s address: A. Razmadze Mathematical Institute Georgian Academy of Sciences 1, M. Aleksidze St., Tbilisi 380093 Georgia