Georgian Mathematical Journal Volume 8 (2001), Number 3, 521–536

ON A TRACE INEQUALITY FOR ONE-SIDED POTENTIALS

AND APPLICATIONS TO THE SOLVABILITY OF NONLINEAR INTEGRAL EQUATIONS

V. KOKILASHVILI AND A. MESKHI

Dedicated to the memory of N. Muskhelishvili

Abstract. Necessary and sufficient conditions, which govern a trace inequality for one-sided potentials in the “diagonal” case, are established. An application to the existence of positive solutions of a certain nonlinear integral equation is presented.

2000 Mathematics Subject Classification: 47B34, 47H50, 45D05. Key words and phrases: Riemann–Liouville and Weyl operators, trace inequality, weight, nonlinear integral equation.

Introduction

It is well-known that the trace inequality for the Riesz potential

∫_{R^{n}} |I_{α}f(x)|^{p}v(x) dx ≤ c ∫_{R^{n}} |f(x)|^{p} dx, 0 < α < n, 1 < p < ∞,

is of great importance for the spectral properties of the Schrödinger operator and has numerous applications to partial differential equations, Sobolev spaces, complex analysis, etc. (see [1]–[7]). The pointwise conditions derived in [3], [6],

[7] turned out to be of particular interest for existence theorems and estimates of solutions of certain semilinear elliptic equations.

The trace inequality for the Riemann–Liouville transform R_{α} in the case where p = 2 and α > 1/2 was derived in [8]. For the extention of this result when 1 < p < ∞ and α > 1/p we refer to [9] (for more general transforms see [10]).

In the present paper criteria for the trace inequality for the Riemann–Liouville and Weyl operators in a more complicated case 0 < α < 1/p, 1 < p < ∞, are established. Some applications solvability problems of certain nonlinear integral equation are presented.

For the basic definitions and auxilliary results concerning fractional integrals on the line we refer to the monorgaph [11].

ISSN 1072-947X / $8.00 / \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag www.heldermann.de

1. Trace Inequalities

In this section we establish necessary and sufficient conditions for the validity of the trace inequality for the Riemann–Liouville and Weyl operators. Twoweighted inequalities are also derived.

Let ν be a locally finite Borel measure on Ω ⊂ R. Denote by L_{ν}^{p}(Ω) (1 < p < ∞) a Lebesgue space with respect to the measure ν consisting of all ν -measurable functions f for which

∥f∥_{L_{ν}^{p}(Ω)} = ( ∫_{Ω} |f(x)|^{p} dν(x))^{1/p} < ∞.

If dν(x) = v(x) dx, with a locally integrable a.e. positive function v on Ω, then we use the notation L_{ν}^{p}(Ω) ≡ L_{v}^{p}(Ω). If dν(x) = dx is a Lebesgue measure, then we assume that L_{ν}^{p}(Ω) ≡ L^{p}(Ω).

Let

R_{α}f(x) = ∫_{0}^{x} f(y)(x − y)^{α−1} dy, x > 0, α > 0,

W_{α}f(x) = ∫_{x}^{∞}f(y)(y − x)^{α−1} dy, x > 0, α > 0

for measurable f : R_{+} → R^{1}.

Theorem 1.1. Let 1 < p < ∞ and let 0 < α < \frac{1}{p}. Then the inequality

∫_{0}^{∞}|R_{α}f(x)|^{p}v(x) dx ≤ c_{0} ∫_{0}^{∞}|f(x)|^{p} dx  (1.1)

holds if and only if W_{α}v ∈ L_{loc}^{p′}(R_{+}) and

W_{α}[W_{α}v]^{p′}(x) ≤ cW_{α}v(x) a.e.  (1.2)

To prove this theorem we need

Proposition 1.1. Let 1 < p < ∞, and let 0 < α < \frac{1}{p}. If (1.1) is fulfilled, then

∫_{x}^{x+h}v(y) dy ≤ c h^{1−αp}  (1.3)

for all positive x and h.

Proof. By the duality argument, (1.1) is equivalent to the inequality

∥W_{α}f∥_{L^{p′}(R_{+})} ≤ c_{0}^{1/p}∥f∥_{L_{v^{1−p′}}^{p′} (R_{+})}.  (1.4)

Replaing here f(y) = χ_{(x,x+h)}(y)v(y) for 0 < h ≤ x, we get

∫_{x−h}^{x} ( ∫_{x}^{x+h}v(z)(z − y)^{α−1} dz)^{p′} dy ≤ c_{0}^{p′−1} ∫_{x}^{x+h}v(y) dy.

Hence (1.3) holds for all x and h with the condition 0 < h ≤ x. Now let 0 < x < h < ∞. Then taking into account the condition 0 < α < \frac{1}{p} we obtain

∫_{x}^{x+h}v(y) dy = ∑_{k=0}^{∞} ∫_{x+ \frac{h}{2^{k+1}}}^{x+ \frac{h}{2^{k}}} v(y) dy

= ∑_{k=0}^{∞} ( ∫_{x+ \frac{h}{2^{k+1}}}^{x+ \frac{h}{2^{k}}} v(y) dy) (\frac{h}{2^{k+1}} )^{αp−1}(\frac{h}{2^{k+1}} )^{1−αp}

≤ sup_{\begin{matrix}t,a \\ t≤a \end{matrix}} (( ∫_{a}^{a+t}v(y) dy)t^{αp−1})h^{1−αp} ∑_{k=0}^{∞} 2^{(k+1)(αp−1)} ≤ c h^{1−αp}.

Therefore that (1.3) holds. Note that c=c_{0}2^{(1−α)p} max \{1,\frac{2^{αp−1}}{1−2^{αp−1}} \} in (1.3).

Proof of Theorem 1.1. Necessity. Let us first show that, from (1.1) it follows that W_{α}v ∈ L_{loc}^{p′}(R_{+}). For f(y) = v(y)χ_{(x,x+h)}(y) (x ∈ R_{+} and h > 0) from (1.1) we have

∫_{x}^{x+h}(W_{α}(vχ_{(x,x+h)}))^{p′}(y) dy ≤ c ∫_{x}^{x+h}v(y) dy.  (1.5)

Let v_{1}(y) = χ_{(x,x+2h)}v(y) and v_{2}(y) = χ_{R_{+}\(x,x+2h)}(y)v(y), where x ∈ R_{+} and h > 0. We have

∫_{x}^{x+h}(W_{α}v)^{p′}(y) dy ≤ c(I_{1}(x) + I_{2}(x)),

where

I_{1}(x) = ∫_{x}^{x+h}(W_{α}v_{1})^{p′}(y) dy and I_{2}(x) = ∫_{x}^{x+h}(W_{α}v_{2})^{p′}(y) dy.

From (1.5) we get

I_{1}(x) ≤ c ∫_{x}^{x+2h}v(y) dy < ∞.

Thus W_{α}v_{1} ∈ L_{loc}^{p′}(R_{+}).

Note that for z > x + 2h and x < y < x + h we have z − x ≤ 2(z − y). Now using (1.3), we come to the estimate

(W_{α}v_{2})(y) = ∫_{x+2h}^{∞} (z − y)^{α−1}v(z) dz ≤ 2^{1−α} ∫_{x+h}^{∞} (z − x)^{α−1}v(z) dz

= 2^{1−α} ∫_{h}^{∞}t^{α−2}( ∫_{x}^{x+t}v(z) dz) dt ≤ c 2^{1−α} ∫_{h}^{∞}t^{α−1−αp} dt < ∞.

Therefore W_{α}v_{2} ∈ L_{loc}^{p′}(R_{+}). Thus W_{α}v ∈ L_{loc}^{p′}(R_{+}). Now we prove that (1.1) yields (1.2). In the sequel the following equality

W_{α}v(x) = (1 − α) ∫_{0}^{∞}τ^{α−1}( ∫_{x}^{x+τ}v(y) dy) \frac{dτ}{τ}  (1.6)

will be used.

Thus

W_{α}[(W_{α}v)^{p′}](x) = (1 − α) ∫_{0}^{∞}τ^{α−1}( ∫_{x}^{x+τ}(W_{α}v)^{p′}(y) dy) \frac{dτ}{τ} .  (1.7)

Let v_{1} and v_{2} be defined as before. By (1.5) we have

∫_{x}^{x+h}(W_{α}v_{1})^{p′}(y) dy ≤ c ∫_{x}^{x+2h}v(y) dy.  (1.8)

Then from (1.7) and (1.8) we derive the estimate

W_{α}[(W_{α}v_{1})^{p′}](x) ≤ c ∫_{0}^{∞}τ^{α−1}( ∫_{x}^{x+2τ}v(t) dt) \frac{dτ}{τ} = c W_{α}v(x).  (1.9)

It is easy to see that for t ∈ (x, x + h)

(W_{α}v_{2})(t) ≤ c ∫_{h}^{∞}r^{α−1}( ∫_{x}^{x+r}v(y) dy) \frac{dr}{r} .

Therefore (1.7) yields

W_{α}[(W_{α}v_{2})^{p′}](x) ≤ c ∫_{0}^{∞}t^{α}( ∫_{t}^{∞}r^{α−1}( ∫_{x}^{x+r}v(y) dy) \frac{dr}{r} )^{p′} \frac{dt}{t} .

Integration by parts on the right-hand side of the last inequality leads to the estimate

W_{α}[(W_{α}v_{2})^{p′}](x)
≤ c ∫_{0}^{∞}r^{α}( ∫_{r}^{∞}τ^{α−1}( ∫_{x}^{x+τ}v(y) dy) \frac{dτ}{τ} )^{p′−1}(r^{α} ∫_{x}^{x+r}v(y) dy) \frac{dr}{r} . (1.10)

Now recall that estimate (1.3) holds by Proposition 1.1. From inequality (1.3), by a simple computation, we obtain

W_{α}[(W_{α}v_{2})^{p′}](x) ≤ c ∫_{0}^{∞}h^{α−1}( ∫_{x}^{x+h}v(y) dy) \frac{dh}{h} .

Thus

W_{α}[(W_{α}v_{2})^{p′}](x) ≤ c(W_{α}v)(x) a.e.  (1.11)

Finally (1.9) and (1.11) imply (1.2).

Remark 1.1. It follows from the proof of necessity of the Theorem 1.1 that if c_{0} is the best constant in (1.2), then for c from (1.3) we have

c = c_{0}^{p′−1}2^{p′−α} + 2^{p′−1}(1 − α)^{p′} \frac{p′c_{1}}{α(αp − α)^{p′−1}} ,

where c_{1} = c_{0}2^{(1−α)p} max \{1, \frac{2^{αp−1}}{1−2^{αp−1}} \}.

Sufficiency of Theorem 1.1. In order to show the sufficiency, we shall need the following lemmas.

Lemma 1.1. Let 1 < p < ∞ and 0 < α < 1. Then there exists a positive constant c such that for all f ∈ L_{loc}^{1}(R_{+}), f ≥ 0, and for arbitrary x ∈ R_{+} the following inequality holds:

(R_{α}f(x))^{p} ≤ c R_{α}((R_{α}f)^{p−1}f)(x)  (1.12)

(for c we have c = 2^{\frac{1}{p−1}} if p ≤ 2 and c = 2^{p(p−1)} if p > 2).

Proof. First we assume that R_{α}f(x) < ∞ and prove (1.12) for such x. We also assume that

V_{α}f(x) ≤ (R_{α}f(x))^{p},

526 V. KOKILASHVILI AND A. MESKHI

where V_{α}f(x) ≡ R_{α}((R_{α}f)^{p−1}f)(x), otherwise (1.12) is obvious for c = 1. Now let us assume that 1 < p ≤ 2. Then we have

(R_{α}f(x))^{p} = ∫_{0}^{x}(x − y)^{α−1}f(y)( ∫_{0}^{x}(x − z)^{α−1}f(z) dz)^{p−1} dy

≤ ∫_{0}^{x}(x − y)^{α−1}f(y)( ∫_{0}^{y}(x − z)^{α−1}f(z) dz)^{p−1} dy

+ ∫_{0}^{x}(x − y)^{α−1}f(y)( ∫_{y}^{x}(x − z)^{α−1}f(z) dz)^{p−1} dy

≡ I_{1}(x) + I_{2}(x).

It is obvious that if z < y < x, then y − z ≤ x − z. Consequently,

I_{1}(x) ≤ ∫_{0}^{x}(x − y)^{α−1}f(y)( ∫_{0}^{y}(y − z)^{α−1}f(z) dz)^{p−1} dy = V_{α}f(x).

Now we use Hölder’s inequality with respect to the exponents \frac{1}{p−1}, \frac{1}{2−p} and measure dσ(y) = (x − y)^{α−1}f(y) dy. We have

I_{2}(x) ≤ ( ∫_{0}^{x}(x − y)^{α−1}f(y) dy)^{2−p}

×( ∫_{0}^{x} ( ∫_{y}^{x}(x − z)^{α−1}f(z) dz)(x − y)^{α−1}f(y) dy)^{p−1}

= (R_{α}f(x))^{2−p}(J(x))^{p−1},

where

J(x) ≡ ∫_{0}^{x} ( ∫_{y}^{x}(x − z)^{α−1}f(z) dz)(x − y)^{α−1}f(y) dy.

Using Tonelli’s theorem we have

J(x) = ∫_{0}^{x}(x − z)^{α−1}f(z)( ∫_{0}^{z}(x − y)^{α−1}f(y) dy) dz.

Further, it is obvious that the following simple inequality

∫_{0}^{z}(x − y)^{α−1}f(y) dy ≤ ( ∫_{0}^{z}(x − y)^{α−1}f(y) dy)^{p−1}(R_{α}f(x))^{2−p}
≤ (R_{α}f(z))^{p−1}(R_{α}f(x))^{2−p}

holds, where z < x.

A TRACE INEQUALITY FOR ONE-SIDED POTENTIALS 527

Taking into account the last estimate, we obtain

J(x) ≤ ( ∫_{0}^{x}(x − z)^{α−1}f(z)(R_{α}f(z))^{p−1} dz)(R_{α}f(x))^{2−p}
= (V_{α}f(x))(R_{α}f(x))^{2−p}.

Thus

I_{2}(x) ≤ (R_{α}f(x))^{2−p}(R_{α}f(x))^{(2−p)(p−1)}(V_{α}f(x))^{p−1}
= (R_{α}f(x))^{p(2−p)}(V_{α}f(x))^{p−1}.

Combining the estimates for I_{1} and I_{2} we derive

(R_{α}f(x))^{p} ≤ V_{α}f(x) + (R_{α}f(x))^{p(2−p)}(V_{α}f(x))^{p−1}.

As we have assumed that V_{α}f(x) ≤ (R_{α}f(x))^{p}, we obtain

V_{α}f(x) = (V_{α}f(x))^{2−p}(V_{α}f(x))^{p−1} ≤ (V_{α}f(x))^{p−1}(R_{α}f(x))^{p(2−p)}. Hence

(R_{α}f(x))^{p} ≤ (V_{α}f(x))^{p−1}(R_{α}f(x))^{p(2−p)} + (V_{α}f(x))^{p−1}(R_{α}f(x))^{p(2−p)}
= 2(V_{α}f(x))^{p−1}(R_{α}f(x))^{p(2−p)}.

Using the fact R_{α}f(x) < ∞ we deduce that

(R_{α}f(x))^{p−1} ≤ 2^{\frac{1}{p−1}} (V_{α}f(x)).

Now we shall deal with the case p > 2. Let us assume again that

V_{α}f(x) ≤ (R_{α}f(x))^{p},

where

V_{α}f(x) ≡ R_{α}[(R_{α}f)^{p−1}f](x).

As p > 2 we have

(R_{α}f(x))^{p} = ∫_{0}^{x} f(y)(x − y)^{α−1}( ∫_{0}^{x}(x − z)^{α−1}f(z) dz)^{p−1} dy

≤ 2^{p−1} ∫_{0}^{x} f(y)(x − y)^{α−1}( ∫_{0}^{y}(x − z)^{α−1}f(z) dz)^{p−1} dy

+ 2^{p−1} ∫_{0}^{x} f(y)(x − y)^{α−1}( ∫_{y}^{x}(x − z)^{α−1}f(z) dz)^{p−1} dy

≡ 2^{p−1}I_{1}(x) + 2^{p−1}I_{2}(x).

It is clear that if z < y < x, then (x − z)^{α−1} ≤ (y − z)^{α−1}. Therefore I_{1}(x) ≤ V_{α}f(x). Now we estimate I_{2}(x). We obtain

( ∫_{y}^{x}(x − z)^{α−1}f(z) dz)^{p−1} = ( ∫_{y}^{x}(x − z)^{α−1}f(z) dz)^{p−2}( ∫_{y}^{x}(x − z)^{α−1}f(z) dz)

≤ (R_{α}f(x))^{p−2} ∫_{y}^{x}(x − z)^{α−1}f(z) dz.

Using Tonelli’s theorem and the last estimate we have

I_{2}(x) ≤ (R_{α}f(x))^{p−2} ∫_{0}^{x} f(y)(x − y)^{α−1}( ∫_{y}^{x}(x − z)^{α−1}f(z) dz) dy

= (R_{α}f(x))^{p−2} ∫_{0}^{x} f(z)(x − z)^{α−1}( ∫_{0}^{z}(x − y)^{α−1}f(y) dy) dz

≤ (R_{α}f(x))^{p−2} ∫_{0}^{x} f(z)(x − z)^{α−1}( ∫_{0}^{z}(z − y)^{α−1}f(y) dy) dz.

Using Hölder’s inequality with respect to the exponents p−1 and \frac{p−1}{p−2} we derive

∫_{0}^{x}(x − z)^{α−1}f(z)( ∫_{0}^{z}(z − y)^{α−1}f(y) dy) dz ≤ ( ∫_{0}^{x}(x − z)^{α−1}f(z) dz)^{\frac{p−2}{p−1}}

×( ∫_{0}^{x} ( ∫_{0}^{z}(z − y)^{α−1}f(y) dy)^{p−1}(x − z)^{α−1}f(z) dz)^{\frac{1}{p−1}}

= (R_{α}f(x))^{\frac{p−2}{p−1}} (V_{α}f(x))^{\frac{1}{p−1}} .

Combining these estimates we obtain

(R_{α}f(x))^{p} ≤ 2^{p−1}V_{α}f(x) + 2^{p−1}(R_{α}f(x))^{\frac{p(p−2)}{p−1}} (V_{α}f(x))^{\frac{1}{p−1}} .

From the inequality V_{α}f(x) ≤ (R_{α}f(x))^{p} it follows that

V_{α}f(x) = (V_{α}f(x))^{\frac{1}{p−1}} (V_{α}f(x))^{\frac{p−2}{p−1}} ≤ (V_{α}f(x))^{\frac{1}{p−1}} (R_{α}f(x))^{\frac{p(p−2)}{p−1}} .

Hence

(R_{α}f(x))^{p} ≤ 2^{p−1}((V_{α}f(x))^{\frac{1}{p−1}} (R_{α}f(x))^{\frac{p(p−2)}{p−1}} + (V_{α}f(x))^{\frac{1}{p−1}} (R_{α}f(x))^{\frac{p(p−2)}{p−1}} )
= 2^{p}(V_{α}f(x))^{\frac{1}{p−1}} (R_{α}f(x))^{\frac{p(p−2)}{p−1}} .

The last estimate yields

(R_{α}f(x))^{p} ≤ 2^{p(p−1)}(V_{α}f(x)),

where 2 < p < ∞.

Next we shall show that (1.12) holds for x satisfying R_{α}f(x) = ∞.

Let k_{n}(x, y) = χ_{(0,x)}(y) min{(x − y)^{α−1}, n}, where n ∈ N. It is easy to verify that (1.12) holds if we replace k(x, y) = χ_{(0,x)}(y)(x − y)^{α−1} by k_{n}(x, y). Let I = (a, b), where 0 < a < b < ∞. Then

∫_{I} k_{n}(x, y)f(y)dy < ∞ and sup_{I,n} ∫_{I} k_{n}(x, y)f(y) dy = ∞.

Taking into account the above arguments we obtain

( ∫_{0}^{x} χ_{I}(y)k_{n}(x, y)f(y) dy)^{p}

≤ c( ∫_{0}^{x} χ_{I}(y)( ∫_{0}^{y} χ_{I}(z)k_{n}(y, z)f(z) dz)^{p−1}f(z)k_{n}(x, z) dy).

(In the last inequality we can assume that f has a support in I. In this case

∫_{0}^{x} k_{n}(x, y)f(y) dy < ∞.) The constant c is defined as follows: c = 2^{\frac{1}{p−1}} if

1 < p ≤ 2 and c = 2^{p(p−1)} if p > 2. Taking the supremum with respect to all I and passing n to +∞, we obtain (1.12) for all x.

Remark 1.2. Let 1 < p < ∞, 0 < α < 1, k_{n}(x, y) = min{n, (x − y)^{α−1}}. Then for all f ∈ L_{loc}^{1}(R_{+}) (f ≥ 0) and for all x ∈ R_{+} we have the inequality

( ∫_{0}^{x} k_{n}(x, y)f(y) dy)^{p} ≤ c ∫_{0}^{x} k_{n}(x, y)( ∫_{0}^{y} k_{n}(x, y)f(z) dz)^{p−1}f(y) dy,

where c is the same as in inequlity (1.12).

Lemma 1.2. Let 0 < α < 1, v be a locally integrable a.e. positive function on R_{+}. Let there exist a constant c > 0 such that the inequality

∥R_{α}f∥_{L_{v_{1}}^{p}(R_{+})} ≤ c_{1}∥f∥_{L^{p}(R_{+})}, v_{1}(x) = [(W_{α}v)(x)]^{p′}  (1.13)

holds for all f ∈ L^{p}(R_{+}). Then

∥R_{α}f∥_{L_{v}^{p}(R_{+})} ≤ c_{2}∥f∥_{L^{p}(R_{+})}, f ∈ L^{p}(R_{+}),

where c_{2} = c_{1}^{1/p′}c^{1/p} and c is the same as in (1.12).

Proof. Let f ≥ 0. Using Lemma 1.1, Tonelli’s theorem and Hölder’s inequality we have

∫_{0}^{∞}(R_{α}f(x))^{p}v(x) dx

≤ c ∫_{0}^{∞}R_{α}[f(R_{α}f)^{p−1}](x)v(x) dx = c ∫_{0}^{∞}(R_{α}f)^{p−1}(y)f(y)(W_{α}v)(y) dy

≤ c( ∫_{0}^{∞}(f(y))^{p} dy)^{\frac{1}{p}} ( ∫_{0}^{∞}(R_{α}f(y))^{p}v_{1}(y) dy)^{\frac{1}{p′}}

= c∥f∥_{L^{p}(R_{+})}∥R_{α}f∥_{L_{v_{1}}^{p}(R_{+})}^{p−1} ≤ c_{1}^{p−1}c∥f∥_{L^{p}(R_{+})}∥f∥_{L^{p}(R_{+})}^{p−1} = c_{1}^{p−1}c∥f∥_{L^{p}(R_{+})}^{p}.

Hence

∥R_{α}f∥_{L_{v}^{p}(R_{+})} ≤ c1^{\frac{1}{p′}} c^{\frac{1}{p}} ∥f∥_{L^{p}(R_{+})}.

Lemma 1.3. Let 1 < p < ∞, 0 < α < 1, W_{α}v ∈ L_{loc}^{p′}, and let

W_{α}(W_{α}v)^{p′}(x) ≤ c_{3}(W_{α}v)(x) a.e.

Then we have

∥R_{α}f∥_{L_{v_{1}}^{p}(R_{+})} ≤ c_{4}∥f∥_{L^{p}(R_{+})}, f ∈ L^{p}(R_{+}),  (1.14)

where v_{1}(x) = [(W_{α}v)(x)]^{p′} and c_{4} = c c_{3} (c is from (1.12)).

Proof. Let f ≥ 0 and let I ⊂ R_{+} be a support of f, where I has a form I = (a, b), 0 < a < b < ∞. Let k_{n}(x, y) = min{(x − y)^{α−1}, n}. Then using Lemma 1.1, Remark 1.1 and Tonelli’s theorem we have

∫_{0}^{∞}( ∫_{0}^{x} k_{n}(x, y)f(y) dy)^{p}v_{1}(x) dx

≤ c ∫_{0}^{∞}( ∫_{0}^{x} k_{n}(x, y)( ∫_{0}^{y} k_{n}(y, z)f(z) dz)^{p−1}f(y) dy)v_{1}(x) dx

= c ∫_{0}^{∞}f(y)( ∫_{0}^{y} k_{n}(y, z)f(z) dz)^{p−1}( ∫_{y}^{∞}k_{n}(x, y)v_{1}(x) dx) dy

≤ c∥f∥_{L^{p}(R_{+})}( ∫_{I} ( ∫_{0}^{y} k_{n}(y, z)f(z) dz)^{p}(W_{α}[(W_{α}v)^{p′}](y))^{p′} dy)^{1/p′}

≤ c c_{3}∥f∥_{L^{p}(R_{+})}( ∫_{I} ( ∫_{0}^{y} k_{n}(y, z)f(z) dz)^{p}[(W_{α}v)(y)]^{p′} dy)^{1/p′}.

The last expression is finite as

∫_{0}^{y} k_{n}(y, z)f(z) dz ≤ n ∫_{I} f(z) dz ≤ n|I|^{\frac{1}{p′}} ∥f∥_{L^{p}(R_{+})} < ∞

A TRACE INEQUALITY FOR ONE-SIDED POTENTIALS 531

and

∫_{I} ((W_{α}v)(y))^{p′} dy < ∞.

Consequently,

( ∫_{I} ( ∫_{0}^{x} k_{n}(x, y)f(y) dy)^{p}v_{1}(x) dx)^{1/p} ≤ c_{3} c∥f∥_{L^{p}(R_{+})},

where c is from (1.12). Finally, we have (1.14).

Combining the above-proved Lemmas we obtain sufficiency of Theorem 1.1. The next theorem concerns the boundedness of W_{α} and is proved just in the

same way as the previous result.

Theorem 1.2. Let 1 < p < ∞ and 0 < α < 1/p. Then the inequality

∥W_{α}f∥_{L_{v}^{p}(R_{+})} ≤ c∥f∥_{L^{p}(R_{+})}, f ∈ L^{p}(R_{+}),

holds if and only if R_{α}v ∈ L_{loc}^{p′}(R_{+}) and

R_{α}[R_{α}v]^{p′}(x) ≤ c R_{α}v(x)

for a.a. x ∈ R_{+}.

Let us consider the Riemann–Liouville and Weyl operators:

R_{α}f(x) = ∫_{−∞}^{x} (x − y)^{α−1}f(y) dy, x ∈ R,
W_{α}f(x) = ∫_{x}^{∞}(y − x)^{α−1}f(y) dy, x ∈ R.

The following statements follow analogously and their proofs are omitted.

Theorem 1.3. Let 1 < p < ∞ and 0 < α < 1/p. Assume that v is a weight on R. Then R_{α} is bounded from L^{p}(R) to L_{v}^{p}(R) if and only if W_{α}v ∈ L_{loc}^{p′}(R) and

W_{α}[W_{α}v]^{p′}(x) ≤ c W_{α}v(x)

for a.a. x ∈ R.

Theorem 1.4. Let 1 < p < ∞ and 0 < α < 1/p. Then the inequality

∥W_{α}f∥_{L_{v}^{p}(R)} ≤ c∥f∥_{L^{p}(R)}, f ∈ L^{p}(R),

holds if and only if R_{α}v ∈ L_{loc}^{p′}(R) and

R_{α}[R_{α}v]^{p′}(x) ≤ c R_{α}v(x)

for a.a. x ∈ R.

532 V. KOKILASHVILI AND A. MESKHI

Let us now consider the case of two weights.

Let μ and ν be locally finite Borel measures on R_{+} and let

R_{α,μ}f(x) = ∫_{[0,x]} f(y)(x − y)^{α−1} dμ(y),
W_{α,ν}f(x) = ∫_{[x,∞)} f(y)(y − x)^{α−1} dν(y),

where x ∈ R_{+} and α ∈ (0, 1).

Theorem 1.5. Let 1 < p < ∞ and 0 < α < 1. Assume that the measures μ and ν satisfy the conditions W_{α,ν}(1) ∈ L_{μ,loc}^{p′}(R_{+}) and

B ≡ sup_{\begin{matrix}x,r \\ x>0,r>0 \end{matrix}} ( ∫_{r}^{∞} \frac{ν(I_{t}(x))}{t^{1−α}} \frac{dt}{t})^{1/p}( ∫_{0}^{r} \frac{μ(I_{t}(x))}{t^{α−1}} \frac{dt}{t})^{1/p′} < ∞,  (1.15)

where I_{r}(x) is the interval of the type [x, x + r). Then the inequality

∫_{0}^{∞}|R_{α,μ}f(x)|^{p} dν(x) ≤ c_{0} ∫_{0}^{∞}|f(x)|^{p} dμ(x), f ∈ L_{μ}^{p}(R_{+}),  (1.16)

holds if and only if

W_{α,μ}[W_{α,ν}(1)]^{p′}(x) ≤ c W_{α,ν}(1)(x)  (1.17)

for μ-a.a. x.

Sufficiency of this theorem follows using the following Lemmas:

Lemma 1.4. Let 1 < p < ∞ and 0 < α < 1. Then there exists a positive constant c such that for all f ∈ L_{μ,loc}^{1}(R_{+}), f ≥ 0, and for arbitrary x ∈ R_{+} the inequality

(R_{α,μ}f(x))^{p} ≤ c R_{α,μ}((R_{α,μ}f)^{p−1}f)(x)  (1.18)

holds (for c we have c = 2^{\frac{1}{p−1}} if p ≤ 2 and c = 2^{p(p−1)} if p > 2).

Lemma 1.5. Let 0 < α < 1. Suppose that there exists a positive constant c_{1} > 0 such that the inequality

∥R_{α,μ}f∥_{L_{ν_{1}}^{p}(R_{+})} ≤ c_{1}∥f∥_{L_{μ}^{p}(R_{+})}, dν_{1}(x) = [(W_{α,ν}(1))(x)]^{p′} dμ(x) (1.19)

holds for all f ∈ L_{μ}^{p}(R_{+}). Then

∥R_{α,μ}f∥_{L_{ν}^{p}(R_{+})} ≤ c_{2}∥f∥_{L_{μ}^{p}(R_{+})}, f ∈ L_{μ}^{p}(R_{+}).

where c_{2} = c_{1}^{1/p′}c^{1/p} and c is the same as in (1.18).

Lemma 1.6. Let 1 < p < ∞, 0 < α < 1, W_{α,ν}(1) ∈ L_{μ,loc}^{p′}(R_{+}), and let

W_{α,μ}(W_{α,ν}(1))^{p′}(x) ≤ c_{3}(W_{α,ν}(1))(x) μ-a.e.

Then we have

∥R_{α,μ}f∥_{L_{ν_{1}}^{p}(R_{+})} ≤ c_{4}∥f∥_{L_{μ}^{p}(R_{+})}, f ∈ L_{μ}^{p}(R_{+}),

where dν_{1}(x) = [(W_{α,ν}(1))(x)]^{p′} and c_{4} = cc_{3} (c is from (1.18)).

These lemmas can be proved in the same way as Lemmas 1.1, 1.2 and 1.3 above.

Taking into account the proof of Theorem 1.1, we easily obtain necessity. Moreover, for the constant c in condition (1.17) we have

c = 2^{p′−1}(c_{0}^{p′−1}2^{1−α} + (1 − α)^{p′}2^{(1−α)p′}B^{p′}p′),

where c_{0} and B are from (1.16) and (1.15), respectively.

Finally we note that the following proposition holds for the Volterra-type integral operator

K_{μ}f(x) = ∫_{[0,x]} f(y)k(x, y) dμ(y),

where μ is a locally finite Borel measure on R_{+} and the kernel k satisfies the condition: there exists a positive constant b such that for all x, y, z, with 0 < y < z < x < ∞, the inequality

k(x, y) ≤ bk(z, y)  (1.20)

is fulfilled.

Theorem 1.6. Let 1 < p < ∞, the kernel k satisfy condition (1.20). Let ν and μ be locally finite Borel measures on R_{+}. Suppose that K′_{ν}(1) ∈ L_{μ}^{p′}(R_{+}), where

K′_{ν}g(x) = ∫_{[x,∞)} g(y)k(y, x) dν(y).

Then the condition

K′_{μ}[K′_{ν}(1)]^{p′}(x) ≤ cK′_{ν}(1)(x), μ-a.e.

implies the boundedness of the operator K_{μ} from L_{μ}^{p}(R_{+}) to L_{ν}^{p}(R_{+}).

The proof of this statement follows in the same way as sufficiency of Theorem 1.5 and therefore is omitted.

2. Application to the Existence of Positive Solutions of Nonlinear Integral Equations

The goal of this section is to give a characterization of the existence of positive solutions of some Volterra-type nonlinear integral equations.

First let us consider the integral equation

φ(x) = ∫_{x}^{∞} \frac{φ^{p}(t)}{(t − x)^{1−α}} dt + ∫_{x}^{∞} \frac{v(t)}{(t − x)^{1−α}} dt, 0 < α < 1  (2.1)

with given non-negative v ∈ L_{loc}(R_{+}).

Theorem 2.1. Let 1 < p < ∞, 0 < α < 1, p′ = \frac{p}{p−1}, and A_{p} = (p′−1)(p′)^{−p}.

(i) If W_{α}v ∈ L_{loc}^{p}(R_{+}) and the inequality

W_{α}[W_{α}v]^{p}(x) ≤ A_{p}W_{α}v(x) a.e.  (2.2)

holds, then (2.1) has a non-negative solution φ ∈ L_{loc}^{p}(R_{+}). Moreover, (W_{α}v)(x) ≤ φ(x) ≤ p′(W_{α}v)(x).

(ii) If 0 < α < \frac{1}{p′} and (2.1) has a non-negative solution in L_{loc}^{p}(R_{+}), then W_{α}v ∈ L_{loc}^{p}(R_{+}) and

W_{α}[(W_{α}v)^{p}](x) ≤ c W_{α}v(x) a.e.  (2.3)

for some constant c > 0.

Proof. We shall use the following iteration procedure. Let φ_{0} = 0, and let for k = 0, 1, 2, …

φ_{k+1}(x) = W_{α}(φ_{k}^{p})(x) + W_{α}v(x).  (2.4)

By induction it is easy to verify that

W_{α}v(x) ≤ φ_{k}(x) ≤ φ_{k+1}(x), k = 0, 1, 2, …  (2.5)

From (2.4) we shall inductively derive an estimate of φ_{k}(x).

Let

φ_{k}(x) ≤ c_{k}W_{α}v(x)  (2.6)

for some k = 0, 1, . . .. It is obvious that c_{1} = 1. Then (2.2), (2.3) and (2.6) yield

φ_{k+1}(x) ≤ (A_{p}c_{k}^{p} + 1)(W_{α}v)(x),

where A_{p} is the constant from (2.2). Thus c_{k+1} = A_{p}c_{k}^{p} + 1 for k = 1, 2, . . .. Now by induction and the definition of A_{p} we deduce that the sequence (c_{k})_{k} is increasing. Indeed, it is obvious that c_{1} < c_{2}. Let c_{k} < c_{k+1}. Then

c_{k+1}(x) = A_{p}c_{k}^{p} + 1 < A_{p}c_{k+1}^{p} + 1 = c_{k+2}.

It is also clear that (c_{k})_{k} is bounded from the above by p′ and consequently it converges. As the equation z = A_{p}z^{p} +1 has only one solution, x = p′, it follows that lim_{k→∞} c_{k} = p′. On the other hand, the sequence (φ_{k})_{k} is nondecreasing and by (2.6) we get

φ(x) = lim_{k→∞} φ_{k}(x) ≤ p′(W_{α}v)(x).

By our assumption W_{α}v ∈ L_{loc}^{p}(R_{+}) and from the preceding estimate we conclude that φ ∈ L_{loc}^{p}(R_{+}). Moreover, (W_{α}v)(x) ≤ φ(x) ≤ p′(W_{α}v)(x).

Now we prove the statement (ii). Suppose (2.1) has a solution φ ∈ L_{loc}^{p}(R^{+}). We have

W_{α}(φ^{p})(x) ≤ φ(x) < ∞ a.e.  (2.7)

Hence W_{α}(φ^{p}) ∈ L_{loc}^{p}(R_{+}). Then from (2.7) we get

W_{α}[W_{α}(φ^{p})(x)]^{p}(x) ≤ W_{α}(φ^{p})(x) a.e.

Applying Theorem 1.1, we deduce that

∥R_{α}f∥_{L_{ρ}^{p′}} ≤ ∥f∥_{L^{p′}},

where ρ(x) = φ^{p}(x). But (W_{α}v)(x) ≤ φ(x). Due to (2.7) we get

∥R_{α}f∥_{L_{ρ_{1}}^{p′}} ≤ c∥f∥_{L^{p′}}

with ρ_{1}(x) = (W_{α}v)^{p}(x). Using Lemma 1.2 we arrive at the inequality

∥R_{α}f∥_{L_{v}^{p′}} ≤ ∥f∥_{L^{p′}}.

Applying Theorem 1.1 we come to condition (2.3).

Analogously, we can prove

Theorem 2.2. Let 1 < p < ∞, 0 < α < 1, and let A_{p} = (p′ − 1)(p′)^{−p}.

(i) If R_{α}v ∈ L_{loc}^{p}(R_{+}) and the inequality

R_{α}[R_{α}v]^{p}(x) ≤ A_{p}R_{α}v(x) a.e.

holds, then the integral equation

φ(x) = R_{α}(φ^{p})(x) + R_{α}(v)(x)  (2.8)

has a non-negative solution φ ∈ L_{loc}^{p}(R_{+}). Moreover, R_{α}v(x) ≤ φ(x) ≤ p′(R_{α}v)(x).

(ii) If 0 < α < \frac{1}{p′} and (2.8) has a non-negative solution in L_{loc}^{p}(R_{+}), then R_{α}v ∈ L_{loc}^{p}(R_{+}) and

R_{α}[(R_{α}v)^{p}](x) ≤ c R_{α}v(x) a.e.

for some constant c > 0.

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(Received 12.02.2001)

Authors’ address: A. Razmadze Mathematical Institute Georgian Academy of Sciences 1, M. Aleksidze St., Tbilisi 380093 Georgia E-mail: kokil@rmi.acnet.ge

meskhi@.rmi.acnet.ge