On the Hardy-type integral operators in Banach function spaces

Publicacions Matem`atiques, Vol 42 (1998), 165–194. ON THE HARDY-TYPE INTEGRAL OPERATORS IN BANACH FUNCTION SPACES Elena Lomakina and Vladimir Stepanov Abstract Characterization of the mapping properties such as boundedness, compactness, measure of non-compactness and estimates of the approximation numbers of Hardy-type integral operators in Banach function spaces are given. 1. Introduction Let Xand Ybe two Banach spaces of measurable functions defined on R+. We consider the Hardy-type integral operator K:X→Ygiven by (1.1) Kf(x)=ϕ(x)Zx 0 k(x, y)ψ(y)f(y)dy, x > 0, where the real functions ϕ(x) and ψ(x) (weights) are measurable and finite almost everywhere on R+, and the kernel k(x, y)≥0, satisfies (1.2) D−1(k(x, z)+k(z,y)) ≤k(x, y) ≤D(k(x, z)+k(z,y)),x≥z≥y≥0, Keywords. Banach function spaces, Hardy-type integral operator, compact operator, approximation number. 1991 Mathematics subject classifications: 46E30, 47B38, 47G60. The research work of the both authors was partially supported by the INTAS project 94-881 and of the second author by the Centre de Recerca Matem`atica at Barcelona and by the RFFR grant 97-01-00604. 166 E. Lomakina, V. Stepanov where the constant D≥1 does not depend on x, y, z. Typical examples of such a kernel are (x−y)α,α≥0; logβµx y¶,β≥0; or µZx y h(s)ds¶γ , γ≥0 with nonnegative h(s), and their various combinations. Introduced by R. Oinarov [13], [14] the condition (1.2) extends in some sense the well-known ∆2-condition for convex functions [10] used in [12], [18] for convolution operators and thus, (1.2) seems to be a balance point between generality of conditions imposed on a kernel and implicitness of a criterion for the boundedness of the Hardy-type operators. A survey of the mapping properties of operators (1.1) with Oinarov’s kernel in Lebesgue and Lorentz spaces can be found in [19]. The paper is devoted to operators of the form (1.1) acting in Banach spaces of Lebesgue-measurable functions on R+(see Definition 1 below). Investigation in this area was recently initiated by E. Berezhnoi [2], [3], who, in particular, characterized weak-type estimates for the operator (1.1) with the kernel k(x, y)≥0 increasing with respect to the first variable and also strong estimates, when k(x, y) = 1 and the spaces X and Ysatisfy an `-condition (see Definition 3 below). E. Berezhnoi [3] has also obtained some necessary and/or sufficient conditions for the boundedness of operators (1.1) with restrictions on k(x, y)≥0, stronger than (1.2). Sections 2 and 3 contain definitions and the statement of the main results and further comments, respectively. Our first result characterizes the boundedness of the operator (1.1) with kernel satisfying (1.2) in the spaces Xand Ysatisfying an `-condition (Theorem 1). This leads to a characterization of the compactness (Theorem 2) and measure of noncompactness (Theorem 3) of the operator. Upper and lower estimates for the behaviour of the approximation numbers of operators (1.1), when k(x, y) = 1, are given in Theorems 6 and 7. Sections 4 and 5 provide the proofs. 2. Definitions Definition 1 [1]. A real normed linear space X={f:kfkX<∞} is called a Banach function space (BFP) if in addition to the usual norm axioms kfkXsatisfies the following conditions: (1) kfkXis defined for every Lebesgue-measurable function fon R+, and f∈Xif, and only if, kfkX<∞;kfkX= 0 if, and only if, f= 0 almost everywhere (a.e.); (2) kfkX=k|f|k Xfor all f∈X; (3) if 0 ≤f≤ga.e., then kfkX≤kgk X; Hardy operators in Banach function spaces 167 (4) if 0 ≤fn↑fa.e., then kfnkX↑kfk X; (5) if mes E<∞, then kχEkX<∞; (6) if mes E<∞, then ZE f(x)dx ≤CEkfkX. Given a BFS X, its associate space X0is defined by X0=½g:Z∞ 0 |fg|<∞for all f∈X¾, and endowed with the associate norm (2.1) kgkX0= sup ½Z∞ 0 |fg|:kfkX≤1¾. X0is also a Banach function space satisfying axioms (1)-(6) and, moreover, X0is a norm fundamental subspace of the dual space X∗, that is the equality (2.2) kfkX= sup ½Z∞ 0 |fg|:kgkX0≤1¾ holds for all f∈X[1]. The spaces X,X0are complete normed linear spaces and X00 =X[1]. The H¨older inequality¯¯¯¯Z∞ 0 fg¯¯¯¯≤kfk Xkgk X 0 holds for all f∈Xand g∈X0and is sharp in both directions on the strength of (2.1) and (2.2). The relationships (2.1) and (2.2) give rise to the following Principle of duality. T:X→Yis a bounded linear operator, that is kTfkY≤CkfkXfor all f∈Xwith a finite positive constant C,if and only if (i) kT0gkX0≤CkgkY0for all g∈Y0, where the conjugate operator T0:Y0→X0is defined by the formulae Z∞ 0 (Tf)g=Z∞ 0 f(T0g), or (ii) ¯¯¯¯Z∞ 0 (Tf)g¯¯¯¯≤CkfkXkgkY0for all f∈Xand g∈Y0, with the same constant C. The least possible constant Cdefines the norm kTkand, thus kTkX→Y=kT0kY0→X0. 168 E. Lomakina, V. Stepanov Xhas absolutely continuous norm (AC-norm), if for all f∈X, kfχEnkX→0 for every sequence of sets {En}⊂R +such, that χEn(x)→0 a.e. We assume throughout the paper that X0and Yhave AC-norms. Let `be a Banach sequence space (BSS), what means that axioms (1)-(6) are satisfied with respect to the count measure and let {ek}denote the standard basis in `. Definition 2 [3]. Given a BFS Xand a BSS `,Xis said to be `-concave, if for any sequence of disjoint intervals {Jk}such that SJk=R+, and for all f∈X (2.3) ° ° ° ° °X k ekkχJkfkX° ° ° ° °` ≤d1kfkX, where d1is a finite positive constant independent of f∈Xand {Jk}. Analogously, BFS Yis said to be `-convex, if for any sequence of disjoint intervals {Ik}, such that SIk=R+and for all g∈Y (2.4) kgkY≤d2° ° ° ° °X k ekkχIkgkY° ° ° ° °` for a finite positive constant d2>0, independent of g∈Yand {Ik}. Definition 3 [3]. (The Berezhnoi `-condition). We say, that Banach function spaces Xand Ysatisfy an `-condition, if there exists a Banach sequence space `such that Xis `-concave and Yis `-convex simultaneously. Let `0denote the associate space. We need the following Lemma 1 [3].Let Ybe an `-convex BFS and suppose (2.4) holds. Then Y0is an `0-concave BFS and (2.5) ° ° ° ° °X k ekkχIkfkY0° ° ° ° °`0 ≤d2kfkY0, for all f∈Y0and {Ik}, such that SIk=R+. Throughout the paper the expressions of the form 0 ·∞,0/0, ∞/∞ are taken equal to zero, the inequality A¿Bmeans A≤cB, where c depends only on D, and possibly on the constants d1and d2of Definition 2; however the relationship A≈Bis interpreted as A¿B¿A or A=cB.χEdenotes the characteristic function (indicator) of a set E⊂R+. Hardy operators in Banach function spaces 169 3. Statement of the main results Put for all t≥0 A0= sup t>0 A0(t) = sup t>0° °χ[t,∞]ϕ° °Y° °χ[0,t](·)k(t, ·)ψ(·)° °X0,(3.1) A1= sup t>0 A1(t) = sup t>0° °χ[t,∞](·)k(·,t)ϕ(·) ° °Y° °χ [0,t]ψ° °X0 (3.2) and let A= max(A0,A 1). Note, that A0=A1,ifk(x, y)=1. Theorem 1. Let Xand Ybe BFS satisfying the Berezhnoi `-condition and let Kbe an integral operator of the form (1.1) with the kernel k(x, y)≥0satisfying (1.2). Then K:X→Yis bounded, if and only if, A<∞. Moreover, (3.3) D−1A≤kKk X→Y≤d 1 d 2 γ(D)A, where γ(D)depends only on D. Remark 1. (i) The boundedness of Kwas characterized in [14], [19] (for Lebesgue spaces) and in [11] (for Lorentz spaces.) The case k(x, y)=1 has been intensively studied for the last few decades by many authors and has led to further developments (sf. [15], [19]). (ii) The Bereznoi `-condition corresponds to the case p≤qin the Lp−Lqsetting and to the case max(r, s)≤min(p, q) in the Lorentz Lrs −Lpq setting, see [6], [11]. If no `-condition holds, then the lower bound in (3.3) is nevertheless valid. Moreover, there exists an operator, for which (3.3) is valid for spaces with no `-condition [17]. Theorem 2. Let the assumptions of Theorem 1 be fulfilled and suppose the spaces X0and Yhave AC-norms. Then K:X→Yis compact, if and only if A<∞and (3.4) lim t→ai Ai(t) = lim t→bi Ai(t)=0; i=0,1, where (3.5) ai= inf{t>0:A i (t)>0},b i= sup{t>0:A i (t)>0};i=0,1. Remark 2. In fact, it follows from the proof of Theorem 2 below, that a0=a1,b0=b1. 170 E. Lomakina, V. Stepanov The condition (3.4) has been formulated by many authors only for a0=a1=0,b 0=b 1=∞. However, it is easy to find a formal counterexample, for which A<∞and (3.4) is valid with a0=a1=0, b 0=b 1=∞, but Kis non-compact. The matter is, that the condition (3.4) has to be formulated for the end-points of the “real” interval of action of operator K(see Remark 4 below for further details). In the non-compact case we estimate the measure of non-compactness of the operator K(or, equivalently, the distance of Kfrom the set of finite rank operators) defined by α(K) = inf{kK−Pk; rank P<∞}. To this end we need additional notation; put for all 0 <a<z<b<∞: (3.6) J0 L(a) = sup 0<t<a ° °χ[t,a]ϕ° °Y° °χ[0,t](·)k(t, ·)ψ(·)° °X0, J1 L(a) = sup 0<t<a ° °χ[t,a](·)k(·,t)ϕ(·) ° °Y° °χ [0,t]ψ° °X0, JL(z) = max(J0 L(z),J1 L(z)),J L = lim z→a0 JL(z); J0 R(b) = sup b<t<∞° °χ[t,∞]ϕ° °Y° °χ[b,t](·)k(t, ·)ψ(·)° °X0, J1 R(b) = sup b<t<∞° °χ[t,∞](·)k(·,t)ϕ(·) ° °Y° °χ [b,t]ψ° °X0, JR(z) = max(J0 R(z),J1 R(z)),J R = lim z→b0 JR(z); J= max(JL,J R). Theorem 3. Let the assumptions of Theorem 2 be valid and K:X→Y be bounded. Then (3.7) D−1J≤α(K)≤d2 1d2 2γ(D)J. Utilizing the scheme from [11] we estimate from above and below the approximation numbers of the Hardy operator of the form (3.8) Hf(x)=ϕ(x)Zx 0 ψ(y)f(y)dy. This part of the paper has been initiated by D. E. Edmunds, W. D. Evans and D. J. Harris in the work [5]. Afterwards the extention for convolution operators with the polynomial kernel was given in [7] and for the Hardy operator in Lorentz spaces in [11]. The statement of the results and proofs of this part are given in Section 5. Hardy operators in Banach function spaces 171 4. Boundedness, compactness and measure of non-compactness We begin with an alternative proof of the criterion for the boundedness of the Hardy operator due to E. Berezhnoi. Then we establish the proof for case in which the kernel satisfies Oinarov’s condition. The basic idea is to apply the principle of duality to obtain the upper bound instead of using direct estimates. Theorem 4 [3].Let Xand Ybe BFS satisfying the `-condition, and let operator Hbe defined by (3.8). Then H:X→Yis bounded if, and only if (4.1) A= sup t>0 A(t) = sup t>0° °χ[t,∞]ϕ° °Y° °χ[0,t]ψ° °X0<∞. Moreover, A≤kHk X→Y≤4d 1 d 2 A. Proof: Necessity: For the lower bound we repeat the Berezhnoi argument [3]. If H:X→Yis bounded, then using axioms (2) and (3) of BFS we find for arbitrary t>0 and for all f∈Xsuch that f(y)ψ(y)≥0 kHkX→YkfkX≥kHfkY=° ° ° °ϕ(x)Zx 0 f(y)ψ(y)dy° ° ° °Y ≥° ° ° °χ[t,∞)(x)ϕ(x)Zx 0 f(y)ψ(y)dy° ° ° °Y ≥° °χ[t,∞)(x)ϕ(x)° °YZt 0 f(y)ψ(y)dy =° °χ[t,∞)ϕ° °YZ∞ 0 χ[0,t](y)f(y)ψ(y)dy. Consequently, applying (2.1), we have kHkX→Y≥A(t) for all t>0 and it follows that kHkX→Y≥A. Sufficency: It follows from the principle of duality that for the upper bound it is sufficient to prove the estimate J≡¯¯¯¯Z∞ 0 ϕFg¯¯¯¯¿AkfkXkgkY0 172 E. Lomakina, V. Stepanov for all f∈Xand g∈Y0, where F(x)=Zx 0 f(y)ψ(y)dy. Suppose, that f(y)ψ(y)6= 0 on a set of positive measure, then we can choose a sequence {xk}⊂R +such, that Zxk 0 |f(y)ψ(y)|dy =2 k ,−∞ <k≤N≤∞, where N= sup{k:Ik=[x k−1 ,x k)6=∅}. Then, applying H¨older’s inequality, (2.3), (2.5) and (4.1), we get J≤Z∞ 0 |ϕFg|≤ X k≤N 2 k+1 ZIk+1 |ϕg| =4X k≤NZI k |fψ|ZIk+1 |ϕg| ≤4X k≤N kχIkfkXkχIkψkX0kχIk+1 ϕkYkχIk+1 gkY0 ≤4AX k≤N kχIkfkXkχIk+1 gkY0 ≤4A° ° ° ° °X k ekkχIkfkX° ° ° ° °`° ° ° ° °X k ekkχIk+1 gkY0° ° ° ° °`0 ≤4Ad1d2kfkXkgkY0. Consequently, kHkX→Y≤4d1d2A. We shall need the following modification of Theorem 4. Theorem 5. Let Xand Ybe BFS satisfying the `-condition and Hωf(x)=ϕ(x)Zω(x) 0 ψ(y)f(y)dy, where y=ω(x)is a differentiable increasing function on R+such that ω(0)=0,ω(∞)=∞and, thus, the inverse function x=ω−1(y)exists. Then (4.2) Aω≤kH ω k X→Y≤4d 1 d 2 A ω , where Aω= sup t>0° °χ[0,t]ψ° °X0° °χ[ω−1(t),∞)ϕ° °Y= sup t>0° °χ[0,ω(t)]ψ° °X0° °χ[t,∞)ϕ° °Y. Hardy operators in Banach function spaces 173 Proof of Theorem 5: Is similar to the proof of Theorem 4. We omit details. Proof of Theorem 1: Necessity: Note that the Oinarov condition (1.2) implies (4.3) k(x, y)≥D−1k(t, y) for all x≥t≥y≥0. Consequently, applying (4.3), we obtain for all f∈Xsuch that f(y)ψ(y)≥0 kKkX→YkfkX≥° ° ° °ϕ(x)Zx 0 k(x, y)f(y)ψ(y)dy° ° ° °Y ≥D−1° °χ[t,∞)ϕ° °YZt 0 k(t, y)f(y)ψ(y)dy and arguing as in the necessity part of Theorem 4 we find, that kKkX→Y≥D−1A0. By the principle of duality kKkX→Y=kK0kY0→X0, where K0g(y)=ψ(y)Z∞ y k(x, y)ϕ(x)g(x)dx. Applying the above argument to the operator K0, we find kK0kX→Y≥ D−1A1and, thus, kKkX→Y≥D−1A. For sufficiency we need the following two lemmas. Lemma 2. Let k0(x, y)≥0,x≥y≥0be nondecreasing and continuous with respect to x. Assume that k0(x, y)≤D0(k0(x, z)+k 0 (z,y)),x≥z≥y≥0 with D0≥1independent of x, z, y.Letf(y)be locally integrable, ψ(y) be bounded and compactly supported and f(y)ψ(y)≥0.LetG 0 (x)= Zx 0 k 0 (x, y)ψ(y)f(y)dy be such, that 0<G 0 (x)<∞for some x>0. For a fixed number δ>0we define ∆k=©x>0:G 0 (x)≥(δ+1) kª, k∈Z,N= max ∆k6=∅k;xk= inf ∆k,k≤N,xN+1 =∞if N<∞.If δ≥D 0 , then 0<···<x k−1<x k<···<x N<∞and the inequality (4.4) (δ+1) k−1 ≤Zx k x k−1 k 0 (x k,y)f(y)ψ(y)dy +D0k0(xk,x k−1)Zx k−1 0 f(y)ψ(y)dy holds for all k≤N. 180 E. Lomakina, V. Stepanov where (4.20) γ(D)=D¡1 + max(2,D2) ¢2¡1 + 2 max(2,D2) ¢. By Fatou’s theorem we obtain (4.19) for f, g, ϕ, ψ with no restriction. Theorem 1 is proved. Remark 4. (i) There are three natural analogues of Theorem 1. The first is a restriction to an interval of real axis, the second deals with the associate operator and the third is concerned the non-Volterra case if the kernel is symmetric with respect to xand y. We omit details. (ii) Note, that if k(x0,y 0)=∞for some ∞>x 0≥y 0>0, then Oinarov’s condition implies k(x, y)=∞,x≥x 0 ≥y 0 ≥y>0. Consequently, the convention 0 ·∞= 0 yields, that A<∞is possible, only if ° °χ[x0,∞)ϕ° °Y+° °χ[0,y0]ψ° °X0=0. Thus, such an operator Kis actually reduced to the interval [y0,x 0], where it coincides with K0f(x)=ϕ(x)Zx y 0 k(x, y)ψ(y)f(y)dy, y0≤x≤x0, being the null-operator outside of the interval. Thus we may and shall assume the kernel to be bounded k(x, y)≤cτ<∞on every domain of the form Ωτ={(x, y): ∞>τ≥x≥y≥0}. Proof of Theorem 2: Necessity: That A= max(A0,A 1)<∞follows from Theorem 1. If f∈X,f(y)ψ(y)≥0 then exploiting Oinarov’s condition, we find ∞>Akfk XÀkKfkY≥° °χ[t,∞)Kf° °Y ≥D−1° °χ[t,∞)ϕ° °YZt a0 k(t, y)ψ(y)f(y)dy. Now, by the principle of duality for an arbitrary fixed γ∈(0,1) we may find a function ft, such that supp ft⊆[a0,t], ft(y)ψ(y)≥0, kftkX=1 and (4.21) ∞>AÀkKftkY≥γD−1° °χ[t,∞)ϕ° °Y° °χ[a0,t]k(t, ·)ψ(·)° °X0. Hardy operators in Banach function spaces 181 Given G∈X0the H¨older inequality and absolute continuity of the norm in X0yield (4.22) ¯¯¯¯Z∞ 0 ftG¯¯¯¯≤° °χ[a0,t]G° °X0→0,t→a 0 . Since K0:Y0→X0is also a compact operator, for any given ε>0 there exists a finite number of functions G1,G 2,... ,G n εsuch that (4.23) min 1≤n≤nε kK0g−GnkX0≤ε for every g∈Y0,kgkY0≤1. Given ε>0 and ftwith kftkX= 1 we find by the principle of duality and (4.23) g∈Y0,kgkY0≤1 and Gn, such that kKftkY≤(1 −ε)¯¯¯¯Z∞ 0 Kftg¯¯¯¯ and kK0g−GnkX0≤ε, respectively, and using (4.21) we obtain kKftkY≤(1 −ε)¯¯¯¯Z∞ 0 ftK0g¯¯¯¯ ≤(1 −ε)ε+(1−ε)¯ ¯ ¯ ¯Z∞ 0 f t G n ¯ ¯ ¯ ¯≤ε, t →a0. Consequently, kKftkY→0, t→a0, and a part of (3.4), namely lim t→a0 A0(t) = 0, now follows from (4.21). Analogously, begining with the inequality ∞>AÀkKfkY≥° °χ[t,∞)Kf° °Y ≥D−1° °χ[t,∞)(·)ϕ(·)k(·,t) ° °YZt a 1 ψ(y)f(y)dy, we prove, that lim t→a1 A1(t) = 0. The dual assertions on infinity follows from the similar observations for the associate operator. For proving sufficiency we need the following result. Lemma 4. Let Xand Ybe BFS on a separable σ-finite measure space and T:X→Ybe an integral operator of the form Tf(α)= ZT(α, β)f(β)dβ. If both X0and Yhave AC-norms and (4.24) AT=° ° °kT(α, ·)kX0° ° °Y<∞, then Tis compact. 182 E. Lomakina, V. Stepanov Proof of Lemma 4: It is sufficient to establish, that the set of the functions of the form i0 X i=1 µi(α)ηi(β) is dense in the space Y[X0] with the norm defined by the right side of (4.24), where µi∈Yand ηi∈X0. On the strength of ([9, Chapter XI, Lemma 2]) it is true if the both spaces X0and Yare “order continuous”. This is fulfilled, when X0and Yhave AC-norms, because of ([1, Chapter 1, Proposition 3.5]), and so the Lemma is proved. We continue the proof of the sufficiency part of Theorem 2. Let us show first, that (4.25) a0=a1,b 0 =b 1 . To this end assume, for instance, that 0 ≤a0<a 1 . Then A0(t)= A 1 (t)=0,t∈[0,a 0] and it follows from the Landau resonance theorem ([1, Lemma 2.6]) and Theorem 1, restricted to the interval [0,a 0], that for a.e. x∈[0,a 0] (4.26) ϕ(x)k(x, y)ψ(y) = 0 for a.e. y∈[0,x]. From (3.5) we find, that A1(t)=° °χ [t,∞)(·)k(·,t)ϕ(·) ° °Y° °χ [0,t]ψ° °X0=0,a 0 <t≤a 1 . If ° °χ[0,t]ψ° °X0=0,a 0 <t≤a 1 , then ψ(y) = 0 for a.e. y∈[0,t] by the first axiom of BFS and, hence, A0(t)=0,t>a 0 , which contradicts the definition of a0.If ° °χ [t,∞)(·)k(·,t)ϕ(·) ° °Y=0,a 0 <t≤a 1 , then for all a0<t≤a 1 ϕ(x)k(x, t) = 0 for a.e. x∈[t, ∞) and for all g∈Y0such, that supp g⊆[a0,a 1], ϕ(x)g(x)≥0 and arbitrary f∈Xsuch, that f(t)ψ(t)≥0, we find Za1 a0 Kf(x)g(x)dx =Za1 a0 ϕ(x)g(x)dx Zx a0 k(x, t)ψ(t)f(t)dt =Za1 a0 ψ(t)f(t)dt Za1 t ϕ(x)k(x, t)g(x)dx =0 Hardy operators in Banach function spaces 183 and, again by the Landau theorem, we get for a.e. x∈[a0,a 1] (4.27) ϕ(x)k(x, y)ψ(y) = 0 for a.e. y∈[a0,x]. Now by (4.26) and (4.27) Za1 0 Kf(x)g(x)dx =Za1 a0 ϕ(x)g(x)dx Za0 0 k(x, t)ψ(t)f(t)dt and by the H¨older inequality and Oinarov’s condition we find ¯¯¯¯Za1 0 Kf(x)g(x)dx¯¯¯¯ ≤DZa1 a0 k(x, a0)|ϕ(x)g(x)|dx Za0 0 |ψ(t)f(t)|dt +DZa1 a0 |ϕ(x)g(x)|dx Za0 0 k(a0,t)|ψ(t)f(t)|dt ≤D³° °χ[a0,a1](·)ϕ(·)k(·,a 0) ° °Y° °χ [0,a0]ψ° °X0 +° °χ[a0,a1]ϕ° °Y° °χ[0,a0](·)k(a0,·)ψ(·)° °X0´° °χ[0,a1]g° °Y0° °χ[0,a1]f° °X ≤D(A0(a0)+A 1 (a 0 )) ° °χ[0,a1]g° °Y0° °χ[0,a1]f° °X=0. Hence, by the principle of duality we obtain kKkX[0,a1]→Y[0,a1]=0, and, in particular, Theorem 1, restricted to the interval [0,a 1], implies A0(t)=0,0≤t≤a 1 . Thus, a0=a1, and by similar arguments it can be proved that b0=b1. For this reason we may and shall assume further for simplicity, that a0=a1=0,b 0=b 1=∞. Let 0 <a<b<∞and put Paf=χ[0,a]f, Qbf=χ[b,∞)f, Pabf=χ[a,b]f. Then we have (4.28) Kf =(P a+P ab +Qb)K(Pa+Pab +Qb)f =PaKPaf+QbKQbf+PabKPabf+QbKPaf+QbKPabf+PabKPaf. By Theorem 1 restricted to the intervals [0,a]or[b, ∞) and (3.4) we have (4.29) kPaKPak≤½sup 0<t<a A0(t) + sup 0<t<a A1(t)¾→0,a→0, kQ b KQbk≤½sup t>b A0(t) + sup t>b A1(t)¾→0,b→∞. 184 E. Lomakina, V. Stepanov It follows from Lemma 4 that the operator QbKPais compact. Indeed, (4.30) AQbKPa≤° ° °° °χ[0,b](·)k(x, ·)ψ(·)° °X0χ[b,∞)(x)ϕ(x)° ° °Y ≤D³° °χ[b,∞)(x)k(x, b)ϕ(x)° °Y° °χ[0,b]ψ° °X0 +° °χ[b,∞)ϕ° °Y° °χ[0,b](·)k(b, ·)ψ(·)° °X0´ ≤D(A0(b)+A 1 (b)) <∞. Anagously, we find (4.31) AQbKPab ≤D(A0(b)+A 1 (b)) <∞, APabKPa≤D(A0(a)+A 1 (a)) <∞. Note, that 0<° °χ[a,∞)ϕ° °Y,° °χ[0,b]ψ° °X0<∞, otherwise A0(a)=A1(b)=0, and k(x, y)≤cb<∞,b≥x≥y≥a. By Remark 4(ii), we may write APabKPab =° ° °° °χ[a,b](·)k(x, ·)ψ(·)° °X0χ[a,b](x)ϕ(x)° ° °Y ≤cb° °χ[a,∞)ϕ° °Y° °χ[0,b]ψ° °X0<∞ and hence, by Lemma 4, operator PabKPab is compact too. Using this and (4.29)-(4.31) we see, that Kis a limit of compact operators. This ends the proof of Theorem 2. Proof of Theorem 3: We assume for simplicity, that a0=a1=0, b 0=b 1=∞. Let 0 <a<b<∞. By Theorem 1 we obtain D−1JL(a)≤kP a KPak≤d 1 d 2 γ(D)J L (a), D −1 J R (b)≤kQ b KQbk≤d 1 d 2 γ(D)J R (b), where the constant γ(D) is defined by (4.20). Now, using (4.28) and taking into account the compactness of the last four components there, we see, that, α(K)≤kP a KPa+QbKQbk. Put S=PaKPaand T=QbKQb. Then kS+Tk= sup f6=0 kSf +TfkY kfkX ≤d2sup   ° ° °kSfkY+kTfkY° ° °` kfkX ;f6=0,fψ≥0,P abf=0   . Hardy operators in Banach function spaces 185 If Pabf= 0, then by the Berezhnoi `-condition kfkX≥d−1 1° ° °kgkX+khkX° ° °`, where g=Pafand h=Qbf. Hence, kS+Tk≤d 1 d 2sup f6=0,f=g+h° ° °kSfkY+kTfkY° ° °` ° ° °kgkX+khkX° ° °` =d1d2sup f6=0,f=g+h° ° ° ° ° ° kSgkY kgkX ·kgkX ° ° °kgkX+khkX° ° °` +kThkY khkX ·khkX ° ° °kgkX+khkX° ° °` ° ° ° ° ° °` ≤d2 1d2 2γ(D)° ° ° ° ° °JL(a)kgkX ° ° °kgkX+khkX° ° °` +JR(b)khkX ° ° °kgkX+khkX° °` ° ° ° ° ° °` ≤d2 1d2 2γ(D)J. To obtain the lower bound, let θ>α(K).If Yhas AC-norm, then Y is separable [1]. Hence, there exists T:X→Ysuch that rankT<∞ and kKf −TfkY≤θkfkXfor all f∈X. Since range of the operator T is formed by a finite number of functions from Y, we can approximate each of them by a bounded function with compact support [1] and, thus, given ε>0, there exist T0:X→Ywith rank T0= rank Tand the numbers 0 <δ<N<∞, such that kT−T0k<εsupp T0f⊂[δ, N] for all f∈X. Hence, kKf −T0fkY≤(θ+ε)kfkXfor all f∈X. Let fbe such that supp f⊂[0,δ]S[N,∞) and fψ ≥0.Then (θ+ε)kfkX≥kKfkY=kKPδf+KQNfkY≥kP δ KPδf+QNKQNfkY, since all the functions involved are non-negative. Thus (θ+ε)kfkX≥kP δ KPδfkY 186 E. Lomakina, V. Stepanov for all f∈Xwith supp f⊂[0,δ] and fψ ≥0 and (θ+ε)kfkX≥kQ NKQNfkY for all f∈Xwith supp f⊂[N,∞) and fψ ≥0. Hence, applying the lower bound from Theorem 1, we obtain (θ+ε)≥D−1JL(δ) and (θ+ε)≥D−1JR(N), Letting θ→α(K), ε→0 and then δ→0, N→∞we establish the lower bound. Theorem 3 is proved. Remark 5. Theorem 3 for Lebesgue spaces was proved in [7], the case k(x, y) = 1 was given in [6]. 5. Approximation numbers We begin with the reminder, that for any positive integer m, the m-th approximation number amof a bounded linear map T:X→Yis defined by (5.1) am(T) = inf{kT−Pk;Pa bounded linear operator and rank P<m}. For further information on the approximation numbers we refer the reader to the monographs [4], [8] and [16]. We consider the operator H:X→Yof the form (3.8) and suppose, that His compact. We also assume for simplicity, that a0=0,b 0=∞for the operator H.By Theorem 2 we get A= sup t>0 A(t) = sup t>0° °χ[t,∞]ϕ° °Y° °χ[0,t]ψ° °X0<∞, lim t→0A(t) = lim t→∞ A(t)=0. Given sufficiently small ε,0<ε<kHk, we choose the numbers 0 = c0< c1<c 2<··· <c N−1<c N<c N+1 =∞and intervals Ik=[c k ,c k+1], k=0,1,... ,N, such that (5.3) A[c1]=A[c N]=ε, where A[c1] = sup 0<t<c1 A(t) = sup 0<t<c1° °χ[t,c1]ϕ° °Y° °χ[0,t]ψ° °X0 A[cN] = sup cN<t<∞ A(t) = sup cN<t<∞° °χ[t,∞]ϕ° °Y° °χ[cN,t]ψ° °X0. Hardy operators in Banach function spaces 187 Lemma 5. Let Xand Ybe BFS satisfying the Berezhnoi `-condition, and suppose Yand Y0have AC-norms. Let 0<a<b<∞,I=(a, b) and (5.4) F(x)=Zx a ψ(y)f(y)dy, a ≤x≤b; FI=1 µ(I)ZI Fdµ, µ(I)=ZI dµ, where dµ(x)=ϕ(x)g(x)dx, and g(x)is a function on Isatisfying the inequality (5.5) (1 −δ)kχ[a,b]ϕkYkχ[a,b]gkY0≤ZI ϕ(x)g(x)dx for a sufficiently small 0<δ≤0,01. Then 3 10 max(B0,B 1)≤sup f6=0 kχ[a,b]ϕ(F−FI)kY kχ[a,b]fkX ≤82 25d2 1d2 2max(B0,B 1), where B0= sup a<x<c kχ[x,c]ψkX0kχ[a,x]ϕkY, B1= sup c<x<b kχ[c,x]ψkX0kχ[x,b]ϕkY. Proof of Lemma 5: Given f∈X[a,b],c∈(a, b)weput Ψ c (x)=         −Zc x ψ(y)f(y)dy, a ≤x<c, Zx c ψ(y)f(y)dy, c ≤x<b and Ψc,I =1 µ(I)ZI Ψcdµ. Then F(x)−FI=Ψ c (x)−Ψ c,I. To obtain the lower bound, we take f∈X[a,b]such that supp f⊆[a, c] and suppose, that the inequality ° °χ[a,b]ϕ(F−FI)° °Y≤C° °χ[a,b]f° °X 188 E. Lomakina, V. Stepanov holds for all f∈X[a,b]with a constant Cindependent of f. Then C° °χ[a,c]f° °X≥° °χ[a,c]ϕ(F−FI)° °Y=° °χ[a,c]ϕ(Ψc−Ψc,I)° °Y ≥° °χ[a,c]ϕΨc° °Y−|Ψ c,I|° °χ[a,c]ϕ° °Y =° °χ[a,c]ϕΨc° °Y−1 µ(I)¯¯¯¯ZI Ψcdµ¯¯¯¯° °χ[a,c]ϕ° °Y ≥° °χ[a,c]ϕΨc° °Y−1 µ(I)ZI |Ψc|dµ ° °χ[a,c]ϕ° °Y =° °χ[a,c]ϕΨc° °Y−1 µ(I)ZI |Ψcϕϕ−1|dµ ° °χ[a,c]ϕ° °Y ≥° °χ[a,c]ϕΨc° °Y−1 µ(I)° °χ[a,c]ϕΨc° °Y° °χ[a,c]g° °Y0° °χ[a,c]ϕ° °Y =µ1−1 µ(I)° °χ[a,c]g° °Y0° °χ[a,c]ϕ° °Y¶° °χ[a,c]ϕΨc° °Y =µ1−V(a, c) µ(I)¶° °χ[a,c]ϕΨc° °Y, where V(a, c)=° °χ [a,c]ϕ° °Y° °χ[a,c]g° °Y0. Because of the absolute continuity of the norms Yand Y0, we can for any fixed β∈(0,1−δ) find a point c∈(a, b) such that V(a, c)=βµ(I), therefore by Theorem 4 restricted to the interval [a, c]wehaveC≥(1 −β)B0. A similar argument applied for all fsuch that supp f⊂[c, b) gives C° °χ[c,b]f° °X≥µ1−W(c, b) µ(I)¶° °χ[c,b]ϕΨc° °Y, where W(c, b)=° °χ [c,b]ϕ° °Y° °χ[c,b]g° °Y0. 1−W(c, b) µ(I)=1 µ(I)(µ(I)−W(c, b)) =1 µ(I)³(1 −δ)° °χ[a,b]ϕ° °Y° °χ[a,b]g° °Y0−W(c, b)´ ≥β(1 −δ)−W(c, b) µ(I). Hardy operators in Banach function spaces 189 If c→b, then β→(1 −δ) and W(c, b)→0, therefore we can choose c∈(a, b) such that W(c, b) µ(I)≤β(1 −δ) 2. By Theorem 4 we get C≥β(1 −δ) 2B1. Now, if we take βsuch that 1−β=β(1 −δ) 2, then β=2 3−δand the required lower bound C≥ 3 10 max(B0,B 1). Sufficiency: Using H¨older’s inequality and the Berezhnoi `-condition, we see that ° °χ[a,b]ϕ(F−FI)° °Y=° °χ[a,b]ϕ(Ψc−Ψc,I)° °Y ≤° °χ[a,b]ϕΨc° °Y+|Ψc,I|° °χ[a,b]ϕ° °Y ≤° °χ[a,b]ϕΨc° °Y +1 µ(I)° °χ[a,b]ϕΨc° °Y° °χ[a,b]g° °Y0° °χ[a,b]ϕ° °Y ≤2 (1 −δ)° °χ[a,b]ϕΨc° °Y ≤2 (1 −δ)d2° ° °° °χ[a,c]ϕΨc° °Y+° °χ[c,b]ϕΨc° °Y° ° °` ≤8d2 1d2 2 1−δmax(B0,B 1)° °χ [a,b]f° °X, and the required result follows. The proof of Lemma 5 is complete. By Lemma 5 the norm of the operator HIf(x)=χ I(x)ϕ(x)(F(x)−FI) depends continuously on the interval I. We choose the intervals Ik= [ck,c k+1], k=1,... ,N −1 so, that (5.6) kHIkk=ε, k =1,... ,N −2, kH N−1k≤ε.