Weighted Hardy's inequalities for negative indices

Publ. Mat. 48 (2004), 423–443 WEIGHTED HARDY’S INEQUALITIES FOR NEGATIVE INDICES Dmitry V. Prokhorov Abstract In the paper we obtain a precise characterization of Hardy type inequalities with weights for the negative indices and the indices between 0 and 1 and establish a duality between these cases. 1. Introduction The Hardy inequality with weights u, v ≥0 Z+∞ 0 u(x)Zx 0 f(y)v(y)dyq dx1 q ≤CZ+∞ 0 f(x)pdx1 p , f ≥0 has been completely characterized for p, q > 0 by G. Talenti [11], B. Muckenhoupt [7], J. S. Bradley [3], V. G. Maz’ja and A. Rozin [6,§1.3], G. Sinnamon [9], G. Sinnamon and V. D. Stepanov [10] and some other authors (see the monographs [8] and [5] for details). An analogous problem for p, q < 1 was studied by P. R. Beesack and H. P. Heinig [1], where sufficient conditions and necessary conditions under some restrictions on the weight functions were given for the inequality (1) Z+∞ 0 f(x)pdx1 p ≤CZ+∞ 0 u(x) Zx 0 f(y)v(y)dyq dx1 q , f ≥0. In the present paper we obtain a precise characterization of (1) for p, q <0 and p, q ∈(0,1). Moreover, we establish a duality between these cases. The paper is organized as follows: Section 2 contains an explicit criterion of (1) for the cases −∞ < q ≤p < 0 (Theorem 1) and −∞ < p < q < 0 (Theorem 2) and similar results with the dual operator. Theorem 3 of Section 3 proves that the Hardy type inequality with negative indices 2000 Mathematics Subject Classification. 26D15, 26D10. Key words. Hardy’s inequality, weight function, negative indices, duality. The research was partially supported by the Russian Foundation for Basic Research (Project 03–01–00017) and by the Russian Academy of Science (Project 04–3–Γ–01– 049). 424 D. V. Prokhorov of integration is equivalent to the same inequality with the dual operator for conjugate indices. As a consequence we are able to characterize (1) for p, q ∈(0,1) in the corollary of Section 4. Other propositions of Section 4 supplement the main results by a number of characterizations for similar inequalities. Throughout this paper A.Band B&Ameans that A≤cB, where the constant cdepends only on p,qand may be different in different places. If both A.Band A&B, then we write A≈B.Nstands for the set of positive integers, Zis the set of all integers and (a, b) is a nonempty interval of the real line (−∞,+∞). The symbol p0:= p p−1denotes the conjugate number of p,q0:= q q−1and the symbol marks the end of proof of any statement. As usual we abbreviate “almost everywhere” by a.e. and “if and only if” by iff. Sometimes for simplicity we also use the notation Rb af:= Rb af(x)dx. Let us mention one more thing before we start. A peculiarity of the inequality (1) caused by the negativity of the indices force us to work with measurable functions having their values in the extended semiaxis [0,+∞] equipped by arithmetic: (2) 0 + (+∞) = a+ (+∞) = a·(+∞) = +∞, a ∈(0,+∞]; 0·(+∞) = 0; (+∞)α= 0−α= +∞,(+∞)−α= 0α= 0, α ∈(0,+∞). Note also simple corollaries of the axioms (2) which we often use: Let α, β ∈(−∞,+∞)\{0}and a, b ∈[0,+∞]. Then 1. (aα)β=aαβ. 2. aαaβ≤aα+β; if (a < +∞and α+β > 0) or (a > 0and α+β < 0), then aαaβ=aα+β. 3. aαbα≤(ab)αwith equality unless α < 0and {a, b}={0,+∞}. 2. The main results Denote by M+(a, b) the class of all measurable functions f: (a, b)→ [0,+∞] and put (I×f)(x) := Zx a f(y)dy, (I×f)(x) := Zb x f(y)dy, x∈(a, b). Hardy’s Inequalities for Negative Indices 425 Theorem 1. Let −∞ < q ≤p < 0,u, v ∈M+(a, b)and I=I×or I=I×. Then the following statement (a)There exists a constant C > 0such that (3) Zb a f(x)pdx!1 p ≤C Zb a u(x) [(I(fv))(x)]qdx!1 q for all f∈M+(a, b) holds is equivalent to A<+∞, where A:= sup a<t<b A(t) := sup a<t<b [(Iu)(t)]−1 qh(Ivp0)(t)i−1 p0 . Moreover, A ≈ Cfor the least possible constant Cin (3). Proof: Let I=I×(the case I=I×can be proved analogously). At first we note that finiteness of the constant Ais equivalent to the condition (a1)Ra∗ au= 0, Rt a∗u < +∞for all t∈(a∗, b∗), A < +∞and Rb∗ a∗u= +∞implies Rb∗ a∗vp0= +∞, where a∗:= sup{t∈[a, b)|Rt avp0= 0}, b∗:= sup{t∈[a, b)|Rt avp0<+∞} and A:= sup a∗<t<b∗ A(t) := sup a∗<t<b∗Zt a∗ u−1 qZt a∗ vp0−1 p0 . Moreover, A=A. If (a1) holds then for t∈(a, a∗] we have Rt au−1 q= 0 and for t∈(b∗, b) we have Rt avp0−1 p0 = 0. For t∈(a∗, b∗] we have A(t) = A(t)≤A. Putting these together shows that A ≤ A < +∞. Conversely, if A<+∞then A(a∗)<+∞ and Ra∗ avp0= 0 so Ra∗ au= 0. The condition that Rt a∗u < +∞ for t∈(a∗, b∗) is evident because Rt avp0∈(0,+∞) there. If Rb∗ a∗u= +∞then we must have Rb∗ a∗vp0−1 p0 = 0 so Rb∗ a∗vp0= +∞. Finally, because Ra∗ avp0= 0 we obtain A≤ A <+∞and we have completed (a1). . 426 D. V. Prokhorov Let (a) hold. Assume that a∗< b∗. Fix an arbitrary t∈(a∗, b∗). Substituting ft(x) = +∞χ(t,b)(x) + v(x)p0−1χ(a,t)(x) into (3), we obtain 0<Zt a vp01 p ≤CZt a u(x)Zx a vp0q dx1 q =C Za∗ a u(x)Zx a vp0q dx+Zt a∗ u(x)Zx a vp0q dx!1 q , since v < +∞a.e. on (a, b∗). It implies Za∗ a u(x)Zx a vp0q dx < +∞and Zt a∗ u(x)Zx a vp0q dx < +∞, otherwise we have a contradiction 0 <0. Since for x∈(a, a∗) it holds that Rx avp0= 0, we have Ra∗ au= 0 and (4) Zt a∗ vp01 p ≤CZt a∗ u1 qZt a∗ vp0, because Rt a∗vp0∈(0,+∞). If Rt a∗u= +∞, then (4) implies that Rt a∗vp0= +∞which contradicts with the definition of b∗. If Rt a∗u=0, then A(t)=0. If Rt a∗u∈(0,+∞), then it follows from (4) that A(t)≤C. Analogously, using the function ftwith t=b∗we obtain (4) with t= b∗and this inequality implies that Rb∗ a∗vp0= +∞in the case Rb∗ a∗u= +∞. Now if a∗=b∗=b, then Rb av= 0 and by putting f= +∞into (3) we get +∞ ≤ C Zb a u(x)·(+∞)dx!1 q . Thus, Rb au= 0 and all the conditions of (a1) hold, as well as, in the case a∗=b∗=a. Let (a1) hold. Since Ra∗ au=Ra∗ avp0= 0, it yields that (3) is equivalent to the inequality (5) Zb a∗ u(x)Zx a∗ v(y)g(y)1 qdyq dx ≤C−q Zb a gp q!q p for all g∈M+(a, b). If a∗=b∗=b, then (5) holds. Let a∗< b∗or a∗=b∗=a. Fix an arbitrary non negative measurable function gon (a, b). If Rb agp q= +∞, Hardy’s Inequalities for Negative Indices 427 then we have (5). Now let Rb agp q<+∞. If b∗< b, then, by H¨older’s inequality with exponents pand p0, we find that for any x∈(b∗, b) Zx a∗ v(y)g(y)1 qdy ≥Zx a∗ vp01 p0Zx a∗ gp q1 p = +∞. Thus, (6) J1:=Zb a∗ u(x)Zx a∗ v(y)g(y)1 qdyq dx=Zb∗ a∗ u(x)Zx a∗ v(y)g(y)1 qdyq dx. In particular, if a∗=b∗=a, then (5) holds. Now let a∗< b∗. Put N:=      inf (k∈Z|k≥log2 Zb∗ a∗ vp0!),if Zb∗ a∗ vp0 <+∞, +∞,otherwise, and construct the sequence {ak}k≤Nby the relations: Rak a∗vp0= 2k, k < N;aN=b∗. Then it follows from (6) that (7) J1=X k<N Zak+1 ak u(x)Zx a∗ v(y)g(y)1 qdyq dx. Note that Rak+1 aku < +∞for all k < N since Rt a∗u < +∞for all t∈ (a∗, b∗) and Rb∗ a∗u= +∞implies that Rb∗ a∗vp0= +∞. By applying the H¨older inequality with exponents p0and p, we have for all x∈(a∗, b∗) Zx a∗ v(y)g(y)1 qdy =Zx a∗ v(y)g(y)1 qZy a∗ vp01 pp0−1 pp0 dy ≥ Zx a∗ v(y)p0 Zy a∗ vp0−1 p dy!1 p0 Zx a∗ g(y)p qZy a∗ vp01 p0 dy!1 p = (p0)1 p0Zx a∗ vp01 p02 Zx a∗ g(y)p qZy a∗ vp01 p0 dy!1 p . 428 D. V. Prokhorov Using this relation and (7) we estimate the left part of (5) as follows: J1.X k<N Zak+1 ak uZak a∗ vp0q p02 Zak+1 a∗ g(y)p qZy a∗ vp01 p0 dy! q p ≤X k<N Zak+1 ak u(2k) q p02 X j≤k"Zaj+1 aj gp q#(2j+1)1 p0  q p . Moreover, by applying Minkowski’s inequality we finally obtain that J1.  X j<N "Zaj+1 aj gp q#(2j+1)1 p0 X j≤k<N Zak+1 ak u(2k) q p02  p q   q p .A−q  X j<N "Zaj+1 aj gp q#(2j)1 p0 X j≤k<N (2k)−q p0p  p q   q p .A−q"Zb a gp q#q p . The proof is complete. Theorem 2. Let −∞ < p < q < 0,1 r:= 1 q−1 p,u, v ∈M+(a, b) and I=I×or I=I×. Then (a)is equivalent to B<+∞, where B:= Zb a [(Iu)(x)]r ph(Ivp0)(x)ir p0 u(x)dx!−1 r . Moreover, B ≈ Cfor the least possible constant Cin (3). Proof: As in Theorem 1 we only consider the case I=I×and note that finiteness of Bis equivalent to the condition (a2)Ra∗ au= 0, Rt a∗u < +∞for all t∈(a∗, b∗), B <+∞and Rb∗ a∗u=+∞ implies Rb∗ a∗vp0= +∞, where a∗,b∗be the same as in Theorem 1 and B:= Zb∗ a∗Zx a∗ ur pZx a∗ vp0r p0 u(x)dx!−1 r . Hardy’s Inequalities for Negative Indices 429 Moreover, B=B. If (a2) holds, then B=B, that is B<+∞. Conversely, let B<+∞. Since B ≥ q r−1 rA, all conditions of (a2), except B < +∞, follow from finiteness of Ain the same way as in the proof of Theorem 1. It implies the equality B=B and (a2) follows. The case a∗=b∗can be proved analogously with the proof of Theorem 1. Therefore, we assume that a∗< b∗. Let (a) hold. All conditions, except B < +∞, follow in the same way as in the proof of Theorem 1. We only need to show that Bis finite. If Rb∗ a∗u= 0, then B= 0 <+∞. Let Rb∗ a∗u > 0. Then there exist numbers t1and t2such that a∗< t1< t2< b∗and Rt2 t1u > 0. Denote ˜u:= uχ(t1,t2)and let ˜ Bbe similar to Bwith ˜uinstead of u. Then ˜ B="Zt2 t1Zx t1 ur pZx a∗ vp0r p0 ˜u(x)dx#−1 r ≤Zt1 a∗ vp0−1 p0Zt2 t1 u−1 q <+∞, ˜ B > 0 and the inequality (3) holds with ˜uinstead of u: Zb a f(x)pdx!1 p ≤C Zb a ˜u(x)Zx a fvq dx!1 q for all f∈M+(a, b). The last inequality is equivalent to (8) Zb∗ a∗ ˜u(x)Zx a∗ v(y)g(y)1 qdyq dx ≤C−q Zb a gp q!q p for all g∈M+(a, b). Put g(y)p q:= v(y)p0 Zb∗ y ˜u(z)Zz a∗ vp0q−1 dz!r q χ(a∗,b∗)(y). 430 D. V. Prokhorov Then Zb a gp q=Zb∗ a∗ v(y)p0 Zb∗ y ˜u(z)Zz a∗ vp0q−1 dz!r q dy =: J2 and since v < +∞a.e. on (a∗, b∗) and Rb∗ y˜u(z)Rz a∗vp0q−1 dz < +∞ for all y∈(a∗, b∗) we have Zb∗ a∗ ˜u(x)Zx a∗ v(y)g(y)1 qdyq dx ≥Zb∗ a∗ ˜u(x)"Zb∗ x ˜u(z)Zz a∗ vp0q−1 dz#r pZx a∗ vp0q−1Zx a∗ v(y)p0 dydx =Zb∗ a∗ v(y)p0 Zb∗ y ˜u(x)Zx a∗ vp0q−1"Zb∗ x ˜u(z)Zz a∗ vp0q−1 dz#r p dx dy =q rJ2, so that, by (8), q rJ2≤C−qJ q p 2. Let {ak}k≤Nbe the same sequence as in the proof of Theorem 1. Recall that Rak+1 ak˜u < +∞for all k < N and we have J2≤X k<N Zak+1 ak vp0 Zb∗ ak ˜u(z)Zz a∗ vp0q−1 dz!r q .X k<N 2k X k≤j<N "Zaj+1 aj ˜u#(2j)q−1  r q . Putting β r q j:= (2j)r p0Raj+1 aj˜ur q, we find that J2.X k<N 2k X k≤j<N βj(2j)−q r  r q .X k<N β r q k by the discrete Hardy inequality (see e.g. [2]) and since for any fixed n∈Z  X k≤n 2k  q r X j≥nh(2j)−q rip q  q p .1 Hardy’s Inequalities for Negative Indices 431 holds. Conversely, J2≥X k<N "Zak ak−1 vp0# Zb∗ ak ˜u(z)Zz a∗ vp0q−1 dz!r q &X k<N 2kZak+1 ak ˜ur q (2k)(q−1) r q=X k<N β r q k. Analogously, ˜ B−r.X k<N Zak+1 ak ˜uZak+1 a∗ ˜ur p (2k)r p0 ≤X k<N (2k)r p0 X j≤kZaj+1 aj ˜u  r q =X k<N (2k)r p0 X j≤k βj(2j)−q p0  r q .X k<N β r q k and ˜ B−r&X k<N Zak+1 ak ˜u(x)Zx a∗ ˜ur p dx(2k)r p0 ≥X k<N (2k)r p0Zak+1 ak ˜u(x)Zx ak ˜ur p dx =q rX k<N β r q k. Thus, ˜ B−r.J2≤C−qJ q p 2.C−q˜ B−rq p, that is ˜ B.C, since ˜ B∈(0,+∞). By now letting t1→a∗and t2→b∗, we conclude that B < +∞and the proof of the implication (a)⇒(a2) is complete. Conversely, assume that (a2) holds. Fix any non-negative measurable function gon (a, b). By arguing similar as in the proof that (a1) 438 D. V. Prokhorov every x∈(a, b∗) by the definition of b∗. Therefore, we have Zb a f(x)pdx!1 p ≤ Zb∗ a f(x)pdx!1 p ≤ Zb∗ a g(x)pv(x)−pdx!1 p = Zb a g(x)pv(x)−pdx!1 p ≤C Zb∗ a u(x) [(Ig)(x)]qdx!1 q =C Zb∗ a u(x) [(I(fv))(x)]qdx!1 q =C Zb a u(x) [(I(fv))(x)]qdx!1 q . The proof of the case I=I×is complete and the case I=I×can be proved analogously. Proposition 3. Let p, q ∈(−∞,0) and I=I×or I=I×. Then the inequality (13) is equivalent to (15) Zb a g(y)pv(y)−pdy!1 p ≤C Zb ahu(x)1 q(Ig)(x)iq dx!1 q for all g∈M+(a, b). Proof: This proposition can be proved in the same way as the proof of Proposition 2. We only note that both (15) and (13) implies u < +∞ a.e. on (a, b) (see the proof of Proposition 1) and, consequently, for almost all x∈(a, b) the equality (Ih)(x) = +∞implies u(x)1 q(Ih)(x) = +∞. Hardy’s Inequalities for Negative Indices 439 Proposition 4. Let p, q ∈(−∞,0) and I=I×or I=I×. Then the inequality (16) Zb a [g(y)v(y)−1]pdy!1 p ≤C Zb a u(x) [(Ig)(x)]qdx!1 q for all g∈M+(a, b) holds, iff mes{x∈(a, b)|v(x) = +∞} >0or (3) holds. Proof: Sufficiency. Let (3) hold. Fix an arbitrary measurable function g≥0. Put f(x) := g(x)v(x)−1into (3). Then we have (16) since f(x)v(x)≤g(x). Let mes{x∈(a, b)|v(x) = +∞} >0. Then Zb a [g(y)v(y)−1]pdy ≥Z{x∈(a,b)|v(x)=+∞} [g(y)v(y)−1]pdy = +∞ for any measurable function g≥0. Hence, the left part of (16) is zero. Necessity. Let mes{x∈(a, b)|v(x) = +∞} = 0 and (16) holds. Fix an arbitrary measurable function f≥0 and let the function f0be integrable on (a, b) and f0(x)∈(0,+∞), x∈(a, b). Denote E:= {x∈ (a, b)|v(x) = 0}and Ec:= (a, b)\E. Put gn=fvχEc+1 nf0χE,n∈N into (16). Then we obtain Zb a f(x)pdx!1 p ≤ZEc f(x)pdx1 p = Zb a [gn(x)v(x)−1]pdx!1 p ≤C Zb a u(x) [(Ign)(x)]qdx!1 q for all n∈N. Since u(x)[(Ign)(x)]q↑u(x)[(I(fv))(x)]qas n→ ∞ for every x∈(a, b) the Monotone Convergence Theorem implies (3). 440 D. V. Prokhorov Proposition 5. Let p, q ∈(−∞,0) and I=I×or I=I×. Then the inequality (17) Zb a [g(y)v(y)−1]pdy!1 p ≤C Zb ahu(x)1 q(Ig)(x)iq dx!1 q for all g∈M+(a, b) holds, iff mes{x∈(a, b)|v(x) = +∞} >0or (13) holds. Proof: Proposition 5 can be proved in a completely similar way as Proposition 4, so we leave out the details. Remark. It is clear that if p, q ∈(0,1), then (14), (15), (16) and (17) are equivalent, and (3) is equivalent to (13). Proposition 6. Let p, q ∈(0,1) and I=I×. Then (16) holds, iff Rb∗ tu= +∞for all t∈(a, a∗),Rb b∗v(x)−pdx = 0 and (18) Zb∗ a∗ f(x)pdx!1 p ≤C Zb∗ a∗ u(x)Zx a∗ fvq dx!1 q for all f∈M+(a∗,b∗) holds, where a∗:= inf{t∈(a, b]|mes{x∈(t, b)|v(x) = 0}= 0}and b∗:= inf{t∈[a∗, b]|Rb tu= 0}. Proof: Necessity. Fix any t∈(a, a∗). There exists γ∈(t, a∗) such that mes{x∈(t, γ)|v(x) = 0}>0. Put g:= χ(t,γ)into (16). Then we obtain +∞=Zγ t v(x)−pdx1 p ≤C Zb a u(x)Zx a gq dx!1 q =C Zb∗ t u(x)Zx t gq dx!1 q ≤C(γ−t) Zb∗ t u!1 q , that is, Rb∗ tu= +∞. Insert g:= χ(b∗,b)into (16) and we find that Rb b∗v(x)−pdx = 0. Hardy’s Inequalities for Negative Indices 441 Now fix any measurable f≥0. If Rb∗ a∗u(x)Rx a∗fvqdx = +∞, then (18) holds. Let Rb∗ a∗u(x)Rx a∗fvqdx < +∞. Denote E:= x∈(a∗,b∗)|u(x)Zx a∗ fvq <+∞. By the definition of b∗for any ξ∈(a∗,b∗) there exists x∈(ξ, b∗)∩E such that u(x)6= 0. Then Rx a∗fv < +∞. Since ξwas taken arbitrary we have fv < +∞a.e. on (a∗,b∗) and mes{x∈(a∗,b∗)|f(x)6= 0, v(x) = +∞} = 0. This relation and v > 0 a.e. on (a∗,b∗) imply that f(y)v(y)v(y)−1=f(y) for almost every y∈(a∗,b∗). It remains only to put g:= fvχ(a∗,b∗) into (16) to obtain (18). Sufficiency. Fix any measurable g≥0. Let Ra∗ ag= 0. Insert f(x) = g(x)v(x)−1,x∈(a∗,b∗), into (18). Since fv ≤gand Rb b∗v(x)−pdx = 0 we obtain Zb a [g(y)v(y)−1]pdy!1 p = Zb∗ a∗ [g(y)v(y)−1]pdy!1 p ≤C Zb∗ a∗ u(x)Zx a∗ gq dx!1 q =C Zb a u(x)Zx a gq dx!1 q . If Ra∗ ag > 0, then there exists a number t∈(a, a∗) such that Rt ag > 0 and Zb a u(x)Zx a gq dx!1 q ≥ Zb∗ t u!1 qZt a g= +∞ by the first condition. Finally, in a similar way we can prove the following statement: 442 D. V. Prokhorov Proposition 7. Let p, q ∈(0,1) and I=I×. Then (16) holds, iff Rt ¯ au= +∞for all t∈(¯ b, b),R¯ a av(x)−pdx = 0 and (19) Z¯ b ¯ a f(x)pdx!1 p ≤C Z¯ b ¯ a u(x)"Z¯ b x fv#q dx!1 q for all f∈M+(¯ a,¯ b) holds, where ¯ b:= sup{t∈[a, b)|mes{x∈(a, t)|v(x) = 0}= 0}and ¯ a:= sup{t∈[a, ¯ b]|Rt au= 0}. Acknowledgement. The author expresses his deep gratitude to Professor Lars-Erik Persson for drawing the author’s attention to this problem and for fruitful discussions. He also thanks Professor Vladimir D. Stepanov for valuable comments and remarks. References [1] P. R. Beesack and H. P. Heinig, Hardy’s inequalities with indices less than 1, Proc. Amer. Math. Soc. 83(3) (1981), 532–536. [2] G. Bennett, Some elementary inequalities. III, Quart. J. Math. Oxford Ser. 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Appl. 160(2) (1991), 434–445. Hardy’s Inequalities for Negative Indices 443 [10] G. Sinnamon and V. D. Stepanov, The weighted Hardy inequality: new proofs and the case p= 1, J. London Math. Soc. (2) 54(1) (1996), 89–101. [11] G. Talenti, Osservazioni sopra una classe di disuguaglianze, Rend. Sem. Mat. Fis. Milano 39 (1969), 171–185. Computing Centre FEB RAS Tikhookeanskaya 153 Khabarovsk 680042 Russia E-mail address:[email protected], [email protected] Primera versi´o rebuda el 19 de novembre de 2003, darrera versi´o rebuda el 9 de mar¸c de 2004.