The classical Hilbert and Hardy integral inequalities are derived within a functional analytic framework. That framework is further used to generate a new inequality, sharper than Hilbert's

Peachey, Tom C

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A Framework for Proving Hilbert's Double Integral Inequality and Related Results

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A FRAMEWORK FOR PROVING HILBERT’S DOUBLEINTEGRAL INEQUALITY AND RELATED RESULTST.C. PEACHEYAbstract. The classical Hilbert and Hardy integral inequalities are derivedwithin a functional analytic framework. That framework is further used togenerate a new inequality, sharper than Hilbert’s.1. IntroductionThe inequality of our title is the following celebrated result.Theorem 1. Suppose p > 1, q = p(p−1) , f ∈ Lp (0,∞) and g ∈ Lq (0,∞) . Then∫ ∞0∫ ∞0f (x) g (y)x+ ydxdy ≤ pisin(pip) ‖f‖p ‖g‖q .(1.1)The above inequality is the integral analogue of Hilbert’s double series theorem. Itwas first proved by Schur [7] for the case p = 2, and as given here by Hardy [2].We are also concerned with the integral version of Hardy’s inequality.Theorem 2. Suppose p > 1, f ∈ Lp (0,∞) , F (x) =∫ x0 f (t) dt. Then∥∥x−1F (x)∥∥p≤ pp− 1 ‖f‖p .(1.2)In [3], Hardy, Littlewood and Polya show how these two inequalities are implied bytheir Theorem 319. The relevant part of that theorem is:Theorem 3. Suppose p > 1, q = pp−1 , that K (x, y) is non-negative and homoge-neous of degree -1, and that∫ ∞0K (x, 1)x−1p dx =∫ ∞0K (1, y) y−1q dy = C.Then∫ ∞0∫ ∞0K (x, y) f (x) g (y) dxdy ≤ C(∫ ∞0|f (x)|p dx) 1p(∫ ∞0|g (y)|q dy) 1q(1.3)Date: June, 1999.1991 Mathematics Subject Classification. Primary 26D15; Secondary 26D10, 46E30.Key words and phrases. Hilbert inequality, Hardy inequality.12 T.C. PEACHEYand ∫ ∞0dy(∫ ∞0K (x, y) f (x) dx)p≤ Cp∫ ∞0|f (x)|p dx.(1.4)The aim of this paper is to show how results essentially equivalent to this theoremmay be derived in a natural way from considerations of a certain inner productspace.The classical results above have been greatly generalized in recent decades [1], [6].The approach presented here may not be capable of proving such generalizations.It does, however, seem conducive to generating some novel results. For example,an inequality that implies (1.1) , but that is sharper, will be derived.2. Framework for the ProofsThe general functions considered in this work are assumed to be real-valued andmeasurable on a domain (0,∞) . For two such functions, u and v, we define theconvolutionu ∗ v (x) =∫ ∞0u(xt)v (t)dtt.(2.1)(The integral here is Lebesgue, as are all the following ones.) It has long been real-ized, [1], that in terms of this convolution, (1.4) is essentially ‖g ∗ f‖p ≤ ‖g‖1 ‖f‖p .The convolution (2.1) satisfies the commutative propertyu ∗ v (x) = v ∗ u (x)(2.2)in the sense that, when one side exists, so does the other, and the two sides agree.Likewise, for the associative property[u ∗ (v ∗ w)] (x) = [(u ∗ v) ∗ w] (x) ,(2.3)which will be considered in more detail later.We also employ an inner product〈u, v〉 =∫ ∞0u (x) v (x)dxx.(2.4)This can be related to the convolution (2.1) by〈u, v〉 = [(Ru) ∗ v] (1) ,(2.5)using the operator R, defined byRu (x) = u(1x).(2.6)We will also have occasion to use the operator Ma defined for a > 0 byMau (x) = u (ax) .(2.7)A FRAMEWORK FOR PROVING SOME INEQUALITIES 3Associated with the inner product (2.4) is the norm∦ u ∦p =[∫ ∞0|u (x)|p dxx] 1p.(2.8)Further, u ∈ L˜p means that ∦ u ∦p < ∞. Note the unusual notation. This is usedto avoid confusion with the more common norm ‖f‖ which lacks the 1x factor inthe integrand. The following relations are simply proved:∦ Ru ∦p = ∦ u ∦p(2.9)and∦Mau ∦p = ∦ u ∦p .(2.10)3. ResultsThis section establishes some relations connecting the definitions of Section 2.Theorem 4. If p > 1, q = pp−1 , u ∈ L˜p, v ∈ L˜q, then u ∗ v (x) exists for all x > 0and|u ∗ v (x)| ≤ ∦ u ∦p ∦ v ∦q .(3.1)Proof. By Ho¨lder’s inequality∫ ∞0∣∣∣u(xt)∣∣∣ |v (t)| dtt=∫ ∞0∣∣∣u(xt)t−1p∣∣∣ ∣∣∣v (t) t− 1q ∣∣∣ dt≤[∫ ∞0∣∣∣u(xt)∣∣∣p dtt] 1p[∫ ∞0|v (t)|q dtt] 1q= ∦ RMxu ∦p ∦ v ∦q= ∦ u ∦p ∦ v ∦q,where the last step uses (2.9) and (2.10) . Existence of u ∗ v (x) and (3.1) follow.Next, we consider a Ho¨lder type inequality for the inner product and a partialconverse.Theorem 5. Suppose p > 1, and q = pp−1 . Then(i)|〈u, v〉| ≤ ∦ u ∦p ∦ v ∦q,(3.2)(ii) if 〈|u| , |v|〉 ≤ U ∦ v ∦q for all v ∈ L˜q, then ∦ u ∦p ≤ U.Proof. Part (i) can be deduced from the previous theorem using (2.5) .Alternatively,defining φ (x) = x−1pu (x) and ψ (x) = x−1q v (x), then ∦ u ∦p = ‖φ‖p , ∦ v ∦q = ‖ψ‖qand ∫ ∞0u (x) v (x)dxx=∫ ∞0φ (x)ψ (x) dx.4 T.C. PEACHEYThen both (i) and (ii) follow from similar results for φ and ψ in conventionalLebesgue spaces. See [3], Theorem 191 for a statement of the required theory.We are now in a position to prove a result equivalent to the first part of Theo-rem 3. This shows that, for the inner product (2.4) , the operator (Ru) ∗ is adjointto u ∗ .Theorem 6. Suppose that p > 1, q = pp−1 , u ∈ L˜p, v ∈ L˜q and k ∈ L˜1. Then〈k ∗ u, v〉 exists with〈k ∗ u, v〉 = 〈k, (Ru) ∗ v〉 = 〈u, (Rk) ∗ v〉(3.3)and|〈k ∗ u, v〉| ≤ ∦ k ∦1 ∦ u ∦p ∦ v ∦q .(3.4)Proof. Consider〈|k| ∗ |u| , |v|〉 =∫ ∞0|v (x)| dxx∫ ∞0∣∣∣k (xt)∣∣∣ |u (t)| dtt=∫ ∞0|v (x)| dxx∫ ∞0|k (s)|∣∣∣u(xs)∣∣∣ dss=∫ ∞0|k (s)| dss∫ ∞0∣∣∣(Ru)( sx)∣∣∣ |v (x)| dxx= 〈|k| , |Ru| ∗ |v|〉 .However, by (2.9) , Ru ∈ L˜p, so by Theorem 4 with u replaced by Ru,|Ru| ∗ |v| ≤ ∦ Ru ∦p ∦ v ∦q = ∦ u ∦p ∦ v ∦qusing (2.9) again. Thus〈|k| ∗ |u| , |v|〉 ≤ 〈|k| , ∦ u ∦p ∦ v ∦q〉= ∦ u ∦p ∦ v ∦q 〈|k| , 1〉= ∦ k ∦1 ∦ u ∦p ∦ v ∦q < ∞.This shows the existence of 〈k ∗ u, v〉 and gives (3.4) . Now, Fubini’s theorem allowsus to repeat the argument above, changing the order of integration, but with |k| ,|u| , and |v| replaced by k, u and v respectively. This gives 〈k ∗ u, v〉 = 〈k, (Ru) ∗ v〉 ,the first part of (3.3) . The remaining part can be shown by changing the order ofintegration without the change of variable.Theorem 7. If p > 1, u ∈ L˜p and k ∈ L˜1, then k ∗ u ∈ L˜p and∦ k ∗ u ∦p ≤ ∦ k ∦1 ∦ u ∦p .(3.5)Proof. Introduce q = pp−1 . Theorem 6 ensures that for all v ∈ L˜q,〈|k| ∗ |u| , |v|〉 ≤ ∦ k ∦1 ∦ u ∦p ∦ v ∦q .Then, Theorem 5 Part (ii) with |u| replaced by |k| ∗ |u| gives |k| ∗ |u| ∈ L˜p, with∦ |k| ∗ |u| ∦p ≤ ∦ k ∦1 ∦ u ∦p . The required results follow.A FRAMEWORK FOR PROVING SOME INEQUALITIES 54. ApplicationsIn Theorem 6, if we replace u (x) by x1p f (x) and v (x) by x1q g (x) , then ∦ u ∦p= ‖f‖p and ∦ v ∦q = ‖g‖q . If we also replace k (x) by x1p(x+1) , we have∦ k ∦1 =∫ ∞0x1px+ 1dxx=∫ 10s1p−1 (x− 1)− 1p ds,which is B(1p , 1− 1p)or pi cosec(pip). Further,〈k ∗ u, v〉 =∫ ∞0∫ ∞0f (x) g (y)x+ ydxdy.(4.1)Then (3.4) becomes the Hilbert inequality (1.1) .To obtain the Hardy inequality we use Theorem 7. There we substitute x1p f (x)for u (x) , as before. In addition, we take k (x) to behµ (x) ={Γ−1 (µ)x1p−µ (x− 1)µ−1 for x > 1,0 for 0 < x ≤ 1,(4.2)where µ is any positive constant. Note that k ∈ L˜1, in fact,∦ k ∦1 =∫ ∞11Γ (µ)x1p−µ−1 (x− 1)µ−1 dx=1Γ (µ)∫ 10s−1p (1− s)µ−1 ds=1Γ (µ)B(1− 1p, µ)=Γ(1− 1p)Γ(µ+ 1− 1p) .Moreover,k ∗ u (x) =∫ x01Γ (µ)(xt− 1)µ−1 (xt) 1p−µt1p f (t)dtt= x1p−µIµf (x) ,whereIµf (x) =1Γ (µ)∫ x0(x− t)µ−1 f (t) dt,the Riemann-Liouville fractional integral of order µ. Then Theorem 7 gives thefollowing result.Theorem 8. If p > 1, µ > 0 and f ∈ Lp (0,∞) , then∥∥x−µIµf (x)∥∥p≤Γ(1− 1p)Γ(µ+ 1− 1p) ‖f‖p .6 T.C. PEACHEYThis is a generalization of Theorem 1.2, due to Knopp [4]; the case µ = 1 is theoriginal Hardy inequality.Alternatively, substituting u (x) = xµ+1p f (x) andk (x) ={Γ−1 (µ)x1p (1− x)µ−1 for 0 < x < 1,0 for x ≥ 1,gives a theorem involving the Weyl fractional integralJµf (x) =1Γ (µ)∫ ∞x(t− x)µ−1 f (t) dt,namely,Theorem 9. If p > 1, µ > 0 and f ∈ Lp (0,∞) , then‖Jµf (x)‖p ≤Γ(1p)Γ(µ+ 1p) ‖xµf (x)‖p .This is also a well-known result. See [3], Theorem 329.5. ExtensionsWe begin with sufficient conditions for the relation (2.3) .Theorem 10. If p > 1, w ∈ L˜p and u, v ∈ L˜1, then[(u ∗ v) ∗ w] (x) = [u ∗ (v ∗ w)] (x)(5.1)for almost all x > 0 and∦ u ∗ (v ∗ w) ∦p ≤ ∦ u ∦1 ∦ v ∦1 ∦ w ∦p .(5.2)Proof. By Theorem 7, v ∗w ∈ L˜p. Then another application of that theorem, withk and u replaced by u and v ∗ w respectively, gives∦ u ∗ (v ∗ w) ∦p ≤ ∦ u ∦1 ∦ v ∗ w ∦pwhich cannot exceed ∦ u ∦1 ∦ v ∦1 ∦ w ∦p ≤ ∞.This proves (5.2) and ensures the existence of [u ∗ (v ∗ w)] (x) for almost all x > 0.For those x, ∫ ∞0∣∣∣u(xt)∣∣∣ dtt∫ ∞0∣∣∣∣v( ts)∣∣∣∣ |w (s)| dss < ∞.Fubini’s theorem now permits the second step in[u ∗ (v ∗ w)] (x) =∫ ∞0u(xt) dtt∫ ∞0v(ts)w (s)dss=∫ ∞0w (s)dss∫ ∞0u(xt)v(ts)dtt=∫ ∞0w (s)dss∫ ∞0u(xys)v (y)dyy= [(u ∗ v) ∗ w] (x) ,A FRAMEWORK FOR PROVING SOME INEQUALITIES 7giving (5.1) .Versions of the index laws for fractional integration can be deduced from The-orem 10. Let p > 1, µ, ν > 0. Replace w (x) by x1p f (x) so f ∈ Lp. Replace v byhµ as defined in (4.2) and replace u (x) by x−µhν (x) . Then u ∗ v (x) = hν+µ (x) .Theorem 10 then gives:Theorem 11. If p > 1, f ∈ Lp thenx−µ−νIνIµf (x) = x−µ−νIν+µf (x) ∈ Lp.If, instead, we swap the definitions of u and v, then we obtainTheorem 12. Under the conditions of Theorem 11:x−µIµx−µ−νIνxµf (x) = x−µ−νIν+µf (x) .This is a version of the “second index law” due to Love [5]. Use of the adjoints(Ru) ∗ and (Rv) ∗ give similar results for the Weyl fractional integral.Returning to more general considerations, we haveTheorem 13. Suppose p > 1, q = pp−1 , u ∈ L˜p, v ∈ L˜q and k1, k2 ∈ L˜1. Then〈(k1 ∗ k2) ∗ u, v〉 = 〈k2 ∗ u, (Rk1) ∗ v〉(5.3)and|〈(k1 ∗ k2) ∗ u, v〉| ≤ ∦ k2 ∗ u ∦p ∦ (Rk1) ∗ v ∦q .(5.4)Proof. By Theorem 10, (k1 ∗ k2) ∗ u = k1 ∗ (k2 ∗ u) in L˜p.So〈(k1 ∗ k2) ∗ u, v〉 = 〈k1 ∗ (k2 ∗ u) , v〉 .(5.5)Since, by Theorem 7, k2 ∗ u ∈ L˜p, Theorem 6 applies to the right hand side andthis yields (5.3) .By (2.9) , Rk1 ∈ L˜1 so that (Rk1 ∗ v) ∈ L˜q, again using Theorem 7. Hence,by Theorem 5|〈(k1 ∗ k2) ∗ u, v〉| = |〈k2 ∗ u, (Rk1) ∗ v〉|≤ ∦ k2 ∗ u ∦p ∦ (Rk1) ∗ v ∦q,which completes the proof.As an application of this result, take k1 (x) = x1p e−x, k2 (x) = x−1q e−1x , u (x) =x1p f (x) and v (x) = x1q g (x) . Then(Rk1) ∗ v (x) = x− 1p∫ ∞0e−tx g (t) dt = x−1pG(1x),8 T.C. PEACHEYwhere G is the (one-sided) Laplace transform of g. Similarly, k2∗u (x) = x− 1qF( 1x),F being the Laplace transform of f. In addition,k1 ∗ k2 (x) = x1px+ 1= k (x)say. So the left side of (5.3) is 〈k ∗ u, v〉 which we have seen in (4.1) is equal toHilbert’s double integral.The right side of (5.3) is〈k2 ∗ u, (Rk1) ∗ v〉 =∫ ∞0x−1qF(1x)x−1pG(1x)dxx=∫ ∞0F (s)G (s) ds.Thus, in this case (5.3) givesTheorem 14. For f ∈ Lp, g ∈ Lq where p > 1 and 1p + 1q = 1,∫ ∞0∫ ∞0f (x) g (y)x+ ydxdy =∫ ∞0F (s)G (s) ds,(5.6)where F and G are the Laplace transforms of f and g.Further, since ∦ k2 ∗ u ∦p = ‖x1− 2pF (x) ‖p and ∦ (Rk1) ∗ v ∦q = ‖x1− 2qG (x) ‖q,then (5.4) gives the following.Theorem 15. Under the conditions of Theorem 14∣∣∣∣∫ ∞0∫ ∞0f (x) g (y)x+ ydxdy∣∣∣∣ ≤ ∥∥∥x1− 2pF (x)∥∥∥p∥∥∥x1− 2qG (x)∥∥∥q.(5.7)This inequality implies Hilbert’s. For∥∥∥x1− 2pF (x)∥∥∥p= ∦ k2 ∗ u ∦p≤ ∦ k2 ∦1 ∦ u ∦p= Γ(1q)‖f‖pand similarly ∥∥∥x1− 2qG (x)∥∥∥q≤ Γ(1p)‖g‖q .AsΓ(1q)Γ(1p)= pi cosec(pip),(5.7) implies (1.1) .To show that (5.7) is a sharper inequality than (1.1) , consider the case f (x) =g (x) = e−x, p = q = 2. Then F (x) = G (x) = 1x+1 and∥∥∥x1− 2pF (x)∥∥∥p=∥∥∥x1− 2qG (x)∥∥∥q= 1A FRAMEWORK FOR PROVING SOME INEQUALITIES 9while‖f‖p = ‖g‖q = p−1p =1√2.The right side of (5.7) is thus equal to 1 while that of (1.1) is pi2 .Note that (5.6) can be obtained directly by changing the order of integration onthe right side, using Hilbert’s inequality to justify the process. However, we canfind no mention of formula (5.6) in the literature.References[1] Erde´ly; A., An extension of a Hardy-Littlewood-Po´lya inequality, Proc. Edinburgh Mathemat-ical Society, 21, (1978), pp. 11-15.[2] Hardy G.H., Note on a theorem of Hilbert concerning series of positive terms, Proc. LondonMathematical Society (2), 23, (1925), Records of Proceedings, xlv-xlvi.[3] Hardy G.H., J.E. Littlewood and G. Polya, Inequalities, (Second Edition) CUP, 1952.[4] Knopp K., U¨ber Reihen mit positiven Gliedern (2te Mitteilung), Journal of the London Math-ematical Society, 5, (1930), pp. 13-21.[5] Love E. R., Two index laws for fractional integrals and derivatives, Journal of the AustralianMathematical Society, 14, (1972), pp. 385-410[6] Opic B. and A. Kufner, Hardy-type inequalities, Longman, Harlow, 1990.[7] Schur I., Bemerkungen zur Theorie der beschra¨nken Bilinearformen mit unendlich vielenVera¨nderlichen, Journal f. Math, 140, (1911), pp. 1-28.School of Communications and Informatics, Victoria University of Technology, POBox 14428, Melbourne City, MC 8001, AustraliaE-mail address: tom@matilda.vu.edu.au

An extension of a Hardy-Littlewood-P olya inequality,

Bemerkungen zur Theorie der beschr anken Bilinearformen mit unendlich vielen Ver anderlichen,

Hardy-type inequalities,

Note on a theorem of Hilbert concerning series of positive terms,

Two index laws for fractional integrals and derivatives,

Uber Reihen mit positiven Gliedern (2te Mitteilung),

A Framework for Proving Hilbert's Double Integral Inequality and Related Results

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