New index transforms with the product of Bessel functions

0.13 MB·English

by  Semyon Yakubovich

NEW INDEX TRANSFORMS WITH THE PRODUCT OF BESSEL FUNCTIONS
S.YAKUBOVICH
5
1
0
2 ABSTRACT. Newindextransformsareinvestigated,whichcontainasthekernelproductsoftheBesselandmod-
ifiedBesselfunctions. MappingpropertiesandinvertibilityinLebesguespacesarestudiedfortheseoperators.
p
RelationshipswiththeKontorovich-LebedevandFouriercosinetransformsareestablished. Inversiontheorems
e
areproved. Asanapplication, asolution oftheinitial value problem forthefourthorderpartial differential
S
equation,involvingtheLaplacianispresented.

5

1. INTRODUCTION AND PRELIMINARY RESULTS
]
A
Themainobjectsofthepresentpaperarethefollowingoperatorsoftheindextransforms[1]
C
1 ¥
h. (Ff)(t )= sinh(pt /2) 0 Kit (2√2x)Im Jit (2√2x) f(x)dx, t ∈R\{0}, (1.1)
t Z h i
a ¥ g(t )
m (Gg)(x)= ¥ Kit (2√2x)Im Jit (2√2x) sinh(pt /2) dt , x∈R+. (1.2)
[ Z− h i
Here i is the imaginary unit and Im denotes the imaginary part of a complex-valued function. The con-

2 vergence of the integrals (1.1), (1.2) will be clarified below. Functions Jm (z),Km (z), m C [2], Vol. II
v are,correspondingly,theBesselandmodifiedBesselorMacdonaldfunctions,whichsatisf∈ythedifferential
9
equations
0
6 z2d2u+zdu+(z2 m 2)u=0, (1.3)
4 dz2 dz −
0
d2u du
1. z2 +z (z2+m 2)u=0, (1.4)
dz2 dz−
0
respectively.Ithastheasymptoticbehaviour
5
1 2 p p 1/2
v: Jm (z)= p zcos z− 4(2m +1) [1+O(1/z)], Km (z)= 2z e−z[1+O(1/z)], z→¥ , (1.5)
i r (cid:16) (cid:17) (cid:18) (cid:19)
X andneartheorigin
ar Jm (z)=O(zm ), zm Km (z)=2m −1G (m )+o(1), z→0, (1.6)
K (z)= logz+O(1), z 0. (1.7)
0
− →
TheMacdonaldfunctioncanberepresentedbytheintegral
¥
Km (z)= e−zcoshucosh(m u)du, Rez>0, m C. (1.8)
0 ∈
Z
Date:September8,2015.
2000MathematicsSubjectClassification. 44A15,33C10,44A05.
Keywordsandphrases. IndexTransforms,Besselfunctions,modifiedBesselfunctions,Kontorovich-Lebedevtransform,Fourier
transform,Mellintransform.
1
2 S.Yakubovich
Concerningthe kernelof the transform(1.1)our key representationwill be in terms of the Mellin-Barnes
integral,givingbyrelation(8.4.23.11)in[3],Vol. III,namely,
sKinith((2pt√2/x2)) Im Jit (2√2x) =−16p1i√p gg+i¥i¥ G (s/2)G ((sG +((i1t )/s2))/G2()(s−it )/2)x−sds, (1.9)
h i Z − −
where x>0, t R 0 , g >0, G (z) is Euler’s gamma -function [2], Vol. I. The theory of the index
∈ \{ }
transformscanbefoundintheauthormonograph[1](seealso[4]).ThefamiliarexampleistheKontorovich-
Lebedevtransform
¥
(KLf)(t )= Kit (x)f(x)dx. (1.10)
0
Z
OurmethodofinvestigationisbasedonthetheoryoftheMellintransformintheLebesguespaces[5]. In
fact,wedefinetheMellintransformas
¥
f (s)= f(x)xs 1dx (1.11)
∗ −
0
Z
anditsinverse,accordingly,
1 n +i¥
f(x)= f (s)x sds. (1.12)
2p i n i¥ ∗ −
Z −
Integrals(1.11),(1.12)areconvergent,forinstance,inmeanintheweightedL -spaces,1< p 2andthe
p
≤
Parsevalequalitytakesplace[5]
¥ 1 n +i¥
f(x)g(x)dx= f (s)g (1 s)ds. (1.13)
0 2p i n i¥ ∗ ∗ −
Z Z −
It is importantfor us for further investigationto obtain an integral representationof the kernel in (1.1) in
termsoftheFouriercosinetransform[5]
2 ¥
(F f)(x)= f(t)cos(xt)dt. (1.14)
c p
r Z0
Precisely,weprovethefollowing
Lemma1. Letx>0,t R 0 .Then
∈ \{ }
sKinith((2pt√2/x2))Im Jit (2√2x) =−p2 0¥ cos(t u)Re K0 4ep i/4√xcoshu du, (1.15)
Z
h i h (cid:16) (cid:17)i
whereRedenotestherealpartofacomplex-valuedfunction.
Proof. Indeed,takingtheFouriercosinetransform(1.14)frombothsidesoftheequality(1.9),wechange
theorderofintegrationowingtoFubini’stheorem.Infact,employingthereciprocalequalities(seeformula
(1.104)in[1])viatheFouriercosinetransform(1.14)
¥ s it s it p G (s)
G + G cos(ut )dt = , Res>0, (1.16)
Z0 (cid:18)2 2(cid:19) (cid:18)2− 2(cid:19) 2s−1coshsu
s it s it G (s) ¥ cos(t u)
G + G = du, (1.17)
(cid:18)2 2(cid:19) (cid:18)2− 2(cid:19) 2s−2Z0 coshsu
onecanintegratetwicebypartsintheintegral(1.17),showingtheuniformestimate
s it s it G (s+1)
G 2+ 2 G 2− 2 ≤ | t 2 |[c1+c2|s|], Res>0, t ∈R\{0}, (1.18)
(cid:12) (cid:18) (cid:19) (cid:18) (cid:19)(cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
IndexTransformsWiththeProductofBesselFunctions 3
wherec ,c areabsolutepositiveconstants. Hencewiththe useoftheStirlingasymptoticformulaforthe
1 2
gamma-function[2],Vol. Iandtheelementaryinequality
s it s it
G + G G (s)B(g /2,g /2) (1.19)
2 2 2− 2 ≤| |
(cid:12) (cid:18) (cid:19) (cid:18) (cid:19)(cid:12)
(cid:12) (cid:12)
where B(a,b) is Euler’s beta-(cid:12)function [2], Vol. I, it gu(cid:12)arantees the absolute convergence of the iterated
(cid:12) (cid:12)
integral
¥ g+i¥ G (s/2)G ((s+it )/2)G ((s it )/2)
− ds dt <¥ .
0 g i¥ G ((1 s)/2)
Z Z − (cid:12) − (cid:12)
Therefore,(1.9)and(1.16)yield (cid:12) (cid:12)
(cid:12) (cid:12)
¥ cos(t u) (cid:12) 1 g+i¥ G (s)(cid:12)G (s/2)
0 sinh(pt /2)Kit (2√2x)Im Jit (2√2x) dt =−8i√p g i¥ G ((1 s)/2)(2xcoshu)−sds. (1.20)
Z h i Z − −
Moreover, appealing to the duplication formula for the gamma-function[2], Vol. I and making a simple
substitution,thelatterMellin-Barnesintegralcanbewrittenas
1 g+i¥ G (s)G (s/2) 1 g+i¥ G 2(s)G (s+1/2)
−8i√p g i¥ G ((1 s)/2)(2xcoshu)−sds=−8p i g i¥ G (1/2 s) (x2cosh2u)−sds. (1.21)
Z − − Z − −
Meanwhile,considering0<g <1/2,wehavethevalueoftherelativelyconvergentintegral(cf.[5],Section
7.3)
G (s) 21 2s ¥ 21 2s(2s 1) ¥
G (1/2 s) = √−p 0 costt2s−1dt=− − √p − 0 sintt2(s−1)dt,
− Z Z
where the latter integral is obtained after the integration by parts and, evidently, convergesabsolutely for
0<g <1/2.Hence,substitutingitintheright-handsideof(1.21)andchangingtheorderofintegration,we
findwiththeaidofrelation(8.4.23.1)in[3],Vol. III
1 g+i¥ G 2(s)G (s+1/2)
(x2cosh2u) sds
−8p i g i¥ G (1/2 s) −
Z − −
= 1 ¥ sint g+i¥ [2G (1+s)G (s+1/2) G (s)G (s+1/2)] 2xcoshu −2sdsdt
4p i√p 0 t2 g i¥ − t
Z Z − (cid:18) (cid:19)
1√2xcoshu ¥ 4xcoshu 4xcoshu sint
= K 1 dt.
√p 0 1/2 t t − t5/2
Z (cid:18) (cid:19)(cid:20) (cid:21)
Since
p
K (z)= e z,
1/2 2z −
r
wegetwiththesimplesubstitution
1 g+i¥ G 2(s)G (s+1/2) 1 ¥ 4xcoshu sint
−8p i g i¥ G (1/2 s) (x2cosh2u)−sds= 2 0 e−4xcosh(u)/t t −1 t2 dt.
Z − − Z (cid:20) (cid:21)
We calculate the latter integral, appealing to relation (2.5.37.1)in [3], Vol. I. In fact, taking in mind the
identityforMacdonald’sfunctions[2],Vol. II
2m
Km +1(z) Km 1(z)= Km (z),
− − z
wededuce
1 ¥ 4xcoshu sint 1
e 4xcosh(u)/t 1 dt= K 4ep i/4√xcoshu +K 4e p i/4√xcoshu
2 0 − t − t2 −2 2 2 −
Z (cid:20) (cid:21) h (cid:16) (cid:17) (cid:16) (cid:17)i
4 S.Yakubovich
+ 1 ep i/4K1 4ep i/4√xcoshu e−p i/4K1 4e−p i/4√xcoshu
4i√xcoshu −
h (cid:16) (cid:17) (cid:16) (cid:17)i
= Re K 4ep i/4√xcoshu .
0
−
Thuscombiningwith(1.20),(1.21),wefindthhev(cid:16)alueoftheindex(cid:17)initegral
0¥ sincho(sp(tt u/)2)Kit (2√2x)Im Jit (2√2x) dt =−Re K0 4ep i/4√xcoshu , x>0, u∈R. (1.22)
Z
h i h (cid:16) (cid:17)i
Inthemeantime,using(1.9),(1.18),onecanverifythatforeachx>0thekernelof(1.1)belongstoL (R )
1 +
L (R ). Since(see(1.5),(1.7),(1.8)) ∩
2 +
¥
K0 4e±p i/4√xcoshu = exp 4e±p i/4√xcoshu cosht dt
(cid:12)(cid:12) (cid:16) ¥ (cid:17)(cid:12)(cid:12) (cid:12)(cid:12)Z0 (cid:16)− (cid:17) (cid:12)(cid:12)
(cid:12) exp 2√2(cid:12)xco(cid:12)(cid:12)shu cosht dt=K0(2√2xcoshu), (cid:12)(cid:12)
≤ 0 −
Z
thesameistruefortheright-handsi(cid:16)deof(1.22). Henceta(cid:17)kingtheinverseFouriercosinetransform[5],we
comeupwithequality(1.15),completingtheproofofLemma1. (cid:3)
Corollary1. Ithas
tli→m0 sKinith((2pt√2/x2))ImhJit (2√2x)i=−p2Z0¥ RehK0(cid:16)4ep i/4√xcoshu(cid:17)idu, x>0.
Corollary2. Letx>0,t R 0 .Thenthekernelin(1.1)satisfiesthefollowinginequality
∈ \{ }
(cid:12)sKinith((2pt√2/x2)) Im Jit (2√2x) (cid:12)≤ p4 e−d |t|K02 cos d2 2xcos(d ) , (1.23)
(cid:12) h i(cid:12) (cid:18) (cid:18) (cid:19)p (cid:19)
whered [0,p /2(cid:12)(cid:12)). (cid:12)(cid:12)
∈ (cid:12) (cid:12)
Proof. Indeed,theintegralintheright-handsideof(1.15)canbewrittenas
1 ¥
eitu K 4ep i/4√xcoshu +K 4e p i/4√xcoshu du
−2p ¥ 0 0 −
Z− h (cid:16) (cid:17) (cid:16) (cid:17)i
andbytheanalyticpropertyoftheintegrandandtheabsoluteconvergenceoftheintegralonecanmovethe
contouralongtheopeninfinitehorizontalstrip(id ¥ , id +¥ )withd [0,p /2),i.e.
− ∈
1 ¥
eitu K 4ep i/4√xcoshu +K 4e p i/4√xcoshu du
−2p ¥ 0 0 −
1 ¥ Z− h (cid:16) (cid:17) (cid:16) (cid:17)i
= eit(id +u) K 4ep i/4 xcosh(id +u) +K 4e p i/4 xcosh(id +u) du, (1.24)
−2p ¥ 0 0 −
wherethemainZ−branchofthhesq(cid:16)uarerooptischosen.Hen(cid:17)cedeno(cid:16)tingby p (cid:17)i
z=cosh(id +u)=cos(d )coshu+isin(d )sinhu= zeiargz,
| |
whereargz=arctan(tan(d )tanhu) ( d , d ).Then
∈ −
argz p argz p
Re 4e±p i/4 xcosh(id +u) =4 xz cos 4 xcos(d )coshu cos
| | 2 ± 4 ≥ 2 ± 4
whencosh argz pp= 1 [cos(argiz/2) psin(argz(cid:16)/2)]>0,i.(cid:17)e tanp(argz/2) <1,which(cid:16)istrue. Mo(cid:17)reover,
2 ± 4 √2 ∓ | |
wehave
(cid:0) (cid:1) argz p d p
cos >cos + .
2 ± 4 2 4
(cid:16) (cid:17) (cid:18) (cid:19)
IndexTransformsWiththeProductofBesselFunctions 5
Therefore,returningto(1.15),(1.24),weobtain
(cid:12)sKinith((2pt√2/x2)) Im Jit (2√2x) (cid:12)≤ 2e−pd |t| 0¥ K0 4cos d2 +p4 xcos(d )coshu du
(cid:12) h i(cid:12) Z (cid:18) (cid:18) (cid:19)p (cid:19)
(cid:12) (cid:12)
(cid:12)(cid:12) 2e−d |t| ¥ K(cid:12)(cid:12) 4cos d +p xcos(d )cosh(u/2) du
≤ p 0 0 2 4
Z (cid:18) (cid:18) (cid:19) (cid:19)
p
4e d t ¥ ¥ d p
= − | | exp 4cos + xcos(d )coshu cosht dudt
p 0 0 − 2 4
Z Z (cid:18) (cid:18) (cid:19) (cid:19)
p
4e d t d
− | |K2 cos 2xcos(d ) ,
≤ p 0 2
(cid:18) (cid:18) (cid:19) (cid:19)
p
completingtheproofofCorollary2.
(cid:3)
AnimmediateconsequenceofLemma1isalso
Corollary3. Letx>0. Thenforallt Rtheinequality
∈
(cid:12)sKinith((2pt√2/x2)) Im Jit (2√2x) (cid:12)≤ p4 K02 √2x (1.25)
(cid:12) h i(cid:12) (cid:16) (cid:17)
(cid:12) (cid:12)
isfulfilled. (cid:12) (cid:12)
(cid:12) (cid:12)
Proof. Wehave
2 ¥ 2 ¥
cos(t u)Re K 4ep i/4√xcoshu du K 2√2xcoshu du
p (cid:12)Z0 0 (cid:12)≤ p Z0 0
(cid:12) h (cid:16) (cid:17)i (cid:12) (cid:16) (cid:17)
(cid:12) 4 ¥ 4 ¥ ¥(cid:12)
(cid:12) K 2√2x coshu du= (cid:12)exp √2x coshu coshv
≤ p 0 0 p 0 0 −
Z Z Z
(cid:16) (cid:17) (cid:16) (cid:17)
4
exp √2x coshu coshv dudv K2 √2x
× − ≤ p 0
(cid:16) (cid:17) (cid:16) (cid:17)
andtheresultfollows.
(cid:3)
Remark1. Inequality(1.25)isaparticularcaseoftheinequality(1.23)withd =0.
EmployingtheMellin-Barnesrepresentation(1.9)ofthekernelin(1.1),whichwedenoteby
Y t (x)= sKinith((2pt√2/x2)) Im Jit (2√2x) ,
h i
wewillderiveanordinarydifferentialequation,whosesolutionisY t (x). Precisely,itisgivenby
Lemma2. ThekernelY t (x)isafundamentalsolutionofthefollowingfourthorderdifferentialequation
withvariablecoefficients
x2d4Y t +5xd3Y t +(4+t 2)d2Y t +16Y t =0. (1.26)
dx4 dx3 dx2
6 S.Yakubovich
Proof. Recalling the Stirling asymptoticformulaforthe gamma-function[2], Vol. I, we see thatfor each
t Rtheintegrandin(1.9)behavesas(s=g +it)
∈
G (s/2)G ((sG +((i1t )/s2))/G2()(s−it )/2) =e−p |t|/2|t|2g−3/2, |t|→¥ .
−
This argumentallows to differentiate repeatedly with respect to x under the integral sign in (1.9). Hence
withthereductionformulaforthegamma-functionweobtain
d d 1 g+i¥ s2G (s/2)G ((s+it )/2)G ((s it )/2)
xdxxdx Y t =−16p i√p Zg−i¥ G ((1−s)/2) − x−sds
1 g+i¥ G (s/2)G (1+(s+it )/2)G (1+(s it )/2)
=−t 2Y t (x)−4p i√p g i¥ G ((1 s)/2) − x−sds
Z − −
1 2+g+i¥ G (s/2 1)G ((s+it )/2)G ((s it )/2)
=−t 2Y t (x)−4p i√p Z2+g−i¥ − G ((3−s)/2) − x2−sds
1 2+g+i¥ G (s/2)G ((s+it )/2)G ((s it )/2)
=−t 2Y t (x)−p i√p Z2+g−i¥ (s−2)(1−s)G ((1−s−)/2) x2−sds.
Differentiatingagaintwicebothsidesofthelatterequalityandusing(1.9),wefind
d2 x d 2Y t = t 2d2Y t 16Y t(x).
dx2 dx − dx2 −
(cid:18) (cid:19)
Thusafterfulfillingthedifferentiationintheleft-handsideofthelatterequality,wearriveat(1.26).Lemma
2isproved.
(cid:3)
Finally in this section we note that the obtained inequalities and integral representations of the kernel
in (1.1) will be used in the sequel to study the boundedness, compactness and invertibility of the index
transforms(1.1),(1.2).
2. BOUNDEDNESS AND COMPACTNESS IN LEBESGUE’SSPACES
Webegin,introducingthefollowingweightedL -space
1
¥
L0 L R ; K2(√2x)dx := f : K2(√2x) f(x)dx<¥ . (2.1)
≡ 1 + 0 0 0 | |
(cid:16) (cid:17) (cid:26) Z (cid:27)
Inparticular,aswewillshowbelow,itcontainsspacesLn ,p(R+)forsomen R, 1 p ¥ withthenorms
∈ ≤ ≤
¥ 1/p
f n ,p= xn p−1 f(x) pdx <¥ , (2.2)
|| || 0 | |
(cid:18)Z (cid:19)
||f||n ,¥ =esssupx≥0|xn f(x)|<¥ .
Whenn = 1 weobtaintheusualnorminL denotedby .
p p ||||p
Lemma3. Let n <1,1 p ¥ , q= p . Thentheembeddingholds
≤ ≤ p 1
−
Ln ,p(R+) L0 (2.3)
⊆
and
2
||f||L0 ≤2−2(1/p+n )"G 1/(q2(q2)q2((11−−n )n )) B(1−n , 1−n )# ||f||n ,p, 1<p≤¥ , (2.4)
IndexTransformsWiththeProductofBesselFunctions 7
||f||L0 ≤sup K02(√2x)x1−n ||f||n ,1. (2.5)
x 0
≥ h i
Proof. Infact,withthedefinitionofthenorm(2.1)andtheHo¨lderinequalityweobtain
¥ ¥ 1/q p
||f||L0 = 0 K02(√2x)|f(x)|dx≤ 0 K02q(√2x)x(1−n )q−1dx ||f||n ,p, q= p 1 (2.6)
Z (cid:18)Z (cid:19) −
and the latter integralvia asymptotic behaviorof the Macdonald function(1.5), (1.6), (1.7) convergesfor
n <1. Henceintegral(1.8)andthegeneralizedMinkowskiinequalityyield
¥ 1/q ¥ ¥ 2q 1/q
K02q(√2x)x(1−n )q−1dx = x(1−n )q−1 e−√2x coshudu dx
(cid:18)Z0 (cid:19) Z0 (cid:18)Z0 (cid:19) !
¥ ¥ 1/q 2
x(1 n )q 1e 2q√2x coshudx du
− − −
≤ Z0 (cid:18)Z0 (cid:19) !
¥ du 2
=22(n 1/p)(2q)4(n 1)G 2/q(2q(1 n )) .
− − − (cid:18)Z0 cosh2(1−n )u(cid:19)
Calculatingthe integralwith thehyperbolicfunctionviarelation(2.4.4.4)in [3], Vol. I,we comeupwith
theestimate(2.4). Forthecase p=1weendupimmediatelywith(2.5),using(2.6),wherethesupremum
isfiniteviatheconditionn <1. Thustheembedding(2.3)isestablishedandLemma3isproved.
(cid:3)
Theorem 1. The index transform (1.1) is well-defined as a bounded operator from L0 into the space
C (R)ofboundedcontinuousfunctionsvanishingatinfinity. Besides,thefollowingcompositionrepresen-
0
tationholds
(Ff)(t )=(F(K f)(cosht))(t ), (2.7)
c 0
wheretheFouriercosinetransformF isdefinedby(1.14)and
c
2 ¥
(K f)(x)= Re K 4ep i/4√xt f(t)dt (2.8)
0 −rp Z0 0
h (cid:16) (cid:17)i
istheoperatoroftheMeijertypeK-transform(cf. [4]).
Proof. Infact,theinequality(1.25)implies
4 ¥ 4
||Ff||C0(R)=tsuRp|(Ff)(t )|≤ p 0 K02(√2x)|f(x)|dx= p ||f||L0 <¥ ,
Z
∈
whichmeansthattheoperator(1.1)iswell-definedandtheintegralconvergesabsolutelyanduniformlywith
respecttot R.Thus(Ff)(t )iscontinuous.Ontheotherhand,recalling(1.15)andCorollary3,wederive
∈
2 ¥ ¥ 4
(Ff)(t ) K 2√2xcosht f(x)dxdt f <¥ .
| |≤ p 0 0 0 | | ≤ p || ||L0
Z Z
(cid:16) (cid:17)
Hence in view of Fubini’s theorem one can invert the order of integration in the corresponding iterated
integral and arrive at the composition (2.7). Moreover, the previous estimate says that (K f)(cosht)
0
L (R). Consequently,(Ff)(t )vanishesatinfinityowingtotheRiemann-Lebesguelemma. (cid:3)∈
1
8 S.Yakubovich
Corollary4. Theoperator(1.1) F :Ln ,p(R+) Lp(R), p 2, n <1isboundedand
→ ≥
p 1/p 1 p
||Ff||Lp(R)≤ 2− q2(n −1)[G (q(1−n ))]2/qB(1−n , 1−n )||f||n ,p, q= p 1. (2.8)
−
Proof. Indeed, taking the composition(2.7), we appealto the Hausdorff-Younginequality for the Fourier
cosinetransform(1.14)(cf.[5],Theorem74)
p
||Fcf||Lq(R)≤(2p )1/q−1/2||f||Lp(R), 1<p≤2, q= p 1,
−
wefind
¥ 1/q
||Ff||Lp(R)≤√2p 1/p−1/2 0 |(K0f)(cosht)|qdt . (2.9)
(cid:18)Z (cid:19)
Hencebythe generalizedMinkowskiandHo¨lderinequalitieswith relation(2.16.2.2)from[3], Vol. IIwe
obtain
¥ 1/q ¥ ¥ 1/q
√2p 1/p 1/2 (K f)(cosht)qdt 2p 1/p 1 f(x) Kq 2√2x cosh(t/2) dt dx
− 0 | 0 | ≤ − 0 | | 0 0
(cid:18)Z (cid:19) Z (cid:18)Z (cid:16) (cid:17) (cid:19)
¥ ¥ ¥ 1/q
21/q+1p 1/p 1 f(x) e 2q√2xcoshtcoshudt dudx
− −
≤ 0 0 | | 0
Z Z (cid:18)Z (cid:19)
¥ ¥
=21/q+1p 1/p 1 f(x) K1/q 2q√2xcoshu dudx
− 0 0 | | 0
Z Z
¥ ¥ (cid:16) (cid:17) 1/q
21/q+1p 1/p−1 f n ,p x(1−n )q−1K0 2q√2x coshu dx du
≤ || || 0 0
Z (cid:18)Z (cid:16) ¥ du(cid:17) (cid:19)
=2n p 1/p−1q2(n −1)[G (q(1−n ))]2/q||f||n ,pZ0 cosh2(1−n )u
=2−n p 1/p−1q2(n −1)[G (q(1 n ))]2/qB(1 n , 1 n ) f n ,p.
− − − || ||
Consequently,combiningwith(2.9),weget(2.8). (cid:3)
Nowweinvestigatethecompactnessoftheoperator(1.1).
Theorem2. Theoperator(1.1) F :Ln ,p(R+) Lq(R), 1<p 2,n <1, q=p/(p 1)iscompact.
→ ≤ −
Proof. Theproofisbasedonapproximationoftheoperator(1.1)byasequenceofcompactoperatorsofa
finiterankwithcontinuouskernelsofcompactsupport. Buttoachievethisgoal,itissufficienttoverifythe
followingHilbert-Schmidt-typecondition
¥ ¥
Y t (x)qx(1−n )q−1dt dx<¥ . (2.10)
0 ¥ | |
Z Z−
Infact,similarlyasabovewerecall(1.15)andthegeneralizedMinkowskiinequalitytodeduce
¥ ¥ 1/q
Y t (x)qx(1−n )q−1dt dx
0 ¥ | |
(cid:18)Z Z− (cid:19)
¥ ¥ 1/(p 1) 1/q
≤21/2+1/pp 1/q−1/2 Z0 x(1−n )q−1(cid:18)Z0 K0p(2√2xcosht)dt(cid:19) − dx!
¥ ¥ q 1/q
21/2+1/pp 1/q 1/2 x(1 n )q 1 K1/p(2p√2xcosht)dt dx
≤ − 0 − − 0 0
(cid:18)Z (cid:18)Z (cid:19) (cid:19)
IndexTransformsWiththeProductofBesselFunctions 9
¥ ¥ 1/q
≤21/2+1/pp 1/q−1/2 0 0 x(1−n )q−1K01/(p−1)(2p√2xcosht)dx dt
Z (cid:18)Z (cid:19)
=2n 3/2p 1/q 1/2p2(n 1)B(1 n , 1 n )
− − −
− −
¥ 1/q
× 0 x2(1−n )q−1K01/(p−1)(x)dx ≤2n −3/2p 1/q−1/2p2(n −1)B(1−n , 1−n )
(cid:18)Z (cid:19)
¥ ¥ p 1 1/p
du x2(1−n )q−1e−xcoshu/(p−1)dx − =2−n +1/2−2/pp 1/q−1/2q2(n −1)G 1/q(2q(1 n ))
× Z0 (cid:18)Z0 (cid:19) ! −
B(1 n , 1 n )B1/p(p(1 n ), p(1 n ))<¥ .
× − − − −
(cid:3)
Anotherrepresentationofthetransform(1.1)canbegivenviatheParsevalequalityfortheMellintrans-
form(1.13)andtheMellin-Barnesintegralrepresentation(1.9). Infact,animmediateconsequenceofThe-
orem87in[5]andStirling’sasymptoticformulaforthegamma-functionis
Theorem3. Let f L1 n ,p(R+), 1<p 2.Thenforallt R
∈ − ≤ ∈
1 n +i¥ G (s/2)G ((s+it )/2)G ((s it )/2)
(Ff)(t )=−16p i√p n i¥ G ((1 s)/2) − f∗(1−s)ds. (2.11)
Z − −
Finallyinthissectionweinvestigatetheexistenceandboundednessoftheoperator(1.2),whichisthead-
jointof(1.1).Infact,itfollowsfromthegeneraloperatortheory.However,wewillproveitdirectly,getting
an explicit estimation of its norm. Assuming g(t ) L (R), 1< p 2 and recalling the asymptotic for-
p
∈ ≤
mula(1.5)fortheMacdonaldfunction,wefindthatforeachx>0thefunctionRe K 4ep i/4√xcosht
0
∈
Lp(R), 1< p 2. Hence via the Parseval theorem for the Fourier transform (chf. [(cid:16)5], Theorem 75)(cid:17)aind
≤
equality(1.15),operator(1.2)canbewrittenas
2 ¥
(Gg)(x)= Re K 4ep i/4√xcosht (Fg)(t)dt, x>0, (2.12)
− p ¥ 0
where(Fg)(t) L (R), q= pristZh−eFourhiert(cid:16)ransformofg (cid:17)i
∈ q p 1
−
1 ¥
(Fg)(x)= g(t)eixtdt (2.13)
√2p ¥
Z−
andtheintegralconvergesintheL -sense.
p
Theorem4. Letg L (R), 1<p 2. Thenoperator(1.2)iswell-definedandforallx>0
p
∈ ≤
1 1
|(Gg)(x)|≤21/(4p)−1p−1/(2p)x−1/(4p)B 4p, 4p ||g||Lp(R). (2.14)
(cid:18) (cid:19)
Proof. Indeed,taking(2.12),werecalltheHo¨lderinequality,theHausdorff-YounginequalityfortheFourier
transform(2.13)[5]andthegeneralizedMinkowskiinequalitytoobtain
2 ¥ 1/p
|(Gg)(x)|≤ p ¥ K0p(2√2xcosh(t/2))dt ||Fg||Lq(R)≤2p 1/q−1||g||Lp(R)
r (cid:18)Z− (cid:19)
¥ ¥ 1/p ¥ du
× 0 ¥ e−p√2xt2 coshudt du= 21−1/(4p)p−1/(2p)x−1/(4p)||g||Lp(R) 0 cosh1/(2p)u
Z (cid:18)Z− (cid:19) Z
1 1
=21/(4p)−1p−1/(2p)x−1/(4p)B 4p, 4p ||g||Lp(R),
(cid:18) (cid:19)
10 S.Yakubovich
whichproves(2.14).
(cid:3)
Theorem5. TheoperatorG:Lp(R) Ln ,r(R+), 1< p 2, q= p/(p 1),r 1, n >0isbounded
→ ≤ − ≥
and
||Gg||n ,r≤2n −1+1/r−2/pp 1/q−1G 1/rr2(n2Gn1r/)pG(22/npp(n)p) B(n , n ) ||g||Lp(R),
where p,rhavenodependence.
Proof. Againwith(2.12),thegeneralizedMinkowski,Ho¨lderinequalitiesandtheHausdorff-Younginequal-
ityfortheFouriertransform(2.13)wefind
2 ¥ ¥ 1/r
||Gg||n ,r≤rp Z−¥ |(Fg)(t)|(cid:18)Z0 xn r−1K0r(2√2xcosh(t/2))dx(cid:19) dt
2 ¥ ¥ p/r 1/p
≤21/prp ||Fg||Lq(R) Z−¥ (cid:18)Z0 xn r−1K0r(2√2xcosht)dx(cid:19) dt!
¥ dt 1/p ¥ 1/r
≤21−3n +1/rp 1/q−1||g||Lp(R) ¥ cosh2n pt 0 x2n r−1K0r(x)dx
(cid:18)Z− (cid:19) (cid:18)Z (cid:19)
¥ 1/r
=21−n +1/r−2/pp 1/q−1B1/p(n p, n p) 0 x2n r−1K0r(x)dx ||g||Lp(R)
(cid:18)Z (cid:19)
¥ ¥ 1/r
≤21−n +1/r−2/pp 1/q−1B1/p(n p, n p) 0 0 x2n r−1e−xrcoshudx du||g||Lp(R)
Z (cid:18)Z (cid:19)
¥ du
=21−n +1/r−2/pp 1/q−1r−2n G 1/r(2n r)B1/p(n p, n p) ||g||Lp(R) 0 cosh2n u
Z
=2n −1+1/r−2/pp 1/q−1r−2n G 1/r(2n r)B1/p(n p, n p)B(n , n ) ||g||Lp(R).
(cid:3)
3. INVERSION THEOREMS
The compositionrepresentation(2.11) and the propertiesof the Fourier and Mellin transformsare key
ingredientstoprovetheinversiontheoremfortheindextransform(1.1)(Ff)(t ). Namely,wehave
Theorem6. Let f(t) L1 n ,p(R+) L1((1,¥ );tdt), 1< p 2,0<n <1, q= p/(p 1),andletthe
Mellintransform(1.11)s∈atisf−ythecond∩ition f (1)=0. If,besid≤es,t ept F(t ) L (R ),t−henforallx>0
∗ 1 +
∈
thefollowinginversionformulaholds
4 d ¥ t (Ff)(t )
f(x)=−p dx 0 cosh(pt /2)Kit 2√2x ReJit 2√2x dt . (3.1)
Z
(cid:16) (cid:17) (cid:16) (cid:17)
Proof. In fact, since f L1 n ,p(R+) then by virtue of Theorem 86 in [5] its Mellin transform f∗(s)
L (1 n i¥ ,1 n +i¥∈). H−enceitisnotdifficulttoverifywiththeuseofHo¨lder’sinequalitythatintegra∈l
q
− − −
(2.11) converges absolutely. Moreover, taking the Fourier cosine transform (1.14) of both sides of this
equality,wechangetheorderofintegrationinitsright-handsidebyFubini’stheorem.Indeed,thisisbecause
the absolute convergence of the corresponding iterated integral can be immediately verified, employing
inequalities(1.18),(1.19).Then,recalling(1.16),wecomeupwiththeequality
¥ 1 n +i¥ G (s/2)G (s)
0 (Ff)(t )cos(t u)dt =−8i√p n i¥ G ((1 s)/2)f∗(1−s)(2coshu)−sds. (3.2)
Z Z − −