New inversion, convolution and Titchmarsh's theorems for the half-Hartley transform

0.09 MB·

by  Semyon Yakubovich

NEW INVERSION, CONVOLUTION AND TITCHMARSH’S THEOREMS FOR THE
HALF-HARTLEY TRANSFORM
S.YAKUBOVICH
4
1
0
2

r ABSTRACT. The generalizedParseval equality for the Mellin transformis employedto provethe in-
a
M versiontheoreminL2 withtherespectiveinverseoperatorrelatedtotheHartleytransformonthenon-
negativehalf-axis(thehalf-Hartleytransform). Moreover,involvingtheconvolutionmethod,whichis

0 basedonthedoubleMellin-Barnesintegrals,thecorrespondingconvolutionandTitchmarsh’stheorems
1 forthehalf-Hartleytransformareestablished. Asanapplication,weconsidersolvabilityconditionsfor
ahomogeneousintegralequationofthesecondkindinvolvingtheHartleykernel.
]
A
C
. 1. INTRODUCTION AND AUXILIARY RESULTS
h
t
a Thefamiliarreciprocal pairoftheHartley transforms
m
1 ¥
[ (H f)(x)= [cos(xt)+sin(xt)]f(t)dt, x R, (1.1)
√2p ¥ ∈
2 Z−
v 1 ¥
3 f(x)= [cos(xt)+sin(xt)](H f)(t)dt (1.2)
4 √2p ¥
1 Z−
is well-known [1] in connection with various applications in mathematical physics. Mapping and
3
. inversion properties of these transforms in L as well as their multidimensional analogs were inves-
1 2
0 tigated, for instance, in [2], [3], [4]. These operators were treated as the so-called bilateral Watson
4
transform. Recently, theauthorfound thepaper[5] (seealso [6], [7]), where theattemptto invertthe
1
: HartleytransformwiththeintegrationoverR+
v
Xi 2 ¥
(H f)(x)= [cos(xt)+sin(xt)]f(t)dt, x R , (1.3)
ar + rp Z0 ∈ +
was undertaken. However,theinversionformulaobtainedby theauthors is dependingon theFourier
transformoftheimageand,indeed,needstobeimproved. Herewewillachievethismaingoal,prov-
ing theinversiontheorem fortransformation(1.3) in L (R ). Moreover, wewill construct and study
2 +
properties of the convolution operator, related to the half-Hartley transform by general convolution
method developed by the author in 1990, and which is based on the double Mellin-Barnes integrals
[8], [9], [4]. Namely,wewill provetheconvolutiontheoremand Titchmarsh’stheorem aboutthe ab-
senceofdivisorsintheconvolutionproduct. Finally,weapplythehalf-Hartleytransform(1.3)tofind
2000MathematicsSubjectClassification. 44A15,44A35,45E05,45E10.
Keywordsandphrases. Hartleytransform,convolutionmethod,Mellintransform,Titchmarshtheorem,homogeneous
integralequation.
1
2 S.YAKUBOVICH
solvability conditions and the form of solutions for a homogeneous integral equation of the second
kind.
WenoteinthissectionthatournaturalapproachwillinvolvetheL -theoryoftheMellintransform
2
[10]
¥ 1
(M f)(s)= f (s)= f(t)ts 1dt, s s = s C,s= +it , (1.4)
∗ −
0 ∈ { ∈ 2 }
Z
where the integral is convergent in the mean square sense with respect to the norm in L (s ). Recip-
2
rocally,theinversionformulatakes place
1
f(x)= f (s)x sds, x>0 (1.5)
2p i s ∗ −
Z
with the convergence of the integral in the mean square sense with respect to the norm in L (R ).
2 +
Furthermore, forany f , f L (R ) thegeneralized Parsevalidentityholds
1 2 2 +
∈
¥ 1
f (xt) f (t)dt = f (s)f (1 s)x sds, x>0 (1.6)
0 1 2 2p i s 1∗ 2∗ − −
Z Z
withParseval’sequalityofsquares ofL -norms
2
¥ ¥ 2
1 1
f(x) 2dx= f +it dt . (1.7)
0 | | 2p ¥ ∗ 2
Z Z− (cid:12) (cid:18) (cid:19)(cid:12)
(cid:12) (cid:12)
2. INVERSION (cid:12)THEOREM (cid:12)
(cid:12) (cid:12)
We begin with the following inversion theorem for the half-Hartley transform (1.3). Precisely, it
has
Theorem1. Thehalf-Hartleytransform(1.3)extendstoaboundedinvertiblemapH :L (R )
+ 2 +
L (R )and foralmostall x R thefollowingreciprocalformulashold →
2 + +
∈
2 d ¥ f(t)
(H f)(x)= [1+sin(xt) cos(xt)] dt, x R , (2.1)
+ rp dxZ0 − t ∈ +
2 ¥
f(x)= [sin(xt)S(xt)+cos(xt)C(xt)](H f)(t)dt, (2.2)
p +
r Z0
where S(x), C(x)areFresnelsin-andcosine-integrals,respectively,
2 √x 2 √x
S(x)= sin(t2)dt, C(x)= cos(t2)dt
p p
r Z0 r Z0
and integral (2.2) converges with respect to the norm in L (R ). Finally, the norm inequalities take
2 +
place
√2 f H f 2 f . (2.3)
|| ||L2(R+) ≤|| + ||L2(R+) ≤ || ||L2(R+)
Proof. Let f belong to the spaceC(2)(R ) of continuously differentiable functions of compact sup-
c +
port,whichisdenseinL (R ). Thenintegratingbypartsin(1.4),wefindthats2f (s)isboundedon
2 + ∗
s and therefore f (s) L (s ) L (s ). Hence mindingtheknownformulas[10]
∗ 2 1
∈ ∩
¥ sint G (s) p s
ts 1dt = cos , s s , (2.4)
−
0 t 1 s 2 ∈
Z − (cid:16) (cid:17)
THEHALF-HARTLEYTRANSFORM 3
¥ 1 cost G (s) p s
− ts 1dt = sin , s s , (2.5)
−
0 t 1 s 2 ∈
wecall thegeneralized ParseZvalequality(1.6)to der−ivefor(cid:16)allx(cid:17)>0
2 ¥ f(t)
[1+sin(xt) cos(xt)] dt
rp Z0 − t
2 1 p s p s x1 s
= G (s) sin +cos f (1 s) − ds. (2.6)
p 2p i s 2 2 ∗ − 1 s
It is easily seen the porssibilitytZo differehntia(cid:16)te th(cid:17)rough w(cid:16)ith r(cid:17)esipect to x in−equality (2.6). Thus com-
biningwith(1.3),wederive(2.1)togetherwiththeequality
2 1 p s p s
(H f)(x)= G (s) sin +cos f (1 s)x sds, (2.7)
+ p 2p i s 2 2 ∗ − −
r Z
h (cid:16) (cid:17) (cid:16) (cid:17)i
which is valid for any f C(2)(R ). Furthermore, from (1.7) one immediately obtains the norm
c +
∈
estimates
1/2
2 1 ¥ pt 1 1 2
H f = cosh2 G +it f +it dt
|| + ||L2(R+) √p 2p ¥ 2 2 ∗ 2 !
Z− (cid:16) (cid:17)(cid:12)(cid:12) (cid:18) (cid:19) (cid:18) (cid:19)(cid:12)(cid:12)
(cid:12) 1/2 (cid:12)
1 ¥ cosh2(pt /2) (cid:12) 1 2 (cid:12)
=2 f +it dt
2p ¥ cosh(pt ) ∗ 2 !
Z− (cid:12) (cid:18) (cid:19)(cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12) 1/2
1 ¥ cosh2(pt /2) (cid:12)1 2 (cid:12)
=2 f +it dt 2 f
2p ¥ 2cosh2(pt /2) 1 ∗ 2 ! ≤ || ||L2(R+)
Z− − (cid:12) (cid:18) (cid:19)(cid:12)
(cid:12) (cid:12)
and plainly (cid:12) (cid:12)
(cid:12) (cid:12)
H f √2 f .
|| + ||L2(R+) ≥ || ||L2(R+)
Thus we proved (2.3) for any f C(2)(R ). Further, since C(2)(R ) is dense in L (R ), there
c + c + 2 +
is a unique extension of H as a∈n invertible continuous map H : L (R ) L (R ). Now, let
+ + 2 + 2 +
→
f L (R ). Thereisasequence f , f C(2)(R )suchthat f f 0, n ¥ . Denoting
∈ 2 + { n} n∈ c + || n− ||L2(R+)→ →
by
2 1 p s p s
h (x)= G (s) sin +cos f (1 s)x sds, (2.8)
n p 2p i s 2 2 n∗ − −
r Z
we observeby virtue of (2.3) that h is ahCau(cid:16)chy s(cid:17)equenc(cid:16)e and(cid:17)iit has a limitin L (R ), which we
n 2 +
{ }
willcall h. Hence, integratingthroughin (2.8), wehave
x 2 x 1 p s p s
h (y)dy= G (s) sin +cos f (1 s)y sds dy. (2.9)
Z0 n rp Z0 (cid:18) 2p iZs 2 2 n∗ − − (cid:19)
h (cid:16) (cid:17) (cid:16) (cid:17)i
In themeantime,by theSchwarz inequality
x
[h (y) h(y)]dy √x h h 0, n ¥
0 n − ≤ || n− ||L2(R+) → →
Z
4 S.YAKUBOVICH
and intheright-handsideof(2.9)onecan changetheorderofintegrationby Fubini’stheorem. Then
passing to the limit when n ¥ under integral signs in the obtained equality due to the Lebesgue
→
dominatedconvergencetheorem,wefind
x 2 1 p s p s x1 s
h(y)dy= G (s) sin +cos f (1 s) − ds. (2.10)
Z0 rp 2p iZs h (cid:16) 2 (cid:17) (cid:16) 2 (cid:17)i ∗ − 1−s
Differentiatingby xin (2.10),wecomeoutwiththeequalityforalmostallx>0
2 1 d p s p s x1 s
h(x) (H f)(x)= G (s) sin +cos f (1 s) − ds, (2.11)
≡ + p 2p idx s 2 2 ∗ − 1 s
r Z h (cid:16) (cid:17) (cid:16) (cid:17)i −
which coincides with (2.7) for any f C(2)(R ). Consequently, appealing to (2.4), (2.5), (1.6) and
c +
∈
(2.6), wecompletetheproofofrepresentation(2.1).
Finally, we establish the inversion formula (2.2). To do this, we denote by h (s) the Mellin trans-
∗
form ofh(t)(1.4)in L andwritereciprocally to(2.11)foralmostall x>0
2
p 1 d p s p s 1 x1 s
f(x)= G 1(1 s) sin +cos − h (1 s) − ds.
2 2p idx s − − 2 2 ∗ − 1 s
r Z h (cid:16) (cid:17) (cid:16) (cid:17)i −
Meanwhile with the supplement formula for gamma-functions and elementary trigonometric manip-
ulationsit becomes
1 1 d sin(p s)[sin(p s/2)+cos(p s/2)] x1 s
f(x)= G (s) h (1 s) − ds
√2p 2p idx s 1+sin(p s) ∗ − 1 s
Z −
1 1 1 d sin(p s/2)+cos(p s/2) x1 s
= (H h)(x) G (s) h (1 s) − ds
2 + −√2p 2p idx s 1+sin(p s) ∗ − 1 s
Z −
1 1 1 G (s)
= (H h)(x) h (1 s)x sds,
2 + −√2p 2p i s sin(p s/2)+cos(p s/2) ∗ − −
Z
1 1 1 s 1 3 s
= (H h)(x) G (s)G + G h (1 s)x sds,
2 + −2p √p 2p i s 2 4 4−2 ∗ − −
Z (cid:18) (cid:19) (cid:18) (cid:19)
wherethedifferentiationundertheintegralsignisallowedviatheabsoluteanduniformconvergence.
But thegeneralized Parsevalidentity(1.6)yields
1 1 ¥
f(x)= (H h)(x) k(xt)h(t)dt, (2.12)
2 + −2p √p 0
Z
where
1 s 1 3 s
k(x)= G (s)G + G x sds, x>0.
2p i s 2 4 4−2 −
Z (cid:18) (cid:19) (cid:18) (cid:19)
Hence employingagain (1.6)and using relations(8.4.2.5), (8.4.3.1)in[11], Vol. 3, thelatterintegral
can bewrittenin theform
¥ e x/t dt ¥ e xt√t
− −
k(x)=2 =2 dt
0 1+t2 √t 0 1+t2
Z Z
and itiscalculated viarelation(2.3.7.10)in [11], Vol. 1,namely,
k(x)=p √2[sinx+cosx] 23/2p [sinxS(x)+cosxC(x)],
−
THEHALF-HARTLEYTRANSFORM 5
where S(x), C(x) are Fresnel sin- and cosine- integrals (see above). Hence, substitutingthe value of
k(x) in (2.12) and making use (2.1), we come out with the inversion formula (2.2) and and complete
theproofofTheorem 1. (cid:3)
Remark 1. Taking into account the value of the integral (2.5.5.1) in [11], Vol. 1 and relation
(7.14.2.75)in[11], Vol. 3, wehave
2 1 x cost
[sin(x)S(x)+cos(x)C(x)]= dt
p p √x t
r Z0 −
1
= ei(x p /4)erf eip /4√x +e i(x+p /4)erfi eip /4√x ,
2√p − −
h (cid:16) (cid:17) (cid:16) (cid:17)i
where erf(z) and erfi(z) are the error function and the error function of imaginary argument, respec-
tively,
2 z 2 z
erf(z)= e t2dt, erfi(z)= et2dt.
√p − √p
0 0
Z Z
Therefore, inversionformula(2.2)can bewrittenas
1 ¥
f(x)= ei(xt p /4)erf eip /4√xt +e i(xt+p /4)erfi eip /4√xt (H f)(t)dt.
2√p − − +
0
Z
h (cid:16) (cid:17) (cid:16) (cid:17)i
3. CONVOLUTION OPERATOR FOR THE HALF-HARTLEY TRANSFORM
In this section we will construct and study mapping properties of the convolution, related to the
transformation(1.3). Followingthegeneral convolutionmethoddeveloped forintegral transforms of
theMellinconvolutiontype(cf. [8], [9], [4])wehave
Definition 1. Let f,g be functions from R into C and f , g be their Mellin transforms (1.4).
+ ∗ ∗
Then thefunction f g beingdefined on R bythedoubleMellin-Barnesintegral
+
∗
1 G (s)G (w) sin(p (s+w)/2)+cos(p (s w)/2)
(f g)(x)= −
∗ (2p i)2 s s G (s+w 1/2) sin(p (s+w)/2)
Z Z −
f (1 s)g (1 w)xs+w 3/2dsdw (3.1)
∗ ∗ −
× − −
iscalledtheconvolutionof f andgrelatedtothehalf-Hartleytransform(1.3)(providedthatitexists).
Theconvolutiontheorem forthehalf-Hartley transformcan bestatedas
Theorem 2. Let f , g betheMellintransformsof f,g,respectively, satisfyingconditionssf (s),
∗ ∗ ∗
sg (s) L (s ). Thentheconvolution(3.1) f gexistsandbelongstoL (R )withthenormestimate
∗ 2 2 +
∈ ∗
2 ¥ 1/2
||f ∗g||L2(R+) ≤4 p ¥ |(1/2+iq )g∗(1/2+iq )|2dq
r (cid:18)Z− (cid:19)
¥ 1/2
(1/2+it )f (1/2+it ) 2dt dq . (3.2)
∗
× ¥ | |
(cid:18)Z− (cid:19)
Moreover, itsMellintransform(M (f g))(1 s)isequalto
∗ −
√2
(M(f g))(1 s)= G (s w+1/2)cos(p (s w)/2)G (w)
∗ − 2p iG (s)sin(p (s+1/2)/2) s − −
Z
6 S.YAKUBOVICH
sin(p (w+1/2)/2)f (1/2 s+w)g (1 w)dw, s s . (3.3)
∗ ∗
× − − ∈
Besides,thefactorizationequalityholds
xp
(H (f g))(x)= (H f)(x)(H g)(x), x>0 (3.4)
+ + +
∗ 2
r
aswell as thegeneralizedParsevaltypeidentity
¥
(f g)(x)= [sin(xt)S(xt)+cos(xt)C(xt)]√t (H f)(t)(H g)(t)dt, (3.5)
+ +
∗ 0
Z
where integral(3.5)converges in theL -sense.
2
Proof. In fact, for s=1/2+it ,w=1/2+iq , (t ,q ) R weobtain
∈
G (s)G (w) sin(p (s+w)/2)+cos(p (s w)/2) G (1/2+it )G (1/2+iq )
− =2
G (s+w 1/2) sin(p (s+w)/2) G (1/2+i(t +q ))
(cid:12) − (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) cosh(pt /2)cosh(pq /2) cosh1/2(p (t +(cid:12)q ))co(cid:12)sh(pt /2)cosh(pq /2) (cid:12)
(cid:12) =2√p (cid:12) (cid:12) (cid:12)
× cosh(p (t +q )/2) cosh1/2(pt )cosh1/2(pq )cosh(p (t +q )/2)
cosh(pt /2)cosh(pq /2)
2√2p 4√2p . (3.6)
≤ cosh1/2(pt )cosh1/2(pq ) ≤
Meanwhile, plainly via conditions sf (s),sg (s) L (s ) we have f (s), g (w) L (s ). Therefore
∗ ∗ 2 ∗ ∗ 1
∈ ∈
by Fubini’s theorem the double integral (3.1) is equal to the corresponding iterated integrals. Then
making in (3.1) the simple substitution z = s+w 1/2, using elementary trigonometric formulas,
−
(1.5), (1.7) and the above estimate, we easily come out with (3.3) and the estimate (3.2). Namely, it
has
1 ¥ 1/2
||f ∗g||L2(R+) = 2p ¥ |(M(f ∗g))(1/2+it )|2dt
(cid:18) Z− (cid:19)
2 ¥ ¥ 1/2 4 ¥ dq 1/2
g (1/2+iq ) f (1/2+i(t q ) 2dt dq
≤ p ¥ | ∗ | ¥ | ∗ − | ≤ p ¥ q 2+1/4
Z− (cid:18)Z− (cid:19) (cid:18)Z− (cid:19)
¥ 1/2 ¥ 1/2
(1/2+iq )g (1/2+iq ) 2dq (1/2+it )f (1/2+it ) 2dt dq
∗ ∗
× ¥ | | ¥ | |
(cid:18)Z− (cid:19) (cid:18)Z− (cid:19)
2 ¥ 1/2 ¥ 1/2
=4 (1/2+iq )g (1/2+iq ) 2dq (1/2+it )f (1/2+it ) 2dt dq ,
p ∗ ∗
¥ | | ¥ | |
r (cid:18)Z− (cid:19) (cid:18)Z− (cid:19)
wheretheSchwarz andgeneralized Minkowskiinequalitiesareemployed. Thefactorization equality
(3.4)comesimmediatelyfrom (3.3)and(1.6)because
2 1
(H (f g))(x)= (M(f g))(1 z)G (z)sin(p (z+1/2)/2)x zdz
+ ∗ √p 2p i s ∗ − −
Z
2 2
= G (z w+1/2)cos(p (z w)/2)G (w)sin(p (w+1/2)/2)
p (2p i)2 s s − −
r Z Z
f (1/2 z+w)g (1 w)x zdwdz
∗ ∗ −
× − −
THEHALF-HARTLEYTRANSFORM 7
2 2√x
= G (s)cos(p ((s 1/2)/2)G (w)sin(p (w+1/2)/2)
p (2p i)2 s s −
r Z Z
xp
f (1 s)g (1 w)x s wdwdz= (H f)(x)(H g)(x).
∗ ∗ − − + +
× − − 2
r
Finally,thegeneralized Parsevalidentity(3.5)isadirect consequenceoftheinversionformula(2.2).
(cid:3)
This section ends with an analog of the Titchmarsh theorem about the absence of divisors of zero
inconvolution(3.1). Wehave
Theorem3. Let f , g satisfyconditionsep s f (s), ep sg (s) L (s ). Thenif(f g)(x)=0, x>
∗ ∗ | | ∗ | | ∗ 1
0, theneither f(x)=0 org(x)=0on R . ∈ ∗
+
Proof. In fact, theintegral
1 G (s)G (w) sin(p (s+w)/2)+cos(p (s w)/2)
F(z)= −
(2p i)2 s s G (s+w 1/2) sin(p (s+w)/2)
Z Z −
f (1 s)g (1 w)zs+w 3/2dsdw
∗ ∗ −
× − −
represents an analytic function in the domain D = z C : argz < p , since under condition of
the theorem it converges uniformly for any z C {: ∈z a|> 0,| argz} < p . Precisely, we have
(s=1/2+it , w=1/2+iq , zs+w 3/2 = z 1/∈2e(t +q )|arg|z≥)via(3.6)| |
− −
| |
G (s)G (w) sin(p (s+w)/2)+cos(p (s w)/2)
−
s s G (s+w 1/2) sin(p (s+w)/2)
Z Z (cid:12) −
(cid:12)
(cid:12) f (1 s)g (1 w)zs+w 3/2dsdw
(cid:12) ∗ ∗ −
× − −
4 2p ¥ ¥ ep [t +q ] f (1/2+it )g (1/2+(cid:12)(cid:12)(cid:12)iq ) dt dq <¥ .
| | | | ∗ ∗
≤ a ¥ ¥ | |
r Z− Z−
Moreover, (3.1) yields that F(x) = (f g)(x). Thus by virtue the uniqueness theorem for analytic
∗
functions F(z)=(f g)(z),z D. Moreover, equality (3.4) holds for z D, where the main branch
∗ ∈ ∈
ofthesquareroot ischosen. Hencecalling(2.7), wededuce
1 p s p s
(H f)(z) G (s) sin +cos f (1 s)z sds
| + |≤ p √2p s 2 2 ∗ − −
Z
(cid:12) h (cid:16) (cid:17) (cid:16) (cid:17)i (cid:12)
1 ¥ cosh(pt /2) f (1/2+(cid:12)(cid:12) it ) ep t dt 1 2 ¥ f (1/2+it ) ep (cid:12)(cid:12)t dt <¥ ,
≤ p √a ¥ cosh1/2(pt )| ∗ | | | ≤ p a ¥ | ∗ | | |
Z− r Z−
whichmeansthat(H f)(z)isanalyticinD. Therefore, if(f g)(x)=0, x>0,thenviatheunique-
+
∗
nesstheorem (f g)(z) 0, z D and (3.4)yields
∗ ≡ ∈
(H f)(z)(H g)(z)=0, z D.
+ +
∈
Since the left-hand side of the latter equality is the product of analytic functions in D, it means that
either(H f)(z) 0,or(H g)(z) 0inD. Finally,weobservethatunderconditionsofthetheorem
+ +
f,g L (R )and≡fromoperational≡propertiesoftheinverseMellintransform(1.5)itfollowsthat f,g
2 +
∈
8 S.YAKUBOVICH
are infinite times differentiable functions. Thus appealing to Theorem 1 and inversion formula (2.2)
wefind thateither f =0 org=0 onR . (cid:3)
+
4. A HOMOGENEOUS INTEGRAL EQUATION INVOLVING THE HALF-HARTLEY TRANSFORM
HerewewillestablishsolvabilityconditionsinL (R ),concerningthefollowingintegralequation
2 +
ofthesecondkind
¥
2
[cos(xt)+sin(xt)]f(t)dt =l f(x), x R , l C. (4.1)
p +
r Z0 ∈ ∈
Themainresultofthesectionis
Theorem 4. Let l <√2. In order to an arbitraryfunction f L (R ) be a solution of integral
2 +
| | ∈
equation(4.1)itis necessaryto havetheform oftheintegralinthemeansquaresense
1 p 1 p s p s
f(x)= l + sec +csc j (s)x sds, x>0 (4.2)
2p i s 2 G (1 s) 2 2 −
Z (cid:20) r − h (cid:16) (cid:17) (cid:16) (cid:17)i(cid:21)
in terms of some function j (s), satisfying condition j (s)=j (1 s), s s , i.e. j (1/2+it ) is even
with respect tot R. Besides,its Mellin’stransform − ∈
∈
p 1 p s p s
f (s)= l + sec +csc j (s), s s (4.3)
∗ 2 G (1 s) 2 2 ∈
(cid:20) r − h (cid:16) (cid:17) (cid:16) (cid:17)i(cid:21)
aswell as j belongto L (s )andthefollowingL -normestimateshold
2 2
1
(2+|l |)−1||f∗||L2(s ) ≤||j ||L2(s ) ≤ √2−|l | − ||f∗||L2(s ). (4.4)
The condition j (s)=j (1 s), j L (s ) and theform(cid:16)of solutio(cid:17)ns(4.2)are also sufficient for any
2
y astheinverseMellintra−nsformo∈fj , satisfyingtheintegralequation
2 ¥ y (t)
(l 2 2)y (x) dt =0, x R , (4.5)
− −p 0 x+t ∈ +
Z
where integral(4.5)converges absolutely.
Proof. Let f beasolutionofequation(4.1). Thenviaformulas(2.1)and (2.6)wederivetheequality
x 2 1 p s p s x1 s
l f(y)dy= G (s) sin +cos f (1 s) − ds. (4.6)
Z0 rp 2p iZs h (cid:16) 2 (cid:17) (cid:16) 2 (cid:17)i ∗ − 1−s
In themeantime,itsleft-hand sideis equalto (see(1.6))
x l x1 s
l f(y)dy= f (s) − ds. (4.7)
0 2p i s ∗ 1 s
Z Z −
Hence, comparing right-hand sides of (4.6), (4.7) and since both integrand are from L (s ) they are
1
equal byvirtueofTh. 32in [10]. Hence
2 p s p s
l f (s)= G (s) sin +cos f (1 s), s s . (4.8)
∗ p 2 2 ∗ − ∈
r
h (cid:16) (cid:17) (cid:16) (cid:17)i
THEHALF-HARTLEYTRANSFORM 9
But 1 s s . Therefore, changings on 1 sin (4.8),it becomes
− ∈ −
2 p s p s
l f (1 s)= G (1 s) sin +cos f (s). (4.9)
∗ − p − 2 2 ∗
r
h (cid:16) (cid:17) (cid:16) (cid:17)i
Subtracting (4.9) from (4.8) and then using the supplement and duplication formulas for gamma -
functions,weobtain
1 1
l [f (s) f (1 s)]=√2p +
∗ − ∗ − G (s/2)G (1 s/2) G ((1 s)/2)G ((1+s)/2)
(cid:20) − − (cid:21)
G ((1+s)/2) G (s/2)
[G (s)f (1 s) G (1 s)f (s)]=2s 1/2f (1 s) +
× ∗ − − − ∗ − ∗ − G (1 s/2) G ((1 s)/2)
(cid:20) − − (cid:21)
G ((1 s)/2) G (1 s/2)
21/2 sf (s) − + − .
− − ∗ G (s/2) G ((1+s)/2)
(cid:20) (cid:21)
Hence
G ((1 s)/2) G (1 s/2)
f (s) l +21/2 s − + −
∗ − G (s/2) G ((1+s)/2)
(cid:20) (cid:20) (cid:21)(cid:21)
G ((1+s)/2) G (s/2)
= f (1 s) l +2s 1/2 + , s s , (4.10)
∗ − − G (1 s/2) G ((1 s)/2) ∈
(cid:20) (cid:20) − − (cid:21)(cid:21)
or, since
p 1 p s p s 2cosh(pt /2)
√2 sec +csc = 2, s=1/2+it , t R, (4.11)
≤ 2 G (s) 2 2 cosh1/2(pt ) ≤ ∈
r (cid:12) (cid:12)
(cid:12) h (cid:16) (cid:17) (cid:16) (cid:17)i(cid:12)
(4.10)undercon(cid:12)dition l <√2 yields (cid:12)
(cid:12) (cid:12)
| |
p 1 p s p s −1
f (s) l + sec +csc
∗ 2 G (1 s) 2 2
(cid:20) r − h (cid:16) (cid:17) (cid:16) (cid:17)i(cid:21)
p 1 p s p s −1
= f (1 s) l + sec +csc =j (s), s s . (4.12)
∗ − 2 G (s) 2 2 ∈
(cid:20) r (cid:21)
h (cid:16) (cid:17) (cid:16) (cid:17)i
Thus we find that j (s) = j (1 s), i.e is even with respect to t R, where s = 1/2+it . So, we
− ∈
established (4.3) and reciprocally in L the representation (4.2). Meanwhile estimates (4.11) yield
2
(4.4).
Now, let us assume for some j L (s ) in (4.3) the condition j (s) = j (1 s), s s . Then
2
∈ − ∈
substitutingthevalueof f (s)in (4.3)intoequation(4.8),weobtain
∗
p 1 p s p s 2 p s p s
l l + sec +csc j (s)= G (s) sin +cos
2 G (1 s) 2 2 p 2 2
(cid:20) r − h (cid:16) (cid:17) (cid:16) (cid:17)i(cid:21) r h (cid:16) (cid:17) (cid:16) (cid:17)i
p 1 p s p s
l + sec +csc j (s), s s ,
× 2 G (s) 2 2 ∈
(cid:20) r (cid:21)
h (cid:16) (cid:17) (cid:16) (cid:17)i
10 S.YAKUBOVICH
oraftersimplecalculationsitdrivesto theequation
2
(l 2 2)j (s) j (s)=0, s s . (4.13)
− −sin(p s) ∈
Taking the inverse Mellin transform (1.5) of both sides of the latter equality, we employ the general-
ized Parsevalequality(1.6)andrelation (8.4.2.5)in[11], Vol. 3 to find
2 ¥ y (t)
(l 2 2)y (x) dt =0, x>0.
− −p 0 x+t
Z
Thus f(x) by formula (4.2) is a solution of integral equation (4.1) for all j (s) under condition
j (s) = j (1 s) such that its inverse Mellin transform satisfies integral equation (4.5). The abso-
−
luteconvergenceofthecorrespondingintegralfollowsfromtheSchwarz inequality.
(cid:3)
Corollary1. Let l ( √2,√2). Then theonlytrivialsolutionsatisfiesintegralequation(4.1).
∈ −
Proof. In fact, since l 2 2 2/sin(p s)<0, s s , we have from (4.13) j (s) 0 on s . Therefore
− − ∈ ≡
from (4.3) it follows f (s) 0 and the inverse Mellin transform implies f = 0, i.e. the solution of
∗
(4.1)istrivial. ≡ (cid:3)
Acknowledgments
Thepresentinvestigationwassupported,inpart,bythe”CentrodeMatema´tica”oftheUniversityof
Porto.
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