Internat. J. Math. & Math. Sci.
Vol. 8 No. 2 (1985) 325-344
325
ANALYTIC REPRESENTATION OF THE DISTRIBUTIONAL FINITE
HANKEL TRANSFORM
O. P. SINGH
Uepa r tmen t of Ma thema t cs
Banaras Hindu University
Vararasi 221005, India
and
RAM S. PATHAK
Department of Mathemat
Faculty of Science
King Saud University
Riyadh, Saudi Arabia
(Received April 30, 1983)
ABSIRACT.
Var!ous
representations of finite Hankel transforFs of generalized
functions :re obtainec. One of the represe,tations is shown to be the limit of a
certain fami|y of regular generalized functions and ths limit is interpreted as a
process of truncation for the generalized functions (distributions). An inversion
thcurem for the: 5ereralized finite Hankel transform is established (in the
Oistributional se,se) which gives a Fourier-Bessel set ies Fepresentat ion of
generalized functions.
KEY WORDS AND PHRASES. Hankel tranc’ons
spaaes, Fnite ila,kel t.ransform.
o’ 9e,.eraized functions,
1980 MATHEMATICS SUBJECT CLASSIFICATION CODE.
Tec’t
’unctiw,,
46F12, 44A20, 44A15.
INTRODUCTION.
Zemanan [1] extended Hankel transformations to the distribution space H’. H’
is the dual of the space of testing functions H
defined as follows" for each real
I.
number
u,
H
y
,.u (@)
-m,...._
let
{
sup
(0,-)
Ixm
(
m
is smooth on
(x -I D) k [x --} (x)]
(0,-) and
<
for each m k
satisfies
0
2
(1.1)
(I 1)
o<x<o=
consists of certain distrbutions of slow growth. Then later [2] he
obtained a
more general result by removing the restriction on the
slow growth of the
distributions. He defined the Hankel transformation of a distribution of rapid
growth
in the space
is the dual of
the strict inductive limit of the testing
Hu
B’.p
B’
B,
O. P. SINGH AND R. S. PATHAK
326
Bp,
b tends to infinity through a
function spaces
b (defined in section 2) as
of
numbers.
positive
monotonically increasing sequence
are identically zero after
We take advantage of the fact that functions in
b
b, to define the finite Hankel transformation for the generalized functions in its
We find that for
dual B’,b
This is done by generalizing Parseval’s equation
>
B’u,b isomorphically onto the
the finite Hankel transform / p maps
The aim of the present
(defined in section 3)
generalized function space Y’
Bp,
-
p=per is to obtain various representations of the generalized functions in
Y’u,b
and
to find an invesion formula for the generalized finite Hanke] transform which also
gives another representation of the members of B’h,b as a Fourier-Bessel series.
We follow the notation and terminology of Schwartz [3] and Zemanidn [4,5]. Here
deotes the open interval (0,). The letters x,y,t and w are used as real
th
derivative of an ordinary or generalized function f(x) is
variab1s on I. The k
k
D
(though the symbol D Xk f(x) is also used)
f(x)
usually denoted by
denotes the space of smooth functions that have compact support on I. The topolog9
of D(!) is that which makes its dual the space D’(1) of Schwartz’s distributiu,s
[3, vol. I, p.65].
TESTING FUNCTION SPACZS g ,b AND Y ,b"
2.
Let b > 0 be a fixed arbitrary real number. Then for u
c
R, where R
is he
nez of re] numbers, we define
B
z(x) is smooth (x) : 0 for x
{
!(x -I D) k Ix -u-; q(x)]l
sup
o<x<.
, for each k
b and
0,1,2
satisfies
( I)}
(2.1)
B ,b is a linear space to hich we assign the topology generated by the countable set
is a sequentially complete countably multi-normed space [2].
of seminorms Yk"
O, the finite Harkel transform of a testing function
Classically, for u +
in
b is defined as
[u,b
B,
@(:L n
o
I b (x)
vn
x J
( n x)dx n=1,2,3
(2 2a)
J
der.otes the Bes3el function of the first kind of order h and
n
(bz)= 0 (arranged in ascending order of
are positive roots of
can be extended to the analytic function of the complex
However @,
where as usual
(n 1,2,3
magnitude).
variable z
J
yiw
by
..
(z)
f
(x)
j (xzdx.
(2.2)
(z) is ar analytic function on the finite z-plane except for a branch
at z
0 [4, p. 145
Henceforth, the finite Hankel transform of a testing
function
in
shall be defined as the na]ytic function @(z) given in (2.2)
and denoted by /
For a given real number b
0 Y
is the space of functions (z) which
satisfy:
is an even entire function of z and for each non-negative integer
k, the quantity
Note that
point
B,b
z--(z)
DISTRIBUTIONAL FINITE HANKEL TRANSFORM
u
b,k ()
e-blwl z 2k-(+1/2)
sup
z
327
(z)l
(2.3)
),k
Y u,b is the one generated by using the
The topology of
is a sequentially complete countably normed
as seminorms. Y
0,1,2,
space. For further properties el these spaces one can look into Zemanian [4, 2].
For a given testing function @ in Y u,b’ consider the function
finite.
is
k
Then for
-,
t,
-
f (y),’
Zemdnian
>-
d
Ip
h, [@].
(xy)dy
(2.4)
[2, Theorem I] has proved"
onto Y ,b"
is an isomorphism from B
},b
shall
Here isomorphism means topological isomorphism. Henceforth, the symbol
b used to denote a testing function in Y u,b whose pre-image is a testig function
For
Theerem 2
in
B, ,b"
,
h
i, Y
the classical inverse of the finite Hdnkel transform
For a given
(2.2a’ is a Fourier-Besse] series of the form, [6,7]
2
n=1
b
Since
’
J(x)n
(x/x n)
j2
t
+1
(bAn)
.
,(x n"
(2.5)
satisfies (2.3), we have
0,1,2
k
],2,3
n
2k-(+1/2)
n
2
is smooth and bounded on
is cc.nstant. Also
where
0
Consequently, the Fourier-Bessel series (2.5) converges
for
nx
O. Let us write
absolutely ano uniformly in x for all x
Ak ,
(x/n)} [Ju(XAn)/Jh+1(bAn)]
-.
Y(x)
2
Ju(x>. n
(x/)
n
j2 (b,kn)
+1
Z
n=
(,)
n
then
l(x
ib)k
x --I
/(x)l =2
E
n =I
(>n
J+k(XXn 2k+p(xX)--k
n
n
+l(bn)
xn
(X n
(2.6)
(XXn)--kJ
s smooth and
+k x
it follows that the right-hand side of (2.6)
for
bounded on (0,=)
0,1,2
Hence
converges absolutely and uniformly fcr all x 0 and for every k
0,1,2
the left-hand side is continuous and bounded on 0
for each
x
Since
is of rapid descent as
_>
-,
and
Xn
Hence
P(T)
Moreover,
(x
ID)k(x -1/2(x))
x
h
k
1/2Lake
X
0
2
+
akl
D..
X
2k-i
k
+
+ a
kk
D
--],
X
d
where the aki denote constants dr, C D
d" So we see that the Fourier-Bessel
series defines an infinitely differentiable function (x) satisfying
for
0
each k
2
But
b
may not be zero for x
may not be in B u,b as
(’)
But
(x)
lira
t-,o+
(x)(x)
c
B ,b’
O. P. SINGH AND R. S. PATHAK
328
(x)
where
0
for
e
is defined as"
b/4,
E(x/2),
0
I,
2
x (x): I-E (x-b+2 ),
2c
b
O,
x
x
<
_
2
2
b-2
x
<
b,
(2.7)
b,
.>_
and
u
o I exp(i/x(x-1))dx
E(u)
(2.8)
oI1exp(I/x(x-1))dx
Note that x (x) is a multiplier in
we have
@ s B
u
)
yk(
e
Now pick X such that 0
sup
x
<
k
(nk) Yn
z
n=o
X < 2.
_<
B ,b
I(x-ID) m A
(xl
-ZDJ n, c
(x)l
0
for each
(@)sup l(x -1 D) kO<x<b
<
<
b/4
since
for any
n;e(X)
Then
(R),
X<x<b
and
sup l(x
o<x<X
where A is a constant.
So we see that
( )
It is easy to see that
y<
im+
0
<- A sup
o<x<X
Also
E
(x)(x)
42 )]I
l(x-1D)m-l[x- lexP(x(2_x
, (x)
is smooth on
(x)
for any
(0,).
in
Hence
},
B
,b"
B ,b"
GENERALIZED FUNCTION SPACES B’
AND YI,b"
,b
The spaces
B’,b and Y’,b are the dual spaces of B ,b and
Y ,b
respectively. We shall use only the weak topology of B’,b’ that is the topology
assigned to it by the seminorms
p
B’u,b"
l<f,>
b, f
3.
B,
(f)
B ,b is a sequentially complete countably normed space, B’
is also
,b
sequentially complete [4, Theorem 1.83]. Similarly, we equip
with the weak
topology generated by the seminorms (F)
Y’,b"
b’ F
Y’,b is a sequentially complete space
We now construct a generalized function in B’,b which is not in D’(1)
Let
{T
a
be
monotone
sequence
of
increasing
numbers
b+l
with
positive
limit
For
n
e B
every
the
formula
,b’
Since
I<F >I
f,
n=l
+( n
is easily seen to define a generalized function
is an arbitrary testing
function
in
Y
D(1),
f
Y’,b
(3 1)
in
B’
then
Zn
On the other hand, if
(Tn) iS in general an
DISTRIBUTIONAL FINITE HANKEL TRANSFORM
329
infinite sum and it need not be convergent.
Note that
(i) B’,b
contains every regular distribution that corresponds to a function
which is Lebesgue integrable on 0
x < b. In this case we have
>
B ,b"
f
o f(x) (x) dx
f
is a tempered distribution whose support is contained in [X, ) for
(i i) If
some X 0 then f B’,b
(iii) Similarly, every regular distribution F, which can be defined by a locally
integrable function F(y) through the equation
fb
< F,
>
f" F(y)(y) dy
belo,gs to Y’,b" Note that F(y) need not be ntegrable
though typically it would be of slow growth, i.e., for some integer
O,
F(y) 0 as y
FINITE HANKEL TRANSFORMATION OF B’
for every
over 0
y
N
4
in
,
Y ,b’
.
y-N
e
Henceforth, we assume that
we define the fnte Hankel transform
-.
For f
Bu ,b and
B’,b’
h f
F
by
< f, >.
Hankel transform
F,
(x-k)
Example i.
equation (4.1)"
<
(x-k), (x)
<
(x-k),
I
oI
,
>,
for 0
k
(kz)
Y ,b’
From Theorem
is given by the
b,
J (xy)dy
(y) / J (ky)dy
This defines a regular distribution F(z)
(kz) J
(kz), for 0 k b.
h (x-k)
Example 2. The finite Ha[kel transform of 6(x-k)
h-iF.
f
onto Y’,b"
The fnite Hankel transfo of the delta function
(z)
u
(4.1)
The above equation also defines the nverse
2.1 one readily obtains:
Theorem 4.1
is an isomorphsm from B’
,b
h a(x-k),
=,z
>,
(using (2.4))
<
in
(kz)1/2J
Y’,b"
Consequently
for k
b
is the "Zero"
0 for all
generalized function in Y’
since <6(x-k) (x)>
in
(k)
b.
u,b
3.
The
finite Hankel transform of the generalized function in B’
defined
Example
,b
by (3.1) is the generalized function defined by the series
Bu,
F
s
n=l
since, for
@
Y
yVTn J
(y)
n
,b’
<
F,@
=n:r1 I (y) vny Ju(ny)dy
z
n--1
(n
f,
>
Suppose f is a regulaF generalized function, corresponding to a Lebesgue
Than the ordinary finite Hankel transform
integrable function over (O,b), in B’
,b"
of f is given by
Example 4.
o
’b
f(x)
/Xn x
J ( n x)dx; n
1,2 3
330
O. P. SINGH AND R. S. PATHAK
(O,b),
Since f is integrable over
the analytic function
olbf(x) J-f J
F(z)
We show that F
(f). Since f
/
f,
<
Since the integrand
0
x
b, 0
y
,
its finite Hankel transform may be extended to
is a regular generalized function,
f,@>
>
o
fb
f(x) q(x)dx
o
Ib
f(x)
(xy);J
f(x)+(y)
the order o
(zx)dx.
(xy)
(o/
@(y)
(us ing (2.4)).
is absolutely integrable over the domain
integration may be changed, and we obtain
of dy (y) olbdx f(x)
/f’@
’z- Ju(xy)dy)dx
(xy) 1/2
Jp(xy)
F,>.
I b f(x)
Note that F( n
o
Hece, for any +
in
converges.
numbers
Y h,b’
(nX) J (nX)dx
equation
Furthermore, if a sequence
{ F( )nl(>, )}
n
n
also converges
gives that
IF(; n )I
is bounded
(2.3) ensures that the series
{
m
converges in
Hence, the sum
2
Y ,b
z
F(An)@(;n
n=1
then the sequence of
F(n)( n
defines a
continuous linear functional on Y’
,b"
Next we investigate a representation for the finite Hankel transform of a
generalized function in B’
,b" Let D(O b) be the space of infinitely differentiable
functions on (O,b) with compact support contained in
(O,b). The topolog. of
D(O,b) is that which makes its dual D’(O,b) of Schwartz’s distribution. Then
h
D(O,b)C
maps D(O b) into a subspace of Y
b and
u,b" Let W be the
subspace of
onto which D(O,b) is mapped. Then we have
Theorem 4.2. For any generalized function f in
b, there exists a continuous
function F(y) of slow growth such that the finite Hankel transform ; f of f
B,
Y,b
B,
restricted to W is equivalent, in the functional sense, to the regular generalized
function F in Y’
,b"
Proof. For a given generalized function f, there exists an integer r
0 and a
continuous function h(x) [5, Theorem 3.4.2] such that
< f, >
Drh, >, for every @ in D(O,b).
We take h 0 outside (O,b). Then using (2.4), we have, for
D(O,b),
f, ;
F,@
(-I) r
h(x),
f,
Dr@(x)
Drh,@
>
DISTRIBUTIONAL FINITE HANKEL TRANSFORM
(-I) r o I b dx h(x)
(-l)r o Ib
(-I) r
r
Z
l
Dr(x)
dx h(x) I
o
b dx
h(x) o I
(-1)i+ra
(,)
i=o
331
dy
)r
(y)@-( x/y d(xy))
(4.2)
r
[(-l)iai(,)yixi-r x/’y a,+i(xy)]
b
(4 3)
i dx xi-rh(x)[ol=dy ,y,yif x a +i (xy)]
dy (y)
11
i=o
,
xi-rh(x),
for each i. If gi-r (x)
where ai() is a constant depending on
0 outside (O,b). Since (y)
then gi_r(X) is continuous on (0,) and gi_r(X)
the order of
0 < xy
(R),
on
bounded
is
and
descent
of
rapid
is
integration in (4.3)may be interchanged. Therefore, (4.3)becomes
r
11
I dy (y)y
F,
(xy)1/2 Ju+i(xY)
i=o
Denote
/+i[gi_r(X)]
by
(-l)i+rai()
+i
Gi_r(y),
r
F.
< S
i=o
hu+i[gi_r(X)].
W,
then for
i+U(y)
(-1)i+rai(p)yiGi_r
(4 4)
(y)>
Clearly, the continuous function
i+ (y)
Gi-r
olb g
i-r
(x)(xy)1/2j u+l (xy)dX
may be extended to an analytic function.
r
z
W
h f
Equation (4.4) gives
i+(y)
(-1)i+rai()yiGi_r
i+u
Finally since,
Gi_ r (y) <, it is obvious that
we know that / a(x-k)
From Example
Example 5
On the other hand, if we define
for x > k,
h(x)
F(y)
F(y)
(kz)1/2J
is of slow growth.
(kz)
for
0
<
k
:0,
x-k, for x
>
k,
we obtain from Theorem 4.2
h 6(x-k)
It is easily seen that
ao(
ao()GU_2(y) al(u)y
F(y)
p
2
, al()
Also
GP2(y)
klb (- k/x2)(xy) J(xy)dx.
1/2
1/2
k
(xy)dx,
G_l(y) klb(l- )(xy)du+l
and
G O+2 (y)
b
1/2
k I (x-k)(xy)
Hence (4.4) gives
Ju+2 (xy)dx
2(+I), and
G
l(y) + a2(u/y
a2()
I.
4o (y).
b.
O. P. SINGH AND R. S. PATHAK
332
(yk)1/2j
F(y)
(yk)
1/2(1 +
)(yb) 1/2 J
--)y(yb)1/2j’(yb).
(yb) + (I
This is another representation of / 6(x-k). It can be shown that this representation
is equivalent to the one given in example 1, since
-
Note.
(1 +
u-)(yb)1/2J
(b)
p
u) ol(R)(y) (yb)1/2j u (yb)dy,
-}(1 + ) (b) O.
Y
-(I +
(yb)
follows from the continuity property.
0
Thus (I
From (2.4) we have
k
B)
<y(yb) 1/2 J’p (yb)
(yb)1/2j(yb)dy.
(xy)1/2d (xy)dy + o ,r y (y)(xy)1/2d (xy)(dy).
(b))1/2j(by)dy + of y (y)(by)1/2j (by)(dy).
2 of(R) (y)
’(x)
Therefore,
’(b)
o
k
) o I"
f(R) (y)
((1
y (y)
So we have
Hence
<
F,
>
<
(yk) 1/2 J (yk), (y)
>.
Thus we get the same result as derived in Example 1.
0 for k
b. This also
Example 6. We have shown in Example 2 that h a(x-k)
0 for x < k and
2 and define h(x)
follows from Theorem 4.2. Take r
O,
(y)
G
h(x) x-k for x > k. It is easily seen that
O.
hence F(y)
Corollary 4.3. For any generalized function f in B’u,b with r h(x) and F(y)
defined as in the proof of Theorem 4.2, we have
+
0
(4.5)
F(y) < D r h(x), X (x) (xy) 1/2 J (xy) > as
Proof. Theorem 4.2 gives
l(y)
GU2(y)_
r
Iz
V
F(y)
flw
i=o
_
(-l)i+rai(u)yiGi_i+(y)
r
for
r
(-l)rofb dx o (-l)iai(p)x r( xy)
(-l)rof b
}1
B’- b
u
f
(xy) 1/2J
v+
(xy)
r
)x
r
(-1)rof b (x)
(-l)r<
Go
h(x),
x/
dv(xy)]
-
h(x) dx
r
xr
xr
x/
[x9
dv(xy)]
(x)]
h(x) dx as
0
+
.s
>
+
as
0
(and h(x)
0 outside (O,b)),
order of differentiation may be interchanged in the preceding equation to give
F(y) (-1) r < h(x),
as
0
E>.(x)(xy)
But since
>,(x)
on
(O,b)
Drx
<
Dxr h(x), e(x) (xy)1/2dv(xy)
+,
J(xy)]
>,
as
the
e
0
+,
differentiation).
(from the definition of distributional
333
DISTRIBUTIONAL FINITE HANKEL TRANSFORM
b
Example 7. While calculating the finite Hankel transform of a(x-k), 0 < k
(Example 5) using the method of Theorem 4.2, it was necessary to evaluate certain
integrals to find F(y). This may be avoided by using the above Corollary. From the
definition of h(x), we see that
D h(x)=
and
D 2 h(x)
I 0:1
0
x
x
>
k,
<
k,
6(x-k).
Hence, (4.7) gives
F(y)
im+
<
(xy)1/2j
6(x-k)
(x)
(k)(ky)1/2j
(ky)
(xy)>
0
im+
(ky)1/2j
(ky)
(k)
since
as 0
+
SOME STRUCTURE THEOREMS
In this section we shall obtain representations for members of B p,b’ B’
and
,b
Y’p,b under suitable conditions. Note that the structure formula given by Theorem 4 2
is valid only when F
Y’p,b is restricted to W, a subspace of Y
Here we will
obtain a more general result, viz., a structure formula shall be established for
FY’p b restricted to a larger subspace than W of Y
This section is very
similar to section 3.4 of Zemanian [5, p. 86-93], consequently the corresponding
results will be stated without proof or, perhaps, with only an indication of the
proof. To begi with, we define certain spaces associated with B
B(1) by
Definition 5.1. We define the spaces
b’
b and
5.
Bp Cp
B op,b
C Ou,b
g(x)
o(x p+1/2) as x
B p,b
(0,=)
g
o(x p+1/2) as x
g
0
+
0
+
(5.1)
(0,(R)), g
is continuous on
0
for x
>
b
(5.2)
},
and
"I’(
B ,b
p+3/2)- as x
o(x
’
B p,b
Bo
0 +}
(5.3)
carries the natural topology induced on it by
Note that
+
0
as x
o(x
b
p+1/2)
Bp,
This is true because
Bp, b.
P
Yo
-p-1/2
Ix
O<x<b
(0): sup
(x)I
<
We prove an interesting property of functions in B p,b
Then
B
Lemma 5 2 Let
q
Proof
B
Let
(t
oxP+5/2j
as x
Now
Dt)[t-u-O(t) t-.-I
,
[_
0
in
+
(. + )
(5.4)
t2].
(5.5)
334
Write
n(t)
n(t)
O,
Hence,
Hence
’
for
( + 1/2)
t
b.
Bu, b.
n(t)
-
O. P. S INGH AND R. S. PATHAK
Clearly
Also,
(r,)
Yk
y
is a smooth function on
+i()
<
,
o(tU+1/2),
n(t)
Therefore,
n(t)
d
u-1/2
d- (t- (t))
O(t)
as
0
t
as
t
0
as
t
0
Interating (5.6), we obtain
t-u-1/2(t) O(t2),
proving Lenna 5.2.
We assign a norm to the space
ljll o
CO
+.
+
(5.6)
+,
(5.7)
by
Ix-U-1/2g(x)l,
sup
0,1,2
k
for each
(0, (R)) and
Cu, b.
g
for each
x(O,b)
(5.8)
We need the following lemma which is
Thus C Ou,b becomes a topological vector space
stated without proof since the proof is identical to the proof of [5, Lemma I, p. 88].
o
o
Lemma 5.3.
b is a dense subset of
bThe followip.g proposition gives an integral representation for the functions in
Bo
Bu,
Cu,
Proposition 5 4
B
Let
Then
(x)
o
fb
satisifies the integral equation
2 t-u-1/2
u(x,t) (t- Dt)(
(t))dt,
(5.9)
xU-1/2u*(x,t),
(5.o)
where
u(x,t)
and
x3
2t3
u*(x,t)
t(t2-b2)’
for
x(x2-b2),
for 0
0
<
x -< t -< b,
t <- x <- b,
(5.11)
elsewhere,
0
for 0 < x (R), 0 < t < -.
Proof. Trivial.
are distributional derivatives
Next we prove that generalized functions in B’
of certain continuous functions.
We start with the following boundedness property of f B’
r and a positive constant
For each f
b, there exists a non-negative integer
A such that for all
(5 12)
<f,>l -< A max YkP () pr () (say)
Bu,
Bu’b’
o<k_<r
Suppose f
e
B’
is such that
(5.12) is satisfied with r
O.
l<f,,>l <-A sup x--1/2 (x)l.
Then
(5.13)
o<x<b
f, satisfying the inequality (5.13), continuously and uniquely onto
C Ou,b" Let g in C Ou,b be arbitrary. Then by Lemma 5 3 there exists a
converges to g in
{n of testing functions in B u,b such that
We now extend
the space
sequence
Cn
DISTRIBUTIONAL FINITE HANKEL TRANSFORM
C o,b"
335
We define <f,g> by
<f,g>
lim
n+
C O,b
on
linear functional
This defines a continuous
(5 14)
<f’n >
satisfying the
inequality
(5.13).
C O,b’
Clearly u(x t)
satisfying (5.13).
For
Definition 5 5
hence the following definition makes sense for
sotisfying the inequality (5.13), define
f e B’
w,b
h(t)
Note:
h(t)
(i)
(ii)
0
t
for
h()l
lh(t)
<
f(x), u(x,t)
<
3AbZlt-l,
0
<
(5.15)
>.
for t e b
-<
b.
t
b, 0
as u(x,t)
b,
0
B, b, ( D t) 2[t-- }(t)j is uniformly
B
2it-P-1/2 (t)]
n(t) (
Lemma 5.6. For
Proof. Let
B’u,b
f
(5.16)
continuous on
(O,b].
0(I) as t 0 + and In (t)l < (R), proving the Lemma 5.6.
For f B’,b satisfying the condition
n(t)
Then
Theorem 5 7
x-u-1/2 (x)I
A sup
<
B ,b’
V
O<x<b
we have
Go,b"
for every
(5.15).
If
<f,>
Here h(t)
Dt[t-IDt(t-lh(t))]
Ib
h(t)(t-lDt)2[t-u-1/2(t)]dt
(5.17)
is the continuous function defined by equation
is Lebesgue integrable over
(O,b),
(5.17) can
be
written as
< f,g>
<t -u-} D t
(t-lDt(t-lh (t)))
(t)>
(5 18)
o
B p,b"
for every
Proof. The proof of (5.17) is very similar to the proof of [5, equation (9),
p. 90-91] and (5.18) follows easily from (5.17).
We now generalize Theorem 5.7 for the case when l<f’>l < PrP(), r > 0 For
this we need the following"
Definition 5.8.
B b’ and H"
B(n)
,b
H(n)
u,b
where
B’
H (n)
For each non-negative integer n, we define the spaces B(n.)
u,b’ ,b’
by
{
lJ
’(x)
{n(X) :(u+)X
(n)(x) Dn(x)
b
b
,b
B p,b
for
(n) (x)
n # 0
h,b
{n(x)
and
)I o(x+))
kJ(x
(p+1/2)x- x
as x
(5 19)
0+}
B(n.
u,b
x
and
H
o( xU+312
(5.2o)
(O)(x)
as x
@(x),
O+
for each k
0
2
},(5 21)
336
O.P. SlNGH AND R. S. PATHAK
Note that H (1) C
B
()
B u,b"
B u,b
and
o
Let
Bo
o
Ib t-u+1/2o(t)dt
be such that
I.
(5.23)
In the subsequent development we shall need the following lend, a, whose proof is
omitted since it is similar to that of [5, Lemma I, p.68].
be a fixed testing function in B
LeFmma 5.9. Let
u,b satisfying (5.23). Then any
B
be
function
in
may
decomposed uniquely according to
testing
u,b
o
where
H (I) and the constant K
u,b
is in
n
K
Suppose f
function
(O,b].
o
Ko
(5.24)
+ n
is given by
Ib t-u+1/2(t)dt.
(5.25)
is a regular generalized function in B’
f such that f is Lebesgue integrable over
Then for
n
H (I) we have
u,b
<f,,>
o
Ib
+
f(x) n (x)dx
[x-U-}@(x)]dx
xU-1/2f(x)
o Ib
b (x-]f(x))
since (b)
Therefore
0
and
generated by a differentiable
(O,b) and f’ is bounded on
o(x u+3/2
x
as
B
(I))
(x)dx,
0+.
x
<x -U-1/2DxU-;
<f,n>
(for some
f(x),-(x)>.
(5.26)
But
x uu+1/2
n
Dx-U-
(x),
therefore,
<f(x), x u-1/2
Dx--}(x) >
Let
L
and
T
Then for any
-u-1/2 Dx u-1/2 f(x),
-(x)>.
x u-1/2 Dx -u-
(5.27)
(5 28)
-u-1/2 Dx-1/2
(5.29)
B’ I)
L
and
<x
-1 D)(x -u’ (x))
x u+1/2 (x
y (L)
We write the above as
The operation
Lenna 5 I0
o
B
For an arbitrary f n B’,b and any
B
Yk+l ()
into
in
B(,
set
337
DISTRIBUTIONAL FINITE HANKEL TRANSFORM
T f,
<
>
From (5.30) we see that T f
x -u-1/2
DxU-1/2
f,
f,
>
’I’I
B h,b"
is a linear functional on
x-U-1/2
DxU-!
(5.30)
-Lu >.
Write
g.
f
We then have formally,
xU+1/2
Dx u-} f
g.
Now define,
x -u+
With this notation (5.26) suggests that
g] (-1)
[xU+1/2
x-U+1/2 [xU+1/2 f](-l),
n
c
Bo
"Ihis can be etended to all of
Assign
x-U+1/2 [xU+1/2 f](-l),
Then for any
(5.32)
f,- >.
>
u+
(1)
Equation (5 32) defines a linear functional on H
B u,b by using Lenmla 5 9
(5.3i)
f.
o
Ko
>
(arbitrary).
(5.33)
we have from Lemma 5.9,
+ n.
= Ko
Hence the operator x-U+1/2 [xU+1/2
Proposition 5.11. Suppose f
f,@
for all
@
in
B
f](-l)
B’
>I
is defined for all of
B
satisfies
-< A sup
o_<k_<r
y (),
(5.34)
r >
Then
I< x -u+ [x +1/2] f(x)](-l)
(x)>
<-A
sup
osk<r-1
y (),
(5.35)
(I)
H ,b"
Now use Lemma
Bo
for all
Proof. Using (5.32) the proof follows trivially for members of
5.9 to complete the proof.
Proposition 5.12 For f B’u,b satisfying
f,
>I
<
A sup
ok<l
y (),
B
V
we have
f,@
>
oI
b
h(t)
(t-lDt)3(t-u-]
(t))dt
in B ’1’I and where h is a continuous function
for each
u,b
O. For
Proof. We know that the theorem is true for r
<
f,
-(u+1/2)
<
x -+1/2 [x }+ 1/2
f](1)
n
n
(u
1
B p,b’
(5 32))
H I)
,b (equation
(5.36)
338
O. P. SINGH AND R. S. PATHAK
-(+1/2) oIb
h(t)(t-lDt)2(t--1/2n(t))dt,
by Theorem (5.7), since
<
x-+1/2[x + f](-l)
n
<
>
and
A You (n), (see Proposition (5 11))
o(x"+1/2),
n(x)
as
x
0/.
Therefore,
f’
>
-(/]) o" b
-o Ib
If we replace both
h(t)(t-iDt)3(t--}(t))dt.
Bu,b and B(1)u,b by Bu,b and H (
5.10 and Propositions 5.11, 5.12 are still true.
For f B’
Theorem 5 13
satisfying
>I
f,
_
for some non-negative integer
f
d
t +1/2
--1/2
(t))] dt
u+1/2 t dt (t
h(t)(t-IDt)2[t-u-1/2
>
<
r,
A
o-<k<rSUp y
by H
,b,
Lemmas 5.9,
Hence by induction on r, we get
(),
(V
e
B
O.,b
we have
(_l)r 0 ib
h(t)(t-IGt)r+2[t-,-1/2(t)]dt
(5.37)
for each @ e B ,b where h is a continuous function.
The above structure theorem helps us to say more about the finite Hankel
transform of elements in B’
,b"
e B
Let
B’,b and
,b and F, be the finite Hankel transforms of f
respectively. Then
< F, >
< f, >
(_l)r-2oIb h(X)ol’(y) 1/2 (y)( __) r x-J,(xy))dydx
-
(5.38)
(using equations (2.4) and (5.37).
Since
(.1)r(1
then
F,
Now let
Since
gr(X)
r (x
Ju(xY)) yrx-u-rJ u+ r (xy),
o Ib h(x) o I" (y)1/2(y)yrx-u-rJ+r(xy)dydx
olyr (y) oIb rh(X)
+r(xy)dxdy.
Xxa
_h.(x)
(5.40)
x+1/2+r"
x J+r(XY)
x+1/2+r
(xy)U+r+1/2
x+1/2+r
asx/
O+
DISTRIBUTIONAL FINITE HANKEL TRANSFORM
339
so that
,b
Ghr+r(Y)
o
is well defined
gr(X)
(xy)
J+r (xy)dx
Thus (5.40) becomes
<
F,
>
o
f(R) yr(y) G +r
r (y)dy
=<Y
Consequently,
yr G+r(y)
r
F
-’IG+r(y)l
r
is of slow growth since
(=
Y’,b’
For any F
Theorem 5 14
r
’-rG+r(y))
<
.
Furthermore,
6.
Y
b
/,
(y)
>
We list the above as
there exists a continuous function
F,@
where
r^+r(y)
(in the functional sense)
of slow growth such that
Y,b’
for
r
FIy ,b
<
G(y)
G(y)
is equivalent to
i.e.
G(y),(y) >,
[B ,b
G(y) may be extended to an analytic function
FURTHER PROPERTIES OF THE FINITE HANKEL TRANSFORM
A
Notation
Let us write f
for any F
members of Y
Also write
Y’,b"
A
fl B,b
for any f
and likewise
B’,b,
to denote the members of
B
",b
FIy b
F
and
^
for the
For any f, we have for some integer r -> 2,
<
^A
f,@
>
(-I) r o f b
where h(x) is a continuous function.
Definition 6.1. We define a new space
H(1)
{
Hu (k)(x)
h(x)( d)r (x --1/2
H(1)
(x))dx,
by
o(x u+3/2) as x
0
+,
for k
(6.1)
0,1,2,...},
(0,-).
where
to be the periodic extension of period b of h(x) on
+
R
Then for any x
hb(X) h(x-nb) for some positive integer n (R)’such
0
x-nb
b. Associated with hb(X) is the regular distribution in H (1)
that
having the value
Now define
hb(X)
(0 b]
<
for any
h b (x), (x)
fb
>
(6.2)
h(x) (x+nb)dx,
n=o
H (I)
The right-hand siae of (6.2) converges, since
by
on
define a functional
fb
Hu(1)
is of rapid descent as
x
.
Now
O.P. SINGH AND R. S. PATHAK
340
fb’
<
of(R) hb(X)( Dx) r (x --1/2 (x))dx
(-l)r[ofh(x-nb)( D x )r (x--1/2,(x))dx]
(-l)r
>
z
n=o
z (-I) r
p=o
nbf( n+l)b h(x-nb)( D x )r(x-U-lo(x))dx
This defines a linear continuous functional on H (1). Also for
have
r
>
D r
(x))dx (since ^ 0
fb,
(-I)
orb h(x)(
x) x--1/2
in
for x
(6 3)
Bu, b,
we
b)
=<f,>.
So we see that (b s a periodic extension of f.
be extended to a periodic linear functional,
Theorem 6.2. Every f in
b may
which is continuous in the topology of
with period b, on H
and each f in B’,b’ the function
in (0 b/4)
For every
Theorem 6 3
Bu,
Hu.
u(1)
(6.4)
J(xy)>
(x) (xy)
f(x)
F(y)
(x) is defined by (2.7) is a smooth function of slow growth and defines a
where
E
regular generalized function in Y’,b"
Proof. The proof of the above theorem is similar to the proof of [1, Lemma 12] given
by Zemanian.
Theorem 6.4. The finite Hankel transform, huf, of a generalized function f in
0
of the family F (z) of regular generalized functions
B’,b is the limit as
defined in Theorem 6.3.
it is sufficient to show that
Proof. Since F(z) is a regular functional in
Y’,b,
<
for each
Bb
in
For each
O.
as
Y,b’
in
F,
>
/pf,
As
given by Equation (2 4)
>
in
O+
(x)
Y,b
there exists a unique
on
(O,b)
Now we have
<
F
,
>
(x)(xy)J
f(x),
<<
f
<
(x)(xy)J
f(x),
<
f(x)
<
f(x),
, (x)
<f(x),(x)
(xy)
Of
(x) (x)
>,
as
>,
(xy)
(y)(xy)1/2j
>
0+.
(y)
>
>
(y)dy
(xy)dy
(by [8 Corollary 5 3 2b p. 121])
341
DISTRIBUTIONAL FINITE HANKEL TRANSFORM
Consequently,
<
F
,
f(x), (x)
<
>
f,
/
<
as
>,
0+,
for each
Y ,b"
(x)(xy)1/2j
Since
(xy)
b)(X)(xy)1/2j
x( 0
(xz), as
0+,
(where X(o b)
is
the characteristic function of the interval (O,b)) and the latter is not a testing
O,
it is not true, in general, that the limit of F (z), as
function in B
p,b’
exists as an ordinary function. Where the limit does exist as an ordinary function,
it will be denoted by F
Corollary 6.5. If f is a regular generalized function in B’p,b’ then the limiting
exists and is equivalent to the finite Hankel transform of f.
function
in (0, b/4),
Proof. If f is regular, then for each
o(z)-
Fo(Z)
ofbf(x)(x)(xy) J(xz)dx.
F(z)
As
0
+,
we obtain
Fo(Z)
of
b
f(x)(xy)1/2j u (xz)dx,
Fo(Z) is
X(O,b)(X), (6.5)
and from Example 4, we see that
Using the function
Fo(Z)
equivalent to / f.
can be written as
X(O,b)(x)f(x),
<
(6.5)
x(x)
(xz)]J
(xz)>,
where >.(x) is a testing function in D(R) such that x(x)
on (O,b]. In this
case, to calculate the finite Hankel transform we merely truncate the regular
b, as would be expected. In a similar way one might
distribution in B’
u,b at x
O+ of the family of the functions F (z) as a process of
interpret the limit as
truncation for distributions in general, for one is replacing
f(x) by the
distributional limit
(x) f(x) as c 0+.
Corollar,v 6.6.
If the generalized function f
B’p,b has support in (0 b]
the function
exists and is equivalent to the finite Hankel transform of
Proof. Let (x) be a testing function in D(1) such that (x)
neighborhood of the support of f. Then
Fo(Z
such that,
lim
/0+
since
(x)
F (z)
<
F(z):
F o (z):
f(x),
on the support of
f
x(x)
f(x)
>,(x)(xz)1/2jl (xz)
;(x)(xz)1/2J
O+
as
(xz)>
But Zemanian [i
proved that
f
for every functional ir,
Example 8.
H
f(x)
<
(x)(xz)1/2J
(xz)
>
having compact support.
For the distribution (x-k),
F(z)
(x-k),
<
g
then
f.
on a
from (6.4) we obtain
(x)(xz)}J(xz)
(k)(kz)1/2ju(kz)
>
Theorem 2] has
O. P. SINGH AND R. S. PATHAK
342
But for 0
k < b,
: (k)
O+
as
And for
k
b, ; (k)
>
0
Therefore,
kV-
Fo(Z
jp(kz),
for 0
k
b
<
for k z b,
O,
/
<
[6(x-k)].
THE FOURIER-BESSEL SERIES
7.
Classically the
FourieroBessel series
finite
inverse
2__
transform
J(x; n
(xlX)3
n j2
u+l(bXn
nZ
b2
Hankel
is
considered
to
be
the
(; n
(7 I)
dx
(7.2)
where
Ib(x)(xX n )1/2j(xX n)
(x n
.
o
[7] of some function
is the finite Hankel transform
In section 2, we showed that for any
B ,b
1/2
J(xX
+1(bXn
and
(x n
given by (7 2)
we
have
z
@(x)
X(x)
n=l
We obtain an
inversion
transforms of members in
x
J
B’,b"
Theorem 7.1 (inversion). Let
where
-1/2. Let F z (f)
convergence in B’,b’ we have
be an arbitrary generalized function in B’
,b’
be the finite Hankel transform
Then in the sense of
f
,,x,:
’ [J,(x;n
Z:
Let >.(x)
-
<
D
2
=IN(n)| j2p+l(bXn
N
hi::1 (nn)1/2
n
Z
n
-(x)1/2 Jh(x; n
Since
<
2
(bXn)]F(>.n)"
)/J
+I
n
be an arbitrary testing function in B
,b"
N+"
Proof.
(7.1) for generalized finite Harkel
theorem similar to
F(;n),
(x)
>
is locally integrable over
J(xX )n
j2p+l (b’
, (x)
F(X n
>
n
fb2_2
o
b
N
z
2
N
z
n=1
-
<
(7.4)
We wish to prove that
f(x), (x) >,
ds
N
(0 b)
F(, n
Jp+1(bXn
F(X n
n:1 O2
n p+l (bn
1/2
(-n)
jp (xX n
(x)dx
(from (2 2))
(R).
DISTRIBUTIONAL FINITE HANKEL TRANSFORM
N
r. 2
2
n=l b
<
(>’ )n
, nj2+I (b)n
im <
/0+
f(x) x c (x)(x s 1/2
343
n )>
(from Theorem 6.4)
N (R).
as
(x)>,
f(x),
We verify this inversion Theorem by means of a numerical example.
Example 9. For 0 < k < b, a(x-k) c E’(1)C
CB’u,b. The finite Hankel transform
of
a (k-x)
H’u
(kz)1/2j(kz)
is,
Iz (x-k)
F(z).
Hence,
F( n )=
(kX)1/2
J(k n
n
n
1,2 3
Now
N
b
n=1
J(x n
--
j2p+l(bAn
2
b
_2
b
=i
J(kkn )’
Jp (k}.)n
N
nZ
(kn)2
(L)1/2
n J+l(b n
(x)>
(k)1/2 o/ b (x;,)n Ju(x; n )(x )dx
J(k
N
n
(k) 1/2 (x n
ns--1 () j2’
n ,+l(bn
(k)
<
This also yields
6(x-k)
6(x-k),
2
, bim
(x)
N
z ,/E- J
n
in the sense of convergence in
(kn)Jp (x)n
j’2u+l(b>,n
B’,b"
ACKNOWLEDGEMENT:
This work was supported by Grant No. 1402/28 from Science Center,
Saud
King
University, Saudi Arabia.
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