SUPPORTS OF FUNCTIONS AND INTEGRAL ... - Semantic Scholar

SUPPORTS OF FUNCTIONS AND INTEGRAL TRANSFORMS

Vu Kim TUAN1

Department of Mathematics and Computer Science Faculty of Science Kuwait University P.O. Box 5969, Safat 13060 Kuwait e-mail: [email protected]

1. Introduction. In this paper we apply a method of spectral theory of linear operators [10] to establish relations between the support of a function f on Rk with properties of its image T f under a linear operator T : Rk → Rk . The classical approach uses analytic continuation of the image T f into some complex domain (theorems of Paley-Wiener type [4, 5, 6, 7]), and therefore, could not apply to functions whose images T f are not analytic in any domain (for example, Fourier transforms of functions vanishing on some ball are in general not analytic anywhere). Our approach can be applied to a quite wide class of linear operators, for example, linear integral operators with theorems of Plancherel type, whose kernels are eigenfunctions of some differential operators (Fourier, Hankel, Kontorovich-Lebedev, Yand H- transforms etc. are having these properties). Here we will apply this method only to the Fourier and Hankel transforms. Fourier transforms of compactly supported functions, functions with polynomial domain supports, functions vanishing on some ball and functions vanishing on a half-line are considered. Some of these cases could not be described by the classical method, and even in classical cases (Fourier transforms of compactly supported functions and functions vanishing on a half-line) the results obtained here are also new. Similar results, obtained by different technique, but only for Fourier transforms of functions with bounded supports are considered in [2, 3]. Hankel transform of compactly supported functions is also considered. Applications to the Y-integral transform will be considered in the forthcoming work [9]. 2. Main Theorem. Assume that P (y) is a non-constant continuous function on Rk . Let U be a subset with relative norm topology of the space Lp (Rk ; du), 1 ≤ p ≤ ∞, of p-integrable functions with respect to the measure du such that P (y)f (y) ∈ U if f ∈ U . Let V be a subset 1

Supported by the Kuwait University research grant SM 112

with relative norm topology of the space Lq (Rk ; dv), 1 ≤ q ≤ ∞, of q-integrable functions with respect to the measure dv. Suppose that T : U → V is a linear homeomorphism and Q : V → V is an operator such that Q(T f )(x) = T (P (y)f (y))(x).

(1)

THEOREM 1. The following relation holds 1/n

lim kQn (T f )kLq (Rk ;dv) =

n→∞

sup |P (y)|.

(2)

y∈supp f

where supp f denotes the support of function f [10]. PROOF. From (1) we have Qn (T f )(x) = T (P n (y)f (y))(x), n = 0, 1, 2, . . . .

(3)

It is sufficient to consider the case f 6≡ 0. Let first supy∈supp f |P (y)| < ∞. From (3) and continuity of T in corresponding relative norm topologies we obtain kQn (T f )kLq (Rk ;dv) ≤ CkP n (y)f (y)kLp (Rk ;du) = CkP n (y)f (y)kLp (supp f ;du) ≤ C sup |P (y)|n kf kLp (supp f ;du) , n = 0, 1, 2, . . . ,

(4)

y∈supp f

where C is a universal constant, that may be distinct in different places. Therefore, 1/n

lim sup kQn (T f )kLq (Rk ;dv) n→∞

≤

1/n

sup |P (y)| lim sup C 1/n kf kLp (supp f ;du) =

y∈supp f

n→∞

sup |P (y)|.

(5)

y∈supp f

In particular, if supy∈supp f |P (y)| = 0 the identity (2) follows at once. Suppose now, therefore, that supy∈supp f |P (y)| > 0. For any , 0 <  < supy∈supp f |P (y)|, there exists a point x0 ∈ supp f such that |P (x0 )| >

sup |P (y)| − /2.

(6)

y∈supp f

For P is a continuous function, there exists a neighborhood Ux0 of x0 such that |P (x)| >

sup |P (y)| − , y∈supp f

x ∈ Ux0 .

(7)

Since x0 ∈ supp f we have kf kLp (Ux0 ;du) > 0. Because the inverse of T is also continuous in corresponding relative norm topologies, we have kQn (T f )kLq (Rk ;dv) ≥ CkP n (y)f (y)kLp (Rk ;du) ≥ CkP n (y)f (y)kLp (Ux0 ;du) ≥ C( sup |P (y)| − )n kf kLp (Ux0 ;du) . y∈supp f

(8)

Hence, 1/n

1/n

lim inf kQn (T f )kLq (Rk ;dv) ≥ ( sup |P (y)|−) lim C 1/n kf kLp (Ux n→∞

n→∞

y∈supp f

0 ;du)

sup |P (y)|−.

=

y∈supp f

(9) For  can be chosen arbitrarily small, from (5) and (9) it follows (2). Suppose now supy∈supp f |P (y)| = ∞. Then for any N > 0 there exists a point x0 ∈ supp f such that |P (x0 )| > 2N . Since P is a continuous function there exists a neighborhood Ux0 of x0 on which |P (x)| > N . Because x0 is in the support of f , we have kf kLp (Ux0 ;du) > 0. Hence, 1/n

1/n

lim inf kQn (T f )kLq (Rk ;dv) ≥ lim inf C 1/n kP n (y)f (y)kLp (Rk ;du) n→∞

n→∞

1/n

1/n

≥ lim inf C 1/n kP n (y)f (y)kLp (Ux

0 ;du)

n→∞

≥ N lim inf C 1/n kf kLp (Ux n→∞

0 ;du)

= N.

(10)

Since N can be chosen arbitrarily large, we have 1/n

lim kQn (T f )kLq (Rk ;dv) = ∞.

n→∞

(11)

Theorem is proved. Put ΩP = {y ∈ Rk : |P (y)| ≤ 1}. If P (y) is a polynomial, ΩP is called a polynomial domain in Rk . A ball is a polynomial domain. A polynomial domain (for example, U = {x : |x1 x2 . . . xn | ≤ 1}) may be unbounded and nonconvex. We have the following COROLLARY 1. A function f ∈ U vanishes outside the domain ΩP if and only if 1/n

lim kQn (T f )kLq (Rk ;dv) ≤ 1.

n→∞

(12)

It follows from the fact that supp f ⊆ ΩP if and only if supy∈supp f |P (y)| ≤ 1. Theorem 1 is especially useful when T is a linear integral operator (T f )(x) =

Z

k(x, y)f (y)dy,

(13)

kT f kL2 (Rk ;dv) = kf kL2 (Rk ;du) ,

(14)

Rk

satisfying a theorem of Plancherel type

with the kernel being an eigenfunction of some differential operator Q = Q(D), D = ( ∂x∂ 1 , . . . , ∂x∂n ), Q(D)k(x, y) = P (y)k(x, y). (15) The Fourier, Hankel, Kontorovich-Lebedev, Y- and H- integral transforms etc. are having these properties. In the next sections we show some applications of Theorem 1

and Corollary 1 to the Fourier and Hankel transforms. 3. Fourier transform of functions with ball supports. Let T be the Fourier transform f (x) = (2π)−k/2

Z Rk

fˆ(y)e−ix.y dy.

(16)

The Fourier transform (16) is unitary in L2 (Rk ) (the Plancherel theorem [4, 7]) and ∂2 ∂2 satisfies equation (1) with P (y) = −|y|2 and Q = ∆ = ∂x 2 + · · · + ∂x2 – the Laplacian. n 1 Let U be the space of all square integrable functions decreasing faster than any polynomial of |y|−1 , V be the space of infinitely differentiable functions belonging to L2 (Rk ) together with all their partial derivatives. Then −|y|2 fˆ(y) ∈ U if fˆ ∈ U , ∆f ∈ V if f ∈ V and the Fourier transform is a homeomorphism from U to V in relative L2 –norm topology. Since supy∈supp fˆ |y|2 = (supy∈supp fˆ |y|)2 we obtain THEOREM 2. Let f ∈ C ∞ (Rk ) such that all partial derivatives of f belong to L2 (Rk ). Then there always exists the limit 1/(2n)

lim k∆n f k2 n→∞

= sup{|y| :

y ∈ supp fˆ}.

(17)

In particular, function f is the Fourier transform of a square integrable function vanishing outside a ball with radius r centered at the origin if and only if 1/(2n)

lim k∆n f k2

n→∞

≤ r.

(18)

Theorem 2 is another version of the classical Paley-Wiener theorem on the Fourier transform of functions with compact supports [4, 6, 7]. 4. Fourier transform of functions with polynomial domain supports. Let U = V = S(Rk ) – the space of infinitely differentiable functions approaching 0 at infinity together with all its partial derivatives faster than any polynomial of |x|−1 [4, 7, 10]. It is known that the Fourier transform is a homeomorphism in S(Rk ) in relative L2 –norm topology. Let P (y) be any non-constant polynomial. If f ∈ S(Rk ) then both P (y)fˆ(y) and P (iD)f (x) belong to S(Rk ). Since d )(y) = P (y)fˆ(y), (P (iD)f

(19)

applying Theorem 1 and Corollary 1 we obtain the following Theorem for the case p = 2:

THEOREM 3. For any function f from S(Rk ) the following relation holds lim kP n (iD)f k1/n = p

n→∞

sup |P (y)|,

1 ≤ p ≤ ∞.

(20)

y∈supp fˆ

In particular, function f ∈ S(Rk ) is the Fourier transform of a function fˆ ∈ S(Rk ) vanishing outside a polynomial domain ΩP if and only if ≤ 1. lim kP n (iD)f k1/n p

(21)

n→∞

PROOF. The case f ≡ 0 is trivial, so we assume that f 6≡ 0. Hereafter let q be the conjugate exponent of p, that means p−1 + q −1 = 1. i) Suppose that 2 ≤ p < ∞. Applying the Hausdorff-Young inequality [7, 8] kP n (iD)f kp ≤ CkP n (y)fˆ(y)kq = CkP n (y)fˆ(y)kLq (supp fˆ) ≤ C sup |P (y)|n kfˆkLq (supp fˆ) ,

(22)

y∈supp fˆ

we obtain lim sup kP n (iD)f k1/n p n→∞

≤

sup y∈supp fˆ

|P (y)| lim sup C 1/n kfˆk1/n = q n→∞

sup |P (y)|.

(23)

y∈supp fˆ

Let now 1 ≤ p < 2. We have kf kpp =

Z Rk

(1 + |x|2 )−kp |(1 + |x|2 )k f (x)|p dx ≤ k(1 + |x|2 )−kp k

2 2−p

k(1 + |x|2 )k f (x)kp2

≤ Ck(1 + |x|2 )k f (x)kp2 = Ck(1 − ∆)k fˆkp2 ,

(24)

where the H¨older inequality and the Plancherel theorem for the Fourier transform [4, 7] are applied. Therefore, kP n (iD)f kp ≤ Ck(1 − ∆)k [P n (x)fˆ(x)]k2 .

(25)

By induction one can show that (1 − ∆)k [P n (x)fˆ(x)] = P n−2k (x)φn (x), with supp φn ⊆ supp fˆ,

n > 2k,

(26)

kφn k2 ≤ Cn2k . Hence,

kP n (iD)f kp ≤ CkP n−2k (x)φn (x)k2 ≤ C( sup |P (y)|n−2k ) kφn k2 ≤ Cn2k sup |P (y)|n−2k . y∈supp fˆ

(27)

y∈supp fˆ

Therefore, ≤ lim sup kP n (iD)f k1/n p n→∞

sup |P (y)|. y∈supp fˆ

(28)

Let now p = ∞. We have −k/2

kf k∞ ≤ (2π)

kfˆk1 = (2π)−k/2

Z Rk 2 k

(1 + |x|2 )−k |(1 + |x|2 )k fˆ(x)|dx

≤ (2π)−k/2 k(1 + |x|2 )−k k2 k(1 + |x| ) fˆ(x)k2 ≤ Ck(1 + |x|2 )k fˆ(x)k2 .

(29)

Hence, kP n (iD)f k∞ ≤ CkP n (x)(1 + |x|2 )k fˆ(x)k2 .

(30)

Consequently, lim supkP n (iD)f k1/n ∞ ≤ n→∞

|P (y)| =

sup y∈supp (1+|x|2 )k fˆ

sup |P (y)|.

(31)

y∈supp fˆ

ii) Since f ∈ S(Rk ), function f and its partial derivatives vanish at infinity, therefore, integration by parts gives Z Rk

P n (−iD)f (x)P n (iD)f (x)dx =

Z Rk

f (x)P 2n (iD)f (x)dx.

(32)

Hence, by the H¨older inequality, kP n (iD)f k22 ≤ kf kq kP 2n (iD)f kp .

(33)

Consequently, applying formula (20) for the proved case p = 2 we obtain sup |P (y)| = y∈supp fˆ

≤

lim kf k1/(2n) lim inf q n→∞ n→∞

1/n

lim kP n (iD)f k2

n→∞

2n

kP (iD)f kp1/(2n) = lim inf kP 2n (iD)f kp1/(2n) . n→∞

(34)

In formula (33) replacing f by P (iD)f we have kP n+1 (iD)f k22 ≤ kP (iD)f kq kP 2n+1 (iD)f kp .

(35)

Since f ∈ S(Rk ), f cannot be a polynomial. Therefore, P (iD)f 6≡ 0, and we have sup |P (y)| = y∈supp fˆ 2/(2n+1)

= n→∞ lim kP n+1 (iD)f k2

≤

1/(n+1)

lim kP n+1 (iD)f k2

n→∞

lim kP (iD)f kq1/(2n+1)

n→∞

lim inf kP 2n+1 (iD)f kp1/(2n+1) = lim inf kP 2n+1 (iD)f kp1/(2n+1) . n→∞ n→∞

(36)

Hence, from (34) and (36) we get lim inf kP n (iD)f k1/n ≥ p n→∞

sup |P (y)|. y∈supp fˆ

Formula (37) together with formulas (23), (28) and (31) give formula (20).

(37)

In particular, Theorem 2 is still valid in Lp (Rn ) norm for 1 ≤ p ≤ ∞ if f ∈ S(Rk ). 5. Fourier transform of function vanishing on a ball Let U = V = S(Rk ) and P (y) = exp(−|y|2 ). Then Q=

∞ X ∆m m=0

(38)

m!

satisfies relation (1) where T is the Fourier transform. Applying Corollary 1 we obtain the case p = 2 of the following THEOREM 4. For any function f ∈ S(Rk ) the following equation holds

∞

X nm ∆m f 1/n

= exp(− inf |y|2 ), lim

n→∞ m! y∈supp fˆ m=0

1 ≤ p ≤ ∞.

(39)

p

In particular, a function f ∈ S(Rk ) is the Fourier transform of a function fˆ ∈ S(Rk ) vanishing on some ball with radius r centered at the origin if and only if

∞

X nm ∆m f 1/n

lim

≤ exp(−r2 ). n→∞

m! m=0

(40)

p

PROOF. The proof is very similar to the proof of Theorem 3. If 2 ≤ p ≤ ∞ applying the Hausdoff–Young inequality

∞

X nm ∆m f

≤ Ck exp(−n|y|2 )fˆ(y)kq ≤ C exp(−n inf |y|2 )kfˆkq ,

m! y∈supp fˆ m=0

(41)

p

we obtain

1/n ∞

X nm ∆m f

lim sup ≤ exp(− inf |y|2 ).

m! n→∞ m=0 y∈supp fˆ

(42)

p

For the case 1 ≤ p < 2 using inequality (24) we get

∞

X nm ∆m f



≤ Ck(1 − ∆)k [exp(−n|x|2 )fˆ(x)]k2 .

m! m=0

(43)

(1 − ∆)k [exp(−n|x|2 )fˆ(x)] = exp(−n|x|2 )φn (x),

(44)

p

One can show that

with supp φn (x) ⊆ supp fˆ and kφn k2 < Cn2k . Hence, inequality (42) follows.

In case p = ∞ using inequality (29) we have

∞

X nm ∆m f



m! m=0

≤ Ck exp(−n|x|2 )(1 + |x|2 )k fˆ(x)k2 .

(45)

∞

Therefore, we get inequality (42). Using the Parseval and Plancherel theorems for the Fourier transform and the H¨older inequality we have

2 ∞

X nm ∆m f



m! m=0

Z

=

Rk m=0

2

= =

Z Rk

Z Rk

Z

exp(−2n|y|2 )|fˆ(y)|2 dy = ∞ X (2n)m ∆m f (x)

f (x)

m!

m=0

dx

2 ∞ X nm ∆m f (x) dx m!

Rk

fˆ(y) exp(−2n|y|2 )fˆ(y)dy

∞

X (2n)m ∆m f

. ≤ kf kq

m! m=0

(46)

p

Similarly,

2

∞ ∞

X

X nm ∆m f

∆m f



≤



m! m=0 m=0 m! 2

q

∞

X (2n − 1)m ∆m f



.

m! m=0

(47)

p

Hence,

∞

∞

X nm ∆m f 1/n

X nm ∆m f 1/n



≥ lim = exp(− inf |y|2 ). lim inf

n→∞

n→∞

m! m! y∈supp fˆ m=0 m=0 p

(48)

2

Theorem is proved. 6. Fourier transform of functions vanishing on a half-line Consider now the Fourier transform only on R. If we choose P (y) = 1 + erf (−y), where erf (y) is the error function [1], then 0 < P (y) ≤ 1 on [0, ∞) and 1 < P (y) < 2 on (−∞, 0). Since ∞ 2 X (−1)m y 2m+1 P (y) = 1 − √ , (49) π m=0 (2m + 1)m! then Q can be chosen to be d Q dx

!

∞ √ X = Id − 2 π

(−1)m d2m+1 , 2m+1 m=0 (2m + 1)m! dx

(50)

where Id is the identity operator. In that case equation (1) holds on S(R). Therefore, applying Corollary 1 we have proved the following theorem for p = 2:

THEOREM 5. A function f ∈ S(R) is the Fourier transform of a function fˆ ∈ S(R) vanishing on (−∞, 0) if and only if

lim

Qn n→∞

where Q



d dx



!

1/n

d f (x)

dx p

≤ 1,

1 ≤ p ≤ ∞,

(51)

is defined by (50).

The general case is proved similar to the proofs of Theorems 3 and 4. 7. Hankel transform of polynomial decreasing functions The Hankel transform is defined by [8] Z ∞ √ f (x) = (Hν g)(x) = xyJν (xy)g(y) dy, 0

x ∈ R+ = (0, ∞), Re ν > −1,

(52)

if the integral converges in some sense (absolutely, improper, mean convergence), where Jν (x) is the Bessel function of the first kind [1]. It is well-known [8] that the Hankel transform is a homeomorphism in L2 (R+ ) (if ν is a real number, then it is even unitary), and its inverse has the symmetric form Z ∞ √ xyJν (xy)f (y) dy, x ∈ R+ , Re ν > −1. (53) g(x) = (Hν f )(x) = 0

In [5] the Hankel transform of infinitely differentiable functions with compact supports has been characterized as entire functions of exponential type. We give here another characterization based on Corollary 1. Let U be the space of square integrable functions on R+ decreasing faster than any polynomial of x−1 . First we have to describe the range V of the Hankel transform on U . THEOREM 6. A function f (x) is the Hankel transform Hν , Re ν ≥ 1/2, of a function g(y) square integrable together with y n g(y), n = 1, 2, . . . , if and only if i) f (x) is infinitely differentiable on R+ ; 1 1 d2 2 n n = 0, 1, . . . , belong to L2 (R+ ); ii) ( dx 2 + x2 ( 4 − ν )) f (x), 2 d 1 1 2 n iii) ( dx2 + x2 ( 4 − ν )) f (x), n = 0, 1, . . . , tend to 0 as x tends to 0 and to infinity; d d2 iv) dx ( dx2 + x12 ( 41 − ν 2 ))n f (x), n = 0, 1, . . . , tend to 0 as x tends to infinity and are bounded at 0; PROOF. a) Let y n g(y) belong to L2 (R+ ) for all n = 0, 1, 2, . . ., then y n g(y) belong to L1 (R+ ) for all n = 0, 1, 2, . . . . Let f (x) be the Hankel transform Hν of g(y). a-i) We have [1] n X dn −n (−1)j J (x) = 2 ν n dx j=0

n j

!

Jν−n+2j (x).

(54)

Hence, n X k X dn √ xyJ (xy)) = ( (−1)n+j−k 2−k (−1/2)n−k ν dxn k=0 j=0

n k

!

k j

!

x1/2+k−n y 1/2+k Jν−k+2j (xy).

(55)

Here (a)n is the Pochhammer symbol defined by (a)n = Γ(a + n)/Γ(a). The Bessel function of the first kind Jν (y) has the asymptotics [1] ( q

h

2 cos πy Re ν



νπ 2

−

π 4



1−4ν 2 8y



sin y −

νπ 2

−

π 4

i

+ O(y −2 ) , ,

y→∞ . O(y ) y→0 (56) ∂n √ 1/2+Re ν xyJ (xy)] as a function of y has the order O(y ) in the neighConsequently, ∂x [ ν n ∂n √ bourhood of 0 and O(y n ) at infinity. Hence, ∂x xyJν (xy)]g(y), Re ν > −1, as a n[ function of y belong to L1 (R+ ) for all n = 0, 1, 2, . . .. Therefore, f (x) is infinitely differentiable on R+ . a-ii) Since Jν (x) satisfies the differential equation [1] Jν (y) =

y−

+

00

0

x2 u + xu + (x2 − ν 2 )u = 0, then

√

(57)

xJν (x) is a solution of the equation 1 x u + x + − ν 2 u = 0. 4 2



00



2

(58)

Therefore, we have "

1 ∂2 + 2 2 ∂x x



1 − ν2 4

# n

√ √ ( xyJν (xy)) = (−y 2 )n xyJν (xy).

(59)

Consequently, "

1 d2 + dx2 x2



1 − ν2 4

# n

f (x) = (−1)n

∞

Z

√

0

xyJν (xy)y 2n g(y)dy,

Re ν > −1.

(60)

For y 2n g(y) ∈ L2 (R+ ) and the Hankel transform is bounded on L2 (R+ ) we obtain that + x12 ( 14 − ν 2 )]n f (x), Re ν > −1, n = 0, 1, . . . , belong to L2 (R+ ). √ a-iii) Because the kernel xyJν (xy) has the asymptotics xν+1/2 as x tends to 0, is uniformly bounded on (0, ∞) if Re ν ≥ −1/2, and y 2n g(y) ∈ L1 (0, ∞), then applying the dominated convergence theorem [10] we have

d2 [ dx 2

d2 1 lim + 2 2 x→0 dx x "

= (−1)n

Z 0

∞



1 − ν2 4

# n

f (x)

√ lim [ xyJν (xy)]y 2n g(y)dy = 0, Re ν > −1/2.

x→0

(61)

Reasoning the same way, for every  > 0 one can choose N large enough so that Z

∞

√

N

xyJν (xy)y

2n

g(y)dy < ,

(62)

uniformly with respect to x ∈ R+ . The Bessel function Jν (y) has the asymptotics (56), therefore, the integral Z bx √ yJν (y)dy, Re ν ≥ −1/2, (63) ax

is uniformly bounded for all non-negative a, b and x. Hence b

Z

a

√

xyJν (xy)dy =

1 Z bx √ yJν (y)dy, x ax

Re ν ≥ −1/2,

(64)

tends to 0 uniformly in a, b for 0 ≤ a < b < ∞ as x tends to infinity. Consequently, applying the generalized Riemann-Lebesgue theorem [8] we get lim

Z

N

x→∞ 0

√

xyJν (xy)y 2n g(y)dy = 0,

0 < N < ∞, Re ν ≥ −1/2.

(65)

Because  can be taken arbitrarily small, from (62) and (65) we have lim x→∞

Z

∞

0

√

xyJν (xy)y 2n g(y)dy = 0,

Re ν ≥ −1/2.

(66)

Hence, d2 1 lim + 2 2 x→∞ dx x "



1 − ν2 4

# n

f (x) = 0,

n = 0, 1, . . . , Re ν ≥ −1/2.

(67)

a-iv) Using the formula [1] 1 ∂ √ [ xyJν (xy)] = ∂x 2

r

y y√ xy(Jν−1 (xy) − Jν+1 (xy)) Jν (xy) + x 2

(68)

we have # n

d2 1 1 1 Z ∞√ n d 2 (−1) + −ν f (x) = xyJν (xy)y 2n g(y)dy dx dx2 x2 4 2x 0 1Z ∞√ 1Z ∞√ 2n+1 + xyJν−1 (xy)y g(y)dy − xyJν+1 (xy)y 2n+1 g(y)dy. 2 0 2 0 "



(69)

As functions of x from a-iii) and especially from (61) and (66) we see that when Re ν ≥ 1/2 the first and the second expressions on the right hand side of (69) tend to 0 at infinity, whereas the third expression tends to 0 both at 0 and infinity. Since (xy)−1/2 Jν (xy) and (xy)1/2 Jν−1 (xy), Re ν ≥ 1/2, are uniformly bounded, the first and the second expressions on the right hand side of (69) are bounded for all x ∈ R+ and, in particular, at d d2 0. Hence, dx [ dx2 + x12 ( 14 −ν 2 )]n f (x), Re ν ≥ 1/2, tends to 0 at infinity and is bounded at 0 .

2

d 1 1 b) Suppose now that f satisfies the conditions i)-iv) of the theorem. Then [ dx 2 + x2 ( 4 − ν 2 )]n f (x), n = 0, 1, . . . , belong to L2 (R+ ). Let gn (y) be their Hankel transforms, that means

gn (y) =

∞

Z

√

0

d2 1 xyJν (xy) + dx2 x2 "



1 − ν2 4

# n

Re ν ≥ 1/2, n = 0, 1, 2, . . . ,

f (x)dx,

(70) where the integral is understood in L2 meaning. Putting gnN (y)

=

Z

N

√

1/N

d2 1 xyJν (xy) + dx2 x2 "



1 − ν2 4

# n

f (x)dx,

n = 0, 1, 2 . . . ,

(71)

we see that gnN (y) tends to gn (y) in L2 norm as N → ∞. Let n ≥ 1. Integrating (71) by parts twice we obtain  x=N  f (x)  x=1/N   x=N " # n−1    ∂ √  1 1 d2 2 − ( xyJν (xy)) + − ν f (x)  ∂x  dx2 x2 4  √

d d2 1 gnN (y) = xyJν (xy) + 2 2  dx dx x "



1 − ν2 4

#n−1

x=1/N

+

Z

N

1/N

"

2

∂ 1 + ∂x2 x2



1 − ν2 4

#

√ ( xyJν (xy))

"

2

d 1 + dx2 x2



1 − ν2 4

#n−1

f (x)dx.

(72)

Using formulas (59) and (68) we get gnN (y)

=

q

1 d d2 + N yJν (N y) dx dx2 x2 "

d d2 1 y Jν (y/N ) + 2 − 2 N dx dx x "

r





1 − ν2 4

1 − ν2 4

#n−1

1 − ν2 4

#n−1

1 − ν2 4

#n−1



d2 1 1 q + 2 + y N yJν+1 (N y) 2 2 dx x



1q d2 1 + N yJν (y/N ) + 2 2 2 dx x



1 − ν2 4

#n−1

1 y d2 1 + y Jν−1 (y/N ) + 2 2 2 N dx x



1 − ν2 4

#n−1

1 y d2 1 − y Jν+1 (y/N ) + 2 2 2 N dx x



1 − ν2 4

#n−1

"

"

"

r

r

"

"

(74)

#n−1

1 q d2 1 − y N yJν−1 (N y) + 2 2 2 dx x

"

f (1/N )

1 − ν2 4



r

(73)

#n−1

d2 1 y Jν (N y) + 2 2 N dx x

1 − 2

f (N )

f (N )

(75)

f (N )

(76)

f (N )

(77)

f (1/N )

(78)

f (1/N )

(79)

f (1/N )

(80)

−y

2

Z

N

√

1/N

Here P



d dx



f (N ) means P

d2 1 xyJν (xy) + 2 2 dx x "



√

d dx





f (x)

x=N



1 − ν2 4

#n−1

f (x)dx.

(81)

. h



2

in−1

d d Since N yJν (N y) is uniformly bounded and dx + x12 14 − ν 2 f (N ) tends to 0 dx2 as N approaches infinity (property iv)), the expression on the right hand side of (73) tends in−1 h 2  d 1 1 d 2 f (1/N ) − ν + to 0 as N approaches infinity. Using property iv) we see that dx dx2 x2 4

is bounded, whereas function

q

y J (y/N ) N ν

has the order O(N −ν−1/2 ) at infinity. Hence, ex-

pression (74) tends to 0 as N approaches infinity. Similarly, function h

d2



1 2

q

y J (N y) N ν

has the

in−1

order O(N −1 ) and dx2 + x12 41 − ν 2 f (N ) is o(1) (property iii)), therefore, expression √ √ 1 (75) is o(1). Functions 2 y N yJν−1 (N y) and 21 y N yJν+1 (N y) both are o(1), and function h

d2 dx2

1 x2



1 4

− ν2

in−1

f (N ) is also o(1) (property iii)), therefore, both expressions (76) q √ and (77) tend to 0 as N tends to infinity. Functions 21 N yJν (y/N ), 12 y Ny Jν−1 (y/N ), and 1 y 2

q

+

y J (y/N ) N ν+1

h

2



in−1

d 1 1 f (1/N ) is o(1) as are bounded, whereas function dx − ν2 2 + x2 4 N tends to infinity (property iii)), hence, all the expressions (78), (79) and (80) tend to 0 as N tends to infinity. Function (81) converges to −y 2 gn−1 (y) as N approaches infinity, hence, gn (y) = −y 2 gn−1 (y), and therefore, gn (y) = (−y 2 )n g0 (y), n = 0, 1, . . . . But if g is the Hankel transform (53) of f , then f is the Hankel transform (52) of g. Therefore, f (x) is the Hankel transform of a function g(y) = g0 (y) such that y 2n g(y) ∈ L2 (R+ ), n = 0, 1, . . . , and Theorem 6 is thus proved.

COROLLARY 2. The Hankel transform Hν , Re ν ≥ 1/2, is a bijection in the space of functions f (x), square integrable together with all xn f (x), n = 1, 2, . . . , and satisfying conditions i)-iv) of Theorem 6. Indeed, if f (x) is square integrable together with all xn f (x), n = 1, 2, . . . , then its Hankel transform satisfies all conditions i)-iv) of Theorem 6. Because f (x) satisfies all conditions i)-iv) of Theorem 6, and the inverse of the Hankel transform has the symmetric form, then its Hankel transform multiplied by xn is square integrable for all n = 0, 1, . . .. Therefore, the Hankel transform maps one-to-one the space of functions f (x), satisfying conditions i)-iv) of Theorem 6 and square integrable together with all xn f (x), n = 1, 2, . . . , into its subspace. But since the Hankel transform has the symmetric inverse in this space (as a subspace of L2 (R+ )), it is a bijection. 8. Hankel transform of functions with compact supports or vanishing at 0 Since the Hankel transform is a homeomorphism on L2 (R+ ) from Theorem 6 it follows that it is also a homeomorphism from U onto V in L2 (R+ ) norm. If we choose P (y) = y 2   1 1 d2 2 and Q = dx2 + x2 4 − ν then P (y) is a multiplier of U and Q is an operator in V and

moreover, "

d2 1 + 2 2 dx x



1 − ν2 4

#

Hν f (x) = −Hν (y 2 f (y))(x),

Re ν > −1.

(82)

Therefore, we can describe now the Hankel transform of square integrable functions with compact supports.

References [1] M. Abramowitz and I.A. Stegun, Handbook of Mathematical Functions, with Formulas, Graphs, and Mathematical Tables. Dover Publication, New York, 1972. [2] H.H. Bang, A property of infinitely differentiable functions. Proc. Amer. Math. Soc. 108(1990), no. 1, p. 73-76. [3] H.H. Bang, Functions with bounded spectrum. Trans. Amer. Math. Soc. 347(1995), no. 3, p. 1067-1080. [4] Yu.A. Brychkov, H.-J. Glaeske, A.P. Prudnikov and Vu Kim Tuan, Multidimentional Integral Transformations. Gordon and Breach, New York - Philadelphia - London - Paris Montreux - Tokyo - Melbourne - Singapore, 1992. [5] Koornwinder T. A new proof of a Paley-Wiener type theorem for the Jacobi transform. Ark. Mat. 13(1975), 145-159. [6] R.E.A.C. Paley and N. Wiener, Fourier Transforms in the Complex Domain. Colloq. Publ. Amer. Math. Soc., 1934. [7] E.M. Stein and G. Weiss, Introduction to Fourier analysis on Euclidean spaces. Princeton Univ. Press, Princeton, 1971. [8] E.C. Titchmarsh, Introduction to the Theory of Fourier Integrals. Chelsea Publishing Company, New York, 1986. [9] Vu Kim Tuan, On the range of the Y-transform. Bull. Australian Math. Soc. 54(1996), no. 2. [10] K. Yosida, Functional Analysis. Sixth Edition, Springer-Verlag, Berlin–Heidelberg–New York, 1980.