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MORE ABOUT MELLIN TRANSFORM IN WEAK FUNCTIONS AND MÜNTS FORMULA
2004,24B(1):1-8
MORE ABOUT MELLIN TRANSFORM IN WEAK FUNCTIONS AND MUNTS FORMULA 1 Ding Xiaqi ( T 1.Ill1)
Luo Peizhu ( 11Jil.J-l- )
Institute of Applied Mathematics, Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing 100080, China
Abstract This paper is a further continuation of the paper [1]. In the present paper Mellin transform and Miints formula of weak functions in complex domain will be treated. Key words
Weak function, Mellin transform, Miints formula, Fourier transform
2000 MR Subject Classification
1
33C45; 42C05
Notations and Known Results
In [1] we noted that the sequence {x-! ?jJn(1og x), n = 0,1,2, ... } forms an orthogonal basis +00), where ?jJn(~) are Hermite functions. of L From the general theory developed in [2], we can construct all the weak functions over (0,00) as 2(0,
L anx-!?jJn(1ogx) = x-h(x) 00
f(x) =
(1.1)
n=O
where an, n = 0,1,2"" are complex numbers. The series (1.1) converges weakly over
K "
{,t,
bnx-I >p" (logx); b.,
n
~ 0,1, ... ,N,
We can define the Mellin transform of f(x) as
Mf(x)
= ~ a nM x - ! ?jJn(1og x ) = ~ an =
~ an
1:
00
1
are complex }.
00
xs-!-l?jJn(logx)dx
e~(S-!)?jJn(Od~.
If f(x) is a function with f(x),f'(x) continuous and bounded over any finite interval and O(x- a ) , O(x- J3 ) , a > CJ > 1, (3 > 1 as x ---.. 00. It was proved in [1] that
((s) 1 Received
roo xS-1f(x)dx= roo xS-1ff(nx)dx
Jo
May 24, 2G02
Jo
n=l
s=CJ+it,CJ>l,
(1.2)
2
ACTA lvIATHElvIATICA SCIENTIA
((s)
roo x s-l j(x)dx = roo x s-l
Jo
Jo
(f
j(nx) -
n=l
Vo1.24 Ser.B
~ roo j(V)dV) dx, Jo
0
< a < 1.
(1.3)
They are called Miints formula. Let S denote the space of rapidly decreasing functions, S' denote its conjugate space. In the following we need Theorem 1.1 below T'heorern 1.1 Suppose cp(x) E S, c > o. For any q and E > 0, there exists ci > 0 such that (1.4) Then the Fourier Transform cp(s)
1 = (j5((j + it) = 27f
1+
00
-00
. e-1sxcp(x)dx
exists and is analytic in It I < c, further for any t in It I < c, (j5( o + it) E S[(j]. On the contrary suppose 'ljJ(s) = 'ljJ((j + it) is analytic for any t in It I < c, 'ljJ((j + it) E S[(j] then there exists cp(x) E S,'ljJ(s) = (j5(x), and for any q.e > 0, there exists Cl, such that (1.4) holds. The proof may be found in [3]. Let K (c) denote all the functions satisfying (1.4), Z (c) denote the Fourier transform of all the functions in K(c);K'(c),Z'(c) denote the conjugate spaces of K(c),Z(c) respectively. This means K'(c) denotes all the weak limits of K(c) functions over K(c), Z'(c) denotes all the weak limits of Z(c) functions over Z(c). In the following we want to generalize the Miints formula to K'(c).
2
Generalizations Firstly in [1] we have proved the modified Miints formula
((s,a)
00 xS-lj(x)dx= 100 xS-lL:j((k+a)x)dx, 00 1o 0
((s, a)
roo x s-l j(x)dx = roo x s-l
Jo
Jo
(f 1
f((k
a > 1,
(2.1)
k=O
+ a)x) -
k=O
~ roo f(V)dV) dx, 0 < a < 1. Jo
(2.2)
00
(((s,ad -((s,az))
=
xS-lf(x)dx
j,~ x'-, (~f«k+ a,)x) - ~ f( (k + a,))x)
We suppose Then by Theorem 1.1, we know that
dx, a > O.
(2.3)
Ding & Luo: MORE ABOUT MELLIN TRANSFORM IN WEAK FUNCTIONS
No.1
is in Z(c) if Now
10" -
~I
3
< c. in
So
(((s, ad - ((s, a2))
1
00
1
10" - 21 < c, as t
x s-1f(x)dx
-+ 00.
is also in Z(c).
We take f(x) to be any finite linear combination of {X-t'lf'n(logX)}, N
fN(e€)
= I>n'lf'n(~)e-! = e-!hN(e€), n=O
then we have
1
00
(((s,al) - ((s,a2))
x s- 1fN(X)dx
~ 1~ X,-1 (t,fN(k + atlx) -
t,fN«k+a,)X))
dz,
a> 0,
(2.4)
by modified Miints formula, we put 00
gN(X) = :~:)fN((k + al)x) - fN((k
+ a2)x)]
k=O
N
= I:[((k + al)x)-t I: an'lf'n(log(k + al)x) 00
k=O
= x-~
N
((k
+ a2)x)-t I: an'lf'n(log(k + a2)x)]
n=O
N
00
n=O
k=O
n=O
I: an I:[(k + al)-~'lf'n(log(k + al)x) -
((k
+ a2)-~'lf'n(log(k + a2)x)]
N
= x-~
I: anPn(x) = x- t jN(X). n=O
It is easy to see that jN(X) is in K(c) for c = ~. So we have
(((s,al) - ((s,ai)) This is
(((s, al) - ((s, a2))
any
100
xs-t-1hN(X)dx =
100
xs-t-1jN(X)dx.
-00+00 ei(t-(".-t)i)€hN(e€)d~ = 1+00 -00 ei(t-(".-t)i)€jN(e€)d~. 1
Now we suppose h(e€) E K'(c) and h(e€) is the weak limit of {hN(e€)} over K(~), i.e., for E K(~), we have
ip
(2.5) So
4
ACTA MATHEMATICA SCIENTIA
Vol.24 Ser.B
By Theorem 1.1
again. So
converges to
This means
(1: ei(t-(T-i)i)~jN(e~)d~, 1: ei(t-(T-i)i)~ep(-~)d~)
converges, so
00
converges,
(J"
> 0, so we have
converges. This means jN(e~) weakly converges over K(~). Put the weak limit to be j(e~), and write
l:= anPn(e~), 00
j(e~) =
n=O
we have So we get (J"
> 0,
or
Here
f(e~)
=
e-h(e~),
g(e~)
=
e-Hj(e~),
h(e~), j(e~) E K(~),
so we have proved Theorem 2.1
If f(x) = x-h(x),h(e~) E K'(~), and (2.5) holds, then we have
1
00
(((s, ci) - ((s, a2)) where g(x)
= x-~j(x),j(e~)
x s - 1f(x)dx =
1
is the weak limit of jN(e~)
00
xS-1g(x)dx,
= 2:;":'=0 anPn(e~)
(J"
> 0,
in K'(~).
(2.6)
Ding & Luo: MORE ABOUT MELLIN TRANSFORM IN WEAK FUNCTIONS
No.1
Moreover, we have in the critical strip 0
< < 1, (J
is uniformly convergent in 0 < al s:: (J s:: a2 < 1, so we have Theorem 2.2 If f(x) = x-h(x),h(e~) E K'(~), then
Proof Sirnilarl to the proof of Theorem 2.1, we can prove
where 00
s" (x) = x-~ i" (x) = x-~ 'L, anP~ (x), n=O K
P~ (x) = 'L,[(k + ad-~1/Jn(log(k + adx) - (k + a2)-~1/Jn(log(k + a2)x)). k=O
TIlls is
-(k + a2)-~
00
'L, an1/Jn(log(k + a2)x))dx n=O
.
So
weakly approaches in Z' ( ~ ). So K
e~ 'L,[f(e~+IOg(k+a,») - f(e~+log(k+a2))) k=O
5
6
ACTA MATHEMATICA SCIENTIA
Vol.24 Ser.B
approaches weakly in K ' (~). Let the limit be 00
e~ 2:)f(e€(k + al)) - f(e€(k + a2))], k=O
so lastly we have
roo x s-l f(x)dx = ioroo x s-l {;[f((k + adx) 00
(((s, al) - ((s, a2)) io
f((k + az)x)]dx.
This is just (2.7).
3
Example We consider
f(x) = sin 7rX, x < sin
(3.1)
e/
e" ~ E L( -00, +(0), so e"€ f( e€) E 8', then by Theorem 2 in [1] we have
1
00
(((S,I)-((S,~))
xS-lf(x)dx
_ roo x s-l (~sin(k + 1)7rX _ ~ sin(k + ~)7rx) dx
- io =
~
(k+l)x
~
k=O
k +1
k=O
roo xs-2 (f sin(k + 1)7rX _ f
io
(3.2)
(k+~)x
sin(2k + 1)¥). dx 2k + 1 '
0< < l. (J
From the identity
~ sin 2kx7r _ [ ] _ ~ ~ kt: - x x+ 2'
k=l
we have
~ sin(k + 1)7rx = 7r ([=:] ~ (k+l) 2 ~ sin(2k + 1)¥
c: k=O =7r
2k + 1
=
7r
_=:
~)
2+2
{~sink7r~ _ ~ Sink7rx}
c: k=l
k7r
c: k=l
2k7r
1 Ix XI) X x ( [-]--+---([-]--+-) 4422222'
and so (2 _ 2 S )( (s ) (_ r(s - 1) cos S7r) 7r s - l 2
= 7r =
7r
1 1 1
00
o
00
o
X
1 - 2 [x Xs-2 {[x] 2 - - -x -] } dx
2
00
=
7r
s-2 { [-] X - -x + -1 - 2[-] x + -x -1 + [-] x - -X+I-} dx 22242 222
x
s-2
{2[~] -
2
2[i] -
4
~} dx.
(3.3)
Ding & Luo: MORE ABOUT MELLIN TRANSFORM IN WEAK FUNCTIONS
Na.1
7
Other Dirichlet Series
4
We consider
(/(S) _ ~ A(n) - ((s) - ~ 7 '
(5
(4.1)
> 1.
For this Dirichlet series we have the Miints formula
But we have
~ A(n)j(nx) so we have
1=
A([u] + l)f(ux)du = x
-~((:] 1= 1= + 1= 1= x s- 1j(x)dx =
xs-1dx
A([u] + l)j'(ux)(u - [u])du,
1=
A([u] + l)f(ux)du
A([u] + l)j'(ux)(u - [u])du.
xSdx
Now we use (4.1) to prove ((s) =I- 0 on We have
((/(S))
-R
1=
((s)
=
(5
f
(4.2)
= 1.
A(n)costlogn
n=l
ti"
,(5
> 1,
so
[ =
R( ( /(S) )] 2 :S ~ A(n) cos2 tlogn ~ A(n) ((s)
.L.J
.L.J n
n=l
~ (~A(n) + ~ A(n)COS2tlogn) ~ A(n) 2
.L.J n
n=l
1 = "2(P
n
n=l
.L.J n
n=l
+ Q)P.
Now
P If ((s) has a zero at s
n
n=l
1
as
rv - -
(5-1
(5
--> 1.
= 1 + if, then -R
((/(s)) ((s)
rv
_1_
(5-1'
as
s=
(5
+ i,--> 1 + if,
so we have
~
(1- E)2 (_1_)2 :S {(I + E)_l_ (5-1 2 (5-1
+
Q} (1 + E)_l_, (5-1
as s --> 1 + i,.
Hence we get
Q> {2(1- E)2 _ (1 +E)} _1_. l+E (5-1
(4.3)
8
ACTA MATHEMATICA SCIENTIA
So
Q ----'
00,
as
8 ----'
Vol.24 Ser.B
1 + i'y.
Hence the point 1 + 2i'y must be a zero of (( 8) again.
R (_ (/(8)) > {2(1- .s) _ (1 + .s)} _1_. ((8) 1+.s (1-1 But if ((8) = (8 - (1 + 2i'y))ng(8), g(1 + 2i'y) -::j:. 0, then
('(8) = _n_ ((8) (1-1
+ 0(1)
as 8 ----' 1 + 2i'y,
so
Q=R(_(/(8)) =O(1) _ _ n- as 8=(I+2i'y----,1+2i'y. ((8) (1-1 This is impossible. Hence ((1
(4.4)
+ i'y) -::j:. 0. References
1 Ding Xiaqi, Luo Peizhu. Mellin Transform in Weak Functions and Miints Formula. Acta Mathematica Scientia, 2003, 23B(3): 413-418 2 Ding Xiaqi, Luo Peizhu. Generalized Expansions in Hilbert Space. Acta Mathematica Scientia, 1999, 19(3): 241-250 3 Feng Kang. On the Theory of Distribution (in Chinese). Mathematical Advance, 1955, 1(3)
MORE ABOUT MELLIN TRANSFORM IN WEAK FUNCTIONS AND MUNTS FORMULA 1 Ding Xiaqi ( T 1.Ill1)
Luo Peizhu ( 11Jil.J-l- )
Institute of Applied Mathematics, Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing 100080, China
Abstract This paper is a further continuation of the paper [1]. In the present paper Mellin transform and Miints formula of weak functions in complex domain will be treated. Key words
Weak function, Mellin transform, Miints formula, Fourier transform
2000 MR Subject Classification
1
33C45; 42C05
Notations and Known Results
In [1] we noted that the sequence {x-! ?jJn(1og x), n = 0,1,2, ... } forms an orthogonal basis +00), where ?jJn(~) are Hermite functions. of L From the general theory developed in [2], we can construct all the weak functions over (0,00) as 2(0,
L anx-!?jJn(1ogx) = x-h(x) 00
f(x) =
(1.1)
n=O
where an, n = 0,1,2"" are complex numbers. The series (1.1) converges weakly over
K "
{,t,
bnx-I >p" (logx); b.,
n
~ 0,1, ... ,N,
We can define the Mellin transform of f(x) as
Mf(x)
= ~ a nM x - ! ?jJn(1og x ) = ~ an =
~ an
1:
00
1
are complex }.
00
xs-!-l?jJn(logx)dx
e~(S-!)?jJn(Od~.
If f(x) is a function with f(x),f'(x) continuous and bounded over any finite interval and O(x- a ) , O(x- J3 ) , a > CJ > 1, (3 > 1 as x ---.. 00. It was proved in [1] that
((s) 1 Received
roo xS-1f(x)dx= roo xS-1ff(nx)dx
Jo
May 24, 2G02
Jo
n=l
s=CJ+it,CJ>l,
(1.2)
2
ACTA lvIATHElvIATICA SCIENTIA
((s)
roo x s-l j(x)dx = roo x s-l
Jo
Jo
(f
j(nx) -
n=l
Vo1.24 Ser.B
~ roo j(V)dV) dx, Jo
0
< a < 1.
(1.3)
They are called Miints formula. Let S denote the space of rapidly decreasing functions, S' denote its conjugate space. In the following we need Theorem 1.1 below T'heorern 1.1 Suppose cp(x) E S, c > o. For any q and E > 0, there exists ci > 0 such that (1.4) Then the Fourier Transform cp(s)
1 = (j5((j + it) = 27f
1+
00
-00
. e-1sxcp(x)dx
exists and is analytic in It I < c, further for any t in It I < c, (j5( o + it) E S[(j]. On the contrary suppose 'ljJ(s) = 'ljJ((j + it) is analytic for any t in It I < c, 'ljJ((j + it) E S[(j] then there exists cp(x) E S,'ljJ(s) = (j5(x), and for any q.e > 0, there exists Cl, such that (1.4) holds. The proof may be found in [3]. Let K (c) denote all the functions satisfying (1.4), Z (c) denote the Fourier transform of all the functions in K(c);K'(c),Z'(c) denote the conjugate spaces of K(c),Z(c) respectively. This means K'(c) denotes all the weak limits of K(c) functions over K(c), Z'(c) denotes all the weak limits of Z(c) functions over Z(c). In the following we want to generalize the Miints formula to K'(c).
2
Generalizations Firstly in [1] we have proved the modified Miints formula
((s,a)
00 xS-lj(x)dx= 100 xS-lL:j((k+a)x)dx, 00 1o 0
((s, a)
roo x s-l j(x)dx = roo x s-l
Jo
Jo
(f 1
f((k
a > 1,
(2.1)
k=O
+ a)x) -
k=O
~ roo f(V)dV) dx, 0 < a < 1. Jo
(2.2)
00
(((s,ad -((s,az))
=
xS-lf(x)dx
j,~ x'-, (~f«k+ a,)x) - ~ f( (k + a,))x)
We suppose Then by Theorem 1.1, we know that
dx, a > O.
(2.3)
Ding & Luo: MORE ABOUT MELLIN TRANSFORM IN WEAK FUNCTIONS
No.1
is in Z(c) if Now
10" -
~I
3
< c. in
So
(((s, ad - ((s, a2))
1
00
1
10" - 21 < c, as t
x s-1f(x)dx
-+ 00.
is also in Z(c).
We take f(x) to be any finite linear combination of {X-t'lf'n(logX)}, N
fN(e€)
= I>n'lf'n(~)e-! = e-!hN(e€), n=O
then we have
1
00
(((s,al) - ((s,a2))
x s- 1fN(X)dx
~ 1~ X,-1 (t,fN(k + atlx) -
t,fN«k+a,)X))
dz,
a> 0,
(2.4)
by modified Miints formula, we put 00
gN(X) = :~:)fN((k + al)x) - fN((k
+ a2)x)]
k=O
N
= I:[((k + al)x)-t I: an'lf'n(log(k + al)x) 00
k=O
= x-~
N
((k
+ a2)x)-t I: an'lf'n(log(k + a2)x)]
n=O
N
00
n=O
k=O
n=O
I: an I:[(k + al)-~'lf'n(log(k + al)x) -
((k
+ a2)-~'lf'n(log(k + a2)x)]
N
= x-~
I: anPn(x) = x- t jN(X). n=O
It is easy to see that jN(X) is in K(c) for c = ~. So we have
(((s,al) - ((s,ai)) This is
(((s, al) - ((s, a2))
any
100
xs-t-1hN(X)dx =
100
xs-t-1jN(X)dx.
-00+00 ei(t-(".-t)i)€hN(e€)d~ = 1+00 -00 ei(t-(".-t)i)€jN(e€)d~. 1
Now we suppose h(e€) E K'(c) and h(e€) is the weak limit of {hN(e€)} over K(~), i.e., for E K(~), we have
ip
(2.5) So
4
ACTA MATHEMATICA SCIENTIA
Vol.24 Ser.B
By Theorem 1.1
again. So
converges to
This means
(1: ei(t-(T-i)i)~jN(e~)d~, 1: ei(t-(T-i)i)~ep(-~)d~)
converges, so
00
converges,
(J"
> 0, so we have
converges. This means jN(e~) weakly converges over K(~). Put the weak limit to be j(e~), and write
l:= anPn(e~), 00
j(e~) =
n=O
we have So we get (J"
> 0,
or
Here
f(e~)
=
e-h(e~),
g(e~)
=
e-Hj(e~),
h(e~), j(e~) E K(~),
so we have proved Theorem 2.1
If f(x) = x-h(x),h(e~) E K'(~), and (2.5) holds, then we have
1
00
(((s, ci) - ((s, a2)) where g(x)
= x-~j(x),j(e~)
x s - 1f(x)dx =
1
is the weak limit of jN(e~)
00
xS-1g(x)dx,
= 2:;":'=0 anPn(e~)
(J"
> 0,
in K'(~).
(2.6)
Ding & Luo: MORE ABOUT MELLIN TRANSFORM IN WEAK FUNCTIONS
No.1
Moreover, we have in the critical strip 0
< < 1, (J
is uniformly convergent in 0 < al s:: (J s:: a2 < 1, so we have Theorem 2.2 If f(x) = x-h(x),h(e~) E K'(~), then
Proof Sirnilarl to the proof of Theorem 2.1, we can prove
where 00
s" (x) = x-~ i" (x) = x-~ 'L, anP~ (x), n=O K
P~ (x) = 'L,[(k + ad-~1/Jn(log(k + adx) - (k + a2)-~1/Jn(log(k + a2)x)). k=O
TIlls is
-(k + a2)-~
00
'L, an1/Jn(log(k + a2)x))dx n=O
.
So
weakly approaches in Z' ( ~ ). So K
e~ 'L,[f(e~+IOg(k+a,») - f(e~+log(k+a2))) k=O
5
6
ACTA MATHEMATICA SCIENTIA
Vol.24 Ser.B
approaches weakly in K ' (~). Let the limit be 00
e~ 2:)f(e€(k + al)) - f(e€(k + a2))], k=O
so lastly we have
roo x s-l f(x)dx = ioroo x s-l {;[f((k + adx) 00
(((s, al) - ((s, a2)) io
f((k + az)x)]dx.
This is just (2.7).
3
Example We consider
f(x) = sin 7rX, x < sin
(3.1)
e/
e" ~ E L( -00, +(0), so e"€ f( e€) E 8', then by Theorem 2 in [1] we have
1
00
(((S,I)-((S,~))
xS-lf(x)dx
_ roo x s-l (~sin(k + 1)7rX _ ~ sin(k + ~)7rx) dx
- io =
~
(k+l)x
~
k=O
k +1
k=O
roo xs-2 (f sin(k + 1)7rX _ f
io
(3.2)
(k+~)x
sin(2k + 1)¥). dx 2k + 1 '
0< < l. (J
From the identity
~ sin 2kx7r _ [ ] _ ~ ~ kt: - x x+ 2'
k=l
we have
~ sin(k + 1)7rx = 7r ([=:] ~ (k+l) 2 ~ sin(2k + 1)¥
c: k=O =7r
2k + 1
=
7r
_=:
~)
2+2
{~sink7r~ _ ~ Sink7rx}
c: k=l
k7r
c: k=l
2k7r
1 Ix XI) X x ( [-]--+---([-]--+-) 4422222'
and so (2 _ 2 S )( (s ) (_ r(s - 1) cos S7r) 7r s - l 2
= 7r =
7r
1 1 1
00
o
00
o
X
1 - 2 [x Xs-2 {[x] 2 - - -x -] } dx
2
00
=
7r
s-2 { [-] X - -x + -1 - 2[-] x + -x -1 + [-] x - -X+I-} dx 22242 222
x
s-2
{2[~] -
2
2[i] -
4
~} dx.
(3.3)
Ding & Luo: MORE ABOUT MELLIN TRANSFORM IN WEAK FUNCTIONS
Na.1
7
Other Dirichlet Series
4
We consider
(/(S) _ ~ A(n) - ((s) - ~ 7 '
(5
(4.1)
> 1.
For this Dirichlet series we have the Miints formula
But we have
~ A(n)j(nx) so we have
1=
A([u] + l)f(ux)du = x
-~((:] 1= 1= + 1= 1= x s- 1j(x)dx =
xs-1dx
A([u] + l)j'(ux)(u - [u])du,
1=
A([u] + l)f(ux)du
A([u] + l)j'(ux)(u - [u])du.
xSdx
Now we use (4.1) to prove ((s) =I- 0 on We have
((/(S))
-R
1=
((s)
=
(5
f
(4.2)
= 1.
A(n)costlogn
n=l
ti"
,(5
> 1,
so
[ =
R( ( /(S) )] 2 :S ~ A(n) cos2 tlogn ~ A(n) ((s)
.L.J
.L.J n
n=l
~ (~A(n) + ~ A(n)COS2tlogn) ~ A(n) 2
.L.J n
n=l
1 = "2(P
n
n=l
.L.J n
n=l
+ Q)P.
Now
P If ((s) has a zero at s
n
n=l
1
as
rv - -
(5-1
(5
--> 1.
= 1 + if, then -R
((/(s)) ((s)
rv
_1_
(5-1'
as
s=
(5
+ i,--> 1 + if,
so we have
~
(1- E)2 (_1_)2 :S {(I + E)_l_ (5-1 2 (5-1
+
Q} (1 + E)_l_, (5-1
as s --> 1 + i,.
Hence we get
Q> {2(1- E)2 _ (1 +E)} _1_. l+E (5-1
(4.3)
8
ACTA MATHEMATICA SCIENTIA
So
Q ----'
00,
as
8 ----'
Vol.24 Ser.B
1 + i'y.
Hence the point 1 + 2i'y must be a zero of (( 8) again.
R (_ (/(8)) > {2(1- .s) _ (1 + .s)} _1_. ((8) 1+.s (1-1 But if ((8) = (8 - (1 + 2i'y))ng(8), g(1 + 2i'y) -::j:. 0, then
('(8) = _n_ ((8) (1-1
+ 0(1)
as 8 ----' 1 + 2i'y,
so
Q=R(_(/(8)) =O(1) _ _ n- as 8=(I+2i'y----,1+2i'y. ((8) (1-1 This is impossible. Hence ((1
(4.4)
+ i'y) -::j:. 0. References
1 Ding Xiaqi, Luo Peizhu. Mellin Transform in Weak Functions and Miints Formula. Acta Mathematica Scientia, 2003, 23B(3): 413-418 2 Ding Xiaqi, Luo Peizhu. Generalized Expansions in Hilbert Space. Acta Mathematica Scientia, 1999, 19(3): 241-250 3 Feng Kang. On the Theory of Distribution (in Chinese). Mathematical Advance, 1955, 1(3)