* Research supported by Office of IJaval ?esearch under contract NG0014-67-A0321-CG2.
NOTES ON CHArACTERISTIC FUNCTIONS - I:
ON DENSITIES SMALL AT INFINITY*
by
l/lalter L. Smith
Department of Statistics
University of North CaroUna at Chapel Hill
I'1Sti tute of Stati sti cs Mi1'1eO Seri es ,;;1'' ;3B
July
107~·
NOTES ON
CW\~I\CT[RISTIC
ON DENSITIES
Sf~Ll
FUNCTIONS
=
I:
AT INFINITY.*
by
1'!al ter L. Smi th
University of P70rth Carolina at C7w.peZ Hill
1.
Introduction
Let
p(x)
be a probability density function (p.d.f.)
corresponding characteristic function (c.f.).
integer
D71d
pt(e)
tile
Suppose that, for some positive
k ,
oo
o ek Ip i' (e) Ide
f
Then the first
zero as
function
x
-+
derivatives of
k
00
•
will exist, be bounded, and tend to
TI1is suggests that if, for some positive and non-decreasing
00
!~(e)
p(x)
<
of
e > 0 ,
(1.1)
then the more rapidly growing
be.
ha.ppens to be; the 1·!smol1ther: 1
Indeed, as is easily seen, if (1.1) holds with
sorle fixed real
z
i-ice)
in the strip -
a > 0 , then
p (z)
\'Jill be
~m
If.(e) = e
p(x)
Hill
as ,say, fer
analytic :function of complex
a < I"z < a
*Research supported by Office of Naval r:.esearch under contract N00014-67A-0321-01)2.
-2In proving certain theorems of renewal theory, probability densities
(see [2]).
In any case it is of interest to know to what extent a p.d.f.
with compact support is restricted in its snoothness.
A famous theorem of
Paley and Yliener settles this issue fairly easily, and we shall prove the
following.
THEOREfii 1.
pex)
A neoeBBaxoy and Buff1.:C1.:ent oondition for the exiBtenoe of a p.d.f.
with oompaot Bupport3 Buch that
(1.1)
(1.2)
<
hoZds iB that
00
•
It might be noted that one can easily arrange
tion which vanishes identically when
Ixl
~
p(x)
to be an even func-
1 , should this prove convenient.
An intriguing question centers on situations where the integral (1.2)
diverges.
For example, suppose
M(a)
Although no associated
~
exp{a/log a} ,as
p(x)
a
+
00
•
with compact support exists for such a case, is
it nevertheless possible that there are densities which are extremely small
outside a compact interval or, more correctly, small at infinity?
pretation of the phrase
statements below.
lI
TIle
int~T-
extremely small tt will be made clear in the detailed
-3-
We shall as a first step, in section 4, deduce the following theorem from
the aforementioned theorem of Paley and Wiener.
THEOREM 2. Let
to
L (_00, +(0)
2
~(x)
be a
reaZ~
non-nulZ and non-negative function
belongir~
and such that
log
~(x) I
l+lxl
~ >
Choose any convenient fixed
a(r)
=
J
Itl>r
dx < 00 •
3
and for large positive
0
r
define
logrl/~(t)] dt
Itl 3
and
b(r)
Then:
(i)
L (_00, +(0)
1
as
x
-7-
If
=~ J
~ ~~Itl~r
log[l/~(t)] dt .
t
2
x2~(x) belongs to 1 1 (_00, +(0) ~ there exists a
t.)ith a Fota'ier tmnsfoY'711
g t (6)
00
g(b(x)) = O(a(x)) .
such that
Ig1" (6) I : :
g(x)
~ (6)
in
, and.?
-4-
If
(ii)
?
x'~ep (x)
L (-00,00) there exists a, g(x) in
2
g io (6) ,[s suoh that IgO( (6) I :: ep (6) 5 and
be Zong.r3 to
x 2ep(x)
It might ~e noted here that should
~elong to L1
L2 then the
taken to be the same
functions
g(x)
function.
This claim will be clear from the proof of the theorem.
TI1eore~
arising in (i) and (ii) above can
n
2 gives us some idea of what can be
~e
ac~ieved
in situations where
the conditions for the classical Theorem XII of Paley and "liener (1] fail to
hold.
Some special examples are illuminating.
Exa~e_~
Suppose
ep(t)
cases (i) and (ii) apply.
and
a(r) = c/r.
= e- c1tl
~~
can
for some constant
c > O.
= 1 anc
rex) with
bCr)
t~<e
Thus Vle can find a
Ig(x)1
=0
choose
r -+ 00
~
.
Suppose
= e
b(r) = (cliT) log log r
Again (.1) and,., (U) apply.
IgiOCt) I
::
= (c/v)log
e- c1tl
r
and
o[e-'~J
logep (t) = -(cltl)/logltl
and get
find
both
[e-'~J
=
Example 2.
~
TI~en
,
for
and
t
~
a(r)
Por (i), say, l'le find
e
'"
We
(r log r)
-1
as
-5-
The second example indicates tl'at ",hen
rapid than exponentially, the function
l'exceedingly small n as
t -+- +
00
b(r)
g (t)
(t)
decreases at rates less
can inr:leec1. be l:-ade to be
It might ge noted that s!1ould
.1 cit
log <p (t)
2
1+t
then
<fJ
<
00
is bounded above by sone finite constant
the conclusion already in the Paley - t'Jiener
be identically zero for all
t
~
theore~,
k, say, and we obtain
namely, that
get)
can
k .
Finally, in section 5, we shall use T'1eorem 2 to treat the original
" smooth.ness l1 question of probability densities l'V"hich are swall at infinity.
!!e shall prove:
THEOREM 3.
Let H(8)
be a real:- non-decreasing fUYI-Otion of 8
and
f
OOl
I
I de
log M(8)
83
<
00
•
Deline
~:(4t)
log
3
t
and" for fixed
A > 0 ,
dt,
~
0 " 1':(0»0,
-6-
b 2 (x)
=p
+
~ IX
1T
Jog t~(4t)
2
dt,
t
A
wheT'e
P
= ~1T
oo
JA
2
log(1+16t) dt
2
.
t
Then if
[3a]
l/M(lxl)
€
L1 '
OT'
[3b]
theT'e exists an even p.d.f.
p(x)
with a
a.f. Pt (8) suah that
(1. 3)
-_._----If (Ja) holds then
(1.4)
and if (3b) 7wlds
3
then
-7(1.5 )
~x
way of illustration, consider an example similar to Example 2, above,
but for ease suppose
out, by (1.4),to be
log
r~(4x)
= ex/log x. Then the density p(x) turns
o[eXl" [ _ 1T I~ L_ e
1Tlx-pIJ]
c
as
displays the main g:eneral finding in this paper.
This exal1lple
If
Mea)
grows "just too
fast" for (1.2), but not as fast as an exponential, then a suitahly smooth
p.d.f.
can be found l'!hicr.., while not havinr: compact support, is
Ixl
extremely snaIl (e.g.o(e-e
)) at infinity.
2.
p(x)
Proof of Theoren 1
T11e necessity part of
tl~e
theorem is easily dispose0. of.
Since M(a)
..t.
positive and non-decreasing, (1.1) implies that
p(x)
pi (8) E L (_00, +00) •
l
is necessarily bounded, and, having compact support, therefore belongs
also to
L (_00, +co) •
Z
Thus
p"r (a)
must also belong to
L (-co, +co) , and \ve
Z
can appeal to Theorem XII of Paley and !:riener [1] to infer that
This implies that
(2.1)
Thus
is
-8-
Ibwever, Jensen's inequality, gives
(2.2)
since (1.1) is assumed to hold.
J
OOo
But
~
M(a)
M(O)
rr.ax{ Ilog H(O)
r~w
by
I,
log M(6)
-=--:2~ d 6 < 00 .
1+6 .
0 for all
>
If \'Ie comhine (2.1) and (/..2) \>Ie (Uscover
e
>
a
Ilog M(a)1
,so
~
log M(6)} , and (1. 2) follows.
suppose (1.2) to be given.
t1ce)(l+a 2 ) •
Let us set, for
TI1en (1.2) also holds with
_00
<
a
<
M(a) replaced
00 ;
1
<1>(6) = - - - - - - 2
Cl+l6e )H(4·1 e I)
Since
a,
M(le!) ~ M(O) >
it follohls tl'at
<P(6)
E
L..,(-oo, +00)
and we can
{~
again appeal to Theorem XII of Paley and Fiener (1] to deduce the existence of
a function
(i)
Cii)
g(x)
g(x) - 0
in
for
the transform
L (_00, +00)
2
x
~
"lith the properties;
1
gt ce )
of
g(x)
is such that
Igt(e)1 -- <p(e) •
-9-
4> (8) , and hence g' (8) , is in L1 (_00,
But it is plain that
Thus
+00)
must be equal almost everywhere to a bounded and continuous function.
g(x)
Fe can thus assume, with no loss of generality, by a suitable finite
of
g(x)
if necessary, that
for some'
e: >
is non-zero in some:' interval
g(x)
I1
shift n
(:"e:, +e:)'"
a
Now define
= Ig(x) 12
u(x)
Then
.
u(x)
belongs to
1. (_00, +00)
1
and has
a Fourier Transform
1
t
u (8)
(2.3)
= 2n
f+oo
t
-t---
g (a) g (a - 8) dx
_00
Thus
Suppose
8
~
O.
Then
lu t (a)1 s 12i f~8
4>(a)4>(a - a)da
_00
+
~i foo
te
In
_00
if we ...r rite
(2.4)
4>(a)4>(a - 8)da •
< a S ~a, Ha - 0) S cP(J,ra)
I
=
f::
in
4> (x)dx , then
~[~-]
n 2 .
]~ S
a <
00,
4>(a)
S 4>(~e)
. Thus,
-10-
It should be clear that (2.4) will also be true if
e
<
0 .
Consider
vex)
= u(x)u(-x) .
The following statements are true of
vex) ;
it is non-null, non-negative,
bounded, and identically zero outside the interval
(-1, +1) .
f1oreover it
has a Fourier transform
An argument similar to that applied to (2.3), using (2.4), ':dll then show
for some finite constant
tl~at,
A,
A
=--,-----.
(l+e--)M(lel)
Thus
(2.5)
The sufficiency part of Theorem 1 follows from (2.5) i f we take
be some suitable multiple of
vex) .
p(x)
to
-11-
3.
Some Lemmas
In this section we establish certain lemmas needed in the proof of
Theorem 2.
~(x)
Suppose
belongs to
L (_00, +00)
2
and takes only non-negative values.
Let us define A to be the set of real points
x
,,,here
(3.1)
and let
if
x
€
while if
B
be the complementary set of reals where (3.1) fails to hold.
A and
x
€
~(x) ~
A and
TIl en
1 ,
~(x)
>
1 ,
Ilog $(x)1 ~ ~(x) .
Thus
JA
110g ¢(x)
l+x
2
I. dx
Both the integrals on the right converge, the second one because
Thus
(3.2)
JA
Ilog $(x) I dx
1+x 2
<
00
•
$(x)
€
L~
-12-
Let us therefore ce£ine
1jJ
(x) =
<P
(x)
1
= --4
1'.c
....
X €
A
if
X
B,
€
l+x
T~1en,
for all real
x,
1/J(x)
<P
(x) + _1_
4
1+x
€
J +~ 11:ewiX)I
+x
dx
<
J ~~x)1
A
_00
dx
l+x'
+ J+OO
_00
~Og(l;X4)
dx
< 00
€
Ll
n
L2 then
1:ly (3.2).
l+x
have therefore proved the following.
benT'''' 3.1_ .T;
<p (x)
another function
is any non·negative function in
1/J (x)
in
+oo
J-co
Furthermore 3 if $(x)
+OO
J
s;
But
LZ also. Indeed, i~ we were given that <I>(x)
also shm'1s that 1/J (x) € L n L ' On the other han~.,
l
Z
so that
~:;e
~ <I> (x),
o s; 1/J (x)
(3.3)
(3.3)
1jJ (x)
_00
xZ¢(x)dx
< co
then
€
L2 such that
IlOg1jJ~x)LdX
<
00
~
ep(x)
then there exists
for aU
x and
•
l+x-'
L n L then
2
l
J::
1/J (x)
L2
2
x 1jJ(x)dx
1jJ(x)
<
00
€
L n L2 aZso: and if
1
-13Note that the final cow.ment in t!1is enunciation £ollo\·/s inmecHate1y from the
inequalities (3.3).
Pe shall also need the
followin~
lemma which helps us in assessing the
relative order of magnitudes of certain integrals t1,at we shall encounter.
If 4> (x)
Lermla 3.2
for. aU l.ajoge
real.
1:8
any real vaZued, non-negative
funat1.~on
in
L2 then
r
(3.4)
If
f::
2
x 4>Cx)dx
any fixed
~,
0 <
(impl.ying
< 00 s
~ <
4>(x) E L1
n
Lx ) then (3.4) shows that for
1 .
(3.5)
Proof.
Since
10g(1/y)
is convex in
(0,00), it follows from Jensen's
inequali ty that
where
y = 2r
2 Joo [4>(X )]2 dx S i2. foo
f.4>(x)]2dx =
3
r
x
r
say,
-14-
and
oCr)
~
0 as
r
Thus
~ 00.
oo
f
log{1/¢(x)} dx > logf.r/o(r)]
r
x3
4r2-'--
and this is a stronger result than the claimed one, (3.4).
To prove the latter part of the lemma we Merely note that
provided
<
00
TI1is completes
•
t~e
rroof of the lemma.
4. Proof of TheoreM 2
Let
¢(x)
EO
1./_
00 ,
+00)
be non-negative, non-null, and
suc~
that
Ilo g¢ (x) '. dx
3
l+lxl
(4.1)
although
+OO
J
(4.2)
_00
Ilo g ¢ (x)
1+x 2
I dx =
00
By Lemma 3.1 we can find a real function
$(x)
~
¢(x) ,all
x, and
IIO g¢ (x) 1_ dx
(4.3)
?
l+xFor large r
¢(x)
>
a define
<
00
•
in
L 2 (_00
'. ,
+00)
such that
-15~r(x)
Then, for
z
=x
= ~(x)
= w(x)
0
+ iy, y >
J
f+oo
_00
Let
Vr(x,y)
Ixl ~ r
for
Ixl
define
1
Ur(x,y) = =IT
(4.4)
for
t~e
> r
harmonic function
y log~r(t)
2 2 dt •
(x-t) +y
be the function conjugate to
Ur(x,y)
ail-d set
(4.5)
Evidently
+OO
f
_ll_O_g~_r~(X)
_00
l+x
2
I
dx <
00
•
Thus the following four results £0110'" £ror:l Paley!!, :"jener (1, pp. 17,18]: -
lim U (x,y)
r
(a)
y+o
= log~
r
(x)
(b)
(c)
such that
if \'Ie define
Ie (e) - 0
r
for
e
~
0 , and
-16-
then
= I'r(x)
l.i.m. hr(x + iy)
.
y-l-O
Thus
IKr (x) I
= ~ r (x)
almost ever~~here.
Next we note that for
u
r
=~ J
(x,y) - U (x,y)
s
~
f
(4.6)
s > r ,
logrw(t~/p~!2l
y.
r<ltl<s
(z) =
rs
--f- J
~l
cit
(x-t) +y
r<ltl<s
= Rf
(z), say, where
rs
!~fJ1/I(t)N(t)] dt
t-z
is analytic everywhere except for the points on the segment
r < pz < s.
s
Further, if
=0
,
Fe infer that
v (x,y) - V (x,y) = If
r
lz
()
rs Z
=
1
-
i
Jr<!tl<s
(t-x)log[1/I(t)/~(t)J dt
(t_x)2+ y 2
.
Ixl < r , we have
lim V.(x,y) - V (x,y)
y-l-O
I
S
=v
rs
(x) , say,
where
(4.7)
Fix
~,
0 <
siI'1ple identity
~
< 1 , and assume that
x
~ ~r.
Then, using in (4.7) the
-17-
"
1
t-x
1
x
x·'
= - + --- + - _.._-
t2
t
t2(t_X)
we find
=-
vrs (x)
(4.8)
~
J
1og[~(t)/~(t)] ct
n r<ltl<s
- !.
J
r<ltl<s
n
t
1ogr~(t)N(t)1
nt
2
.'
t
rs (x). say,
+ P
where
Ip
(4.9)
(x) I
-~ J
<
(1-~)
rs
r<ltl<s
1og(W(t)/~(t)]
Itl 3
Let us now set, for any convenient choice of a fixed
a(r)
(4.10)
=~ J
1og[W(t)/~(t)]
n ~~Itl~r
t
at .
~
>1
0 ,
dt
and
(4.11)
Then, from (4.8), (4.10), (4.11) we derive
(4.12)
Vrs(x)
Let us next set
= [a(r)
- a(s)]
+ x~B(r)
- 8(s)]
+
Prs(x) .
-18-
= e- ia (r)-iX(3(r)K
(4.13)
Then
r
IGr(x)
I = epr(x)
form of Gr(x)
(x)
almost everywhere, and if
gr(8)
is the Fourier Trans-
in the sense that (a.e.):
and
+A ·8
J-Ae:t xGr (x)dx
,
then (4.13) shows that (a.e.)
= e- ia (r)l:
(4.14 )
r
(e -
(3(r)) .
From (3.3), for Ixl < ~r ,
(4.15 )
lu
r
(x v) - U (x y)\
,s'
::;!
~
Jr<ltl<s 109[\jJ(t){<!J(t)]
(l-~
t
Let us, for typographical ease, write
E(r,s,x) =
e
e
-ia (r) -is (r)x
-ia(s) -is(s)x
Then define
(4.16)
mrs(x,y)
={
!1r(X+i Y)}
hs(x+iy) E(r,s,x) - 1 .
dt
= I(r,s)
, say.
From (4.5) we have
=e
(4.17)
A
rs (X,y) , say,
where
Thus, by (4.14), for all
y > 0 and all
<_ 1
Imrs (x,y ) I _
(4.18)
Furthermore, for all
Ixl ~ sr
<
+
Ixl ~ sr ,
e I (r, s) <
y+O
•
r , Ne have from (a) th.at
lim U (x,y) - Us(x,y)
(4.19)
00
r
=0
that is,
lin RA
y+o
rs
(x,y)
=0
•
Additionally, (4.16) and (4.5) show
lA rs (x,y) = Vr(x,y) - Vs(x,y) - [a(r) - a(s)] - xfS(r) - S(s)] ,
and so,
(4.20)
using (4.i2),
lim IArs(x,y)
y+O
= Vrs(x)
- [a(r)-a(s)] - x[B(r)-B(s)]
= p rs (x)..
-20-
Thus (4.17), (4.19), and (4.20) show that, as
(4.21)
Mrs(X,y) ~ e
0 ,
y
~
+
iY)IJx
ex)
ip
rs
- 1 .
t1e know from (d) that
(4.22)
and, since
we also have
(4.23)
From (4.13),
<
-
f~r
-~r
IK
(x) - h
s'
l';r
+
f -z;;r
z;r
+
J-z;r
S
(x
-21-
By (4.23) we can replace the first t\iO terms on the right of this
inequality by
and
l'
T;;r
$
J -~r1'
e(y) , a bounded function which tends to zero as
fixed (recalls> 1').
5
IKs(x)\'lm s(x,y)ldx +
l'
term is
~he ren~ining
J1;r
In 5 (x
-1;1'
+
iy) -
y + 0 , for
l(
5
I 1m1'5 (x,yJldx.
(x)
Thus, if we use (4.18),
[1 + e
e(y)
1m
f1;T
+
-1;1'
where
e (y)
's)
] f1;r
-1;1'
f1;r 1m1'5 (x,y)
-1;1'
+
$
1(1'
(x,y)
1'5
Iv. 5 (x)
- h s ex
+
iy)ldx
I IK S (x) Idx ,
I~(x)dx
,
in the last inequality includes the first two terms on the rig:1'i::
of the preceeding inequality.
Then, by
dominate~
convergence and a further
appeal to (4.18), we have from (4.21)
1;r
Ie
J
ip
rs
(x)
- ll~(x)dx
$
A
J1;1'
-;;1'
-7;1'
for some absolute constant
<
-
A.
The last integral, in tUTTI is, by (4.9),
J
log[¢(t)~~(t)] dt
~(1-1;) .r<ltl<s
Itl
A
x
J1;:r
-1;1'
x2¢(x)dx.
-22Let us set
and let us proceed on the assumption
IGr(x) I
But
= ¢r(x)
TrillS it transpires that
T.
(x)
and
~
5
ex) ),
J
{G ex)}
functions and hence there exists
~
S is finite.
a.e., so we may infer that
+
of
tJ~at
r
G(x)
Itl>~r
is a
€
L1
~
Gr(x)
(x)dx +
r
funda~ental
suc~
Tten
is in
J
ltl>~r
L
~
and that
l
(x)dx .
s
sequence of
that (USiiig the
L1 -
definition~
-23-
I+
(4.24 )
oo
IG (x) - G(x) /dx
r
_00
~
f
AS
'IT(1-Z;)
+ 2
+
The terms involving
I
logfw(t)/$(t)] dt
Itl>r
Itl 3
Jz;r</tl<r ~(t)dt
[$(t)
It/>r
+
w(t)]dt .
W on the right of this inequality give
(4.25)
+
r
W(t)dt
-r
If we appeal to Lemma 3.2, applied to
to be ultiwately negative.
The
sa~e
W(x) , then (3.4) and (3.5) show (4.25)
lemma applied to the terms involving
on the ri.ght of (4.24) then shows t!lat we can have a simplier inequality
(4.26)
for all large r,where
Let
is in
gee)
L1
...
,
\'m
A?
is some new constant.
be the Fourier Transform of
have that for almost all
TIlerefore, from (4.26), for all 12rge
e,
r,
G(x)
~
-24-
Eowever, (4.14) shows that
gr(e):: 0 when
e
~
S(r).
Thus we discover th?t
(4.27)
Note that, cy (4.11), and (3.4)
a~p1ied
to
~(x)
, for all large
r
we
shall have
B(r)
S
1. fr 10 g[l/4>(t)] = b (r)
'IT
~e
1
t2
.
then have from (4.27) the more convenient result
g(b(r)) = O(a(r)) ,
where
aCr)
=f
Itl>r
10g[1/~(t)] dt .
It I
This proves one part of Theorem 2 , the pnrt based on the assumption
2
X ~(x)
E
L1 .
To prove the other part, on the assum?tion
proceed as follows.
Then let us note that
For any function
f, say, in
?
x~~(x) E
L2(-~r, ~r)
t'-~at
L2 ' we
t~ite
-25-
+
and
IE(r,s,x)
I
=
1.
!Ie
h (x
s
iy) - K (x)
+
s
+
h (x
s
+
iy)m(x,y) ,
rs
then have
11Kr (x) - hr (x
+
From (4.22) we see that, as
II r r (x)
y
Ilh s (x
+
+
iy) I I
iy)mrz'_(x,y)l
I
+0 ,
- h (x +
r
iy) II
-+
0
and
II KS (x)
- h (x + iy)
5
II
-+ 0 •
Also,
~
!!{hs (x
+
iy) -
Ks (x)}mrs (x,Y)11
-:. 11K s (x)mrs (x,y)11 .
-26-
By (4.18) we see that
+
!'10re:->ver
IK5 (x) I
=
<p
IKs (x) I
a. e., and thus
(x)
S
Y ~ 0 , since
0 as
= Hx)
I(r,s) <
11hen
00
Ix I
$
S •
hence by dominated convergence and (4.21),
II \
as
y
~
O.
rs (x,y)
(x)m
If we write
,I<p(x){e ip rs (x) _
nIl
II
II cP (x){e
+
<
A{
J
- l} II
10gljJ(t)/Ht) dt
r<ltl<s
= AI
J
x
2
<I>
(x)
} I Ix2</l(x) II
It\3
10gtj!(t)~<P(t) dt ,
r<ltl<s
13:1t (4.13) sho\'ls
(x)
rs
A,Ai, ... and so on for various constants, (4.9) shoHs
-
since "Ie may now suppose
ip
€
L2 ·
It\
Thus we
FlaY
concl..:je
-27-
TI1US,
rather as before, we can deduce that
{Gr(x)}
L2 this time. There exists an L~ function
tends to G(x) ,as S + 00 , in L2 ' and
in
[{~(t)}2
Joo
+
r
is a fundamental sequence,
{~(t)}2]dt
+
such that
'S(x)
GS (x)
.
Now
so, by Lemma 3.2 (especially (3.4)), we find
A similar result holds \'lith
~
(t)
in place of
<p
(t) , and
in the earlier case, to the result
(4.29)
J
But, by
Parsev~lis
+OO
IG (x)
_00
r
2
.:1
- G( xJ... 1 uX
theorem,
{t.. : J
I t I >r
lo~! 1/3.1-C!-1l
(t
dt}2
"'8
are led,
~uc~
as
-28-
Since
The
g (6) :: 0
rer~ining
€J ~ 13 (1') ,
in
substitution of
VIe can hence infer from (4.29) the result
b(r)
for
S(r)
t~en
completes the proof of
Theorem 2.
5.
Proof of Theorem 3
Little need be said concerning this proof since an argUlI1Cnt almost
identical \'lith the sufficiency part of the proof of
needed, after an appeal to Theorem 2.
He)
€
(a)
if [3a] holds,
(b)
if [3b] holds,
¢(6)
1 is all that is
2
= 1/(1
+ l6e )M(4Iel).
r1en
L1 n 1.
and:
2
Suppose [3a] holds.
bounded
g (x)
inte:rv~_l
gt(0)
Set
Tl~eore!a
in
Then 111.eorem 2 shoNs the existence of a cO:1tinuous and
L
(,..E> ",1';)"
suc~ that
Ig(x)cc-x)12
Z
which is identically zero in
for
19t(t))I
,SQme.
==
€
> 0 , aTtd with
~
1 , non-,zero in som.e
a Fourier Transform
He) . t,s before, one can tako vex) =
~ 0 and deduce that v-r-(e)
00.
x
But
= O(¢(e/4))
, so that
vex) = o(lg(lxl)12) , and, by Theorem 2,
-29-
= o(aclxl))
g(bclxl))
C5.1)
, where
aClxl)
and
bclxl)
=
!
'IT
f
lx
\
A
logCl+16t2)~~ dt
--'~-----:;2
t
as defined in the enunciation of T!1eoreJ:l 3.
where
a2C!x\)
5: P + b 2 cl x
\) ,
Fe a1 so have, from CS .1), t!19.t
is also defined in that enunciation.
But, under [3a],
1,
1/(M(4It I) f2
large
Thus
Ix I
belongs to
Thus \:'e can deduce from Lenma 3.2 that for all
L .
2
,
aClxl)
= o(a2clxl))
, and part of Theorem 3 is proved.
clear that the remaining part, when
[.~h]
It should ~e
holds, fo11o'l:5 fro!'! Theorem 2 in a
very similar way.
i'le end with an example.
Intuitively one
mig'~t
monoton0 and increasing sufficiently fast to make
lY'l
J
IlogM(X)
---''<':;2;-'.' -
x
I
dx =
co
feel that if
M(x)
were
-30-
Then
[M(lxl)]-l
would necessarily belong to
example, 'vould be unnecessary.
be dropped.
{F, }
n
Define
and
Condition [3aJ, for
I'o\iJever this intuition is false and [3a) camlOt
To see this, we shall construct an increasinp sequence of reals
as follows.
F,n
Ll ' say.
M(x)
= X in F,n
-+00
as
n
-+
00
and
F,l = 1
~uppose
:0:; X
have reen defined.
n 2 F,
,..
n
'1'~~en
l; I = e
F,n+l > r, n
n+
F,2,F,3" .. ,F,
and set
< 7.F,n
, as desired.
Also
dx
- = loe 2
x
so that
Further, define
ensures that
M(x)
H(x)
=e
2l;
n
in
is non-c1.ecrea.sin.r;,-.
2F,n
:0:;
x < l; n+_1
•
It is clear that ttis
But
as
n
-+
00
•
Thus
OO
f1
J10gM(x)
I dx =
2
x
and the demonstration is apparent.
ne~d
for [3b].
00
Clearly a similar example will shm'!
t:;{~
-31-
References
[1]
PCl,ley, R. E. A. C. & ~'Tiener, 1-1. (1934), Fourier transforms in the co!".pl ex
domain. American Hath. Boc. Colloquium Pu,bUcation8.; Vol. XIX, Providence.
[2]
Smit!1, ~·r. L. (1967), ft.. theorer.-l on functions of' d,aracterlstic functions
and its application to some renewal theoretic random wdk problems,
Proceedings of the Fifth Berkeley Symposium on'~thematical Statistics
and Probability, Vol. II, Part 2, pp. 265-309, Berkeley, University of
California Press.