On Possible Deterioration of Smoothness under the
Operation of Convolution
A. Muhammed Uludag*
˘
Department of Mathematics, Bilkent Uni¨ersity, 06533 Bilkent, Ankara, Turkey Submitted by N. H. Bingham
Received March 8, 1996
We give some sufficient conditions of deterioration of smoothness under the operation of convolution. We show that the convolution of two probability densities which are restrictions to ⺢ of entire functions can possess infinite essential supremum on each interval. 䊚 1998 Academic Press
1. INTRODUCTION
It is known that, as a rule, the operation of convolution improves
w x
smoothness. This rule was mentioned by Paul Levy in his book 1, p. 91 . In
´
w x order to elaborate the domain of applicability of this rule, D. Raikov 2 constructed two probability densities p , p on1 2 ⺢ which are restrictions to ⺢ of entire functions, but their convolution⬁
p x
Ž .
s p ) pŽ
1 2. Ž .
x sH
p1Ž
xy s p s ds,.
2Ž .
xg ⺢,y⬁
although infinitely differentiable, is not analytic everywhere on ⺢. We show that the deterioration of smoothness under convolution can be much greater than in Raikov’s example. We prove this by a method different from Raikov’s. Nevertheless, Raikov’s method permits us to obtain some conditions of deterioration presented in this article.
* Current address for correspondence: A. Muhammed Uludag, Institut Fourier, B.P. 74,˘
38402 Saint Martin-d’Heres Cedex, France. E-mail: uludag@ujf-grenoble.fr.`
335
0022-247Xr98 $25.00 Copyright䊚 1998 by Academic Press All rights of reproduction in any form reserved.
2. NOTATION
Ž . We shall adopt the following notation for some subsets of L1⺢ :
䢇 Lq1 is the set of all nonnegative functions on⺢ belonging to L ⺢1Ž . and not equivalent to 0;
䢇 ELq is the set of all functions of Lq which are restrictions to⺢ of
1 1
entire functions;
䢇 E L⬁ q is the subset of ELq consisting of the restrictions to ⺢ of
1 1
< < 4 entire functions bounded in each strip z: Im z F r , r ) 0;
䢇 E L q is the subset of ELq consisting of the functions which are
1 1
restrictions to⺢ of entire functions of order not exceeding ;
䢇 ULq is the set of all functions fg L possessing the followingq
1 1
x w
property: for any nonempty interval ␣,  , the equality
ess sup f x
Ž .
s ⬁Ž .
1x w
xg␣ ,  is valid.
The set Lq1 consists of functions equal to a probability density up to a positive constant factor. The sets ELq1, E⬁Lq1, E L q1 can be viewed as subsets of Lq1 consisting of functions with ‘‘extremely good smoothness.’’ The set ULq1 can be viewed as a subset of Lq1 consisting of functions with ‘‘extremely bad smoothness.’’
We define the operators S: Lq1ª Lq1 by the equality ⬁
Sf x s f xq t f t dt, xg ⺢. 2
Ž
. Ž .
H
Ž
. Ž .
Ž .
y⬁
Ž .
We accept the agreement that S is defined by 2 e¨erywhere on⺢. Note
Ž . that Sf is an even function, and S is the operator of convolution of f x
Ž .
and f yx . We call Sf the symmetrization of f.
A standard characterization of growth of a function analytic in the disc z: z< <- R , R F ⬁, is4
< <
M r , f
Ž
.
[ max f z ,Ž .
0F r - R.< <zFr
< < 4 Ž .
If f is analytic in the strip z: Im z - R , we shall use, besides M r, f , the characteristic
< <
H r , f
Ž
.
[ sup f z ,Ž .
0F r - R.Ž . Ž .
Evidently, M r, f F H r, f , for 0 F r - R. If R s ⬁, i.e., f is an entire function, then, besides its order, defined by
log log M r , f
Ž
.
w x
f [ lim sup , log r rª⬁ w xwe shall consider another characteristic f , defined by log log H r , f
Ž
.
w x
f [ lim sup . log r rª⬁ w x w x Evidently, 0F f F f F ⬁. q w xIf fg L , we define the quantity h f as1
w x
h f [ sup r ) 0: f is the restriction to ⺢ of a function
< <
4
analytic and bounded in the strip z : Im z F r .
4
⬁ q w x
If fg E L , we define h f s ⬁. If there is no function whose restric-1 < < 4 tion to ⺢ is f and analytic and bounded in some strip z: Im z F r , we
w x define h f s 0.
3. STATEMENT OF RESULTS
As we have mentioned, the set ULq1 consists of functions with extremely bad smoothness. For example, if fg ULq1, then f cannot coincide almost everywhere with a continuous function in any interval.
q q Ž q.
THEOREM1. There exists fg EL such that Sf g UL , i.e., S EL l1 1 1
ULq1/ ⭋.
In the proof of this theorem, we use a theorem of T. Carleman on ‘‘touching’’ approximation by entire functions on ⺢. By the help of the generalization of this theorem due to Keldysh, it is possible to prove the following refinement of Theorem 1:
THEOREM 2. There is a function fg E L3 q1 such that Sfg ULq1, i.e.,
Ž q. q
S E L3 1 l UL / ⭋.1
Now we give some conditions of deterioration of smoothness obtained by use of Raikov’s method. The basic result in this direction is the next theorem.
THEOREM3. Let fg Lq1. Then
Ž .i h Sfw xG h f , and M r, Sf s H r, Sf F f H r, fw x Ž . Ž . 5 51 Ž . for r
-w x
h f .
Ž .ii If Sf is analytic in the disc z: z < <- R , then h Sf G R, h f G4 w x w x
Rr2, and the following inequality is¨alid:
1r2 1 5 5 5 5 M r , Sf
Ž
.
s H r, Sf F f H r, f F fŽ
.
1Ž
.
1½
M 2 rŽ
Ž
q h , Sf.
.
5
, h r) 0, h ) 0, 2 r q h F R.Ž
.
Ž .
3 ⬁ q w xSince fg E L m h f s ⬁, the following corollary is immediate:1 COROLLARY 1. In order that fg E⬁Lq1 it is necessary and sufficient that
Sfg EL1q. Moreo¨er, Sfg ELq1 implies Sfg E⬁Lq1. If fg E⬁Lq1 then the
w x w x w x w x
relation f F f s Sf s Sf is¨alid.
Ž w x w x. ⬁ q
Now we describe the possible pairs f , f for f g E L :1
Ž .
THEOREM4. Let , be a pair of numbers such that 1 F F F ⬁.
⬁ q w x w x
There exists a function fg E L such that1 f s , f s .
⬁ q w x w x
Therefore, if fg E L is of fixed order1 f s , then the order Sf
⬁ q w x w x
of Sf can be arbitrarily large. Now let f, gg E L . If1 f - g , then it
w x w x
is natural to consider f as ‘‘smoother’’ than g. Since Sf s f , the functions constructed in Theorem 4 can be interpreted as examples of deterioration of smoothness under convolution.
Ž . w x
From Theorem 3 ii , it also follows that if h f s 0, then Sf cannot be w x
analytic at the origin. This is Raikov’s result 2 . To show that convolution Ž . can deteriorate smoothness, he then considered the function f x s
w yxx4 q w x
drdx exp 1 y exp e , which belongs to EL , but h f1 s 0. However, although not analytic at the origin, Sf is infinitely differentiable on⺢, and
w x
f s ⬁. We construct the following examples:
q w x
THEOREM5. There exists fg E L with h f s 0, i.e., Sf is not analytic1 1
at the origin.
THEOREM 6. For each n, there exists an fg E1q1r nLq1 such that Sf is
Ž .
not 2 nq 2 -times differentiable at the origin.
Theorem 6 is proved by the help of the following theorem, which is obtained by a refinement of Raikov’s method.
THEOREM7. If fg Lq1 is not n-times differentiable, or if it is but not all
Ž .
of the n deri¨ati¨es are bounded, then Sf is not 2 nq 2 -times differentiable at the origin.
Note that, by Theorem 7, for any function fg ELq1 unbounded on⺢, Sf is not twice differentiable at the origin.
4. PROOF OF THEOREM 1
We begin by the construction of a continuous function gg Lq1 such
Ž .Ž . Ž .Ž . 5 5
that Sg x s ⬁ for any x g ⺡. Note that Sg 0 s g . However, there2
q Ž .
are continuous functions gg L such that g f L ⺢ . This is the basic1 2 fact in this construction.
For 2F n g ⺞ denote by s the function continuous on ⺢, equal to zeron
x y3 y3w w y3x
for xf n y n , n q 2n , equal to n for xg n, n q n , and linear
w y3 x w y3 y3x
for xg n y n , n , and for x g n q n , n q 2n . Define ⬁
qs
Ý
s .nŽ .
4ns2
5 5 y2
Since the supports of s s do not overlap, andn sn 1s 2n , it is easy to verify that q is continuous on⺢ and belongs to Lq1. For any nonnegative integer a, we have ⬁ ⬁ ⬁ Sq a s s t s tq a dt
Ž
. Ž .
H
½
Ý
nŽ .
5 ½
Ý
nŽ
.
5
y⬁ ns2 ns2 ⬁ ⬁ sH
Ý
s t snŽ .
nqaŽ
aq t dt.
y⬁ ns2 ⬁ y3 ⬁ n Ž . nq nqa GÝ
H
n nŽ
q a dt G.
Ý
2 s ⬁. nŽ
nq 2.
ns2 ns2 Set ⬁ 1 2 g xŽ .
sÝ
q kxŽ
y k ..
Ž .
5 k ks1 Ž .Since each summand of 5 is continuous on⺢ and, moreover, the support
w w
of the k th summand is contained in k,⬁ , the series converges everywhere and g is continuous on⺢. Since
⬁ 1 ⬁ 1
2
5 5g 1F
Ý
k q kxŽ
y k.
1s q5 51Ý
k2,we have gg Lq1. Let x be a non-negative rational number; set xs arb, 4 where ag ⺞ j 0 , b g ⺞. We have ⬁ ⬁ ⬁ 1 2 1 2 Sg x s q kty k q ktq kx y k dt
Ž
. Ž .
H
½
Ý
Ž
.
5 ½
Ý
Ž
.
5
k k y⬁ ks1 ks1 ⬁ 1 2 1 2 GH
½
q btŽ
y b.
5 ½
q btŽ
q a y b.
5
dt b b y⬁ ⬁ 1 s 3H
q s q sŽ . Ž
q a ds.
b y⬁ 1 s 3Ž
Sq. Ž .
a s ⬁. b Ž .Ž .Since Sg is an even function, we conclude that Sg x s ⬁ for any x g ⺡.
Thus, the function g with the properties mentioned at the beginning of the proof has been constructed. In order to construct the desired function
q w x
fg EL , we need the following theorem by Carleman 4 .1
Ž . Ž .
THEOREM Carleman . Let g be a complex-¨alued continuous function
Ž . w w
on ⺢. Let ⑀ s ⑀ r be a positi¨e decreasing continuous function on 0,⬁ . There exists an entire function f such that
< <
g x
Ž .
y f x -Ž .
⑀ x ,Ž
.
xg ⺢.Ž .
6 We shall use the following corollary to this theorem.COROLLARY 2. If g is assumed to be real ¨alued on ⺢, then f can be
Ž . Ž .
chosen real¨alued and such that f x ) g x on ⺢.
To derive the corollary, note that, by Carleman’s theorem, there exists an entire function f such that1
1 < < 1 < <
g x
Ž .
q2⑀ x y f x - ⑀ x ,Ž
.
1Ž .
4Ž
.
xg ⺢.1
Ž . Ž . 4
It is easy to see that the function f z s2 f z1 q f z1
Ž .
is entire,Ž . Ž . Ž .
satisfies f x ) g x on ⺢, and 6 is valid. Now we can construct the function fg ELq1 such that Sfg ULq1. Let g be the function defined by Ž .5 . By the corollary to Carleman’s theorem, there exists an entire function
f positive on⺢ and satisfying the condition
y< x <
g x
Ž .
y f x - eŽ .
, xg ⺢.q q Ž . Ž .
Hence, fg EL . It remains to show that Sf g UL . From f x ) g x G1 1
Ž .Ž . Ž .Ž . Ž .Ž .
0, it follows that Sf x G Sg x . Since Sg x s ⬁ for x g ⺡, we
conclude that
Sf x s ⬁, xg ⺡. 7
Ž . q
In order to derive from 7 that Sfg UL , we shall use the following two1 lemmas.
LEMMA1. If f, g are continuous nonnegati¨e functions, then the con¨
olu-tion f) g is lower semicontinuous.
Proof. The function f) g can be represented as the pointwise limit of
the nondecreasing sequence of continuous functions ⬁
n
f x
Ž
y t g t dt.
Ž .
.H
½
yn5
ns1Since the limit of a nondecreasing sequence of continuous functions is lower semicontinuous, so is f) g.
Ž .
LEMMA 2. If f is a lower semicontinuous function such that f x s q⬁
for x in a dense subset M of ⺢, then f possesses infinite essential supremum in any inter¨al.
Ž . 4 x w4
Proof. By the lower semicontinuity of f, the set x: f x ) C l ␣, 
x w
is open for any C) 0 and for any interval ␣,  . By the condition of the lemma, this set is nonempty. Since any nonempty open set has a positive Lebesgue measure, we obtain, for any set E with meas Es 0,
Ž . Ž .
supxg x␣ ,  w_ E f x ) C. Hence ess supxg x␣ ,  w f x ) C. Using the
arbitra-riness of C, we get the desired result.
We are now ready to complete the proof of Theorem 1. By Lemma 1, Sf Ž .Ž .
is a lower semicontinuous function. Since Sf x s ⬁ for x g ⺡, Sf g
q
UL1 according to Lemma 2.
5. PROOF OF THEOREM 2
Now we proceed to show that there exists an entire function f of order F 3 such that Sf g ULq. In order to construct the function f, we shall use
1
w x
a refinement of the Carleman theorem due to Keldysh 5 . For a detailed w x
exposition of this theorem see 6 .
Ž . Ž .
THEOREM Keldysh . Let g be a complex-¨alued differentiable function on⺢. Put
log r
Ž .
< <
w x
r [ max g⬘ x
Ž .
Ž .
and g [ lim sup . log r < <xFr rª⬁Then for each ⑀ G 0 there exists an entire function f whose order does not
w x < Ž . Ž .<
The following corollary is easily derived by imitating the proof of Corollary 2.
COROLLARY 3. Let g be a real-¨alued differentiable function on⺢. Then
w x
there exists an entire function f whose order does not exceed g q 1 which is
Ž . Ž .
real¨alued on⺢ and satisfies 0 - f x y g x - 1.
Now we start with the construction of the function f whose existence is asserted by Theorem 2.
Step 1. For 3F n g ⺞ denote by s the function continuous on ⺢,n
x y1 y3 y1 y3 w 3
equal to zero for xf n y n log n, n q 2n log n , equal to n log n
w y1 y3 x w y1 y3 x
in the interval n, nq n log n , and linear for n y n log n, n and
w y1 y3 y1 y3 x
for nq n log n, n q 2n log n . One can make the edges of sn
smoother, so that it becomes a differentiable function. Define ⬁ q x
Ž .
sÝ
snŽ .
xŽ .
8 ns3 and set ⬁ q kxŽ
y k!.
g xŽ .
sÝ
2 .Ž .
9 k ks3Since the supports of s s do not overlap, q is differentiable onn ⺢. Likewise,
Ž . w w
the support of the function q kxy k! is contained in k y 1, ⬁ , so that the series defining g converges everywhere and g is also differentiable. We will approximate this function by an entire function according to the
Ž .
corollary to Keldysh’s theorem. Let us first calculate r for the function Ž .
g. If x- 1, then g⬘ x is identically 0; so it suffices to consider x ) 1 only.
Ž .
So assume that 1- x - r. Then, since the function q kx y k! vanishes for
Ž . Ž . 4
kxy k!- 1, only the finite number n r [ 噛 k: k y 1 !- r of terms
Ž . Ž .
contributes to g. Note that by Stirling’s formula n r s O log r as r ª ⬁. Hence Ž . n r q kxy k!
Ž
.
g xŽ .
sÝ
2 , 1F x F r, k ks3 Ž . Ž . n r q⬘ kx y k! n rŽ
.
<g⬘ x FŽ .
<Ý
FÝ
q⬘ kx y k! ,Ž
.
1F x F r. k ks3 ks3< Ž .< Ž .2 6Ž .
Now clearly we have q⬘ x F x q 1 log x q 1 . Inserting this in the above inequality, we get, as rª ⬁,
Ž . n r 2 6 <g⬘ x F
Ž .
<Ý
Ž
kxy k!q 1 log kx y k!q 1.
Ž
.
ks3 F n r x2log6xs O r2log7 r .Ž .
Ž
.
Ž . Ž 2 7 .Therefore r s O r log r as r ª ⬁, and hence log r
Ž .
w x
g s lim sup F 2.
log r
rª⬁
By Corollary 3, we conclude that there is an entire function f , real valued0
w x Ž . Ž . Ž .
and nonnegative on ⺢ with f F 3, satisfying f x s g x q ␦ x ,0 0 Ž .
where 0-␦ x - 1. The desired function will be obtained by ‘‘shrinking’’
f0 by multiplying with the function h we describe in the lemma below, whose proof is rather technical and will be given at the end of this section.
q w x
LEMMA 3. There exists a function hg EL1 such that h s 1 and
Ž . Ž 2 . Ž y3r2.
h x s 1r x log x q O x as xª ⬁ in ⺢.
Ž . Ž . Ž .
Put f z [ f z h z . We claim that f is a function with desired0 properties.
q w x w x
Step 2. Now we shall prove that fg E L . Since3 1 h s 1, f F 3,0
w x
f is entire and f F 3. Clearly f is nonnegative on ⺢, and it remains
only to show that it is integrable. Put ␦ [ f y g. Then we have0 5 5f 1s g q5
Ž
␦ h F ␦ h q gh ..
51 5 51 5 515 5
Since␦ is bounded and h is integrable, ␦ h - ⬁. On the other hand, by1 Ž .9 we have
⬁ 1
5 5gh 1F
Ý
25q kxŽ
y k! h x. Ž .
51.k ks3
By the change of variable kxy k!s y we obtain
1 yq k!
5q kx
Ž
y k! h x. Ž .
51s q y hŽ .
ž
/
.k k 1
5 5 5 Ž . ŽŽ
Hence, in order to show that gh 1- ⬁, it suffices to show that q y h y
. .5 Ž . Ž . q k! rk 1s O k . Indeed, by 9 we have ⬁ yq k! yq k! q y h
Ž .
ž
/
FÝ
snŽ .
y hž
/
. k 1 ns3 k 1x y1 y3
Since the support of sn is contained in ny 2n log n, nq
y1 y3 w Ž . 3
2 n log n , and s xn F n log n, we obtain
yq k! sn
Ž .
y hž
/
k 1 yq k! yq k! 3 Ž . nq2r n log n FH
hž
/
dyF 4 max hž
/
. 3 k w x k Ž . yg ny1, nq1 ny2r n log n Ž . Ž 2 . Ž . Ž . Ž y3r2.Now define r x [ 1r log x . From h x s r x q O x as xª ⬁ it
Ž . Ž .
follows that h x F Cr x , x G 2 with some positive constant C. The function r is decreasing, so we have, for kG 3
yq k! ny 1 q k!
max h
ž
/
F rž
/
.k k
w x
yg ny1, nq1
On the other hand, recall the formula
⬁ ⬁
f k
Ž
.
F f k qŽ
.
f y dy,Ž .
Ý
0H
k0
ksk0
which is valid if f is a decreasing function. Using this formula, we get ⬁ yq k! ny 1 q k! q y h
Ž .
ž
/
FÝ
4Crž
/
k 1 ns3 k ⬁ 2q k! yy 1 q k! F 4Crž
/
q 4CH
rž
/
dy. k 3 k Ž .Substitute xs y y 1 q k! rk in the integral to get, for some C, ⬁ yq k! 2q k! q y h
Ž .
ž
/
F Crž
/
q CkH
r x dxŽ .
k 1 k Ž2qk! rk. ⬁ 2q k! Ck s Crž
/
y s O kŽ
.
as kª ⬁. k log x Ž2qk! rk. We conclude that f is integrable.q Ž .Ž .
Step 3. We shall prove Sfg UL . Let us first show Sf x s ⬁ for1
xg ⺡. We shall, for b g ⺞,
⬁ q kt
Ž
y k!.
q btŽ
y b!.
Sfs S hg q h␦ G S hg s S h
Ž
.
Ž
.
ž
Ý
2/
G S hž
2/
.k b
Therefore, ⬁ 1 Sf x G h t h xq t q bt y b! q bx q bt y b! dt.
Ž
. Ž .
4H
Ž . Ž
. Ž
. Ž
.
b y⬁Now let x be a nonnegative rational number; set xs arb, where a g 4
⺞ j 0 , b g ⺞. Upon the change of variable bt y b!s y and recalling that bxs a, the above inequality becomes
⬁ 1 yq b! yq a q b! Sf x G h h q y q aq y dy.
Ž
. Ž .
5H
ž
/ ž
/
Ž . Ž
.
b b b y⬁ Ž . Ž . Ž y3r2. Ž . Ž .Recall that h x s r x q O x as xª q⬁, so that h y G r y r2 for y) y . If we increase the lower limit of the above integral to by , the0 0 inequality will be preserved:
⬁ 1 yq b! yq a q b! Sf x G r r q y q aq y dy.
Ž
. Ž .
5H
ž
/ ž
/
Ž . Ž
.
b b 4 b by0Since r is a decreasing function we have ⬁ 1 yq a q b! 2 Sf x G r q y q aq y dy.
Ž
. Ž .
5H
ž
b/
Ž . Ž
.
4 b by0Now we insert the series defining q into the last inequality:
⬁ ⬁ ⬁ 1 yq a q b! 2 Sf x G r s y s aq y dy
Ž
. Ž .
5H
ž
b/
½
Ý
nŽ .
5 ½
Ý
nŽ
.
5
4 b by0 ns3 ns3 ⬁ ⬁ 1 yq a q b! 2 G 4 b5H
rž
b/
Ý
snŽ .
y saqnŽ
aq y dy..
by0 ns3 w xPut n0[ a q by q 1. Simply by eliminating the terms for which n - n0 0 above, we obtain ⬁ ⬁ 1 yq a q b! 2 Sf x G r s y s aq y dy
Ž
. Ž .
4 b5H
ž
b/
Ý
nŽ .
aqnŽ
.
by0 nsn0⬁ 1 nq1r nqa log nqaŽ . 3Ž . yq a q b!
2 G
Ý
5H
rž
/
b 4 b n nsn0 ⬁ =Ý
snŽ .
y saqnŽ
aq y dy..
nsn0⬁ w Ž The last inequality follows from the fact that Dnsn n, nq 1r n q
0
. 3Ž .4x w x Ž . 3
a log nq a ; by , ⬁ . On the other hand, note that s y s n log n,0 n
Ž . Ž . 3Ž . 2Ž 4 . 2Ž
saqn aq y s a q n log a q n , and r yq a q b! rb G r 2n q
4 . w Ž . 3Ž .4x
aq b! rb for y g n, n q 1r n q a log n q a . Using these, we
ob-tain y2 ⬁ 1 2 nq b q b! 2 nq b q b! 3 y4 Sf x G n log n log
Ž
. Ž .
5Ý
ž
/
ž
/
b b b nsn 0 s ⬁. Ž .Ž . qHence, the result Sf x s ⬁ for all x g ⺡ is proved. Since Sf is an
Ž .Ž .
even function, we conclude that Sf x s ⬁ for all x g ⺡. Since Sf is
q
lower semicontinuous to Lemma 1, Sfg UL by Lemma 2.1
Ž q. q
We believe that S E L1 1 l UL / ⭋, but we have failed to prove it.1 The latter condition cannot be improved since E L q1s ⭋ for - 1 by the Phragmen
´
᎐Lindelof theorem.¨
Proof of Lemma 3. For the construction of a function with properties
described in Lemma 3, we will use the following theorem based on an idea w x
first used by Mittag᎐Leffler in 1903 7 . This theorem will be used extensively throughout this paper.
Ž . < < 4
THEOREM 8 Mittag-Leffler . Denote by G the angle z: arg z - ,
0- - , and let g be a function analytic in G for some ␥, satisfying␥ < <y2 < <
g z
Ž .
s O zŽ
.
as z ª ⬁ in G _ G ,␥ ␣Ž
10.
Ž .
where 0-␣ - ␥. For z g int ⺓ _ G , define␦
1 g
Ž
d.
f z
Ž .
[ yH
Ž
11.
2 i ⭸ G␦ y z
for some ␦ such that ␣ - ␦ - ␥. Then
Ž .i The function f does not depend on ␦ g ␣, ␥ and can be contin-Ž .
ued to ⺓ as an entire function.
Ž .ii The following asymptotic formulas are¨alid for any ns 0, 1, 2, . . . :
< <yny1 < < O z
Ž
.
, as z ª ⬁ in ⺓ _ G ,␦ Ž n. f s yny1Ž
12.
Ž n.½
gŽ .
z q O zŽ
< <.
, as z< <ª ⬁ in G .␦ Ž . Ž . Ž .Proof. i By virtue of 10 , integral 11 converges uniformly on every
Ž . Ž .
< < 4 Ž .
put G␦, R[ z g G : z G R . If z g int ⺓ _ G , then, by the Cauchy␦ ␦ Ž .
theorem, the integral 11 does not change if we replace ⭸ G by ⭸ G␦ ␦ , R for
Ž .
any R) 0. Since the integral along ⭸ G␦, R is analytic in zg int ⺓ y G␦ , R , and, moreover, R is arbitrary, we conclude that f can be analytically continued into ⺓. According to this rule of continuation, for z g G we␦ have the representation
1 g
Ž
d.
f z
Ž .
s yH
,Ž
13.
2 i ⭸ G␦ , R y z
< < Ž .
where R) z . The function f does not depend on ␦ g ␣, ␥ since, for
Ž . Ž .
zg int ⺓ _ G , the integral 11 does not change if we replace␥ ⭸ G by␦
Ž .
⭸ G for any ␦ ⬘ g ␣, ␥ : This follows from the Cauchy theorem and␦ ⬘ Ž .
condition 10 .
Ž .ii For zg int ⺓ _ G , we chooseŽ . ␦ ⬘ g ␣, ␦ and represent fŽ . Ž n. ␦ in the form n! g
Ž
d.
Ž n. f s y2 iH
nq1.Ž
14.
⭸ G␦ ⬘Ž
y z.
Since< y z G z sin ␦ y ␦ ⬘< < <
Ž
.
for g ⭸ G and z g int ⺓ _ G , 15␦ ⬘Ž
␦.
Ž
.
Ž . Ž . Ž .
by using 14 and 10 , we obtain the first part of 12 . On the other hand, Ž .
for zg⭸ G , by the Cauchy theorem, the representation 13 gives␦
g
Ž
d.
gŽ
d.
gŽ
d.
y
H
qH
s yH
s 2 ig z .Ž .
y z y z Ž . y z
⭸ G␦ , R ⭸ G␦ ⭸ G _G␦ ␦ , R
Ž .
Hence, for zg int G we have the representation␦ 1 g
Ž
d.
f z
Ž .
s g z yŽ .
H
.Ž
16.
2 i ⭸ G␦ ⬘ y z
Ž . Ž .
Now choose␦ ⬘ g ␦, ␥ . By 16 , for z g G we have␦
n! g
Ž
d.
Ž n. Ž n.
f
Ž .
z s gŽ .
z yH
nq1.Ž
17.
2 i ⭸ G␦ ⬘
Ž
y z.
Ž . Ž . Ž .
Using 15 and 10 , we obtain the second part of 12 .
Now we are ready to prove Lemma 3. We apply the Mittag-Leffler yi z
'
Ž . Ž .
i 4
region G0[ z s re : yr2 - - 3r2; z / 1, 0 , with the branch of the logarithm real on⺢q.
1. Let us first look at the asymptotic behaviour of g in the subset of
G lying in the lower half-plane. Let0 ␦ be such that 0 - ␦ - r2. Then if
arg z- y␦ or arg z ) q ␦, we have
ey< z <sin␦ y< z <sin␦ <g z
Ž .
<s F e for z< <) 3.'
< z log z< 4Denote G[ z: yr4 - arg z - 5r4 . By Theorem 8, the function given in⺓ _ G by
1 g
Ž
d.
f z
Ž .
[ yH
2 i ⭸ G y z can be continued analytically to⺓ and satisfies
< <y1 < < O z
Ž
.
, as z ª ⬁ in ⺓ _ G, f zŽ .
s½
y1 < < < < g zŽ .
q O zŽ
.
, as z ª ⬁ in G. w x Ž . Ž . Ž < <. < <Clearly, f s 1 and f x s g x q O 1r x as x ª ⬁ in ⺢. Hence we have
¡
eyi x y1 q O xŽ
.
, as xª q⬁,'
x log x~
f xŽ .
s yi xŽ
18.
e y1 < < q O xŽ
.
, as xª y⬁.¢
i'
< <x log xŽ
q i.
2 Ž . Ž . 42. Consider the function t z [ f z q f z
Ž .
, which is also entire, < Ž .< and nonnegative on ⺢. From the estimate for f we obtain t x FŽ< < 2< <.
2r x log x . Therefore t is integrable. On the other hand, as x ª ⬁, by Ž18 we have. 2 yi x i x 2 e e 4 cos x y1 y3r2 t x
Ž .
s½
q q O xŽ
.
5
s 2 q O xŽ
.
.'
x log x'
x log x x log xŽ . Ž . 3. Finally, define the entire function h of order 1 as h z [ t z q
Ž .4 q
t zqr2 r4. Then h is the desired function. Clearly, h g EL . As1 xª q⬁ we have cos2 x sin2 x y3r2 h x
Ž .
s 2 q 2 q O xŽ
.
x log xŽ
xqr2 log x q r2.
Ž
.
1 y3r2 s x log x2 q O xŽ
.
.6. PROOF OF THEOREM 3
Before proving Theorem 3, we recall some of Raikov’s results cited w x
in 3 .
Ž .
THEOREM 9 Raikov . Let g be the Fourier transform of the function
ˆ
q < < 4
gg L . If g is analytic in the disc z: z - R then1
ˆ
⬁ r x
e g x dx
Ž .
F ⬁, yR - r - R.H
y⬁
w x
Moreo¨er, h g
ˆ
G R and the following representation is¨alid in the strip z:<Im z<- R :4 ⬁ i z x g z
Ž .
s e g x dx.Ž .
ˆ
H
y⬁ $ˆ
Ž .Now note that f
Ž
yx t s f t . Thus, we have.
Ž .
$ 2
ˆ
ˆ
<ˆ
< Sf t s f t f t s f t G 0.Ž
. Ž .
Ž . Ž .
Ž .
Hence, the transform of Sf is always nonnegative. For such functions, the following fact is valid:
q
ˆ
Ž . Ž .
THEOREM10 Raikov . Let fg L be continuous at 0. If f t G 0, then1
ˆ
q fg L and1 ⬁ 1 yi t xˆ
f xŽ .
sH
e f t dt.Ž .
2 y⬁COROLLARY4.$ If Sf is continuous at 0, then Theorem 9 is applicable for
Ž .
gs Sf and g s Sfr 2 .
ˆ
Now we pass to the proof of Theorem 3.
Ž .i First, we shall prove the following lemma:
w x w x
LEMMA 4. Let f, g be functions such that gg L . Then h f ) g G h f ,1
Ž . 5 5 Ž . w x
and the inequality H r, f) g F g H r, f is satisfied for r - h f .1
< < w x < Ž .< Ž< < .
Proof. Clearly, for Im z - h f we have f z y t F H Im z , f .
< < Hence the convolution integral converges uniformly in the strip z: Im z
w x4
F r - h f , and f ) g is analytic in this strip. Moreover, ⬁ 5 5 H r , f
Ž
) g s sup.
H
f zŽ
y t g t dt F g H r, f ,.
Ž .
1Ž
.
y⬁ <Im z<-r w xw x w x Ž . 5 5 Ž .
Now, by Lemma 4, h Sf G h f and H r, Sf F f H r, f . On the1 < < w x
other hand, by Corollary 4 we have, for Im z - h f , ⬁ 1 2 yi zt
ˆ
<Ž
Sf. Ž .
z <sH
e <f tŽ .
< dt 2 y⬁ ⬁ 1 2 Im Ž z t .<ˆ
< FH
e f tŽ .
dts Sf i Im z ,Ž
. Ž
.
2 y⬁ Ž . Ž .which shows that H r, Sf s M r, Sf . Ž .ii We want to show that the integral
⬁ r< t <
ˆ
< < e f tŽ .
dtH
y⬁is finite for 0F r - Rr2. Let r - r⬘ - Rr2. Then
⬁ ⬁ ⬁
r t<
ˆ
< r< t <<ˆ
< Ž ryr⬘.< t < r⬘< t <<ˆ
<e f t
Ž .
dtF e f tŽ .
dts e e f tŽ .
dt.H
H
H
y⬁ y⬁ y⬁
By Schwarz’s inequality, it follows that ⬁ Ž ryr ⬘.< t < r ⬘< t <
ˆ
< < e e f tŽ .
dtH
y⬁ 1r2 ⬁ 2Ž ryr⬘.< t < ⬁ 2 r⬘< t < 2ˆ
< < F½
H
e dtH
e f tŽ .
dt5
.Ž
19.
y⬁ y⬁ Ž .For the first integral in the right-hand side of 19 we have ⬁ 2Ž ryr⬘.< t < 1
e dts .
H
r⬘ y ry⬁ For the second integral,
⬁ 2 r⬘< t < 2 ⬁ 2 r⬘t 2 ⬁ y2 r⬘t 2
ˆ
ˆ
ˆ
< < < < < <
e f t
Ž .
dtF e f tŽ .
dtq e f tŽ .
dt.H
H
H
y⬁ y⬁ y⬁
< < 4
Now assume that Sf is analytic in the disc z: z - R . Then, both of the last two integrals are finite by Corollary 4, and
⬁ 2 r⬘t 2
ˆ
< < e f tŽ .
dts 2 Sf 2ir⬘ F 2 M 2r⬘, Sf ,Ž
. Ž
.
Ž
.
H
y⬁ ⬁ y2 r ⬘t 2ˆ
< < e f tŽ .
dts 2 Sf y2ir⬘ F 2 M 2r⬘, Sf .Ž
. Ž
.
Ž
.
H
y⬁Hence ⬁ 2 r⬘< t < 2
ˆ
< e f tŽ .
dtF 4 M 2r⬘, Sf ,Ž
.
H
y⬁ and we finally have1r2 ⬁ r< t < 4 M 2r⬘, Sf
Ž
.
ˆ
< < e f tŽ .
dtF F ⬁.H
½
r⬘ y r5
y⬁It follows that the integral
⬁ 1 yi zt
ˆ
e f t dtŽ .
H
2 y⬁ < < 4converges uniformly in the strips z: Im z F r for r - Rr2, and f is
< < 4
analytic in the strip z: Im z - Rr2 . On the other hand ⬁ 1 r< t <
ˆ
< < < < H r , fŽ
.
s sup f z FŽ .
H
e f tŽ .
dt 2 y⬁ <Im z<-r 1r2 1 4 M 2r⬘, SfŽ
.
F½
5
,Ž
20.
2 r⬘ y r Ž . w xwhich means that H r, f - ⬁ for r - Rr2; i.e., h f G r - Rr2. Now put Ž .
r⬘ y r s h and substitute in 20 to get
1r2 ⬁ 1 r< t < M 2 r
Ž
Ž
q h , Sf.
.
ˆ
< < H r , fŽ
.
FH
e f tŽ .
dtF½
5
. 2 y⬁ h Ž . Ž . Ž . 5 5 Ž .By part i , M r, Sf s H r, Sf F f H r, f . Joining this with the above1 inequality, we obtain the desired result.
⬁ q w x Ž .
Proof of Corollary 1. If fg E L , then h f s ⬁ and by Theorem 3 i1
w x ⬁ q q q
it follows that h Sf s ⬁, i.e., Sf g E L ; EL . Similarly if Sf g EL ,1 1 1
Ž . w x w x
then by Theorem 3 ii it follows that both h Sf s ⬁ and h f s ⬁, i.e.,
⬁ q ⬁ q Ž .
Sfg E L , f g E L . Substituting h s 1 in inequality 3 one has1 1
w x
w x
w x
Sf s Sf F f 1r2 log log M 2 1Ž
Ž
q r , Sf.
.
4
w x
F lim sup s Sf . log r rª⬁ w x w x w x w x w xHence Sf s Sf s f . Since the inequality f F f is always valid, we get the desired result.
7. PROOF OF THEOREM 4 w x
To calculate the quantity f for the functions we shall construct, the following lemma will be helpful.
< <
LEMMA 5. For 0- - r2, define the set A [ z: Im z - r,r,
<arg z<- or arg z y -  , and for the entire function f put H r, f [< < <4 Ž . < Ž .<
supzg Ar , f z . Then the inequality
w x
w x
w x
f s max f , fŽ
.
is¨alid, where log log HŽ
r , f.
w x
f [ lim sup . log r rª⬁ w x Ž w x w x.Proof. Evidently, f G max f , f . On the other hand, put
< < 4 Ž . < Ž .<
Br,[ z: Im z - r _ A , and let B r, f [ supr,  zg Br , f z . Define
log log B
Ž
r , f.
w x
b f [ lim sup . log r rª⬁ Ž . Ž Ž . Ž .. w x Ž w xSince H r, f s max B r, f , H r, f , we have  f F max b f ,
w x. w x w x f . Finally, b f F f since  < < B
Ž
r , f.
F sup f zŽ .
s M rrsinŽ
 , f ,.
< <zFrrsin w x Ž w x w x. so that f F max f , f .Now we pass to the proof of the theorem.
Case 1. 1- - - ⬁. For 0 - - y 1 - , consider the function
g z
Ž .
[ exp yizŽ
y z.
,where the branches of the power functions are taken to be positive on⺢q.
i 4
It is analytic in the region zs re : y - - . For y - - - 1 2
Ž . i 4
denote by G , 1 2 the angle zs re : - - ; and put ␥ [1 2
Ž .
min 2r, . 1. We have
< i <
log g re
Ž
.
s r sin y r cos ,Ž
21.
Ž .
which is majorized by the term r sin since ) . Hence, g z s
Ž Ž .. i Ž . Ž
. q
q ␦, y␦ , ␦ being sufficiently small. Note that, for x g ⺢ , one has
Ž . Ž .
g x s O 1rx .
Fix an ␣ satisfying r - ␣ - ␥. By Theorem 8, the function f given in
Ž Ž .. int⺓ _ G yr2 , ␣ by g
Ž
d.
f zŽ .
[ yH
y z Ž . ⭸ G yr2 , ␣can be continued analytically to⺓ and satisfies < <y1 < < O z
Ž
.
, as z ª ⬁ in ⺓ _ G yr2 , ␣ ,Ž
.
Ž n. f s yny1 Ž n.½
gŽ .
z q O zŽ
< <.
, as z< <ª ⬁ in G yr2 , ␣ .Ž
.
22Ž
.
w x w x Ž .2. Clearly, f s . Now let us show that f s r y q 1 .
Ž .
There exists an angle  - r 3 such that f is bounded in the angle z: < 4
y arg z -  . Moreover, f is bounded on the lower half-plane. By
Ž .
Lemma 5 we want to estimate H r, f . It suffices to consider the angle z:
4 Ž .
0- arg z - only. In order to estimate H r, f , we shall find the
< < q 4
supremum of f on the lines ly[ z s x q iy: 0 - arg z - . By the
construction of f we have
< <y1 < < q
f z
Ž .
s g z q O zŽ .
Ž
.
as z ª ⬁, z g l ,yŽ .
so that by 21 it follows that
< < y1 i q
log f z
Ž .
s r sin y r cos q O rŽ
.
as rª ⬁, z s re g l .ySubstitute sin s yrr, and use the estimates
2 F sin F for 0F F , 2 2 1 F cos F 1, for 0F F , 2 3 to get, for zs x q iy s reig lq, as rª ⬁, y 2 y y 1 y1 y1 < < y1 y1 r y r q O r
Ž
.
F log f z FŽ .
r y r q O rŽ
.
. 2 2Hence, for r large enough, we have
2 y y1 y y1 1
< <
r y r y 1 F log f z F
Ž .
r y r q 1.Ž
23.
2 2
For any fixed y0) 0, the largeness of r can be taken uniformly in y with 0- y F y . In the estimation of f from above, we see that the dominant0
Ž .
term as rª ⬁ for fixed y is yr r2 since G y 1 , so that the function
q Ž .
is bounded on ly and H y, f - ⬁ for all y ) 0. Calculation of the
Ž .
maximum of both sides of 23 by the usual method of differentiation gives
rŽyq1. < < rŽyq1.
K y1 F sup log f z F log H y, f F K y
Ž .
Ž
.
2Ž
24.
q
zgly
for y large enough with some positive constants K , K . On the other1 2
Ž . q < Ž .<
hand, 23 tells us that supzg ly log f z is bounded in 0- y - y for any0
Ž . rŽyq1.
y . Hence log H0  y, f F K y2 for large y. We conclude that
w x Ž . Ž
f s r y q 1 . Hence, for given , if we substitute s y
. Ž . w x
1r y 1 , then f s .
2 ⬁ q
Ž . Ž . 4 w x w x
3. Put h z [ f z q f z . Then h g E L ,
Ž .
1 h s , and hŽ . Ž < <. < < Ž .
s. Indeed, f x s O 1r x as x ª ⬁ in ⺢ by construction, so h x s Ž 2. < <
O 1rx as x ª ⬁ on ⺢, and h is an integrable function. Being
nonnega-tive on⺢, we conclude that h g ELq1. On the other hand, f is bounded on Ž< Ž .< .2 the lower half-plane, say by the constant C, therefore f z y C F
< Ž .<h z F f z q CŽ< Ž .< .2 if Im zG 0. Applying the same argument for Im z F 0 we obtain 2 2 M r , f
Ž
.
y C F M r, h F M r, f q C ,Ž
.
Ž
.
Ž
.
Ž
.
2 2 H r , fŽ
.
y C F H r, h F H r, f q C .Ž
.
Ž
.
Ž
.
Ž
.
w x w x w x w x Hence, h s f s and h s f s .For the remaining cases, we shall give a sketch of proof.
Case 2. 1- - s ⬁. For ) 1, apply the procedure in Case 1 to
Ž . Ž 2 y1 3 .
the function g z [ exp yiz log z y z log z .
Ž . Ž .
Case 3. 1s - - ⬁. Put s 2 r y 1 , and consider the
func-Ž . Ž 2 . < <
tion g z [ exp yiz log z y log z , r2 - arg z - 3r2; z ) 1. De-fine the entire function by the Cauchy-type integral along the contour
< < 4 < <
L[ z: z s 2, yr4 - arg z - 5r4 j z: z G 3, arg z s yr4 or
4
5r4 . Then apply the same procedure in Case 1.
Ž . Ž 2
Case 4. 1s - s ⬁. Consider the function g z [ exp yiz log
2 . i
zy log z log log z , which is analytic in the region G s z s re : r ) 1;
4
Case 5. 1F s F ⬁. The desired functions are Sf, where f is one of the functions constructed in previous cases. Indeed, we have exhibited
w x
functions f with given f G 1. By Corollary 1 for each of these
func-⬁ q w x w x w x
tions we have Sfg E L and1 f s Sf s Sf .
Proof of Theorem 5. Theorem 3 can be considered as a test for
w x analyticity of Sf. For example, it immediately implies that if h f - ⬁, then Sf/ ELq1. For each h) 0, there exists a function f g ELq1 such that
w x Ž . w Ž Ž ..x
h f s h; for example, consider the function f x s exp ycosh xr 2h .
Hence, symmetrizations Sf of these functions cannot be entire. As we have already mentioned, another immediate consequence of Theorem 3 is that
w x
if h f s 0, then Sf cannot be analytic at 0. Now we shall show the
q w x
existence of a function E L1 1 with h f s 0. From the function
eyi z log2z 3
g z
Ž .
[ , wherey - arg z - ,z 2 2
construct the entire function f as in the proof of Theorem 4, Case 3, and
2 q
˜
Ž . 4˜
˜
put f[ f z q f z . Then f is the desired function. Clearly, f g EL
Ž .
1˜
w x w x
and f s 1. Now let us show h f s 0. We have
< i < 2 2
log g re
Ž
.
s r log r sin q 2r log r cos y r sin y log r,< Ž i.< so that for 0F F r2 and for sufficiently large r one has log g re G
r log2 r sin y r2 sin y log r. Now assume y ) 0. For z s reig l [
y
zs x q iy: x ) 0 , one has sin4 s yrr. Hence, for z s re g l we havei y
y
i 2
< <
log g re
Ž
.
G y log y y y log r ª ⬁ as rª ⬁. 2Ž . Ž . Ž < <. Ž .
Since f z s g z q O 1r z in the upper half-plane, we get H y, f s ⬁.
Ž .
On the other hand, note that H f z1
Ž .
- ⬁ since f z is bounded in theŽ .
upper half-plane. Hence
2 2
˜
H y, f
Ž
.
G H y, f y H y, f zŽ
.
Ž
Ž .
.
4
G H y, f y H 1, f zŽ
.
Ž
Ž .
.
4
s ⬁.˜
w x Since y) 0 is arbitrary, we conclude that h f s 0.
However, symmetrization of the function constructed in Theorem 5, and
Ž . w yxx4
the symmetrization of Raikov’s example, f x s drdx exp 1 y exp e , are infinitely differentiable by the following theorem.
THEOREM 11. Assume fg Lq1 is a bounded function with continuous,
bounded deri¨ati¨es up to the order n. Then Sf has continuous, bounded
Proof. For kF n we have
Žk. Ž k . Ž k .
<
Ž
Sf.
Ž .
x <s f ) f F f< < 5 5 5 5⬁ f 1- ⬁. It follows that the convolution integrals⬁
Ž k .
f
Ž
xq t f t dt. Ž .
H
y⬁
are uniformly convergent, and Sf has n bounded derivatives.
By the refinement of Raikov’s method, a statement of the converse type can also be proved. This is the content of Theorem 7.
8. PROOF OF THEOREM 7
Ž w x.
We begin with a theorem of Raikov cited in 3 , which is actually the first step of the proof of Theorem 9.
Ž . q
THEOREM12 Raikov . Let g be the transform of g
ˆ
g L . If g is 2n-times1ˆ
differentiable on⺢, then⬁ m
< <x g x dx - ⬁,
Ž .
ms 0, 1, . . . , 2n.H
y⬁
Moreo¨er, g is 2 n-times differentiable on
ˆ
⺢, and these deri¨ati¨es can be represented by the integrals⬁
Ž m . m m i t x
g
Ž .
t s i x e g x dx,Ž .
ms 0, 1, . . . , 2n.ˆ
H
y⬁
Since the transform of Sf is nonnegative, by Theorem 10 we have the following corollary:
COROLLARY5.$If Sf is continuous at 0, then Theorem 11 is applicable to
Ž . Ž .
gs Sf and g s Sf r 2 .
ˆ
Now we are ready to prove Theorem 7. We have
ˆ
< < ⬁ ␣ ⬁ f tŽ .
 ␣ˆ
<f tŽ .
< < <t dtsŽ
1q t< < < <.
t dt.H
H
< < 1q t y⬁ y⬁By Schwarz’s inequality, this is bounded above by 1r2 ⬁ 2  2 2␣ ⬁ 1
ˆ
<f tŽ .
<Ž
1q t< <.
< <t dt dt .H
H
2½
y⬁ y⬁ < <5
1q tŽ
.
The second integral above is finite whenever  ) 1r2. For the first
ˆ
2 < <integral, we can consider f as the transform of Sf. Suppose that Sf is 2 n-times differentiable at the origin. By Corollary 5 the first integral above is finite whenever 2␣ q 2 F 2n, that is, when ␣ - n y 1r2. Hence the
⬁
ˆ
Ž . yi t xŽ .kintegral Hy⬁ f t e it dt converges uniformly for kF n y 1. Therefore
Ž .
f is ny 1 -times differentiable on ⺢, and since these derivatives tends to
0 at infinity by the Riemann᎐Lebesgue theorem, they are bounded.
Proof of Theorem 6. From the function
eyi zlog3z
g z
Ž .
s ,'
z log zconstruct the entire function f as in the proof of Theorem 4, Case 1, and
Ž . Ž . 4 w x
put h z [ f z q f z
Ž .
again. Clearly, h is entire, h s , and it is Ž . Ž < < 2< <. < <nonnegative on ⺢. Since h x s O 1r x log x as x ª ⬁ in ⺢, h is integrable and hg E L q1. Now let us show that h⬘ is unbounded on ⺢.
Ž . Ž . Ž . Ž 2.
Indeed, by Theorem 8 ii , we have f⬘ x s g⬘ x q O 1rx as x ª q⬁ on⺢. Note that, as x ª q⬁, we have
2 cos xlog3 x
Ž
.
y1 f xŽ .
q f x sŽ .
q O xŽ
.
'
x log x and y3r2 2 3 y3r2f⬘ x q f⬘ x s y2
Ž .
Ž .
x log x sin x log xŽ
.
q O xŽ
log x ..
Hence
h⬘ x s y4 x
Ž .
y2log x sin 2 xŽ
log3 x.
q O xŽ
y2.
, so that h⬘ is unbounded if G 2.Note that, if G 1 q 1rn in the above construction, then hŽ n. is
Ž .
unbounded and Sh is not 2 nq 1 -times differentiable at 0.
ACKNOWLEDGMENTS
I express my gratitude to Professor I. V. Ostrovskii for supervising this work and to Prof. A. Kerimov for his valuable critiquing.
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Ž . Ž .