Spaces of holomorphic almost periodic functions on a strip

0.16 MB·English

by  S. Favorov

Spaces of holomorphic almost periodic functions on a strip
Favorov S.Yu., Udodova O.I.
Karazin Kharkov National University
7
Department of Mechanic and Mathematic
0
0 Sq. Svobody 4, Kharkov,61077,Ukraine
2 E-mail: [email protected],[email protected]

n
The notions of almost periodicity in the sense of Weyl and Besicovitch of the order p 1 are extended
a
≥
J to holomorphic functions on a strip. We prove that the spaces of holomorphic almost periodic functions in

8 the sense of Weyl for various orders p are the same. These spaces are considerably wider than the space of
2 holomorphic uniformly almostperiodic functions and considerablynarrowerthan the spaces ofholomorphic
almostperiodicfunctionsinthesenseofBesicovitch. Besidesweconstructexamplesshowingthatthespaces
]
V of holomorphic almost periodic functions in the sense of Besicovitch for various orders p are all different.
2000 Mathematics Subject Classification 42A75, 30B50.
C
.
h A continuous function f on a strip Π = z =x+iy C:a y b is called almost periodic if for
[a,b]
{ ∈ ≤ ≤ }
t any ε>0 the set of ε-almost periods
a
m
τ R:dU (f(z),f(z+τ))<ε
[ ∈ [a,b]

n o
1 is relatively dense on R, i.e., its intersection with any segment of the length L=L(ε) is nonempty. Here
v
2 dU (g,h)= sup g(z) h(z) (1)
1 [a,b] z∈Π[a,b]| − |
8
1 is the standard uniform metric on Π[a,b].
0 We denote by U the space of almost periodic functions on Π . By the Approximation Theorem,
[a,b] [a,b]
7 U is equal to the closure of the set of all finite exponential sums
[a,b]
0
h/ N
t cn(y)eiλnx, λn R, cn(y) C[a,b] (2)
a ∈ ∈
n=1
m X
v: with respect to the metric dU[a,b].
Note that any continuous periodic function on a strip Π with real periods is almost periodic since
i [a,b]
X any period is an ε-almost period for every ε > 0, and the set of periods forms a dual-sided arithmetical
r progression.
a We can replace the uniform metric dU by either the Stepanov metric
[a,b]
1
1 p
dSp (g,h)= sup g(z+t) h(z+t)pdt , p 1 (3)
[a,b]  | − |  ≥
z∈Π[a,b] Z0
 
or the Weyl metric
1
T p
1
dWp (g,h)= lim sup g(z+t) h(z+t)pdt , p 1 (4)
[a,b] T→∞z∈Π[a,b]2T ZT | − |  ≥
 − 
or the Besicovitch metric
1
T p
1
dBp (g,h)= lim sup g(t+iy) h(t+iy)pdt , p 1. (5)
[a,b] T a y b2T | − |  ≥
→∞ ≤ ≤ ZT
 − 
1
We suppose that the functions g(z) and h(z) are measurable and g(x+iy)p and h(x+iy)p are locally
| | | |
integrable with respect to the variable x for fixed y.
The closure ofthe setof sums (2) inmetrics (3) - (5)will be calledthe spaceof almostperiodic functions
inthesenseofStepanovoforderp,thespaceofalmostperiodicfunctionsinthesenseofWeyloforderp,the
spaceofalmostperiodicfunctions inthe senseofBesicovitchoforderp,and, respectively,willbe denoteby
p p p
S ,W and B . (6)
[a,b] [a,b] [a,b]
In the casea=b=0, i.e., for functions defined onthe realaxis,these spacesare well-known(see [2],[3],
[6], p. 189-247.) Note that we can also define the spaces (6) by using the concept of ε-almost period. But
for the space of B this definition is much more complicated (see. [2], p. 91-104).
[a,b]
It is clear that every sum (2) is uniformly bounded on Π , and therefore an almost periodic function
[a,b]
f(z) from the space U (or from spaces (6)) is bounded in the corresponding metric, i.e. dU (f,0)<
[a,b] [a,b] ∞
(or dSp (f,0) < , dWp(f,0) < , dBp (f,0) < ). Observe that in the definitions of spaces (6) we can
[a,b] ∞ [a,b] ∞ [a,b] ∞
replace sums (2) by functions from U .
[a,b]
It is easy to see that the elements of metric spaces (6) are equivalent classes of functions with the
corresponding zero distances. For example the equivalent class for the space Sp consists of the functions
[a,b]
which coincide a. e. on every horizontal straight line. Note that every two functions with the difference
e (x+iy)2 always belong to the same class in the spaces Wp and Bp , since dWp (e (x+iy)2,0)=0 for all
− [a,b] [a,b] [a,b] −
p < and a b. Nevertheless we will use the notation f Wp (or f Bp ); this means that the
∞ ≤ ∈ [a,b] ∈ [a,b]
equivalent class of the function f belongs to the corresponding space.
The Ho¨lder inequality implies that for all p<p
′
dSp (g,h) dSp′ (g,h), dWp (g,h) dWp′(g,h), dBp (g,h) dBp′(g,h).
[a,b] ≤ [a,b] [a,b] ≤ [a,b] [a,b] ≤ [a,b]
Besides for all p 1
≥ dBp (g,h) dWp (g,h) dSp (g,h) dU (g,h)
[a,b] ≤ [a,b] ≤ [a,b] ≤ [a,b]
(the inequality between dWp (g,h) and dSp (g,h) follows from the relation
[a,b] [a,b]
L [L] j+1
1 1 [L]+1
f(x+t) g(x+t)pdt f(x+t) g(x+t)pdt dSp (f,g))
2L | − | ≤ 2L | − | ≤ L [a,b]
−ZL j=−X[L]−1 Zj
Hence we have
Sp′ Sp ,Wp′ Wp ,Bp′ Bp (7)
[a,b] ⊂ [a,b] [a,b] ⊂ [a,b] [a,b] ⊂ [a,b]
and
U Sp Wp Bp . (8)
[a,b] ⊂ [a,b] ⊂ [a,b] ⊂ [a,b]
Herealltheinclusionsmeanthateachequivalentclassina”narrower”spaceiscontainedinsomeequivalent
class in a ”wider” space. All the inclusions are strict even in the case a=b=0: there exists the equivalent
class in a ”wider” space containing no class from a ”narrower”space (see [4]).
Note that there exists the mean value
T
1
(M f)(y)= lim f(t,y)dy (9)
t
T 2T
→∞ Z
T
−
uniformly in y [a,b] for the before mentioned functions. Indeed, the mean value of finite exponential
∈
sums (2) equals the coefficient at ei0t; in the general case, the existence of limits (9) follows easily from the
definition of almost periodicity. By the same way, we can prove the relation
T
1
lim sup f(x+t,y)dt (M f)(y) =0 (10)
(cid:12) t (cid:12)
T→∞(cid:12)(cid:12)ax∈Rymb, 2T ZT − (cid:12)(cid:12)
(cid:12) ≤ ≤ − (cid:12)
(cid:12) (cid:12)
for the space W1 and for all ”narro(cid:12)(cid:12)wer”spaces. (cid:12)(cid:12)
Let Π = z =x+iy C: a<y <b be an open strip, maybe infinite width. A func-
(a,b)
{ ∈ −∞≤ ≤∞}
tion f(z) is uniformly almost periodic on Π (or almost periodic in the sense of Stepanov, Weyl, and
(a,b)
2
Besicovitch), if the restriction f(z) to every strip Π for a<α<β <b is almost periodic in metrics (1),
[α,β]
(3) - (5). We denote by HU , HSp , HWp , HBp the correspondingspaces of holomorphic almost
(a,b) (a,b) (a,b) (a,b)
periodic functionsonΠ . Theinclusionssimilarto(7)and(8)holdforthese spaces. ThespacesHBp
(a,b) (a,b)
were studied earlier in [7]. These spaces were also defined in [1] as sets of holomorphic functions on a strip
with the following property: the restrictions to each straight line in the strip are almost periodic functions
in the sense of Besicovitch of order p. However as it follows from [1], theorem 3.4, for functions growing as
O(ee|z|N) on a strip these definitions coincide.
The following Linfoot’s theorem is well-known:
Theorem L. (see. [2], p. 146)
Spaces HU and HSp coincide for all p 1.
(a,b) (a,b) ≥
Here we obtain the similar result:
Theorem 1.
The spaces HWp coincide for all p 1.
(a,b) ≥
The proof of this theorem is based on the following proposition.
Proposition 1.
Each function f HW1 is uniformly bounded on every substrip Π , a<α<β <b.
∈ (a,b) [α,β]
Next, we prove that inclusions HU HW1 , HW1 HBp , HBp′ HBp , p > p 1,
(a,b) ⊂ (a,b) (a,b) ⊂ (a,b) (a,b) ⊂ (a,b) ′ ≥
are strict in the same sense as inclusions (7) and (8)).
Theorem 2.
Thereexistsafunctionf HW1 suchthateveryfunctiong equivalenttof inanyspaceHW1 ,
∈ ( , ) [ H,H]
does not belong to HU for all−H∞>∞0. −
[ H,H]
Theorem 3. −
There exists a function f HBp such that every function g equivalent to f in any space
∈ ( , )
p 1 −∞∞
HBp , p 1, does not belong tT≥o HW1 for all H >0.
[ H,H] ≥ [ H,H]
T−heorem 4. −
For all p > p 1 there exists a function f HBp such that every function g equivalent to f in
′ ≥ ∈ ( , )
any space HBp , does not belong to HBp′ for−a∞ll H∞>0.
[ H,H] [ H,H]
The proof−of Proposition 1. −
Suppose a<α <α<β <β <b. Since dW1 (f,0)< , we have
′ ′ [α′,β′] ∞
T0
1
f(z+u)du C
2T | | ≤
0 Z
−T0
for some C < , T < and all z Π .
0 [α′,β′]
∞ ∞ ∈
Fix r<min T , α α, β β .
0 ′ ′
{ − − }
Since the function f(z) is holomorphic, we have for all z Π
0 [α,β]
∈
1
f(z ) f(z)dxdy
| 0 |≤ πr2 | |
Z
|z−z0|<r
and
T0 T0
1 1 2T C
0
f(z ) f(z +u+iv)dudv sup f(z+u)du .
| 0 |≤ πr2 | 0 | ≤πr2 | | ≤ πr2
−ZT0|v|Z<r z∈Π[α,β]−ZT0
Proposition 1 is proved.
Consider the sums
K(q)(t)= knqe−iλnt,
n
X
whicharecalledBochner-Fejerkernels(see,forexample,[6],p.66-71). Hereλ runsoverthelinearenvelope
n
of the countable set Λ R over the field Q. The following properties of Bochner-Fejer kernels are fulfilled:
⊂
1) 0 kq 1;
≤ n ≤
2) there is only a finite number of nonzero coefficients kq for every fixed q;
n
3) kq 1 as q for every fixed n;
n → →∞
3
4) K(q)( t)=K(q)(t);
−
5) K(q)(t) 0 for all t R;
≥ ∈
6) M K(q)(t) =1.
t
Clearly, M e iλteiµt = 0 for λ = µ. Hence for any finite sum Q(x,y) of type (2) with all exponents
(cid:8) t −(cid:9) 6
λ Λ, we have
n
∈ (cid:8) (cid:9)
N
(Q K(q))(x,y)=M Q(x+t,y)K(q)(t) = c (y)kqeiλnx.
∗ t n n
n o nX=1
It follows from condition 3) that this sum convergesto Q(x,y) as q uniformly on the strip Π .
[a,b]
→∞
Take a function f W1 . Fix ε > 0 and take any sum Q(x,y) of type (2) such that dW1 (f,Q) < ε.
∈ [a,b] [a,b]
Then choose q such that
sup Q(x,y) (Q K(q))(x,y) <ε.
− ∗
(x,y)∈Π[a,b](cid:12) (cid:12)
(cid:12) (cid:12)
Put (f K(q))(x,y) = M f(x+t,y)K(q)(cid:12)(t) . It follows from (10)(cid:12)that the mean value does not change
t
∗
under shift on x R. Since
∈ (cid:8) (cid:9)
M f(x+t,y)e iλt =M f(x+t,y)e iλ(x+t) eiλx,
t − t −
n o
(cid:8) (cid:9)
we see that f K(q) (x,y) is a finite sum of type (2). Next, we have
∗
(cid:8) (cid:9)
T
1
(f K(q))(x,y) lim f(x+t,y)K(q)(t)dt
| ∗ |≤ 2T | |
T
→∞ ZT
−
for all x R, y K. By Fatou’s lemma, we obtain
∈ ∈
X T X
1 1 1
(f K(q))(x+τ,y)dx lim f(x+t+τ,y)K(q)(t)dtdx
2X | ∗ | ≤ 2T 2X | | ≤
ZX T→∞ ZT ZX
− − −
T X
1 1
lim K(q)(t) sup f(x+t,y)dx dt
T→∞2T ZT t∈R,y∈[a,b]2X ZX | | 
− −
for every τ R and X < . The expressionin the curly brackets does not exceed dW1 (f,0)+ε for large
∈ ∞ [a,b]
X. Consequently, passing to the limit as X , and then ε 0, gives the inequality
→∞ →
T
1
dW1 (f K(q),0) dW1 (f,0) lim K(q)(t)dt=dW1 (f,0).
[a,b] ∗ ≤ [a,b] T 2T [a,b]
→∞ Z
T
−
Replace f by f Q to get
−
dW1 (f K(q),f) dW1 ((f Q) K(q),0)+dW1 (Q K(q),Q)+dW1 (f Q,0) 3ε,
[a,b] ∗ ≤ [a,b] − ∗ [a,b] ∗ [a,b] − ≤
thus the exponential sums f K(q) approximate the function f with respect to the metric dW1 as well.
∗ [a,b]
The proof of Theorem 1.
Letf HW1 ,[α,β] (a,b). AccordingwithProposition1wehave sup f(z) =C < ,therefore
∈ (a,b) ⊂ | | ∞
z∈Π[α,β]
for any K(q)
f K(q) (z) = M f(z+t)K(q)(t) CM K(q)(t) =C.
t t
∗ ≤
(cid:12)(cid:16) (cid:17) (cid:12) (cid:12) n o(cid:12) n o
Hence for any p>1 (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
T
p 1
dWp (f K(q),f) = lim sup f(z+t) (f K(q))(z+t)pdt
(cid:16) [α,β] ∗ (cid:17) T→∞z∈Π[α,β] 2T ZT | − ∗ |
−
4
T
1
lim sup f(z+t) (f K(q))(z+t) f(z+t) (f K(q))(z+t)p 1dt
−
≤T→∞z∈Π[α,β] 2T ZT | − ∗ |·| − ∗ |
−
≤(2C)p−1dW[α,1β](f ∗K(q),f).
Thus the exponential sums f K(q) approximate f in the metric dWp , as claimed.
∗ [α,β]
The proof of Theorem 2.
Consider the function f(z) = e 4(z n)2, where I = n=3l 1(3k+1), k Z, l N . For any
− − −
∈ ∈
n I
z = x + iy C we have f(z) P∈e4y2 e 4(x n)2 e4y(cid:8)2 e 4(x n)2. The series ∞(cid:9) e 4(x n)2
− − − − − −
∈ | | ≤ n I ≤ n Z n=
converges for every x R and is a periodiPc∈function with the peP∈riod 1, therefore f(z) is bouPn−d∞ed in every
∈
strip z =x+iy :x R, y <H andf(z)isanentirefunction. Inparticular,f(z)isuniformlycontinuous
{ ∈ | | }
on each strip.
Let us check that f(z) HW1 . Put
∈ ( , )
−∞∞
ϕ (z)= e 4(z n)2
l − −
n=3l−X1(3k+1),
k Z
∈
. The function f (z)= ϕ (z) is the sum of the functions with periods 3l, l m, therefore f (z) is a
m l m
≤
1 l m
periodic function with the≤Pr≤ealperiod3m, and belongs to the space HU . It is sufficient to provethat
( , )
−∞∞
dWp (f,f ) 0 (11)
[−H,H] m →
as m for each H < . We have
→∞ ∞
f(z) f (z) = ∞ e 4(z 3ln 3l−1)2 ∞ e 4(x 3mn)2e4H2
m − − − − −
| − | ≤
l=Xm+1nX∈Z(cid:12)(cid:12) (cid:12)(cid:12) n=X−∞
(cid:12)T (cid:12)
and dW1 (f,f ) e4H2 lim sup 1 ∞ e 4(x+t 3mn)2dt.
Pu[−tH,H] m ≤ T→∞x∈R2T n=P−∞−RT − −
1
E1 = n Z:n 3−mx 3−mT , n1 =supE1,
∈ ≤ − − 2
(cid:26) (cid:27)
1 1
E = n Z:3 mx 3 mT <n<3 mx+3 mT + ,
2 − − − −
∈ − − 2 2
(cid:26) (cid:27)
1
E = n Z:n 3 mx+3 mT + , n =infE
3 − − 2 3
∈ ≥ 2
(cid:26) (cid:27)
for any fixed x R, T > 0. Denote by cardE the number of elements of the set E. Note that cardE
2
∈ ≤
2 3 mT +2.
−
·
For n E and t [ T,T]
1
∈ ∈ −
1
(x+t 3mn) 3m n + n
1
− ≥ 2 −
(cid:18) (cid:19)
and
e−4(x+t−3mn)2 e−32m(2(n1−n)+1)2.
≤
Similarly,
1
(3mn x t) 3m n n +
2
− − ≥ − 2
(cid:18) (cid:19)
and
e−4(x+t−3mn)2 e−32m(2(n−n2)+1)2
≤
for n E , t [ T,T]. Consequently,
3
∈ ∈ −
T
1 e 4(x+t 3mn)2dt ∞ e (3mn)2 ∞e (3mu)2du= √π ,
2T − − ≤ − ≤ − 2 3m
nX∈E1 ZT nX=1 Z0 ·
−
5
and the same estimate holds for the sum
T
1
e−4(x+t−3mn)2dt.
2T
nX∈E3 ZT
−
Next, we have
T
∞ √π
e 4(x+t 3mn)2dt e 4u2du=
− − −
≤ 2
Z Z
T
− −∞
and
T
1 1 √π 1 √π
e 4(x+t 3mn)2dt cardE 3 m2T +2
− − 2 −
2T ≤ 2T · 2 ≤ 2T · 2
nX∈E2 ZT (cid:0) (cid:1)
−
for n E . Thus dW1 (f,f ) 3√π 3 me4H2, and we obtain (11).
∈ 2 [ H,H] m ≤ 2 · −
Choose H < .−Let us check that dW1 (f,g) = 0 for no function g HU . We will prove the
strongerresult: t∞hereareno almostperio[−dHic,Hfu]nctions g(x)inthe senseofS∈tepan(o−v∞o,∞n)R withthe property
dW1(f,g)=0.
0
{ }We need some auxiliary lemmas:
Lemma 1.
√π
sup f(x) <1 inf f(x). (12)
x∈Z\I ≤ 2 ≤x∈I
The proof of Lemma 1.
We have
∞ √π
f(x)= ϕ (z) e 4n2 e 4t2dt=
l − −
≤ ≤ 2
lX∈N n∈XZ\{0} −Z∞
for any x Z I, and f(x) e−4(x−n0)2 =1 for x=n0 I.
∈ \ ≥ ∈
Lemma 2.
For all q Z 0 there exists a two-sided arithmetical progression I(q) I
∈ \{ } ⊂
such that (I(q)+q) I = .
∩ ∅
The proof of Lemma 2.
Every positive integer q has the form q =3r 1(3m+1) or q =3r 1(3m 1), r N, m Z. In the first
− −
− ∈ ∈
case take n =3r 1(3j+1), j Z, because
j −
∈
3r 1(3m+1)+3r 1(3j+1)=3r 1(3(m+j+1) 1) / I.
− − −
− ∈
In the second case take n =3r 1(3(3j m 1)+1), j Z, because n +q =3r(3j 1) / I for any j Z.
j − j
− − ∈ − ∈ ∈
Lemma 3.
There exists γ > 0 such that for each τ R, τ 1 there is a relatively dense set I(τ) R, with the
∈ | | ≥ ⊂
property
inf f(x+τ) f(x) >γ. (13)
x I(τ)| − |
∈
The proof of Lemma 3.
Since f(x) is uniformly continuous on R, there exists N < such that
∞
1 √π
f(x) f(t) < 1 (14)
| − | 2 − 2
(cid:18) (cid:19)
whenever x t 1.
| − |≤ N
Let τ be an arbitrary real number, τ 1. We show that inequality (13) takes place with γ =
| | ≥
1 1 √π for all points from some relatively dense set in R.
2N − 2
(cid:16) (cid:17)
6
Since the fractional parts of numbers 0,τ, 2τ, ..., Nτ belong to the half-open interval [0,1), there are
two numbers kτ and k τ, 0 k < k N, such that the distance between their fractional parts is at most
′ ′
≤ ≤
1, i.e. for some q Z 0 the inequality
N ∈ \{ }
1
kτ k′τ q
| − − |≤ N
holds. For M =k k we obtain
′
− 1
Mτ q . (15)
| − |≤ N
Let L be the difference of the arithmetic progressionI(q) from Lemma 2. Fix a real number a R, and
∈
n I(q) [a,a+L). Taking into account Lemma 1 and 2, we see that
∈ ∩
√π
f(n) f(n+q) 1 . (16)
| − |≥ − 2
On the other hand,
M 1
−
f(n+q) f(n) f(n+q) f(n+Mτ) + f(n+kτ) f(n+(k+1)τ). (17)
| − |≤| − | | − |
k=0
X
By(14)and(15),wehave f(n+q) f(n+Mτ) 1 1 √π . Henceinequalities(16)and(17)imply
| − |≤ 2 − 2
that (cid:16) (cid:17)
1 √π
f(n+k′′τ) f(n+(k′′+1)τ) 1
| − |≥ 2N − 2
(cid:18) (cid:19)
for some k , 0 k M N.
′′ ′′
≤ ≤ ≤
Thusthereexiststhepointx=n+k τ, x [a,a+L+Nτ]suchthat f(x+τ) f(x) >γ. Thelemma
′′
∈ | − |
is proved.
We continue the proof of Theorem 3. Let γ be the constant from Lemma 3. Take δ > 0 such that the
inequality
γ
f(x) f(t) , (18)
| − |≤ 5
holds whenever x, t R, x t <δ. Let us check that an arbitraryfunction g(x) from the equivalent class
∈ | − |
of the function f(x) in the space W1 satisfies the inequality
0
{ }
x+1
γδ
dS1 (g(t+τ),g(t))= sup g(t+τ) g(t)dt (19)
{0} x R | − | ≥ 10
∈ Zx
for arbitrary τ R, τ 1. Then the set of ε-almost periods for the function g in the Stepanov metric for
ε< γδ is contai∈ned in| |th≥e segment [ 1,1], and so g(x) / S1 .
10 − ∈ 0
In order to prove (19) for fixed τ R put { }
∈
γ
F = x R: g(x) f(x) ,
1
∈ | − |≥ 5
n o
γ
F = x R: g(x+τ) f(x+τ) .
2
∈ | − |≥ 5
n o
Take L < such that the set I(τ) from Lemma 3 has nonempty intersection with every interval of the
∞
length L. Since
nL
1 5 5
mes(F [ nL, nL]) f(x) g(x) dx f(x) g(x) dx,
1
2nL ∩ − ≤ 2nLγ | − | ≤2nLγ | − |
Z Z
F1∩[−nL,nL] −nL
and f, g belong to the same equivalent class in the space W1 , we get
0
{ }
1
lim mes(F [ nL, nL]) dW1 f,g =0.
n 2nL 1∩ − ≤ {0}{ }
→∞
7
The same equality holds for the set F . Hence
2
n 1
1 − mes (F1 F2) [kL,(k+1)L] 1
lim { ∪ ∩ } = lim mes (F F ) [ nL,nL] =0.
1 2
n 2n L n 2nL { ∪ ∩ − }
→∞ k= n →∞
X−
Therefore for n sufficiently large there exists an interval [k L,(k +1)L] such that
0 0
δ
mes (F F ) [k L,(k +1)L] < . (20)
1 2 0 0
{ ∪ ∩ } 2
Take a real number x [k L,(k +1)L] I(τ), where I(τ) is defined in Lemma 3. By (18) we get the
0 0
∈ ∩
inequality
3γ
f(t+τ) f(t) f(x+τ) f(x) f(x+τ) f(t+τ) f(x) f(t)
| − |≥| − |−| − |−| − |≥ 5
for each point t (x δ,x+δ).
∈ −
Note thatthe lengthofthe interval(x δ,x+δ) (k L,(k +1)L)is atleastδ. Itfollowsfrom(20)that
0 0
the measure of the set (x δ,x+δ) [F −F ] is at∩least δ. Next, for t (x δ,x+δ) [F F ] we have
− \ 1∪ 2 2 ∈ − \ 1∪ 2
γ
g(t+τ) g(t) f(t+τ) f(t) f(t+τ) g(t+τ) f(t) g(t) .
| − |≥| − |−| − |−| − |≥ 5
Thus,
x+δ
γδ
g(t+τ) g(t) dt g(t+τ) g(t) dt .
| − | ≥ | − | ≥10
Z Z
x−δ [x−δ,x+δ]\(F1∪F2)
The last inequality implies (19). The theorem is proved.
For the proof of other theorems we need following lemmas:
Lemma 4.
Any collection of functions æ (x)=e 4(x 3n)2 satisfy the inequality
n − −
p
∞ ∞
æk(x) 2p−1 æk(x) .
! ≤ !
k=1 k=1
X X
The proof of Lemma 4.
Fix x R and put n = x + 1 . Since 3n 3 x 3n 1, we obtain
∈ 0 3 2 0− 2 ≤ ≤ 0− 2
n0 e(cid:2)−4(x−3(cid:3)n)2 n0 e−(6(n0−1−n)+3)2 ∞ e−(3n)2 √π
≤ ≤ ≤ 6
n= n= n=1
X−∞ X−∞ X
and
∞ e−4(x−3n)2 ∞ e−(6(n−n0−1)+3)2 ∞ e−(3n)2 √π,
≤ ≤ ≤ 6
n=Xn0+1 n=Xn0+1 nX=1
hence ∞ e 4(x 3n)2 √π <1. The statement of the lemma follows from the inequality
n∈Z\{n0} − − ≤ 3
P
(a+b)p 2p−1(a+b)
≤
with a=æ (x) and b= æ (x).
n0 n
The proof of TheorneP6=mn03.
For z =x+iy C, l N put
∈ ∈
ϕ(z)= e 4(z n)2,
l − −
n=3l−X1(3k+1),
k Z
∈
∞
f(z)= lϕ(z). (21)
l
l=1
X
8
First of all, for y H and any x R we have
| |≤ ∈
ϕ (z) e4H2 e 4(x 3l−1(3k+1))2 e4H2 e 4(x n)2. (22)
l − − − −
| |≤ ≤
k Z n Z
X∈ X∈
Hence each termof sum (21) is uniformly bounded onevery horizontalstrip. If x 3l0−2 for some l0 N,
| |≤ ∈
then for all k Z, l l we have
0
∈ ≥
x 3l−1(3k+1) 3l−1 3k+1 3l−2 3l−2(4k +1)
− ≥ | |− ≥ | |
and for z =x+iy, x 3(cid:12)(cid:12)l0−2, y H (cid:12)(cid:12)
| |≤ | |≤
ϕ (z) e4H2 e 4(3l−2(4k+1))2 e4H22 ∞ e 4(3l−2n)2 e4H22 ∞e 4(3l−2u)2du= e4H2√π,
| l |≤ − | | ≤ − ≤ − 2 3l 2
k Z n=1 Z · −
X∈ X 0
so all terms of series (21) with indices l l are majorized by the terms of the convergentseries
0
≥
∞ 9√πe4H2 l
.
2 · 3l
l=1
X
Thus series (21) uniformly converge on every compact set in C, and f(z) is an entire function.
m
Next, ϕ(z) is an entire function with the period 3l, and the sum lϕ(z) is a periodic function with
l l
l=1
the period 3m, therefore this sum belongs to the space HU( , ). HenPce if
−∞∞
1
T p p m
lim 1 sup ∞ lϕ(z)dx =dBp lϕ(z),f(z) 0 (23)
T→∞2T |y|≤H ZT (cid:12)(cid:12)l=Xm+1 l (cid:12)(cid:12)  [−H,H] Xl=1 l !→
 − (cid:12)(cid:12) (cid:12)(cid:12) 
as m for all H >0, then f(z) (cid:12) HBp , p(cid:12) 1.
→∞ ∈ ( , ) ≥
Fix T >0 and consider the integral −∞∞
T T
ϕl(x)dx= e−4(x−3l−1(3k+1))2dx.
Z k ZZ
T X∈ T
− −
Put
1
E1 = n Z:n 31−lT , n1 =supE1,
∈ ≤− − 2
(cid:26) (cid:27)
1 1
E = n Z: 31 lT <n<31 lT + 0 ,
2 − −
∈ − − 2 2 \{ }
(cid:26) (cid:27)
1
E = n Z:n 31 lT + , n =infE .
3 − 2 3
∈ ≥ 2
(cid:26) (cid:27)
Note that
T T
e 4(x 3l−1(3k+1))2dx e 4(x 3l−1n)2dx+
− − − −
≤
kX∈ZZT nX∈E1 ZT
− −
T T
e 4(x 3l−1n)2dx+ e 4(x 3l−1n)2dx.
− − − −
nX∈E2 ZT nX∈E3 ZT
− −
Note also that the number cardE is at most 2T 31 l.
2 −
·
For n E and x [ T,T] we have
1
∈ ∈ −
1
x 3l 1n 3l 1 n + n
− − 1
− ≥ 2 −
(cid:18) (cid:19)
(cid:0) (cid:1)
9
and
e−4(x−3l−1n)2 e−32(l−1)(2(n1−n)+1)2,
≤
by the same way, for n E , x [ T,T], we have
3
∈ ∈ −
1
3l 1n x 3l 1 n n +
− − 2
− ≥ − 2
(cid:18) (cid:19)
(cid:0) (cid:1)
and
e−4(x−3l−1n)2 e−32(l−1)(2(n−n2)+1)2.
≤
Therefore
T
1 e 4(x 3l−1n)2dx ∞ e 32(l−1)n2 ∞e (3l−1u)2du= √π .
2T − − ≤ − ≤ − 2 3l 1
nX∈E1 ZT nX=1 Z0 · −
−
The same estimate is true for the sum over n E . Next, for n E we get
3 2
∈ ∈
T
∞ √π
e 4(x 3l−1n)2dx e 4u2du=
− − −
≤ 2
Z Z
T
− −∞
and
T
1 √π √π
e 4(x 3l−1n)2dx cardE 31 l2T.
− − 2 −
2T ≤ 4T ≤ 4T
nX∈E2 ZT
−
T T
Thus 21T T ϕl(x)dx=k Z21T T e−4(x−3l−1(3k+1))2dx≤ 3√π2·31−l.
Apply−iRng Lemma 4 tP∈o the−fRunctions e 4(x 3l−1(3k+1))2, k Z, we obtain
− −
∈
T T
1 1
ϕp(x)dx 2p ϕ (x)dx 2p−19√π3−l. (24)
2T l ≤ 2T l ≤
Z Z
T T
− −
The Ho¨lder inequality implies
1 1
∞ lϕ(x) ∞ l2pϕp(x) p ∞ 1 q ,
(cid:12) l (cid:12)≤ l ! lq!
(cid:12)l=Xm+1 (cid:12) l=Xm+1 l=Xm+1
(cid:12) (cid:12)
(cid:12) (cid:12)
where 1 + 1 =1. Therefore(cid:12)for sufficientl(cid:12)y large m we have
p q
T p T
1 ∞ ∞ 1 ∞ l2p
2T lϕl(x)! dx≤ l2p2T ϕpl(x)dx≤2p−19√π 3l .
ZT l=Xm+1 l=Xm+1 ZT l=Xm+1
− −
If we combine the letter with the inequality
sup ∞ lϕ(z) e4H2 ∞ lϕ(x),
l l
y H(cid:12) (cid:12)≤
| |≤ (cid:12)l=Xm+1 (cid:12) l=Xm+1
(cid:12) (cid:12)
(cid:12) (cid:12)
we get (23). (cid:12) (cid:12)
Let us check that anarbitraryfunction g(x) fromthe equivalentclass off(x) in the space Bp does not
0
belong to the space W1 . Then it follows that g does not belong to the space HW1 for al{l H} >0.
0 [ H,H]
Assume the contrar{y.}Then −
T
1 p
lim sup g(x+t)pdt= dWp(0,g) < ,
x→∞2T x∈RZT | | (cid:16) {0} (cid:17) ∞
−
10