Oscillation of Hölder Continuous Functions

0.15 MB·English

by  José González Llorente

Oscillation of H¨older Continuous Functions
Jos´e Gonz´alez Llorente and Artur Nicolau
3
1
0 Abstract
2
Local oscillation of a function satisfying a Ho¨lder condition is considered and it is proved
n
a that its growth is governedby a version of the Law of the Iterated Logarithm.
J

0
2 1 Introduction

]
A
For 0 < α < 1, let Λ (R) be the class of functions f: R R for which there exists a constant C =
α
C →
C(f) > 0 such that f(x) f(y) C x y α for any x,y R. The infimum of such constants C
h. | − | ≤ | − | ∈
is denoted by f . For b > 1, G. H. Hardy proved in [H] that the Weierstrass function
t α
a k k
m ∞
[ fb(x) = b−jαcos(bjx), x R,
∈
j=1
1 X
v
is in Λ (R) and exhibits the extreme behaviour
1 α
7
6 fb(x+h) fb(x)
limsup| − | > 0
4 hα
h→0
. | |
1
0 for any x R. However fixed x R one may expect many changes of sign of fb(x+h) fb(x)
∈ ∈ −
3 as h 0. Next definition provides a way of quantifying it. Given a function f Λ (R) and
1 α
→ ∈
v: 0 < ε< 1/2, consider
i
X 1 f(x+h) f(x h)dh
ar Θε(f)(x) = Zε h−α − h , x ∈ R. (1.1)
It is clear that Θ (f) 2α f log(1/ε). Moreover this uniform estimate can not be improved
ε ∞ α
k k ≤ k k
astheelementary examplef(x)= x αsign(x)shows. However atalmosteverypointx,theuniform
| |
estimate can be substantially improved. The main result of the paper is the following
Theorem 1. Fix 0 < α < 1. For f Λ (R) and 0 < ε < 1/2, let Θ (f)(x) be given by (1.1).
α ε
∈
Then, there exists a constant c(α) > 0, independent of ε and f, such that
(a) For any interval I R, I = 1, one has
⊂ | |
Θ (f)(x)2dx c(α)(log1/ε) f 2.
| ε | ≤ k kα
ZI
0Both authors are partially supported bythe grants MTM2011-24606 and 2009SGR420.
1
(b) At almost every point x R, one has
∈
Θ (f)(x)
ε
limsup | | c(α) f .
α
ε→0+ log(1/ε)logloglog(1/ε) ≤ k k
p
Themain technical step in the proof is the following estimate which provides the right subgaus-
sian decay: there exists a constant c = c(α) > 0 such that for any t > 0 one has
x [0,1] : Θ∗(f)(x) > t log(1/ε) f ce−t2/c. (1.2)
|{ ∈ | ε | k kα}| ≤
p
HereΘ∗(f)isthemaximalfunctiongivenbyΘ∗(f)(x) = sup Θ (f)(x) :1/2 δ ε . Theorem1
ε ε {| δ | ≥ ≥ }
follows from this subgaussian estimate by standard arguments. Our proof of (1.2) is organized in
twosteps. Firstwestateandproveadyadicversionof (1.2)andlaterweuseanaveragingprocedure
due to J. Garnett and P. Jones ([GJ]). Theorem 1 is sharp up to the value of the constant c(α).
Moreover there exists f Λ (R) for which there exists a constant c = c(f) > 0 such that for any
α
∈
0 < ε< 1/2 one has
1 f(x+h) f(x h) dh
| − − | > clog(1/ε)
hα h
Zε
at almost every x R. So, Theorem 1holds dueto certain cancellations which occur in theintegral
∈
defining Θ (f)(x).
ε
Subgaussian estimates and Law’s of the Iterated Logarithm play a central role in the boundary
behavior of martingales and have also appeared in function theory. For instance, in the relation
between the boundary behaviour of a harmonic function in an upper half space and the size of
its area function ([W], [CWW], [BM]) or in differentiability properties of functions defined in
the euclidean space ([AP], [SV]). Our result is inspired in the nice work of Y. Lyubarskii and
E. Malinnikova ([LM]) who studied the oscillation of harmonic functions in the Koremblum class.
Related results can be found in [E], [EM] and [EMM].
Thepaper is organized as follows. Section 2 is devoted to thedyadic version of Theorem 1. The
averaging procedure which is used to prove the results in the continuous setting from their dyadic
counterparts, is given in Section 3. Section 4 contains the proof of the subgaussian estimate (1.2)
as well as the proof of Theorem 1. In Section 5, the sharpness of the results is discussed. Finally,
Section 6 provides a higher dimensional analogue of Theorem 1.
The letters c and c(α) will denote a constant and a constant depending on the parameter α
whose value may change from line to line.
It is a pleasure to thank Eugenia Malinnikova for several sharp remarks on a first version of
this paper.
2 Dyadic Model
For 1 ρ 2, let = (ρ) be the collection of intervals of the form [j2−kρ,(j +1)2−kρ), where
≤ ≤ D D
j Z and k = 0,1,2,... . Let = (ρ) be the collection of intervals of of length 2−kρ and
k k
∈ D D D
2
let = (ρ) be the σ-algebra generated by the intervals of . In the rest of this section the
k k k
F F D
number 1 ρ 2 is fixed. A dyadic martingale is a sequence of functions S defined in [0,ρ]
k
≤ ≤ { }
such that for any k = 0,1,2,... the following two conditions hold: (a) S is adaptated to ;
k k
F
(b) the conditional expectation of S respect to is S . In other words: S is constant in each
k+1 k k k
F
interval of and
k
D 1
(S (x) S (x))dx =0
k+1 k
I −
| |ZI
for any I , k = 0,1,2,.... Given a dyadic martingale S , its quadratic variation S is
k n n
∈ D { } h i
defined as
n
S 2(x)= (S (x) S (x))2, n= 1,2,... .
h in k − k−1
k=1
X
It is well known that the quadratic variation governs the boundary behaviour of the martingale.
More concretely, the sets x [0,ρ] : lim S (x) exists and x [0,ρ] : S (x) < coincide
n ∞
{ ∈ n→∞ } { ∈ h i ∞}
except at most for a set of Lebesgue measure 0. Moreover there exits a universal constant c > 0
such that
S (x)
n
limsup | | c,
n→∞ hSi2n(x)logloghSin(x) ≤
at almost every point x where S ∞(xp) = . We also mention that an elementary orthogonality
h i ∞
argument gives that
ρ ρ
S (x)2dx = S 2(x)dx, n = 1,2,... .
| n | h in
Z0 Z0
Fix 0 < β < 1. Let S be a dyadic martingale satisfying S 2nβ, n = 0,1,2,... . For
n n ∞
{ } k k ≤
N = 1,2,..., consider
N
Γ (x) = Γ ( S )(x) = 2−kβS (x).
N N n k
{ }
k=1
X
Itis clear that Γ N. Moreover this uniformestimate is bestpossible. Actually, if theinitial
N ∞
k k ≤
martingale S satisfies, S 0, S = 2nβ and S (x) = 2kβ for some x R and any k N;
n 0 n ∞ k
{ } ≡ k k ∈ ≤
then Γ ( S ) = N. However, as nextresultshows, this uniformestimate can besubstantially
N n ∞
k { } k
improved at almost every point. Parts (b) and (c) are the discrete analogues of Theorem 1.
Theorem 2. Fix 0 < β < 1 and C > 0. Let S be a dyadic margingale with respect (ρ) with
n
{ } D
S 0 and S C2nβ, n = 1,2,... . For N = 1,2,..., consider
0 n ∞
≡ k k ≤
N
Γ (x) = 2−kβS (x),
N k
k=1
X
Γ∗ (x) = sup Γ (x).
N | k |
k≤N
Then, there exists a constant c= c(β,C) > 0 such that
(a) For any λ > 0 and any N = 1,2,..., one has
ρ
exp(λΓ∗ (x)) dx cecλ2N.
N ≤
Z0
3
(b) For any N = 1,2,..., one has
ρ
Γ∗ (x)2dx cN.
| N | ≤
Z0
(c) For almost every x [0,ρ] one has
∈
Γ (x)
N
limsup | | c.
√N loglogN ≤
n→∞
Proof. We can assume C = 1. Although Γ is not a dyadic martingale, we will show that its
N
{ }
size is comparable to the size of a dyadic martingale with bounded differences. Actually, consider
the dyadic martingale T defined by T 0 and
n 0
{ } ≡
n
S S
k k−1
T = − , n = 1,2,... .
n 2kβ
k=1
X
The subgaussian estimate (see [BM, p. 69]) gives that
x [0,ρ] : T∗(x) > t 2exp( t2/2 T 2 ),
|{ ∈ n }|≤ − kh ink∞
for any t > 0. Here T∗(x) = sup T (x) : 1 k n . Hence
n {| k | ≤ ≤ }
ρ ∞ ∞
exp(T∗(x)) dx = et x [0,ρ] : T∗(x) > t dt 2 exp t t2/2 T 2 dt
n |{ ∈ n }| ≤ − kh ink∞
Z0 Z0 Z0
(cid:0) (cid:1)
We deduce that
ρ
exp(T∗(x)) dx 2√2π T exp T 2 /2 , n = 1,2,...
n ≤ kh ink∞ kh ink∞
Z0
(cid:0) (cid:1)
Since T T 1+2−β for any n, one has T 2 n(1+2−β)2 for n = 1,2,... . We
k n+1 − nk∞ ≤ kh ink∞ ≤
deduce that for any λ > 0, one has
ρ λ2
exp(λT∗(x)) dx 2(1+2−β)√2πnλexp n(1+2−β)2 , n = 1,2,...,.
n ≤ 2
Z0 (cid:18) (cid:19)
On the other hand, summation by parts gives that
T = (1 2−β)Γ +2−nβS .
n n−1 n
−
Hence
Γ∗ (1 2−β)−1(T∗ +1) (2.1)
n ≤ − n+1
We deduce that for any n = 1,2,..., and any λ > 0, one has
ρ 1+2−β 1 1+2−β 2
exp(λΓ∗(x)) dx 2 2π(n+1)λexp λ(1 2−β)−1 exp λ2(n+1) .
n ≤ 1 2−β − 2 1 2−β
Z0 − p (cid:16) (cid:17) (cid:18) − (cid:19) !
Hence, the trivial estimate λ(1 2−β)−1 λ2/2+(1 2−β)−2/2 finishes the proof of (a).
− ≤ −
4
The estimate (2.1) gives
ρ ρ
Γ∗(x)2dx 2(1 2−β)−2 T∗ (x)2dx+2ρ(1 2−β)−2.
| n | ≤ − | n+1 | −
Z0 Z0
Since by Doob’s maximal inequality ([S, p.493])
ρ ρ
T∗ (x)2dx c T 2 (x)dx c(n+1),
| n+1 | ≤ h in+1 ≤
Z0 Z0
(b) follows. Finally, the Law of the Iterated Logarithm applied to T gives
n
{ }
T (x)
n
limsup | | c a.e. x.
√nloglogn ≤
n→∞
We deduce
Γ (x)
limsup | n | c(1 2−β)−1 a.e. x
n→∞ √nloglogn ≤ −
which finishes the proof.
3 Averaging
An averaging procedure due to J. Garnett and P. Jones ([GJ]) will be used to go from the discrete
situation of Theorem 2 to the continuous one of Theorem 1.
Given x R, let Iρ(x) be the unique interval of (ρ) of length 2−kρ which contains x. Given
∈ k D
a function f: R R and an interval I = [a,b) we denote ∆f(I) = f(b) f(a) and consider the
→ −
dyadic martingale with respect to the filtration (ρ) given by
D
(ρ)
∆f(I (x))
S(ρ)(f)(x) = k , k = 0,1,2,... .
k 2−kρ
If f Λ (R), we have S(ρ)(f) (2k/ρ)β f , k = 0,1,... where β = 1 α. As in Section 2,
∈ α k k k∞ ≤ k kα −
consider
n n (ρ)
∆f(I (x))
Γ(ρ)(f)(x) = Γ(ρ)( S(ρ) )(x) = 2−kβρβS(ρ)(f)(x) = k . (3.1)
n n { k } k (2−kρ)α
k=1 k=1
X X
The main purpose of this section is to describe an averaging argument with respect both ρ [1,2]
∈
and translates of the dyadic net (ρ). We start with a preliminary result.
D
Lemma 3.1. Let f: R R be a locally integrable function. For s R let f be the function defined
s
→ ∈
by f (x) = f(x s), x R. Then for any x R and any k = 1,2,..., one has
s
− ∈ ∈
ρ 2−kρ
∆f (I(ρ)(x+s))ds = 2k (f(x+t) f(x t))dt.
s k − −
Z0 Z0
5
Proof. Fix x R and k = 1,2,... . Let I(ρ)(x) = [a,b). Fix an integer j with 0 j 2k 1 and
∈ k ≤ ≤ −
consider [2−kjρ,2−k(j +1)ρ) = J K where J = J(x) = [2−kjρ,2−kjρ+b x) and K = K(x) =
∪ −
[2−kjρ+b x,2−k(j +1)ρ). Note that for s J one has I(ρ)(x+s)= [a+2−kjρ,b+2−kjρ) and
− ∈ k
b−x
∆f (I(ρ)(x+s))ds = (f(b+2−kjρ s) f(a+2−kjρ s))ds = (f(x+t) f(x+t 2−kρ))dt.
s k − − − − −
ZJ ZJ Z0
For s K one has I(ρ)(x+s) =[a+2−k(j +1)ρ,b+2−k(j +1)ρ) and
∈ k
∆f (I(ρ)(x+s))ds = (f(b+2−k(j +1)ρ s) f(a+2−k(j +1)ρ s))ds
s k − − −
ZK ZK
2−kρ
= (f(x+t) f(x+t 2−kρ))dt.
− −
Zb−x
Thus
2−k(j+1)ρ 2−kρ 2−kρ
∆f (I(ρ)(x+s))ds = (f(x+t) f(x+t 2−kρ))dt = (f(x+t) f(x t))dt.
Z2−kjρ s k Z0 − − Z0 − −
Adding on j = 0,...,2k 1, one finishes the proof.
−
We now state the main result of this section.
Proposition 3.2. Fix 0 < α 1. Let f be a locally integrable function. For s R let f be the
s
≤ ∈
function defined by f (x) = f(x s), x R. For n = 1,2,..., consider Γ(ρ)(f ) as defined in (3.1).
s n s
− ∈
Then for any x R, one has
∈
2 ρ dρ 1 1 f(x+t) f(x t)
Γ(ρ)(f )(x+s)ds = − − dt+A (f)(x)
Z1 Z0 n s ρ2 1+αZ2−n t1+α n
where
1 2−n
A (f)(x) c(α) f(x+t) f(x t) dt+c(α)2n(1+α) f(x+t) f(x t) dt.
n
| | ≤ Z2−n| − − | Z0 | − − |
In particular if f Λ (R), one has sup A (f)(x) < C(α) f
∈ α n,x| n | k kα
Proof. For k = 1,2,..., consider
2 ρ ∆(f )(I(ρ)(x+s)) dρ
B = s k ds .
k (2−kρ)α ρ2
Z1 Z0
Lemma 3.1 gives that
2 2−kρ f(x+t) f(x t) dρ
B = 2k − − dt .
k (2−kρ)α ρ2
Z1 Z0
The change of variables h =2−kρ gives
2−k+1 1 h
B = (f(x+t) f(x t))dtdh.
k Z2−k h2+α Z0 − −
6
Adding on k = 1,...,n, one deduces
2 ρ dρ 1 1 h
Γ(ρ)(f )(x+s)ds = (f(x+t) f(x t))dtdh.
Z1 Z0 n s ρ2 Z2−n h2+α Z0 − −
Applying Fubini’s Theorem one deduces
2 ρ dρ 1 1 f(x+t) f(x t)
Γ(ρ)(f )(x+s)ds = − − dt
Z1 Z0 n s ρ2 1+α Z2−n t1+α
1 1
(f(x+t) f(x t))dt
− 1+α Z2−n − −
2n(1+α) 1 2−n
+ − (f(x+t) f(x t))dt
1+α − −
Z0
which finishes the proof.
4 Continuous setting
Inthissection, theresultsofthedyadicmodelofSection 2andtheaveraging procedureofSection 3
will be used to prove Theorem 1.
Given f Λ (R) and 0 < ε < 1, pick an integer N such that 2−N−1 ε < 2−N. Observe that
α
∈ ≤
Θε(f)(x) Θ2−N(f)(x) 2 f α. Hence the estimates of Θ2−N(f)(x) can be easily transferred to
| − | ≤ k k
Θε(f)(x). The main technical step in proving the relevant subgaussian estimate of Θ2−N(f)(x) is
stated in next result.
Proposition 4.1. Let f Λ ([ 1,2]) with f 1. For x [0,1] and N = 1,2,..., consider
α α
∈ − k k ≤ ∈
1 f(x+h) f(x h)dh
Θ2−N(f)(x) = Z2−N h−α − h ,
Θ∗2−N(f)(x) = sup|Θ2−k(f)(x)|.
k≤N
Then, there exists a constant c(α) > 0 such that for any λ > 0 and any N = 1,2,..., one has
1
exp λΘ∗ (f)(x) dx c(α)exp c(α)λ2N .
2−N ≤
Z0
(cid:0) (cid:1) (cid:0) (cid:1)
Proof. Consider the set A= (ρ,s) :1 ρ 2, 0 s ρ and the measure dµ defined as
{ ≤ ≤ ≤ ≤ }
dρ
µ(E) = ds ,E R2
ρ2 ⊂
ZETA
For any k = 1,2,..., Proposition 3.2 gives that
(ρ)
Θ2−k(f)(x) = (1+α) Γk (fs)(x+s)dµ(ρ,s)+Ak(f)(x).
ZA
7
Moreover there exists a constant C = C(α) such that sup A (x) C. Here is where the normal-
k
| | ≤
k,x
ization f 1 is used. Hence if k and N are integers with k N we deduce
α
k k ≤ ≤
|Θ2−k(f)(x)| ≤ (1+α) (ΓN(ρ))∗(fs)(x+s)dµ(ρ,s)+C.
ZA
Here (Γ(ρ))∗(f )(x) = sup Γ(ρ)(f )(x) :k N . Hence for any N = 1,2,..., one has
N s {| k s | ≤ }
Θ∗ (f)(x) (1+α) (Γ(ρ))∗(f )(x+s)dµ(ρ,s)+C.
2−N ≤ N s
ZA
Now, Jensen’s inequality and Fubini’s Theorem give that
1 1
exp λΘ∗ (f)(x) dx exp(λC) exp λ(α+1)(Γ(ρ))∗(f )(x+s) dxdµ(ρ,s). (4.1)
2−N ≤ N s
Z0 ZAZ0 (cid:16) (cid:17)
(cid:0) (cid:1)
(ρ) (ρ)
Recall that Γ (f ) is defined via the formula (3.1) from the martingale S (f ) which is given
N s k s
(ρ) (ρ)
by S (f )(x) = (f (b) f (a))/(b a), where x I (x) = [a,b) (ρ). The normalization
k s s − s − ∈ k ∈ D
(ρ)
f 1 gives that there exists an absolute constant c > 0 such that S (f ) c for any
k kα ≤ 1 | 0 s | ≤ 1
(ρ,s) A. Recall that if f 1, the martingale S(ρ) satisfies S(ρ) (2k/ρ)1−α. According
∈ k kα ≤ k k k k∞ ≤
to (a) of Theorem 2, there exists a constant c (α) > 0 such that
1
1
exp λ(1+α)(Γ(ρ))∗(f )(x+s) dx c (α)exp c (α)(c λ+λ2N) .
N s ≤ 1 1 1
Z0 (cid:16) (cid:17)
(cid:0) (cid:1)
The trivial estimate 2λ λ2+1 shows that there exists a constant c (α) > c (α) such that
2 1
≤
1
exp λ(1+α)(Γ(ρ))∗(f )(x+s) dx c (α)ec2(α)λ2N.
N s ≤ 2
Z0 (cid:16) (cid:17)
By (4.1) one deduces
1
exp λΘ∗ (f)(x) dx c (α)exp(Cλ)exp c (α)λ2N .
2−N ≤ 2 2
Z0
(cid:0) (cid:1) (cid:0) (cid:1)
Again the trivial estimate 2λ λ2+1 finishes the proof.
≤
Now the subgaussian estimate follows easily.
Corollary 4.2. Let f Λ ([ 1,2]) with f 1. Then there exists a constant c(α) > 0 such
α α
∈ − k k ≤
that for any N > 0 and any t > 0 one has
x [0,1] :Θ∗ (f)(x) > √Nt c(α)exp t2/c(α) .
|{ ∈ 2−N }|≤ −
(cid:0) (cid:1)
Proof. Let E = x [0,1] : Θ∗ (f)(x) > √Nt . Previous Proposition 4.1 and Chebyshev
{ ∈ 2−N }
inequality gives that for any λ >0 one has exp λ√Nt E c(α)exp c(α)λ2N , that is,
| | ≤
(cid:16) (cid:17)
(cid:0) (cid:1)
E c(α)exp(c(α)λ2N λ√Nt).
| |≤ −
We take λ=t/2c(α)√N and deduce E c(α)exp( t2/4c(α))which finishes the proof.
| |≤ −
8
We can now prove Theorem 1.
Proof of Theorem 1. Intheproofofpart(a)wecanassumethatI istheunitintervaland f = 1.
α
k k
Given0< ε < 1/2,pickanintegerN suchthat2−N−1 ≤ ε < 2−N. Since|Θε(f)(x)−Θ2−N(f)(x)| ≤
2, Corollary 4.2 gives that
x [0,1] : Θ (f)(x) >√Nt c(α)e−t2/c(α),
ε
|{ ∈ | | }| ≤
for any t > 0 such that t√N > 2. Now (a) follows easily from
1 ∞
Θ (f)(x)2dx = 2 λ x [0,1] : Θ (f)(x) > λ dλ.
ε ε
| | |{ ∈ | | }|
Z0 Z0
TheLawoftheIteratedLogarithmofpart(b)followsfromthesubgaussianestimateofCorollary4.2
via an standard Borel-Cantelli argument. Consider the set A of points x [0,1] for which
∈
Θ∗ (f)(x) > 2c N loglogN
2−N
p
for infinitely many N 0. Here c = c (α) is a constant which will be chosen later. Let N = 2m.
0 m
≥
If Θ∗ (f)(x) > 2c√N loglogN and N < N N then
2−N m−1 ≤ m
Θ∗ (f)(x) Θ∗ (f)(x) > 2c N loglogN c N loglogN .
2−Nm ≥ 2−N ≥ m m
p p
Thus A A where
⊂ ∩k ∪m≥k Nm
A = x : Θ∗ f(x)> c N loglogN .
Nm { 2−Nm m m}
p
Now Corollary 4.2 with t = c√loglogN = c(log(mlog2))1/2 gives A c(α)(mlog2)−c2/c(α)
m | Nm| ≤
and for c2 > c(α) the Borel-Cantelli lemma gives A = 0. Thus for almost every x [0,1] one has
| | ∈
Θ∗ (f)(x)
limsup | 2−N | 2c
√N loglogN ≤
N→∞
and the proof is completed.
5 Sharpness
In this section the sharpness of our results is discussed.
5.1 Sharpness of Theorem 1
Both parts (a) and (b) in Theorem 1 as well as Proposition 4.1 and its Corollary 4.2, are sharp up
to the value of the constants c(α). Since Theorem 1 follows from Corollary 4.2, it is sufficient to
construct a function f Λ (R) for which
α
∈
1
Θ (f)(x)2dx c(log(1/ε)), 0< ε < 1/2 (5.1)
ε
| | ≥
Z0
9
and
Θ (f)(x)
ε
limsup > c, a.e. x [0,1] (5.2)
ε→0 log(1/ε)logloglog(1/ε) ∈
for a certain constant c = c(αp) > 0. Fix 0 < α < 1. As it is usual in this kind of questions, the
function f will be given by a lacunary series. More concretely, consider
∞
f(x)= 2−jαsin(2π2jx).
j=0
X
Then,
1 f(x+h) f(x h)dh
Θ2−N(f)(x) = Z2−N h−α − h
∞ 1 sin(2j2πh)
= 2 2−jα dh cos(2j2πx)
j=0 (cid:18)Z2−N hα+1 (cid:19)
X
∞
= 2 c cos(2j2πx),
j,N
j=0
X
where
2j sin(2πt)
c = dt.
j,N Z2j−N tα+1
Integrating by parts one shows that there exists a constant c (α) > 0 such that
1
c c (α)2−(j−N)(α+1), j = 1,2,..., N = 1,2,... .
j,N 1
| |≤
Hence
c 2c (α). (5.3)
j,N 1
| | ≤
j≥N
X
On the other hand, using the estimate sint t, we have
| |≤
N 2j−N sin2πt 2π N
dt 2(j−N)(1−α) c (α). (5.4)
Xj=0(cid:12)(cid:12)Z0 tα+1 (cid:12)(cid:12) ≤ 1−α Xj=0 ≤ 2
(cid:12) (cid:12)
(cid:12) (cid:12)
Using (5.3) and (5.4) one(cid:12)deduces that (cid:12)
N
Θ2−N(f)(x)= bjcos(2j2πx)+EN(x), (5.5)
j=0
X
where E (x) c (α) = 2c (α)+c (α) for any x R and any N = 1,2,... and
N 3 1 2
| | ≤ ∈
2j sin2πt
b = 2 dt.
j tα+1
Z0
Consider
∞ sin2πt
A(α) = lim b = 2 dt
j→∞ j Z0 tα+1
10