Universal function for a weighted space L^1_u[0,1]

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by  Artsrun Sargsyan

Positivity manuscript No.
(will be inserted by the editor)
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Universal function for a weighted space L [0,1]
µ
Artsrun Sargsyan · Martin Grigoryan
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Received: date/Accepted: date
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n Abstract Itisshownthatthereexistsuchafunctiong ∈L1[0,1]andaweight
a function 0<µ(x)≤1 that g is universal for the weighted space L1[0,1] with
J µ
respect to signs of its Fourier–Walsh coefficients.
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2 Mathematics Subject Classification (2010) MSC 42C10 · MSC 43A15

]
A
F 1 Introduction
.
h
t Historically,thefirsttypeofauniversalfunctionwasconsideredbyG.Birkhoff
a
[1] in 1929. He proved, that there exists an entire function g(z), which is
m
universal with respect to translations, i.e. for every entire function f(z) and
[
for each number r > 0 there exists a growing sequence of natural numbers

1 {nk}∞k=1, so that the sequence {g(z+nk)}∞k=1 uniformly convergesto f(z) on
v thedisk|z|≤r.In1952G.MacLane[2]provedasimilarresultforanothertype
6 of universality,namely, there exists an entire function g(z), which is universal
7
with respect to derivatives, i.e. for every entire function f(z) and for each
7
number r > 0 there exists a growing sequence of natural numbers {n }∞ ,
5 k k=1
0 so that the sequence {g(nk)(z)}∞ uniformly converges to f(z) on the disk
k=1
. |z| ≤ r. Further, in 1975 S. Voronin [3] proved the universality theorem for
1
0 the Riemann zeta function ζ(s), which states that any nonvanishing analytic
7 function can be approximated uniformly by certain purely imaginary shifts
1
: A.Sargsyan
v
YSU,AlexManoogian1,0025,Yerevan,Armenia/CANDLESRI,Acharyan31,0040,Yere-
i
X van,Armenia
Tel.:[phone]
r
a E-mail:[e-mail]
M.Grigoryan
YSU,AlexManoogian1,0025,Yerevan,Armenia
Tel.:[phone]
E-mail:[e-mail]
2 ArtsrunSargsyan,MartinGrigoryan
of the zeta function in the critical strip, namely, if 0 < r < 1 and g(s) is a
4
nonvanishing continuous function on the disk |s| ≤ r, that is analytic in the
interior, then for any ε>0, there exists such a positive real number τ that
max g(s)−ζ(s+3/4+iτ) <ε.
|s|≤r
(cid:12) (cid:12)
(cid:12) (cid:12)
In 1987 K. Grosse–Erdman[4] provedthe existence of infinitely differentiable
functionwith universalTaylorexpansion,namely,there existssucha function
g(x) ∈ C∞(R) with g(0) = 0 that for every function f(x) ∈ C(R) with
f(0)=0andforeachnumberr >0thereexistsagrowingsequenceofnatural
numbers {n }∞ , so that the sequence
k k=1
nk g(m)(0)
S (g,0)= xm
nk m!
m=1
X
uniformly converges to f(x) on |x|≤r.
There are also many works devoted to the existence of universal series
(in the common sense, with respect to rearrangements,partial series, signs of
coefficientsandetc.)invariousclassicalorthogonalsystems.Themostgeneral
results were obtained by D. Menshov [5], A. Talalyan [6], P. Ulyanov [7] and
their disciples (see [8]–[19]).
Theresultspresentedinthecurrentpaperareanadditiontothisattractive
area of mathematical research.
Let |E| be the Lebesgue measure of a measurable set E ⊆ [0,1], χ (x)–
E
its characteristic function, Lp[0,1] (p > 0) – the class of all those measur-
able functions on [0,1] that satisfy the condition 1|f(x)|pdx < ∞, Lp[0,1]
0 µ
(weighted space) – the class of all those measurable functions on [0,1] that
R
satisfy the condition 1|f(x)|pµ(x)dx < ∞, where 0 < µ(x) ≤ 1 is a weight
0
function, and {ϕ } – a complete orthonormal system in L2[0,1].
k
R
Definition 1. We say that a function g ∈ L1[0,1] is universal for a class
Lp[0,1], p>0, with respect to signs of its Fourier coefficients (signs–subseries
of its Fourier series) by the system {ϕ }, if for each function f ∈Lp[0,1] one
k
canchoosesuchnumbersδ =±1(δ =0,±1)thattheseries ∞ δ c (g)ϕ (x),
k k k=0 k k k
where c (g)= 1g(x)ϕ (x)dx, converges to f in Lp[0,1] metric, i.e.
k 0 k P
R 1 m p
lim δ c (g)ϕ (x)−f(x) dx=0.
k k k
m→∞Z0 (cid:12)(cid:12)kX=0 (cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
Definition 2. Let µ(x) be a weight function defined on [0,1]. We say that a
function g ∈ L1[0,1] is universal for a weighted space Lp[0,1] with respect to
µ
signs of its Fourier coeffcients (signs–subseries of its Fourier series) by the
system {ϕ }, if for each function f ∈ Lp[0,1] one can choose such numbers
k µ
UniversalfunctionforaweightedspaceL1[0,1] 3
µ
δ = ±1 (δ = 0,±1) that the series ∞ δ c (g)ϕ (x) converges to f in
k k k=0 k k k
Lp[0,1] metric, i.e.
µ P
1 m p
lim δ c (g)ϕ (x)−f(x) µ(x)dx=0.
k k k
m→∞Z0 (cid:12)(cid:12)kX=0 (cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
Let us recallthe definition of the Walsh orthonormalsystem {W (x)}∞ .
n n=0
FunctionsoftheWalshsystemaredefinedbymeansofRademacher’sfunctions
R (x)= sign(sin2nπx), x∈[0,1], n=1,2,...,
n
in the following way (see [20]): W (x)≡1 and for n≥1
0
p
W (x)= R (x),
n ki+1
i=1
Y
where n=2k1 +2k2 +···+2kp (k1 >k2 >···>kp).
Remark. Theredoes notexista functiong ∈L1[0,1]whichis universalfor a
certain class Lp[0,1], p≥1, neither with respect to signs of its Fourier–Walsh
coefficients nor with respect to signs–subseries of its Fourier–Walshseries.
Indeed,ifsuchuniversalfunctionexistedthenforthefunctionk c (g)W (x),
0 k0 k0
where k >1 is any natural number with condition c (g)6=0, one could find
0 k0
such numbers δ =±1 or δ =0,±1 that
k k
1 m p
lim δ c (g)W (x)−k c (g)W (x) dx=0,
m→∞Z0 (cid:12)(cid:12)kX=0 k k k 0 k0 k0 (cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
which simply leads to(cid:12) contradiction: δ =k >1. (cid:12)
k0 0
Itturnsout,however,thatthesituationchangeswhenconsideringweighted
spaces Lp[0,1], p ≥ 1, or classes Lp[0,1], p ∈ (0,1). For the latter case in
µ
[21] the authorshaveconstructeda universalfunction with respectto signs of
Fourier–Walshcoefficients.ForthecaseLp[0,1],p≥1in[22]theauthorscould
µ
constructauniversalfunctionwithrespecttosigns–subseriesofFourier–Walsh
series.As a next step,the existence ofa universalfunction for spaces Lp[0,1],
µ
p ≥ 1 with respect to signs of Fourier–Walsh coefficients was considered. We
wouldliketonotethattheuniversalitywithrespecttosignsofFourier–Walsh
coefficients is much harder to acheive than the universality with respect to
signs–subseriesofFourier–Walshseries.Therefore,ithasbeenpossibletocon-
struct such a function only for the p = 1 case yet. The question is open for
spaces Lp[0,1], p>1.
µ
The following theorem is true for the Walsh system:
4 ArtsrunSargsyan,MartinGrigoryan
Theorem. There exist a function g ∈ L1[0,1] and a weight function 0 <
µ(x)≤1, so that g is universal for the weighted space L1[0,1] with respect to
µ
signs of its Fourier-Walsh coefficients.
Moreover,it will be shown that the measure of the set on which µ(x)=1
can be made arbitrarily close to 1, and the function g(x) can be choosen to
have strictly decreasing Fourier–Walsh coefficients and converging to it by
L1[0,1] norm Fourier–Walsh series.
Note that, considering generalities of many results obtained for the Walsh
and trigonometric systems, an interest arises to find out whether the proved
theorem is true for the trigonometric system.
2 Main lemmas
Forthesakeofsimplicitytheproofofthemaintheoremisdividedintoseveral
steps which are given in the form of lemmas. Let us start from the known
properties of the Walsh system {W (x)}∞ . It is known (see [20]) that for
k k=0
each natural number m we have
2m−1 2m, when x∈[0,2−m),
(1.a) W (x)=
k (0, when x∈(2−m,1],
k=0
X
and, consequently,
2m+1−1 2m, when x∈[0,2−m−1),
(1.b) W (x)= −2m, when x∈(2−m−1,2−m),
k 
kX=2m 0, when x∈(2−m,1].
Obviously,foranynaturalnumberM ∈[2m,2m+1)andnumbers{a }2m+1−1
k k=2m
one gets
1
1 M 1 2m+1−1 2 2
(2) a W (x) dx≤ a W (x) dx .
We will usZe0als(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)koX=t2hme fkollokwing(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) lemmaZ[203](cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12): kX=2m k k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) 
Lemma 1. For each dyadic interval ∆= l ,l+1 , l ∈[0,2K), and for every
2K 2K
such natural number M > K that M−K is a whole number, there exists a
2 (cid:2) (cid:3)
polynomial in the Walsh system
2M+1−1
H(x)= a W (x),
k k
kX=2M
so that
1) |ak|=2−M+2K, when k ∈[2M,2M+1),
UniversalfunctionforaweightedspaceL1[0,1] 5
µ
2) H(x)=−1, if x∈E , |E |= 1|∆|,
1 1 2
3) H(x)=1, if x∈E , |E |= 1|∆|,
2 2 2
4) H(x)=0, if x6∈∆.
where E and E are finite unions of dyadic intervals.
1 2
Now let us proceed to main lemmas of the paper:
Lemma 2 Let number n ∈N and dyadic interval ∆= l ,l+1 , l∈[0,2K)
0 2K 2K
be given. Then for any numbers ε ∈ (0,1), γ 6= 0 and natural number q
(cid:2) (cid:3)
there exist a measurable set E ⊂ ∆ with measure |E | = (1−2−q)|∆| and
q q
polynomials
2nq−1 2nq−1
P (x)= a W (x) and H (x)= δ a W (x), δ =±1,
q k k q k k k k
k=X2n0 k=X2n0
in the Walsh system, so that
1) 0<a ≤a <ε when k ∈[2n0,2nq −1),
k+1 k
2) |γχ (x)−H (x)|dx<ε and |H (x)|dx<ε,
∆ q q
ZEq Z[0,1]\∆
1 M
3) max δ a W (x) dx<3|γ||∆|+ε,
k k k
2n0≤M<2nqZ0 (cid:12)(cid:12)k=X2n0 (cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
1 M
4) max a W (x) dx<ε.
k k
2n0≤M<2nqZ0 (cid:12)(cid:12)k=X2n0 (cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
Proof of lemma2 Theproofisperformedbyusingthemathematicalinduc-
tion method with respect to the number q. We choose such a natural number
K >K that
1
(3) |γ|2−K12+1 < ε,
2
and present the interval ∆ in the form of disjoint dyadic intervals’ union
N1
∆= ∆(1)
i
i=1
[
with ∆(i1) =2−K1−1, i∈[1,N1]. Obviously, N1 =2K1−K+1.
By denoting K(1) ≡n −1, for each natural number i∈[1,N ] we choose
(cid:12) (cid:12) 0 0 1
sucha(cid:12) natu(cid:12)ralnumberK(1) >K(1) K(1) >K thatthefollowingconditions
i i−1 1 1
take place:
(cid:0) (cid:1)
6 ArtsrunSargsyan,MartinGrigoryan
a) Ki(1)−K1−1 is a whole number,
2
b) (Ki(1)−Ki(−1)1)|γ|2−Ki(1)+2K1+1 < 4Nε1,
c) 2|γ|2−Ki(12)+1 < ε.
2
It immediately follows from (3) that
(4) |γ|2−K1(1)+2K1+1 <ε.
By successively applying lemma 1 for each interval ∆(1) (i ∈ [1,N ]) and
i 1
corresponding number K(1) we can find polynomials in the Walsh system
i
2Ki(1)+1−1
(5) H(1)(x)= a˜ W (x), i∈[1,N ],
i k k 1
k=X2Ki(1)
e
so that
(6) |a˜k|=|γ|2−Ki(1)+2K1+1 when k∈ 2Ki(1),2Ki(1)+1 ,
(cid:2) (cid:1)
−γ, for x∈E (1) ⊂∆(1), E (1) = ∆(i1) ,
i i i 2
(7) Hi(1)(x)=γ0,, ffoorr xx∈∈/ E∆fi((11)).⊂∆(i1), (cid:12)(cid:12)Efi(1)(cid:12)(cid:12)= (cid:12)(cid:12)(cid:12)∆2(i1)(cid:12)(cid:12)(cid:12),
e fi (cid:12)f (cid:12) (cid:12) (cid:12)
f (cid:12)f (cid:12)
Hence, by denoting
N1
(8) H (x)= H(1)(x),
1 i
i=1
X
e e
we get
−γ, for x∈E ⊂∆, E = |∆|,
1 1 2
(9) H (x)= γ, for x∈∆\E ,
1  1 (cid:12) (cid:12)
0, for x∈/ ∆e. (cid:12)e (cid:12)
e e
AsthepolynomialH(1)(x)isalinearcombinationofWalshfunctionsfrom
i
K(1)group,itisclear,thatthesetE canbepresentedasaunionofcertainN
i 1 2
amount of disjoint dyadeic intervals ∆(i2) with measure ∆(i2) =2−KN(11)−1, i∈
e
[1,N ]:
2 (cid:12) (cid:12)
N2 (cid:12) (cid:12)
E = ∆(2).
1 i
i=1
[
e
UniversalfunctionforaweightedspaceL1[0,1] 7
µ
After making the following definitions
(10) E =∆\E
1 1
and e
(11) a˜k =|γ|2−Ki(1)+2K1+1, when k ∈ 2Ki(−1)1+1,2Ki(1) , i∈[1,N1],
ak =|a˜k|, δk = |aa˜˜kk|, when k ∈(cid:2)2n0,2KN(11)+1 ,(cid:1)
(cid:2) (cid:1)
let us verify that the set E and polynomials
1
2KN(11)+1−1 2KN(11)+1−1
P (x)= a W (x), H (x)= δ a W (x)
1 k k 1 k k k
k=X2n0 k=X2n0
satisfy all lemma 2 statements for q = 1. Indeed, from (9) and (10) it imme-
diately follows that |E | = (1−2−1)|∆|. The statement 1) follows from (4),
1
(6), (11) and from monotonicity of numbers K(1) (i ∈ [1,N ]). For the proof
i 1
of statements 2) and 3) we present the polynomial H (x) in the form of
1
N1 2Ki(1)−1
(12) H (x)=H (x)+ a W (x)=
1 1 k k
Xi=1k=2XKi(−1)1+1
e
N1 Ki(1)−Ki(−1)1−1
=H (x)+ Q(1)(x),
1 i,j
i=1 j=1
X X
e
where
2Ki(−1)1+j+1−1
(13) Q(1)(x)= a W (x)
i,j k k
k=2XKi(−1)1+j
δk =1 when k ∈ 2Ki(−1)1+1,2Ki(1) ,i∈[1,N1] .
(cid:0) Sinceallcoeffic(cid:2)ientsakareequ(cid:1)alwhenk ∈(cid:1)2Ki(−1)1+1,2Ki(1) , i∈[1,N1](see
(11)) then considering (1.b), (9)–(13) and b) condition for numbers K(1) (i∈
(cid:2) (cid:1) i
[1,N ]) we obtain
1
|γχ (x)−H (x)|dx≤ γχ (x)−H (x) dx+
∆ 1 ∆ 1
ZE1 ZE1(cid:12) (cid:12)
(cid:12)(cid:12) e (cid:12)(cid:12)
+ N1 Ki(1)−Ki(−1)1−1 1 Q(i,1j)(x) dx= N1 Ki(1)−Ki(−1)1−1 |γ|2−Ki(1)+2K1+1 <ε
i=1 j=1 Z0 i=1
X X (cid:12) (cid:12) X(cid:0) (cid:1)
(cid:12) (cid:12)
8 ArtsrunSargsyan,MartinGrigoryan
and
N1 Ki(1)−Ki(−1)1−1 1
|H (x)|dx≤ Q(1)(x) dx<ε.
1 i,j
Z[0,1]\∆ i=1 j=1 Z0
X X (cid:12) (cid:12)
(cid:12) (cid:12)
Toprovestatements3)and4)wepresentthenaturalnumberM ∈ 2n0,2KN(11)+1
in the form of M = 2n˜ +s, s ∈ [0,2n˜), where n˜ ∈ K(1) ,K(1) for some
m−1 m (cid:2) (cid:1)
m∈[1,N ]. From (8), (12) and (13) it follows that
1
(cid:0) (cid:3)
1 M 1 m−1
(1)
δ a W (x) dx≤ H (x) dx+
k k k i
Z0 (cid:12)(cid:12)k=X2n0 (cid:12)(cid:12) Z0 (cid:12)(cid:12)Xi=1 (cid:12)(cid:12)
(cid:12) (cid:12) (cid:12) e (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) m−1Ki(1)−Ki(−1(cid:12))1−1 1 (cid:12) (cid:12)
+ Q(1)(x) dx+
i,j
i=1 j=1 Z0
X X (cid:12) (cid:12)
(cid:12) (cid:12)
n˜−Km(1−)1−1 1 1 2n˜+s
+ Q(1) (x) dx+ δ a W (x) dx.
m−1,j (cid:12) k k k (cid:12)
Xj=1 Z0 (cid:12) (cid:12) Z0 (cid:12)(cid:12)kX=2n˜ (cid:12)(cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
By using (1.b)–(3), (5)–(8) and (11) we get (cid:12) (cid:12)
(cid:12) (cid:12)
1 m−1 1 2n˜+s
H(1)(x) dx+ δ a W (x) dx≤
m−Z10 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)1Xi=1 ei (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Z01(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)k2Xn=˜+2n˜s k k k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)2 21
(1)
≤ H (x) dx+ δ a W (x) dx ≤
i  (cid:12) k k k (cid:12) 
Xi=1 Z0 (cid:12)(cid:12)(cid:12)e (cid:12)(cid:12)(cid:12) Z0 (cid:12)(cid:12)(cid:12)(cid:12)kX=2n˜ (cid:12)(cid:12)(cid:12)(cid:12)  1
N1 1 1 2(cid:12)n˜+1−1 (cid:12)2 2
≤ H(1)(x) dx+ δ a W (x) dx =
i  (cid:12) k k k (cid:12) 
Xi=1Z0=(cid:12)(cid:12)(cid:12)|γe||∆|+(cid:12)(cid:12)(cid:12)|γ|2−Km(1Z)+02K1(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)+k1X=22n2˜n˜<3|γ||∆|+(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)ε, 
2
thus, taking
m−1Ki(1)−Ki(−1)1−1 1 n˜−Km(1−)1−1 1
Q(1)(x) dx+ Q(1) (x) dx≤
i,j m−1,j
i=1 j=1 Z0 j=1 Z0
X X (cid:12) (cid:12) X (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
≤ N1 Ki(1)−Ki(−1)1−1 1 Q(i,1j)(x) dx= N1 Ki(1)−Ki(−1)1−1 |γ|2−Ki(1)+2K1+1 < 4ε
i=1 j=1 Z0 i=1
X X (cid:12) (cid:12) X(cid:0) (cid:1)
(cid:12) (cid:12)
into account we verify the validity of the statement 3).
UniversalfunctionforaweightedspaceL1[0,1] 9
µ
Further, for each natural number n ∈ n ,K(1) we denote b = a , k ∈
0 N1 n k
[2n,2n+1) (coefficients a of Walsh functions from n–th group are equal in
k (cid:2) (cid:3)
H (x)). It follows from (1.b)–(3), (6), (11) and b) condition for numbers
1
K(1) (i∈[1,N ]) that
i 1
KN(11) bn = N1 Ki(1) bn = N1 Ki(1)−Ki(−1)1 |γ|2−Ki(1)+2K1+1 < 4ε
nX=n0 Xi=1n=KXi(−1)1+1 Xi=1(cid:0) (cid:1)
and
1 M n˜−1 1 2n˜+s
a W (x) dx≤ b + a W (x) dx≤
k k n (cid:12) k k (cid:12)
Z0 (cid:12)(cid:12)k=X2n0 (cid:12)(cid:12) nX=n0 Z0 (cid:12)(cid:12)kX=2n˜ (cid:12)(cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
≤ KN(11) bn(cid:12)(cid:12) + 1(cid:12)2n˜+1−(cid:12)(cid:12)1bn˜Wk(x)(cid:12)221 < 4ε(cid:12)(cid:12)+|γ|2−Km(1)+2K(cid:12)(cid:12)1+12n2˜ <ε,
nX=n0 Z0 (cid:12)(cid:12) kX=2n˜ (cid:12)(cid:12) 
(cid:12) (cid:12)
 (cid:12) (cid:12) 
which proves the state(cid:12)ment 4). (cid:12)
Assume that for q >1 natural numbers
K(1) <···<K(1) <···<K(q−1) <···<K(q−1),
1 N1 1 Nq−1
sets
E ⊂∆ and E =∆\E
q−1 q−1 q−1
and polynomials
e e
2KN(qq−−11)+1−1 2KN(qq−−11)+1−1
P (x)= a W (x), H (x)= δ a W (x), δ =±1
q−1 k k q−1 k k k k
k=X2n0 k=X2n0
are already chosen to satisfy the following conditions:
a′) Ki(ν)−KN(νν−−11)−1 is a whole number K(0) ≡K ,
2 N0 1
b′) Ki(ν)−Ki(−ν)1 2ν−1|γ|2−Ki(ν)+K2N(νν−−1(cid:0)1)+1 < 2ν+ε1N(cid:1)ν,
c′) (cid:0)2ν|γ|2−Ki(ν2)+1(cid:1)< ε,
2
(14) ak =2ν−1|γ|2−Ki(ν)+K2N(νν−−11)+1 for k∈ 2Ki(−ν)1+1,2Ki(ν)+1 ,
(cid:2) (cid:1)
K(ν−1), if ν >1,
K(ν) ≡ Nν−1
0
(n0−1, if ν =1,
10 ArtsrunSargsyan,MartinGrigoryan
for any natural numbers i∈[1,N ] and ν ∈[1,q−1]. Besides,
ν
K(q−1)
Nq−1 q−1 ε
(15) b < , where b ≡a , k ∈[2n,2n+1),
n 2k+1 n k
nX=n0 Xk=1
q−1 Nν 2Ki(ν)−1
(16) H (x)=H (x)+ a W (x),
q−1 q−1 k k
Xν=1Xi=1k=2XKi(−ν)1+1
e
−(2q−1−1)γ, for x∈E ,
q−1
(17) H (x)= γ, for x∈E ,
q−1  q−1
0, for x∈/ ∆e,
e

(18) E =2−q+1|∆| and E = 1−2−q+1 |∆|
q−1 q−1
(cid:12) (cid:12) (cid:12) (cid:12) (cid:0) (cid:1)
andthe setE(cid:12)q−e1 ca(cid:12)nbe presentedasau(cid:12)niono(cid:12)fcertainNq amountofdisjoint
dyadic interveals ∆i(q) with measure ∆i(q) =2−KN(qq−−11)−1, i∈[1,Nq]:
(cid:12) (cid:12)
(cid:12) Nq(cid:12)
E = ∆(q).
q−1 i
i=1
[
e
For each natural number i ∈ [1,N ] we choose such a natural number
q
(q) (q) (q) (q−1)
K >K K ≡K that the following conditions hold:
i i−1 0 Nq−1
K(q)−K(cid:0)(q−1)−1 (cid:1)
a′′) i Nq−1 is a whole number,
2
b′′) Ki(q)−Ki(−q)1 2q−1|γ|2−Ki(q)+K2N(qq−−11)+1 < 2q+ε1Nq,
c′′) (cid:0)2q|γ|2−Ki(q2)+1(cid:1)< ε.
2
By a successive application of lemma 1 for each interval ∆(q) ⊂E (i∈
i q−1
[1,N ])andcorrespondingnumberK(q) wecanfindpolynomialsintheWalsh
q i
system e
2Ki(q)+1−1
(19) H(q)(x)= a˜ W (x), i∈[1,N ],
i k k q
k=X2Ki(q)
e
so that
(20) |a˜k|=2q−1|γ|2−Ki(q)+K2N(qq−−11)+1 when k ∈ 2Ki(q),2Ki(q)+1 ,
(cid:2) (cid:1)