c Allerton Press, Inc., 2016. ISSN 1066-369X, Russian Mathematics (Iz. VUZ), 2016, Vol. 60, No. 1, pp. 42–59.  c V.I. Shcherbakov, 2016, published in Izvestiya Vysshikh Uchebnykh Zavedenii. Matematika, 2016, No. 1, pp. 49–68. Original Russian Text 

Divergence of the Fourier Series by Generalized Haar Systems at Points of Continuity of a Function V. I. Shcherbakov1* 1

Moscow Technical University of Communication and Information Science, ul. Narodnogo Opolcheniya 32, Moscow, 123995 Russia Received May 26, 2014

Abstract—We obtain a connection between the Dirichlet kernels and partial Fourier sums by generalized Haar and Walsh (Price) systems. Based on this, we establish an interrelation between convergence of the Fourier series by generalized Haar and Walsh (Price) systems. For any unbounded sequence we construct a model of continuous function on a group (and even on a segment [0, 1]), whose Fourier series by generalized Haar system generated by this sequence, diverges at some point. DOI: 10.3103/S1066369X16010059 Keywords: Abelian group, modified segment [0, 1], continuity on modified segment [0, 1], systems of characters, Price’s systems, generalized Haar’s systems.

INTRODUCTION. BASIC DEFINITIONS Let p0 = 1, {pn }∞ n=1 be an integer sequence with pn ≥ 2, mn =

n 

pk , n = 0, 1, 2, . . . Any natural

k=0

number n can be uniquely decomposed by the formula n=

s 

ak mk = as ms + n ,

(1)

k=0

where ak , s and n are integers with 0 ≤ ak ≤ pk+1 − 1, 1 ≤ as ≤ ps+1 − 1, i.e., ms ≤ n ≤ ms+1 − 1, and 0 ≤ n ≤ ms − 1. Note that if n < p1 = m1 , then s = 0, as = a0 = n and n = 0. Any real number x ∈ [0, 1] can be represented in the form ∞  xk , where xk are integer numbers, 0 ≤ xk ≤ pk − 1. x= mk

(2)

k=1

If x is {pn }-irrational, as well as equal zero or one, then its representation by formula (2) is unique. For x = mln , l = 1, 2, . . . , mn − 1, n = 1, 2, . . . , there exist its two decompositions in the form of equality (2), one of which is finite, i.e., xk = 0 for k ≥ n + 1, which we denote as mln , and the other is infinite, xk = pk − 1 for k > n, mln −. Thus, the segment [0, 1] maps to a certain set of integer sequences G = {{xn }∞ n=1 | xn = 0, 1, 2, . . . , pn − 1}, in which {pn }-rational points (except zero and one) “bifurcated”. Assuming mln − < mln , with [0, 1] on G one can extend the notion of order of points, then on G a segment [a, b] (a, b ∈ G) is defined. *

E-mail: [email protected].

42

DIVERGENCE OF THE FOURIER SERIES BY GENERALIZED HAAR SYSTEMS +

43

+

On the set G, let us define operation · of componentwise addition modulo pn : {xn } · {yn } = −

{(xn + yn ) mod pn }, with respect to which the set G becomes an Abelian group, let · be the inverse operation. The neighborhoods of zero in G are the subgroups   Gn = 0, m1n − = {{xk }∞ k=1 ∈ G | x1 = x2 = · · · = xn = 0}, n = 0, 1, 2, . . . , G0 = G. +

Thus, in G a topology is defined (a system of neighborhoods of point x ∈ G is co-sets x · Gn ), with respect to which in group G the limit and continuity of function are defined, a function is a mapping of the group G to the set of complex numbers C. This continuity is equivalent to ◦ ordinary continuity at {pn }-irrational points, ◦ right-continuity at ◦ left-continuity at

l mn ,

l mn −.

We can extend the notions of measure and Lebesgue integral, orthogonal and orthonormal systems of functions, as well as Fourier series by these systems and classes of functions Lp (G) without any change from [0, 1] to G. Also assume x 0 = 0 and n  xk . x n = mk

(3)

k=1

xn + Then x ∈ [ xn , (

1 mn )−]

+

and x · Gn = [ xn , ( xn +

1 mn )−].

1. GENERALIZED HAAR AND WALSH SYSTEMS  l  2iπxk+1 = exp p2iπl , if x∈ mk+1 , ml+1 − , l = 0, 1, 2, . . . , mk+1 − 1; Let ψ0 (x)≡1, ψmk (x)= exp pk+1 k+1 k+1 s  (ψmk (x))ak , where ak and s are defined k = 0, 1, 2, . . . , and xk be given by equality (2), ψn (x) = k=0

by (1). Then {ψn (x)}∞ n=0 is a complete orthonormal system of continuous on group G functions, possessing the characters property: +

ψn (x · y) = ψn (x)ψn (y).

(4)

For pn ≡ 2 the collection {ψn (x)}∞ n=0 is a Walsh system [1] in Paley enumeration [2], for pn ≡ p is a Chrestenson (or Chrestenson–Levi) system [3], in the general case {ψn (x)}∞ n=0 is a Price system [4] or a Vilenkin system on a modified segment [0, 1]. N. Ya. Vilenkin [5] considered even more general case, when {ψn (x)}∞ n=0 is a character system of zero-dimensional compact Abelian group. Consider another complete orthonormal system of continuous on group G piecewise constant functions {χn (x)}∞ n=0 : χ0 (x) ≡ 1, ⎧√  n   n n +1  2iπlas l n l+1 ⎪ ⎨ ms exp ps+1 , if x ∈ ms + ms+1 , ( ms + ms+1 ) − ⊂ ms , ms ; χn (x) = χas ms +n (x) = l = 0, 1, . . . , ps+1 − 1; ⎪ ⎩ 0 for other x, as and n are defined by (1). On the group G, we can also define the function √ s xs+1 ms exp 2iπa , ps+1 χn (x) = χas ms +n (x) = 0 x s is from (3). RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 60 No. 1 2016

if n = x s ms ; s ms , for n = x

(5)

44

SHCHERBAKOV

In case pn ≡ 2 the system {χn (x)}∞ n=0 is the Haar system [6]. By equality (1) for n < p1 = m1 we get (6)

χn (x) = ψn (x).

Unlike the Price system, {χn (x)}∞ n=0 does not possess property (4). Indeed, if the function f (t) +

possesses property (4), i.e., f (x · t) = f (x)f (t), and for some x0 we have f (x0 ) = 0, then for any x ∈ G −

−

+

f (x) = f ((x · x0 ) · x0 ) = f (x · x0 )f (x0 ) = 0, i.e., if a function, possessing the character property (4), equals zero at some point, then it turns to zero everywhere. However, the function χn (x) for n ≥ m1 is not of such a kind, due to equality (5) there are points x0 , where χn (x0 ) = 0, and there are points y0 ∈ G, where χn (y0 ) = 0.

2. MAIN RESULT The main goal of this paper is to study the interrelation between the partial Fourier sums by systems ∞ ∞ {ψn (x)}∞ n=0 and {χn (x)}n=0 . By this interrelation, for any sequence {pn }n=1 with sup pn = ∞ we n

will construct an example of continuous on group G (also on segment [0, 1]) function, depending on ∞ sequence {pn }∞ n=1 , whose Fourier series by generalized Haar system {χn (x)}n=0 is divergent at some point. The respective theorems will be formulated and proved in Section 4, and the contrary instances will be constructed in Sections 5–8 of this paper.

3. DIRICHLET KERNELS BY GENERALIZED HAAR AND WALSH SYSTEMS Recall that an n-th Dirichlet kernel by orthonormal system of functions {ϕn (x)}∞ n=0 is evaluated by n−1 2 1 ϕk (x)ϕk (t). Since |ψn (x)| ≡ 1, it follows ψk (t) = |ψψkk(t)| the formula Dn (x) = (t) = ψk (t) , and using k=0

equality (4), for Price system we have Dn (x, t) = may assume Dn (x) =

n−1

n−1

ψk (x)ψk (t) =

k=0

n−1

−

−

ψk (x · t) = Dn (x · t), here we

k=0

ψk (x). It is known that (see, e.g., [7]; [5], 2.2; [8], P. 267, Lemma 3; [9], P. 134,

k=0

equalities (2) and (3))

   mn , if x ∈ Gn = 0, m1n − ; Dmn (x) =   0 for x ∈ G \ Gn = m1n , 1 ,

(7)

Dn (x) = Das ms (x) + (ψms (x))as Dn (x),

(8)

and where s, as , and

n

are defined by (1). Also, if j, 1 ≤ j ≤ pn+1 − 1, is integer, then Djmn (x) = Dmn (x)

1 − (ψmn (x))j 1 − ψmn (x)

and since for x ∈ Gn+1 and k < mn+1 ψk (x) = 1, we have that Djmn (x) =

(9) jm n −1

ψk (x) = jmn . We

k=0

have

⎧ ⎪ ⎪ ⎨jmn ,

  1 if x ∈ Gn+1 = 0, mn+1 − ;  1  j mn (x)) 1 Djmn (x) = mn 1−(ψ , if x ∈ G \ G = , − ; n n+1 mn+1 mn 1−ψmn (x) ⎪  1  ⎪ ⎩0 in case x ∈ G \ Gn = mn , 1 .

(10)

RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 60 No. 1 2016

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45

Now consider the Dirichlet kernels for generalized Haar systems. By equality (6), if n ≤ m1 = p1 , −

then Dn (x, t) = Dn (x · t) and, in particular,    − m1 , if x · t ∈ G1 = 0, m11 − ; Dm1 (x, t) = 0 for other x and t.

(11)

Let now n = as ms + n > m1 . Unlike in (1), here we assume ms + 1 ≤ n ≤ ms+1 and 1 ≤ n ≤ ms . Then the number n = ms for s ≥ 2 will be written as n = ms = (ps − 1)ms−1 + ms−1 c as−1 = ps − 1 and n = ms−1 , and n = as ms will be decomposed as n = as ms = (as − 1)ms + ms as ms +n −1 as m s −1 with n = ms . Let us find Dn (x, t) = Das ms +n (x, t) = χk (x)χk (t) = χk (x)χk (t) + as ms +n −1

k=0

k=0

χk (x)χk (t), i.e., we get the formula

k=as ms

Dn (x, t) = Das ms (x, t) +

 −1 n

(12)

χas ms +k (x)χas ms +k (t).

k=0

Note that the summation index in equality (12) k ≤ n − 1 ≤ ms − 1, and hence functions χas ms +k (x) can be evaluated by formula (5). Consider the following cases. ∞ s − tk tk  / Gs , i.e., x s =  ts (by formulas (2) and (3) t = and t = 1. Let x · t ∈ s mk mk ). Then either k=1

k=1

ts ms and χas ms +k (t) = 0. Hence by (12) for k= x s ms and by equality (5) χas ms +k (x) = 0, or k =  −

x · t ∈ G \ Gs we have (13)

Dn (x, t) = Das ms +n (x, t) = Das ms (x, t). −

s =  ts . Then if k < x s ms =  ts ms , then by (5) we have χas ms +k (x) = 2. Let now x · t ∈ Gs , i.e., x − s ms =  ts ms . If k = x s ms =  ts ms , χas ms +k (t) = 0. Hence equality (13) is valid for x · t ∈ Gs , but n ≤ x then by formula (5) −

−ts+1 ) as s xs+1 s ts+1 ) exp(−i 2πaps+1 )=ms (exp 2iπ(xs+1 ) =ms (ψms (x · t))as , χas ms +k (x)χas ms +k (t)=ms exp(i 2πaps+1 ps+1

s ms =  ts ms we obtain the where ψms is a respective function from the Price system. Thus, for k = x equality −

χas ms +k (x)χas ms +k (t) = ms (ψms (x · t))as .

(14)

ts ms , then χas ms +k (x) = χas ms +k (t) = 0. Hence for n > x s ms =  ts ms under the If k > x s ms =  summation sign of the right-hand side of formula (12) only one summand will be non-zero, where the ts ms , then by (12)–(14) we get summation index k = x s ms =  ⎧ − ⎪ ⎪ in case x · t ∈ / Gs ; ⎨Das ms (x, t) − (15) Dn (x, t) = Das ms (x, t), if x · t ∈ Gs and n ≤ x s ms =  ts ms ; ⎪ ⎪ − − ⎩D (x, t) + m (ψ (x · t))as for x · t ∈ G and n > x  m = tm , as ms

s

ms

s

s

s

s

s

numbers s, as , and n are defined by (1) with the corrections stated right after equality (11). Assuming s to be n, as = an to be j − 1, then 2 ≤ j ≤ pn+1 and assuming n = mn = ms , we get  − − D(j−1)mn (x, t) + mn (ψmn (x · t))j−1 for x · t ∈ Gn ; Djmn (x, t) = D(j−1)mn +mn (x, t) = − for x · t ∈ G \ Gn , D(j−1)mn (x, t) = Dmn (x, t) (16) RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 60 No. 1 2016

46

SHCHERBAKOV −

if x · t ∈ / Gn , then D(j−1)mn (x, t) can also be presented by formula (16): Djmn (x, t) = D(j−1)mn (x, t) = D(j−2)mn (x, t) up to D2mn (x, t), Djmn (x, t) = D(j−1)mn (x, t) = D(j−2)mn (x, t) = · · · = Dmn (x, t)). −

For x · t ∈ Gn let us show that Djmn (x, t) = Dmn (x, t) + mn

j−1 

−

(ψmn (x · t))l .

(17)

l=1

Let us prove equality (17) by mathematical induction by j (starting from j = 2). −

Substituting j = 2 into (16), we get D2mn (x, t) = Dmn (x, t) + mn ψmn (x · t), that proves formula (17) in case j = 2. Let now equality (17) be fulfilled for j = k. By formula (16), let us find D(k+1)mn (x, t), where instead −

of Dkmn (x, t) we substitute its value from equality (17): D(k+1)mn (x, t) = Dkmn (x, t) + mn (ψmn (x · k−1 k − − − (ψmn (x · t))l + mn (ψmn (x · t))k = Dmn (x, t) + mn (ψmn (x · t))l and t))k = Dmn (x, t) + mn l=1

l=1

the fulfillment of equality (17) was shown for j = k + 1. Formula (17) is proved. −

−

If x · t ∈ Gn \ Gn+1 , then ψmn (x · t) = 1, and let us transform the sum in the right-hand side of j−1 − (ψmn (x · t))l = equality (17) in the following way (we add and subtract a summand for l = 0): j−1

−

(ψmn (x · t))l − 1 =

l=0 −

1−(ψmn

−

(x · t))j −

1−ψmn (x · t)

− 1 or mn

l=1

−

For x · t ∈ Gn+1 ψmn (x · t) = 1 and

j−1

l=1

j−1

−

−

(ψmn (x · t))l = mn 1−(ψmn (x ·−t)) − mn . j

1−ψmn (x · t)

−

(ψmn (x · t))l = j − 1, i.e., mn

l=1

j−1

−

(ψmn (x · t))l = jmn −

l=1

mn . Hence the equality is fulfilled ⎧ − ⎪ D (x, t) + jmn − mn , if x · t ∈ Gn+1 ; ⎪ ⎪ ⎨ mn − j − Djmn (x, t) = Dmn (x, t) + mn 1−(ψmn (x ·−t)) − mn , if x · t ∈ Gn \ Gn+1 ; ⎪ 1−ψ (x · t) mn ⎪ ⎪ − ⎩ if x · t ∈ G \ Gn , Dmn (x, t),

(18)

j, 1 ≤ j ≤ pn+1 is integer, the validity of case j = 1 in formula (18) is verified by straightforward

 n+1 pn+1 = 1, and then, assuming substitution. Note that for x ∈ Gn \ Gn+1 (ψmn (x))pn+1 = exp 2iπx pn+1 in formula (18) j = pn+1 we get the following recurrent relation for Dirichlet kernels Dmn (x, t) by generalized Haar system {χn (x)}∞ n=0 : ⎧ − ⎪ ⎪Dmn (x, t) + mn+1 − mn , if x · t ∈ Gn+1 ; ⎨ − (19) Dmn+1 (x, t) = Dmn (x, t) − mn , if x · t ∈ Gn \ Gn+1 ; ⎪ ⎪ − ⎩D (x, t), if x · t ∈ G \ G . mn

n

From (19) and (11) by mathematical induction let us obtain a general formula for Dirichlet kernels Dmn (x, t) by generalized Haar systems:  − mn if x · t ∈ Gn ; (20) Dmn (x, t) = − 0, if x · t ∈ G \ Gn . Indeed, the induction base is formula (11). Induction step. −

1. For x · t ∈ Gk+1 we have Dmk+1 (x, t) = Dmk (x, t) + mk+1 − mk = mk + mk+1 − mk = mk+1 . −

2. If x · t ∈ Gk \ Gk+1 , then Dmk+1 (x, t) = Dmk (x, t) − mk = mk − mk = 0. RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 60 No. 1 2016

DIVERGENCE OF THE FOURIER SERIES BY GENERALIZED HAAR SYSTEMS

47

−

3. In case x · t ∈ G \ Gk we get Dmk+1 (x, t) = Dmk (x, t) = 0. Formula (20) is proved. Then by equalities (18), (20) we have ⎧ − ⎪ m + jmn − mn = jmn , if x · t ∈ Gn+1 ; ⎪ ⎪ ⎨ n − j − if x · t ∈ Gn \ Gn+1 ; Djmn (x, t) = mn 1−(ψmn (x ·−t)) , ⎪ 1−ψ (x · t) mn ⎪ ⎪ − ⎩ 0, if x · t ∈ G \ Gn .

(21)

As we compare formulas (10) and (21), we see that Dirichlet kernels for Price systems and generalized Haar systems with numbers jmn , j = 1, 2, . . . , pn+1 − 1, coincide: −

Djmn (x, t) = Djmn (x · t).

(22)

By equality (15) we get that by generalized Haar systems {χn (x)}∞ n=0 either Dn (x, t) = Das ms +n (x, t) = Das ms (x, t), or Dn (x, t) = Das ms +n (x, t) = D(as +1)ms (x, t), to be more precise (numbers as , ms , and n are defined by (1)), ⎧ − ⎪ ⎪ if x · t ∈ Gs and n ≤ x s ms =  ts ms ; ⎨Das ms (x, t), −  (23) Dn (x, t) = D(as +1)ms (x, t), if x · t ∈ Gs and n > x s ms = ts ms ; ⎪ ⎪ − ⎩D (x, t) = 0, if x · t ∈ G \ G . as ms

s

The second row in (23) can be obtained by comparing the first row of formula (16) with the third row in equality (15) for n = s and j = as + 1. 4. FOURIER SERIES BY GENERALIZED HAAR AND WALSH SYSTEMS It is known that an n-th partial Fourier sum of the function f (t) at x by orthonormal system 1 (ϕ) f (t)Dn (x, t)dt, where Dn (x, t) are Dirichlet {ϕn (t)}∞ n=0 can be found by the formula Sn (x, f ) = kernels. Then for Price systems (χ)

(Ψ) Sn (x, f )

alized Haar systems Sn (x, f ) =



=



0

−

f (t)Dn (x · t)e dt =

G



−

f (x · t)Dn (t)dt, and for gener-

G

f (t)Dn (x, t)dt.

G

By equalities (22) and (23) (s, as , and n are defined by (1)) we get  (χ) Sas ms (x, f ), if n ≤ x s ms ; (χ) Sn(χ) (x, f ) = Sas ms +n (x, f ) = (χ)  s ms . S(as +1)ms (x, f ), if n > x Indeed, Sn(χ) (x, f )

 =

 f (t)Dn (x, t)dt =

 f (t)Dn (x, t)dt +

−

G

{t|x · t∈G\Gs }

s ms these integrals equal By equality (15) for n ≤ x   f (t)Dn (x, t)dt = −

f (t)Das ms (x, t)dt

−

{t|x · t∈Gs }

{t|x · t∈Gs }



 f (t)Dn (x, t)dt = −

{t|x · t∈G\Gs }

f (t)Dn (x, t)dt. −

{t|x · t∈Gs }

and

(24)

f (t)Das ms (x, t)dt. −

{t|x · t∈G\Gs }

RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 60 No. 1 2016

(25)

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SHCHERBAKOV

Substituting the latter two equalities into (25), we get   Sn(χ) (x, f ) = f (t)Das ms (x, t)dt + f (t)Das ms (x, t)dt −

−

{t|x · t∈Gs }

{t|x · t∈G\Gs }

 f (t)Das ms (x, t)dt = Sa(χ) (x, f ), s ms

= G

and in case

n

n

≤x s ms equality (24) is proved. If > x s ms , then by formula (23) we obtain   f (t)Dn (x, t)dt = f (t)D(as +1)ms (x, t)dt −

−

{t|x · t∈Gs }

and

{t|x · t∈Gs }



 f (t)Dn (x, t)dt = 0 = −

f (t)D(as +1)ms (x, t)dt −

{t|x · t∈G\Gs }

{t|x · t∈G\Gs }

(see also the third row in (21)). Substituting the latter equalities into the right-hand side of (25), we get   (χ) f (t)Dn (x, t)dt + f (t)Dn (x, t)dt Sn (x, f ) = −

−

{t|x · t∈Gs }

{t|x · t∈G\Gs }





f (t)D(as +1)ms (x, t)dt + 0 =

= −

f (t)D(as +1)ms (x, t)dt −

{t|x · t∈Gs }

{t|x · t∈Gs }





f (t)D(as +1)ms (x, t)dt =

+ −

(χ)

f (t)D(as +1)ms (x, t)dt = S(as +1)ms (x, f ), G

{t|x · t∈G\Gs }

s ms . Formula (24) is proved. i.e., equality (24) is obtained for n > x (χ)

(χ)

(χ)

Thus, partial Fourier sums Sn (x, f ) coincide either with Sas ms (x, f ), or with S(as +1)ms (x, f ). By equality (22) we find (χ)

(Ψ)

(26)

Sjn mn (x, f ) = Sjnmn (x, f ) (χ)

jn , 1 ≤ jn ≤ pn+1 − 1, are integers. Indeed, Sjn mn (x, f ) =

 G

f (t)Djn mn (x, t)dt =



−

f (t)Djn mn (x ·

G

(Ψ)

t)dt = Sjn mn (x, f ). Further, due to formula (26), for partial sums Sjmn (x, f ), we will not state by which system they are evaluated. Let us denote by Jn a subset of the set of natural numbers Jn = {mn , 2mn , . . . , (pn+1 − 1)mn } ∞

and assume J = ∪ Jn . Then the increasingly ordered set J defines a subsequence of the sequence n=1

of natural numbers. / J. Set J can Note that J does not coincide with the set of all natural numbers. For example, p1 + 1 ∈ also be defined as the set of natural numbers, such that in their decomposition by formula (1) ak = 0 for all k = s, as well as n = 0. Equalities (24) and (26) show the validity of RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 60 No. 1 2016

DIVERGENCE OF THE FOURIER SERIES BY GENERALIZED HAAR SYSTEMS

49

Theorem 1. A Fourier series of function f (t) ∈ L(G) by generalized Haar system {χn (x)}∞ n=0 converges at x ∈ G to a (generally speaking, complex) number S if and only if at this point (Ψ) subsequence {Sn (x, f )}n∈J of partial Fourier sums by Price system {ψn (x)}∞ n=0 converges to S. In Section 9 we will say that in case pn ≡ 2 Theorem 1 was known earlier. Theorem 2. Let some c > 2 be given. Assume Jnc = {mn , 2mn , . . . , min([c], pn+1 −1))mn } and ∞

J c = ∪ Jnc , where [c] denotes the integer part of c. Then if function f (t) ∈ L(G) is continuous n=1

at x ∈ G, then sequence {Sk (x, f )}k∈J c converges to f (x). Proof. Since k ∈ J c , i.e., k ∈ Jnc for some n, then k = jmn , where j, 1 ≤ j ≤ c, is integer. Then, by continuity of f (t), at x for any ε > 0 there exists N such that for all n > N and t ∈ Gn ε − (27) |f (x · t) − f (x)| < . 2c  However, Djmn (t)dt = 1. Hence for n > N we have G

     −  Djmn (t)dt |Sk (x, f ) − f (x)| = |Sjmn (x, f ) − f (x)| =  f (x · t)Djmn (t)dt − f (x) G

G

    −  =  (f (x · t) − f (x))Djmn (t)dt ≤



−

|f (x · t) − f (x)| |Djmn (t)|dt −

G

{t|x · t∈Gn+1 }

  |f (x · t) − f (x)| |Djmn (t)|dt + 





−

+ −

  (f (x · t) − f (x))Djmn (t)dt. −

−

{t|x · t∈Gn \Gn+1 }

{t|x · t∈G\Gn }

summand here turns to zero. Using further (10) and (27), measure μ(Gn+1 ) =  By (10),  the latter 1 1 , μ(Gn \ Gn+1 ) < m1n , and condition |ψn (x)| = 1, μ 0, mn+1 = mn+1 |Dn (x)| ≤

n−1 

|ψk (x)| = n,

(28)

k=0

by the latter inequality we get   ε ε ε ε 1 1 + j < ε. + jmn < |Sk (x, f ) − f (x)| = Sjmn (x, f ) − f (x) < jmn 2c mn+1 2c mn 2c 2c



Then by Theorems 1 and 2, as well as equality (24) it is easy to obtain Corollary 1. If subsequence of partial Fourier sums {Snk (x, f )}∞ k=1 by generalized Haar systems of nk n = , m is defined by (1), are bounded, then Snk (x, f ) converge function f (t) ∈ L(G) is such that m s ms s to f (t) at each of its points of continuity x. 

n ≥ as , i.e., numbers Indeed, if we divide all parts of formula (1) by ms , then we get mns = as + m s nk n as ≤ ms = ms are bounded, and then by Theorem 2 we get that Sas ms (x, f ) and S(as +1)ms (x, f ) (χ)

(χ)

converge to f (x). Hence, by equality (24), Sn (x, f ) = Snk (x, f ) are also converging to f (x). Corollary 2. If sup pn < ∞, then Fourier series by generalized Haar system of function f (t) ∈ L(G) n

converges to it at each point of continuity x of function f (t). Note that in case sup pn = ∞ there may be no convergence of Fourier series by generalized Haar n

system even for continuous functions, that we will show in the next Section. RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 60 No. 1 2016

50

SHCHERBAKOV

5. FIRST CONTRARY INSTANCE Let us give an example of a continuous on group G function with diverging, at some point, Fourier series by generalized Haar system in case when the sequence {pn }∞ n=1 contains an infinite subsequence, consisting solely of odd numbers. pn+1 −1 . 2

Then, using equalities 1 − e2iα = −2 sin αei(α+π/2) and iπ(4l+1)pn+1   − 4l+2 = i, as well as formula (21), by generalized Haar systems for x · t ∈ m4l+1 ; − ⊂ e 2pn+1 n+1 mn+1  pn+1  Gn \ Gn+1 , i.e., l, 0 ≤ l ≤ 4 , is integer, we will find Let pn+1 be odd, and jn =

−

Djn mn (x, t) = mn

1 − (ψmn (x · t))jn −

1 − ψmn (x · t) = mn

= mn

n+1 −1) sin π(4l+1)(p 2pn+1

sin π(4l+1) pn+1 π 2 π(4l+1) 2pn+1

sin(2πl + = mn

2 sin

= mn

1 − exp 2iπ(4l+1) pn+1

n n sin π(4l+1)j exp(i π(4l+1)j + π2 ) pn+1 pn+1

π(4l+1) π sin π(4l+1) pn+1 exp(i pn+1 + 2 )

n −1) i π(4l+1)(j p

e

n 1 − exp 2iπ(4l+1)j pn+1

n+1

= mn

π(4l+1) −3iπ(4l+1) 2pn+1 ) ie 2pn+1 π(4l+1) cos 2pn+1

−

n+1 sin( π(4l+1)p − 2pn+1

2 sin π(4l+1) 2pn+1 cos

= mn

π(4l+1) π(4l+1)(pn+1 −3) 2pn+1 ) i 2pn+1 e π(4l+1) 2pn+1

cos π(4l+1) 2pn+1 2 sin

π(4l+1) 2pn+1

cos

π(4l+1) 2pn+1

=

ie

−3iπ(4l+1) 2pn+1

mn 2 sin

π(4l+1) 2pn+1

ie

−3iπ(4l+1) 2pn+1

.

We have shown that the equality is fulfilled, Djn mn (x, t) = −

for x · t ∈



4l+1 4l+2 mn+1 ; mn+1

mn 2 sin

ie π(4l+1)

−3iπ(4l+1) 2pn+1

(29)

2pn+1

 − ; l = 0, 1, 2, . . . , [ pn+1 8 ].

Let now {pn }∞ n=1 be an unbounded sequence that has an infinite increasing subsequence, consisting of odd numbers, which we will denote as {pnk +1 }∞ k=0 . Without loss of generality, we may assume pnk +1 ≥ 5. Then numbers jk =

pnk +1 − 1 2

(30)

are integer. Consider the function ⎧ 3iπ(4l+1)  4l+1 4l+2  − pnk +1 ⎪ 2pn +1 ⎪ √ −i ⎪ ⎨ ln pnk +1 e k , if u = x · t ∈ mnk +1 , mnk +1 − , l = 1, 2, . . . , [ 8 ] − f (t) = (i. e. u = x · t ∈ Gnk \ Gnk +1 , k = 0, 1, 2, . . . ); ⎪ ⎪ ⎪ ⎩0 for other t. Note that f (t) is a continuous on group G function, since lim |f (t)| = lim √ln p1 t→x

k→∞

nk +1

= 0 = f (x). Let

us find its partial Fourier sum Sjk mnk (x, f ) at x by generalized Haar system   f (t)Djk mnk (x, t) dt = f (t)Djk mnk (x, t)dt Sjk mnk (x, f ) = G

−

{t|x · t∈Gnk +1 }

RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 60 No. 1 2016

DIVERGENCE OF THE FOURIER SERIES BY GENERALIZED HAAR SYSTEMS





+

f (t)Djk mnk (x, t)dt + −

51

f (t)Djk mnk (x, t)dt. −

{t|x · t∈Gnk \Gnk +1 }

{t|x · t∈G\Gnk }

By formula (21), the third summand in the right-hand side of the latter equality turns to zero. Thus, we have shown that   f (t)Djk mnk (x, t)dt + f (t)Djk mnk (x, t)dt. (31) Sjk mnk (x, f ) = −

−

{t|x · t∈Gnk +1 }

   

{t|x · t∈Gnk \Gnk +1 }

By formulas (21) and (28), the first summand in (31) will be estimated as      f (t)Djk mnk (x, t)dt ≤ jk mnk |f (t)|dt −

−

{t|x · t∈Gnk +1 }

{t|x · t∈Gnk +1 }



mn +1 < k ln pnk +1

dt =  −

1 , (32) ln pnk +1

{t|x · t∈Gnk +1 }

the measure μ(Gnk +1 ) =

1 mnk +1 .

The second summand in (31) will be estimated, using (29):  pn +1  k   8  f (t)Djk mnk (x, t)dt = l=0   −  − {t|x · t∈Gn \Gn +1 } t  x · t∈ 4l+1 k

k

−i = ln pnk +1

, 4l+2 − mn +1 mn +1 k k

 pn +1  k 8  l=0



  −  t  x · t∈

mn =  k 2 ln pnk +1

4l+1 , 4l+2 mn +1 mn +1 k k

 pn +1  k 8  l=0



e

3iπ(4l+1) 2pn +1 k



f (t)Djk mnk (x, t)dt

imnk 2 sin

e π(4l+1)

−3iπ(4l+1) 2pn +1 k

dt

2pnk +1



1 sin π(4l+1) 2pnk +1   −  t  x · t∈

4l+1 , 4l+2 − mn +1 mn +1 k k



dt

 pn +1  k 8  2pnk +1 mnk  ≥ π(4l + 1) 2mnk +1 ln pnk +1 l=0  pn +1  k  pn +1  8  ln( k8 − 1) 1 2mnk pnk +1   > . > 2πmnk +1 ln pnk +1 l=0 4(l + 1) 4π ln pnk +1

pnk +1   pnk +1 + 1 , and hence Here we will also use that the argument of sine (since l ≤ 8 ]≤ 8

pn +1  k π +4 πpn +1 π(4l+1) 2 < 4pn k+1 + pn2π+1 < π4 + 23 π < π (since pnk +1 ≥ 5 (see condition on the se2pn +1 < 2pn +1 k

k

k

k

π(4l+1) quence {pnk +1 }∞ k=1 before formula (30))) and then sin pn +1 > 0. The right-hand side of the latter k

equality tends to infinity as k → ∞. Thus, by the latter equality, as well as formulas (31) and (32) we get (x, f ) = ∞, i.e., the Fourier series by generalized Haar system is diverging (even that lim S pnk +1 k→∞

2

mnk

boundedly) at x. RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 60 No. 1 2016

52

SHCHERBAKOV

Similarly to the inference of the latter inequality, we can show that  mnk +1 (b − a)  f (t)Djk mnk (x, t)dt ≥ . 4π(4l + 1) ln pnk +1   −    t  x · t∈[a,b]⊆ 4l+1 , 4l+2 − ⊂Gn \Gn +1 k

mn +1 mn +1 k k

(33)

k

6. SECOND CONTRARY INSTANCE Let us give an example of continuous on group G function with diverging, at some point, Fourier series by generalized Haar system in case of unbounded sequence {pn }∞ n=1 that does not satisfy the pn+1 condition of Section 5. So, let the number pn+1 be even, and jn = 2 . Let us find Djn mn (x, t)     − , 4l+2 − ⊂ Gn \ Gn+1 , i.e., l = 0, 1, 2, . . . , pn+1 . Using equality (21), as well as for x · t ∈ m4l+1 4 n+1 mn+1 π

formula 1 − e2iα = −2 sin αei(α+ 2 ) , we have −

−

Djn mn (x, t) = mn (1 − (ψmn (x · t))jn )/(1 − ψmn (x · t))      2iπ(4l + 1)jn 2iπ(4l + 1) 1 − exp = mn 1 − exp pn+1 pn+1        iπ(4l+1)jn iπ −iπ(4l+1) iπ π(4l+1)jn  π(4l+1) exp − 2 sin + − − 2 sin = mn exp pn+1 2 pn+1 2 pn+1 pn+1      iπ(4l + 1)(jn − 1) π(4l + 1) π(4l + 1)pn+1 sin sin = mn exp pn+1 2pn+1 pn+1      iπ(4l + 1)(pn+1 − 2) π(4l + 1) π(4l + 1) sin sin = mn exp 2pn+1 2 pn+1       iπ(4l + 1)(pn+1 − 2) π(4l + 1) π sin sin 2πl + = mn exp 2pn+1 2 pn+1     iπ(4l + 1)(pn+1 − 2)  π(4l + 1) sin . mn exp 2pn+1 pn+1 We obtain the equality Djn mn (x, t) = −

for x · t ∈



4l+1 4l+2 mn+1 , mn+1

mn sin

e π(4l+1)

iπ(4l+1)(pn+1 −2) 2pn+1

imn

= sin

pn+1



− 2iπ(4l+1) 2 +1

e π(4l+1)

(34)

pn

pn +1

− ⊂ Gn \ Gn+1 , i.e., l is integer with 0 ≤ l ≤

 pn+1  4

and jn =

pn+1 2 .

So, let unbounded sequence {pn }∞ n=0 not satisfy the condition of Section 5. This means that any infinite subsequence (it must be, otherwise sup pn = ∞) of sequence {pn }∞ n=0 may consist of at most n

finite number of odd numbers. As we drop them, we get infinite subsequence, entirely consisting of even numbers. Passing in it, if necessary, to a subsequence, we construct an infinite increasing subsequence {pnk +1 }∞ k=1 satisfying the condition pnk +1 ≥ 12 and in which all numbers are even. Consider a function ⎧ iπ(4l+1)(pn +1 −2) k   ⎪ − − ⎪√ 1 2pn +1 ⎪ k , if u = x · t ∈ m4l+1 , 4l+2 − ⊂ Gnk \ Gnk +1 ⎨ ln pn +1 e nk +1 mnk +1 k

 pnk +1  f (t) = ⎪ i. e. l = 0, 1, 2, . . . , , k = 1, 2, . . . ; ⎪ 8 ⎪ ⎩ 0 for other t. Since lim |f (t)| = lim |f (t)| = lim √ln p1 −

Assume

x · t→0

t→x

k→∞

nk +1

jk =

= 0, the function f (t) is continuous on group G.

pnk +1 . 2

(35)

RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 60 No. 1 2016

DIVERGENCE OF THE FOURIER SERIES BY GENERALIZED HAAR SYSTEMS

First consider



  −  t  x · t∈[a,b]⊆

4l+1 , 4l+2 − mn +1 mn +1 k k



⊂Gnk \Gnk +1



53

f (t)Djk mnk (x, t)dt. Passing in the integral

−

to a new integration variable u = x · t, and, using equality (34), we get  f (t)Djk mnk (x, t)dt      − t  x · t∈[a,b]⊆ 4l+1 , 4l+2 − ⊂Gn \Gn +1 k

mn +1 mn +1 k k

k

1 mnk = ln pnk +1 sin π(4l+1) p nk +1

≥

   − t  x · t∈[a,b]⊆

e

ln pnk +1 π(4l+1) pnk +1 a

iπ(4l+1)(pn +1 −2) k 2pn +1 k

−

e

iπ(4l+1)(pn +1 −2) k 2pn +1 k

du =

mnk pnk +1 mnk +1 (b − a)   (b − a) = . π(4l + 1) ln pnk +1 π(4l + 1) ln pnk +1



4l+1 , 4l+2 − mn +1 mn +1 k k



du

a

b

mnk

We obtain the inequality

b

⊂Gnk \Gnk +1



f (t)Djk mnk (x, t)dt ≥

mnk +1 (b − a)  . π(4l + 1) ln pnk +1

(36)

Inequality (36), in particular, means that its left-hand side is always a real number. Now consider   f (t)Djk mnk (x, t)dt Sjk mnk (x, f ) = f (t)Djk mnk (x, t)dt = −

G

{t|x · t∈Gnk +1 }



 f (t)Djk mnk (x, t)dt. (37)

f (t)Djk mnk (x, t)dt +

+ −

−

{t|x · t∈Gnk \Gnk +1 }

{t|x · t∈G\Gnk }

By (21), the latter summand in (37) turns to zero. Let us show that   f (t)Djk mnk (x, t)dt + Sjk mnk (x, f ) = −

(38)

−

{t|x · t∈Gnk +1 }

   

f (t)Djk mnk (x, t)dt.

{t|x · t∈Gnk \Gnk +1 }

Let us estimate the first summand in (38) by equality (32) (jk are defined by (35)):         f (t)Djk mnk (x, t)dt = jk mnk  f (t)dt −

−

{t|x · t∈Gnk +1 }

{t|x · t∈Gnk +1 }



pn +1 mnk ≤ k 2

mn +1 |f (t)|dt ≤  k ln pnk +1

−

{t|x · t∈Gnk +1 }

−

mn +1 1 1 dt ≤  k = , ln pnk +1 mnk +1 ln pnk +1

{t|x · t∈Gnk +1 }

since the measure μ(Gnk +1 ) = μ([0, mn1 +1 −]) = k

We have shown that



   

 −

1 mnk +1 .

  1 f (t)Djk mnk (x, t)dt ≤  . ln pnk +1

{t|x · t∈Gnk +1 }

RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 60 No. 1 2016

(39)

54

SHCHERBAKOV

Using the definition of the function f (t), consider the second integral in the right-hand side of (38):  pn +1  k   8  f (t)Djk mnk (x, t)dt. f (t)Djk mnk (x, t)dt =     l=0 − − {t|x · t∈Gn \Gn +1 } t  x · t∈ 4l+1 , 4l+2 − k

mn +1 mn +1 k k

k

Let us estimate the right-hand side of the latter relation by inequality (36), a = hence b − a = 

1 mnk +1 .

4l+1 mnk +1

and b =

4l+2 mnk +1 ,

We get

f (t)Djk mnk (x, t)dt −

{t|x · t∈Gnk \Gnk +1 }

=

 pn +1  k 8 



  −  t  x · t∈

l=0

4l+1 , 4l+2 − mn +1 mn +1 k k

 pn +1  k 8 

mnk +1  ≥ πmnk +1 ln pnk +1

l=0



f (t)Djk mnk (x, t)dt

1 1 ≥  4l + 1 π ln pnk +1

Thus, the second summand in equality (38) is estimated as       f (t)Djk mnk (x, t)dt ≥  −

 pn +1  k 8  l=0

pn

pn

+1

ln[ k ] 1 ≥  8 . 4(l + 1) 4π ln pnk +1

+1

ln[ k8 ]  . 4π ln pnk +1

(40)

{t|x · t∈Gnk \Gnk +1 }

Then, substituting inequalities (39) and (40) into (38), we get       f (t)Djk mnk (x, t)dt |Sjk mnk (x, f )| ≥  −

{t|x · t∈Gnk \Gnk +1 }

  − 

 pn +1  ln[ k8 ] 1  f (t)Djk mnk (x, t)dt ≥  − . 4π ln pnk +1 ln pnk +1

 −

{t|x · t∈Gnk +1 }

Since the subsequence {pnk +1 }∞ k=1 is infinite, the right-hand side of the latter inequality tends to infinity as k → ∞. Hence lim Sjk mnk (x, f ) = ∞, i.e., partial sums Sn (x, f ) by generalized Haar k→∞

system is diverging (even unboundedly). Thus, the divergence of Fourier series by generalized Haar systems for continuous on group G functions is possible for any {pn }∞ n=1 with sup pn = ∞. n

7. THIRD CONTRARY INSTANCE Let us give an example of continuous on segment [0, 1] function with diverging, at some point, Fourier series by generalized Haar system in case when the sequence {pn }∞ n=1 satisfies the condition of Section 5. ∞ So, let {pnk +1 }∞ k=1 be an infinite increasing subsequence of the sequence {pn }n=0 , consisting solely of odd numbers and pnk +1 ≥ 12, and jk =

pnk +1 −1 . 2

Consider the function

RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 60 No. 1 2016

DIVERGENCE OF THE FOURIER SERIES BY GENERALIZED HAAR SYSTEMS

h(t) = g(u) ⎧ 3iπ(4l+1) ⎪ 2pn +1 ⎪ k e , ⎪ √ln−i ⎪ pnk +1 ⎪ ⎪ 3iπ(4l+1) ⎪ ⎪ ⎪ 4l+1 √ −i 2pn +1 ⎪ k 10m (u − ) e , n +1 ⎪ k mnk +1 ⎪ ln pnk +1 ⎪ ⎨ 3iπ(4l+1) = 10mnk +1 ( m4l+2 − u) √ −i e 2pnk +1 , nk +1 ln pnk +1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 0 

−

 4l+1.1



−





−

 4l+1.9

if u = x · t ∈ if u = x · t ∈ if u = x · t ∈

4l+1.9 mnk +1 , mnk +1 4l+1 4l+1.1 mnk +1 , mnk +1

55

; ;



4l+2 mnk +1 , mnk +1 ,  pn +1 −4  k ; k = 1, 2, 3, . . . l = 0, 1, . . . , 8 − (i. e. u = x · t ∈ Gnk \ Gnk +1 ); −

for other u = x · t.

Note that g(u) is a continuous on segment [0, 1] function, since at the ends of segments    4l+1.9 4l+2 , − it is joined by continuity. and m n +1 mn +1

4l+1 4l+1.1 mnk +1 , mnk +1

k

k

For example, if we denote A = √ln−i p

e

2iπ(4l+1) 2pn +1 k

nk +1

A=

lim

4l+1.9 u→ m −0 nk +1

g(u), while

lim

u→ m4l+2 −0 nk +1

We can similarly show that lim

u→ m4l+1 −0

g(u). Also for u ∈



, we get

lim

4l+1.9 u→ m +0

g(u) = 10Amnk +1

lim

4l+1.1 u→ m +0 nk +1

4l+1 4l+1.1 mnk +1 , mnk +1

g(u) = 10Amnk +1

nk +1

4l+2 mnk +1

−

4l+2 mnk +1



=0=

4l+2−(4l+1.9)  mnk +1

lim

u→ m4l+2 −0

=

g(u).

nk +1

g(u) = A = 

lim

4l+1.1 u→ m −0

g(u) and

nk +1

lim

u→ m4l+1 +0

g(u) = 0 =

nk +1

we have

nk +1

4l + 1  4l + 1  10mnk +1 4l + 1.1 1 10mnk +1 − u− ≤ = |g(u)| =  mnk +1 mnk +1 ln pnk +1 ln pnk +1 mnk +1 ln pnk +1  4l+1.9 4l+2  and for u ∈ mn +1 , mn +1 k k   4l + 1.9 1 10mnk +1 0.1 10mnk +1 4l + 2 − = . = |g(u)| ≤  mnk +1 ln pnk +1 mnk +1 ln pnk +1 mnk +1 ln pnk +1

(41)

Thus, lim g(u) = 0 = g(0), i.e., g(u) is continuous on segment [0, 1] function. Almost verbatim u→0

repeating the inference of equality (32), we show that      1  g(u)Djk mnk (x, t)dt ≤  .  ln pnk +1

(42)

−

{t|u=x · t∈Gnk +1 } 4l+2 mn +1 k

Note that in



4l+1 mn +1 k

0, 1, 2, . . . ,

− g(u)Djk mnk (x, x · u)du the integrand function is real-valued and positive l =

 4l+1.9 4l+2    pnk +1 −4  − , , k = 1, 2, . . . . For example, for u = x · t ∈ m by (29) we get 8 n +1 mn +1 k

k

  −3iπ(4l+1) 3iπ(4l+1)  mnk 10mnk +1 4l + 2 − 2pn +1 2pn +1 k k  ie −u − ie g(u)Djk mnk (x, x · u) = mnk +1 ln pnk +1 2 sin π(4l+1) 2pnk +1   4l + 2 5mnk mnk +1 − u > 0, =  sin π(4l+1) ln pn +1 mnk +1 2pnk +1

RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 60 No. 1 2016

k

56

SHCHERBAKOV π(4l+1) 2pnk +1

since

≤

π(

pn −4 k +1) 2 2pnk +1

<

π 4

< π2 , i.e., sin π(4l+1) 2pn +1 > 0. Hence k

4l+2 mn +1 k

4l+1.9 mn +1 k



4l+1.9 mn +1 k



−

g(u)Djk mnk (x; x · u)du ≥ 4l+1 mn +1 k



−

−

g(u)Djk mnk (x; x · u)du = 4l+1.1 mn +1 k

−

f (x · u)Djk mnk (x; x · u)du, 4l+1.1 mn +1 k

where f (u) is the function defined in Section 5. Then by inequalities (43) and (33) for a = b=

4l+1.9 mnk +1 ,

b−a =

0.8 mnk +1 ,

4l+1.1 mnk +1

(43) and

we get

4l+2 mn +1 k

4l+1.9 mn +1 k





−

−

g(u)Djk mnk (x, x · u)du ≥ 4l+1 mn +1 k

−

f (x · u)Djk mnk (x, x · u)du 4l+1.1 mn +1 k

1 0.8mnk +1   = . (44) 4π(4l + 1)mnk +1 ln pnk +1 5π(4l + 1) ln pnk +1

≥ −

Then (u = x · t) from (44) we have  g(u)Djk mnk (x, t)dt =



−

g(u)Djk mnk (x, x · u)du Gnk \Gnk +1

−

{t|x · t∈Gnk \Gnk +1 }

=

 pn +1 −4  m4l+2 k nk +1 8  l=0

1 ≥  5π ln pnk +1

 pn +1 −4  k 8  l=0

−

g(u)Djk mnk (x, x · u)du

4l+1 mn +1 k

1 1 >  4l + 1 5π ln pnk +1

 pn +1 −4  k 8  l=0

pn

+1 −4

ln[ k 8 ] 1  > . (45) 4(l + 1) 20π ln pnk +1

−

Now consider (h(t) = g(x · t)):  Sjk mnk (x, h) = h(t)Djk mnk (x, t)dt =



−

g(x · t)Djk mnk (x, t)dt −

G



{t|x · t∈Gnk +1 }



−

−

g(x · t)Djk mnk (x, t)dt +

+ −

{t|x · t∈Gnk \Gnk +1 }

g(x · t)Djk mnk (x, t)dt. −

{t|x · t∈G\Gnk }

By (21), the third integral on the right-hand side of the latter equality turns to zero. Then, using inequalities (45) and (42), by the latter formula we get       − −  g(x · t)Djk mnk (x, t)dt −  g(x · t)Djk mnk (x, t)dt |Sjk mnk (x, h)| ≥ −

{t|x · t∈Gnk \Gnk +1 }

−

{t|x · t∈Gnk +1 } pn

+1 −4

ln[ k 8 ] 1  − . (46) > 20π ln pnk +1 ln pnk +1 RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 60 No. 1 2016

DIVERGENCE OF THE FOURIER SERIES BY GENERALIZED HAAR SYSTEMS

57

The right-hand side of inequality (46) tends to infinity as k → ∞. Hence the Fourier series of continuous on the segment [0, 1] function h(t) is diverging (even unboundedly) at x.

8. FOURTH CONTRARY INSTANCE Let us give an example of continuous on segment [0, 1] function with diverging, at some point, Fourier series by generalized Haar system in case when unbounded sequence {pn }∞ n=1 does not satisfy the condition of Section 5. If the sequence {pn }∞ n=1 with sup pn = ∞ does not satisfy the conditions of Section 5, then, as it n

was shown in Section 6, there exists an infinite increasing subsequence {pnk +1 }∞ k=1 , consisting solely of even numbers. Without loss of generality, we may assume pnk +1 > 12. pnk +1 2

Let jk =

(see (35)). Consider the function

−

h(t) = g(x · t)

⎧ 2iπ(4l+1) ⎪ −i 2pn +1 ⎪ √ k e , ⎪ ⎪ ln pnk +1 ⎪ ⎪ 2iπ(4l+1) ⎪

 ⎪ ⎪ 4l+1 √ −i 2pn +1 ⎪ k e , u − 10m n +1 ⎪ k mnk +1 ⎪ ln pnk +1 ⎪ ⎨ 2iπ(4l+1) 

= g(u) = 10mnk +1 m4l+2 − u √ −i e 2pnk +1 , n +1 ln pnk +1 ⎪ k ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 0

 4l+1.1

 4l+1.9 , mnk +1 mnk +1 ;   − if u = x · t ∈ m4l+1 , 4l+1.1 ; nk +1 mnk +1  4l+1.9 4l+2  − if u = x · t ∈ m , ; n +1 mnk +1  pnk +1k−4  ; k = 1, 2, . . . l = 0, 1, . . . 8 −

if u = x · t ∈

−

(i.e., u = x · t ∈ Gnk \ Gnk +1 ); −

for other u = x · t.

Similarly to Section 7 we show that g(u) is a continuous on the segment [0, 1] function. Note that   4l+2 , ⊂ Gnk \ Gnk +1 the inequality is fulfilled, for u ∈ m4l+1 m n +1 n +1 k

k

−

g(u)Djk mnk (x, x · u) ≥ 0,

(47)

in particular, its left-hand side is real-valued. For example, for u ∈



4l+1 4l+1.1 mnk +1 , mnk +1



by (34) we get

  −iπ(4l+1) −i 4l + 1 pn +1 e k 10mnk +1 u − g(u)Djk mnk (x, x · u) =  mnk +1 ln pnk +1 −

×

imnk sin π(4l+1) pn +1

e

−iπ(4l+1) pn +1 k

k

since sin π(4l+1) pn +1 k

 10mnk mnk +1 u − m4l+1 nk +1 = > 0,  sin π(4l+1) ln pnk +1 pn +1 k

p −4  > 0 (since, by definition of the function g(u), l ≤ nk +1 , and hence 8

π(4l+1) pnk +1

< π2 ).

−

Almost verbatim repeating the inference of inequality (39), we get (t = x · u)      1 −  g(t)Djk mnk (x, x · u)du ≤  .  ln pnk +1 Gnk +1

RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 60 No. 1 2016

(48)

58



In view of (47), on the segment 4l+2 mn +1 k

SHCHERBAKOV 4l+1 4l+2 mnk +1 , mnk +1



⊂ Gnk \ Gnk +1 consider

4l+1.9 mn +1 k



4l+1.9 mn +1 k



−

g(u)Djk mnk (x, x · u)du≥ 4l+1 mn +1 k



−

−

g(u)Djk mnk (x, x · u)du= 4l+1.1 mn +1 k

−

f (x · u)Djk mnk (x, x · u)du, 4l+1.1 mn +1 k

(49) −

u = x · t, and f (x) is the function defined in Section 6. Using further inequality (36), by (49) for p  4l+1.1 4l+1.9 and b = mn , b − a = mn0.8+1 , and l, 0 ≤ l ≤ nk8+1 , is integer, we get a= m n +1 k +1 k

k

4l+2 mn +1 k



−

g(u)Djk mnk (x, x · u)du ≥ 4l+1 mn +1 k

mnk +1 0.8  m π(4l + 1) ln pnk +1 nk +1 =

4 4 1   > . (50) 5π 4(l + 1) ln pnk +1 5π(4l + 1) ln pnk +1

Then by formula (50) let us estimate 

−

g(u)Djk mnk (x, x · u)du =

 pn +1 −4  m4l+2 k nk +1 8  l=0

Gnk \Gnk +1

−

g(u)Djk mnk (x, x · u)du

4l+1 mn +1 k

1 ≥  5π ln pnk +1

 pn +1 −4  k 8  l=0

pn

+1 −4

ln[ k ] 1 >  8 . (51) l+1 5π ln pnk +1

−

Consider (u = x · t)    − − Sjk mnk (h, x) = h(t)Djk mnk (x, t)dt = g(x · t)Djk mnk (x, t)dt = g(u)Djk mnk (x, x · u)du 

G



−

G

G −

g(u)Djk mnk (x, x · u)du +

=

g(u)Djk mnk (x, x · u)du +

Gnk \Gnk +1

Gnk +1



−

g(u)Djk mnk (x, x · u)du.

G\Gnk

Due to formula (21), the third integral in the rightmost part of the latter equality turns to zero. Then, using inequalities (48) and (51), by the latter formula we get  − g(u)Djk mnk (x, x · u)du |Sjk mnk | ≥ Gnk \Gnk +1

   −  Gnk +1

 pn +1 −4  ln[ k ] 1 g(u)Djk mnk +1 (x, x · u)du ≥  8 − . (52) 5π ln pnk +1 ln pnk +1 −

The right-hand side of inequality (52) tends to infinity as k → ∞. Hence the Fourier series of continuous on the segment [0, 1] function h(t) by generalized Haar system is diverging (even unboundedly) at x. RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 60 No. 1 2016

DIVERGENCE OF THE FOURIER SERIES BY GENERALIZED HAAR SYSTEMS

59

9. EARLIER RESULTS The generalization of Haar systems (by analogy with the generalization of Walsh systems), apparently, was first studied by B. I. Golubov and A. I. Rubinshtein [10] (for sup pn < ∞), as well as by n

B. I. Golubov [11]. In [11], in particular, it was shown that if pn+1 > emn , then the Fourier series of function f (t) ≡ t is diverging at x = 0. However, the behavior of partial Fourier sums for any {pn }∞ n=1 with sup pn = ∞ was not investigated in [11]. n

Note that in case pn ≡ 2 the convergence of {Sn (x, f )}n∈J coincides with the convergence of Smn (x, f ) = S2n (x, f ) (n → ∞). This was known earlier (see, e.g., [12], P. 213). In [11] and [13] (χ) (Ψ) B. I. Golubov and S. F. Lukomskii showed that Smn (x, f ) = Smn (x, f ) for any sequences {pn }∞ n=0 , i.e., they obtained formula (20). However, there are no equalities of type (21) in these articles. ACKNOWLEDGMENTS The author wishes to express his gratitude to D. V. Dubnov for valuable help. REFERENCES 1. Walsh, J. L. A Closed Set of Normal Orthogonal Functions, Amer. J. Math. 45, No. 1, 5–24 (1923). 2. Paley, R. E. A. C. A Remarkable Series of Orthogonal Functions. I, Proc. London Math. Soc. 34, 241–264 (1932). 3. Paley, R. E. A. C. A Remarkable Series of Orthogonal Functions. II, Proc. London Math. Soc. 34, 265– 279 (1932). 4. Chrestenson, H. E. A Class of Generalized Walsh Functions, Pacific J. Math. 5 (1), 17–31 (1955). 5. Price, J. J. Certain Group of Orthonormal Step Functions, Canadian J. Math. 9, No. 3, 417–425 (1957). 6. Vilenkin, N. Ya. On a Class of Complete Orthonormal Systems, Izv. Akad. Nauk SSSR Ser. Mat. 11, No. 4, 363–400 (1947) [in Russian]. 7. Haar A. Zur Theorie der orthogonalen Functionensysteme, Math. Ann. 69, No. 3, 331–371 (1910). 8. Shcherbakov, V. I. On the Pointwise Convergence of Fourier Series with Respect to Multiplicative Systems, Vestnik MSU, Ser. Matem., Mekh. No. 2, 37–42 (1983) [in Russian]. 9. Onneweer, C. W., Waterman, D. Uniform Convergence of Fourier Series on Groups, Michigan Math. J. 18, No. 3, 265–273 (1971). 10. Shcherbakov, V. I. Dini–Lipschitz Test and Convergence of Fourier Series with Respect to Multiplicative Systems, Analysis Math. 10, No. 1, 134–150 (1984). 11. Golubov, B. I., Rubinshtein, A. I. A Class of Convergence Systems, Matem. Sb. 71, No. 1, 96–115 (1966) [in Russian]. 12. Golubov, B. I. On a Certain Class of Complete Orthonormal Systems, Sib. Matem. Zh. 9, No. 2, 297–314 (1968) [in Russian]. 13. Golubov, B. I., Efimov, A. V., Skvortsov V.A. Walsh Series and Transforms. Theory and Applications (Nauka, Moscow, 1987) [in Russian]. 14. Lukomskii, S. F. Haar Series on Compact Zero-Dimensional Abelian Group, Izv. Saratov Univ. (N. S.), Ser. Math. Mech. Inform. 9, No. 1, 24–29 (2009) [in Russian].

Translated by P. A. Novikov

RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 60 No. 1 2016