Journal of Mathematical Sciences, Vol. 155, No. 1, 2008

ON UNIFORMLY CONVERGENT REARRANGEMENTSOF TRIGONOMETRIC FOURIER SERIES

S. V. Konyagin UDC 517.5

Abstract. We show that if the module of continuity ω(f, δ) of a 2π-periodic function f ∈ C(T) iso(1/ log log 1/δ) as δ → 0+, then there exists a rearrangement of the trigonometric Fourier series of fconverging uniformly to f .

Introduction. P. L. Ul’yanov has set the following problem (see [8]). Is it true that the trigonometricFourier series of an arbitrary 2π-periodic continuous function f may be rearranged so that the obtainedseries converges uniformly to f? This problem is still open. Sz. Gy. Revesz has proved that there isa uniformly convergent subsequence of partial sums of the rearranged Fourier series; in general, thisrearrangement and this subsequence depend on f (see [6]). In [7], this result was generalized to multipleFourier series of continuous functions. In this paper, we give an affirmative answer to Ul’yanov’squestion if the module of continuity ω(f, δ) of the function f is o(1/ log log 1/δ) as δ → 0+.

Let T = R/2πZ be a one-dimensional torus and C(T) be the space of complex functions which arecontinuous on T. To each function f ∈ C(T) assign its Fourier series in the complex and in the realform:

f ∼∑

k∈Z

ckeikx, f ∼

∞∑

k=0

dk cos(kx + ϕk).

For brevity, set Ak(x) = dk cos(kx + ϕk). As usual, the function

ω(f, δ) = maxx,y∈T,|x−y|≤δ

|f(x) − f(y)|, δ ≥ 0,

is the module of continuity of f in C(T). For f ∈ C(T) put ‖f‖ = maxx∈T

|f(x)|. Denote by C1, C2, . . .

absolute positive constants.

Theorem 1. Assume that f ∈ C(T) and ω(f, δ) = o(1/ log log 1/δ) as δ → 0+. Then there exists apermutation σ of the integers such that

limn→∞

∥∥∥∥∥f − d0 −n∑

k=1

Aσ(k)

∥∥∥∥∥ = 0.

Theorem 1 deals with rearrangements of Fourier series in the real form. The analogous statementabout Fourier series in the complex form follows from Theorem 1, whose conclusion can be rewrittenin the form ∥∥∥∥∥f − c0 −

n∑

k=1

(cσ(k)e

ikx + c−σ(k)e−ikx

)∥∥∥∥∥→ 0 (n → ∞).

Theorem 1 follows immediately from the next two theorems, which we shall prove.

Theorem 2. Let B(u) be a nondecreasing positive function such that(1) B(n2) = O(B(n)) for n ∈ N;

Translated from Sovremennaya Matematika. Fundamental’nye Napravleniya (Contemporary Mathematics.Fundamental Directions), Vol. 25, Theory of Functions, 2007.

1072–3374/08/1551–0081 c© 2008 Springer Science+Business Media, Inc. 81

(2) for any trigonometric polynomial T (x) =n∑

k=0

Ak(x) there is a permutation τ of the set {0, . . . , n}such that ∥∥∥∥∥

m∑

k=0

Aτ(k)

∥∥∥∥∥ ≤ B(n)‖T‖

for all m = 0, . . . , n.Then for any function f ∈ C(T) with the property that ω(f, δ) = o(1/B(1/δ)) as δ → 0+ there existsa permutation σ of the integers such that

limn→∞

∥∥∥∥∥f − d0 −n∑

k=1

Aσ(k)

∥∥∥∥∥ = 0.

Theorem 3. The function B(u) = C1 log log(u + 3) satisfies the hypothesis of Theorem 2.

Given any module of continuity ω(δ) = o(1/ log log 1/δ) as (δ → 0+), the proof of Theorem 1 allowsus to construct a sequence εn → 0 as (n → ∞) such that for any function f ∈ C(T) with the propertythat ω(f, δ) ≤ ω(δ) for δ ≥ 0 there exists a permutation σ : N → N satisfying the condition

∥∥∥∥∥f − d0 −n∑

k=1

Aσ(k)

∥∥∥∥∥ ≤ εn (n ∈ N).

For instance, if ω(δ) = O(ω(δ2)), then we can set εn = O(ω(f, 1/n) log log(n+3)). However, we do notknow if this is true for an arbitrary module of continuity ω(δ). On the other hand, it is not unlikelythat for any function f ∈ C(T) there exists a permutation σ : N → N such that

∥∥∥∥∥f − d0 −n∑

k=1

Aσ(k)

∥∥∥∥∥ = O (ω (f, 1/n))

for all n ∈ N.The results of the paper have been announced in [4].

1. Proof of Theorem 2. The proof is based on the notions of [6]. Given f ∈ C(T) and positiveintegers n and m, by Sn(f) denote the partial sum

Sn(f) =n∑

k=1

Ak

of the Fourier series for f and by Vn,m(f) denote the de la Vallee-Poussin sum

Vn,m(f) =1m

n+m−1∑

k=n

Sk(f).

Let us restate [6, Lemma 2] for the case of complex functions.

Lemma 1. Assume that f ∈ C(T), η > 0, n > 7, m ≤ n. Ifn+m∑

k=n+1

|dk|2 <η

log n, (1)

then there is a collection (ω1, . . . , ωm) ∈ {0, 1}m such that∥∥∥∥∥Sn(f) +

m∑

k=1

ωkAn+k − Vn,m(f)

∥∥∥∥∥ < 12√

η. (2)

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In [6], in the proof of the corresponding proposition for real functions f the collection (ω1, . . . , ωm) ∈{0, 1}m was considered as random; it has been proved that the probability of the event

∥∥∥∥∥Sn(f) +m∑

k=1

ωkAn+k − Vn,m(f)

∥∥∥∥∥ ≥ 8√

η

is less than 26n−2 provided an appropriate probabilistic space is chosen. Therefore, for a complexfunction f , each of the events

∥∥∥∥∥Sn(Re f) +m∑

k=1

ωk Re An+k − Vn,m(Re f)

∥∥∥∥∥ ≥ 8√

η,

∥∥∥∥∥Sn(Im f) +m∑

k=1

ωk Im An+k − Vn,m(Im f)

∥∥∥∥∥ ≥ 8√

η

has probability less than 26n−2. Then for n > 7 both of these two events do not occur with positiveprobability ≥ 1 − 52n−2, so (2) holds.

Lemma 2. For any f ∈ C(T), there exist subsets N0, N1, . . . of the integers such that N0 = ∅,

{1, . . . , 22l−1} ⊂ Nl ⊂ {1, . . . , 22l},∥∥∥∥∥∥f − d0 −

∑

k∈Nl

Ak

∥∥∥∥∥∥≤ C2ω

(f, 1/22l−1)

for all l ∈ N.

Proof. For l = 1 and l = 2, the sets Nl may be chosen arbitrarily, for instance Nl = {1, . . . , 22l−1}. Itcan easily be checked that the conclusion of the lemma is valid for such l. Now assume that l ≥ 3.For δ ≥ 0 the equality

ω(f, δ)2 = sup0≤h≤δ

⎛

⎝∫

T

|f(x + h) − f(x)|2 dx

⎞

⎠1/2

defines the module of continuity of f in L2(T). Applying the Jackson theorem in L2(T) (see, e.g., [9])and taking into account that the best approximations in L2(T) are Fourier sums, we obtain

∑

k>22l−1

|dk|2 ≤ C3ω(f, 1/22l−1)2

2≤ 2πC3ω(f, 1/22l−1

)2.

Thus there exists j ∈ {0, . . . , 2l−1 − 1} such that2n∑

k=n+1

|dk|2 ≤ 4πC3ω(f, 1/22l−1

)2

2l≤ C4

ω(f, 1/22l−1)2

log n

for n = 22l−1+j . Applying Lemma 1 to the case of m = n, η = C4ω(f, 1/22l−1)2 and putting

Nl = {k ∈ N : k ≤ n} ∪ {n + k : 0 < k ≤ n, ωk = 1},we obtain ∥∥∥∥∥∥

Vn,n(f) − d0 −∑

k∈Nl

Ak

∥∥∥∥∥∥≤ C5ω(f, 1/22l−1

). (3)

Further, owing to the Jackson theorem in C(T) (see [9]) and to the estimates of the approximation byde la Vallee-Poussin’s sums (see [9]) we obtain

‖f − Vn,n(f)‖ ≤ C6ω(f, 1/n) ≤ C6ω(f, 1/22l−1). (4)

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Thus, the conclusion of Lemma 2 follows from (3) and (4).

Note that from the conditions on the sets Nl of Lemma 2 it immediately follows that

N0 ⊂ N1 ⊂ · · · ,⋃

l∈N

Nl = N.

We now turn directly to the proof of Theorem 2. Assume that f ∈ C(T) and ω(f, δ) = o(1/B(1/δ))as δ → 0+. Put

Tl =∑

k∈Nl\Nl−1

Ak;

then

f = d0 +∑

l

Tl. (5)

The series on the right-hand side of (5) converges uniformly because of the fact that by Lemma 2∥∥∥∥∥f − d0 −

L∑

l=1

Tl

∥∥∥∥∥ = O(ω(1/22L−1

)). (6)

Further, by condition (1) of Theorem 2 we have

‖Tl‖ ≤ C2

(ω(f, 1/22l−1

) + ω(f, 1/22l))

= o

(1

B(22l−1)

)= o

(1

B(22l)

). (7)

By nl denote the number of elements of the set Nl. By the hypothesis of Theorem 2, noting thatthe degree of the polynomial Tl is less than 22l

there exists a bijective mapping σl of {nl−1 +1, . . . , nl}onto Nl \ Nl−1 such that

∥∥∥∥∥∥

m∑

k=nl−1+1

Aσl(k)

∥∥∥∥∥∥≤ B(22l

)‖Tl‖,

for any m = nl−1 + 1, . . . , nl. From this and from (7) it follows that

maxnl−1<m≤nl

∥∥∥∥∥∥

m∑

k=nl−1+1

Aσl(k)

∥∥∥∥∥∥= o(1) (l → ∞). (8)

Let us construct a permutation σ of the set N setting σ(k) = σl(k) for k ∈ {nl−1 + 1, . . . , nl}. Thenfor any n ∈ N, choosing L such that n ∈ {nL−1 + 1, . . . , nL}, we have

∥∥∥∥∥f − d0 −n∑

k=1

Aσ(k)

∥∥∥∥∥ =

∥∥∥∥∥∥f − d0 −

L−1∑

l=1

Tl −n∑

k=nL−1+1

Aσ(k)

∥∥∥∥∥∥≤∥∥∥∥∥f − d0 −

L−1∑

l=1

Tl

∥∥∥∥∥+

∥∥∥∥∥∥

n∑

k=nL−1+1

Aσ(k)

∥∥∥∥∥∥;

by virtue of (6) and (8)∥∥∥∥∥f − d0 −

n∑

k=1

Aσ(k)

∥∥∥∥∥→ 0 (n → ∞).

The theorem is proved.

84

2. Proof of Theorem 3. We must prove that if

T (x) =n∑

k=0

Ak(x) =n∑

k=0

dk cos(kx + ϕk) n ∈ N

is a trigonometric polynomial of degree less than n and

‖T‖ ≤ 1, (9)

then there exists a permutation τ of the set {0, . . . , n} such that∥∥∥∥∥

m∑

k=0

Aτ(k)

∥∥∥∥∥ ≤ C1 log log(n + 3) (10)

for any m = 0, . . . , n.The central idea of the proof is that choosing a prime p we split the terms Ak of the polynomial

T into packs putting together all the Ak with k ≡ ±j (mod p) to the jth pack (j = 0, . . . , (p − 1)/2).We sum the packs in the natural order while the order of summands within each pack is arbitrary. Itwill be proved that the obtained order is that of interest provided a suitable choice of a prime p.

First we estimate norms of packs and of sums over packs. Condition (9) will be assumed throughoutthis paper.

Lemma 3. Let p be an odd prime number. Then(1) for any j = 0, . . . , (p − 1)/2 ∥∥∥∥∥∥

∑

k≡±j (mod p)

Ak

∥∥∥∥∥∥≤ 2;

(2) for any J = 0, . . . , (p − 1)/2∥∥∥∥∥∥

J∑

j=0

∑

k≡±j (mod p)

Ak

∥∥∥∥∥∥≤ C7 log p.

Proof. With real packs we shall consider complex packs

Tj(x) =∑

k≡j (mod p)

ckeikx (|j| ≤ (p − 1)/2),

where ck are the coefficients of the polynomial T expansion in the complex form

T (x) =n∑

k=−n

ckeikx.

We have∑

k≡0 (mod p)

Ak = T0, (11)

∑

k≡±j (mod p)

Ak = Tj + T−j (j = 1, . . . , (p − 1)/2). (12)

Fix x0 ∈ T and denote xl = x0 + 2πl/p (|l| ≤ (p − 1)/2). It is clear that the expansion

T =∑

j

Tj

85

induces a discrete Fourier series expansion of the polynomial T on the grid {xl : |l| ≤ (p− 1)/2}. Here

Tj(x0) =1p

∑

|l|≤(p−1)/2

e−2πijl/pT (xl) (|j| ≤ (p − 1)/2). (13)

Then |Tj(x0)| ≤ 1; owing to the arbitrariness of x0 ∈ T we obtain ‖Tj‖ ≤ 1. The first conclusion ofthe lemma follows now from this, (11), and (12).

The second one may be proved in the usual way using discrete Dirichlet kernels. Taking intoaccount (11), (12), and (13), for any J = 0, . . . , (p − 1)/2 we have∣∣∣∣∣∣

J∑

j=0

∑

k≡±j (mod p)

Ak(x0)

∣∣∣∣∣∣=

∣∣∣∣∣∣

J∑

j=−J

Tj(x0)

∣∣∣∣∣∣=

∣∣∣∣∣∣1p

∑

|l|≤(p−1)/2

⎛

⎝J∑

j=−J

e−2πijl/p

⎞

⎠T (xl)

∣∣∣∣∣∣

≤ 1p

∑

|l|≤(p−1)/2

∣∣∣∣∣∣

J∑

j=−J

e−2πijl/p

∣∣∣∣∣∣≤ 1

p

⎛

⎝2J + 1 +∑

1≤|l|≤(p−1)/2

1sin(π|l|/p)

⎞

⎠ ≤ C7 log p.

Lemma 3 shows that norms of sums over packs are bounded by a fixed value of order log log(n + 3)if p is less than a fixed power of log(n+3). The following lemma allows us to choose a prime number p.

Lemma 4. There exists an odd prime p ≤ 2 log3(n + 3) such that∑

k1 �=k2,k1≡±k2 (mod p)

|dk1 |2|dk2 |2 ≤ C8

log2(n + 3).

Proof. Note that condition (9) implies ∑

k

|dk|2 ≤ 2. (14)

Let P be the set of all odd primes which are less than 2 log3(n + 3). By (14) we have∑

p∈P

∑

k1 �=k2,k1≡±k2 (mod p)

|dk1 |2|dk2 |2 ≤∑

k1 �=k2

∣∣{p ∈ P : k21 − k2

2 ≡ 0 (mod p)}∣∣ · |dk1 |2|dk2 |2

≤ maxk1 �=k2

∣∣{p ∈ P : k21 − k2

2 ≡ 0 (mod p)}∣∣(∑

k

|dk|2)2

≤ 4 maxk1 �=k2

∣∣{p ∈ P : k21 − k2

2 ≡ 0 (mod p)}∣∣.

If m is the number of distinct prime divisors of k21 − k2

2 for k1 �= k2, 0 ≤ k1 < n, 0 ≤ k2 < n, thenm! ≤ |k2

1 − k22| < n2, whence

m ≤ C9log(n + 3)

log log(n + 3),

∑

p∈P

∑

k1 �=k2,k1≡±k2 (mod p)

|dk1 |2|dk2 |2 ≤ 4C9log(n + 3)

log log(n + 3).

On the other hand, |P | ≥ C10 log3(n + 3)/ log log(n + 3) (see, e.g., [5, p. 27]). Therefore there existsp ∈ P such that

∑

k1 �=k2,k1≡±k2 (mod p)

|dk1 |2|dk2 |2 ≤ 4C9

C10 log2(n + 3).

The lemma is proved.

86

Fix a number p in accordance with Lemma 4 and choose an arbitrary number

j ∈ {0, . . . ,min((p − 1)/2, n)}.

Among all the numbers k, 0 ≤ k ≤ n, satisfying the inequality k ≡ ±j (mod p), let us choose a numberk(j) such that |dk(j)| is maximal. Put

Nj ={k : k ≡ ±j (mod p), k �= k(j)

}.

We have⎛

⎝∑

k∈Nj

|dk|2⎞

⎠2

=∑

k∈Nj

(|dk|2)2 +∑

k1 �=k2,k1∈Nj ,k2∈Nj

2|dk1 |2|dk2 |2 ≤∑

k∈Nj

|dk|2|dk(j)|2 +∑

k1 �=k2,k1∈Nj ,k2∈Nj

2|dk1 |2|dk2 |2,

whence by Lemma 4∑

k∈Nj

|dk|2 ≤√

2C8

log(n + 3). (15)

We now putnj = |Nj |, Uj =

∑

k∈Nj

Ak.

Note that, by virtue of (14), |dk(j)| ≤√

2, therefore, the first conclusion of Lemma 3 implies

‖Uj‖ ≤ 2 +√

2. (16)

Lemma 5. There exists a permutation τj = {k1, . . . , knj} of the set Nj such that

‖Ak1 + · · · + Akm‖ ≤ C11

for any m ∈ {1, . . . , nj}.Theorem 3 easily follows from Lemmas 3–5. The required ordering in the set {0, . . . , n} is con-

structed as follows: {{τ0}, k0, {τ1}, k1, . . . , {τj}, kj , . . .};

continuing until j ≤ min((p − 1)/2, n). For any m = 0, . . . , n we have∥∥∥∥∥

m∑

k=0

Aτ(k)

∥∥∥∥∥ ≤ C7 log p + C11 ≤ C7 log(2 log3(n + 3)

)+ C11,

whence the proof of Theorem 3 follows.Thus, it remains to prove Lemma 5. Let ξ be a random vector whose components are independent

random variables ξk (k ∈ Nj) each of which is +1 or −1 with probability 1/2. Define the randompolynomial

Uj,ξ =∑

k∈Nj

ξkAk.

By the Chobanian theorem (see [1, 2]) there exists a permutation τj = {k1, . . . , knj} such that

‖Ak1 + · · · + Akm‖ ≤ 9(E‖Uj,ξ‖ + ‖Uj‖

)(17)

for all m ∈ {1, . . . , nj}.We use the following inequality (see [3]) to estimate E‖Uj,ξ‖:

P

⎛

⎜⎝‖Uj,ξ‖ ≥⎛

⎝C12 log n∑

k∈Nj

|dk|2⎞

⎠1/2⎞

⎟⎠ ≤ 1/n2.

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Furthermore, by (14) the inequality ‖Uj,ξ‖ ≤ √2n holds for any ξ. This implies that

E‖Uj,ξ‖ ≤⎛

⎝C12 log n∑

k∈Nj

|dk|2⎞

⎠1/2

+1n2

√2n;

taking into account (15) we obtain

E‖Uj,ξ‖ ≤ (2C8)1/4C1/212 +

√2.

This and (17) and (16) prove the assertion of Lemma 5.

Acknowledgment. This work was supported by the Russian Foundation for Basic Research (projectNo. 96-01-00378) and by the program “Leading scientific schools” (project No. 96-15-96072).

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S. V. KonyaginMoscow State University, Moscow, Russia

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