Trigonometric series with monotone coefficients

Karlstads universitet 651 88 Karlstad Faculty of Technology and Science

Jona Ekström

Trigonometric series

With monotone coefficients

Mathematics C-level thesis

Date/Term: 2006-11-26 Supervisor: Viktor Kolyada Examiner: Håkan Granath

Contents

1 Introduction 2

2 General series 2

3 Trigonometric series 4

3.1 The Dirichlet and Fej´er kernels . . . . 4

4 Fourier series 7

5 Trigonometric series with monotone

coefficients 9

6 Uniform convergence 20

References 23

1 Introduction

This work is devoted to trigonometric series with monotone coefficients. The main problem is to study conditions under which a given series is the Fourier series of an integrable (or continuous) function.

It is easy to see that whenever a trigonometric series is uniformly conver- gent it is the Fourier series of its sum. In this case the series converges to a continuous function. However, the Fourier series of a continuous function is not necessarily uniformly convergent.

One of the main theorems studied in this work asserts that if the sum of trigonometric series with monotone coefficients is integrable, then this series is the Fourier series of its sum. We observe that this statement is true for any trigonometric series that converges everywhere, but the proof is much more complicated.

The main objective of this work was to give a complete and self-contained exposition of some basic results concerning trigonometric series with mono- tone coefficients. These results are contained in the book by Hardy and Rogosinski and the book by Zygmund (see the list of references). However, some proofs in these books are omitted and some of them are too concise. In this work we fill in the details that have been omitted in the books.

2 General series

Theorem 1 (Abel transform). Let u_k and v_k, (k = 1, 2, ..., n) be real numbers and

U_n=

n

X

k=1

u_k. Then

n

X

k=1

u_kv_k=

n−1

X

k=1

U_k(v_k− v_k+1) + U_nv_n Proof.

n−1

X

k=1

U_k(v_k− v_k−1) + U_nv_n

= U₁(v₁− v₂) + U₂(v₂− v₃) + . . . + U_n−1(v_n−1− v_n) + U_nv_n

= v1U1+ v2(U2− U1) + v3(U3− U2) + . . . + vn(Un− Un−1)

=

n

X

k=1

u_kv_k

For series of functions, u_k(x), we have the following theorem.

Theorem 2. Let {v_k} be a sequence of real numbers and let u_k(x), x ∈ A be a sequence of real-valued functions. If the sums

U_n(x) =

n

X

k=1

u_k(x)v_k

are uniformly bounded on A and the sequence {v_k} decreases monotonically to zero, then the series

∞

X

k=1

u_k(x)v_k (1)

converges uniformly on A.

Proof. Let

|U_k(x)| ≤ M ∀k ∈ N, ∀x ∈ A.

Using the Abel transform and the triangle inequality we get 

q

X

k=p

u_k(x)v_k     

=     

q−1

X

k=1

Uk(x)(vk− vk+1) + Uq(x)vq−

p−2

X

k=1

Uk(x)(vk− vk+1) − Up−1(x)vp−1

=     

q−1

X

k=p−1

U_k(x)(v_k− v_k+1) + U_q(x)v_q− U_p−1(x)v_p−1     

≤

q−1

X

k=p−1

M |v_k− v_k−1| + M v_q+ M v_p−1

Since {v_k} is monotonically decreasing, v_k− v_k+1 ≥ 0, and 

q

X

k=p

uk(x)vk

≤ 2M vp−1+ M vq ≤ 3M vp−1 ∀x ∈ A.

Thus, the series (1) is uniformly convergent on A by the Cauchy criterion.

3 Trigonometric series

Trigonometric series are series of the form a₀

2 +

∞

X

k=1

a_kcos kx + b_ksin kx

In this work we assume that the coefficients a_k and b_k are real numbers.

3.1 The Dirichlet and Fej´er kernels

Definition 1. The Dirichlet kernel is defined as the partial sum D_n(x) = 1

2 +

n

X

k=1

cos kx and the conjugate Dirichlet kernel as

Den(x) =

n

X

k=1

sin kx

By multiplying D_n(x) with 2 sin¹₂x we get D_n(x)2 sin1

2x

= sin1 2x +

n

X

k=1

2 sin 1

2x cos kx

= sin1 2x +

n

X

k=1

sin(1

2− k)x + sin(1 2 + k)x

= sin1 2x +

n

X

k=1

− sin(k − 1

2)x + sin(k + 1 2)x

= sin(n + 1 2)x.

We get that

D_n(x) = sin(n +¹₂)x

2 sin¹₂x , (2)

and similarly for eD_n(x)

De_n(x) = cos¹₂x − cos(n + ¹₂)x

2 sin¹₂x . (3)

Lemma 1. Let 0 <  < π. Then D_n(x) and eD_n(x) are uniformly bounded on the interval [, 2π − ]

Proof. By (2) we have

|D_n(x)| =    

sin(n + ¹₂)x 2 sin¹₂x

≤ 1

2 sin¹₂x (4)

On the interval [, 2π − ], sin¹₂x is positive and attains its minimum at  and 2π −  . Thus, we get

|D_n(x)| ≤ 1

2 sin¹₂ ∀n ∈ N, ∀x ∈ [, 2π − ]

In the same way we get from (3) that 

 De_n(x)

  =

cos¹₂x − cos(n + ¹₂)x 2 sin¹₂x

≤ 1

sin¹₂x. (5) Thus, on the interval [, 2π − ], eD_n(x) is bounded by

  Den(x)

≤ 1

sin¹₂ ∀n ∈ N, ∀x ∈ [, 2π − ].

Lemma 2.

1 π

Z π

−π

D_n(x) dx = 1 Proof.

1 π

Z π

−π

D_n(x) dx = 1 π

Z π

−π

1

2dx + 1 π

n

X

k=1

Z π

−π

cos kx dx = 1 since

1 π

Z π

−π

cos kx dx = 0 ∀k 6= 0

Lemma 3. Let {a_k} be a sequence of positive real numbers that decreases monotonically to zero. For 1 ≤ p < q < ∞ we have,

q

X

k=p

a_ksin kx     

≤ 2a_p

sin¹₂x, x ∈ (0, π] (6)

and 

q

X

k=p

a_kcos kx     

≤ a_p

sin¹₂x, x ∈ (0, π]. (7)

Proof. Using Theorem 1, we get 

q

X

k=p

a_ksin kx     

=     

q−1

X

k=p

(a_k− a_k+1) eD_k(x) + a_qDe_q(x) − a_pDe_p−1(x)      . Since {a_k} is decreasing, a_k− a_k+1 > 0. Applying the triangle inequality and estimate (5), we get

q

X

k=p

a_ksin kx     

≤

q−1

X

k=p

(ak− ak+1)   Dek(x)

  + aq

  Deq(x)

  + ap

Dep−1(x)   

≤a_p− a_q

sin¹₂x + a_q

sin¹₂x + a_p sin¹₂x

≤ 2a_p

sin¹₂x, x ∈ (0, π].

The proof of (7) is nearly identical.

Definition 2. The Fej´er kerner is defined as Fn(x) = 1

n + 1

n

X

k=0

Dk(x) Lemma 4.

F_n(x) = 1 − cos(n + 1)x 4(n + 1) sin^{2 1}₂x Proof.

F_n(x) = 1 n + 1

n

X

k=0

D_k(x)

= 1

n + 1

n

X

k=0

sin(n +¹₂)x 2 sin¹₂x

= 1

n + 1

n

X

k=0

2 sin¹₂x sin(n + ¹₂)x (2 sin¹₂x)²

= 1

n + 1

n

X

k=0

cos kx − cos(k + 1)x (2 sin¹₂x)²

= 1 − cos(n + 1)x 4(n + 1) sin^{2 1}₂x

Remark. Since 1 − cos x ≥ 0 for all x, F_n(x) ≥ 0 for all x.

Lemma 5.

1 π

Z π

−π

F_n(x) dx = 1 Proof. From lemma 2 we have

1 π

Z π

−π

D_n(x) dx = 1 so for the Fej´er kernel we get

1 π

Z π

−π

F_n(x) dx = 1 n + 1

n

X

k=0

1 π

Z π

−π

D_k(x) dx = 1

Lemma 6. Let 0 <  < π. Then the sequence nFn(x) is uniformly bounded on the interval [, 2π − ].

Proof. For any integer n and all x in [, 2π − ] we have, nF_n = n1 − cos(n + 1)x

4(n + 1) sin^{2 1}₂x ≤ 2n

4(n + 1) sin^{2 1}₂x ≤ 1 2 sin^{2 1}₂.

Remark. Since nFn(x) is bounded, Fn(x) → 0 uniformly on [, 2π − ] for 0 <  < π.

4 Fourier series

Definition 3. Let f (x) be an integrable function on [−π, π]. The numbers α_k = 1

π Z π

−π

f (x) cos kxdx

β_k= 1 π

Z π

−π

f (x) sin kxdx are called the Fourier coefficients of f , and the series

1 2α₀ +

∞

X

k=1

α_kcos kx + β_ksin kx is called the Fourier series of f .

A special case where a trigonometric series is a Fourier series, is when it is uniformly convergent. In this case it is the Fourier series of its sum. Namely, we have the following theorem.

Theorem 3. If the series 1 2a₀+

∞

X

k=1

a_kcos kx + b_ksin kx (8)

converges uniformly to f (x) on the interval [−π, π], then it is the Fourier series of f .

Proof. Since the series are uniform convergent, they can be integrated termwise.

We get the Fourier coefficients α_m = 1

π Z π

−π

f (x) cos mx dx

= 1 π

Z π

−π

(1 2a₀+

∞

X

k=1

a_kcos kx +

∞

X

k=1

b_ksin kx) cos mx dx

= 1 π

Z π

−π

1

2a₀cos mx dx + 1 π

∞

X

k=1

a_k Z π

−π

cos kx cos mx dx

+ 1 π

∞

X

k=1

bk

Z π

−π

sin kx cos mx dx.

These sums can be reduced to a single term since Z π

−π

cos kx cos mx dx =

(0 (k 6= m)

π (k = m 6= 0) (9)

Z π

−π

sin kx sin mx dx =

(0 (k 6= m)

π (k = m 6= 0) (10)

and Z π

−π

sin kx cos mx dx = 0 (11)

We get

α_m = 1 π

Z π

−π

1

2a₀cos mx dx + 1 π

∞

X

k=1

a_mcos kx cos mx = a_m.

In the same way we get β_m = 1

π Z π

−π

g(x) sin mx dx = b_m.

We get that for any uniformly convergent trigonometric serie, its coefficients is equal to the Fourier coefficients of its sum, thus it is the Fourier series of its sum.

Remark. The converse is not true. Even if (8) is a Fourier series of a con- tinuous function, it may not be uniformly convergent. A counter example was first given by du Bois-Reymond, who showed that there is a continuous function whose Fourier series diverges at a point, [2, Ch. VIII, Th. 1.1].

Lebesgue later showed that there is a continuous function whose Fourier series converges everywhere, but not uniformly, [2, Ch. VIII, Th. 1.13].

5 Trigonometric series with monotone coefficients

When the coefficients a_k decreases monotonically to zero we will denote the series

1 2a₀+

∞

X

k=1

a_kcos kx (C)

and ∞

X

k=1

a_ksin kx (S)

by (C) and (S).

Theorem 4. Let 0 <  < π. The series (C) and (S) converge uniformly on the interval [, 2π − ].

Proof. The Dirichlet kernels are uniformly bounded on [, 2π − ] by lemma 1, Thus, the theorem follows from theorem 2.

At the point x = 0

∞

X

k=1

a_ksin kx = 0 so (S) converges pointwise everywhere.

For (C) we have at x = 0 1

2a₀+

∞

X

k=1

a_kcos kx = 1 2a₀+

∞

X

k=1

a_k

so (C) converges pointwise everywhere if and only if

∞

X

k=1

a_k < ∞ (12)

As we showed in theorem 3, uniform convergence on [−π, π] is a sufficient condition for an arbitrary trigonometric series to be the Fourier series of its sum. We will now look at some weaker conditions for the series (C) and (S) Since (C) is even and (S) is odd any result that is proved on the interval [0, π] can be extended to [−π, π].

The sums of (C) and (S) are continuous on (0, π]. In the next theorem we assume that the sum is Lebesque integrable.

Theorem 5. If the sum of (C) or (S) is integrable on [−π, π], then this series is the Fourier series of its sum. [1, Th. 46]

Proof. We will first show that, for any fixed m, the series

∞

X

k=1

a_ksin kx sin mx (13)

is uniformly convergent on the interval [0, π]. For any 1 ≤ p < q < ∞ we have,

q

X

k=p

a_ksin kx sin mx     

≤ mx     

q

X

k=p

a_ksin kx     

, x ∈ (0, π].

By Lemma 3

mx     

q

X

k=p

a_ksin kx     

≤ mx 2a_p

sin¹₂x, x ∈ (0, π].

On the interval [0, π] we have,

sin1 2x ≥ x

π. (14)

We get that     

q

X

k=p

a_ksin kx sin mx     

≤ 2πma_p, x ∈ [0, π].

Since a_p → 0, we have that (13) is uniformly convergent on [0, π] by the Cauchy Criterion.

We know that (S) converges everywhere to some integrable odd function g(x). The Fourier coefficients of this function are

α_m = 1 π

Z π

−π

g(x) sin mxdx = 1 π

Z π

−π

∞

X

k=1

a_ksin kx sin mxdx.

As we have seen, this series is uniformly convergent. So, we can integrate it termwise

α_m = 1 π

Z π

−π

∞

X

k=1

a_ksin kx sin mxdx = 1 π

∞

X

k=1

Z π

−π

a_ksin kx sin mxdx

In this sum every term but k = m is zero (by Eq. (10)). We get the Fourier coefficients of g

α_m = 1 π

Z π

−π

a_msin mx sin mxdx = a_m

thus (S) is the Fourier series of its sum g(x).

The series (C) converges to some integrable function f (x) for all x ∈ [−π, π] except possibly at x = 0. We first show that, for any fixed m, the series

1

2a₀(1 − cos mx) +

∞

X

k=1

a_kcos kx(1 − cos mx)

converges uniformly on [0, π]. On this interval we have 1 − cos mx ≤ 1

2m²x².

For 1 ≤ p < q < ∞ we get an upper bound for the partial series 

q

X

k=p

a_kcos kx(1 − cos mx)     

≤ 1 2m²x²

q

X

k=p

a_kcos kx      .

By lemma 3 we have 1

2m²x²     

q

X

k=p

a_kcos kx     

≤ 1

2m²x² a_p

sin¹₂x, x ∈ (0, π].

Using the inequality (14) we get, 

q

X

k=p

a_kcos kx(1 − cos mx)     

≤ 1

2m²πxa_p, x ∈ [0, π].

Since a_p → 0, the series is uniformly convergent by the Cauchy criterion.

Further, using uniform convergence, 1

π Z π

−π

f (x) dx − 1 π

Z π

−π

f (x) cos mx dx

=1 π

Z π

−π

f (x)(1 − cos mx) dx

=1 π

Z π

−π

1

2a₀(1 − cos mx) +

∞

X

k=1

a_kcos kx(1 − cos mx)

!

dx (15)

=1 π

Z π

−π

a₀

2 dx − 1 π

Z π

−π

cos mx dx

+ 1 π

∞

X

k=1

a_k Z π

−π

cos kx dx − 1 π

∞

X

k=1

a_k Z π

−π

cos kx cos mx dx

=a₀− a_m

a_m → 0 and by the Riemann-Lebesgue theorem

m→∞lim 1 π

Z π

−π

f (x) cos mx dx = 0.

Thus, if we let m → ∞ in (15) we get a0 = 1

π Z π

−π

f (x) dx = α0

For m 6= 0 we get

a_m = 1 π

Z π

−π

f (x) cos mx dx = α_m

Thus the coefficients a_m are the Fourier coefficients of f and (C) is the Fourier series of its sum f .

Remark. The theorem holds for a general trigonometric series that converges to a finite and integrable function, but the proof is much more complicated, [1, Th. 100].

A Fourier series with coefficients that decreases monotonically to zero converges almost everywhere, this is not true for an arbitrary Fourier series.

Kolmogorov gave in 1926 an example of an integrable function f (x), whose Fourier series is divergent everywhere, [2, Ch VIII, Th. 4.1]. However, if we require f (x) to be continuous its Fourier series will be convergent almost everywhere, this was shown by Carleson in 1966.

Definition 4. Let {a_k} be a sequence of real numbers and s_n be the partial sums

sn =

n

X

k=0

ak.

Let σ_n denote the mean of the n + 1 first partial sums,

σn = 1 n + 1

n

X

k=0

sn

If σ_n→ S as n → ∞, we say that the series

∞

X

k=0

ak.

is (C,1)-summable to S.

By a theorem of Lebesgue, the Fourier series of f (x) is (C,1)-summable almost everywhere to f (x), [2, Ch III, Th. 3.9]. Any convergent sequence s_n that is (C,1)-summable to a number s converge to this number. Thus, we have the following proposition

Proposition 1. If the series 1 2a₀+

∞

X

k=1

a_kcos kx + b_ksin kx

is the Fourier series of a function f (x) and this series converges almost everywhere, then its sum is integrable and equal to f (x) almost everywhere.

The next theorem gives a condition on the coefficients a_n in (C) and (S) for the series to be Fourier series of their respective sums.

Theorem 6. Let

A =

∞

X

k=1

a_k k < ∞.

Then the sum of (C) and (S) are integrable in [−π, π] and the series are the Fourier series of their sums. [1, Th. 47]

Proof. Let

A_k= 1

2a₀ + a₁+ a₂+ . . . + a_k. We will first show that

∞

X

k=1

A_k

k(k + 1) = 1

2a0+ A

∞

X

k=1

A_k k(k + 1)

=1 2a0+

∞

X

k=1

1 k(k + 1)

k

X

m=1

am

=1 2a₀+ 1

2(a₁) + 1

6(a₁ + a₂) + 1

12(a₁+ a₂+ a₃) + · · ·

=1

2a₀+ a₁(1 2 +1

6 + 1

12+ · · · ) + a₂(1 6 + 1

12 + · · · ) + a₃( 1

12+ · · · )

=1 2a₀+

∞

X

m=1

a_m

∞

X

k=m

1 k(k + 1)

=1 2a₀+

∞

X

m=1

a_m

∞

X

k=m

 1

k − 1 k + 1



=1 2a₀+

∞

X

m=1

a_m m

=1

2a₀+ A

Since (C) is even and (S) is odd it is enough to prove the theorem for the interval [0, π]. Both (C) and (S) converges pointwise on [0, π], (C) with the possible exception of x = 0. Let the limits be f (x) and g(x) and define

h(x) = f (x) + ig(x) = 1 2a₀+

∞

X

n=1

a_ne^inx

We need to find an upper bound for the sum

∞

X

n=k

a_ne^inx (16)

By Lemma 3 we have 

q

X

n=p

a_ne^inx     

≤     

q

X

n=p

a_ncos nx     

+     

q

X

n=p

a_nsin nx     

≤ 3a_p sin¹₂x. Using the inequality (14), we get the upper bound

q

X

n=p

a_ne^inx     

≤ 3πa_p x .

This inequality holds for any q and for all x in (0, π) and gives us an upper bound for the sum (16).

For π/(k + 1) ≤ x < π/k we have the following upper bound on h

|h| =      1 2a₀+

k−1

X

n=1

a_ne^inx+

∞

X

n=k

a_ne^inx     

≤1 2a₀+

k−1

X

n=1

a_ne^inx     

+     

∞

X

n=k

a_ne^inx     

≤1 2a₀+

k−1

X

n=1

a_n+ 6πa_k x

≤A_k+ 6(k + 1)a_k.

On the interval [π/(k +1), π/k], both series (C) and (S) are uniformly conver- gent for any k. Since the sum of any uniformly convergent series of continuous functions is continuous, both f and g are continuous on these intervals and thus they are integrable. We have

Z ^π_k

π k+1

|h(x)| dx ≤ Z ^π_k

π k+1

A_k+ 3(k + 1)a_kdx = π

 Ak

k(k + 1)+ 3ak

k



Adding these integrals for all k we get Z π

0

|h(x)| dx =

∞

X

k=1

Z ^π_k

π k+1

|h(x)| dx

≤

∞

X

k=1

π

 A_k

k(k + 1)+ 3a_k k



=π 1

2a₀+ A + 3A



= π 1

2a₀+ 4A



≤ ∞

thus |h| is integrable which implies that h is integrable, this in turn implies that the f and g, the real and imaginary part of h is integrable. By theorem 5, (C) and (S) then are the Fourier series of their sums.

For (S) the condition in Theorem 6 is also necessary. We have the follow- ing theorem.

Theorem 7. Let

∞

X

k=1

aksin kx

be the Fourier series of some integrable function g, then

∞

X

k=1

a_k k converges.

Proof. We have

sn =

n

X

k=1

a_k k =

n

X

k=1

1 kπ

Z 2π 0

g(x) sin kx dx = 1 π

Z 2π 0

g(x)

n

X

k=1

sin kx k Consider the series

∞

X

k=1

sin kx

k . (17)

This series is the Fourier series of the function f (x) =

(₁

2(π − x), x ∈ (0, 2π)

0, x = 0

The series (17) converges to f (x) boundedly, and uniformly on any closed interval free from multiples of 2π, [1, Th. 42]. Since the series converges boundedly, we can apply the Dominated convergence theorem.

n→∞lim s_n = lim

n→∞

1 π

Z 2π 0

g(x)

n

X

k=1

sin kx k

=1 π

Z 2π 0

g(x) lim

n→∞

n

X

k=1

sin kx k

= 1 2π

Z 2π 0

g(x)(π − x) dx Thus, {sn} has a finite limit and

∞

X

k=1

ak

k = 1 2π

Z 2π 0

g(x)(π − x) dx.

Remark. For (C) the condition is not necessary, in the next theorem we give an alternative condition for (C) to be the Fourier series of its sum.

Definition 5. Let {an} be a sequence of real numbers, we denote the differ- ence (a_n− a_n+1) by ∆a_n. The numbers {∆a_n} form a new sequence and we denote the difference (∆a_n− ∆a_n+1) by ∆²a_n. If ∆²a_n ≥ 0 for every n we say that the sequence {an} is convex.

Lemma 7. If a_n→ 0 then

∞

X

k=0

∆ak = a0

and ∞

X

k=0

(k + 1)∆²a_k = a₀ Proof. For the first sum we get the telescopic sum

∞

X

k=0

∆a_k = lim

N →∞

N

X

k=0

∆a_k

= lim

N →∞(a₀− a₁) + (a₁− a₂) + (a₂− a₃) + . . . + (a_N − a_{N +1})

= lim

N →∞a₀− a_{N +1}= a₀

and for the second sum

∞

X

k=0

(k + 1)∆²a_k

=(∆a₀− ∆a₁) + 2(∆a₁− ∆a₂) + 3(∆a₂− ∆a₃) + . . .

=

∞

X

k=0

∆a_k= a₀

Theorem 8. (C) is convergent on [−π, π] except possibly at x = 0 to some function f (x). If {a_n} is convex, then f (x) is non-negative and integrable, and (C) is the Fourier series of f (x). [1, Th. 48]

Proof. Denote the partial sum of (C) by C_n. Using the Abel transform twice we get

C_n =1 2a₀+

n

X

k=1

a_kcos kx

=

n−1

X

k=0

∆akDk(x) + anDn(x)

=

n−2

X

k=0

∆²a_k(k + 1)F_k(x) + ∆a_n−1nF_n−1(x) + a_nD_n(x).

The Dirichlet kernels are uniformly bounded on [, π] for 0 <  < π and since a_n→ 0 we get

n→∞lim a_nD_n(x) = 0 ∀n ∈ N, ∀x ∈ [, π].

Since {a_n} is a convex, ∆a_n → 0, this together with lemma 6 gives us that for 0 <  < π

n→∞lim ∆a_n−1nF_n−1(x) = 0 ∀n ∈ N, ∀x ∈ [, π].

We get that

f (x) = lim

n→∞C_n(x) =

∞

X

k=0

∆²a_k(k + 1)F_k(x)

which is a sum of non-negative terms and we have the first part of the theo- rem, f (x) is non-negative.

Since every term is non-negative the partial sums C_n are monotonically increasing and we can apply the monotone convergence theorem and integrate the sum termwise.

Z π 0

f (x) dx = Z π

0

∞

X

k=0

∆²a_k(k + 1)F_k(x) dx

=

∞

X

k=0

∆²a_k(k + 1) Z π

0

F_k(x) dx

=

∞

X

k=0

∆²a_k(k + 1)π 2

=1 2πa0

where we have used lemma 5 and lemma 7.

Example 1. We consider the series (C) and (S) with coefficients a_k = 1

ln(n + 1)

This sequence decreases monotonically to zero so both series converges almost everywhere. We have that

∆a_k= 1

ln k − 1

ln(k + 1) = Z k+1

k

1 x ln²xdx and thus,

∆²a_k = Z k+1

k

1

x ln²xdx − Z k+2

k+1

1 x ln²xdx.

Since the function

f (x) = 1 x ln²x

is decrasing, we have ∆²a_k ≥ 0. Thus, the sequence {a_k} is convex, so by Theorem 8

∞

X

k=1

cos kx ln(n + 1)

is a Fourier series and since it convergent almost everywhere it is the Fourier series of its sum.

But since

∞

X

n=1

1

n ln(n + 1) = ∞, we have that

∞

X

k=1

sin kx ln(k + 1) is not a Fourier series by Theorem 7.

6 Uniform convergence

Theorem 9. For (S) the following conditions are equivalent.

(1). (S) converges uniformly (2). The sum of (S) is continuous

(3). (S) is the Fourier series of a continuous function (4). ka_k → 0 when k → ∞

Proof. Assume (S) converges uniformly to a function f (x). Each term in (S) is continuous and since the series converges uniformly, its sum will be continuous. This shows that (1) =⇒ (2).

Next, assume that the sum of (S) is continuous, it is then integrable and by theorem 5 it is the Fourier series of its sum. This shows that (2) =⇒ (3).

To show that (3) =⇒ (4), consider the (C,1)-mean of (S)

σ_n(x) = s₀+ s₁+ · · · + s_n

n + 1 =

n

X

k=1

a_k



1 − k n + 1

 sin kx

Assume (S) is the Fourier series of a continuous function, f (x). Then f (0) = 0, so for every  > 0, there is a number δ > 0 such that

|f (x)| < , ∀x ∈ [−δ, δ].

σn(x) converges uniformly to f (x), thus, there is an integer N such that

|f (x) − σn(x)| < , ∀n > N, ∀x ∈ [−π, π].

In particular, the above inequality holds on [−δ, δ].

|σ_n(x)| < 2, ∀n > N, ∀x ∈ [−δ, δ].