Notes regarding classical Fourier series

. A survey of some classical results from the theory of trigonometric series is presented, especially the case of Fourier series. Some new proofs are presented, and Riemann’s theory of trigonometric series is given special attention. MSC. 42A20, 42A24.


INTRODUCTION
The subject of trigonometric series [1,2,3,4,5] comes up in many areas of approximation theory, as well as the study of infinite series, the theory of representation of functions and also in constructing solutions of partial differential equations and eigenvalue problems [6,7] function of a real variable into a trigonometric series we refer to as Fourier's Theorem. It is possible to state certain sufficient conditions under which a function admits a trigonometric series [8,9,10,11].
Let f (t) be defined arbitrarily when −π ≤ t ≤ π and is defined for all other real values by means of the periodicity condition (1.1) f (t + 2π) = f (t).
The state f (t) is a periodic function with period 2π. Suppose f (t) has a Riemann integral over [−π, π] that exists, and if it is improper, suppose it is absolutely convergent.
Theorem 1.1. Define constants a n and b n for n = 0, 1, 2, . . . by If x is an interior point of any interval (a, b) in which f (t) has limited total variation, the trigonometric series The representation of a function by means of a Fourier series can be extended to more general intervals other than (−π, π) as well.

FEJÉR'S THEOREM
There is a theorem due to Fejér which concerns the summability of the Fourier series associated with the function f (t) which is introduced here.
Proof. Let a n , b n denote the Fourier constants (1.2) of f (t) and define (2.2) A 0 = a 0 2 , A n (x) = a n cos(nx)+b n sin(nx), S n (x) = n j=0 A j (x).
It must be proved that provided the limit on the right exists. Note first that
Hence for ϵ an arbitrary positive number, by choosing m sufficiently large, we can enforce the inequality By definition of limit, this leads to (2.10), Consequently, using (2) and (2.8) in (2.9), the Theorem follows. □

THE HURWITZ-LIAPOUNOFF THEOREM
The following Lemma which involves Fourier constants is very useful in what follows.
(a n cos(nx)+b n sin(nx))(a p cos(px)+b p sin(px)) dx Proof. Partition the interval (−π, π) into 4N subintervals such that each subinterval has length δ such that 4N δ = 2π or δ = π/2N . The partition points are y 0 = −π, y 1 = −π + δ, . . . , y k = −π + kδ, . . . , y 4N = −π + 2π = π. Consider the set of subintervals defined as I 0 = (−π, −π +δ), I k = ((2k −1)δ − π, (2k + 2)δ − π), k = 1, . . . , 2N − 1. Let U k , L k be the upper and lower bounds of f (x) on I k and |f (t)| bounded above by constant C for all x ∈ (−π, π). Denote a sample point in interval I k as x * k to be used in the x integration. Choose η k arbitrarily but such that In fact, to carry out the sum, it suffices to fix η k = δ for each k and so as part of the integrand in the x integration, Take the absolute value on both sides of (3.3) and split up the integral over The square of (3.4) can be given in the following way The right-hand side corresponds to the sample point x * k when the Riemann sum is formed for the integration over x. The right side is an upper bound for that integral. Consequently, · δ can be made arbitrarily small by taking N sufficiently large. Given that N and δ have been designated, such a value choose m = m * so that 4N/m * π sin 2 (δ/2) < ϵ/2. The expression on the right side of (3.5) is made arbitrarily small by letting m have any value greater than m * . Hence the expression on the left side of the inequality approaches zero as m → ∞. □ Theorem 3.2. Let f (t) be bounded in the interval (−π, π) and let π −π f (t) dt exist so that the Fourier coefficients a n , b n of f (t) exist. Then the series is convergent and its sum is 1 Proof. Lemma 3.1 and Theorem 3.1 prove Theorem 3.2 by means of the following approach. Begin with the identities By means of (3.7), it is deduced that The results in Lemma 3.1 can be used now Since the integral on the left approaches zero by Theorem 3.1 as m → ∞, and since (3.9) shows it equals the sum of two positive terms, it follows that each of these expressions must tend to zero as well. In particular, Expanding the bracket, the left side is equal to As this expression must go to zero as m → ∞, it follows that as m → ∞,   Proof. The function S m (x) can be expressed directly as an integral as Integrating the equation sin(2m+1) θ sin θ = 1 + 2 cos 2θ + 2 cos 4θ + · · · + 2 cos(2mθ) with respect to θ ∈ (0, π/2), we arrive at Using (4.2), we can form the difference In order to prove S m (x) approaches (1.4) as m → ∞, it is sufficient to prove that The function φ ± (θ)·θ ·csc(θ) is a function with limited total fluctuation on an interval for which θ = 0 is an end point, so In (4.5) χ 1,2 (θ) are bounded positive increasing functions of θ such that χ 1 (+0) = χ 2 (+0) = 0. Given an arbitrary positive number ϵ, a positive number δ can be chosen such that 0 ≤ χ 1 (θ) < ϵ and 0 ≤ χ 2 (θ) < ϵ whenever 0 ≤ θ ≤ δ/2. The integral in (4.4) can be split up The modulus of the first integral can be made less than ϵ by taking m sufficiently large. This follows from the Riemann-Lebesgue lemma since σ ± (θ) csc(θ) has an integral which converges absolutely in (δ/2, π/2). From the second mean value theorem, it follows that there is a number ζ between 0 and δ such that, It is known that ∞ 0 (sin t/t) dt converges, it follows that | ∞ β sin t/t dt| has an upper bound γ which is independent of β. Hence it is clear that The third integral can be treated in a similar way. By taking m sufficiently large By definition of limit, this implies that However, it has been seen that this is a sufficient condition for the limit of S m (x) to equal [f (x + 0) + f (x − 0)]/2 as m approaches infinity. So we have therefore established the convergence of a Fourier series under the conditions stated. □ The condition that x should be an interior point of the interval in which f (t) has total limited variation is merely a sufficient condition for convergence of the Fourier series. It could be replaced by any condition which satisfies the condition (4.10) lim m→∞ π/2 0 sin(2m+1)θ sin θ φ ± (θ) dθ = 0.

THEORIES OF TRIGONOMETRIC SERIES
The theory of Fourier series due to Dirichlet is directed towards series which represent given functions. Riemann made advances in this regard and considered properties of functions defined by series of the form (1.3), where it is assumed that lim n→∞ (a n cos(nx) + b n sin(nx)) = 0. Some properties which lead up to Riemann's theorem are introduced. This states essentially that if two trigonometric series converge and are equal at all points of the interval (−π, π) with the possible exception of a finite number of points, corresponding coefficients of the two series are equal.
Let the sum of (1.3) at any point x where it converges be denoted f (x) and define a function F (x) to be An(x) To prove the theorem here, two significant results are needed. There is a theorem attributed to Hardy; and the following important Lemma which was first introduced by Cantor.
Lemma 5.1. If lim n→∞ A n (x) = 0 for all values of x such that a ≤ x ≤ b, then a n → 0 and b n → 0 as n → ∞. Proof. If it is assumed that the series which defines f (x) converges at all points of a certain interval of the real axis, it is the case by Lemma 5.1 that a n , b n → 0. Then for all real values of x |a n cos(nx) + b n sin(nx)| ≤ (a 2 n + b 2 n ) 1/2 → 0 and the right side is O(1/n). By the result of Hardy, the series (5.1) converges absolutely and uniformly for all real values of x and so F (x) is continuous for all real x. □

PROPERTIES OF THE FUNCTION F (x)
Lemma 6.1 (Riemann). Define Then lim a→0 G(x, a) = f (x) provided that ∞ n=0 A n (x) converges for the value of x under consideration.
Proof. Since the series which define F (x) and F (x±2a) converge absolutely, terms may be rearranged by first noticing that Substituting into (6.1), the function G(x, a) can be calculated 1 n 2 (a n sin 2 (na) cos(nx) + b n sin 2 (nx)sin(nx)) n 2 a n cos(nx) + b n sin(nx) · sin 2 (nx).
Therefore, recalling (2.1), we arrive at, The series converges uniformly with respect to the variable a for all values of a provided that ∞ n=1 A n (x) converges. To this end it should be recalled that if a series of continuous functions of variable x is uniformly convergent for all values of x in a closed interval, the sum is a continuous function there. So for a ̸ = 0, and f (0) = 1 when a = 0, then f n (x) is a continuous function for all values of a. Consequently, if G(x, a) is a continuous function of a and then the limit a → 0 has to exist To prove that the series which defines G(x, a) converges uniformly, the following result due to Hardy is recalled: where k, k ′ are independent of n and x, and if ∞ n=1 α n is a convergent series independent of x, then ∞ n=1 a n ω n (x) converges uniformly when a ≤ x ≤ b.
In this instance, ω n (x) = f n (x) given in (6.4), clearly |f n (x)| ≤ 1. It remains to show that ∞ n=1 |f n+1 (a) − f n (a)| < K and K is independent of a. Hence if ∞ n=0 A n (x) converges, the series which defines G(x, a) converges uniformly with respect to a for all values of a, and so the limit can be computed For the proof of the following results, the well known sum is needed for a > 0 Proof. Since A n (x) → 0 as n → ∞, the first series on the right telescopes The second series can be written as n 2 a A n (x). (6.10) Substitute (6.9) and (6.10) in to (6.8) on the right side, the left side is obtained. □ Lemma 6.3. If a n , b n → 0 in A n (x) then for all values of x.
Proof. It is the case by (5.1) that (6.12) By Lemma 6.2 on the right of (6.12), using Hardy's theorem (H) for uniform convergence, this series converges uniformly with respect to a for all a greater than or equal to zero. Moreover, This limit is the value of the function when a = 0, and the value is zero since lim n→∞ A n (x) = 0. By symmetry, it can be seen the right and left hand limits are the same, so the result is zero when a → −∞. □ Suppose there are two trigonometric series satisfying the given conditions, and let their difference of these trigonometric series be Then f (x) = 0 at all points of the interval (−π, π) with a finite number of exceptions. Let ξ 1 , ξ 2 be a consecutive pair of these exceptional points, and let F (x) be the Riemann's associated function.
If the first term of ϕ(x) is not zero on the interval, there will be some point x = c at which ϕ(x) is not zero. Pick the sign of θ so that the first term is positive at c, and then take h sufficiently small so that ϕ(x) is still positive. As ϕ(x) is continuous, it attains its upper bound which must be positive since ϕ(c) > 0. Let ϕ(x) attain this upper bound at x = β so β ̸ = ξ 1 and β ̸ = ξ 2 . By Riemann's first lemma However, ϕ(β + a) ≤ ϕ(β), ϕ(β − a) ≤ ϕ(β), so this limit must be negative or zero. Hence, by assuming the first term of ϕ(x) is not everywhere zero in (ξ 1 , ξ 2 ), a contradiction has been reached, so it is zero.. Consequently, F (x) is a linear function of x over (ξ 1 , ξ 2 ). □ An immediate consequence of the next theorem is that a function of the type considered cannot be expressed as any trigonometric series in (−π, π) other than its Fourier series. This quotient in the limit is the difference of the slopes of the two segments meeting at a point whose x value is τ . Therefore, the two segments are continuous in direction, so the equation y = F (x) represents a single line, which we write as F (x) = mx + b. Then it follows that m and b have the same values for all values of x. Thus, The right-hand side of (6.16) is periodic with period 2π. This means the lefthand side of this equation must be periodic with period 2π as well, and this implies these three results: An(x) The series (6.17) (iii) is uniformly convergent. Thus we can multiply by cos(nx) or sin(nx) and integrate on both sides to produce two more results (6.18) Therefore, all the coefficients vanish, so the two trigonometric series whose difference is A 0 + ∞ n=1 A n (x) have corresponding coefficients equal as required. □

UNIFORM CONVERGENCE AND SOME EXAMPLES
let f (t) be continuous in the interval a ≤ t ≤ b. Since continuity implies uniform continuity there, the choice of δ corresponding to any value of x in (a, b) is independent of x, and the upper bound of |f (x ± 0)|, that is, |f (x)| is also independent of x so (7.1) and the upper bound of the last expression is independent of x.
Since the choice of m which makes This is independent of x, consequently, approaches the limit f (x) as m → ∞ uniformly throughout a ≤ x ≤ b.
Proof. Let h(t) be a function defined to be equal to f (t) on a ≤ t ≤ b and equal to zero for t outside this interval but in (−π, π). Suppose α n , β n are the Fourier coefficients of h(t) and S (2) n (x) the sum of the first m + 1 terms of the Fourier series associated with h(t). It follows from the results above such as (7.2) that is uniformly summable throughout (a + δ, b − δ). Moreover, there is an xindependent upper bound (7.4) |α n cos(nx) + β n sin(nx)| ≤ (α 2 n + β 2 n ) 1/2 and by Lemma 5.1 it is O(1/n). It follows from Hardy's convergence theorem that (7.3) converges uniformly to the sum h(x) which is equal to f (x). Thus write Choose ϵ > 0 arbitrarily and then enclose the points at which f (t) is unbounded in a set of intervals δ 1 , . . . , δ p such that p i=1 δ i |f (t)| dt < ϵ. Let C be the upper bound of |f (t)| outside these intervals so we have , where the selection of n depends only on a, b and the form of f (t). By a choice of m independent of x, we can assume |S m (x) − S (2) m (x)| is arbitrarily small so S m (x) − S (2) m (x) tends to zero uniformly. □ Let us now finish with a few Examples to illustrate these ideas in a more applied form.