Introduction to Linear Algebra with Mathematica

# Preface

One of the main issues of harmonic analysis is a possibility of restoring a function from its Fourier coefficients. It turns out that this problem is an ill-posed problem. Therefore, practical applications of Fourier series may require a regularization, which is related to the scrutiny of convergence of Fourier series. This topic is known as classical harmonic analysis, a branch of pure mathematics---it will be discussed further in next section and in section devoted to Cesàro summation.

Historically, Fourier series was the first example of expansion of a function with respect to eigenfunctions corresponding to a Sturm--Liouville problem. Before Fourier's discovery, Taylor series were mostly in use; however, their expansions require an utilization of uniform and absolute convergence inside the disk, and pointwise convergence on its boundary. Taylor's series coefficients are defined by derivatives evaluated at one point---the center of convergence, so only infinitesimal information is used for their determination. In opposite, eigenfunction expansions are based on the integration over whole interval. This caused development of another methods: 𝔏² convergence, summability, and the the Cesàro mean.

This section is about some conditions that guarantee pointwise convergence of (formal) Fourtier series S[f] to f(x) either pointwise or uniformly. There are known many sufficient conditions that guarantee convergence of Fourier series to the generating it function. However, we discuss only two important tests, known as Dirichlet and Dini, that provide pointwise convergence. If a function satisfies a Hölder condition, then its Fourier series converges uniformly.

# Finite sums of Fourier Series

The Fourier series of a function f is its decomposition with respect to trigonometric or exponential functions:

$$\label{EqConverge.1} f(x) \sim S[f] = \frac{a_0}{2} + \sum_{k\ge 0} \left[ a_k \cos \frac{k\pi x}{\ell} + b_k \sin \frac{k\pi x}{\ell} \right] = \mbox{P.V.} \sum_{n=-\infty}^{\infty} \hat{f}(n)\, e^{n{\bf j} \pi x/\ell} ,$$
where S[f] denotes a formal Fourier series, whose Fourier coefficients $$\hat{f}(k)$$ and 𝑎n, bn are determined through Euler--Fourier formulas
$$\label{EqFourier.2} \hat{f}(n) = \frac{1}{2\ell} \int_{-\ell}^{\ell} f(x)\, e^{-k{\bf j} \pi x/\ell} \,{\text d} x = \frac{1}{T} \int_{0}^{T} f(x)\, e^{-2k{\bf j} \pi x/T} \,{\text d} x , \qquad k=0, \pm 1, \pm 2, \ldots ;$$
and
$$\label{EqFourier.5T} \left[ \begin{array}{c} a_k \\ b_k \end{array} \right] = \frac{2}{T} \int_{0}^{T} f(x) \left[ \begin{array}{c} \cos \left( \frac{2\pi kx}{T} \right) \\ \sin \left( \frac{2\pi kx}{T} \right) \end{array} \right] {\text d} x , \qquad k= 0, 1, 2, 3, \ldots .$$
Here T = 2ℓ is the period and «P.V.» abbreviates the Cauchy principle value. As usual, j denotes the imaginary unit vector in the positive vertical direction on complex plane ℂ, so j² = −1. These coefficients are related via the Euler formula:
$\hat{f}(n) = \frac{1}{2} \left( a_n - {\bf j} b_n \right) , \quad n \ge 0,\qquad \mbox{and} \qquad \hat{f}(-n) = \overline{\hat{f}(n)} = \hat{f}(n)^{\ast} = \frac{1}{2} \left( a_n + {\bf j} b_n \right) , \quad n > 0.$
There are two standard notations to represent a complex conjugate: either with overline or with asterisk, so we utilize both. Series \eqref{EqConverge.1} is an example of an expansion of f(x) with respect to eigenfunctions corresponding to a Sturm--Liouville problem (see section for detail). Since this problem contains periodic boundary conditions, this expansion reflects this property. Therefore, we will always assume that a function f(x) is periodically expanded to ℝ with the same period T = 2ℓ to match the boundary condition: f(x) = f(x + T).

Example 1: We consider a 2π-periodic function whose Fourier series diverges. This function was constructed by Lipót Fejér in 1900.

$f(x) = \sum_{p\ge 1} \frac{1}{p^2} \,\sin \left[ \left( 2^{p^3} +1 \right) \frac{x}{2} \right] \tag{1.1}$
This function is continuous because its general term decays as 1/p². Since function f(x) is odd, its Fourier series becomes
$f(x) = \sum_{n\ge 1} b_n \sin (nx) ,$
where
$b_n = \frac{2}{\pi} \int_0^{\pi} f(x)\,\sin (nx)\,{\text d} x = \sum_{p\ge 1} \frac{1}{p^2} \cdot \frac{2}{1 + 2^{p^3} -2n} . \tag{1.2}$
Using Mathematica, we get
Integrate[Sin[(m + 1)*x/2]*Sin[n*x], {x, 0, Pi}]*2/Pi
(2 (-(Cos[1/2 (m + 2 n) $Pi]]/(1 + m + 2 n)) + Sin[1/2 (1 + m - 2 n) \[Pi]]/(1 + m - 2 n)))/\[Pi] 1. du Bois-Reymond, P., Ueber die fourierschen reihen, Königliche Gesellschaft der Wissenschaften zu Göttingen, 1873, Vol. 21, pp. 571--582. 2. Castillo, A., Chavez, J., Kim, H., A note on divergent Fourier series and λ-permutations, The Australian Journal of Mathematical Analysis and Applications, 2017, Volume 14, Issue 1, Article 3, pp. 1--9. 3. Kolmogorov, A., Une serie de Fourier--Lebesgue divergente presque partout, Fundamenta Mathematicae, 1923, 4, 324--328. ■ Since Fourier series S[f] of a function f(x) is an example of infinite series, its convergence (or justification) depends on a rule how its partial sums $$\label{EqFourier.4} S_N (f; x) = \frac{a_0}{2} + \sum_{k=1}^N \left[ a_k \cos \left( \frac{k\pi x}{\ell} \right) + b_k \sin \left( \frac{k\pi x}{\ell} \right) \right] = \sum_{n=-N}^N \hat{f}(n)\,e^{{\bf j}n \pi x/\ell}$$ approach a limit. Therefore, we need a compact formula for finite partial sums SN(f;, x); fortunately, such formula is known and it is based on Fourier secomposition of the Dirac delta function. We expand the delta-function into Fourier series, Using Euler--Fourier formulas, we get \[ \delta (x) = \frac{1}{2\ell} + \frac{1}{\ell} \left[ \cos \left( \frac{\pi x}{\ell} \right) + \cos \left( \frac{2\pi x}{\ell} \right) + \cos \left( \frac{3\pi x}{\ell} \right) + \cdots \right] = \frac{1}{2\ell} + \frac{1}{\ell} \sum_{\k\ge 1} \cos \left( \frac{k\pi x}{\ell} \right) .$
This series cannot truly converge because its terms don’t approach zero. However, this series converges in weak sense ,eaning that its partial sums
$\delta_N (x) = \frac{1}{2\ell} + \frac{1}{\ell} \left[ \cos \left( \frac{\pi x}{\ell} \right) + \cos \left( \frac{2\pi x}{\ell} \right) + \cos \left( \frac{3\pi x}{\ell} \right) + \cdots + \cos \left( \frac{N\pi x}{\ell} \right) \right] = \frac{1}{2\ell} + \frac{1}{\ell} \sum_{\k+ 1}^N \cos \left( \frac{k\pi x}{\ell} \right)$
possess the property
$\lim_{N\to +\infty} \int_{-\ell}^{\ell} \delta_N (x) f(x)\,{\text d}x = f(0)$
for a smooth function f(x). Actually there is a neat formula for the partial sum δN(x). It is called the Dirichlet kernel after the German scientist who first dicovered it in 1824.
Lemma 1: The following trigonometric identity holds:
$$\label{EqFourier.5} D_n (x) = \frac{1}{2} + \cos x + \cos 2x + \cdots + \cos nx = \frac{\sin \left( n + \frac{1}{2} \right) x}{2\,\sin \frac{x}{2}} .$$
Since
$\sin \left( k + \frac{1}{2} \right) x - \sin \left( k - \frac{1}{2} \right) x = 2\,\sin \frac{x}{2}\,\cos kx ,$
we have
\begin{align*} 2\,\sin\frac{x}{2}\,\sum_{k=0}^n \cos kx = \sum_{k=0}^n \left[ \sin \left( k + \frac{1}{2} \right) x - \sin \left( k - \frac{1}{2} \right) x \right] \\ &= \sin \left( n + \frac{1}{2} \right) x - \sin \left( - \frac{1}{2} \right) x , \end{align*}
or,
$\frac{1}{2} + \sum_{k=1}^n \cos kx = \frac{\sin \left( n + \frac{1}{2} \right) x}{2\,\sin\frac{x}{2}} .$

# Spaces of Functions

Before we get into the topic of convergence, we need to define some classes of functions that are sufficient for Fourier expansions and wide enough to include the majority of practical applications. First, we extend the set of continuous functions on the interval [−ℓ, ℓ], which is usually denoted as C[−ℓ, ℓ], that have an additional condition f(−ℓ) = f(ℓ). We need to consider piecewise continuous functions that consist of finite number of continuous pieces. We say that f(x) has a jump discontinuity at x = 𝑎 if the limit of the function from the left, $$\lim_{t\uparrow a} f(t) ,$$ denoted $$f(a-0) \ \mbox{ or }\ f(a^{-})$$ and the limit of the function from the right, $$\lim_{t\downarrow a} f(t) ,$$ denoted $$f(a+0) \ \mbox{ or }\ f(a^{+}) ,$$ both exist and $$f(a+0) \ne f(a-0) .$$ A function f(x) is called piecewise continuous on an interval [𝑎, b], if it is continuous on [𝑎, b] except for at most finitely many points $$x_1 , x_2 , \ldots , x_n$$ at each of them the function has both the left-hand and right-hand limits: $$f(x_k -0) \ \mbox{ and }\ f(x_k +0) , \quad k=1,2,\ldots , n .$$ Now we give the formal definition.

Let 𝑎 and b be real numbers such that 𝑎 < b. A function $$f\, : \, [a,b] \mapsto \,\mathbb{R}$$ is said to be piecewise continuous on $$[a,b]$$ if the following conditions are satisfied:
• there exists a finite set $$\{ x_1 , x_2 , \ldots , x_n \} \subset (a, b)$$ such that $$x_1 < x_2 < \cdots < x_n$$ and f is continuous and monotone on each subinterval
$\left( a , x_1 \right) , \quad \left( x_k , x_{k+1} \right) , \quad k=1,2,\ldots , n-1, \quad \left( x_n , b \right) ;$
• all the following one-sided limits exist
$\lim_{x\downarrow a} f \left( x \right) , \quad \lim_{x\uparrow x_k} f\left( x \right) , \quad \lim_{x\downarrow x_k} f\left( x \right) ,\quad k=1,2,\ldots , n-1, \quad \lim_{x \uparrow b} f\left( x \right) .$
A function $$f\, : \, \mathbb{R} \mapsto \,\mathbb{R}$$ is piecewise continuous on $$\mathbb{R} ,$$ if it is piecewise continuous on every finite subinterval of $$\mathbb{R} .$$

Example 5: The function

$f(x) = \begin{cases} 2x , & \ \mbox{ if } 0 < x < 1, \\ 1, & \ \mbox{ if } 1 < x < 2, \end{cases}$
is piecewise continuous on [0,2], but not continuous on [0,2].
f[x_]=Piecewise[{{2*x, 0<x<1},{1,1<x<2}}] Integrate[f[x], {x, 0, 2}]
2
Simplify[Integrate[f[x]*Cos[n*Pi*x], {x, 0, 2}], Assumptions -> Element[n, Integers]]
(2 (-1 + (-1)^n))/(n^2 $Pi]^2) Simplify[Integrate[f[x]*Sin[n*Pi*x], {x, 0, 2}], Assumptions -> Element[n, Integers]] -((1 + (-1)^n)/(n \[Pi])) Therefore, we get the following Fourier series for function x²: \[ f(x) = 1 + \sum_{n\ge 1} \left[ \frac{(-1)^n -1}{n^2\pi^2 /2} \,\cos \left( n\pi x \right) - \frac{(-1)^n +1}{n\pi} \,\sin \left( n\pi x \right) \right] .$
Next, we plot partial sums along with the given function.
Fourier approximation with 10 terms    Fourier approximation with 20 terms    Fourier approximation with 100 terms

The first sufficient conditions for the pointwise convergence of the Fourier series were discovered by a German mathematician Johann Peter Gustav Lejeune Dirichlet (1805--1859). A function that satisfies the Dirichlet conditions is also called piecewise monotone. Such functions must have right and left limits at each point of discontinuity. We prove the corresponding theorem in convergence section.

Theorem 5: Assume that
$F(x) \,=\, \frac{a_0}{2} + \sum_{k= 1}^{\infty} a_k \cos \left( \frac{k \pi x}{\ell} \right) + b_k \sin \left( \frac{k \pi x}{\ell} \right)$
is the Fourier series for a piecewise continuous function f(x) over the interval [−ℓ, ℓ]. If its derivative f'(x) is piecewise continuous on the interval $$[- \ell , \ell ]$$ and has both a left- and right-hand derivative at each point in this interval, then F(x) is pointwise convergent for all x ∈ [−ℓ, ℓ]. The relation $$F(x) = f(x)$$ holds at all points x ∈ [−ℓ, ℓ], where f(x) is continuous. If x = 𝑎 is a point of discontinuity of f, then
$F(x) \,=\, \frac{f(a+0)+ f(a-0)}{2} ,$
where $$f(a-0) = f(a^{-}) = \lim_{t\uparrow a} f(t)$$ and $$f(a+0) = f(a^{+}) = \lim_{t\downarrow a} f(t)$$ denote the left- and right-hand limits, respectively.    ⧫

Since the class of continuous functions C[−ℓ, ℓ] contains functions that are different from their corresponding sum-Fourier series, we consider its subclass containing so called piecewise smooth functions.

We say that f(x) is piecewise smooth on some interval if the interval can be broken up into a finite number of pieces (or sections) such that in each piece the function f(x) is continuous and its derivative df/dx is also continuous. The function f(x) may not be continuous, but the only kind of discontinuity allowed is a finite number of jump discontinuities.

If derivatives in the condition above are replaced by weaker condition so that function f is piecewise monotonic, then such function is said to satisfy the Dirichlet conditions. A function $$f\, : \, \mathbb{R} \mapsto \,\mathbb{R}$$ is piecewise smooth on $$\mathbb{R} ,$$ if it is piecewise smooth on every finite subinterval of $$\mathbb{R} .$$ The tangent function is not piecewise smooth because its discontinuity is of infinite jump. On the other hand, the function sin(1/x) is not piecewise continuous because it has infinite number of maxima and minima in the neighborhood of the origin.

A real-valued function f defined on an interval [𝑎, b] is called piecewise smooth on it if it is piecewise continuous and its derivative is piecewise continuous.

Example 6: Let f(x) = |sin(x)| be the absolute value of the sine function on the interval |−π, π|. This function is continuous on any subinterval of the real axis ℝ. Its derivative is piecewise continuous:

$f'(x) = \begin{cases} \phantom{-}\cos x , & \ \mbox{ if }\quad 0 < x < \pi , \\ -\cos x , & \ \mbox{ if }\quad -\pi < x < 0 . \end{cases}$
 Using Mathematica FourierTrigSeries[Abs[Sin[x]], x, 10] Integrate[Abs[Sin[x]]*Cos[2*n*x], {x, -Pi, Pi}] -((4 Cos[n $Pi]]^2)/(-1 + 4 n^2)) we expand the given function into Fourier series: \[ \left\vert \sin x \right\vert = \frac{2}{\pi} - \frac{4}{\pi} \sum_{n\ge 1} \frac{1}{4n^2 -1} \,\cos \left( 2nx \right) .$ We plot the function |sin(x)| and its derivative with Mathematica: f[x_] = Piecewise[{{Cos[x], 0 < x < Pi}, {-Cos[x], -Pi < x < 0}}]; Plot[{Abs[Sin[x]], f[x]}, {x, -Pi, Pi}, PlotStyle -> Thickness[0.01]] Function |sin(x)| and its derivative. Mathematica code

However, $$f(x) = |x|^{1/2} = \sqrt{|x|}$$ is continuous on [-1,1], but it is not piecewise smooth on [-1,1] because we can break the interval [-1,1] into two subintervals [-1, 0} and (0, 1] on each of these the given function has no a bounded derivative.     ■

# Pointwise Convergence of Fourier Series

In this section we analyze a pointwise convergence of partial Fourier sums
$$\label{EqConverge.4} S_N (f; x) = \frac{a_0}{2} + \sum_{k=1}^N a_k \cos \frac{k\pi x}{\ell} + b_k \sin \frac{k\pi x}{\ell} = \sum_{k=-N}^N \hat{f}(k)\, e^{{\bf j} k\pi x/\ell} ,$$
as N → ∞ for every fixed x∈[−ℓ, ℓ]. It turns out that not every integrable (in Riemann sense, 1855) function possesses a convergent Fourier series. in 1923, Andrew Kolmogorov gave an example of a function from 𝔏¹ with an almost everywhere divergent Fourier series.Soon after, in 1926, he proved another example of Fourier series that diverges at every point. Some function have convergent Fourier series that have limiting values different from generating Fourier series. We denote the relation between a (periodic) function f(x) and its Fourier series as
$f(x) \,\sim \,\frac{a_0}{2} + \sum_{n\ge 1} \left[ a_n \cos \frac{n\pi x}{\ell} + b_n \sin \frac{n\pi x}{\ell} \right] = \mbox{P.V.} \sum_{n=-\infty}^{\infty} \hat{f}(n)\,e^{{\bf j}\pi nx/\ell} .$
This relation indicates that the Fourier coefficients are evaluated according to the Euler--Fourier formulas.

The first convergence theorem was proved in 1829 by the German mathematician Peter Gustav Lejeune Dirichlet (1805--1859). Its proof is based on application of the second mean value theorem. It provides sufficient conditions for Fourier series to comverge pointwise without requirement on existence of a derivative of function f(x).

Second mean value Theorem: Let g(x) be a bounded, increasing real valued function defined on the interval [𝑎,b], that is continuous at 𝑎 and b; and f(x) a Riemann integrable function on [𝑎,b]. Then there exists a real number c∈[𝑎,b] such that
$\int_a^b g(x)\,f(x)\,{\text d} x = g(a+0) \int_a^c f(x)\,{\text d} x + g(b-0) \int_c^b f(x)\,{\text d} x .$
There are known other sufficient conditions that guarantee pointwise convergence of a Fourier series; however, we start with a classical statement credited to Dirichlet.
Theorem (Dirichlet): Let f(x) be a periodic function with period T = 2ℓ and satisfy the following conditions (that are usually referred to as the Dirichlet conditions):
• f(x) must be absolutely integrable over a period;
• f(x) must be of bounded variation (which means that the function has finite number of local maximum and minimum points and it monotonically increases/decreases between them) in any given bounded interval;
• f(x) must have a finite number of discontinuities in any given bounded interval, and the discontinuities cannot be infinite.
Then for all x, the Fourier series
$\frac{a_0}{2} + \sum_{k\ge 1} \left[ a_k \cos \frac{k\pi x}{\ell} + b_k \sin \frac{k\pi x}{\ell} \right] = \sum_{n=-\infty}^{\infty} \alpha_n e^{n\pi{\bf j}x/\ell} = \frac{1}{2} \left[ f(x+0) + f(x-0) \right]$
converges to the mean values of the function f at that point. Here
$f(x+0) = f(x^{+}) = \lim_{\varepsilon \downarrow 0} f(x+\varepsilon ) , \qquad f(x-0) = f(x^{-}) = \lim_{\varepsilon \downarrow 0} f(x-\varepsilon ) , \qquad \varepsilon > 0,$
denotes the right/left limits of f(x).     ⧫
The Dirichlet theorem assures that the Fourier series $$\frac{a_0}{2} + \sum_{k\ge 1} a_k \cos \frac{k\pi x}{\ell} + b_k \sin \frac{k\pi x}{\ell}$$ converges to
1. f(x) at all points where f is continuous inside the interval (−ℓ, ℓ);
2. $$\displaystyle \frac{f(x+0) + f(x-0)}{2} = \frac{f(x^{+}) + f(x^{-})}{2} = \lim_{\varepsilon\downarrow 0} \frac{f(x+\varepsilon ) + f(x- \varepsilon )}{2}$$
at points of discontinuity;
3. $$\displaystyle \frac{f(-\ell +0) + f(\ell -0)}{2}$$
at endponts.

The absolutely integrability condition is a sufficient condition to guarantee existence of Fourier coefficients. The last two conditions of Dichlet's theorem are imposed to apply the second mean value theorem:

In section on Fourier series, a class of functions, called piecewise smooth, was introduced. Such functions satisfy the Dirichlet conditions.

Lemma 2: If f is integrable on the interval [-ℓ,ℓ], then
$\sum_{k=-n}^n \hat{f}(k)\, e^{{\bf j}kx\pi /\ell} = \frac{1}{\ell} \,\int_{-\ell}^{\ell} f(x-t)\, D_n (t)\,{\text d} t ,$
where
$D_n (t) = \frac{1}{2}\,\csc \left( \frac{\pi t}{2\ell} \right) \sin \left( \frac{(2n+1)\pi t}{2\ell} \right) , \qquad t\ne 0,$
is called the Dirichlet kernel. At t = 0, we have Dn(0) = n + ½.
Note that for the complex Fourier series, the Dirichlet kernel is the same:
$D_n (x) = \sum_{n=-N}^N e^{{\bf j} nx} = \frac{\sin \left( \left( N + \frac{1}{2} \right) x \right)}{2\,\sin \left( x/2 \right)} .$
The denominator in the Dirichlet kernel \eqref{EqFourier.4} can vanish and hence this expression holds at the points, where it is not well-defined, according to L'Hôpital's rule.
Lemma 3: The Dirichlet kernel possesses the following properties.
1. Dn(x) is 2π-periodic.
2. Dn(x) is an even function.
3. $$\displaystyle \frac{1}{\pi} \int_{-\pi}^{\pi} D_n (x) \,{\text d} x = 1 .$$
The first two properties are obvious. The last property follows from its series representation:
$\int_{-\pi}^{\pi} D_n (x)\,{\text d}x = \int_{-\pi}^{\pi} \left[ \frac{1}{2} + \sum_{k=1}^n \cos kx \right] {\text d}x = \pi + \sum_{k=1}^n \int_{-\pi}^{\pi} \cos kx \,{\text d}x .$
Since
$\int_{-\pi}^{\pi} \cos kx \,{\text d}x = 0 ,$
Integrate[(Cos[k*x]) , {x, -Pi, Pi}]
(2 Sin[k $Pi]])/k we get the required property.  Here is Mathematica code for n = 9: Plot[Sin[(9 + 1/2)*x]/2/Sin[x/2]/Pi, {x, -2, 2}, PlotStyle -> {Thick, Blue}, PlotRange -> All] Plot of Dirichlet's kernel. Mathematica code Riemann-Lebesgue lemma: Let f(x) ∈ 𝔏[𝑎, b] be absolutely integrable function on the interval [𝑎. b]. Then \[ \lim_{n\to\infty} \int_a^b f(x)\,\cos nx\,{\text d}x = \lim_{n\to\infty} \int_a^b f(x)\,\sin nx\,{\text d}x = 0.$
For differentiable function, the Riemann-Lebesgue lemma follows by integration by parts
$\int_a^b f(x)\,\cos nx\,{\text d}x = \frac{1}{n} \left[ f(b) \sin nb - f(a)\,\sin an - \int_a^b f'(x)\,\sin nx\,{\text d}x \right] .$
Hence,
$\left\vert \int_a^b f(x)\,\cos nx\,{\text d}x \right\vert \le \frac{1}{n} \left[ |f(b)| + |f(a)| + \int_a^b \left\vert f(x) \right\vert {\text d}x . \right]$
Allowing n → ∞, we obtain the stated limit. A similar argument holds for the sine.

Suppose now that f(x) ∈ 𝔏[𝑎. b]. Given ε > 0, we can find a polynomial p(x) such that $$\int_a^b \left\vert f(x) - p(x) \right\vert {\text d}x \le \varepsilon.$$ Now,

$\left\vert \int_a^b f(x)\,\cos nx \,{\text d}x \right\vert = \left\vert \int_a^b \left[ f(x) - p(x) \right] \cos nx \,{\text d}x + \int_a^b p(x)\,\cos nx \,{\text d}x \right\vert \le \int_a^b \left\vert f(x) - p(x) \right\vert {\text d} x + \left\vert \int_a^b p(x)\,\cos nx \,{\text d}x \right\vert .$
Since the limit for the polynomial
$\lim_{n\to\infty} \int_a^b p(x)\,\cos nx\,{\text d}x = 0$
is zero, we get
$\lim_{n\to\infty} \sup \left\vert \int_a^b f(x)\,\cos nx\,{\text d}x \right\vert \le \varepsilon .$
Because &epsilon is arbitrary small, we derive the required statement.
Lemma 5: Let $$S_n (x) = \frac{a_0}{2} + \sum_{k=1}^n a_k \cos kx + b_k \sin kx$$ be the partial sum of the Fourier series of f(x). Then
$S_n (x) - f(x) = \frac{1}{2\pi} \int_0^{\pi} \left[ f(x+t) + f(x-t) - 2\,f(x) \right] \frac{\sin \left( n + \frac{1}{2} \right) t}{\sin\frac{t}{2}} \,{\text d}t .$
We have
$S_n (x) = \int_{-\pi}^{\pi} D_n (x-t)\,f(t)\,{\text d}t = \frac{1}{2\pi}\int_{-\pi}^{\pi} \frac{\sin \left( n + \frac{1}{2} \right) \left( x - t \right)}{\sin\frac{1}{2} \left( x-t \right)}\,f(t)\,{\text d}t .$
Let t = x + τ. Then
$S_n (x) = \frac{1}{2\pi} \int_{-\pi -x}^{\pi -x} D_n (x-t)\,f(\tau + x)\,{\text d}\tau = \frac{1}{2\pi} \int_{-\pi -x}^{\pi -x} f(\tau + x)\, \frac{\sin \left( n + \frac{1}{2} \right) \tau}{\sin\frac{1}{2} \tau}\,{\text d}\tau .$
Assuming that f(x) is extended periodically, we get
$S_n (x) = \frac{1}{2\pi} \int_{0}^{\pi} f(t + x) \cdot \frac{\sin \left( n + \frac{1}{2} \right) t}{\sin\frac{1}{2} t}\,{\text d}t + \frac{1}{2\pi} \int_{-\pi}^{0} f(t + x) \cdot \frac{\sin \left( n + \frac{1}{2} \right) t}{\sin\frac{1}{2} t}\,{\text d}t .$
In the latter integral, we set τ = −t, and obtain
$S_n (x) = \frac{1}{2\pi} \int_{0}^{\pi} \left[ f(t + x) + f(x-t) \right] \frac{\sin \left( n + \frac{1}{2} \right) t}{\sin\frac{1}{2} t}\,{\text d}t .$
Since
$\frac{\sin \left( n + \frac{1}{2} \right) t}{\sin\frac{1}{2} t} \to \frac{1}{2}\,\delta (t) \qquad \mbox{as} \qquad n\to \infty .$
we obtain
$f(x) = \frac{1}{2\pi} \int_{0}^{\pi} 2f(x) \cdot \frac{\sin \left( n + \frac{1}{2} \right) t}{\sin\frac{1}{2} t}\,{\text d}t .$
Lemma 6: Let f(x) ∈ 𝔏[−π, π] be absolutely integrable function on the interval [−π, π]. For any δ such that 0 < δ < π, we have
$S_n (x) - f(x) = \frac{1}{2\pi} \int_0^{\delta} \frac{f(x+t) + f(x-t) - 2\,f(x)}{\sin\frac{t}{2}}\,\sin \left( n+\frac{1}{2} \right) t\,{\text d}t + \varepsilon_n ,$
where εn → 0     as n → ∞.
We write the integral for the partial sum as the sum of two integrals:
$S_n (x) = \frac{1}{2\pi} \int_0^{\pi} \left[ f(x+t) + f(x-t) - 2\,f(x) \right] \frac{\sin \left( n + \frac{1}{2} \right) t}{\sin\frac{t}{2}} \,{\text d}t = \int_0^{\delta} + \int_{\delta}^{\pi} .$
Notice that in the integral over [δ, π], the integrand is an integrable function and hence the Riemann-Lebesgue lemma guarantees that it approaches zero as n → ∞.

Lemma 7: If function ψ(x) is monotonic in an interval [0, δ], for some positive δ, then
$\lim_{m\to\infty} \int_0^{\delta} \psi (z) \,\frac{\sin (mz)}{z}\,{\text d}z = \frac{\pi}{2}\,\psi (+0) .$
Since the function $$\displaystyle \frac{\sin (mz)}{z}$$ is integrable on the interval [0, δ] for any δ, we can apply the second mean value theorem and obtain
$\int_0^{\delta} \psi (z) \,\frac{\sin (mz)}{z}\,{\text d}z = \psi (+0) \int_0^c \frac{\sin (mz)}{z}\,{\text d}z + \psi (\delta -0) \int_c^{\delta} \frac{\sin (mz)}{z}\,{\text d}z .$
The latter integral tends to zero as m → ∞ because of Riemann--Lebesgue lemma
$\int_c^{\delta} \frac{\sin (mz)}{z}\,{\text d}z \ \to\ 0 \qquad \mbox{as} \qquad m \to \infty .$
In the former integral, we change the variable of integration t = m z, which yields
$\int_0^c \frac{\sin (mz)}{z}\,{\text d}z = \int_0^{mc} \frac{\sin t}{t}\, {\text d}t \ \to \, \int_0^{+\infty} \frac{\sin t}{t}\, {\text d}t = \frac{\pi}{2} \qquad\mbox{as} \qquad m \to \infty .$
Without any loss of generality, we can consider the main interval to be (−π, π) instead of (−ℓ, ℓ).

The finite Fourier sum of function f(x) can be written as

$S_n (f; x) = \frac{a_0}{2} + \sum_{k=1}^n a_k \cos kx + b_k \sin kx = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \left[ \frac{1}{2} + \sum_{k=1}^n \cos k \left( t-x \right) \right] {\text d}t .$
Using explicit formula for the Dirichlet kernel, we find
$S_n (f; x) = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t)\, \frac{\sin \frac{(2n+1)\left( t-x \right)}{2}}{2\,\sin \frac{t-x}{2}}\,{\text d}t .$
Next, we expand function f(x) periodically from the interval [−π, π] into ℝ. This allows us to rewrite the latter as
$S_n (f; x) = \frac{1}{\pi} \int_{x-\pi}^{x+ \pi} f(t)\, \frac{\sin \frac{(2n+1)\left( t-x \right)}{2}}{2\,\sin \frac{t-x}{2}}\,{\text d}t .$
Upon setting t = x −2z in the integral from Lemma 3, we get
$S_n (f; x) = \frac{1}{\pi} \int_{0}^{\pi /2} f(x-2z)\, \frac{\sin (2n+1)\,z}{\sin z}\,{\text d}z + \frac{1}{\pi} \int_{0}^{\pi /2} f(x+2z)\, \frac{\sin (2n+1)\,z}{\sin z}\,{\text d}z .$
Assuming that function f(x) is continuous at x, we subtract it from the Fourier partial sum using the identity
$1 = \frac{2}{\pi} \int_{0}^{\pi /} \frac{\sin (2n+1)\,z}{\sin z}\,{\text d}z ,$
to obtain
$S_n (x) - f(x) = \frac{1}{\pi} \int_0^{\pi /2} \left[ f(x+2z) - f(x) \right] \frac{\sin \left( 2n +1 \right) z}{\sin z} \,{\text d}z + \frac{1}{\pi} \int_0^{\pi /2} \left[ f(x-2z) - f(x) \right] \frac{\sin \left( 2n +1 \right) z}{\sin z} \,{\text d}z .$
We rewrite the latter as
$S_n (x) - f(x) = \frac{1}{\pi} \int_0^{\pi /2} \frac{f(x-2z) - f(x)}{-2z} \cdot \frac{-2z}{\sin z} \,\sin \left( 2n + 1 \right) {\text d}z + \frac{1}{\pi} \int_0^{\pi /2} \frac{f(x+2z) - f(x)}{2z} \cdot \frac{2z}{\sin z} \,\sin \left( 2n + 1 \right) {\text d}z$
In order to prove the Dirichlet theorem, we need to show that the difference Sn(x) − f(x) approaches zero as n → ∞. Let us fix x and consider the function
$\psi (t) = f(x+t) + f(x-t) - 2\,f(x) .$
Based on Lemma 5, we have
$2 \int_0^{\delta} \psi (t)\,\frac{\sin \left( n + \frac{1}{2} \right) t}{t} \,{\text d}t = \int_0^{\delta} \psi (t)\,\frac{\sin \left( n + \frac{1}{2} \right) t}{\sin\frac{t}{2}} \,{\text d}t + \int_0^{\delta} \psi (t) \left( \frac{2}{t} - \frac{1}{\sin\frac{t}{2}} \right) \sin \left( n + \frac{1}{2} \right) t\,{\text d}t$
Since the difference
$\frac{2}{t} - \frac{1}{\sin\frac{t}{2}}$
has no singulariry in a neighborhood of the origin, it is integrable over [0, δ]. Hence, by the Riemann--Lebesgue lemma, the latter integral → 0 as n → ∞. Application of the second mean value theorem provides the required result because ψ(0) = 0.

Lemma 8: Let f be a continuous periodic function. If its Fourier series converges absolutely, then the Fourier series converges to f uniformly.    ; ⧫

Theorem 3: For piecewise smooth f(x), the Fourier series of f(x) is continuous and converges to f(x) for −ℓ ≤ x ≤ ℓ if and only if f(x) is continuous and f(−ℓ) = f(ℓ).     ⧫

Theorem 4: A Fourier series for a 2ℓ-periodic function f(x) that is continuous can be differentiated term by term if its derivative f'(x) satisfies the Dirichlet conditions.     ⧫
Theorem 5 (Riemann principle of localization): Let function f∈𝔏¹ be absolutely integrable on an interval of length T = 2ℓ and have a Fourier series \eqref{EqConverge.1}. The convergence of this series to f(x) at a fixed point x depends only upon the behavior of f(x) in an arbitrary small neighborhood of x.
From Lemma 5, it follows that the convergence to f(x) depends upon
$\lim_{n\to +\infty} \int_0^{\delta} \left[ f(x+t) + f(x-t) - 2\,f(x) \right] \frac{\sin \left( n+ \frac{1}{2} \right) t}{t}\, {\text d}t .$
Now this integral utilizes the values of f(x) only in the interval (x −δ), x + δ). ﹡ ⁎ ✱ ✲ ✳ ✺ ✻ ✼ ✽ ❋
A function f of bounded variation on a finite interval [𝑎, b] is a function having bounded total variation
$V_a^b (f) = \sup_P \sum_{i=0}^{n-1} \left\vert f( x_{i+1}) - f(x_i ) \right\vert ,$
where the supremum is taken over every partition of interval [𝑎, b] such that xixi+1
If f is differentiable and its derivative is Riemann-integrable, its total variation is the vertical component of the arc-length of its graph, that is,
$V_a^b (f) = \int_a^b \left\vert f' (x) \right\vert {\text d} x .$
According to Boris Golubov, a bounded variation functions of a single variable were first introduced by Camille Jordan, in the paper (1881) dealing with the convergence of Fourier series. Using this definition, he improved the Dirichlet theorem.
Theorem 6 (Dirichlet--Jordan criterion): Suppose that f(x) is of bounded variation over interval [−ℓ, ℓ]. Then
1. at every point x0, the Fourier series S[f] converges to the value $$\displaystyle \tfrac{1}{2} \left[ f(x_0 +0) + f(x_) -0) \right]$$ ; in particular, the Fourier series S[f] converges to f(x) at every point of continuity of f;
2. if further f is continuous at every point of a closed interval [𝑎, b], then its Fourier series S[f] converges uniformly in [𝑎, b].
The proof of this theorem follows from the Dirichlet theorem and the following lemma.
Lemma (Jordan decomposition): Every function of bounded variation on a finite closed interval is a difference of two bounded monootonic (nondecreasing) functions.
Recall that thetotal variation of functionf is defined to be
$V_a^b (f) = \sup_P \sum_{i=0}^{n-1} \left\vert f( x_{i+1}) - f(x_i ) \right\vert , \qquad P = \left\{ a = x_0 < x_1 < \cdots < x_{n-1} < x_n = b \right\} .$
Thus, f is of bounded variation on [𝑎, b] if and only if Vb𝑎(f) is bounded on [𝑎, b]. Furthermore, given any function of bounded variation f on [𝑎, b], we may rewrite it as a difference of two monotonic functions:
$f(x) = \frac{1}{2} \left[ V_a^x (f) + f(x) \right] - \frac{1}{2} \left[ V_a^x (f) - f(x) \right] .$
Observe that $$f^{+} (x) = \tfrac{1}{2} \left[ V_a^b (f)(x) + f(x) \right]$$ and $$f^{-} (x) = \tfrac{1}{2} \left[ V_a^b (f)(x) - f(x) \right]$$ are both monotone increasing functions. That is, a function is of bounded variation if and only if it is the difference of two monotonicaly increasing functions.
Since the Direchlet kernel is an even function (as sume of cosine functions), the partial Fourier sum can be written as
$S_N (f; x) = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x-y)\,D_N (y)\,{\text d} y = \frac{1}{2\pi} \int_{0}^{\pi} \left[ f(x-y) + f(x+y) \right] D_N (y)\,{\text d} y$
As every bounded variation functionf is the difference of two monotonic functions, it suffices to show that
$\frac{1}{2\pi} \int_{0}^{\pi} g(y)\,D_N (y)\,{\text d} y \,\to \frac{g(0+0)}{2} = \frac{g(0^{+})}{2} \qquad\mbox{as} \quad N \to \infty ,$
where g is monotone, since then we can take g(y) = f(x+y) and g(y) = f(x−y) to complete the result. For G(y) = g(y) − g(0+0), we have
$\frac{1}{2\pi} \int_{0}^{\pi} G(y)\,D_N (y)\,{\text d} y \,\to \frac{G(0+0)}{2} = 0 \qquad\mbox{as} \quad N \to \infty$
if and only if
$\frac{1}{1\pi} \int_{0}^{\pi} \left[ g(y) - g(0+0) \right] D_N (y)\,{\text d} y \,\to \frac{G(0+0)}{2} = 0 \qquad\mbox{as} \quad N \to \infty$
because
$\frac{1}{1\pi} \int_{0}^{\pi} D_N (y)\,{\text d} y = \frac{1}{2} .$
Therefore, without loss of generality, suppose that g(0+0) = 0 and that g is monotone increasing. We now use the Second mean value Theorem to prove
$\frac{1}{2\pi} \int_{0}^{\pi} g(y) \, D_N (y)\,{\text d} y \,\to 0 \qquad\mbox{as} \quad N \to \infty .$
As g(0+0) = 0, for every positive ε there exists δ > 0 such that g(x) < ε whenever 0 ≤ x <. Then δ. Then
$\frac{1}{2\pi} \int_{0}^{\pi} g(y)\, D_N (y)\,{\text d}y = \underbrace{\frac{1}{2\pi} \int_{0}^{\delta} g(y)\, D_N (y)\,{\text d}y}_{= I_1} + \underbrace{\frac{1}{2\pi} \int_{\delta}^{\pi} g(y)\, D_N (y)\,{\text d}y}_{= I_2} .$
Now
$I_2 = \frac{1}{2\pi} \int_{\delta}^{\pi} \frac{g(y)}{\sin (y/2)}\,\sin \left( N + \tfrac{1}{2} \right) y\, {\text d} y = \underbrace{\frac{1}{2\pi} \int_{0}^{\pi} \frac{g(y)}{\sin (y/2)}\,\chi_{[\delta , \pi ]} (y)}_{ \in 𝔏¹} \cdot \sin \left( N + \tfrac{1}{2} \right) y\, {\text d} y\,\to 0 \qquad\mbox{as} \quad N \to \infty$
by the Riemann–Lebesgue lemma. By the mean value formula, taking h = g and φ = DN, we have that there exists a constant C ∈ (0, δ) such that
\begin{align*} I_1 &= \frac{1}{2\pi} \int_{0}^{\delta} g(y)\,D_N (y)\,{\text d} y = g(\delta -0) \,\frac{1}{2\pi} \int_{C}^{\delta} D_N (y)\,{\text d} y \\ &\le \frac{\varepsilon}{2\pi}\,\sup_{C, \delta , N} \int_{C}^{\delta} D_N (y)\,{\text d} y . \end{align*}
As long as the sup is finite, we can send ε → 0 and we are done, so
\begin{align*} \left\vert \int_{0}^{\delta} D_N (y)\,{\text d} y \right\vert &= \left\vert \int_{C}^{\delta} \sin \left( N + \tfrac{1}{2} \right) y \left[ \frac{1}{\sin (y/2)} - \frac{1}{y/2}\right] {\text d} y \right\vert + \left\vert \frac{\sin \left( N + \tfrac{1}{2} \right) y}{y/2}\,{\text d} y \right\vert \\ & \le K + 2\,\sup \left\vert \int_0^{\infty} \frac{\sin t}{t}\,{\text d}t \right\vert = K + \pi \end{align*}
because
$\int_C^{\delta} \frac{\sin y}{y}\,{\text d}y = \int_{C \left( N + \tfrac{1}{2} \right)}^{\delta \left( N + \tfrac{1}{2} \right)} \frac{\sin t}{t}\,{\text d}t \le \int{0}^{\infty} \frac{\sin t}{t}\,{\text d}t = \frac{\pi}{2} .$
Note that a constant K is independenyt on C, N, and δ. This completes the proof

The following theorem was proved in 1880 by an Italian mathematician Ulisse Dini (1845--1918).

Theorem 6 (Dini's criterion): Suppose that a function f(x) ∈ 𝔏[−ℓ, ℓ] is absolutely integrable, and let ψ(t) = f(x + t) + f(xt) −2f(x) for a fixed x. If the integral
$\int_0^{\delta} \left\vert \frac{\psi (t)}{t} \right\vert {\text d} t < \infty$
converges, then Fourier series of f(x) converges at x to f(x)

Theorem 2 (Dini’s convergence theorem): Suppose that for a periodic absolutely integrable function f(x) there exists a positive constant δ such that
$\int_{-\delta}^{\delta} \left\vert \frac{f(x-t) - f(x)}{t} \right\vert {\text d}t < + \infty .$
Then the Fourier series S[f](x) converges to f(x) for every x from the whole interval.
Recall that if f has a continuous derivative, then
$\lim_{t\to 0} \left\vert \frac{f(x-t) - f(x)}{t} \right\vert$
exists and is bounded for all x; in this case, taking δ = π,
$\int_{|t| < \pi} \left\vert \frac{f(x-t) - f(x)}{t} \right\vert {\text d}t \le 2\pi \| f' \|_{\infty} = 2\pi \max | f' (x) | < +\infty .$

For arbitrary integrable function f, we compute

\begin{align*} \left\vert S_N (f; x) - f(x) \right\vert &= \frac{1}{2\pi} \left\vert \int f(x-y)\,D_N (y) \,{\text d}y - \int f(x)\, D_N (y)\,{\text d} y \right\vert \\ &= \frac{1}{2\pi} \left\vert \int \frac{f(x-y) - f(x)}{|y|} \,\sin \left( N + \frac{1}{2} \right) y \cdot \frac{|y|}{\sin (y/2)}\,{\text d} y \right\vert \\ &= \frac{1}{2\pi} \left\vert \frac{1}{2{\bf j}}\int_{-\pi}^{\pi} \frac{f(x-y) - f(x)}{|y|} \frac{|y|}{\sin (y/2)} \cdot \left( e^{{\bf j}y/2} e^{{\bf j} Ny} - e^{-{\bf j}y/2} e^{-{\bf j} Ny} \right) {\text d} y \right\vert \\ &= \frac{1}{2\pi} \cdot \frac{1}{2} \left\vert \int_{-\pi}^{\pi} \, \underbrace{\frac{f(x-y) - f(x)}{|y|} \frac{|y|}{\sin (y/2)} \cdot e^{{\bf j}y/2}}_{= g(y)} e^{{\bf j} Ny} {\text d} y - \int_{-\pi}^{\pi} \, \underbrace{\frac{f(x-y) - f(x)}{|y|} \frac{|y|}{\sin (y/2)} \cdot e^{-{\bf j}y/2}}_{= h(y)} e^{-{\bf j} Ny} {\text d} y \right\vert . \end{align*}
To complete the proof, it suffices to show that g, h ∈ 𝔏¹[−ℓ, ℓ]. For g, we need to show that
$\int_{-\pi}^{\pi} \frac{|f(x-y) - f(x)|}{|y|} \frac{|y|}{|\sin (y/2)|} \cdot \left\vert e^{{\bf j}y/2} \right\vert {\text d} y < + \infty .$
We split the integral into the regions where |y| < δ and δ < |y| < 2π. For latter domain of integration, we have $$\frac{1}{|\sin (y/2) |} \le M$$ for some positive constant M. Then
$\int_{\delta < |y| < \pi} \frac{|f(x-y) - f(x)|}{|\sin (y/2)|} \left\vert e^{{\bf j}y/2} \right\vert {\text d}y \le M \int_{\delta < |y| < \pi} |f(x-y) - f(x)| \, {\text d}y \le 2M\, \| f \|_1 .$
For domain |y| < δ, we observe that    $$\frac{|y|}{|\sin (y/2)|} \le 2 ,$$ so that
$\int_{|y| < \delta} \frac{|f(x-y) - f(x)|}{|y|} \frac{|y|}{|\sin (y/2)|} \cdot \left\vert e^{{\bf j}y/2} \right\vert {\text d} y \le \int_{|y| < \delta} \frac{|f(x-y) - f(x)|}{|y|} \cdot 2 \,{\text d} y < + \infty$
by assumprion.

h because |h| = |g|.
Observe that for Dini’s theorem to hold, it is in fact enough to have that there existconstants H > 0 and α ∈ (0, 1] such that that whenever |y ≤ 2ℓ, we have
$| f(x-y) - f(y)| \le H \left\vert y \right\vert^{\alpha} .$
Such functions are called α-Hölder continuous. Note that if function f has a continuous derivative, then f is automatically 1-Hölder (also known as Lipschitz).

# Uniform Convergence of Fourier Series

It is well known that the Fourier series of a continuous function is not necessarily uniformly convergent in an interval of continuity of the sum-function. However, if, in addition to being continuous, the function is also of bounded variation then it is known that the Fourier series is uniformly convergent in the interval of periodicity. This is the so-called Jordan criterion for uniform convergence,

A second criterion for uniform convergence is the Dini- Lipschitz condition. If for 0 < |t| < 1 $| f(x+t) - f(x) | < \frac{K}{\left( -\log |t| \right)^{1+\alpha}} , \qquad K, \alpha > 0 ,$ then the Fourier series of f ( x ) converges uniformly.

Corollary: Let f(x) be continuous and periodic in [−ℓ, ℓ] and have Fourier coefficients 𝑎k, bk according to Eq.\eqref{EqFourier.5T}. If $$\displaystyle \sum_{k\ge 1} \left( \left\vert a_k \right\vert + \left\vert b_k \right\vert \right)$$ converges, then the Fourier series of f converges absolutely and uniformly to f(x), x∈[−ℓ, ℓ]. Similarly, if $$\displaystyle \sum_{\nu =-\infty}^{+\infty} \left\vert \hat{f}(\nu ) \right\vert < \infty ,$$ then the Fourier series S[f] converges absolutely and uniformly.

The smoothness of the function drastically affects the speed of convergence of the Fourier series: the smoother the function the more rapid is the decrease of the Fourier coefficients. The study of convergence of Fourier series is largely the study of the interplay between assumptions of smoothness and conditions about convergence.

Theorem 7: Let f(x) ∈ Cn[−ℓ, ℓ], that it has n continuous derivatives, and have period 2ℓ, that is, f(−ℓ) = f(ℓ), f'(−ℓ) = f'f(n)(−ℓ) = f(n)(ℓ). Suppose further that its n-th derivative is a Lipschitz function of order α, 0 < α ≤ 1. The the Fourier coefficients of f satisfy the inequalities
$\left\vert a_k \right\vert , \quad \left\vert b_k \right\vert \le \frac{\mbox{constant}}{k^{n+\alpha}} , \qquad k=1,2,\ldots ,$
and the Fourier series converges uniformly to f(x), x ∈ [−ℓ, ℓ].
Theorem 2: Suppose f is a periodic function of period 2ℓ that belong to the class Cm. Then its Fourier coefficients
$\alpha_n (f) = \frac{1}{2\ell} \int_{-\ell}^{\ell} f(x) \,e^{-n{\bf j} \pi x/\ell} {\text d} x = O \left( \frac{1}{|n|^m} \right) \qquad\mbox{as} \quad |n| \to \infty .$
This notation means that there exists a constant K such that |αn| ≤ K/|n|m.
Let f be a periodic function on [−ℓ, ℓ], let t be some point and let δ be a positive number. We define the local modulus of continuity at the point t by
$\omega_f (\delta ) = \max_{|\varepsilon | \le \delta} \left\vert f(t) - f(t+\varepsilon ) \right\vert .$
Theorem 2 (Dini's test): Suppose that f is continuous in a closed interval [𝑎, b] and let ω(δ) be its modulus of continuity there. If ω(δ)/δ is integrable near δ = 0, and if the integrals
$\int_0^{\pi} \frac{\left\vert f(a) - f(a-t) \right\vert}{t}\,{\text d}t , \qquad \int_0^{\pi} \frac{\left\vert f(b+t) - f(b) \right\vert}{t}\,{\text d}t$
are finite, then the Fourier series S[f] converges uniformly to f(x) in [𝑎, b].
In particular, any function of a α-Hölder class for α > ½, has a uniform convergent Fourier series.

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