DIVERGENCE OF THE FOURIER SERIES OF CONTINUOUS FUNCTIONS WITH A RESTRICTION ON THE FRACTALITY OF THEIR GRAPHS

We consider certain classes of functions with a restriction on the fractality of their graphs. Modifying Lebesgue’s example, we construct continuous functions from these classes whose Fourier series diverge at one point, i.e. the Fourier series of continuous functions from this classes do not converge everywhere.

Let f be a 2π-periodic integrable function, and let where a k = 1 π π −π f (t) cos kt dt, b k = 1 π π −π f (t) sin kt dt, be the trigonometric Fourier series of the function f . Denote by S n (f, x) the nth partial sum of (1). It is known (see [1,Ch. 1,Sect. 39]) that if f has bounded variation on the period (f ∈ BV ), then its Fourier series converges everywhere on R, and if, in addition, f is continuous on R, then the Fourier series converges to f uniformly on R. Salem [2] (see also [1,Ch. 4, Sect. 5]) considered the classes BV p of functions of bounded p-variation and proved that if f ∈ BV p , then the Fourier series of f also converges everywhere on R. (Further generalizations of these results see in [3]). The author [4] studied relations between the classes BV p and classes of continuous functions with a restriction on the fractality of their graphs.
Definition 1. Let f : R → R be a bounded 2π-periodic function. By the modulus of fractality of the function f , we call the function ν(f, ε) which, for all ε > 0, gives the minimal number of closed squares with sides of length ε parallel to the coordinate axes that cover the graph of the function f on [−π, π].
Then there exists a continuous function F µ whose Fourier series does not converge everywhere.
P r o o f of Theorem 1. We will require that By (3), the former inequality holds on an interval (0, δ) and, changing the function µ on the interval ( δ 2 , 1), we will obtain the same class F µ . The latter inequality can only reduce the class F µ . Thus, if we find a required function in the narrower class, it will belong to the wider class immediately.
To obtain a function f ∈ F µ with divergent Fourier series, we modify Lebesgue's example from [1, Ch. 1, Sect. 46]. We start with defining an increasing sequence of natural numbers {a k } as follows. Let a 0 = 1. Suppose that the first k elements a 0 , a 1 , . . . , a k−1 have been already defined.
From inequalities (4), it follows that Then, by continuity, there exists the smallest number a such that As a k , we take the largest integer such that a k a and the fraction a k /a k−1 is integer. It is not hard to understand that a k belongs to [a − a k−1 , a], and, in view of the inequalities we conclude that a k > a k−1 .
The definition of a k implies the inequality The definition of a k , inequalities (5), and condition (3) imply that a k a k−1 → +∞, k → +∞.
Consider the half-open intervals Let {k i } ∞ i=0 , k 0 = 1, be an increasing sequence, on which, in what follows, two additional conditions will be imposed. Let . Finally, we define the function f on the interval [−π, π]: We extend the function f to R periodically. The resulting function is continuous on each I k and, since a k /a k−1 is integer, is continuous and vanishes at the points ±π/a k . Thus, the function f is continuous on [−π, π].
Consider now the sequence of partial sums of the Fourier series of f at the point x = 0. As is known [1, Ch. 1, Sect. 32, formula (32.5)], for the function f, we have hence, for x = 0, The function f is even; therefore, Let us show that, after an appropriate choice of {k i }, Then S a k i (0, f ) → +∞ as i → +∞, i.e., the Fourier series of f diverges at x = 0.
To estimate J i , we divide it into three terms: We have sin a k i t t a k i .

Hence, |J
Suppose that k 1 , . . . , k i−1 have been already defined. Then the function f (t)/t is defined, bounded, and continuous on π/a k i −1 , π . Extending this function by zero to [−π, π] and assuming that k i are large enough (this is the first of two conditions on k i ), we can make the Fourier coefficient a k i of the obtained function small enough; more precisely, It remains to estimate J ′′ i . We have According to the second mean value theorem, taking into account that the function 1/t is positive and monotone, we find that Combining (8), (9), (10), and (11), and taking into account (7), we conclude that Let us now estimate the modulus of fractality ν(f, ε). Denote by ν(f, ε) [a,b] the minimal number of squares with sides of length ε parallel to the coordinate axes that cover the graph of the function f on [a, b].