GEORGIAN MATHEMATICAL JOURNAL: Vol. 4, No. 4, 1997, 333-340

ON THE ABSOLUTE CONVERGENCE OF FOURIER SERIES

T. KARCHAVA

Abstract. The necessary and sufficient conditions of the absolute convergence of a trigonometric Fourier series are established for continuous 2π-periodic functions which in [0, 2π] have a finite number of intervals of convexity, and whose nth Fourier coefficients are O(ω(1/n; f)/n), where ω(δ; f) is the continuity modulus of the function f.

Let ω be an arbitrary modulus of continuity, i.e., a nondecreasing function continuous on [0, 1], ω(0) = 0 and ω(δ_{1} + δ_{2}) ≤ ω(δ_{1}) + ω(δ_{2}). As usual, denote by H^{ω} the class of all functions f continuous on [0, 2π] for which

ω(δ; f) = sup_{|x_{1}−x_{2}|≤δ} |f(x_{1}) − f(x_{2})| = O(ω(δ)), 0 ≤ δ ≤ 1

(see, for instance, [5, Ch. 3, pp. 150, 157]).

Let M be the class of all continuous 2π-periodic functions f for which there exists a partitioning of the segment [0, 2π] by the points 0 = x_{1}(f) < ⋯ < x_{m+1}(f) = 2π such that f is convex or concave on each segment [x_{k}(f), x_{x+1}(f)], k = 1, … m.

The Fourier coefficients of a function f with respect to the trigonometric system will be denoted by a_{n} = a_{n}(f), b_{n} = b_{n}(f).

Problems pertaining to the absolute convergence of Fourier series have been studied quite completely (see, for instance, the monographs of Bari [2, Ch. 9], Zygmund [3, Ch. 6], Kahane [4, Ch. 2], and the survey by Guter and Ulyanov [5, p. 391]).

This paper deals with some problems of the absolute convergence of trigonometric Fourier series of a function from the class M.

The following facts are well known:

333

1991 Mathematics Subject Classification. 42A28, 42A16.

Key words and phrases. Periodic funcctions with a finite number of intervals of convexity, absolute convergence of Fourier series, estimates of Fourier coefficients.

1072-947X/97/0700-0333$12.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1997 Plenum Publishing Corporation

(1) The Fourier series of any 2π-periodic continuous even function, convex on [0, 2π], converges absolutely ([4, Ch. 2]).

(2) Let f be an odd function convex on [0, +∞). Then f ∈ A^{loc} if and only if ∫_{0}^{1} f(t)\frac{dt}{t} < ∞ [4, Ch. 2]. Here A^{loc} is the set of all functions f, continuous on (−∞, +∞), for which every point can be encircled by an interval on which f = g, where g is a function, continuous on [0, 2π], whose Fourier series converges absolutely.

We have obtained the following results:

Theorem 1. If f ∈ M, then for the absolute convergence of the Fourier series of the function f it is necessary and sufficient that

∑_{n=1}^{∞} |f(x_{k}(f) + \frac{1}{n}) − f(x_{k}(f) − \frac{1}{n})|\frac{1}{n} < +∞, k = 1, …, m.

Theorem 2.

(a) Let f ∈ M; then

a_{n}(f) = O(ω(\frac{1}{n}; f)\frac{1}{n}), b_{n}(f) = O(ω(\frac{1}{n}; f)\frac{1}{n}).

(b) If ∑_{n=1}^{∞} ω(\frac{1}{n})\frac{1}{n} = +∞, then in the class H^{ω} ∩ M there exists a a function whose Fourier series does not converge absolutely.

Theorem 3. Let f ∈ M and at least one of the following conditions be fulfilled:

(1) for any adjacent intervals (x_{k}(f), x_{k+1}(f)) and (x_{k+1}(f), x_{k+2}(f)), the function f is convex on one of them and concave on the other;

(2) each point x_{k}(f) can be encircled by an interval where the function f is monotonous;

(3) for any x_{k}(f), at least one of the two series

∑_{n=1}^{∞} |f(x_{k}(f) + \frac{1}{n}) − f(x_{k}(f))|\frac{1}{n} and ∑_{n=1}^{∞} |f(x_{k}(f) − \frac{1}{n}) − f(x_{k}(f))|\frac{1}{n}

converges.

Then the convergence of the series ∑_{n=1}^{∞} ω(\frac{1}{n}; f)\frac{1}{n} is the necessary and sufficient condition for the Fourier series of the function f to converge absolutely.

Proof of Theorem 1. Let f_{1}, f_{2}, f be continuous 2π-periodic functions defined as follows: f_{1} is convex or concave on a segment [0, π], f_{1}(0) = f_{1}(π) = 0, linear on [1, π], f_{1}(x) = 0, for x ∈ [−π, 0]; f_{2} is convex or concave on [−π, 0], f_{2}(−π) = f_{2}(0) = 0, linear on [−π, −1], f_{1}(x) = 0 for x ∈ [0, π]; f = f_{1} + f_{2}.

The theorem will be proved by showing that for the Fourier series of f to converge it is necessary and sufficient that

∑_{n=1}^{∞} |f(\frac{1}{n}) − f( − \frac{1}{n})|\frac{1}{n} < +∞.

This follows from Wiener’s theorem and from the following facts: If the function f is convex or concave on a segment [a, b], then f ∈ Lip 1 on any segment [c, d] entirely lying inside [a, b], and the Fourier series of the functions f(x) and f(x + c) simultaneously converge or diverge absolutely.

The function f_{1} is convex on [0, π] and continuous, which means that it is absolutely continuous so that one can apply integration by parts and the Newton–Leibniz formulas to obtain a_{n}(f) = a_{n}(f_{1}) + a_{n}(f_{2}).

a_{n}(f_{1}) = \frac{1}{π} ∫_{0}^{2π}f_{1}(t) cos nt dt = \frac{1}{π} ∫_{0}^{2π}f_{1}(t)d\frac{sin nt}{n} = \frac{1}{π} (f_{1}(t)\frac{sin nt}{n})_{0}^{2π} −

−\frac{1}{πn} ∫_{0}^{2π}f′_{1}(t) sin nt dt = \frac{−1}{πn} ∫_{0}^{2π}f′_{1}(t) sin nt dt = \frac{−1}{πn} ∫_{0}^{π} f′_{1}(t) sin nt dt =

= −\frac{1}{πn} ∫_{0}^{1/n}f′_{1}(t) sin nt dt − \frac{1}{πn} ∫_{1/n}^{1} f′_{1}(t) sin nt dt − \frac{1}{πn} ∫_{1}^{π} f′_{1}(t) sin nt dt.

The derivative f′ of the convex or concave function f is monotonous and therefore, applying the second theorem of the mean value, we obtain

|\frac{1}{πn} ∫_{1/n}^{1} f′_{1}(t) sin nt dt| = |\frac{1}{πn}f′_{1}(\frac{1}{n} + 0) ∫_{1/n}^{ξ} sin nt +

+\frac{1}{πn}f′_{1}(1 − 0) ∫_{ξ}^{1} sin nt dt| ≤ \frac{1}{πn^{2}} |f′_{1}(\frac{1}{n} + 0)| + \frac{1}{πn^{2}} |f′_{1}(1 − 0)|

with 1/n < ξ < 1.

Wherever we come across expressions of the form f′(x ± 0), the left and right limits are considered with respect to the set at whose points the derivative f′ exists.

For the convex (concave) function f we have the relation

\frac{f(x_{2}) − f(x_{1})}{x_{2} − x_{1}} ≥ f′(x_{2} ± 0) ≥ \frac{f(x_{3}) − f(x_{2})}{x_{3} − x_{2}}

(\frac{f(x_{2}) − f(x_{1})}{x_{2} − x_{1}} ≤ f′(x_{2} ± 0) ≤ \frac{f(x_{3}) − f(x_{2})}{x_{3} − x_{2}})

where x_{1} < x_{2} < x_{3}. Therefore

|f′_{1}(\frac{1}{n} + 0)| ≤ \frac{f_{1}(\frac{1}{n}) − f_{1}(\frac{1}{n+1})}{1/n − 1/(n + 1)} ≤ (n + 1)^{2}(f_{1}(\frac{1}{n}) − f_{1}(\frac{1}{n + 1})).

Hence

∑_{n=1}^{∞} |f′_{1}(\frac{1}{n} + 0)|\frac{1}{n^{2}} ≤ 2 ∑_{n=1}^{∞} (f_{1}(\frac{1}{n}) − f_{1}(\frac{1}{n + 1})) < +∞.

Since f_{1} ∈ Lip 1 on the segment [ε, π], we have |f′_{1}(1 − 0)| ≤ M and ∑_{n=1}^{∞} |f′_{1}(1 − 0)|/n^{2} ≤ ∑_{n=1}^{∞} M/n^{2} < +∞.

The function f_{1} is linear on the segment [1, π], i.e., f′_{1}(t) = cos nt = c, so that 1/n| ∫_{1}^{π} f′_{1}(t) sin nt| ≤ c/n^{2}.

Finally, a_{n}(f_{1}) = −\frac{1}{πn} ∫_{0}^{1/n} f′_{1}(t) sin nt dt + γ_{n}, where ∑_{n=1}^{∞} |γ_{n}| < +∞. If we introduce the notation I_{n} = \frac{−1}{πn} ∫_{0}^{1/n} f′_{1}(t) sin nt dt, then a_{n}(f_{1}) =

I_{n} + γ_{n}, I_{n} = a_{n}(f_{1}) − γ_{n}.

Since the function f_{1} has a bounded variation, we have

f_{1}(x) = \frac{a_{n}(f_{1})}{2} + ∑_{n=1}^{∞} a_{n}(f_{1}) cos nx + b_{n}(f_{1}) sin nx.

By substituting here x = 0 we obtain ∑_{n=1}^{∞} a_{n}(f_{1}) < ∞. Therefore ∑_{n=1}^{∞} I_{n} = ∑_{n=1}^{∞}(a_{n}(f_{1}) − γ_{n}) < ∞.

One can easily verify that the values I_{n} do not change their sign for sufficiently large n. Thus ∑_{n=1}^{∞} |I_{n}| < +∞. Since |a_{n}(f_{1})| ≤ |I_{n}| + |γ_{n}|, we obtain ∑_{n=1}^{∞} |a_{n}(f_{1})| < +∞.

In a similar manner we shall show that ∑_{n=1}^{∞} |a_{n}(f_{2})| < ∞. We have |a_{n}(f)| = |a_{n}(f_{1}) + a_{n}(f_{2})| ≤ |a_{n}(f_{1})| + |a_{n}(f_{2})| and ∑_{n=1}^{∞} |a_{n}(f)| < +∞.

Now consider the coefficients b_{n}(f). We have b_{n}(f) = b_{n}(f_{1}) + b_{n}(f_{2}),

b_{n}(f_{1})= \frac{1}{π} ∫_{0}^{2π}f_{1}(t) sin nt dt= \frac{1}{π} ∫_{0}^{2π}f_{1}(t)d\frac{cos nt}{n} = \frac{1}{π}(f_{1}(t)\frac{cos nt}{n})|_{0}^{2π} −

−\frac{1}{πn}∫_{0}^{2π}f′_{1}(t) cos nt dt=−\frac{1}{πn}∫_{0}^{2π}f′_{1}(t) cos nt dt=−\frac{1}{πn}∫_{0}^{π}f′_{1}(t) cos nt dt=

= −\frac{1}{πn} ∫_{0}^{1/n}f′_{1}(t) cos nt dt− \frac{1}{πn} ∫_{1/n}^{1} f′_{1}(t) cos nt dt− \frac{1}{πn}∫_{1}^{π}f′_{1}(t) cos nt dt.

ON THE ABSOLUTE CONVERGENCE OF FOURIER SERIES 337

The function f_{1} is linear on the segment [1, π], i.e., f′_{1}(t) = const = C, so that

\frac{1}{n}| ∫_{1}^{π} f′_{1}(t) cos nt dt| ≤ \frac{C}{n^{2}} .

Again applying the theorem of the mean, we obtain (with 1/n < ξ < 1)

|\frac{1}{n} ∫_{1/n}^{1} f′_{1}(t) cos nt dt|= \frac{1}{n}|f′_{1}(\frac{1}{n} + 0) ∫_{1/n}^{ξ} cos nt dt +

+f′_{1}(1 − 0)∫_{ξ}^{1} cos nt dt|≤ \frac{1}{n^{2}} |f′_{1}(\frac{1}{n} + 0)| + \frac{1}{n^{2}} |f′_{1}(1 − 0)| < +∞.

Therefore b_{n}(f_{1}) = −\frac{1}{πn} ∫_{0}^{1/n} f′_{1}(t) cos nt dt + γ_{n}, ∑_{n=1}^{∞} |γ_{n}| < +∞.

−\frac{1}{πn} ∫_{0}^{1/n}f′_{1}(t) cos nt dt = \frac{1}{πn} ∫_{0}^{1/n}f′_{1}(t)(1 − cos nt − 1)dt =

= \frac{−1}{πn} ∫_{0}^{1/n}f′_{1}(t)dt + \frac{1}{πn} ∫_{0}^{1/n}f′_{1}(t)(1 − cos nt)dt =

= \frac{−1}{πn}f_{1}(\frac{1}{n}) + \frac{1}{πn} ∫_{0}^{1/n}f′_{1}(t)2 sin^{2} \frac{nt}{2}dt,

|\frac{1}{πn} ∫_{0}^{1/n}f′_{1}(t)2 sin^{2} \frac{nt}{2}dt| ≤ \frac{2}{πn} ∫_{0}^{1/n}|f′_{1}(t)| | sin nt|dt = 2|I_{n}|.

As we have seen above, ∑_{n=1}^{∞} |I_{n}| < +∞ and therefore

b_{n}(f_{1}) = −\frac{1}{πn}f_{1}(\frac{1}{n}) + C_{n} = \frac{−1}{πn}f(\frac{1}{n}) + C_{n},

where ∑_{n=1}^{∞} |C_{n}| < +∞. In a similar manner it will be shown that

b_{n}(f_{2}) = \frac{1}{πn}f( − \frac{1}{n}) + P_{n}, where ∑_{n=1}^{∞} |P_{n}| < +∞.

Since b_{n}(f) = b_{n}(f_{1}) + b_{n}(f_{2}), we have

b_{n}(f) = \frac{−1}{πn}\{f(\frac{1}{n}) − f(\frac{−1}{n})\} + γ_{n}, ∑_{n=1}^{∞} |γ_{n}| < +∞.

Proof of Theorem 2.

(a) It is the well-known fact that estimates of Fourier coefficients can be derived using the integral modulus of continuity (see, for instance [3, Ch. 2])

|a_{n}| ≤ sup_{|h|≤ \frac{1}{n}} \frac{1}{π} ∫_{0}^{2π}|f(x + h) + f(x − h) − 2f(x)|dx.

Applying the above inequality, we obtain

|a_{n}| ≤ sup_{|h|≤ \frac{1}{n}} \frac{1}{π} ∫_{0}^{2π}|f(x + h) + f(x − h) − 2f(x)|dx =

= sup_{|h|≤ \frac{1}{n}} \frac{1}{π} ∫_{0}^{2π}|f(x + kh) + f(x + (k − 2)h) − 2f(x + (k − 1)h)|dx =

= \frac{1}{πn} sup_{|h|≤ \frac{1}{n}} ∫_{0}^{2π}∑_{k=1}^{n} |f(x + kh) + f(x + (k − 2)h) −

−2f(x + (k − 1)h)|dx = \frac{1}{πn} sup_{|h|≤ \frac{1}{n}} ∫_{0}^{2π}( ∑_{k=1}^{n} |u_{k} − u_{k−1}|)dx,

where u_{k} = f(x + kh) − f(x + (k − 1)h).

Convexity (concavity) of a function f on some segment [a, b] implies that

f(x+kh)+f(x−(k−2)h)−2f(x+(k−1)h) = u_{k}−u_{k−1} ≤ 0 (u_{k}−u_{k−1} ≥ 0)

for x+(k−2)h, x+(k−1)h, x+kh ∈ [a, b]. Therefore on the segment [0, 2π] the values u_{k} − u_{k−1} change their sign a finite number of times. Thus

|a_{n}(f)| ≤ \frac{1}{πn} sup_{|h|≤ \frac{1}{n}} ∫_{0}^{2π}( ∑_{k=1}^{n} |u_{k} − u_{k−1}|)dx ≤

≤ \frac{1}{πn} sup_{|h|≤ \frac{1}{n}} ∫_{0}^{2π}| ∑_{k=1}^{n} u_{k} − u_{k−1}|dx + \frac{C(f)}{n}ω(\frac{1}{n}, f) ≤ \frac{C_{1}(f)}{n}ω(\frac{1}{n}, f).

The proof of the estimate for b_{n}(f) is similar.

(b) Let ∑_{n=1}^{∞} ω(\frac{1}{n})\frac{1}{n} = +∞. By Stechkin’s lemma (see [6]) there exists a convex modulus of continuity ω′(δ) such that H^{ω} = H^{ω′}. Hence ω(δ) can be regarded as convex function.

Consider a continuous function

f_{0}(t) = \begin{cases}ω(t), & t ∈ [0, 1], \\ linear for & [t ∈ [1, 2π], f_{0}(t + 2π) = f_{0}(t). \end{cases}

Clearly, f_{0} ∈ H^{ω} ∩ M.

On the interval [1, 2π] the function f_{0}(t) is linear, f(2π) = f(0) = 0. Therefore f_{0}( − \frac{1}{n}) = f(2π − \frac{1}{n}) = \frac{c_{0}}{n}, where c_{0} is some number . We have

∑_{n=1}^{∞} |f_{0}(\frac{1}{n}) − f_{0}(\frac{−1}{n})|\frac{1}{n} ≥ ∑_{n=1}^{∞} |f_{0}(\frac{1}{n})| − |f_{0}(\frac{−1}{n})|\frac{1}{n} =

= ∑_{n=1}^{∞} |f_{0}(\frac{1}{n})|\frac{1}{n} − ∑_{n=1}^{∞} |f_{0}( − \frac{1}{n})|\frac{1}{n} = ∑_{n=1}^{∞} ω(\frac{1}{n})\frac{1}{n} − ∑_{n=1}^{∞} \frac{C_{0}}{n^{2}} = +∞.

By virtue of Theorem 1 we see that the necessary condition for Fourier series to be absolutely convergent is not fulfilled. The Fourier series of f_{0} does not converge absolutely.

Proof of Theorem 3. The sufficiency follows from part (a) of the proof of Theorem 2.

First note that if the function is convex (concave) on [a, b], then its modulus of continuity on [a, b] equals

max {|f(a + δ) − f(a)|, |f(b − δ) − f(b)|}.

Therefore

ω(1/n, f) ≤ max_{1≤k≤m} {|f(x_{k} + 1/n) − f(x_{k})| + |f(x_{k} − 1/n) − f(x_{k})|},

where x_{k} ≡ x_{k}(f).

Hence it follows that if ∑_{n=1}^{∞} ω(\frac{1}{n}, f)\frac{1}{n} = +∞, then for some x_{k} we have

∑_{n=1}^{∞} |f(x_{k} + \frac{1}{n}) − f(x_{k})|\frac{1}{n} = +∞ or ∑_{n=1}^{∞} |f(x_{k} − \frac{1}{n}) − f(x_{k})|\frac{1}{n} = +∞.

For convenience we assume that

∑_{n=1}^{∞} |f(x_{k} + \frac{1}{n}) − f(x_{k})|\frac{1}{n} = +∞.

If the conditions (2) are fulfilled, then

∑_{n=1}^{∞} |f(x_{k} + \frac{1}{n}) − f(x_{k} − \frac{1}{n})|\frac{1}{n} ≥ ∑_{n=1}^{∞} |f(x_{k} + \frac{1}{n}) − f(x_{k})|\frac{1}{n} = +∞,

i.e., by Theorem 1 the Fourier series of the function f is not absolutely convergent.

If the conditions (3) are fulfilled, then

∑_{n=1}^{∞} |f(x_{k} + \frac{1}{n}) − f(x_{k} − \frac{1}{n})|\frac{1}{n} =

= ∑_{n=1}^{∞} |f(x_{k} + \frac{1}{n}) − f(x_{k}) + f(x_{k}) − f(x_{k} − \frac{1}{n})|\frac{1}{n} ≥

≥ ∑_{n=1}^{∞} |f(x_{k} + \frac{1}{n}) − f(x_{k})|\frac{1}{n} − ∑_{n=1}^{∞} |f(x_{k} − \frac{1}{n}) − f(x_{k})|\frac{1}{n} = +∞.

which proves the theorem under conditions (3).

If , however, the conditions of (1) are fulfilled, then, as one can easily verify, the conditions of (2) or (3) are fulfilled too.

References

1. V. K. Dziadik, Introduction to the theory of uniform approximation of function by polynomials. (Russian) Nauka, Moscow, 1977.

2. N. K. Bari, Trigonometric series. (Russian) Gos. Isd. Fiz-Mat. Lit., Moscow, 1961.

3. A. Zygmund, Trigonometric series, v. 1. Cambridge University Press, 1959; Russian translation: Mir, Moscow, 1965.

4. J.-P. Kahane, Séries de Fourier absolument convergentes. Springer-Verlag, Berlin, 1970; Russian translation: Mir, Moscow, 1976.

5. R. S. Guter and P. L. Ulyanov, On the new results in the theory of orthogonal series. (Russian) Supplement to the translation from German into Russian of the book S. Caczmarz and H. Steinhaus, Theorie der Orthogonalreihen (Warszawa–Lwow, 1935): Gos. Izd. Fiz.-Mat. Lit., Moscow, 1958.

6. A. V. Efimov, Linear methods of approximation of continuous periodic functions. (Russian) Mat. Sbornik 54(1961), No. 1, 51–90.

(Received 02.06.1995)

Authors’ address: Faculty of Physics and Mathematics Sukhumi Branch of I. Javakhishvili Tbilisi State University 1, I. Chavchavadze ave., Tbilisi 380028 Georgia