ON THE RATE OF CONVERGENCE FOR THE CHEBYSHEV SERIES

Kamel Al–Khaled

Abstract. Let f(x) be a function of bounded variation on [−1, 1] and S_{n}(f; x) the nth partial sum of the expansion of f(x) in a Chebyshev series of the second kind. In this note we give the estimate for the rate of convergence of the sequence S_{n}(f; x) to f(x) in terms of the modulus of continuity of the total variation of f(x).

1. Introduction. Let U_{n}(x) be the Chebyshev polynomial of the second kind [4]. Let f(x) be a function of bounded variation on [−1, 1] and S_{n}(f; x) the nth partial sum of the expansion of f(x) in a Chebyshev series of the second kind:

∑_{n=0}^{∞} a_{n}U_{n}(x)

with

a_{n} = \frac{2}{π} ∫_{−1}^{1}(1 − y^{2})^{1/2}f(y)\frac{sin(n + 1) arccos y}{sin arccos y} dy, (n = 0, 1, …).  (1.1)

According to the equiconvergence theorem for Jacobi series [4], we know that

lim_{n→∞} S_{n}(f; x) = \frac{1}{2}(f(x + 0) + f(x − 0)), x ∈ (−1, 1).

In this note we shall find an estimate for the rate of convergence of the sequence S_{n}(f; x) to f(x). Results of this type for Fourier series of 2π–periodic functions of bounded variation were proved by Bojanic [2].

2. Preliminary Results. Before proving the main theorem we shall state a preliminary result. Al–Khaled [1] has studied the behavior of Chebyshev series for functions of bounded variation on [−1, 1] and he proved the following Theorem.

\underline{Theorem 2.1}. If f(x) is a function of bounded variation on [−1, 1]. Let

A_{x}(y) = \{ \begin{matrix}f(y) − f(x − 0), & −1 ≤ y < x \\ 0, & y = x \\ f(y) − f(x + 0), & x < y ≤ 1. \end{matrix}

Then for every x ∈ (−1, 1) and n ≥ 2 we have

|S_{n}(f; x) − \frac{1}{2}(f(x + 0) + f(x − 0))| ≤ \frac{9}{n\sqrt{1 − x^{2}}} [\frac{1}{1 + x} ∑_{k=1}^{n} V_{x−(1+x)/k}^{x}(A_{x})

+ \frac{1}{1 − x} ∑_{k=1}^{n} V_{x}^{x+(1−x)/k}(A_{x})] + \frac{4}{nπ\sqrt{1 − x^{2}}} |f(x + 0) − f(x − 0)|  (2.1)

where V_{a}^{b}A_{x} is the total variation of A_{x} on [a, b]. Since A_{x}(y) is continuous at y = x, the right–hand side of (2.1) converges to zero. For Theorem 2.1, we can make a rough estimate.

\underline{Corollary 2.2}. Under the assumption of Theorem 2.1, we have

|S_{n}(f; x) − \frac{1}{2}(f(x + 0) + f(x − 0))| ≤ \frac{18}{n(1 − x^{2})^{3/2}} ∑_{k=1}^{n} V_{x−(1+x)/k}^{x+(1−x)/k}(A_{x})

+ \frac{4}{nπ\sqrt{1 − x^{2}}} |f(x + 0) − f(x − 0)|, x ∈ (−1, 1), n ≥ 2.  (2.2)

\underline{Proof}. For the quantities in equation (2.1), we note that

\begin{aligned}\frac{1}{1 + x} ∑_{k=1}^{n} V_{x−(1+x)/k}^{x}(A_{x}) & = \frac{1 − x}{(1 − x^{2})} ∑_{k=1}^{n} V_{x−(1+x)/k}^{x}(A_{x}) \\ & ≤ \frac{2}{1 − x^{2}} ∑_{k=1}^{n} V_{x−(1+x)/k}^{x}(A_{x}). \end{aligned}

Similarly,

\begin{aligned}\frac{1}{1 − x} ∑_{k=1}^{n} V_{x}^{x+(1−x)/k}(A_{x}) & = \frac{1 + x}{(1 − x^{2})} ∑_{k=1}^{n} V_{x}^{x+(1−x)/k}(A_{x}) \\ & ≤ \frac{2}{1 − x^{2}} ∑_{k=1}^{n} V_{x}^{x+(1−x)/k}(A_{x}). \end{aligned}

Combining the above two inequalities we get the required result.

\underline{Example 2.1}. For a fixed x ∈ (−1, 1), consider a function g of bounded variation on [−1, 1], i.e.,

g(y) = |y − x|, y ∈ (−1, 1).

Now we have g(x + 0) = g(x − 0) = g(x) = 0 and A_{x}(y) = g(y), furthermore,

V_{x−(1+x)/k}^{x}(A_{x}) = \frac{1 + x}{k}, V_{x}^{x+(1−x)/k}(A_{x}) = \frac{1 − x}{k},

so, equation (2.1) becomes

|S_{n}(g; x)| ≤ \frac{9}{n\sqrt{1 − x^{2}}} (\frac{1}{1 + x} ∑_{k=1}^{n} \frac{1 + x}{k} + \frac{1}{1 − x} ∑_{k=1}^{n} \frac{1 − x}{k}) = \frac{18}{n\sqrt{1 − x^{2}}} ∑_{k=1}^{n} \frac{1}{k} .

But,

∑_{k=1}^{n} \frac{1}{k} = 1 + \frac{1}{2} + ⋯ + \frac{1}{n} = ln n + γ + O(1).

Therefore, by Theorem 2.1 we get an estimate

S_{n}(g; x) − g(x) = O(\frac{ln n}{n\sqrt{1 − x^{2}}} ).  (2.3)

Hereafter, the bounds of the terms “O” are independent of n and x. If we apply the above corollary from V_{x−(1+x)/k}^{x+(1−x)/k}(A_{x}) = 2/k, we shall obtain another estimate

S_{n}(g; x) − g(x) = O(\frac{ln n}{n(1 − x^{2})\sqrt{1 − x^{2}}} ).  (2.4)

Comparing (2.3) with (2.4), we see that when |x| → 1, the estimate (2.3) is more exact than (2.4).

3. The Main Result. Now we state and prove our main result.

\underline{Theorem 3.1}. If f(x) is a continuous function of bounded variation on [−1, 1] and ω_{v(f)}(δ) is the modulus of continuity of the total variation V_{−1}^{t}(f), then for x ∈ (−1, 1), n ≥ 2 we have

|S_{n}(f; x) − f(x)| ≤ \frac{9}{n\sqrt{1 − x^{2}}} \{\frac{1}{1 + x}ω_{v(f)}(1 + x) + \frac{1}{1 − x}ω_{v(f)}(1 − x)\}

+ \frac{9}{n\sqrt{1 − x^{2}}} ∫_{1/n}^{1} \{\frac{ω_{v(f)}((1 − x)u)}{1 − x} + \frac{ω_{v(f)}((1 + x)u)}{1 + x}\}\frac{du}{u^{2}},  (3.1)

especially, when V_{−1}^{t}(f) belongs to the class Lip α (α ∈ (0, 1)),

S_{n}(f; x) − f(x) = O(\frac{1}{n^{α}(1 − x^{2})^{3/2−α}} ).  (3.2)

Further, for the Cesaro mean (c, λ), λ ∈ (0, 1):

σ_{n}^{λ}(f; x) = \frac{1}{(λ)_{n}} ∑_{k=0}^{n}(λ − 1)_{n−k}S_{k}(f; x)  (3.3)

where in general

(β)_{n} = \frac{Γ(β + n + 1)}{Γ(β + 1)Γ(n + 1)}.

We have also

σ_{n}^{λ}(f; x) − f(x) = O(\frac{1}{n^{γ}(1 − x^{2})^{3/2−α}} ),  (3.4)

where γ = min{α, 1 − λ}.

\underline{Proof}. Since f(x) is a continuous function, we have A_{x}(y) = f(y) − f(x), so

V_{x}^{x+(1−x)/k}(A_{x}) = V_{x}^{x+(1−x)/k}(f) − V_{x}^{x}(f) ≤ ω_{v(f)}(\frac{1 − x}{k}), 2 ≤ k ≤ n

and

V_{x−(1+x)/k}^{x}(A_{x}) ≤ ω_{v(f)}(\frac{1 + x}{k}), 2 ≤ k ≤ n

and

V_{−1}^{x}(A_{x}) ≤ ω_{v(f)}(1 + x), V_{x}^{1}(A_{x}) ≤ ω_{v(f)}(1 − x).

Thus,applying Theorem 2.1, then for x ∈ (−1, 1), n ≥ 2 we obtain

\begin{matrix}|S_{n}(f; x) − f(x)| ≤ \frac{9}{n\sqrt{1 − x^{2}}} \{\frac{1}{1 + x}ω_{v(f)}(1 + x) + \frac{1}{1 − x}ω_{v(f)}(1 − x)\} \\ + \frac{9}{n\sqrt{1 − x^{2}}} \{\frac{1}{1 − x} ∑_{k=2}^{n} ω_{v(f)}(\frac{1 − x}{k}) + \frac{1}{1 + x} ∑_{k=2}^{n} ω_{v(f)}(\frac{1 + x}{k})\}. \end{matrix}

From this and noting that

∑_{k=2}^{n} ω_{v(f)}(\frac{1 − x}{k}) ≤ ∫_{1/n}^{1} ω_{v(f)}((1 − x)u)u^{−2}du

and

∑_{k=2}^{n} ω_{v(f)}(\frac{1 + x}{k}) ≤ ∫_{1/n}^{1} ω_{v(f)}((1 + x)u)u^{−2}du

we have formula (3.1).

When V_{−1}^{t}(f) ∈ Lip α (0 < α < 1), we have

ω_{v(f)}((1 − x)u) = O((1 − x)^{α}u^{α}) and ω_{v(f)}((1 + x)u) = O((1 + x)^{α}u^{α}).

From (3.1), we get (3.2). Now by (3.3) and (β)_{n} = O(n^{β}), we know for x ∈ (−1, 1), n ≥ 2 that

σ_{n}^{λ}(f; x) − f(x) = \frac{1}{(λ)_{n}} ∑_{k=0}^{n}(λ − 1)_{n−k}(S_{k}(f; x) − f(x))

= \frac{1}{(λ)_{n}} ∑_{k=2}^{n−1}(λ − 1)_{n−k}(S_{k}(f; x) − f(x))

+ O(1/n) + O(\frac{1}{n^{α+λ}(1 − x^{2})^{3/2−α}} ).  (3.5)

According to formula (3.2), we get

\frac{1}{(λ)_{n}} ∑_{k=2}^{n−1}(λ − 1)_{n−k}(S_{k}(f; x) − f(x)) = O(\frac{1}{n^{λ}(1 − x^{2})^{3/2−α}} ) ∑_{k=2}^{n−1} \frac{1}{k^{α}(n − k)^{1−λ}} .

(3.6)

Let γ = min{α, 1 − λ}. By the inequality

(a + b)^{γ} ≤ 2^{γ}(a^{γ} + b^{γ}), a > 0, b > 0,

we see that

∑_{k=2}^{n−1} \frac{1}{k^{α}(n − k)^{1−λ}} ≤ ∑_{k=2}^{n−1} \frac{1}{k^{γ}(n − k)^{γ}}

= ∑_{k=2}^{n−1} \frac{1}{n^{γ}} (\frac{1}{k} + \frac{1}{n − k} )^{γ}

= O(\frac{1}{n^{γ}} )\{ ∑_{k=2}^{n−1} \frac{1}{k^{γ}} + ∑_{k=2}^{n−1} \frac{1}{(n − k)^{γ}} \}

= O(\frac{1}{n^{2γ−1}} ).

From this and (3.5), (3.6), we obtain the formula (3.4). This completes the proof of Theorem 3.1.

A result of the type of equality in Theorem 3.1 for 2π–periodic continuous function of bounded variation on [−π, π] was proved by Natanson [3].

\underline{References}

1. K. Al–Khaled, “An Estimate for the Rate of Approximation of Functions by Chebyshev Polynomials,” Revista Colombiana de Matematicas, (to appear).

2. R. Bojanic, “An Estimate of the Rate of Convergence for Fourier Series of Functions of Bounded Variation,” Publ. Inst. Math. (Belgrade) 26 (1979), 57– 60.

3. G. I. Natanson, “On Fourier Series of Continuous Functions of Bounded Variation,” (in Russian), Vest, Leningrad, Univ. 7 (1972), 154—155.

4. G. Szego, Orthogonal Polynomials, American Mathematical Society Colloqium Publ., 1939.

Kamel Al–Khaled Department of Mathematics and Statistics Jordan University of Science and Technology P.O. Box 3030 Irbid 22110, Jordan email: applied@just.edu.jo