Applied Mathematics E-Notes, 1(2001), 143-148 \begin{matrix}c \\ ◯ \end{matrix} Available free at mirror sites of http://math2.math.nthu.edu.tw/∼amen/

On the Rate of Pointwise Summability of Fourier Series ^{∗}

Włodzimierz Łenski^{†}

Received 17 April 2000

Abstract We introduce a local modulus of continuity as a measure of pointwise summa-

bility by Abel and (C,2) means. The (C,1) summability is also considered.

1 Introduction

Let L^{p} (C) be the class of all 2π–periodic real functions integrable in the Lebesgue sense with p–th power (respectively continuous functions) over Q = [−π, π]. Let X = X^{p} where X^{p} = L^{p} when 1 ≤ p < ∞ or X^{p} = C when p = ∞. Let us define the norm of f ∈ X^{p} as

∥f∥_{X^{p}} := ∥f(·)∥_{X^{p}} = \{ \begin{matrix}(∫_{Q} | f(x) |^{p} dx)^{1/p} & 1 ≤ p < ∞ \\ sup_{x∈Q} | f(x) | & p = ∞ \end{matrix} .

Let

Sf(x) := \frac{a_{0}(f)}{2} + ∑_{k=1}^{∞}(a_{k}(f) cos kx + b_{k}(f) sin kx) := ∑_{k=0}^{∞} C_{k}f(x)

and let S_{k}f, σ_{k}^{α}f and A_{r}f be the partial sum, the (C, α) mean and the Abel mean of the trigonometric Fourier series Sf respectively. Thus

σ_{n}^{0}f = S_{n}f, n = 0, 1, 2, ...,
σ_{n}^{α}f = \frac{1}{A_{n}^{α}} ∑_{k=0}^{n} A_{n−k}^{α}C_{k}f, α > −1, n = 0, 1, 2, ...,

A_{k}^{α} = C_{k+α}^{k} = ( \begin{matrix}k + α \\ k \end{matrix} )

and

A_{r}f = ∑_{k=0}^{∞} r^{k}C_{k}f, r ∈ (0, 1) .

∗Mathematics Subject Classifications: 42A10, 42A24

†Institute of Mathematics, University of Zielona Góra, 65-069 Zielona Góra, pl. Słowiański 9, Poland

The pointwise characteristic

\overline{w}_{x}f_{p}(δ) = sup_{0<h≤δ} \{\frac{1}{h} ∫_{0}^{h} |φ_{x} (t)|^{p} dt\}^{1/p} ,

where

φ_{x} (t) = f (x + t) + f (x − t) − 2f (x) ,

was used as a measure of approximation by Aljančič et al. [1]. This characteristic was very often used, but it appears that such approximation cannot be comparable with the norm approximation when X ≠ C. In [5] we introduced the modified quantity

w_{x}f_{p}(δ) := \{\frac{1}{δ} ∫_{0}^{δ} |φ_{x} (t)|^{p} dt\}^{1/p} .

In view of the monotonicity of the product δw_{x}f_{1}(δ) with respect to δ, by the methods in [5], we may easily obtain the following estimation

|σ_{n}^{1}f (x) − f (x)| ≤ \frac{3/2}{n + 1} ∑_{k=0}^{n} w_{x}f_{1} (\frac{π}{k + 1}) , n = 0, 1, 2, ... .

Based on the points of differentiability of the indefinite integral of f (D-points), the quantity

\overline{w}_{x}^{∗}f(δ) := sup_{0<h≤δ} |\frac{1}{h} ∫_{0}^{h} φ_{x} (t) dt|

is considered in [7].

Here we introduce yet another modified quantity

w_{x}^{∗}f(δ) := sup_{0<h≤δ} |\frac{1}{δ} ∫_{0}^{h} φ_{x} (t) dt| .

We can observe that for p ∈ [1, ∞),

w_{x}f_{p} (δ) ≤ \overline{w}_{x}f_{p} (δ) ≤ ωf_{C} (δ) ,

and

w_{x}^{∗}f(δ) ≤ \overline{w}_{x}^{∗}f(δ) ≤ ωf_{C} (δ) ,

and also, for p̃ ∈ [p, ∞], by the Minkowski inequality,

∥w_{·}^{∗}f(δ)∥_{X_{p}} ≤ ∥w_{·}f_{p}(δ)∥_{X_{p}} ≤ ωf_{X_{p}}(δ) ,

where ωf_{X} is the modulus of continuity of f in the space X = X_{p} defined by the formula

ωf_{X} (δ) := sup_{0<|h|≤δ} ∥f(· + h) − f(·)∥_{X} .

It is well-known (see [2,3,4]) that the (C, 1) means do not tend to f at the D-points of f but the Abel means as well the (C, 2) means do. In this paper these facts will be presented in the approximation version with the quantity w_{x}^{∗}f as a measure of such approximation (cf. [6]).

By K we shall designate either an absolute constant or a constant depending on the indicated parameters, not necessarily the same in each occurrence.

2 Results

We start with two theorems for pointwise approximation.

THEOREM 1. If f ∈ L^{1}, then

|σ_{n}^{2}f (x) − f (x)| ≤ K \frac{1}{n + 1} ∑_{k=0}^{n} w_{x}^{∗}f (\frac{π}{k + 1}) ,

for all real x and every positive integer n.

THEOREM 2. If f ∈ L^{1}, then

|A_{r}f (x) − f (x)| ≤ K (w_{x}^{∗}f(π) (1 − r) + (1 − r) ∫_{1−r}^{π} \frac{w_{x}^{∗}f(t)}{t^{2}}dt)

for all real x and every r ∈ (0, 1).

Now using the Minkowski inequality, we can derive from these theorems a corollary. COROLLARY 1. If f ∈ X = X^{p} (p ∈ [1, ∞]), then

∥σ_{n}^{2}f − f∥_{X} ≤ K \frac{1}{n + 1} ∑_{k=0}^{n} ωf_{X} (\frac{π}{k + 1})

and

∥A_{r}f − f∥_{X} ≤ K ((1 − r) ωf_{X} (π) + (1 − r) ∫_{1−r}^{π} \frac{ωf_{X} (t)}{t^{2}}dt)

for every integer n and r ∈ (0, 1) respectively.

From our theorems the results of Fatou [3] and Lebesgue [4] also follow.

COROLLARY 2. If f ∈ L and x is a D-point of f, then σ_{n}^{2}f (x) − f (x) = o_{x} (1) as n → ∞ and A_{r}f (x) − f (x) = o_{x} (1) as r → 1^{−}.

To prove Theorem 1, let us observe that

σ_{n}^{2}f (x) − f (x) = \frac{1}{A_{n}^{2}} ∑_{ν=0}^{n} A_{ν}^{1} (σ_{ν}^{1}f (x) − f (x)) = \frac{1}{2πA_{n}^{2}} ∫_{0}^{π} φ_{x} (t) ∑_{ν=0}^{n} (\frac{sin \frac{ν+1}{2}t}{sin \frac{1}{2}t})^{2} dt.

Putting

G_{n} (0) = ∑_{ν=0}^{n} (ν + 1)^{2}
G_{n} (t) = ∑_{ν=0}^{n} (\frac{sin \frac{ν+1}{2}t}{sin \frac{1}{2}t})^{2} , t ≠ 0,

146 Pointwise Summability and integrating by parts we obtain

σ_{n}^{2}f (x) − f (x) = \frac{1}{2πA_{n}^{2}} \{[∫_{0}^{t} φ_{x} (u) duG_{n} (t)]_{0}^{π} − ∫_{0}^{π} (∫_{0}^{t} φ_{x} (u) du) \frac{d}{dt}G_{n} (t) dt\} .

An easy computation yields

∑_{ν=0}^{n} sin^{2} \frac{ν + 1}{2}t = \frac{2n + 3}{4} − \frac{sin \frac{2n+3}{2}t}{4 sin \frac{1}{2}t},

and therefore

G_{n} (t) = \frac{2n + 3}{4 sin^{2} \frac{1}{2}t} − \frac{sin \frac{2n+3}{2}t}{4 sin^{3} \frac{1}{2}t},

whence

G_{n} (π) = \frac{2n + 3}{4} − \frac{(−1)^{n+1}}{4} .

Thus

|σ_{n}^{2}f (x) − f (x)|
≤ \frac{2n + 3 + (−1)^{n}}{4 (n + 1)^{2}} w_{x}^{∗}f(π) + \frac{1}{π (n + 1)^{2}} (∫0^{\frac{π}{n+1}} + ∫ π_{\frac{π}{n+1}} ) tw_{x}^{∗}f(t) |\frac{d}{dt}G_{n} (t)| dt.

Let S_{1}, S_{2} and S_{3} denote respectively the first, second and the third sum in the right hand side of the above inequality. Immediately we have

S_{1} ≤ \frac{1}{n + 1}w_{x}^{∗}f(π) ≤ \frac{1}{n + 1} ∑_{k=0}^{n} w_{x}^{∗}f (\frac{π}{k + 1}) .

Differentiating G_{n} we obtain

\frac{d}{dt}G_{n} (t) = \frac{− (2n + 3) cos \frac{1}{2}t − \frac{2n+3}{2} cos \frac{2n+3}{2}t}{4 sin^{3} \frac{1}{2}t} + \frac{3 cos \frac{1}{2}t sin \frac{2n+3}{2}t}{8 sin^{4} \frac{1}{2}t},

whence

|\frac{d}{dt}G_{n} (t)| ≤ \frac{4.5 (2n + 3)}{4 sin^{3} \frac{1}{2}t} ≤ \frac{9π^{2}}{4} \frac{n + 1}{t^{3}}, 0 < |t| ≤ π,

and

\frac{d}{dt}G_{n} (t) = \frac{d}{dt} (2 ∑_{ν=0}^{n} ∑_{k=0}^{ν} (\frac{1}{2} + ∑_{μ=1}^{k} cos μt)) = −2 ∑_{ν=0}^{n} ∑_{k=0}^{ν} ∑_{μ=1}^{k} μ sin μt,

W. Łenski 147 Thus, for all t,

|\frac{d}{dt}G_{n} (t)| ≤ 2 ∑_{ν=0}^{n} ∑_{k=0}^{ν} ∑_{μ=1}^{k} μ^{2} sin t ≤ 2n^{5} sin t ≤ 2 (n + 1)^{5} t.

These imply

S_{2} ≤ \frac{1}{π (n + 1)^{2}} \frac{π}{n + 1}w_{x}^{∗}f (\frac{π}{n + 1}) 2 (n + 1)^{5} ∫0^{\frac{π}{n+1}} tdt ≤ 2π^{2}w_{x}^{∗}f (\frac{π}{n + 1}) ,

and

S_{3} ≤ \frac{9π^{2}}{4π (n + 1)^{2}} ∫ π_{\frac{π}{n+1}} tw_{x}^{∗}f(t)\frac{n + 1}{t^{3}}dt = \frac{9π}{4} \frac{1}{n + 1} ∫ π_{\frac{π}{n+1}} \frac{w_{x}^{∗}f(t)}{t^{2}}dt.

Finally, by the monotonicity of tw_{x}^{∗}f(t),

w_{x}^{∗}f(\frac{π}{n + 1}) = \frac{π}{n + 1}w_{x}^{∗}f(\frac{π}{n + 1})\frac{4π}{n + 1} ∫ π_{\frac{π}{n+1}} t^{−3}dt ≤ \frac{4π}{n + 1} ∫ π_{\frac{π}{n+1}} \frac{w_{x}^{∗}f(t)}{t^{2}}dt

and

∫ π_{\frac{π}{n+1}} \frac{w_{x}^{∗}f(t)}{t^{2}}dt = \frac{1}{π} ∫_{1}^{n+1} w_{x}^{∗}f(\frac{π}{u})du = \frac{1}{π} ∑_{k=1}^{n} ∫_{k}^{k+1} u sup_{0<h≤π/u} |∫_{0}^{h} φ_{x} (t) dt| du
≤ \frac{1}{π} ∑_{k=1}^{n} (k + 1) sup_{0<h≤π/k} |∫_{0}^{h} φ_{x} (t) dt|
≤ 2 ∑_{k=1}^{n} w_{x}^{∗}f(\frac{π}{k}) ≤ 2 ∑_{k=0}^{n} w_{x}^{∗}f(\frac{π}{k + 1}).

These estimates imply our result with constant K = 1 + 16π^{3} + 9π/2. The proof is complete.

Next, we prove Theorem 2. First of all, it is clear that

A_{r}f (x) − f (x) = \frac{1}{π} ∫_{0}^{π} φ_{x} (t) P (r, t) dt,

where

P (r, t) = \frac{1 − r^{2}}{2Δ (r, t)}

and

Δ (r, t) = 1 − 2r cos t + r^{2} = (1 − r)^{2} + 4r sin^{2} \frac{1}{2}t.

Integration by parts gives

A_{r}f (x) − f (x)

= [\frac{1}{π} ∫_{0}^{t} φ_{x} (u) duP (r, t)]_{t=0}^{π} − \frac{1}{π} ∫_{0}^{π} (∫_{0}^{t} φ_{x} (u) du) \frac{∂}{∂t}P (r, t) dt
= \frac{1 − r}{2 (1 + r)} \frac{1}{π} ∫_{0}^{π} φ_{x} (u) du + \frac{1}{π} ∫_{0}^{π} (∫_{0}^{t} φ_{x} (u) du) \frac{r (1 − r^{2}) sin t}{Δ^{2} (r, t)}dt.

Hence,

|A_{r}f (x) − f (x)| ≤ (1 − r) w_{x}^{∗}f(π) + \frac{1}{π} ∫_{0}^{π} tw_{x}^{∗}f(t)\frac{r (1 − r^{2}) sin t}{Δ^{2} (r, t)}dt
= (1 − r) w_{x}^{∗}f(π) + \frac{1}{π} (∫_{0}^{1−r} + ∫_{1−r}^{π}) tw_{x}^{∗}f(t)\frac{r (1 − r^{2}) sin t}{Δ^{2} (r, t)}dt
= (1 − r) w_{x}^{∗}f(π) + I_{1} + I_{2}.

Using the estimate Δ (r, t) ≥ (1 − r)^{2} we can see that

I_{1} ≤ \frac{1}{π} (1 − r) w_{x}^{∗}f(1 − r)\frac{r (1 − r^{2}) (1 − r)}{(1 − r)^{4}} ∫_{0}^{1−r} dt ≤ \frac{2}{π} w_{x}^{∗}f(1 − r).

By the estimate Δ (r, t) ≥ \frac{4r}{π^{2}} t^{2} we obtain

I_{2} ≤ \frac{1}{π} ∫_{1−r}^{π} tw_{x}^{∗}f(t)\frac{r (1 − r^{2}) t}{\frac{16r^{2}}{π^{4}} t^{4}}dt = \frac{π^{3}}{8r} (1 − r) ∫_{1−r}^{π} \frac{w_{x}^{∗}f(t)}{t^{2}}dt.

Finally, because tw_{x}^{∗}f_{1}(t) is a non-decreasing function of t,

w_{x}^{∗}f(1 − r) ≤ \frac{π^{2} − 1}{2π^{2}}w_{x}^{∗}f(1 − r) (1 − r)^{2} ∫_{1−r}^{π} t^{−3}dt ≤ \frac{π^{2} − 1}{2π^{2}}(1 − r) ∫_{1−r}^{π} \frac{w_{x}^{∗}f(t)}{t^{2}}dt,

which completes our proof.

References

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