Degree of approximation of the conjugate of signals (functions) belonging to $\mathrm{Lip}(\alpha ,r)$-class by $(C,1)(E,q)$ means of conjugate trigonometric Fourier series

Abstract

In this paper, we determine the degree of approximation of the conjugate of 2π-periodic signals (functions) belonging to $Lip(\alpha ,r)$ ($0<\alpha \le 1$, $r\ge 1$)-class by using Cesàro-Euler $(C,1)(E,q)$ means of their conjugate trigonometric Fourier series. Our result generalizes the result of Lal and Singh (Tamkang J. Math. 33(3):269-274, 2002).

MSC:41A10.

1 Introduction

Let ${\sum }_{n=0}^{\mathrm{\infty }}{u}_n$ be a given infinite series with $\{s_n\}$, the sequence of its n th partial sum. The sequence-to-sequence transform

$$C_n^1=\frac{1}{n+1}\sum _{k=0}^n{s}_k,n=0,1,2,\dots ,$$

(1.1)

defines the Cesàro means of order one of $\{s_n\}$. The series ${\sum }_{n=0}^{\mathrm{\infty }}{u}_n$ is said to be $(C,1)$ summable to s, if ${lim}_{n\to \mathrm{\infty }}{C}_n^1=s$. The sequence-to-sequence transform

$$E_n^q=\frac{1}{{(1+q)}^n}\sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)q^{n-k}{s}_k,q>0,n=0,1,2,\dots ,$$

(1.2)

defines the Euler means of order $q>0$ of $\{s_n\}$. By super imposing the $(C,1)$ means on $(E,q)$ means of $\{s_n\}$, we get $(C,1)(E,q)$ means of $\{s_n\}$ denoted by $C_n^1{E}_n^q$ and defined by

$$C_n^1{E}_n^q=\frac{1}{n+1}\sum _{k=0}^n{E}_k^q=\frac{1}{n+1}\sum _{k=0}^n{(q+1)}^{-k}\sum _{v=0}^k\left(\begin{array}{c}k\\ v\end{array}\right)q^{k-v}{s}_v.$$

(1.3)

The series ${\sum }_{n=0}^{\mathrm{\infty }}{u}_n$ is said to be $(C,1)(E,q)$ summable to s, if ${lim}_{n\to \mathrm{\infty }}{C}_n^1{E}_n^q=s$.

For a given 2π-periodic Lebesgue integrable signal (function), let

$$s_n(f;x)=\frac{a_0}{2}+\sum _{k=1}^n(a_{k}coskx+b_{k}sinkx)$$

(1.4)

denote the $(n+1)$th partial sum, called trigonometric polynomial of degree n (or order n), of the Fourier series of $f\in L_1[-\pi ,\pi ]$.

The conjugate of Fourier series of f is defined by

$$\sum _{k=1}^{\mathrm{\infty }}(b_{k}coskx-a_{k}sinkx),$$

(1.5)

and its n th partial sum is defined as

$$\widetilde{s_n}(f;x)=\sum _{k=1}^n(b_{k}coskx-a_{k}sinkx)$$

(1.6)

The conjugate of f denoted by $\tilde{f}$ is defined by

$$2\pi \tilde{f}(x)=-\underset{ϵ\to 0}{lim}{\int }_{ϵ}^{\pi }\psi (t)cot(t/2)dt,$$

where $\psi (t)=f(x+t)-f(x-t)$ [[1], p.131].

A function $f\in Lip\alpha$, if

$$|f(x+t)-f(x)|=O\left(t^{\alpha }\right),$$

and $f\in Lip(\alpha ,r)$ if

$${\left({\int }_0^{2\pi }{|f(x+t)-f(x)|}^{r}dx\right)}^{1/r}=O\left(t^{\alpha }\right),0<\alpha \le 1,r\ge 1.$$

The $L_r$-norm for $f\in L_r[-\pi ,\pi ]$ is defined by

$${\parallel f\parallel }_r={\left({\int }_0^{2\pi }{|f(x)|}^{r}dx\right)}^{1/r},r\ge 1.$$

The $L_{\mathrm{\infty }}$-norm is defined by

$${\parallel f\parallel }_{\mathrm{\infty }}=sup\{|f(x)|:x\in R\}.$$

The degree of approximation $E_n(f)$ of a function $f\in Lip(\alpha ,r)$ by trigonometric polynomials $T_n(x)$ of degree n is given by

$$E_n(f)=\underset{T_n}{min}{\parallel f-T_n\parallel }_r.$$

This method of approximation is called trigonometric Fourier approximation (tfa). We also write

$$K_n(t)=\frac{1}{n+1}\sum _{k=0}^n\frac{1}{{(1+q)}^k}\sum _{v=0}^k\left(\begin{array}{c}k\\ v\end{array}\right)q^{k-v}\frac{cos(v+1/2)t}{sin(t/2)}$$

and $\tau =[1/t]$, the integral part of $1/t$.

2 Known result

Various investigators such as Dhakal [2], Lal and Singh [3], Mittal et al. [4, 5], Nigam [6], Qureshi [7, 8] have studied the degree of approximation in various function spaces such as Lipα, $Lip(\alpha ,r)$, $Lip(\xi (t),r)$ and weighted $(L_r,\xi (t))$ by using triangular matrix summability and product summability $(C,1)(E,1)$, $(N,p_n)(E,1)$. Recently, Lal and Singh [3] have determined the degree of approximation of the conjugate of $f\in Lip(\alpha ,r)$ by $(C,1)(E,1)$ means of conjugate Fourier series. Lal and Singh [3] have proved the following.

Theorem 1 [3]

If $f:R\to R$ is a 2π-periodic and $Lip(\alpha ,r)$ function, then the degree of approximation of its conjugate function $\tilde{f}(x)$ by the $(C,1)(E,1)$ product means of conjugate series of Fourier series of f satisfies, for $n=0,1,2,\dots$ ,

$$M_n(\tilde{f})=Min{\parallel {(CE)}_n^1-\tilde{f}\parallel }_r=O\left(n^{-\alpha +1/r}\right),$$

(2.1)

where

$${(CE)}_n^1=\frac{1}{n+1}\sum _{k=0}^n\left(\frac{1}{2^k}\sum _{i=0}^k\left(\begin{array}{c}k\\ i\end{array}\right)S_i\right),$$

is $(C,1)(E,1)$ means of series (1.5).

3 Main result

Recently, Nigam and Sharma [9] have studied the degree of approximation of functions belonging to $Lip(\xi (t),r)$-class through $(C,1)(E,q)$ means of their Fourier series. In this paper, we use the $(C,1)(E,q)$ means of conjugate Fourier series of $f\in Lip(\alpha ,r)$ to determine the degree of approximation of the conjugate of f, which in turn generalizes the result of Lal and Singh [3]. More precisely we prove

Theorem 2 Let $f(x)$ be a 2π-periodic, Lebesgue integrable function and belong to the $Lip(\alpha ,r)$-class with $r\ge 1$ and $\alpha r\ge 1$. Then the degree of approximation of $\tilde{f}(x)$, the conjugate of $f(x)$ by $(C,1)(E,q)$ means of its conjugate Fourier series is given by

$${\parallel C_n^1{E}_n^q-\tilde{f}\parallel }_r=O\left(n^{-\alpha +1/r}\right),n=0,1,2,\dots ,$$

(3.1)

provided

(3.2)

(3.3)

where δ is an arbitrary number such that $(\alpha +\delta )s+1<0$ and $1/r+1/s=1$ for $r>1$.

Remark 1 The authors have used conditions ${({\int }_0^{\pi /(n+1)}{|\frac{t\psi (t)}{t^{\alpha }}|}^{r}dt)}^{1/r}=O(1)$ implied by (3.2) and (3.3), but not mentioned in the statement of Theorem 1 [[3], pp.271-272].

4 Lemmas

We need the following lemmas for the proof of our theorem.

Lemma 1 $|K_n(t)|=O(1/t)+O((n+1)t)$ for $0\le t\le \pi /(n+1)\le \pi /(v+1)$.

Proof

$$\begin{array}{rcl}|K_n(t)|& \le & \frac{1}{2\pi (n+1)}\left|\sum _{k=0}^n\frac{1}{{(1+q)}^k}\sum _{v=0}^k\left(\begin{array}{c}k\\ v\end{array}\right)q^{k-v}\frac{cos(v+1/2)t}{sin(t/2)}\right|\\ =& \frac{1}{2\pi (n+1)}\left|\sum _{k=0}^n\frac{1}{{(1+q)}^k}\sum _{v=0}^k\left(\begin{array}{c}k\\ v\end{array}\right)q^{k-v}\frac{cos(v+1-1/2)t}{sin(t/2)}\right|\\ \le & \frac{1}{(n+1)}\sum _{k=0}^n\frac{1}{{(1+q)}^k}\sum _{v=0}^k\left(\begin{array}{c}k\\ v\end{array}\right)q^{k-v}\left|\frac{cos(v+1)tcos(t/2)+sin(v+1)tsin(t/2)}{sin(t/2)}\right|\\ =& \frac{1}{(n+1)}\sum _{k=0}^n\frac{1}{{(1+q)}^k}\sum _{v=0}^k\left(\begin{array}{c}k\\ v\end{array}\right)q^{k-v}[O(1/t)+O(sin(v+1)t)]\\ =& O\left[\frac{1}{(n+1)t}\sum _{k=0}^n\frac{1}{{(1+q)}^k}\sum _{v=0}^k\left(\begin{array}{c}k\\ v\end{array}\right)q^{k-v}\right]\\ +O[\frac{1}{(n+1)}\sum _{k=0}^n\frac{1}{{(1+q)}^k}\sum _{v=0}^k\left(\begin{array}{c}k\\ v\end{array}\right)q^{k-v}(v+1)t]\\ =& O[\frac{1}{(n+1)t}(n+1)]+O[\frac{1}{(n+1)}(n+1)(n+1)t]\\ =& O(1/t)+O((n+1)t),\end{array}$$

in view of $sin(v+1)t\le (v+1)t$ for $0\le t<\pi /(v+1)$ and ${(sin(t/2))}^{-1}<\pi /t$ for $0<t\le \pi$ [[10], p.247]. □

Lemma 2 $|K_n(t)|=O(1/t)+O(1)$ for $\pi /(v+1)\le t\le \pi$.

Proof

$$\begin{array}{rcl}|K_n(t)|& \le & \frac{1}{2\pi (n+1)}\left|\sum _{k=0}^n\frac{1}{{(1+q)}^k}\sum _{v=0}^k\left(\begin{array}{c}k\\ v\end{array}\right)q^{k-v}\frac{cos(v+1/2)t}{sin(t/2)}\right|\\ =& \frac{1}{2\pi (n+1)}\left|\sum _{k=0}^n\frac{1}{{(1+q)}^k}\sum _{v=0}^k\left(\begin{array}{c}k\\ v\end{array}\right)q^{k-v}\frac{cos(v+1-1/2)t}{sin(t/2)}\right|\\ \le & \frac{1}{(n+1)}\sum _{k=0}^n\frac{1}{{(1+q)}^k}\sum _{v=0}^k\left(\begin{array}{c}k\\ v\end{array}\right)q^{k-v}\left|\frac{cos(v+1)tcos(t/2)+sin(v+1)tsin(t/2)}{sin(t/2)}\right|\\ =& \frac{1}{(n+1)}\sum _{k=0}^n\frac{1}{{(1+q)}^k}\sum _{v=0}^k\left(\begin{array}{c}k\\ v\end{array}\right)q^{k-v}[O(1/t)+O(1)]\\ =& O\left[\frac{1}{(n+1)t}\sum _{k=0}^n\frac{1}{{(1+q)}^k}\sum _{v=0}^k\left(\begin{array}{c}k\\ v\end{array}\right)q^{k-v}\right]\\ +O\left[\frac{1}{(n+1)}\sum _{k=0}^n\frac{1}{{(1+q)}^k}\sum _{v=0}^k\left(\begin{array}{c}k\\ v\end{array}\right)q^{k-v}\right]\\ =& O[\frac{1}{(n+1)t}(n+1)]+O[\frac{1}{(n+1)}(n+1)]\\ =& O(1/t)+O(1),\end{array}$$

in view of $|sin(v+1)t|\le 1$ and ${(sin(t/2))}^{-1}\le \pi /t$ for $0<t\le \pi$ [[10], p.247]. □

5 Proof of Theorem 2

The integral representation of $\widetilde{s_n}(f;x)$ is given by

$$\widetilde{s_n}(f;x)=-\frac{1}{\pi }{\int }_0^{\pi }\psi (t)\frac{cos(t/2)-cos(n+1/2)t}{2sin(t/2)}dt.$$

Therefore, we have

$${\tilde{s}}_n(f;x)-\tilde{f}(x)=\frac{1}{2\pi }{\int }_0^{\pi }\psi (t)\frac{cos(n+1/2)t}{sin(t/2)}dt.$$

Now, denoting $(C,1)(E,q)$ transform of $\widetilde{s_n}(f;x)$ by $C_n^1{E}_n^q$, we write

$$\begin{array}{rcl}{C}_n^1{E}_n^q-\tilde{f}& =& \frac{1}{2\pi (n+1)}[\sum _{k=0}^n\frac{1}{{(1+q)}^k}{\int }_0^{\pi }\frac{\psi (t)}{sin(t/2)}\sum _{v=0}^k\left(\begin{array}{c}k\\ v\end{array}\right)q^{k-v}cos(v+1/2)tdt]\\ =& [{\int }_0^{\pi /(n+1)}+{\int }_{\pi /(n+1)}^{\pi }]\psi (t)K_n(t)dt=I_1+I_2,\text{say}.\end{array}$$

(5.1)

Using Lemma 1, Hölder’s inequality, condition (3.2) and Minkowiski’s inequality, we have

$$\begin{array}{rcl}|I_1|& =& {\int }_0^{\pi /(n+1)}|\psi (t)||K_n(t)|dt\\ \le & {\left[{\int }_0^{\pi /(n+1)}{(|\psi (t)|/t^{\alpha })}^{r}dt\right]}^{1/r}{\left[\underset{ϵ\to 0}{lim}{\int }_{ϵ}^{\pi /(n+1)}{\left(t^{\alpha }|K_n(t)|\right)}^{s}dt\right]}^{1/s}\\ =& O\left({(n+1)}^{-1}\right){\left[\underset{ϵ\to 0}{lim}{\int }_{ϵ}^{\pi /(n+1)}{(t^{\alpha -1}+(n+1)t^{\alpha +1})}^{s}dt\right]}^{1/s}\\ =& O\left({(n+1)}^{-1}\right)[{\left(\underset{ϵ\to 0}{lim}{\int }_{ϵ}^{\pi /(n+1)}{t}^{(\alpha -1)s}dt\right)}^{1/s}+{(\underset{ϵ\to 0}{lim}{\int }_{ϵ}^{\pi /(n+1)}(n+1)t^{(\alpha +1)s}dt)}^{1/s}]\\ =& O\left({(n+1)}^{-1}\right)[{(n+1)}^{-\alpha +1-1/s}+(n+1){(n+1)}^{-\alpha -1-1/s}]\\ =& O\left({(n+1)}^{-1}\right)[{(n+1)}^{-\alpha +1/r}+(n+1){(n+1)}^{-\alpha -1-1+1/r}]\\ =& O[{(n+1)}^{-\alpha +1/r-1}+{(n+1)}^{-\alpha -2+1/r}]\\ =& O\left({(n+1)}^{-\alpha -1+1/r}\right).\end{array}$$

(5.2)

Now, we consider

$$|I_2|\le {\int }_{\pi /(n+1)}^{\pi }|\psi (t)||K_n(t)|dt.$$

Using Lemma 2, Hölder’s inequality, condition (3.3) and Minkowiski’s inequality, we have

$$\begin{array}{rcl}|I_2|& \le & {\left[{\int }_{\pi /(n+1)}^{\pi }{\left(\frac{t^{-\delta }|\psi (t)|}{t^{\alpha }}\right)}^{r}dt\right]}^{1/r}{\left[{\int }_{\pi /(n+1)}^{\pi }{\left(\frac{t^{\alpha }|K_n(t)|}{t^{-\delta }}\right)}^{s}dt\right]}^{1/s}\\ =& O\left({(n+1)}^{\delta }\right){\left[{\int }_{\pi /(n+1)}^{\pi }{\left(\frac{t^{\alpha }}{t^{-\delta }}(O(1/t)+O(1))\right)}^{s}dt\right]}^{1/s}\\ =& O\left({(n+1)}^{\delta }\right){\left[{\int }_{\pi /(n+1)}^{\pi }{(t^{\alpha +\delta -1}+t^{\alpha +\delta })}^{s}dt\right]}^{1/s}\\ =& O\left({(n+1)}^{\delta }\right)[{\left({\int }_{\pi /(n+1)}^{\pi }{t}^{(\alpha +\delta -1)s}dt\right)}^{1/s}+{\left({\int }_{\pi /(n+1)}^{\pi }{t}^{(\alpha +\delta )s}dt\right)}^{1/s}]\\ =& O\left({(n+1)}^{\delta }\right)[{(n+1)}^{(-\alpha -\delta +1)-1/s}+{(n+1)}^{(-\alpha -\delta )-1/s}]((\alpha +\delta )s+1<0)\\ =& O[{(n+1)}^{-\alpha +1-1/s}+{(n+1)}^{-\alpha -1/s}]\\ =& O[{(n+1)}^{-\alpha +1/r}+{(n+1)}^{-\alpha -1+1/r}]=O\left[{(n+1)}^{-\alpha +1/r}(1+{(n+1)}^{-1})\right]\\ =& O\left({(n+1)}^{-\alpha +1/r}\right).\end{array}$$

(5.3)

Combining (5.1)-(5.3), we have

$$|C_n^1{E}_n^q-\tilde{f}|=O\left({(n+1)}^{-\alpha +1/r}\right).$$

Hence,

$${\parallel C_n^1{E}_n^q-\tilde{f}\parallel }_r={\left({\int }_0^{2\pi }{|C_n^1{E}_n^q-\tilde{f}(x)|}^{r}dx\right)}^{1/r}=O\left(n^{-\alpha +1/r}\right).$$

This completes the proof of Theorem 2.

Remark 2 The proof of Theorem 2 for $r=1$, i.e., $s=\mathrm{\infty }$, can be written by using sup norm while using Hölder’s inequality.

6 Corollaries

Corollary 1 When $q=1$, then $(C,1)(E,q)$ means reduces to $(C,1)(E,1)$ means.

Hence, Theorem  2 reduces to Theorem  1.

Corollary 2 If $f:R\to R$ is a 2π-periodic, Lebesgue integrable and belonging to the Lipα ($0<\alpha \le 1$) class, then the degree of approximation of $\tilde{f}(x)$, the conjugate of $f(x)\in Lip\alpha$, with $0<\alpha \le 1$ by $(C,1)(E,q)$ means of its Fourier series is given by

$${\parallel C_n^1{E}_n^q-\tilde{f}\parallel }_{\mathrm{\infty }}=O\left(n^{-\alpha }\right)\mathit{\text{for}}n=0,1,2,\dots .$$

Proof If $r\to \mathrm{\infty }$ in Theorem 2, then for $0<\alpha <1$,

$${\parallel C_n^1{E}_n^q-\tilde{f}(x)\parallel }_{\mathrm{\infty }}=O\left(n^{-\alpha }\right).$$

For $\alpha =1$, we can write an independent proof by using $\alpha =1$ and $\psi (t)=O(t)$ in $I_1$ and $I_2$. □