Approximation of Functions of Class $\mathrm {Lip} (\alpha , { p})$ in $L_{p}$-Norm

Abstract

Mittal and Rhoades (Int. J. Math. Game Theory Algebra 9(4), 259–267, 1999 [9]; J. Comput. Anal. Appl. 2(1) 1–10, 2000 [10]) and Mittal et al. (J. Math. Anal. Appl. 326(1) 667–676, 2007 [7]; Appl. Math. Comput. 217(9), 4483–4489, 2011 [8]) initiated the studies of error estimates $E_{n}(f)$ through trigonometric-Fourier approximation (tfa) for situations in which the summability matrix T does not have monotone rows. In this paper, we extend the results of Mittal et al. (Appl. Math. Comput. 217(9), 4483–4489, 2011 [8]) to a more general $C_{\lambda }$-method in view of Armitage and Maddox (Analysis 9, 195–204, 1989 [1]), which in turn generalizes the several previous known results due to Mittal and Singh (Int. J. Math. Math. Sci., Art. ID 267383, 1–6, 2014 [11]), De${\tilde{\mathrm{g}}}$er et al. (Proc. Jangjeon Math. Soc. 15(2), 203–213, 2012 [4]), Leindler (J. Math. Anal. Appl. 302, 129–136, 2005 [6]), Chandra (J. Math. Anal. Appl. 275, 13–26, 2002 [3]) and Quade (Duke Math. J. 3(3), 529–543, 1937 [15]).

1 Introduction

For a given function $f \in L_{p}:= L_{p}[0, 2\pi ],\, p \ge 1$, let

$$\begin{aligned} s_{n}(f):= s_{n}(f;x)= \frac{a_0}{2}+\sum _{k=1}^{n} \left( a_k \cos kx+b_k \sin kx\right) = \sum _{k=0}^{n} u_{k}(f;x) \end{aligned}$$

(1)

denote the partial sums, called trigonometric polynomials of degree (or order) n, of the first $(n+1)$ terms of the Fourier series of f at a point x.

A positive sequence $\mathbf c :=\{c_n\}$ is called almost monotone decreasing (increasing) if there exists a constant $K:=K(\mathbf c )$, depending on the sequence $\mathbf c$ only, such that for all $n \ge m$, $c_n \le Kc_m (Kc_n \ge c_m).$ Such sequences will be denoted by $\mathbf c \in$ AMDS and $\mathbf c \in$ AMIS respectively. A sequence which is either AMDS or AMIS is called almost monotone sequence and will be denoted by $\mathbf c \in$ AMS.

Let $\mathbb {F}$ be an infinite subset of $\mathbb {N}$ and $\mathbb {F}$ the range of strictly increasing sequence of positive integers, say $\mathbb {F}=\{\lambda (n)\}_{n=1}^{\infty }$. The Cesàro submethod $C_{\lambda }$ is defined as

$$\begin{aligned} (C_{\lambda } x)_n = \frac{1}{\lambda (n)} \sum _{k=1}^{\lambda (n)} x_{k}, ( n = 1, 2, 3,...), \end{aligned}$$

where $\{x_k\}$ is a sequence of real or complex numbers. Therefore, the $C_{\lambda }$-method yields a subsequence of the Cesàro method $C_1$, and hence it is regular for any ${\lambda }$. Matrix-$C_{\lambda }$ is obtained by deleting a set of rows from Cesàro matrix. The basic properties of $C_{\lambda }$-method can be found in [1, 14].

Define

$$\begin{aligned} \tau _{n}^{\lambda } (f) = \tau _{n}^{\lambda } (f; x) = \sum _{k=0}^{\lambda (n)} a_{\lambda (n),k} s_{k}(f; x) , \forall n \ge 0. \end{aligned}$$

The trigonometric Fourier series of the signal f is said to be $T^\lambda$-summable to s if $\tau _{n}^{\lambda } (f) \rightarrow s$ as $n \rightarrow \infty$.

Throughout $T \equiv (a_{n,k})$, a linear operator, will denote an infinite lower triangular matrix with nonnegative entries and row sums 1. Such a matrix T is said to have monotone rows if , $\forall n$, $\{a_{n,k}\}$ is either nonincreasing or nondecreasing in $k, 0 \le k \le n$. A linear operator T is said to be regular if it is limit-preserving over the space of convergent sequences.

We write

$$\begin{aligned} s_{n}(f; x) = \frac{1}{\pi } \int _{0}^{2 \pi }f(x+t) D_{n}(t)\, {\text {d}}t ,\,\,\,\, D_{n}(t) = (\sin (n+{1/2})t) / {2 \sin (t/2)}, \end{aligned}$$

$$\begin{aligned} A_{\lambda (n),k} = \sum _{r=k}^{\lambda (n)} a_{\lambda (n),r},\,\,\,\, A_{\lambda (n),0} \equiv 1, \forall n \ge 0. \end{aligned}$$

The notation [x] means the greatest integer contained in x.

2 Known Results

Chandra [3] proved three theorems on the trigonometric approximation using Nörlund and Riesz matrices. Some of them give sharper estimates than the results proved by Quade [15], Mohapatra and Russell [12] and himself earlier [2]. Similar results were proved by Khan [5] for generalized $N_{p}$-mean and Mohapatra et al. [13] for Taylor mean. Leindler [6] extended the results of Chandra [3] without the assumption of monotonicity on the generating sequence $\{p_{n}\}$. Leindler [6] proved the following:

Theorem 1

([6]) If $f \in Lip(\alpha , p)$ and $\{p_{n}\}$ be positive. If one of the conditions

(i) $p>1, 0<\alpha <1$ and $\{p_{n}\} \in$ AMDS,

(ii) $p>1, 0<\alpha <1$ and $\{p_{n}\} \in$ AMIS and

$$\begin{aligned} (n+1)p_{n}= O(P_{n}) \, holds, \end{aligned}$$

(2)

(iii) $p>1, \alpha =1$ and $\sum _{k=1}^{n-1} k| \triangle p_{k} | = O(P_{n})$,

(iv) $p>1, \alpha =1, \sum _{k=0}^{n-1} | \triangle p_{k} | = O({P_{n}}/{n})$ and (2) holds,

(v) $p=1,0<\alpha <1$ and $\sum _{k=-1}^{n-1} | \triangle p_{k} | = O({P_{n}}/{n})$,

maintains, then

$$\begin{aligned} ||f-N_{n}(f)||_{p}= O(n^{-\alpha }). \end{aligned}$$

(3)

Theorem 2

([6]) Let $f \in Lip(\alpha , 1), 0< \alpha < 1$. If the positive $\{p_n\}$ satisfies conditions (2) and $\sum _{k=0}^{n-1} | \triangle p_{k} | = O({P_{n}}/{n})$ hold, then

$$\begin{aligned} ||f-R_{n}(f)||_{1}= O(n^{-\alpha }). \end{aligned}$$

Mittal et al. [7, 8] extended the work of Chandra to general matrices.

Mittal et al. [8] proved the following:

Theorem 3

([8]) Let $f \in Lip(\alpha , p)$ and let $T=(a_{n, k})$ be an infinite regular triangular matrix.

(i) If $p > 1, 0 < \alpha <1, \{ a_{n, k} \} \in$ AMS in k and satisfies

$$\begin{aligned} (n+1) \mathrm{{max}} \{a_{n,0},a_{n,r}\}= O(1). \end{aligned}$$

(4)

where $r:=[n/2]$ then

$$\begin{aligned} ||f-\tau _{n}(f)||_p = O(n^{- \alpha }). \end{aligned}$$

(5)

(ii) If $p>1, \alpha =1$ and $\sum _{k=0}^{n-1} (n-k)|\triangle _k a_{n, k}|=O(1)$, or

(iii)If $p>1, \alpha =1$ and $\sum _{k=0}^{n} |\triangle _k a_{n, k}|=O(a_{n, 0})$, or

(iv) If $p=1, 0 < \alpha < 1$ and $\sum _{k=0}^{n} |\triangle _k a_{n, k}|=O(a_{n, 0})$,

and also $(n+1) a_{n, 0}= O(1)$, holds then (5) is satisfied.

Recently, De${\tilde{\mathrm{g}}}$er et al. [4] extended the results of Chandra [3] to more general $C_{\lambda }$-method in view of Armitage and Maddox [1]. De${\tilde{\mathrm{g}}}$er et al. [4] proved:

Theorem 4

([4]) Let $f \in Lip(\alpha , p)$ and $\{p_{n}\}$ be positive such that

$$\begin{aligned} (\lambda (n)+1)p_{\lambda (n)}= O(P_{\lambda (n)}), \end{aligned}$$

(6)

If either (i) $p>1, 0 < \alpha \le 1$ and $\{p_{n}\}$ is monotonic or (ii) $p=1,0<\alpha <1$ and $\{p_{n}\}$ is nondecreasing then

$$\begin{aligned} ||f-N_{n}^{\lambda }(f)||_{p}= O(n^{-\alpha }). \end{aligned}$$

Theorem 5

([4]) Let $f \in Lip(\alpha , 1), 0 < \alpha < 1$. If the positive $\{p_{n}\}$ satisfies condition (6) and nondecreasing, then $||f-R_{n}^{\lambda }(f)||_{1}= O(n^{-\alpha }).$

Very recently, in [11], the authors of this paper generalized two theorems of De${\tilde{\mathrm{g}}}$er et al. [4], by dropping the monotonicity on the elements of the matrix rows. These results also generalize the results of Leindler [6] to more general $C_{\lambda }$-method.

Theorem 6

([11]) If $f \in Lip(\alpha , p)$ and $\{p_{n}\}$ be positive. If one of the following conditions

(i) $p>1, 0<\alpha <1$ and $\{p_{n}\} \in$ AMDS,

(ii) $p>1, 0<\alpha <1$ and $\{p_{n}\} \in$ AMIS and (6) holds,

(iii) $p>1, \alpha =1$ and $\sum _{k=1}^{\lambda (n)-1} k| \triangle p_{k} | = O(P_{\lambda (n)})$,

(iv) $p>1, \alpha =1, \sum _{k=0}^{\lambda (n)-1} | \triangle p_{k} | = O\left( \frac{P_{\lambda (n)}}{\lambda (n)}\right)$ and (6) holds,

(v) $p=1,0<\alpha <1$ and $\sum _{k=-1}^{\lambda (n)-1} | \triangle p_{k} | = O\left( \frac{P_{\lambda (n)}}{\lambda (n)}\right)$,

maintains, then

$$\begin{aligned} ||f-N_{n}^{\lambda }(f)||_{p}= O\left( (\lambda (n))^{-\alpha }\right) . \end{aligned}$$

(7)

Theorem 7

([11]) Let $f \in Lip(\alpha , 1), 0< \alpha < 1$. If the positive $\{p_n\}$ satisfies (6) and the condition $\sum _{k=0}^{\lambda (n)-1} | \triangle p_{k} | = O\left( \frac{P_{\lambda (n)}}{\lambda (n)}\right)$ holds, then

$$\begin{aligned} ||f-R_{n}^{\lambda }(f)||_{1}= O\left( (\lambda (n))^{-\alpha }\right) . \end{aligned}$$

(8)

3 Main Results

Mittal and Rhoades [9, 10] initiated the studies of error estimates through trigonometric-Fourier approximation (tfa) for situations in which the summability matrix T does not have monotone rows. In continuation of Mittal and Singh [11], in this paper, we generalize Theorem 3 of Mittal et al. [8] using more general $C_{\lambda }$-method. We prove the following:

Theorem 8

Let $f \in Lip(\alpha , p)$ and let $T=(a_{n, k})$ be an infinite regular triangular matrix.

(i) If $p > 1, 0 < \alpha <1, \{ a_{n, k} \} \in$ AMS in k and satisfies

$$\begin{aligned} (\lambda (n)+1) \mathrm{{max}} \{a_{\lambda (n),0},a_{\lambda (n),r}\}= O(1), \end{aligned}$$

(9)

where $r:=[\lambda (n)/2]$ then

$$\begin{aligned} ||f-\tau _{n}^{\lambda }(f)||_p = O((\lambda (n))^{- \alpha }). \end{aligned}$$

(10)

(ii) If $p>1, \alpha =1$ and

$$\begin{aligned} \sum _{k=0}^{\lambda (n)-1} (\lambda (n)-k)|\triangle _k a_{\lambda (n), k}|=O(1), \text {or} \end{aligned}$$

(11)

(iii) If $p>1, \alpha =1$ and

$$\begin{aligned} \sum _{k=0}^{\lambda (n)} |\triangle _k a_{\lambda (n), k}|=O(a_{\lambda (n), 0}), \text {or} \end{aligned}$$

(12)

(iv) If $p=1, 0 < \alpha < 1$ and

$$\begin{aligned} \sum _{k=0}^{\lambda (n)} |\triangle _k a_{\lambda (n), k}|=O(a_{\lambda (n), 0}), \end{aligned}$$

(13)

and also

$$\begin{aligned} (\lambda (n)+1) a_{\lambda (n), 0}= O(1), \end{aligned}$$

(14)

holds then (10) is satisfied.

Remarks (1) If $\lambda (n)=n$, then our Theorem 8 generalizes Theorem 3.

(2) If $T \equiv (a_{n, k})$ is a Nörlund $N_p$ (or weighted $R_p$) matrix then-

(a) If $\lambda (n)=n$, then condition (9) (or (14)) reduces to (2) while the conditions (11), (12), (13) reduce to conditions in (iii), (iv) and (v) of Theorem 1 respectively. Thus our Theorem 8 generalizes Theorems 1 and 2.

(b) De${\tilde{\mathrm{g}}}$er et al. [4] used the monotone sequences $\{p_n\}$ in Theorems 4 and 5 while our Theorem 8 claims less than the requirement of their theorems. For example, condition (11) of Theorem 8 is automatically satisfied if $\{p_n\}$ is nonincreasing sequence, i.e., L.H.S. of (11) gives

$$\begin{aligned} \sum _{k=0}^{\lambda (n)-1} (\lambda (n)-k) \left| \frac{\triangle _k p_{\lambda (n)-k}}{P_{\lambda (n)}} \right| =\,&\frac{1}{P_{\lambda (n)}}\sum _{k=0}^{\lambda (n)-1} (\lambda (n)-k) |p_{\lambda (n)-k} - p_{\lambda (n)-k-1}| \\ =\,&\frac{P_{\lambda (n)-1} - \lambda (n) p_{\lambda (n)}}{P_{\lambda (n)}} = O(1) = R.H.S., \end{aligned}$$

while the condition (12) is always satisfied if $\{p_n\}$ is nondecreasing, i.e.,

$$\begin{aligned} \sum _{k=0}^{\lambda (n)} \left| \frac{\triangle _k p_{\lambda (n)-k}}{P_{\lambda (n)}}\right| =\,&\frac{1}{P_{\lambda (n)}} \sum _{k=0}^{\lambda (n)} |p_{\lambda (n)-k} - p_{\lambda (n)-k-1}| \\ =\,&\frac{1}{P_{\lambda (n)}}[ p_{\lambda (n)} - p_{\lambda (n)-1}+ p_{\lambda (n)-1} - p_{\lambda (n)-2}+...+ p_0 -p_{-1}]\\ =\,&O\left( \frac{p_{\lambda (n)}}{P_{\lambda (n)}}\right) . \end{aligned}$$

Further, condition (9) (or (14)) of Theorem 8 reduces to (6) of Theorem 4. Thus our Theorem 8 generalizes the Theorems 4 and 5 of De${\tilde{\mathrm{g}}}$er et al. [4] under weaker assumptions and gives sharper estimate because all the estimates of De${\tilde{\mathrm{g}}}$er et al. [4] are in terms of n while our estimates are in terms of $\lambda (n)$ and $(\lambda (n))^{-\alpha } \le n^{-\alpha }$ for $0 < \alpha \le 1$.

(c) Also, Theorem 8 extends Theorems 6 and 7 of Mittal, Singh [11] where two theorems of De${\tilde{\mathrm{g}}}$er et al. [4] were generalized by dropping the monotonicity on the elements of matrix rows.

4 Lemmas

We shall use the following lemmas in the proof of our Theorem:

Lemma 1

([15]) If $f \in Lip(1, p)$, for $p>1$ then

$$\begin{aligned} ||\sigma _n (f)-s_{n}(f)||_{p}= O(n^{-1}),\,\, \forall n > 0. \end{aligned}$$

Lemma 2

([15]) If $f \in Lip(\alpha , p)$, for $0< \alpha \le 1$ and $p>1$. Then

$$\begin{aligned} ||f-s_{n}(f)||_{p}= O(n^{-\alpha }) ,\,\, \forall n > 0. \end{aligned}$$

Note: We are using sums upto $\lambda (n)$ in the nth partial sums $s_n$ and $\sigma _n$ and writing these sums $s_{n}^{\lambda }$ and $\sigma _{n}^{\lambda }$, respectively, in the above lemmas for our purpose in this paper.

Lemma 3

Let T have AMS rows and satisfy (4). Then, for $0 < \alpha < 1$,

$$\begin{aligned} \sum _{k=0}^{\lambda (n)} a_{\lambda (n), k}\,\, (k+1)^{- \alpha } = O \left( (\lambda (n)+1)^{-\alpha } \right) . \end{aligned}$$

Proof

Suppose that the rows of T are AMDS. Then there exists a $K > 0$ such that

$$\begin{aligned} \sum _{k=0}^{\lambda (n)} a_{\lambda (n), k} \,(k+1)^{- \alpha }&= \sum _{k=0}^{\lambda (n)} K a_{\lambda (n), 0} (k+1)^{- \alpha } = K a_{\lambda (n), 0} \sum _{k=0}^{\lambda (n)} (k+1)^{- \alpha }\\&= O(a_{\lambda (n), 0}(\lambda (n)+1)^{1 - \alpha }) =O((\lambda (n)+1)^{-\alpha }). \end{aligned}$$

A similar result can be proved if the rows of T are AMIS.

5 Proof of the Theorem 8

Case I. $p>1, 0 < \alpha < 1$. We have

$$\begin{aligned} \tau _{n}^{\lambda }(f) - f = \sum _{k=0}^{\lambda (n)} a_{\lambda (n), k} s_{k}(f) - f = \sum _{k=0}^{\lambda (n)} a_{\lambda (n), k} (s_{k}(f) - f) \end{aligned}$$

(15)

Thus in view of Lemmas 2 and 3 we have

$$\begin{aligned} ||\tau _{n}^{\lambda }(f)-f||_{p} \le \sum _{k=0}^{\lambda (n)} a_{\lambda (n), k} ||s_{k}(f)-f||_{p}&= \sum _{k=0}^{\lambda (n)} a_{\lambda (n), k} O((k+1)^{-\alpha })\\&= O\left( (\lambda (n)+1)^{- \alpha }\right) . \end{aligned}$$

Case III. $p>1, \alpha = 1$. We have

$$\begin{aligned} ||\tau _{n}^{\lambda }(f) - f||_p \le ||\tau _{n}^{\lambda }(f) - s_{n}^{\lambda }(f)||_p + ||s_{n}^{\lambda }(f) - f||_p. \end{aligned}$$

Again using the Lemma 2, we get

$$\begin{aligned} ||\tau _{n}^{\lambda }(f) - f||_p \le ||\tau _{n}^{\lambda }(f) - s_{n}^{\lambda }(f)||_p + O\left( (\lambda (n))^{-1}\right) . \end{aligned}$$

(16)

So, it remains to show that

$$\begin{aligned} ||\tau _{n}^{\lambda }(f) - s_{n}^{\lambda }(f)||_p = O\left( (\lambda (n))^{-1}\right) . \end{aligned}$$

(17)

Since $A_{\lambda (n), 0}=1$, we have

$$\begin{aligned} \tau _{n}^{\lambda }(f)-s_{n}^{\lambda }(f)&= \sum _{k=1}^{\lambda (n)}(A_{\lambda (n),k}- A_{\lambda (n),0}) u_{k}(f)= \sum _{k=1}^{\lambda (n)} \left( \frac{A_{\lambda (n),k}- A_{\lambda (n),0}}{k} \right) (ku_{k}(f)). \end{aligned}$$

Thus using Abel’s transformation, we get

$$\begin{aligned} ||\tau _{n}^{\lambda }(f;x)-s_{n}^{\lambda }(f;x)||_p \le&\sum _{k=1}^{\lambda (n)-1} \left| \triangle _{k}\left( \frac{A_{\lambda (n),k}- A_{\lambda (n),0}}{k} \right) \right| . || \sum _{j=1}^{k}ju_{j}(f)||_p \nonumber \\&\,\,\,\,\,\,\,\,+ \left| \frac{A_{\lambda (n),\lambda (n)}- A_{\lambda (n),0}}{\lambda (n)} \right| . ||\sum _{j=1}^{\lambda (n)}ju_{j}(f)||_p. \end{aligned}$$

(18)

Let $\sigma _{n}(s)$ denote the nth term of the (C, 1) transform of the sequence s, then

$$\begin{aligned} s_{n}^{\lambda }(f)-\sigma _{n}^{\lambda }(f)=\frac{1}{(\lambda (n)+1)} \sum _{j=1}^{\lambda (n)}j u_{j}(f). \end{aligned}$$

Using Lemma 1, we get

$$\begin{aligned} ||\sum _{j=1}^{\lambda (n)}j u_{j}||_{p} = (\lambda (n)+1)||s_{n}^{\lambda }(f)-\sigma _{n}^{\lambda }(f)||_{p} =(\lambda (n)+1)O\left( (\lambda (n))^{-1}\right) = O(1). \end{aligned}$$

(19)

Note that

$$ \left| \frac{A_{\lambda (n),0}-A_{\lambda (n),\lambda (n)}}{\lambda (n)} \right| \le ({\lambda (n)})^{-1} A_{\lambda (n),0}= O\left( (\lambda (n))^{-1}\right) . $$

Thus

$$\begin{aligned} \left| \frac{A_{\lambda (n),0}-A_{\lambda (n),\lambda (n)}}{\lambda (n)} \right| .||\sum _{j=1}^{\lambda (n)}ju_{j}(f)||_p = O\left( (\lambda (n))^{-1}\right) . \end{aligned}$$

(20)

Now

$$\begin{aligned} \triangle _{k} \left( \frac{A_{\lambda (n), k}-A_{\lambda (n), 0}}{k} \right)&= \frac{1}{k} \triangle _{k} (A_{\lambda (n),k}-A_{\lambda (n),0})+ \frac{A_{\lambda (n),k+1}-A_{\lambda (n),0}}{k(k+1)} \nonumber \\&= \frac{1}{k(k+1)} \left[ (k+1) \triangle _k A_{\lambda (n), k} + \sum _{r=k+1}^{\lambda (n)} a_{\lambda (n),r}- \sum _{r=0}^{\lambda (n)} a_{\lambda (n),r} \right] \nonumber \\&= \frac{1}{k(k+1)} \left[ (k+1) a_{\lambda (n),k}- \sum _{r=0}^{k} a_{\lambda (n),r} \right] . \end{aligned}$$

(21)

Next we claim that $\forall k \in N$,

$$\begin{aligned} |\sum _{r=0}^{k} a_{\lambda (n),r} - (k+1)a_{\lambda (n),k}| \le \sum _{r=0}^{k-1}(r+1)|a_{\lambda (n),r}-a_{\lambda (n),r+1}| , \end{aligned}$$

(22)

If $k=1$, then the inequality (22) reduces to

$$ |\sum _{r=0}^{1} a_{\lambda (n),r} - 2 a_{\lambda (n),1}| = |a_{\lambda (n),0}-a_{\lambda (n),1}|. $$

Thus (22) holds for $k=1$. Now let us assume that (22) is true for $k= m$, i.e.,

$$\begin{aligned} |\sum _{r=0}^{m} a_{\lambda (n),r} - (k+1)a_{\lambda (n),m}| \le \sum _{r=0}^{m-1}(r+1)|a_{\lambda (n),r}-a_{\lambda (n),r+1}|. \end{aligned}$$

(23)

Let $k=m+1$, using (23), we get

$$\begin{aligned}&|\sum _{r=0}^{m+1} a_{\lambda (n),r} - (m+2)a_{\lambda (n),m+1}| \\&= |\sum _{r=0}^{m} a_{\lambda (n),r} - (m+1)a_{\lambda (n),m}+(m+1)a_{\lambda (n),m}-(m+1)a_{\lambda (n),m+1}|\\&\le \sum _{r=0}^{m-1}(r+1)|a_{\lambda (n),r}-a_{\lambda (n),r+1}|+(m+1)|a_{\lambda (n),m}-a_{\lambda (n),m+1}| \\&= \sum _{r=0}^{(m+1)-1}(r+1)|a_{\lambda (n),r}-a_{\lambda (n),r+1}|. \end{aligned}$$

Thus (22) is true $\forall k$. Using (12), (14), (21), (22), we get

$$\begin{aligned} \sum _{k=1}^{\lambda (n)} |\triangle _{k} \left( \frac{A_{\lambda (n), k}-A_{\lambda (n), 0}}{k} \right) |&=\sum _{k=1}^{\lambda (n)} \frac{1}{k(k+1)} \left| (k+1) a_{\lambda (n),k}- \sum _{r=0}^{k} a_{\lambda (n),r} \right| \nonumber \\&\le \sum _{k=1}^{\lambda (n)} \frac{1}{k(k+1)} \sum _{m=0}^{k-1}(m+1)|a_{\lambda (n),m}-a_{\lambda (n),m+1}| \nonumber \\&= \sum _{k=1}^{\lambda (n)} \frac{1}{k(k+1)} \sum _{m=1}^{k} m |a_{\lambda (n),m-1}-a_{\lambda (n),m}| \nonumber \\&\le \sum _{m=1}^{\lambda (n)} m |\triangle _{m} a_{\lambda (n),m-1}| \sum _{k=m}^{\infty } \frac{1}{k(k+1)} \nonumber \\&= \sum _{k=0}^{\lambda (n)-1} |\triangle _{k} a_{\lambda (n),k}| = O(a_{\lambda (n),0})=O\left( (\lambda (n))^{-1}\right) . \end{aligned}$$

(24)

Combining (18), (19), (20) and (24) yields (17). From (17) and (16), we get

$$ ||\tau _{n}^{\lambda }(f)-f||_{p}= O\left( (\lambda (n))^{-1}\right) . $$

Case II. $p > 1, \alpha =1$. For this we first prove that the condition $\sum _{k=0}^{\lambda (n)-1}(\lambda (n)-k) | \triangle _k a_{\lambda (n), k} | = O(1)$ implies that

$$\begin{aligned} \sum _{k=1}^{\lambda (n)} \left[ \triangle _{k} \left( \frac{A_{\lambda (n), k}- A_{\lambda (n), 0}}{k} \right) \right] = O\left( (\lambda (n))^{-1}\right) . \end{aligned}$$

(25)

As in case (iii), using (22) and taking $r:=[\lambda (n)/2]$ throughout the case, we have

$$\begin{aligned} \sum _{k=1}^{\lambda (n)} \left| \triangle _{k} \left( \frac{A_{\lambda (n), k}-A_{\lambda (n), 0}}{k} \right) \right|&=\sum _{k=1}^{\lambda (n)} \frac{1}{k(k+1)} \left| (k+1) a_{\lambda (n),k}- \sum _{m=0}^{k} a_{\lambda (n),m} \right| \\&= \sum _{k=1}^{\lambda (n)} \frac{1}{k(k+1)} \sum _{m=0}^{k-1}(m+1)|a_{\lambda (n),m}-a_{\lambda (n),m+1}| \\&= \left( \sum _{k=1}^{r}+ \sum _{k=r+1}^{\lambda (n)}\right) k^{-1}(k+1)^{-1} \sum _{m=1}^{k} m |\triangle _m a_{\lambda (n),m-1}| \\&:= B_1 + B_2, say. \end{aligned}$$

Now interchanging the order of summation and using (11), we get

$$\begin{aligned} B_1&=\sum _{k=1}^{r} k^{-1}(k+1)^{-1} \sum _{m=1}^{k} m |\triangle _m a_{\lambda (n),m-1}| \le \sum _{m=1}^{r} m |\triangle _m a_{\lambda (n),m-1}| \sum _{k=m}^{\infty } k^{-1}(k+1)^{-1} \nonumber \\&= \sum _{m=1}^{r} |\triangle _m a_{\lambda (n),m-1}| = \sum _{m=\lambda (n)-r+1}^{\lambda (n)} |\triangle _{\lambda (n)-m} a_{\lambda (n),\lambda (n)-m}|\nonumber \\&= \sum _{m=r-1}^{\lambda (n)} |\triangle _{\lambda (n)-m} a_{\lambda (n),\lambda (n)-m}|.\left( \frac{m}{r-1} \right) \nonumber \\&\le \frac{1}{r-1} \sum _{m=1}^{\lambda (n)} m |\triangle _{\lambda (n)-m} a_{\lambda (n),\lambda (n)-m}|= \frac{1}{r-1} \sum _{k=0}^{\lambda (n)-1} (\lambda (n)-k) |\triangle _{k} a_{\lambda (n),k}| \nonumber \\&=\frac{1}{r-1} O(1)= O\left( (\lambda (n))^{-1}\right) . \end{aligned}$$

(26)

$$\begin{aligned} \text{ Now }\,\,\,\,B_2&= \sum _{k=r}^{\lambda (n)} k^{-1} (k+1)^{-1} \sum _{m=1}^{k} m | \triangle _m a_{\lambda (n),m-1}| \\&\le \sum _{k=r}^{\lambda (n)} k^{-1} (k+1)^{-1} \left[ \left( \sum _{m=1}^{r} + \sum _{m=r}^{k}\right) m |\triangle _m a_{\lambda (n),m-1}| \right] := B_{21} + B_{22}, say. \end{aligned}$$

Furthermore, using again our assumption, we get

$$\begin{aligned} B_{21}&= \sum _{k=r}^{\lambda (n)} k^{-1} (k+1)^{-1} \sum _{m=1}^{r} m | \triangle _m a_{\lambda (n),m-1}| \nonumber \\&\le r^{-1} \sum _{k=r}^{\lambda (n)} (k+1)^{-1} \sum _{m=1}^{\lambda (n)} m | \triangle _{\lambda (n)-m} a_{\lambda (n),\lambda (n)-m}| \nonumber \\&= r^{-1} \sum _{k=r}^{\lambda (n)} (k+1)^{-1} \sum _{k=0}^{\lambda (n)-1} (\lambda (n)-k) | \triangle _{k} a_{\lambda (n),k}| \nonumber \\&=O(r^{-1})\sum _{k=r}^{\lambda (n)} (k+1)^{-1}=O\left( (\lambda (n))^{-1}\right) . \end{aligned}$$

(27)

Again interchanging the order of summation and using (11), we get

$$\begin{aligned} B_{22}&= \sum _{k=r}^{\lambda (n)} k^{-1} (k+1)^{-1} \sum _{m=r}^{k} m | \triangle _m a_{\lambda (n),m-1}| \le \sum _{k=r}^{\lambda (n)} (k+1)^{-1} \sum _{m=r}^{k} | \triangle _{m} a_{\lambda (n),m-1}| \nonumber \\&\le \sum _{m=r}^{\lambda (n)} | \triangle _{m} a_{\lambda (n),m-1}| \sum _{k=m}^{\lambda (n)}(k+1)^{-1} \le (r+1)^{-1} \sum _{m=r}^{\lambda (n)} | \triangle _{m} a_{\lambda (n),m-1}| \sum _{k=m}^{\lambda (n)} 1 \nonumber \\&= (r+1)^{-1} \sum _{m=r}^{\lambda (n)}(\lambda (n)-m+1) | \triangle _{m} a_{\lambda (n),m-1}| \nonumber \\&= (r+1)^{-1} \sum _{k=r-1}^{\lambda (n)-1}(\lambda (n)-k) | \triangle _{k} a_{\lambda (n),k}| \nonumber \\&=(r+1)^{-1} O(1)=O\left( (\lambda (n))^{-1}\right) . \end{aligned}$$

(28)

Summing up our partial results (26), (27), (28) we verified (25). Thus (16), (18), (19), (25) and Lemma 2, again yield

$$\begin{aligned} ||f-\tau _{n}^{\lambda }(f)||_p = O\left( (\lambda (n))^{-1}\right) . \end{aligned}$$

Case IV. $p=1, 0 < \alpha < 1$.

Using Abel’s transformation, conditions (13), (14), convention $a_{n, n+1}=0$ and the result of Quade [15], we obtain

$$\begin{aligned}&||\tau _{n}^{\lambda }(f)-f||_{1} = || \sum _{k=0}^{\lambda (n)} a_{\lambda (n),k} s_{k}(f) - f ||_{1} = || \sum _{k=0}^{\lambda (n)} a_{\lambda (n),k} (s_{k}(f) - f) ||_1 \\&= || \sum _{k=0}^{\lambda (n)-1} \left( \triangle _k a_{\lambda (n),k} \right) \sum _{r=0}^{k}(s_{r}(f) - f) + ( a_{\lambda (n), \lambda (n)}-a_{\lambda (n),\lambda (n)+1}) \sum _{r=0}^{\lambda (n)}( s_{r}(f) - f ) ||_{1} \\&=||\sum _{k=0}^{\lambda (n)} \left( \triangle _k a_{\lambda (n),k}\right) \sum _{r=0}^{k}(s_{r}(f) - f)||_1 =||\sum _{k=0}^{\lambda (n)} \left( \triangle _k a_{\lambda (n),k} \right) (k+1)(\sigma _{k}(f) - f)||_{1} \\&\le \sum _{k=0}^{\lambda (n)} (k+1) | \triangle _k a_{\lambda (n),k} |.||\sigma _{k}(f) - f) ||_1 = O \left( \sum _{k=0}^{\lambda (n)} (k+1)^{1-\alpha } | \triangle _k a_{\lambda (n),k} | \right) \\&= O \left( {\lambda (n)}^{1-\alpha } \right) \sum _{k=0}^{\lambda (n)} | \triangle _k a_{\lambda (n),k} | = O \left( {\lambda (n)}^{1-\alpha } \right) O \left( a_{\lambda (n),0} \right) = O\left( (\lambda (n))^{-\alpha }\right) . \end{aligned}$$

This completes the proof of case (iv) and hence the proof of Theorem 8 is complete.