Degree of Approximation of $f\in L[0,\infty )$  by Means of Fourier–Laguerre Series

Abstract

In this paper, we determine the degree of approximation of functions belonging to $L[0,\infty )$ by the Hausdorff means of its Fourier–Laguerre series at $x=0.$ Our theorem extends some of the recent results of Nigam and Sharma [A study on degree of approximation by (E, 1) summability means of the Fourier–Laguerre expansion, Int. J. Math. Math. Sci. (2010), Art. ID 351016, 7], Krasniqi [On the degree of approximation of a function by (C, 1)(E, q) means of its Fourier–Laguerre series, International Journal of Analysis and Applications 1 (2013), 33–39] and Sonker [Approximation of Functions by (C, 2)(E, q) means of its Fourier–Laguerre series, Proceeding of ICMS-2014 ISBN 978-93-5107-261-4:125–128.] in the sense that the summability methods used by these authors have been replaced by the Hausdorff matrices.

1 Introduction

Let f be a function belonging to $L[0,\infty )$ in the sense that f is Labesgue integrable in the interval $[0,\infty )$. The Fourier–Laguerre expansion of f is given by

$$\begin{aligned} f(x)\sim \sum _{n=0}^{\infty }a_{n}L_{n}^{(\alpha )}(x), \end{aligned}$$

(1)

where

$$\begin{aligned} \varGamma (\alpha +1)\left( \begin{array}{c} n+\alpha \\ n \\ \end{array} \right) a_{n}=\int _{0}^{\infty }e^{-x}x^{\alpha }f(x)L_{n}^{(\alpha )}(x)dx \end{aligned}$$

(2)

and $L_{n}^{(\alpha )}(x)$ denotes the nth Laguerre polynomial of order $\alpha > -1,$ defined by the generating function

$$\begin{aligned} \sum _{n=0}^{\infty }L_{n}^{(\alpha )}(x)\omega ^{n}=(1-\omega )^{-\alpha -1}\exp \left( \frac{-x\omega }{1-\omega }\right) . \end{aligned}$$

(3)

When $x=0,$

$$ L_{n}^{(\alpha )}(0)= \left( \begin{array}{c} n+\alpha \\ n \\ \end{array} \right) ~[9]. $$

The $n{\text{ th }}$ partial sums of (1) are defined by

$$\begin{aligned} s_{n}(f;x)= \sum _{k=0}^{n}a_{k}L_{k}^{(\alpha )}(x). \end{aligned}$$

(4)

The Ces$\grave{a}$ro means of order $\lambda$ of the Fourier–Laugerre series are defined by

$$C^{\lambda }_{n}(f;x)=\frac{1}{\left( \begin{array}{c} n+\lambda \\ n \\ \end{array} \right) }\sum _{k=0}^{n}\left( \begin{array}{c} \lambda +n-k-1 \\ n-k \\ \end{array} \right) s_{k}(f;x). $$

The Euler means of the Fourier–Laugerre series are defined by

$$E^{q}_{n}(f;x)=\frac{1}{(1+q)^{n}}\sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) q^{n-k}s_{k}(f;x),\;\;q>0. $$

The Hausdorff matrix $H\equiv (h_{n, k})$ is an infinite lower triangular matrix defined by

$$h_{n, k}= \left\{ \begin{array}{ll} \left( \begin{array}{c} n \\ k \\ \end{array} \right) \triangle ^{n-k}\mu _k &{}, 0\le k\le n, \\ 0 &{}, \;\, k>n,\\ \end{array} \right. $$

where $\triangle$ is the forward difference operator defined by $\triangle \mu _n= \mu _n-\mu _{n+1}$ and $\triangle ^{k+1} \mu _n=\triangle ^k (\triangle \mu _n).$ If H is regular, then $\{\mu _n \},$ known as moment sequence, has the representation

$$\mu _n= \int _0^1 u^n d\gamma (u),$$

where $\gamma (u)$, known as mass function, is continuous at $u=0$ and belongs to $\textit{BV}[0, 1]$ such that $\gamma (0)=0, \gamma (1)=1;$ and for $0< u< 1,$ $\gamma (u)=[\gamma (u+0) + \gamma (u-0)]/2$ [11].

The Hausdorff means of the Fourier–Laugerre series are defined by

$$\begin{aligned} {H}_n(f; x):=\sum _{k=0}^n h_{n, k} {s}_k(f; x), \;\;n=0,1,2,... \end{aligned}$$

(5)

The Fourier–Laugerre series is said to be summable to s by the Hausdorff means, if ${H}_n(f;x)\rightarrow s~as~n\rightarrow \infty$, [3].

For the examples of Hausdorff matrices, one can see [7, 8, 11] and references therein.

In this paper, the class of all regular Hausdorff matrices with moment sequence $\{ \mu _n\}$ associated with mass function $\gamma (u)$ having constant derivative, is denoted by $H_1$.

We also write

$$\varphi (y)=\frac{e^{-y}y^{\alpha }(f(y)-f(0))}{\varGamma (\alpha +1)},$$

and

$$g(u,y)=\sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) u^{k}(1-u)^{n-k}L_{k}^{(\alpha +1)}(y).$$

2 Known Results

Gupta [2] obtained the degree of approximation of $f\in L[0,\infty )$ by Ces$\grave{a}$ro means of order k of the Fourier–Laguerre series at the point $x=0$, where $k > \alpha +1/2$. Nigam and Sharma [5] have used (E, 1) means of the Fourier–Laguerre series for $-1 < \alpha < 1/2$ which is more appropriate range from the application point of view. The authors have proved the following result:

Theorem A If

$$E^{1}_{n}=\frac{1}{2^{n}}\sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) s_{k}\;\;\;\rightarrow \infty \;\;\mathrm{as}\;\;n\rightarrow \infty , $$

then the degree of approximation of Fourier–Laguerre expansion at the point $x=0$ by (E, 1) means $E_{n}^1$ is given by

$$\begin{aligned} E_{n}^{1}(0)-f(0)=o(\xi (n)), \end{aligned}$$

(6)

provided that

$$\begin{aligned} \varPhi (t)= \int _{0}^{t}|\varphi (y)|dy=o\left( t^{\alpha +1}\xi (1/t)\right) ,\;\;t\rightarrow 0, \end{aligned}$$

(7)

$$\begin{aligned} \int _{\delta }^{n}e^{y/2}y^{-((2{\alpha }+3)/4)}|\varphi (y)|dy=o\left( n^{-((2{\alpha }+1)/4)}\xi (n)\right) , \end{aligned}$$

(8)

$$\begin{aligned} \int _{n}^{\infty }e^{y/2}y^{-1/3}|\varphi (y)|dy=o(\xi (n)),\;\;n\rightarrow \infty , \end{aligned}$$

(9)

where $\delta$ is a fixed positive constant and $\alpha \in (-1,-1/2),$ and $\xi (t)$ is a positive monotonic increasing function of t such that $\xi (n)\rightarrow \infty$ as $n\rightarrow \infty .$

Following, Nigam and Sharma [5], Krasniqi [4] has used the (C, 1)(E, q) means of the Fourier–Laguerre series to obtain the degree of approximation of $f\in L[0,\infty )$ at point $x=0$ and has proved the following result:

Theorem B The degree of approximation of the Fourier–Laguerre expansion at the point $x=0$ by the $[(C,1)(E,q)]_{n}$ means is given by

$$\begin{aligned}{}[(C,1)(E,q)]_{n}(0)-f(0)=o(\xi (n)), \end{aligned}$$

(10)

provided that the conditions (7)–(9) given in Theorem A are satisfied.

Recently, Sonker [10] has also proved the same result using $[(C,2)(E,q)]_{n}$ means of the Fourier–Laguerre series for the same range of $\alpha$ as follows:

Theorem C The degree of approximation of the Fourier–Laguerre expansion at the point $x=0$ by the $[(C,2)(E,q)]_{n}$ means is given by

$$\begin{aligned}{}[(C,2)(E,q)]_{n}(0)-f(0)=o(\xi (n)), \end{aligned}$$

(11)

provided that the conditions (7)–(9) given in Theorem A are satisfied.

Remark 1 We observe that Krasniqi [4, p. 37] has optimized $\sum _{k=0}^{v}\left( \begin{array}{c} v \\ k \\ \end{array} \right) q^{k}k^{(2\alpha +1)/4}$ by its maximum value $(1+q)^{v}v^{(2\alpha +1)/4}$. This is possible only when $\alpha > -1/2$. But the author has used $-1 < \alpha < 1/2$ [4, p. 35, Theorem 2.1]. Similar error can also be seen in [5, 10].

3 Main Results

In this paper, we extend the above results using the Hausdorff means, which is a more general summability method, for an appropriate range of $\alpha$. More precisely, we prove the following:

Theorem 1

The degree of approximation of $f\in L[0,\infty )$ at the point $x=0$ by the Hausdorff means of the Fourier–Laguerre series generated by $H\in H_{1}$ is given by

$$\begin{aligned} H_{n}(f;0)-f(0)=o(\xi (n)) \end{aligned}$$

(12)

where $\xi (t)$ is a positive increasing function such that $\xi (t)\rightarrow \infty$ as $t\rightarrow \infty$ and satisfies the following conditions

$$\begin{aligned} \varPhi (y)=\int _{0}^{t}|\varphi (y)|dy=o\left( t^{\alpha +1}\xi (1/t)\right) ,\;\;t\rightarrow 0, \end{aligned}$$

(13)

$$\begin{aligned} \int _{\delta }^{n}e^{y/2}\,y^{-((2{\alpha }+3)/4)}|\varphi (y)|dy=o\left( n^{-((2{\alpha }+1)/4)}\xi (n)\right) , \end{aligned}$$

(14)

and

$$\begin{aligned} \int _{n}^{\infty }e^{y/2}\,y^{-1/3}|\varphi (y)|dy=o(\xi (n)),\;\;n\rightarrow \infty , \end{aligned}$$

(15)

where $\delta$ is a fixed positive constant and $\alpha > -1/2.$

For the proof of our theorem, we need the following lemmas:

Lemma 1

[9, p. 177]. Let $\alpha$ be an arbitrary real number, c and $\delta$ be fixed positive constants. Then

$$\begin{aligned} L_{n}^{(\alpha )}(x)=\left\{ \begin{array}{ll} O\left( n^{(\alpha )}\right) ,&{}\;\;\text{ if }\;\; 0 \le x \le \frac{c}{n},\\ O\left( x^{-(2{\alpha }+1)/4}n^{(2{\alpha }-1)/4}\right) ,&{}\;\; \text{ if }\;\; \frac{c}{n} \le x \le \delta , \end{array}\right. \end{aligned}$$

(16)

as $n \rightarrow \infty$.

Lemma 2

[9, p. 240]. Let $\alpha$ be an arbitrary real number, $\delta >0$ and $0 < \eta < 4$. Then

$$\begin{aligned} \max e^{-x/2}x^{(\alpha /2+1/4)} |L_{n}^{(\alpha )}(x)|=\left\{ \begin{array}{cl} O\left( n^{(\alpha /2-1/4)}\right) ,&{}\;\;\text{ if }\;\; \delta \le x \le (4-\eta )n,\\ O\left( n^{({\alpha }/2-1/12)}\right) ,&{}\;\; \text{ if }\;\; x \ge \delta , \end{array}\right. \end{aligned}$$

(17)

as $n \rightarrow \infty$.

Lemma 3

For $0< u< 1$ and $0\le y \le \delta ,$

$$\begin{aligned} \left| \int _{0}^{1}g(u,y)d\gamma (u)\right| =\left\{ \begin{array}{lc} O\left( n^{(\alpha +1)}\right) ,&{}\;\;\text{ if }\;\; 0 \le y \le \frac{1}{n},\\ O\left( y^{-(2{\alpha }+3)/4}n^{(2{\alpha }+1)/4}\right) ,&{}\;\; \text{ if }\;\; \frac{1}{n} \le y \le \delta , \end{array}\right. \end{aligned}$$

(18)

as $n \rightarrow \infty$.

Proof

The g(u, y) can be written as

$$g(u,y)=(1-u)^{n}\sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) {\left( \frac{u}{1-u}\right) }^{k}L_{k}^{(\alpha +1)}(y).$$

Then

$$\begin{aligned} \left| \int _{0}^{1}g(u,y)d\gamma (u)\right|= & {} \left| \int _{0}^{1}(1-u)^{n}\sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) {\left( \frac{u}{1-u}\right) }^{k}L_{k}^{(\alpha +1)}(y)d\gamma (u)\right| \\= & {} \left| M\int _{0}^{1}(1-u)^{n}\sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) {\left( \frac{u}{1-u}\right) }^{k}L_{k}^{(\alpha +1)}(y)du\right| \end{aligned}$$

Now, using Lemma 1 for $0 \le y \le \frac{1}{n}$ (taking $\alpha +1$ for $\alpha$ and $c=1$), we have

$$\begin{aligned} \left| \int _{0}^{1}g(u,y)d\gamma (u)\right|= & {} \int _{0}^{1}(1-u)^{n}\sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) {\left( \frac{u}{1-u}\right) }^{k}O(k^{\alpha +1})du\nonumber \\= & {} O\left( n^{\alpha +1}\int _{0}^{1}(1-u)^{n}\sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) {\left( \frac{u}{1-u}\right) }^{k}du\right) \nonumber \\= & {} O\left( n^{\alpha +1}\int _{0}^{1}(1-u)^{n}du\right) \nonumber \\= & {} O\left( n^{\alpha +1}\right) . \end{aligned}$$

(19)

Again, using Lemma 1 for $\frac{1}{n}\le y \le \delta$, we have

$$\begin{aligned} \left| \int _{0}^{1}g(u,y)d\gamma (u)\right|= & {} \int _{0}^{1}(1-u)^{n}\sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) {\left( \frac{u}{1-u}\right) }^{k}O\left( y^{-(2{\alpha }+3)/4}k^{(2{\alpha }+1)/4}\right) du\nonumber \\= & {} O\left( y^{-(2{\alpha }+3)/4}n^{(2{\alpha }+1)/4}\int _{0}^{1}(1-u)^{n}\sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) {\left( \frac{u}{1-u}\right) }^{k}du\right) \nonumber \\= & {} O\left( y^{-(2{\alpha }+3)/4}n^{(2{\alpha }+1)/4}\right) . \end{aligned}$$

(20)

Collecting (19) and (20), the proof of Lemma 3 is completed.

Lemma 4

For $0< u< 1,$

$$\begin{aligned} \left| \int _{0}^{1}g(u,y)d\gamma (u)\right| =\left\{ \begin{array}{cl} O\left( e^{y/2}y^{-(2{\alpha }+3)/4}n^{(2\alpha +1)/4}\right) ,&{}\;\;\text{ if }\;\; \delta \le y \le {n},\\ O\left( e^{y/2}y^{-(3{\alpha }+5)/6}n^{({\alpha }+1)/2}\right) ,&{}\;\; \text{ if }\;\; y\ge \delta , \end{array}\right. \end{aligned}$$

(21)

as $n \rightarrow \infty$.

Proof

Following the Lemma 3, we have

$$\begin{aligned} \left| \int _{0}^{1}g(u,y)d\gamma (u)\right|= & {} \left| \int _{0}^{1}(1-u)^{n}\sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) {\left( \frac{u}{1-u}\right) }^{k}L_{k}^{(\alpha +1)}(y)du\right| \end{aligned}$$

Now, using Lemma 2 for $\delta \le y \le n$ (taking $\alpha +1$ for $\alpha$ and $\eta =3$), we have

$$\begin{aligned} \left| \int _{0}^{1}g(u,y)d\gamma (u)\right|= & {} \left| \int _{0}^{1}e^{(y/2)}y^{-(2\alpha +3)/4}(1-u)^{n}\sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) {\left( \frac{u}{1-u}\right) }^{k}e^{-(y/2)}y^{(2\alpha +3)/4}L_{k}^{(\alpha +1)}(y)du\right| \nonumber \\= & {} \,\int _{0}^{1}e^{y/2}y^{-(2{\alpha }+3)/4}(1-u)^{n}\sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) {\left( \frac{u}{1-u}\right) }^{k}O\left( k^{(2{\alpha }+1)/4}\right) du\nonumber \\= & {} O\left( e^{y/2}y^{-(2{\alpha }+3)/4}n^{(2{\alpha }+1)/4}\right) . \end{aligned}$$

(22)

Again, using Lemma 2 for $y \ge n$, we have

$$\begin{aligned} \left| \int _{0}^{1}g(u,y)d\gamma (u)\right|= & {} \left| \int _{0}^{1}e^{(y/2)}y^{-(3\alpha +5)/6}(1-u)^{n}\sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) {\left( \frac{u}{1-u}\right) }^{k}e^{-(y/2)}y^{(3\alpha +5)/6}L_{k}^{(\alpha +1)}(y)du\right| \nonumber \\= & {} \,\int _{0}^{1}e^{y/2}y^{-(3{\alpha }+5)/6}(1-u)^{n}\sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) {\left( \frac{u}{1-u}\right) }^{k}O\left( k^{({\alpha }+1)/2}\right) du\nonumber \\= & {} O\left( e^{y/2}y^{-(3{\alpha }+5)/6}n^{({\alpha }+1)/2}\right) . \end{aligned}$$

(23)

Collecting (22) and (23), the proof of Lemma 4  is completed.

Proof of Theorem 1

We have

$$\begin{aligned} s_{n}(0)= & {} \sum _{k=0}^{n}a_{k}L_{k}^{(\alpha )}(0) \\= & {} \,\sum _{k=0}^{n}\frac{1}{\varGamma (\alpha +1)\left( \begin{array}{c} n+\alpha \\ n \\ \end{array} \right) } \left( \int _{0}^{\infty }e^{-y}y^{\alpha }f(y)L_{k}^{(\alpha )}(y)dy\right) L_{k}^{(\alpha )}(0) \\= & {} \, \frac{1}{\varGamma (\alpha +1)}\int _{0}^{\infty }e^{-y}y^{\alpha }f(y)\sum _{k=0}^{n}L_{k}^{(\alpha )}(y)dy \\= & {} \,\frac{1}{\varGamma (\alpha +1)}\int _{0}^{\infty }e^{-y}y^{\alpha }f(y)L_{n}^{(\alpha +1)}(y)dy, \end{aligned}$$

so that

$$\begin{aligned} H_{n}(f;0)= & {} \sum _{k=0}^{n}h_{n,k}s_{k}(0)\\= & {} \,\sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) \varDelta ^{n-k}\mu _{k} \frac{1}{\varGamma (\alpha +1)}\int _{0}^{\infty }e^{-y}y^{\alpha }f(y)L_{k}^{(\alpha +1)}(y)dy. \end{aligned}$$

Thus

$$\begin{aligned} H_{n}(f;0)-f(0)= & {} \sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) \varDelta ^{n-k}\mu _{k}\left( \frac{1}{\varGamma (\alpha +1)}\int _{0}^{\infty }e^{-y}y^{\alpha }f(y)L_{k}^{(\alpha +1)}(y)dy-f(0)\right) \\= & {} \,\sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) \varDelta ^{n-k}\mu _{k}\frac{1}{\varGamma (\alpha +1)}\int _{0}^{\infty }e^{-y}y^{\alpha }(f(y)-f(0))L_{k}^{(\alpha +1)}(y)dy \\= & {} \,\sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) \varDelta ^{n-k}\mu _{k}\int _{0}^{\infty }\varphi (y)L_{k}^{(\alpha +1)}(y)dy\\= & {} \,\int _{0}^{\infty }\varphi (y)\left( \sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) \varDelta ^{n-k}\mu _{k}L_{k}^{(\alpha +1)}(y)\right) dy\\= & {} \, \int _{0}^{\infty }\varphi (y)\left( \sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) \int _{0}^{1}u^{k}(1-u)^{n-k}d\gamma (u)L_{k}^{(\alpha +1)}(y)\right) dy \nonumber \\= & {} \,\int _{0}^{\infty }\varphi (y)\left( \int _{0}^{1}\sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) u^{k}(1-u)^{n-k}L_{k}^{(\alpha +1)}(y)d\gamma (u)\right) dy\\= & {} \,\int _{0}^{\infty }\varphi (y)\left( \int _{0}^{1}g(u,y)d\gamma (u)\right) dy \end{aligned}$$

and

$$\begin{aligned} \left| H_{n}(f;0)-f(0)\right|= & {} \left| \int _{0}^{\infty }\varphi (y)\left( \int _{0}^{1}g(u,y)d\gamma (u)\right) dy\right| \nonumber \\\le & {} \, \int _{0}^{\infty }|\varphi (y)|\left| \int _{0}^{1}g(u,y)d\gamma (u)\right| dy\nonumber \\= & {} \,\left( \int _{0}^{1/n}+\int _{1/n}^{\delta }+\int _{\delta }^{n}+\int _{n}^{\infty }\right) \left( |\varphi (y)|\left| \int _{0}^{1}g(u,y)d\gamma (u)\right| dy\right) \nonumber \\= & {} I_{1}+I_{2}+I_{3}+I_{4}. \end{aligned}$$

(24)

Now, using Lemma 3 for $0 \le y \le \frac{1}{n}$, we have

$$\begin{aligned} I_{1}= & {} \int _{0}^{1/n}|\varphi (y)|\left| \int _{0}^{1}g(u,y)d\gamma (u)\right| dy \nonumber \\= & {} \,O\left( n^{\alpha +1}\right) \int _{0}^{1/n}|\varphi (y)|dy\nonumber \\= & {} \,O (n^{(\alpha +1)})o\left( \left( \frac{1}{n}\right) ^{\alpha +1}\xi (n)\right) \nonumber \\= & {} o(\xi (n)), \end{aligned}$$

(25)

in view of condition (13).

Further, using Lemma 3 for $\frac{1}{n} \le y \le \delta$, we have,

$$\begin{aligned} I_{2}= & {} \int _{1/n}^{\delta }|\varphi (y)|O\left( y^{-(2{\alpha }+3)/4}n^{(2{\alpha }+1)/4}\right) dy \\= & {} \,O\left( n^{(2{\alpha }+1)/4}\right) \left( \int _{1/n}^{\delta }y^{-(2{\alpha }+3)/4}|\varphi (y)|dy\right) . \end{aligned}$$

Following [5, p. 6], we have

$$\begin{aligned} I_{2}= o(\xi (n)), \end{aligned}$$

(26)

in view of condition (13).

Now, using Lemma 4  for $\delta \le y \le n$, we have

$$\begin{aligned} I_{3}= & {} \,\int _{\delta }^{n}|\varphi (y)|\left| \int _{0}^{1}g(u,y)d\gamma (u)\right| dy \nonumber \\= & {} \,\int _{\delta }^{n}O\left( e^{y/2}y^{-((2{\alpha }+3)/4)}n^{(2\alpha +1)/4}\right) |\varphi (y)| dy\nonumber \\= & {} \,O\left( n^{(2\alpha +1)/4}\right) \left( \int _{\delta }^{n}e^{y/2}y^{-((2{\alpha }+3)/4)}|\varphi (y)| dy\right) \nonumber \\= & {} \,O\left( n^{(2\alpha +1)/4}\right) o \left( (n^{-(2\alpha +1)/4})\xi (n)\right) \nonumber \\= & {} \, o(\xi (n)), \end{aligned}$$

(27)

in view of condition (14).

Further, using Lemma 4, we have

$$\begin{aligned} I_{4}= & {} \int _{n}^{\infty }|\varphi (y)|\left| \int _{0}^{1}g(u,y)d\gamma (u)\right| dy \nonumber \\= & {} \,\int _{n}^{\infty }|\varphi (y)|O\left( e^{y/2}y^{-(3{\alpha }+5)/6}n^{({\alpha }+1)/2}\right) dy\nonumber \\= & {} \, O\left( n^{(\alpha +1)/2}\right) \left( \int _{n}^{\infty }\frac{e^{y/2}y^{-1/3}|\varphi (y)|}{y^{(\alpha +1)/2}}dy\right) \nonumber \\= & {} \,o\left( (\xi (n))n^{(\alpha +1)/2}\left( n^{-(\alpha +1)/2}\right) \right) \nonumber \\= & {} \, o(\xi (n)), \end{aligned}$$

(28)

in view of condition (15).

Collecting (24)–(28), we have

$$ H_{n}(f;0)-f(0)=o(\xi (n)).$$

Hence the proof of Theorem 1 is completed.

4 Corollaries

The following corollaries can be derived from our Theorem 1.

Corollary 1

As discussed in [7, p. 306, Lemma 1] and [11, p. 38], if we take the mass function $\gamma (u)$ given by

$$\gamma (u)=\left\{ \begin{array}{cc} 0,&{}\;\;\;0\le u \le a,\\ 1,&{}\;\;\;a\le u \le 1, \end{array}\right. $$

where $a=\frac{1}{(1+q)},q>0$, the Hausdorff matrix H reduces to Euler matrix $(E,q),q>0$ and defines the corresponding (E, q) means given by

$$ E^{n}_{q}(f;x)= \frac{1}{(1+q)^{n}}\sum _{k=0}^{n}\left( \begin{array}{c} n \\ k \\ \end{array} \right) q^{n-k}s_{k}(f;x),\;\;q>0.$$

Hence the Theorem 1 reduces to Theorem A (result proved by Nigam and Sharma [5, p. 3, Theorem 2.1]).

Corollary 2

As discussed in [1, p. 400] and [6, p. 2747], the Ces$\grave{a}$ro matrix of order $\lambda$, is also a Hausdorff matrix obtained by mass function $\gamma (u)=1-(1-u)^\lambda$ and the corresponding Ces$\grave{a}$ro means are given by

$$C^{\lambda }_{n}(f;x)=\frac{1}{\left( \begin{array}{c} n+\lambda \\ n \\ \end{array} \right) }\sum _{k=0}^{n}\left( \begin{array}{c} \lambda +n-k-1 \\ n-k \\ \end{array} \right) s_{k}(f;x). $$

Further, Rhoades [7, p. 308] and Rhoades et al. [8, p. 6869] has mentioned that the product of two Hausdorff matrices is again a Hausdorff matrix. Hence the Theorem B and Theorem C (results proved by Krasniqi [4, p. 35, Theorem 2.1] and Sonker [10, p. 126, Theorem 1]) are also particular cases of our Theorem 1.

Remark 2 This is an open problem to associate the above discussed results with the $L^{p}$-spaces.