Functional approximation in Besov space using generalized Nörlund–Hausdorff product matrix

Abstract

In the present work, a best approximation of a function f in Besov space using the generalized Nörlund–Hausdorff $(N_{pq}\triangle _{H})$ product means, for the two different cases $1<\sigma <\infty, \rho \geq 1, 0\leq \eta <\nu <2$ and $\sigma =\infty, 0\leq \eta < \nu <2$, has been obtained. Our theorem generalizes the results of (Kyungpook Math. J. 50:545–556, 2010; Int. J. Appl. Math. 24(4):479–490, 2011; Nepal J. Sci. Technol. 14(2):117–122, 2013; Ultra Sci. Phys. Sci. 14(1): 53–58, 2002). Thus, the results of (Kyungpook Math. J. 50:545–556, 2010; Int. J. Appl. Math. 24(4):479–490, 2011; Nepal J. Sci. Technol. 14(2):117–122, 2013; Ultra Sci. Phys. Sci. 14(1): 53–58, 2002) become particular cases of our theorem. We also obtain some useful corollaries from our theorem.

1 Introduction

Besov space serves to generalize more elementary functional spaces like Sobolov spaces, Lipschitz spaces, Hölder spaces, and generalized Hölder spaces. It is important to note that Besov space is effective at measuring regularity properties of functions.

Several researchers like those of [3, 5,6,7,8,9] have obtained a degree of approximation of certain functions in different functional spaces such as Lipschitz space and Hölder spaces using single and product summability means. Therefore, in the present work, we obtain the degree of approximation of a function in a Besov space using generalized Nörlund–Hausdorff $(N_{pq}\triangle _{H})$ product means, which provide a more general estimate than those of [1,2,3,4].

2 Preliminaries

From [10] we define the following:

Let $C_{2\pi }:=C[0,2\pi ]$ denote the Banach space of all 2π-periodic continuous function defined on $[0,2\pi ]$ under the supremum norm and

$$ L_{\rho }:=L^{\rho }[0,2\pi ] := \biggl\{ f:[0,2\pi ]\to \mathbb{R}; \int _{0}^{2\pi } \bigl\vert f(y) \bigr\vert ^{\rho } \,dy < \infty, \rho \geq 1 \biggr\} $$

be the space of all 2π-periodic integrable functions.

The $L_{\rho }$-norm of a function f is defined by

$$ \Vert f \Vert _{\rho } = \textstyle\begin{cases} \{ \frac{1}{2\pi }\int _{0}^{2\pi } \vert f(y) \vert ^{\rho } \,dy \}^{\frac{1}{\rho }} & \text{for } 1\leq \rho < \infty; \\ \operatorname{ess} \sup_{f\in (0,2\pi )} \vert f(y) \vert & \text{for } \rho =\infty. \end{cases} $$

The modulus of continuity of a function f in $L_{\rho }$ space is defined by

$$ w(f;l)=\mathop{\sup_{y, y+h \in [0,2\pi ]}}_{ \vert h \vert < l} \bigl\vert f(y+h)-f(y) \bigr\vert . $$

The kth order modulus of smoothness of a function f in $L_{\rho }$ space is defined by

$$\begin{aligned} &w_{k}(f,l)_{\rho }=\sup_{0< h\leq l} \bigl\Vert \triangle _{h}^{k}(f,\cdot) \bigr\Vert _{\rho },\quad l>0, \\ &\triangle _{h}^{k}(f,y)=\sum _{i=0}^{k}(-1)^{k-i}\binom{k}{i}f(y+ih),\quad k\in \mathbb{N}. \end{aligned}$$

Remark 1

• (i) For $\rho =\infty, k=1$ and a continuous function f, the modulus of smoothness $w_{k}(f,l)_{\rho }$ reduces to the modulus of continuity $w(f,l)$.

• (ii) For $0<\rho <\infty$, $k=1$ and a continuous function f, $w_{k}(f,l)_{\rho }$ becomes the integral modulus of continuity of first order $w(f,l)_{\rho }$.

Remark 2

If a function f belongs to $C_{2\pi }$ and $w(f,l)=O(l^{\nu })$, for $0<\nu \leq 1$, then the function f belongs to Lipν. If the function f belongs to $L_{\rho }, 0<\rho < \infty$, and $w(f,l)_{\rho }=O(l^{\nu })$, $0<\nu \leq 1$, then the function f belongs to $\operatorname{Lip}(\nu,\rho )$.

If $\rho =\infty$ in class $\operatorname{Lip}(\nu,\rho )$ then $\operatorname{Lip}(\nu,\rho )$ class reduces to the class Lipν. Thus,

$$ \operatorname{Lip} \nu \subseteq \operatorname{Lip}(\nu, \rho ). $$

(1)

Consider $\nu >0$, $k>\nu$ i.e., $k=[\nu ]+1$, where k is the smallest integer.

For $f\in L_{\rho }$, if

$$ w_{k}(f,l)_{\rho }=O\bigl(l^{\nu }\bigr),\quad l>0, $$

(2)

then the function $f\in \operatorname{Lip}^{*}(\nu,\rho )$ (generalized Lipschitz class) and in this case the seminorm is given by

$$ \vert f \vert _{\operatorname{Lip}^{*}}=\sup_{l>0} \bigl(l^{-\nu }w_{k}(f,l)_{\rho }\bigr). $$

Thus,

$$ \operatorname{Lip}(\nu,\rho )\subseteq \operatorname{Lip}^{*}(\nu,\rho ) . $$

(3)

Remark 3

We are not representing here the definition of well-known Hölder spaces $H_{\nu }$ and $H_{\nu,\rho }$. The reader can consult [11] for detailed work on these spaces, It can be noted that [11, 12]:

• (1) $H_{\nu } \subseteq H_{\eta }\subseteq C_{2\pi }$ for $0<\eta \leq \nu \leq 1$ ($H_{\nu }$ is a Banach space),

• (2) $H_{\nu,\rho } \subseteq H_{\eta,\rho } \subseteq L_{ \rho }$ for $0<\eta \leq \nu \leq 1$, $H_{\nu,\rho }$ is a Banach space for $\rho \geq 1$ and a complete ρ-normed space for $0<\rho <1$.

Let $\nu >0$ be given, and let $k=[\nu ]+1$. For $0<\rho, \sigma \leq \infty$, the Besov space $B_{\sigma }^{\nu }(L_{\rho })$ is a collection of all the functions (2π-periodic) $f\in L_{\rho }$ such that

$$\begin{aligned} \vert f \vert _{B_{\sigma }^{\nu }(L_{\rho })} &:= \bigl\Vert w_{k}(f,\cdot) \bigr\Vert _{\nu,\sigma }= \textstyle\begin{cases} ( \int _{0}^{\pi }[l^{-\nu }w_{k}(f,l)_{\rho }]^{\sigma } \frac{dl}{l} )^{\frac{1}{\sigma }}, & 0< \sigma < \infty; \\ \sup_{l>0}(l^{-\nu }w_{k}(f,l)_{\rho }), & \sigma =\infty, \end{cases}\displaystyle \end{aligned}$$

(4)

is finite [13].

Note 1

From (2) and (3) it is observed that, for $\sigma = \infty$, $B_{\infty }^{\nu }(L_{\rho })=\operatorname{Lip}^{*}(\nu,\rho )$. Then the following cases are obtained:

• (i) If we take $0<\nu \leq 1, 0<\rho <\infty$, then $\operatorname{Lip}^{*} {(\nu,\rho )}$ reduces to the $\operatorname{Lip}{(\nu, \rho)}$ class.

• (ii) If we take $\rho \to \infty$ then $\operatorname{Lip}(\nu,\rho )$ reduces to Lipν class.

It is observed that (4) is a seminorm if $1\leq \rho, \sigma \leq \infty$ but a quasi-seminorm in other cases [10]. In this way, the quasi-norm for Besov space $B_{\sigma }^{\nu }(L_{\rho })$ is given by

$$\begin{aligned} \Vert f \Vert _{B_{\sigma }^{\nu }{(L_{\rho })}} &:= \Vert f \Vert _{\rho }+ \vert f \vert _{B_{q_{1}}^{\nu }(L_{\rho })}= \Vert f \Vert _{\rho }+ \bigl\Vert w_{k}(f,\cdot) \bigr\Vert _{\nu,\sigma }. \end{aligned}$$

(5)

Remark 4

• 1. If $0<\nu <1$, the space $B_{\infty }^{\nu }(L_{\rho })$ reduces to the space $H_{\nu,\rho }$ [14].

• 2. If $\rho =\infty =\sigma$ and $0<\nu <1$, the Besov space reduces to the space $H_{\nu }$ [15].

The δ-order error of approximation of a function $f\in C_{2 \pi }$ is defined by

$$ E_{\delta }(f)= \inf_{t_{\delta }} \Vert f-t_{\delta } \Vert , $$

where $t_{\delta }$ is a trigonometric polynomial of degree δ [16].

If $E_{\delta }(f)\to 0$ as $\delta \to \infty$, the $E_{\delta }(f)$ is said to be the best approximation of f [16].

Let $\sum_{\delta =0}^{\infty }u_{\delta }$ be an infinite series such that $s_{j}=\sum_{i=0}^{j}u_{i}$.

The δth partial sum of the Fourier series (F. S.) is denoted by $s_{\delta }(f;y)$ and is given by [16]

$$ s_{\delta }(f;y)-f(y)=\frac{1}{2\pi } \int _{0}^{\pi }\phi (y,l) \frac{ \sin (\delta +\frac{1}{2})l}{\sin (\frac{l}{2})} \,dl. $$

A Hausdorff matrix is a lower triangle matrix with entries

$$ h_{\delta,m}=\binom{\delta }{m}\triangle ^{\delta -m}\mu _{m}, $$

where $\triangle \mu _{m}=\mu _{m}-\mu _{m+1}$ and $\triangle ( \triangle ^{\delta }\mu _{m})=\triangle ^{\delta +1}\mu _{m}$.

If $t_{\delta }^{\triangle _{H}}=\sum_{j=0}^{\delta }h_{\delta,m}s _{j}$ as $\delta \to \infty$, then the series $\sum_{\delta =0}^{ \infty } u_{\delta }$ is said to be summable to the sum s by the Hausdorff method ($\triangle _{H}$ means).

The Hausdorff matrix H is regular, i.e., H preserves the limit of each convergent sequence iff

$$ \int _{0}^{1} \bigl\vert d\bigl(\nu (z)\bigr) \bigr\vert < \infty, $$

where the mass function $\nu \in BV[0,1]$, $\nu (0+)=\nu (0)=0$, and $\nu (1)=1$. In this case, $\mu _{\delta }$ has the representation [17]

$$ \mu _{\delta }= \int _{0}^{1}z^{\delta }\,d\nu (z). $$

Considering the two sequences $\{p_{\delta }\}$ and $\{q_{\delta }\}$, we write

$$ t_{\delta }^{N_{pq}}=\frac{1}{R_{\delta }}\sum _{k=0}^{\delta }p_{ \delta -k}q_{k}s_{k};\qquad R_{\delta }=\sum_{k=0}^{\delta } p_{k}q_{ \delta -k}\neq 0\quad\text{for all } \delta, $$

then the generalized Nörlund means $(N_{p,q})$ of the sequence $\{s_{\delta }\}$ is denoted by the sequence $t_{\delta }^{pq}$. If $t_{\delta }^{pq}\to s,\text{ as } \delta \to \infty$ then the series $\sum_{\delta =0}^{\infty }u_{\delta }$ is said to be summable to s by $N_{p,q}$ method and is denoted by $s_{\delta }\to s(N_{p,q})$ [18].

The necessary and sufficient conditions for a $N_{p,q}$ method to be regular are

$$ \sum_{k=0}^{\delta } \vert p_{\delta -k}q_{k} \vert =O\bigl( \vert R_{\delta } \vert \bigr) \quad\text{and}\quad p_{\delta -k}=o\bigl( \vert R_{\delta } \vert \bigr) \quad\text{as } \delta \to \infty $$

for every fixed $k\geq 0$ for which $q_{k}\neq 0$ [19].

The $N_{p,q}$ transform of the $t_{\delta }^{\triangle _{ H}}$ transform defines the $N_{pq}\triangle _{H}$ product transform and its δth partial sum is denoted by $t_{\delta }^{N_{pq} \triangle _{H}}$. Thus,

$$\begin{aligned} t_{\delta }^{N_{pq} \triangle _{H}} &= \frac{1}{R_{\delta }} \sum _{k=0} ^{\delta } p_{\delta -k}q_{k} t_{k}^{\triangle _{H}} \\ &=\frac{1}{R_{\delta }}\sum_{k=0}^{\delta }p_{\delta -k}q_{k} \sum_{i=0}^{k}h_{k,i}s_{i}. \end{aligned}$$

If $t_{\delta }^{N_{pq} \triangle _{H}}\to s$ as $\delta \to \infty$, then $\sum_{\delta =0}^{\infty } u_{\delta }$ is summable by $N_{pq} \triangle _{H}$ product means to s. We have

$$\begin{aligned} s_{\delta }\to s & \implies t_{\delta }^{ \triangle _{H}}\to s\quad \text{as } \delta \to \infty, \triangle _{H} \text{ method is obtained as regular} \\ &\implies N_{pq}\bigl(t_{\delta }^{\triangle _{H}} \bigr)=t_{\delta }^{N_{pq} \triangle _{H}}\to s,\quad \text{as } \delta \to \infty, N_{pq} \text{ means is obtained as regular} \\ &\implies N_{pq}\triangle _{H} \quad\text{is obtained as regular.} \end{aligned}$$

Note 2

• (i) △H means reduces to $C^{\alpha }$ means if $\nu (z)=\varPi _{k=1}^{\alpha }z^{k},\alpha \geq 1$.

• (ii) △H means reduces to $E^{q}$ if $h_{\delta,m}= \binom{\delta }{m} \frac{q^{\delta -m}}{(1+q)^{\delta }},0\leq m \leq \delta$.

• (iii) $N_{p,q}$ reduces to $N_{p}$ means if $q=1$.

Remark 5

We define the following particular cases of the product means $N_{pq}\triangle _{H}$:

• (i) $N_{p,q} \triangle _{H}$ means reduces to $(N,p,q)(C,\alpha )$ or $N_{pq}C^{\alpha }$ means in view of Note 2(i).

• (ii) $N_{p,q} \triangle _{H}$ means reduces to $(N,p,q)(E^{q})$ or $N_{pq}E^{q}$ means in view of Note 2(ii).

• (iii) $N_{p,q} \triangle _{H}$ means reduces to $N_{p} \triangle _{H}$ means in view of Note 2(iii).

Note 3

• (i) Above particular case (i) in remark 5 is further reduced to $N_{p,q}C ^{1}$ for $\alpha =1$.

• (ii) Above particular case (ii) in remark 5 is further reduced to $N_{p,q}E^{1}$ for $q=1$.

• (iii) Above particular case (iii) in remark 5 is further reduced to $N_{p}C^{\alpha }$ in view of Note 2(i) and then to $N_{p}C^{1}$ for $\alpha =1$.

• (iv) Above particular case (iii) in remark 5 is further reduced to $N_{p}E^{q}$ in view of Note 2(ii) and then to $N_{p}E^{1}$ for $q=1$.

We write

$$\begin{aligned} & T_{\delta}(y)= t_{\delta}^{N_{pq}\Delta H}(y) -f(y)=\int_{0}^{\pi} \phi_{y}(u) M_{\delta}(u) \,du, \\ \text{where} \\ & M_{\delta }(u)=\frac{1}{2\pi R_{\delta }}\sum_{k=0}^{\delta }p_{ \delta -k}q_{k} \sum_{v=0}^{k} \int _{0}^{1}\binom{k}{v}z^{v}(1-z)^{k-v} \,d \nu (z) \frac{\sin (v+\frac{1}{2})u}{\sin (\frac{u}{2})}; \\ &\phi _{y}(u)=f(y+u)+f(y-u)-2f(y); \\ &\Phi (y,l,u)= \textstyle\begin{cases} \phi _{y+l}(u)-\phi _{y}(u), & 0< \nu < 1, \\ \phi _{y+l}(u)+\phi _{y-l}(u)-2\phi _{y}(u), &1\leq \nu < 2; \end{cases}\displaystyle \\ &\varUpsilon _{\delta }(y,l)= \textstyle\begin{cases} T_{\delta }(y+l)-T_{\delta }(y), & 0< \nu < 1, \\ T_{\delta }(y+l)+T_{\delta }(y-l)-2T_{\delta }(y), &1\leq \nu < 2. \end{cases}\displaystyle \end{aligned}$$

(6)

Remark 6

We prove the following additional results that will be used in the proof of our theorem.

$$\begin{aligned} &\mathrm{(i)}\quad \varUpsilon _{\delta }(y,l)= \int _{0}^{\pi } M_{\delta }(u) \Phi (y,l,u) \,du, \end{aligned}$$

(7)

$$\begin{aligned} & \mathrm{(ii)}\quad w_{k}(T_{\delta },l)_{\rho}= \bigl\Vert \varUpsilon _{\delta }(\cdot,l) \bigr\Vert _{\rho}. \end{aligned}$$

(8)

Proof

(i) We have

$$\begin{aligned} \varUpsilon _{\delta }(y,l) &= \textstyle\begin{cases} T_{\delta }(y+l)-T_{\delta }(y), & 0< \nu < 1, \\ T_{\delta }(y+l)+T_{\delta }(y-l)-2T_{\delta }(y), & 1\leq \nu < 2, \end{cases}\displaystyle \\ &= \textstyle\begin{cases} \int _{0}^{\pi } [\phi _{y+l}(u)-\phi _{y}(u)] M_{\delta }(u) \,du, & 0< \nu < 1, \\ \int _{0}^{\pi } [\phi _{y+l}(u)+\phi _{y-l}(u)-2\phi _{y}(u)] M_{\delta }(u) \,du, & 1\leq \nu < 2, \end{cases}\displaystyle \\ &= \int _{0}^{\pi } \Phi (y,l,u) M_{\delta }(u) \,du. \end{aligned}$$

 □

Proof

(ii) By definition of $w_{k}(f,l)_{\rho }$, we have

$$\begin{aligned} w_{k}(T_{\delta },l)_{\rho } & =\sup_{0< h\leq l} \bigl\Vert \triangle _{h} ^{k}(T_{\delta },\cdot) \bigr\Vert _{\rho } \\ &= \textstyle\begin{cases} \sup_{0< h\leq l} \Vert T_{\delta }(\cdot+h)-T_{\delta }(\cdot) \Vert _{\rho }, &0< \nu < 1, \\ \sup_{0< h\leq l} \Vert T_{\delta }(\cdot+h)+T_{\delta }(\cdot-h)-2T_{\delta }(\cdot) \Vert _{\rho }, & 1\leq \nu < 2, \end{cases}\displaystyle \\ &= \bigl\Vert \varUpsilon _{\delta }(\cdot,l) \bigr\Vert _{\rho }. \end{aligned}$$

 □

3 Main theorem

Theorem 3.1

For a function f (2π-periodic and Lebesgue integrable) for $0\leq \eta <\nu <2$, the best error approximation of f in the Besov space $B_{\sigma }^{\nu }(L_{\rho }), \rho \geq 1,1<\sigma \leq \infty$ by $N_{pq} \triangle _{H}$ transform of its FS is given by

$$ \bigl\Vert T_{\delta }(\cdot) \bigr\Vert _{B_{\sigma }^{\eta }(L_{ \rho })}=O(1) \textstyle\begin{cases} (\delta +1)^{-1}, &\nu -\eta -\sigma ^{-1} >1, \\ (\delta +1)^{-\nu +\eta +\sigma ^{-1}}, &\nu -\eta -\sigma ^{-1}< 1, \\ (\delta +1)^{-1}[\log (\delta +1)\pi ]^{1-\sigma ^{-1}}, &\nu -\eta -\frac{1}{ \sigma }=1. \end{cases} $$

4 Lemmas

Lemma 4.1

If $\{p_{\delta }\}$ and $\{q_{\delta }\}$ are monotonic increasing and monotonic decreasing, respectively, then

$$ (\delta +1)p_{\delta }q_{0}=O(R_{\delta }). $$

Proof

$$\begin{aligned} &R_{\delta }=\sum_{k=0}^{\delta } p_{\delta -k}q_{k}= p_{\delta }q _{0}+p_{\delta -1}q_{1}+ \cdots+p_{0}q_{\delta } \\ &\phantom{R_{\delta }}\geq p_{\delta }q_{0}+p_{\delta }q_{0}+ \cdots+p_{\delta }q_{0} \\ &\phantom{R_{\delta }}=(\delta +1)p_{\delta }q_{0}, \\ &(\delta +1)p_{\delta }q_{0}=O(R_{\delta }). \end{aligned}$$

 □

Lemma 4.2

$M_{\delta }(u)=O (\delta +1 )$ for $0< u\leq \frac{1}{ \delta +1}$.

Proof

For $0< u\leq \frac{1}{\delta +1}$, $\sin (\frac{u}{2})\geq \frac{u}{ \pi }$, $\sin (v+\frac{1}{2})u \leq (v+\frac{1}{2})u$ and $\sup_{0\leq z\leq 1}|\nu ^{\prime }(z)|=N$ we have

$$\begin{aligned} \bigl\vert M_{\delta }(u) \bigr\vert &= \Biggl\vert \frac{1}{2\pi R_{\delta }} \sum_{k=0}^{\delta } p_{\delta -k} q_{k} \sum_{v=0}^{k} \int _{0}^{1} \binom{k}{v}z^{v}(1-z)^{k-v} \frac{\sin (v+\frac{1}{2})u}{\sin ( \frac{u}{2})} \,d\nu (z) \Biggr\vert \\ &\leq \frac{1}{2\pi R_{\delta }} \Biggl\vert \sum_{k=0}^{\delta } p_{ \delta -k} q_{k} \sum_{v=0}^{k} \int _{0}^{1}\binom{k}{v}z^{v}(1-z)^{k-v} \frac{\sin (v+\frac{1}{2})u}{\frac{u}{\pi }} \,d\nu (z) \Biggr\vert \\ &\leq \frac{1}{2R_{\delta }u} \sum_{k=0}^{\delta } p_{\delta -k}q_{k} \sum_{v=0}^{k} \int _{0}^{1}\binom{k}{v} \frac{z^{v}}{(1-z)^{v}}(1-z)^{k} \sin \biggl(v+\frac{1}{2}\biggr)u \bigl\vert d\nu (z) \bigr\vert \\ &\leq \frac{N}{4R_{\delta }u}\sum_{k=0}^{\delta }p_{\delta -k}q_{k} \sum_{v=0}^{k} \int _{0}^{1}\binom{k}{v} A^{v} (2v+1)\frac{u}{2} (1-z)^{k}\,dz, \quad\text{where } A= \frac{z^{v}}{(1-z)^{v}} \\ &=\frac{N}{4\pi }\sum_{k=0}^{\delta } p_{\delta -k} q_{k} \int _{0} ^{1} (1-z)^{k} \,dz \sum _{v=0}^{k} \binom{k}{v} A^{v} (2v+1) \\ &= \frac{N}{4\pi }\sum_{k=0}^{\delta } p_{\delta -k} q_{k} \int _{0} ^{1} (1-z)^{k} \,dz \Biggl\{ 2 \sum_{v=0}^{k}v \binom{k}{v} A^{v} + \sum_{v=0}^{k} \binom{k}{v} A^{v} \Biggr\} \\ &= \frac{N}{4\pi }\sum_{k=0}^{\delta } p_{\delta -k} q_{k} \int _{0} ^{1} (1-z)^{k} \,dz \Biggl\{ 2 \sum_{v=1}^{k} k \binom{k-1}{v-1} A^{v} + (1+A)^{k} \Biggr\} \\ &= \frac{N}{4\pi }\sum_{k=0}^{\delta } p_{\delta -k} q_{k} \int _{0} ^{1} (1-z)^{k} \,dz \bigl\{ 2k(1+A)^{k-1} + (1+A)^{k} \bigr\} \\ &= \frac{N}{4\pi }\sum_{k=0}^{\delta } p_{\delta -k} q_{k} \int _{0} ^{1} \bigl[2k(1-z)+1\bigr] \,dz \quad\text{(substituting the value of $A$)} \\ &= \frac{N}{4\pi }\sum_{k=0}^{\delta } p_{\delta -k} q_{k} (k+1) \\ &\leq \frac{N(\delta +1)}{4\pi }\sum_{k=0}^{\delta } p_{\delta -k} q _{k} \\ &=O(\delta +1). \end{aligned}$$

 □

Lemma 4.3

If $\{p_{\delta }\}$ and $\{q_{\delta }\}$ are monotonic increasing and monotonic decreasing sequences, respectively, then

$$ M_{\delta }(u)=O \biggl(\frac{1}{(\delta +1)u^{2}} \biggr)\quad \textit{for } \frac{1}{\delta +1}< u\leq \pi. $$

Proof

For $\frac{1}{\delta +1}< u\leq \pi$, $\sin ^{2}\delta u \leq 1$, $\sin (\frac{u}{2})\geq \frac{u}{\pi }$ and $\sup_{0\leq z \leq 1}| \nu ^{\prime }(z)|=N$

$$\begin{aligned} \bigl\vert M_{\delta }(u) \bigr\vert &= \Biggl\vert \frac{1}{2\pi R_{\delta }}\sum_{k=0} ^{\delta } p_{\delta -k} q_{k} \sum_{v=0}^{k} \int _{0}^{1}\binom{k}{v} z^{v}(1-z)^{k-v} \frac{\sin (v+\frac{1}{2})u}{\sin (\frac{u}{2})} \,d \alpha (z) \Biggr\vert \\ &\leq \frac{N}{2R_{\delta }u} \Biggl\vert \sum_{k=0}^{\delta }p_{\delta -k}q _{k} \operatorname{Im}\sum_{v=0}^{k} \int _{0}^{1}\binom{k}{v}z^{v}(1-z)^{k-v}e ^{\iota (v+\frac{1}{2})u} \,dz \Biggr\vert . \end{aligned}$$

(9)

First, we solve

$$\begin{aligned} &\operatorname{Im} \sum_{v=0}^{k} \int _{0}^{1}\binom{k}{v}z^{v}(1-z)^{k-v} e ^{i(v+\frac{1}{2})u} \,dz \\ & \quad\leq \operatorname{Im} e^{i\frac{u}{2}} \int _{0}^{1} \sum_{v=0}^{k} \binom{k}{v} z^{v}(1-z)^{k-v}e^{ivu}\,dz \\ &\quad= \operatorname{Im} e^{i\frac{u}{2}} \int _{0}^{1} \sum_{v=0}^{k} \binom{k}{v}(1-z)^{k-v}\bigl(ze ^{iu} \bigr)^{v} \,dz \\ &\quad= \operatorname{Im}e^{i\frac{u}{2}} \int _{0}^{1} \bigl\{ 1+z\bigl(e^{iu}-1 \bigr) \bigr\} ^{k}\,dz \\ &\quad=\operatorname{Im}\frac{e^{i(k+1)u}-1}{(k+1)(e^{i\frac{u}{2}}-e^{-i\frac{u}{2}})} \\ &\quad= \operatorname{Im}\frac{e^{i(k+1)u}-1}{(k+1)2i\sin \frac{u}{2}} \\ &\quad=\operatorname{Im}\frac{\cos (k+1)u+i\sin (k+1)u-1}{2i(k+1)\sin \frac{u}{2}} \\ &\quad= \frac{\sin ^{2}(k+1)\frac{u}{2}}{(k+1) \sin \frac{u}{2}}. \end{aligned}$$

(10)

From (9) and (10), we have

$$\begin{aligned} \bigl\vert M_{\delta }(u) \bigr\vert &\leq \frac{N}{2R_{\delta }u} \Biggl\vert \sum_{k=0}^{\delta }{p_{\delta -k}} {q_{k}}\frac{\sin ^{2}(k+1)\frac{u}{2}}{(k+1) \sin \frac{u}{2}} \Biggr\vert \\ &\leq \frac{N\pi }{2 R_{\delta } u^{2}} \Biggl\vert \sum_{k=0}^{ \delta } {p_{\delta -k}} {q_{k}} \frac{1}{k+1} \Biggr\vert . \end{aligned}$$

Using Abel’s lemma and Lemma 4.1, we have

$$\begin{aligned} &\bigl\vert M_{\delta }(u) \bigr\vert \leq \frac{N \pi }{2 R_{\delta } u^{2}} \Biggl\vert \sum_{k=0}^{\delta -1} (p_{\delta -k}q_{k}-p_{\delta -k-1} q_{k+1}) \sum_{v=0}^{k} \frac{1}{v+1}+p_{0}q_{n}\sum _{k=0}^{\delta } \frac{1}{k+1} \Biggr\vert \\ &\phantom{\bigl\vert M_{\delta }(u) \bigr\vert }\leq \frac{N\pi }{2 R_{\delta } u^{2}} \Biggl[\sum_{k=0}^{v-1} \vert p_{\delta -k} q_{k}-p_{\delta -k-1} q_{k+1} \vert +p_{0}q_{\delta } \Biggr] \max_{0\leq k\leq m} \Biggl\vert \sum_{k=0}^{m} \frac{1}{k+1} \Biggr\vert \\ &\phantom{\bigl\vert M_{\delta }(u) \bigr\vert }\leq \frac{N\pi }{2 R_{\delta } u^{2}} \bigl[ \vert p_{\delta } q_{0}-p_{0} q_{\delta } \vert + p_{0} q_{\delta } \bigr] \\ &\phantom{\bigl\vert M_{\delta }(u) \bigr\vert }\leq \frac{N\pi }{2R_{\delta } u^{2}} (p_{\delta }q_{0}+2p_{0}q_{ \delta }) \\ &\phantom{\bigl\vert M_{\delta }(u) \bigr\vert }\leq \frac{3p_{\delta }q_{o} N\pi }{2R_{\delta }u^{2}}, \\ &M_{\delta }(u)=O \biggl(\frac{1}{(\delta +1)u^{2}} \biggr). \end{aligned}$$

 □

Lemma 4.4

Let $1\leq \rho \leq \infty$ and $0<\nu <2$. If $f\in L_{\rho }$ then for $0< l, u\leq \pi$

• (i) $\Vert \Phi (\cdot,l,u)\Vert _{\rho } \leq 4w_{k}(f,l)_{ \rho }$,

• (ii) $\Vert \Phi (\cdot,l,u)\Vert _{\rho } \leq 4w_{k}(f,u)_{ \rho }$,

• (iii) $\Vert \Phi _{.}(u)\Vert _{\rho }\leq 2w_{k}(f,u)_{ \rho }$,

where $k=[\nu ]+1$.

Proof

The proof of above lemma can be obtained along the same lines of the proofs of Lemma 2 in [20]. □

Lemma 4.5

Let $0 \leq \eta <\nu <2$. If $f\in B_{\sigma }^{\nu }(L_{\rho }),\rho \geq 1, 1<\sigma <\infty$, then

$$\begin{aligned} &\mathrm{(i)}\quad \int _{0}^{\pi } \bigl\vert M_{\delta }(u) \bigr\vert \biggl( \int _{0}^{u}\frac{ \Vert \Phi (\cdot,l,u) \Vert _{\rho }^{\sigma }}{l^{\eta \sigma }} \frac{dl}{l} \biggr)^{\frac{1}{\sigma }} \,du=O(1) \biggl\{ \int _{0}^{\pi }(u ^{\nu -\eta }| \bigl(M_{\delta }(u) \bigr)^{\frac{\sigma }{\sigma -1}} \,du\biggr\} ^{1-\frac{1}{\sigma }}, \\ &\mathrm{(ii)}\quad \int _{0}^{\pi } \bigl\vert M_{\delta }(u) \bigr\vert \biggl( \int _{u}^{\pi }\frac{ \Vert \Phi (\cdot,l,u) \Vert _{\rho }^{\sigma }}{l^{\eta \sigma }} \frac{dl}{l} \biggr)^{\frac{1}{\sigma }} \,du\\ &\phantom{\textit{(ii)}\quad}\quad=O(1) \biggl\{ \int _{0}^{ \pi }\bigl(u^{\nu -\eta +\frac{1}{\sigma }}\bigl|M_{\delta } (u)\bigr| \bigr)^{\frac{ \sigma }{\sigma -1}}\,du\biggr\} ^{1-\frac{1}{\sigma }}. \end{aligned}$$

Proof

The part of above lemma can be established along the same lines of the proofs of Lemma 2 in [20]. □

Lemma 4.6

([20])

Let $0\leq \eta <\nu <2$ and if $f\in B_{\sigma } ^{\nu }(L_{\rho }),\rho \geq 1,\sigma =\infty$, then

$$ \sup_{0< l,u\leq \pi }\bigl(l^{-\eta } \bigl\Vert \Phi ( \cdot,l,u) \bigr\Vert _{\rho }\bigr)=O\bigl(u ^{\nu -\eta }\bigr). $$

5 Proof of the main theorem

5.1 Case I: for $1<\sigma <\infty, \rho \geq 1, 0 \leq \eta < \nu <2$

Proof

Following [16], we have

$$ s_{\delta }(f;y)-f(y)=\frac{1}{2\pi } \int _{0}^{\pi } \phi _{y}(l) \frac{ \sin (\delta +\frac{1}{2})l}{\sin (\frac{l}{2})} \,dl. $$

Denoting the Hausdorff matrix summability transform of $s_{\delta }(y)$ by $t_{\delta }^{\triangle _{H}}(y)$, we get

$$\begin{aligned} t_{\delta }^{\triangle _{H}}(y)-f(y) &= \sum_{m=0}^{\delta } h_{\delta ,m}\bigl[s_{m}(y)-f(y)\bigr] \\ &=\sum_{m=0}^{\delta }\binom{\delta }{m} \triangle ^{\delta -m}\mu _{m} \biggl\{ \frac{1}{2\pi } \int _{0}^{\pi }\phi _{y}(l) \frac{\sin (m+ \frac{1}{2})l}{\sin (\frac{l}{2})} \,dl \biggr\} \\ &=\frac{1}{2\pi } \int _{0}^{\pi }\phi _{y}(l)\sum _{m=0}^{\delta }\binom{ \delta }{m}\triangle ^{\delta -m} \biggl( \int _{0}^{1}z^{m}\,d\nu (z) \biggr) \frac{ \sin (m+\frac{1}{2})l}{\sin (\frac{l}{2})}\,dl \\ &=\frac{1}{2\pi } \int _{0}^{\pi }\phi _{y}(l)\sum _{m=0}^{\delta } \int _{0}^{1}\binom{\delta }{m}z^{m}(1-z)^{\delta -m} \,d\nu (z)\frac{\sin (m+ \frac{1}{2})l}{\sin \frac{l}{2}}\,dl. \end{aligned}$$

The $N_{pq}$ transform of $t_{\delta }^{\triangle _{H}}(y)$, denoted by $t_{\delta }^{N_{pq} \triangle _{H}}(y)$, is given by

$$\begin{aligned} &t_{\delta }^{N_{pq} \triangle _{H}}(y)-f(y) \\ &\quad=\frac{1}{R_{\delta }}\sum_{m=0}^{\delta }p_{\delta -k}q_{m} \Biggl(\frac{1}{2 \pi } \int _{0}^{\pi }\phi _{y}(l)\sum _{v=0}^{m} \int _{0}^{1}\binom{m}{v}z ^{v}(1-z)^{m-v}\,d\nu (z)\frac{\sin (v+\frac{1}{2})l}{\sin \frac{l}{2}}\,dl \Biggr). \end{aligned}$$

(11)

Replacing l by u

$$\begin{aligned} &= \int _{0}^{\pi }\phi _{y}(u) \frac{1}{2\pi R_{\delta }}\sum_{m=0}^{ \delta }p_{\delta -m}q_{m} \sum_{v=0}^{m} \int _{0}^{1}\binom{m}{v}z^{v}(1-z)^{m-v} \,d \nu (z)\frac{\sin (v+\frac{1}{2})u}{\sin (\frac{u}{2})}\,du \\ &= \int _{0}^{\pi }\phi _{y}(u) M_{\delta }(u)\,du. \end{aligned}$$

(12)

Let

$$ T_{\delta }(y)=t_{\delta }^{N_{pq} \triangle _{H}}(y)-f(y)= \int _{0} ^{\pi }\phi _{y}(u) M_{\delta }(u)\,du. $$

(13)

Using the definition of the Besov norm given by (5), we have

$$\begin{aligned} &\bigl\Vert T_{\delta }(\cdot) \bigr\Vert _{B_{\sigma }^{\eta }(L_{\rho })}= \bigl\Vert T _{\delta }(\cdot) \bigr\Vert _{\rho }+ \bigl\Vert w_{k}(T_{\delta },\cdot) \bigr\Vert _{\eta, \sigma }. \end{aligned}$$

(14)

Now using (6) and Lemma 4.4(iii)

$$\begin{aligned} \bigl\Vert T_{\delta }(\cdot) \bigr\Vert _{\rho } &\leq \int _{0}^{\pi } \bigl\Vert \phi _{.}(u) \bigr\Vert _{\rho } \bigl\vert M_{\delta }(u) \bigr\vert \,du \\ &\leq \int _{0}^{\pi }2w_{k}(f,u)_{\rho } \bigl\vert M_{\delta }(u) \bigr\vert \,du. \end{aligned}$$

(15)

Using Hölder’s inequality and definition of Besov space given in (4), we get,

$$\begin{aligned} \bigl\Vert T_{\delta }(\cdot) \bigr\Vert _{\rho } &\leq 2 \biggl\{ \int _{0}^{\pi } \bigl( \bigl\vert M_{\delta }(u) \bigr\vert u^{\nu +\frac{1}{\sigma }} \bigr)^{\frac{ \sigma }{\sigma -1}} \,du \biggr\} ^{1-\frac{1}{\sigma }} \biggl\{ \int _{0}^{\pi } \biggl( \frac{w_{k}(f,u)_{\rho }}{u^{\nu +\frac{1}{\sigma }}} \biggr)^{\sigma }\,du \biggr\} ^{\frac{1}{\sigma }} \\ ={}&O(1) \biggl\{ \int _{0}^{\pi } \bigl( \bigl\vert M_{\delta }(u) \bigr\vert u^{\nu +\frac{1}{\sigma }} \bigr)^{\frac{ \sigma }{\sigma -1}} \,du \biggr\} ^{1-\frac{1}{\sigma }} \\ ={}&O\biggl[ \biggl\{ \int _{0}^{\frac{1}{\delta +1}} \bigl( \bigl\vert M _{\delta }(u) \bigr\vert u^{\nu +\frac{1}{\sigma }} \bigr)^{\frac{\sigma }{ \sigma -1}} \,du \biggr\} ^{1-\frac{1}{\sigma }} \\ &{}+ \biggl\{ \int _{\frac{1}{ \delta +1}}^{\pi } \bigl( \bigl\vert M_{\delta }(u) \bigr\vert u^{\nu +\frac{1}{ \sigma }} \bigr)^{\frac{\sigma }{\sigma -1}} \,du \biggr\} ^{1-\frac{1}{ \sigma }} \biggr] \\ ={}&R+S. \end{aligned}$$

(16)

Now using Lemma 4.2, we have

$$\begin{aligned} R &= O\biggl\{ \int _{0}^{\frac{1}{\delta +1}} \bigl( \bigl\vert M_{\delta }(u) \bigr\vert u^{\nu +\frac{1}{\sigma }} \bigr)^{\frac{\sigma }{\sigma -1}} \,du \biggr\} ^{1-\frac{1}{\sigma }} \\ &=O \biggl[ \int _{0}^{\frac{1}{\delta +1}} \bigl\{ (\delta +1) u^{ \nu +\frac{1}{\sigma }} \bigr\} ^{\frac{\sigma }{\sigma -1}} \,du \biggr] ^{1-\frac{1}{\sigma }} \\ &=O \biggl\{ (\delta +1)^{\frac{\sigma }{\sigma -1}} \int _{0}^{\frac{1}{ \delta +1}} u^{\frac{\nu \sigma }{\sigma -1}+\frac{1}{\sigma -1}} \,du \biggr\} ^{1-\sigma ^{-1}} \\ &=O \biggl\{ \frac{1}{(\delta +1)^{\nu }} \biggr\} . \end{aligned}$$

(17)

Using Lemma 4.3, we have

$$\begin{aligned} S &= O\biggl\{ \int _{\frac{1}{\delta +1}}^{\pi } \bigl( \bigl\vert M_{\delta }(u) \bigr\vert u^{\nu +\frac{1}{\sigma }} \bigr)^{\frac{\sigma }{\sigma -1}} \,du \biggr\} ^{1-\frac{1}{\sigma }} \\ &=O \biggl\{ \int _{\frac{1}{\delta +1}}^{\pi } \biggl( \frac{1}{(\delta +1)u^{2}} u^{\nu +\frac{1}{\sigma }} \biggr)^{ \frac{\sigma }{\sigma -1}} \,du \biggr\} ^{1-\frac{1}{\sigma }} \\ &=O \biggl\{ \int _{\frac{1}{\delta +1}}^{\pi } \biggl(\frac{1}{\delta +1} \times u^{\nu +\frac{1}{\sigma }-2} \biggr)^{ \frac{\sigma }{\sigma -1}} \,du \biggr\} ^{1-\frac{1}{\sigma }} \\ &=O(1) \textstyle\begin{cases} (\delta +1)^{-1}, &\nu >1, \\ (\delta +1)^{-\nu }, &\nu < 1, \\ (\delta +1)^{-1}[\log (\delta +1)\pi ]^{1-\sigma ^{-1}}, &\nu =1. \end{cases}\displaystyle \end{aligned}$$

(18)

Combining (16)–(18), we have

$$ \bigl\Vert T_{\delta }(\cdot) \bigr\Vert _{\rho } =O(1) \textstyle\begin{cases} (\delta +1)^{-1}, &\nu >1, \\ (\delta +1)^{-\nu }, &\nu < 1, \\ (\delta +1)^{-1}[\log (\delta +1)\pi ]^{1-\sigma ^{-1}}, &\nu =1. \end{cases} $$

(19)

Using the generalized Minkowski inequality [21] repeatedly and Lemma 4.5, we get

$$\begin{aligned} \bigl\Vert w_{k}(T_{\delta },\cdot) \bigr\Vert _{\eta,\sigma }={} &\biggl[ \int _{0}^{ \pi } \biggl(\frac{w_{k}(T_{\delta },l)_{\rho }}{l^{\eta }} \biggr) ^{\sigma } \frac{dl}{l} \biggr]^{\frac{1}{\sigma }} \\ ={}& \biggl[ \int _{0}^{\pi } \biggl(\frac{ \Vert \varUpsilon _{\delta }(\cdot,l) \Vert _{\rho }}{l^{\eta }} \biggr)^{\sigma } \frac{dl}{l} \biggr] ^{\frac{1}{\sigma}} \\ \leq{}& \int _{0}^{\pi } \bigl\vert M_{\delta }(u) \bigr\vert \,du \biggl( \int _{0}^{\pi } \frac{ \Vert \Phi (\cdot,l,u) \Vert _{\rho }^{\sigma }}{l^{\eta \sigma }} \frac{dl}{l} \biggr)^{\sigma ^{-1}} \\ \leq{}& \biggl[ \int _{0}^{\pi } \bigl\vert M_{\delta }(u) \bigr\vert \,du \biggl\{ \int _{0}^{u}\frac{ \Vert \Phi (\cdot,l,u) \Vert _{\rho }^{\sigma }}{l^{\eta \sigma }} \frac{dl}{l} \biggr\} ^{\sigma ^{-1}} \biggr] \\ &{}+ \biggl[ \int _{0}^{\pi } \bigl\vert M_{\delta }(u) \bigr\vert \,du \biggl\{ \int _{u}^{\pi }\frac{ \Vert \Phi (\cdot,l,u) \Vert _{\rho }^{\sigma }}{l^{\eta \sigma }} \frac{dl}{l} \biggr\} ^{\sigma ^{-1}} \biggr] \\ ={}&O(1) \biggl\{ \int _{0}^{\pi }\bigl(u^{\nu -\eta } \bigl\vert M_{\delta }(u) \bigr\vert \bigr)^{\frac{ \sigma }{\sigma -1}} \,du \biggr\} ^{1-\frac{1}{\sigma }} \\ &{}+O(1) \biggl\{ \int _{0}^{\pi }\bigl(u^{\nu -\eta +\frac{1}{\sigma }} \bigl\vert M _{\delta }(u) \bigr\vert \bigr)^{\frac{\sigma }{\sigma -1}} \,du \biggr\} ^{1-\frac{1}{\sigma }} \\ ={}&O(1) (R_{1}+S_{1}). \end{aligned}$$

(20)

Since $(a+b)^{\rho} \leq a^{\rho}+b^{\rho}$ for positive a, b and $0 < \rho \leq 1$ for $\rho=1-\frac{1}{\sigma} <1$, then

$$\begin{aligned} R_{1}={}& \biggl\{ \int _{0}^{\pi }\bigl(u^{\nu -\eta } \bigl\vert M_{\delta }(u) \bigr\vert \bigr)^{\frac{ \sigma }{\sigma -1}} \,du \biggr\} ^{1-\sigma ^{-1}} \\ \leq{}& \biggl\{ \int _{0}^{\frac{1}{\delta +1}}\bigl(u^{\nu -\eta } \bigl\vert M_{\delta }(u) \bigr\vert \bigr)^{\frac{ \sigma }{\sigma -1}}\,du \biggr\} ^{1-\sigma ^{-1}} \\ &{}+ \biggl\{ \int _{\frac{1}{\delta +1}}^{\pi }\bigl(u^{\nu -\eta } \bigl\vert M_{\delta }(u) \bigr\vert \bigr)^{\frac{\sigma }{\sigma -1}} \,du \biggr\} ^{1-\sigma ^{-1}} \\ ={}&R_{11}+R_{12}. \end{aligned}$$

(21)

Using Lemma 4.2, we have

$$\begin{aligned} R_{11} &= O\biggl\{ \int _{0}^{\frac{1}{\delta +1}} \bigl( (\delta +1)u^{\nu - \eta } \bigr)^{\frac{\sigma }{\sigma -1}} \,du \biggr\} ^{1-\sigma ^{-1}} \\ &=O\bigl\{ (\delta +1)^{-\nu +\eta +\frac{1}{\sigma }} \bigr\} . \end{aligned}$$

(22)

Using Lemma 4.3, we have

$$\begin{aligned} R_{12} &= O\biggl\{ \int _{\frac{1}{\delta +1}}^{\pi } \biggl(u^{\nu -\eta } \frac{1}{( \delta +1)u^{2}} \biggr)^{\frac{\sigma }{\sigma -1}} \,du \biggr\} ^{1-\sigma ^{-1}} \\ &=O(1) \textstyle\begin{cases} (\delta +1)^{-1}, &\nu -\eta -\sigma ^{-1} >1, \\ (\delta +1)^{-\nu +\eta +\sigma ^{-1}}, &\nu -\eta -\sigma ^{-1}< 1, \\ (\delta +1)^{-1}[\log (\delta +1)\pi ]^{1-\sigma ^{-1}}, &\nu -\eta -\frac{1}{ \sigma }=1. \end{cases}\displaystyle \end{aligned}$$

(23)

Now from (21) to (23), we get

$$ R_{1}=O(1) \textstyle\begin{cases} (\delta +1)^{-1}, &\nu -\eta -\sigma ^{-1} >1, \\ (\delta +1)^{-\nu +\eta +\sigma ^{-1}}, &\nu -\eta -\sigma ^{-1}< 1, \\ (\delta +1)^{-1}[\log (\delta +1)\pi ]^{1-\sigma ^{-1}}, &\nu -\eta -\frac{1}{ \sigma }=1. \end{cases} $$

(24)

Since $(a+b)^{\rho} \leq a^{\rho}+b^{\rho}$ for positive a, b and $0 < \rho \leq 1$ for $\rho=1-\frac{1}{\sigma} <1$, then

$$\begin{aligned} S_{1}={}& \biggl\{ \int _{0}^{\pi }\bigl(u^{\nu -\eta +\sigma ^{-1}} \bigl\vert M_{\delta }(u) \bigr\vert \bigr)^{\frac{ \sigma }{\sigma -1}} \,du \biggr\} ^{1-\sigma ^{-1}} \\ \leq {}& \biggl\{ \int _{0}^{\frac{1}{\delta +1}}\bigl(u^{\nu -\eta +\sigma ^{-1}} \bigl\vert M_{\delta }(u) \bigr\vert \bigr)^{\frac{\sigma }{\sigma -1}} \,du \biggr\} ^{1-\sigma ^{-1}} \\ &{}+ \biggl\{ \int _{\frac{1}{\delta +1}}^{\pi }\bigl(u^{\nu -\eta +\sigma ^{-1}} \bigl\vert M _{\delta }(u) \bigr\vert \bigr)^{\frac{\sigma }{\sigma -1}} \,du \biggr\} ^{1-\sigma ^{-1}} \\ ={}&S_{11}+S_{12} \quad\text{say.} \end{aligned}$$

(25)

Using Lemma 4.2, we have

$$\begin{aligned} S_{11} &= O\biggl\{ \int _{0}^{\frac{1}{\delta +1}} \bigl(u^{\nu -\eta + \sigma ^{-1}} \bigl\vert M_{\delta }(u) \bigr\vert \bigr)^{\frac{\sigma }{\sigma -1}} \,du \biggr\} ^{1-\sigma ^{-1}} \\ &= O\biggl\{ \int _{0}^{\frac{1}{\delta +1}} \bigl(u^{\nu -\eta +\sigma ^{-1}} (\delta +1) \bigr)^{\frac{\sigma }{\sigma -1}} \,du \biggr\} ^{1-\sigma ^{-1}} \\ &=O\bigl\{ (\delta +1)^{-\nu +\eta }\bigr\} . \end{aligned}$$

(26)

Using Lemma 4.3, we have

$$\begin{aligned} S_{12} &= O\biggl\{ \int _{\frac{1}{\delta +1}}^{\pi } \bigl(u^{\nu - \eta +\frac{1}{\sigma }} \bigl\vert M_{\delta }(u) \bigr\vert \bigr)^{\frac{\sigma }{ \sigma -1}} \,du \biggr\} ^{1-\sigma ^{-1}} \\ &= O\biggl\{ \int _{\frac{1}{\delta +1}}^{\pi } \biggl(u^{\nu -\eta +\frac{1}{ \sigma }} \frac{1}{(\delta +1)u^{2}} \biggr)^{ \frac{\sigma }{\sigma -1}} \,du \biggr\} ^{1-\sigma ^{-1}} \\ &=O(1) \textstyle\begin{cases} (\delta +1)^{-1}, &\nu -\eta >1, \\ (\delta +1)^{-\nu +\eta }, &\nu -\eta < 1, \\ (\delta +1)^{-1}[\log (\delta +1)\pi ]^{1-\sigma ^{-1}}, &\nu -\eta =1. \end{cases}\displaystyle \end{aligned}$$

(27)

Now, from (25)-(27), we get

$$\begin{aligned} S_{1}=O(1) \textstyle\begin{cases} (\delta +1)^{-1}, &\nu -\eta >1, \\ (\delta +1)^{-\nu +\eta }, &\nu -\eta < 1, \\ (\delta +1)^{-1}[\log (\delta +1)\pi ]^{1-\sigma ^{-1}}, &\nu -\eta =1. \end{cases}\displaystyle \end{aligned}$$

(28)

Combining (20), (24) and (28), we get

$$ w_{k}(T_{\delta },\cdot)\Vert _{\eta,\sigma }=O(1) \textstyle\begin{cases} (\delta +1)^{-1}, &\nu -\eta -\sigma ^{-1} >1, \\ (\delta +1)^{-\nu +\eta +\sigma ^{-1}}, &\nu -\eta -\sigma ^{-1}< 1, \\ (\delta +1)^{-1}[\log (\delta +1)\pi ]^{1-\sigma ^{-1}}, &\delta - \eta -\sigma^{-1}=1. \end{cases} $$

(29)

From (14), (19) and (29), we get

$$ \bigl\Vert T_{\delta }(\cdot) \bigr\Vert _{B_{\sigma }^{\eta }(L_{\rho })}=O(1) \textstyle\begin{cases} (\delta +1)^{-1}, &\nu -\eta -\sigma ^{-1} >1, \\ (\delta +1)^{-\nu +\eta +\sigma ^{-1}}, &\nu -\eta -\sigma ^{-1}< 1, \\ (\delta +1)^{-1}[\log (\delta +1)\pi ]^{1-\sigma ^{-1}}, &\nu -\eta -\frac{1}{ \sigma }=1. \end{cases} $$

(30)

Case II: For $\sigma=\infty,0\leq \eta <\nu <2$.

$$\begin{aligned} \bigl\Vert T_{\delta }(\cdot) \bigr\Vert _{B_{\infty }^{\eta }(L_{\rho })}&= \bigl\Vert T _{\delta }(\cdot) \bigr\Vert _{\rho }+ \bigl\Vert w_{k}(T_{\delta },\cdot) \bigr\Vert _{\eta, \infty }. \end{aligned}$$

(31)

Using (2) in (15),

$$\begin{aligned} \bigl\Vert T_{\delta }(\cdot) \bigr\Vert _{\rho } &\leq \int _{0}^{\pi }2w_{k}(f,u)_{ \rho } \bigl\vert M_{\delta }(u) \bigr\vert \,du \\ &=O(1) \biggl\{ \int _{0}^{\frac{1}{\delta +1}} \bigl\vert M_{\delta }(u) \bigr\vert u^{\nu } \,du + \int _{\frac{1}{\delta +1}}^{\pi } \bigl\vert M_{ \delta }(u) \bigr\vert u^{\nu } \,du \biggr\} \\ &=O(1)[R_{2}+S_{2}]. \end{aligned}$$

(32)

Using Lemma 4.2, we get

$$\begin{aligned} R_{2} &= \int _{0}^{\frac{1}{\delta +1}} u^{\nu } \bigl\vert M_{\delta }(u) \bigr\vert \,du \\ &\leq \int _{0}^{\frac{1}{\delta +1}} u^{\nu }(\delta +1) \,du=( \delta +1)^{-\nu }. \end{aligned}$$

(33)

Using Lemma 4.3, we get

$$\begin{aligned} S_{2} &= \int _{\frac{1}{\delta +1}}^{\pi } u^{\nu } \bigl\vert M_{\delta }(u) \bigr\vert \,du \\ &\leq \frac{1}{\delta +1} \int _{\frac{1}{\delta +1}}^{\pi } u^{ \nu -2}\,du \\ &= \textstyle\begin{cases} (\delta +1)^{-1}, &\nu >1, \\ (\delta +1)^{-\nu }, &\nu < 1, \\ (\delta +1)^{-1}\log( \delta +1)\pi, & \nu =1. \end{cases}\displaystyle \end{aligned}$$

(34)

From (32) and (34), we get

$$ \bigl\Vert T_{\delta}(\cdot) \bigr\Vert _{\rho}=O(1) \textstyle\begin{cases} (\delta +1)^{-1}, &\nu >1, \\ (\delta +1)^{-\nu }, &\nu < 1, \\ (\delta +1)^{-1}\log(\delta +1)\pi, & \nu =1. \end{cases} $$

(35)

Using the generalized Minkowski inequality [21] and Lemma 4.6, we get

$$\begin{aligned} \bigl\Vert w_{k}(T_{\delta },\cdot) \bigr\Vert _{\eta,\infty } &=\sup_{l>0}\bigl(l^{- \eta }w_{k}(T_{\delta },l)_{\rho } \bigr) \\ &=\sup_{l>0}\bigl(l^{-\eta } \bigl\Vert \varUpsilon _{\delta }(\cdot,l) \bigr\Vert _{\rho }\bigr) \\ &=\sup_{l>0} \biggl[l^{-\eta } \biggl( \frac{1}{2\pi } \int _{0}^{2 \pi } \biggl\vert \int _{0}^{\pi } \bigl\vert M_{\delta }(u) \bigr\vert \Phi (y,l,u) \,du \biggr\vert ^{\rho } \,dy \biggr)^{\frac{1}{\rho }} \biggr] \\ &\leq \sup_{l>0} \biggl[l^{-\eta } \biggl( \frac{1}{2\pi } \biggr) ^{\frac{1}{\rho }} \int _{0}^{\pi } \biggl\{ \int _{0}^{2\pi } \bigl\vert M_{ \delta }(u) \bigr\vert ^{\rho } \bigl\vert \Phi (y,l,u) \bigr\vert ^{\rho } \,dy \biggr\} ^{\frac{1}{ \rho }} \,du \biggr] \\ &=\sup_{l>0} \biggl[l^{-\eta } \int _{0}^{\pi } \bigl\Vert \Phi (\cdot,l,u) \bigr\Vert _{\rho } \bigl\vert M_{\delta }(u) \bigr\vert \,du \biggr] \\ &= \int _{0}^{\pi }\Bigl(\sup_{l>0} l^{-\eta } \bigl\Vert \Phi (\cdot,l,u) \bigr\Vert _{\rho } \Bigr) \bigl\vert M_{\delta }(u) \bigr\vert \,du \\ &= O(1) \int _{0}^{\pi } u^{\nu -\eta } \bigl\vert M_{\delta }(u) \bigr\vert \,du \\ &= O(1) \biggl[ \int _{0}^{\frac{1}{\delta +1}} u^{\nu -\eta } \bigl\vert M_{ \delta }(u) \bigr\vert \,du + \int _{\frac{1}{\delta +1}}^{\pi } u^{\nu -\eta } \bigl\vert M_{\delta }(u) \bigr\vert \,du \biggr] \\ &=O(1)[R_{3}+S_{3}]. \end{aligned}$$

(36)

Using Lemma 4.2, we get

$$\begin{aligned} R_{3} &= \int _{0}^{\frac{1}{\delta +1}} u^{\nu -\eta } \bigl\vert M_{\delta }(u) \bigr\vert \,du \\ &=O\bigl((\delta +1)^{\eta -\nu }\bigr). \end{aligned}$$

(37)

Using Lemma 4.3, we get

$$\begin{aligned} S_{3} &= \int _{\frac{1}{\delta +1}}^{\pi } u^{\nu -\eta } \bigl\vert M_{\delta }(u) \bigr\vert \,du \\ &=O(1) \frac{1}{\delta +1} \int _{\frac{1}{\delta +1}}^{\pi } u^{ \nu -\eta -2} \,du \\ &= O(1) \textstyle\begin{cases} (\delta +1)^{-1}, &\nu -\eta >1, \\ (\delta +1)^{-\nu +\eta }, &\nu -\eta < 1, \\ (\delta +1)^{-1} \log (\delta +1)\pi, &\nu -\eta =1. \end{cases}\displaystyle \end{aligned}$$

(38)

From (36) to (38), we get

$$ \bigl\Vert w_{k}(T_{\delta },\cdot) \bigr\Vert _{\eta,\infty }= O(1) \textstyle\begin{cases} (\delta +1)^{-1}, &\nu -\eta >1, \\ (\delta +1)^{-\nu +\eta }, &\nu -\eta < 1, \\ (\delta +1)^{-1} \log (\delta +1)\pi, &\nu -\eta =1. \end{cases} $$

(39)

Combining (31), (35) and (39) we obtain

$$ \bigl\Vert T_{\delta }(\cdot) \bigr\Vert _{B_{\sigma }^{\eta }(L_{ \rho })}=O(1) \textstyle\begin{cases} (\delta +1)^{-1}, &\nu -\eta -\sigma ^{-1} >1, \\ (\delta +1)^{-\nu +\eta +\sigma ^{-1}}, &\nu -\eta -\sigma ^{-1}< 1, \\ (\delta +1)^{-1}[\log (\delta +1)\pi ]^{1-\sigma ^{-1}}, &\nu -\eta -\frac{1}{ \sigma }=1. \end{cases} $$

 □

6 Corollaries

Corollary 6.1

The error approximation of a function $f \in B_{\sigma}^{\nu}(L_{p})$, $\rho \geq 1$, $1<\sigma \leq \infty$ by $N_{p,q}C^{\alpha}$ means of its F. S is given by

$$ \bigl\Vert T_{\delta }(\cdot) \bigr\Vert _{B_{\sigma }^{\eta }(L_{ \rho })}=O(1) \textstyle\begin{cases} (\delta +1)^{-1}, &\nu -\eta -\sigma ^{-1} >1, \\ (\delta +1)^{-\nu +\eta +\sigma ^{-1}}, &\nu -\eta -\sigma ^{-1}< 1, \\ (\delta +1)^{-1}[\log (\delta +1)\pi ]^{1-\sigma ^{-1}}, &\nu -\eta -\frac{1}{\sigma }=1. \end{cases} $$

Corollary 6.2

The error approximation of a function $f \in B_{\sigma }^{\nu }(L_{\rho }), \rho \geq 1, 1<\sigma \leq \infty$ by $N_{pq} E^{q}$ means of its FS is given by

$$ \bigl\Vert T_{\delta }(\cdot) \bigr\Vert _{B_{\sigma }^{\eta }(L_{ \rho })}=O(1) \textstyle\begin{cases} (\delta +1)^{-1}, &\nu -\eta -\sigma ^{-1} >1, \\ (\delta +1)^{-\nu +\eta +\sigma ^{-1}}, &\nu -\eta -\sigma ^{-1}< 1, \\ (\delta +1)^{-1}[\log (\delta +1)\pi ]^{1-\sigma ^{-1}}, &\nu -\eta -\frac{1}{ \sigma }=1. \end{cases} $$

Remark 7

Corollaries 6.1, 6.2 can be further reduced for $N_{p,q} C^{1}$ and $N_{p,q} E^{1}$ means, respectively in view of Note 3(i), (ii).

Corollary 6.3

The error approximation of $f \in B_{\sigma } ^{\nu }(L_{\rho }), \rho \geq 1, 1<\sigma \leq \infty$ by $N_{p} \triangle _{H}$ means of its F. S is given by

$$ \bigl\Vert T_{\delta }(\cdot) \bigr\Vert _{B_{\sigma }^{\eta }(L_{ \rho })}=O(1) \textstyle\begin{cases} (\delta +1)^{-1}, &\nu -\eta -\sigma ^{-1} >1, \\ (\delta +1)^{-\nu +\eta +\sigma ^{-1}}, &\nu -\eta -\sigma ^{-1}< 1, \\ (\delta +1)^{-1}[\log (\delta +1)\pi ]^{1-\sigma ^{-1}}, &\nu -\eta -\frac{1}{ \sigma }=1. \end{cases} $$

Remark 8

Corollary 6.3 can be further reduced for $N_{p}C^{ \alpha }, N_{p}C^{1}, N_{p}E^{q},N_{p}E^{1}$ in view of Note 3(iii), (iv).

7 Particular cases

• 7.1. Using Note 1(ii) and Note 3(iv) and by putting $\eta =0$ in our result, our theorem becomes a particular case of main theorem of [4].

• 7.2. Using Note 1(i) and Note 3(iii) and by putting $\eta =0$ in our result, our main theorem becomes a particular case of main theorem of [1].

• 7.3. Using Note 1(i) and Note 3(i) by putting $\eta =0$ in our result, our main theorem becomes a particular case of main theorem of [3].

• 7.4. If $\xi (t) = t^{\alpha }$ then $\operatorname{Lip}( (\xi (t),r)$ class reduces to $\operatorname{Lip}(\alpha,r)$ class, where $\xi(t)$ is a positive increasing function and $r\geq 1$. Further as $r\to \infty$ in $\operatorname{Lip}(\alpha,r)$ class reduces to Lipα class. Thus, using this argument in [2] and putting $\eta =0$ in our result, our main theorem becomes a particular case of [2].

8 Conclusion

In the review literature, it has been observed that many results have been obtained by the researchers on the degree of approximation of certain functions in different functional spaces like Lipschitz space, Hölder spaces etc. using the trigonometric Fourier approximation method. Since the Besov space generalizes more elementary functions as mentioned above and this space is very effective in measuring regularity properties of the function, this space has a wide range of applications in different areas of engineering and in mathematics in general and in analysis in particular.

Motivated by the usefulness of the Besov space in approximating the error of a certain function, in the present work we estimate the error of a function f in Besov space using a generalized Nörlund–Hausdorff ($N_{pq}\triangle _{H}$) product matrix, our result generalizes several previously known results obtained by using a Lipschitz space. Thus, the results of [1,2,3,4] become particular cases of our theorem. Some useful results are also deduced in the form of corollaries from our theorem.

Some other studies regarding the modulus of the smoothness of functions using different function spaces may be performed in future work.