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