**1. Introduction**

Approximation theory rudimentarily deals with the approximation of functions by simpler functions or more easily calculated functions. Broadly, it is divided into theoretical and constructive approximation. In 1912, S.N. Bernstein [1] was the first to construct a sequence of positive linear operators to provide a constructive proof of the prominent Weierstrass approximation theorem [2] using a probabilistic approach. One can find a detailed monograph about the Bernstein polynomials in [3,4]. Cárdenas et al. [5] in 2011, defined the Bernstein type operators by *Bn*(*f oτ*<sup>−</sup><sup>1</sup>)*oτ* and showed that its Korovkin set is *e*0, *τ*, *τ*2 instead of {*<sup>e</sup>*0,*e*1,*e*2}. These operators present an interesting byproduct sequence of positive linear operators of polynomial type with nice geometric shape preserving properties, which converge to the identity, which in a certain sense improves *Bn* in approximating a number of increasing functions, and which, apart from the constant functions, fixes suitable polynomials of a prescribed degree. The notion of convexity with respect to *τ* plays an important role. Recently, Aral et al. [6] in 2014 defined a similar modification of Szász-Mirakyan type operators obtaining approximation properties of these operators on the interval [0, <sup>∞</sup>).

Very recently motivated by the above work, ˙Ilarslan et al. [7] introduced a new modification of Lupa¸s operators [8] using a suitable function *ρ*, which satisfies the following properties:


The generalized Lupa¸s operators are defined as

$$\mathcal{L}\_m^\rho(f; u) = 2^{-m\rho(u)} \sum\_{l=0}^\infty \frac{(m\rho(u))\_l}{2^l l!} \left( fo\rho^{-1} \right) \left( \frac{l}{m} \right), \tag{1}$$

for *m* ≥ 1, *u* ≥ 0, and suitable functions *f* are defined on [0, <sup>∞</sup>). If *ρ*(*u*) = *u* then (1) reduces to the Lupa¸s operators defined in [8].

˙ Ilarslan et al. [7] gave uniform convergence results on a weighted space, where the weight function is *φ*(*u*) = 1 + *ρ*<sup>2</sup>(*u*) satisfying the conditions (*ρ*1) and and (*ρ*2) given above, in the sense of Gadjiev's results [9,10]. For the rate of convergence, the authors used a weighted modulus of continuity stated by Holho¸s in [11] using the weight function. They obtained a Voronovskaya-type result and monotonicity of the sequence of operators <sup>L</sup>*ρm*(*f* ; *<sup>u</sup>*). Moreover, they obtained some quantitative type theorems on weighted spaces.

The purpose of this paper is to define the *q*-analogue of operators (1) which depend on *ρ*.

Before proceeding further, let us recall some basic definitions and notations of quantum calculus [12]. For any fixed real number *q* > 0, the *q*-integer [*l*]*<sup>q</sup>*, for *l* ∈ N (set of natural numbers) are defined as

$$[l]\_q := \begin{cases} \frac{(1-q^l)}{(1-q)}, & q \neq 1\\ l, & q = 1, \end{cases}$$

and the *q*-factorial by

$$[l]\_q! := \begin{cases} [l]\_q [l-1]\_q \dots [1]\_q & l \ge 1\\ l & l = 0. \end{cases}$$

The *q*-Binomial expansion is

$$(\mu + y)\_q^m := (\mu + y)(\mu + qy)(\mu + q^2y)\cdots(\mu + q^{m-1}y)\_r$$

and the *q*-binomial coefficients are as follows:

$$\left[ \begin{array}{c} m \\ l \end{array} \right]\_q := \frac{[m]\_q!}{[l]\_q! [m-l]\_q!}.$$

The Gauss-formula is defined as:

$$(\mu + y)\_q^m = \sum\_{j=0}^m \left[ \begin{array}{c} m \\ l \end{array} \right]\_q q^{j(j-1)/2} y^j \mu^{m-j} \dots$$

After development of *q*-calculus, Lupa¸s [13] introduced the *q*-Lupa¸s operator (rational) as follows:

$$L\_{m,q}(f;u) = \sum\_{l=0}^{m} \frac{f\left(\frac{[l]\_q}{[m]\_q}\right) \left[\begin{array}{c} m\\l \end{array}\right]\_q q^{\frac{l(l-1)}{2}} u^l (1-u)^{m-l}$$

$$\prod\_{j=1}^{m} \{(1-u) + q^{j-1}u\} \tag{2}$$

and studied its approximation properties.

Similarly, Phillips [14] constructed another *q*-analogue of Bernstein operators (polynomials) as follows:

$$B\_{m,q}(f;u) = \sum\_{l=0}^{m} \begin{bmatrix} m \\ l \end{bmatrix}\_q u^l \prod\_{s=0}^{m-l-1} (1 - q^s u) \; f\left(\frac{[l]\_q}{[m]\_q}\right), \; u \in [0, 1] \tag{3}$$

where *Bm*,*<sup>q</sup>* : *C*[0, 1] → *C*[0, 1] defined for any *m* ∈ N and any function *f* ∈ *C*[0, 1], where *C*[0, 1] denotes the set of all continuous functions on [0, 1].

The basis of these operators have been used in Computer Aided Geometric Design (CAGD) to study curves and surfaces. Then it became an active area of research in approximation theory as well *Mathematics* **2020**, *8*, 68

as CAGD [15–17]. In the recent past, *q*-analogues of various operators were investigated by several researchers (see [18–23]).

The *q*-analogue is a very interesting idea. It can also be used in statistical and biological physics, multi- type directed scalefree percolation, and modeling epidemic spread with awareness and heterogeneous transmission rates in networks.

Persuaded by the above mentioned work, we introduce the *q*-analogue of operators (1) which depends on a suitable function *ρ*, as follows:

**Definition 1.** *Let* 0 < *q* < 1 *and m* ∈ N*. For f* : [0, ∞) → R, *we define q-analogue of generalized Lupa¸s operators as*

$$\mathcal{L}\_{m,q}^{\rho}(f;\mathfrak{u}) = 2^{-[m]\_q \rho(\mathfrak{u})} \sum\_{l=0}^{\infty} \frac{([m]\_q \rho(\mathfrak{u}))\_l}{2^l [l]\_q!} \left( f o \rho^{-1} \right) \left( \frac{[l]\_q}{[m]\_q} \right), \tag{4}$$

*where* ([*m*]*qρ*(*u*))*<sup>l</sup> is the rising factorial defined as:*

$$\begin{aligned} ([m]\_\mathfrak{q}\rho(\mathfrak{u}))\_0 &= 1, \\ ([m]\_\mathfrak{q}\rho(\mathfrak{u}))\_l &= ([m]\_\mathfrak{q}\rho(\mathfrak{u}))([m]\_\mathfrak{q}\rho(\mathfrak{u})+1)([m]\_\mathfrak{q}\rho(\mathfrak{u})+2)\cdots([m]\_\mathfrak{q}\rho(\mathfrak{u})+l-1), \quad l \ge 0. \end{aligned}$$

The operators (4) are linear and positive. For *q* = 1, the operators (4) turn out to be generalized Lupa¸s operators defined in (1). Next, we prove some auxiliary results for (4).

**Lemma 1.** *Let* <sup>L</sup>*ρ<sup>m</sup>*,*<sup>q</sup> be given by (4). Then for each u* ≥ 0 *and m* ∈ N *we have*

$$\begin{array}{ll} (i) & \mathcal{L}\_{\mathsf{m},q}^{\rho}(1;u) = 1, \\ (ii) & \mathcal{L}\_{\mathsf{m},q}^{\rho}(\rho;u) = \rho(u), \\ (iii) & \mathcal{L}\_{\mathsf{m},q}^{\rho}(\rho^2;u) = \rho^2(u) + \frac{(1+q)}{[m]\_q}\rho(u), \\ (iv) & \mathcal{L}\_{\mathsf{m},q}^{\rho}(\rho^3;u) = \rho^3(u)q^3 + \frac{(3q^3+q^2+2q)}{[m]\_q}\rho^2(u) + \frac{(2q^3+q^2+2q+1)}{[m]\_q^2}\rho(u), \\ (v) & \mathcal{L}\_{\mathsf{m},q}^{\rho}(\rho^4;u) = \rho^4(u)q^6 + \frac{(6q^6+q^3+2q^4+3q^3)}{[m]\_q}\rho^3(u) + \frac{(11q^6+3q^5+6q^4+1aq^3+3q^2+3q)}{[m]\_q^2}\rho^2(u) \\ & + \frac{(6q^6+2q^5+4q^4+7q^3+3q^2+3q+1)}{[m]\_q^3}\rho(u). \end{array}$$

**Proof.** By taking into account the recurrence relation ([*m*]*qρ*(*u*))<sup>0</sup> = 1,([*m*]*qρ*(*u*))*<sup>l</sup>* = ([*m*]*qρ*(*u*)) ([*m*]*qρ*(*u*) + <sup>1</sup>)*<sup>l</sup>*−1, *l* ≥ 1, we have

(i)

$$\mathcal{L}\_{m,q}^{\rho}(1;\mu) = 2^{-[m]\_{q\mathbb{P}}(\mu)} \sum\_{l=0}^{\infty} \frac{([m]\_{q}\rho(\mu))\_{l}}{\mathfrak{L}^{l}[l]\_{q}!} \quad = 1.$$

(ii)

$$\begin{aligned} \mathcal{L}\_{\mathfrak{m},q}^{\rho}(\rho;\mu) &= \ \_ {\mathfrak{L}} ^ {\ [m]} \_ {\!\rho \!\!/ } \_ {\!\rho \!\!/ } ^ {\!\rho \!\/ } \_ {\!\rho \!\/ } ^ {\!\rho \!\/ } \_ {\!\rho \!\/ } ^ {\!\rho \!\/ } \_ {\!\rho \!\/ } ^ {\!\rho \!\/ } \_ {\!\rho \!\/ } ^ {\!\rho \!\/ } \_ {\!\rho \!\/ } \_ {\!\rho \!\/ } ^ {\!\rho \!\/ } \_ {\!\rho \!\/ } \_ {\!\rho \!\/ } ^ {\!\rho \!\/} \_ {\!\rho \!\/ } \_ {\!\rho \!\/ } \_ {\!\rho \!\/ } \_ \_ {\!\rho \!\/ } \_ \_ {\!\rho \!\/ } \_ \_ {\!\rho \!\/ } \_ \_ \times \_ \begin{aligned} &= \ \ \_ {\!\sigma \!\/ } \_ ^ {\!\rho \!\/} \_ {\!\rho \!\/} \_ {\!\rho \!\/} \_ {\!\rho \!\/} \_ {\!\rho \!\/} \_ {\!\rho \!\/} \_ {\!\rho \!\/} \_ \_ \times \begin{aligned} &= \ \ \_ {\!\sigma \!\/} \_ \end{aligned} &= \ \ \_ {\!\sigma \!\/} \_ {\!\rho \!\/} \_ {\!\sigma \!\/} \_ {\!\rho \!\/} \_ {\!\sigma \!\/} \_ {\!\sigma \!\/} \_ \_ \times \begin{aligned} &= \ \ \_ {\!\sigma \!\/} \_ {\!\sigma \!\/} \_ \end{aligned} &= \ \ \ \ \ \end$$

(iii)

$$\begin{split} \mathcal{L}\_{m,q}^{\rho}(\rho^2;u) &= \ 2^{-\lfloor m \rfloor\_{q}\rho(u)} \sum\_{l=0}^{\infty} \frac{(\lfloor m \rfloor\_{q}\rho(u))\_{l}}{[l]\_{q}!2^{l}} \frac{[l]\_{q}^{2}}{[m]\_{q}^{2}} \\ &= \ \ \ \frac{2^{-\lfloor m \rfloor\_{q}\rho(u)}}{[m]\_{q}^{2}} \sum\_{l=0}^{\infty} \frac{(\lfloor m \rfloor\_{q}u)\_{l}}{[l]\_{q}!2^{l}} [l]\_{q}^{2} \\ &= \ \ \ \frac{2^{-\left(\lfloor m \rfloor\_{q}\rho(u)+1\right)}[m]\_{q}\rho(u)}{[m]\_{q}^{2}} \sum\_{l=0}^{\infty} \frac{(\lfloor m \rfloor\_{q}\rho(u)+1)\_{l}}{[l]\_{q}!2^{l}} [l]\_{q} \\ &= \ \ \rho^{2}(u) + \frac{(1+q)}{[m]\_{q}}\rho(u). \end{split}$$

(iv)

$$\begin{array}{rcl} \mathcal{L}\_{m,q}^{\rho}(\rho^3;u) &=& 2^{-\left[m\right]\_{\mathbb{P}}\rho(u)}\sum\_{l=0}^{\infty}\frac{([m]\_{q}\rho(u))\_{l}}{[l]\_{q}!2^{l}}\frac{[l]\_{q}^{3}}{[m]\_{q}^{3}}.\end{array}$$

Now by using [*l* + 1]*q* = (1 + *<sup>q</sup>*[*l*]*q*) and shifting *l* to *l* + 1, we have

$$\begin{split} \mathcal{L}\_{m,q}^{\rho}(\rho^{3};u) &= \quad \frac{2^{-\left(\lceil m \right\rfloor\_{q}\rho(u)+1\right)}[m]\_{q}\rho(u)}{[m]\_{q}^{3}}\sum\_{l=0}^{\infty}\frac{([m]\_{q}\rho(u)+1)\_{l}}{[l]\_{q}!2^{l}}[l+1]\_{q}^{2} \\ &= \quad \frac{2^{-\left([m]\_{q}\rho(u)+1\right)}[m]\_{q}\rho(u)}{[m]\_{q}^{3}}\sum\_{l=0}^{\infty}\frac{([m]\_{q}\rho(u)+1)\_{l}}{[l]\_{q}!2^{l}}(1+q[l]\_{q})^{2} \\ &= \quad \frac{2^{-\left([m]\_{q}\rho(u)+1\right)}[m]\_{q}\rho(u)}{[m]\_{q}^{3}}\sum\_{l=0}^{\infty}\frac{([m]\_{q}\rho(u)+1)\_{l}}{[l]\_{q}!2^{l}} \\ &+ \quad \frac{2^{-\left([m]\_{q}\rho(u)+1\right)}[m]\_{q}\rho(u)}{[m]\_{q}^{3}}\sum\_{l=0}^{\infty}\frac{([m]\_{q}\rho(u)+1)\_{l}}{[l]\_{q}!2^{l}}q^{2}[l]\_{q}^{2} \\ &+ \quad \frac{2^{-\left([m]\_{q}\rho(u)+1\right)}[m]\_{q}\rho(u)}{[m]\_{q}^{3}}\sum\_{l=0}^{\infty}\frac{([m]\_{q}\rho(u)+1)\_{l}}{[l]\_{q}!2^{l}}2q[l]\_{q} \\ &= \quad A+B+C(\text{say}). \end{split}$$

Now, let us calculate the values of *A*, *B*, and *C*

$$\begin{aligned} A\_- &= \frac{2^{-\left(\left[m\right]\_q \rho\left(\mu\right) + 1\right)} \left[m\right]\_q \rho\left(\mu\right)}{\left[m\right]\_q^3} \sum\_{l=0}^\infty \frac{\left(\left[m\right]\_q \rho\left(\mu\right) + 1\right)\_l}{\left[l\right]\_q! 2^l}, \\ &= \frac{\rho\left(\mu\right)}{\left[m\right]\_q^2}. \end{aligned}$$

$$\begin{split} B &=& \frac{2^{-(\lfloor m \rfloor\_{q}\rho(u)+1)}[m]\_{q}\rho(u)}{[m]\_{q}^{3}}\sum\_{l=0}^{\infty}\frac{([m]\_{q}\rho(u)+1)\_{l}}{[l]\_{q}!2^{l}}q^{2}[l]\_{q}^{2}}{[m]\_{q}!2^{l}}\\ &=& \frac{2^{-(\lfloor m \rfloor\_{q}\rho(u)+2)}([m]\_{q}\rho(u)+1)\rho(u)}{[m]\_{q}^{2}}q^{2}\sum\_{l=0}^{\infty}\frac{([m]\_{q}\rho(u)+2)\_{l}}{[l]\_{q}!2^{l}}[l+1]\_{q}\\ &=& \frac{2^{-(\lfloor m \rfloor\_{q}\rho(u)+2)}([m]\_{q}\rho(u)+1)\rho(u)}{[m]\_{q}^{2}}q^{2}\sum\_{l=0}^{\infty}\frac{([m]\_{q}\rho(u)+2)\_{l}}{[l]\_{q}!2^{l}}(1+q[l]\_{q})\\ &=& \rho^{3}(u)q^{3}+\frac{3\rho^{2}(u)q^{3}}{[m]\_{q}}+\frac{2\rho(u)q^{3}}{[m]\_{q}^{2}}+\frac{\rho^{2}(u)q^{2}}{[m]\_{q}}+\frac{\rho(u)q^{2}}{[m]\_{q}^{2}}.\end{split}$$

Also,

$$\begin{aligned} \mathcal{C} &= \frac{2^{-\left(\left[m\right]\_q \rho(u) + 1\right)} [m]\_q \rho(u)}{[m]\_q^3} \sum\_{l=0}^\infty \frac{\left(\left[m\right]\_q \rho(u) + 1\right)\_l}{[l]\_q! 2^l} 2q[l]\_{q'}, \\ &= \frac{2\rho^2(u)q}{[m]\_q} + \frac{2\rho(u)q}{[m]\_q^2}. \end{aligned}$$

On adding *A*, *B*, and *C* we have,

$$\mathcal{L}\_{m,q}^{\rho}(\rho^3;u) = \rho^3(u)q^3 + \frac{(3q^3+q^2+2q)}{[m]\_q}\rho^2(u) + \frac{(2q^3+q^2+2q+1)}{[m]\_q^2}\rho(u).$$

(v)

$$\begin{array}{rcl} \mathcal{L}^{\rho}\_{m,q}(\rho^{4};u) &=& 2^{-[m]\_{q}\rho(u)} \sum\_{l=0}^{\infty} \frac{([m]\_{q}\rho(u))\_{l}}{[l]\_{q}!2^{l}} \frac{[l]\_{q}^{4}}{[m]\_{q}^{4}}. \end{array}$$

Now, by using [*l* + 1]*q* = (1 + *<sup>q</sup>*[*l*]*q*) and shifting *l* to *l* + 1, we have

<sup>L</sup>*ρ<sup>m</sup>*,*<sup>q</sup>*(*ρ*4; *u*) = <sup>2</sup>−([*m*]*qρ*(*u*)+1)[*m*]*qρ*(*u*) [*m*]<sup>4</sup>*q* ∞∑*l*=0 ([*m*]*qρ*(*u*) + <sup>1</sup>)*l* [*l*]*q*!2*<sup>l</sup>* [*l* + 1]3*q* = <sup>2</sup>−([*m*]*qρ*(*u*)+1)*ρ*(*u*) [*m*]<sup>3</sup>*q* ∞ ∑ *l*=0 ([*m*]*qρ*(*u*) + <sup>1</sup>)*l* [*l*]*q*!2*<sup>l</sup>* (1 + *<sup>q</sup>*[*l*]*q*)<sup>3</sup> = <sup>2</sup>−([*m*]*qρ*(*u*)+1)*ρ*(*u*) [*m*]<sup>3</sup>*q* ∞ ∑ *l*=0 ([*m*]*qρ*(*u*) + <sup>1</sup>)*l* [*l*]*q*!2*<sup>l</sup>* + <sup>2</sup>−([*m*]*qρ*(*u*)+1)*ρ*(*u*) [*m*]<sup>3</sup>*q* ∞ ∑ *l*=0 ([*m*]*qρ*(*u*) + <sup>1</sup>)*l* [*l*]*q*!2*<sup>l</sup> <sup>q</sup>*<sup>3</sup>[*l*]<sup>3</sup>*q* + <sup>2</sup>−([*m*]*qρ*(*u*)+1)*ρ*(*u*) [*m*]<sup>3</sup>*q* ∞ ∑ *l*=0 ([*m*]*qρ*(*u*) + <sup>1</sup>)*l* [*l*]*q*!2*<sup>l</sup>* <sup>3</sup>*q*<sup>2</sup>[*l*]<sup>2</sup>*q* + <sup>2</sup>−([*m*]*qρ*(*u*)+1)*ρ*(*u*) [*m*]<sup>3</sup>*q* ∞ ∑ *l*=0 ([*m*]*qρ*(*u*) + <sup>1</sup>)*l* [*l*]*q*!2*<sup>l</sup>* <sup>3</sup>*q*[*l*]*<sup>q</sup>*, = *D* + *E* + *F* + *<sup>G</sup>*(*Say*).

Now, let us calculate the values of *D*, *E*, *F*, and *G*

$$\begin{aligned} D &=& \frac{2^{-(\lceil m \rceil\_q \rho(u) + 1)} \rho(u)}{[m]\_q^3} \sum\_{l=0}^\infty \frac{(\lceil m \rceil\_q \rho(u) + 1)\_l}{[l]\_q! 2^l}, \\ &=& \frac{\rho(u)}{[m]\_q^3} .\end{aligned}$$

*Mathematics* **2020**, *8*, 68

*E* = <sup>2</sup>−([*m*]*qρ*(*u*)+1)*ρ*(*u*) [*m*]<sup>3</sup>*q* ∞∑*l*=0 ([*m*]*qρ*(*u*) + <sup>1</sup>)*l* [*l*]*q*!2*<sup>l</sup> <sup>q</sup>*<sup>3</sup>[*l*]<sup>3</sup>*q* = <sup>2</sup>−([*m*]*qρ*(*u*)+2)*ρ*(*u*)([*m*]*qρ*(*u*) + 1) [*m*]<sup>3</sup>*q q*3 ∞ ∑ *l*=0 ([*m*]*qρ*(*u*) + <sup>2</sup>)*l* [*l*]*q*!2*<sup>l</sup>* [*l* + 1]2*q* = <sup>2</sup>−([*m*]*qρ*(*u*)+2)*ρ*(*u*)([*m*]*qρ*(*u*) + 1) [*m*]<sup>3</sup>*q q*3 ∞ ∑ *l*=0 ([*m*]*qρ*(*u*) + <sup>2</sup>)*l* [*l*]*q*!2*<sup>l</sup>* (1 + *<sup>q</sup>*[*l*]*q*)<sup>2</sup> = <sup>2</sup>−([*m*]*qρ*(*u*)+2)*ρ*(*u*)([*m*]*qρ*(*u*) + 1) [*m*]<sup>3</sup>*q q*3 ∞ ∑ *l*=0 ([*m*]*qρ*(*u*) + <sup>2</sup>)*l* [*l*]*q*!2*<sup>l</sup>* (1 + *<sup>q</sup>*<sup>2</sup>[*l*]<sup>2</sup>*q* + <sup>2</sup>*q*[*l*]*q*) = *ρ*4(*u*) + <sup>6</sup>*ρ*<sup>3</sup>(*u*) [*m*]*q* + <sup>11</sup>*ρ*<sup>2</sup>(*u*) [*m*]<sup>2</sup>*q* + <sup>6</sup>*ρ*(*u*) [*m*]<sup>3</sup>*q q*6 + *ρ*3(*u*) [*m*]*q* + <sup>3</sup>*ρ*<sup>2</sup>(*u*) [*m*]<sup>2</sup>*q* + <sup>2</sup>*ρ*(*u*) [*m*]<sup>3</sup>*q q*5 + <sup>2</sup>*ρ*<sup>3</sup>(*u*) [*m*]*q* + <sup>6</sup>*ρ*<sup>2</sup>(*u*) [*m*]<sup>2</sup>*q* + <sup>4</sup>*ρ*(*u*) [*m*]<sup>3</sup>*q q*4 + *ρ*2(*u*) [*m*]<sup>2</sup>*q* + *ρ*(*u*) [*m*]<sup>3</sup>*q q*3.

Similarly,

$$\begin{aligned} F &=& \frac{2^{-(\lfloor m \rfloor\_q \rho(u) + 1)} \rho(u)}{[m]\_q^3} \sum\_{l=0}^\infty \frac{([m]\_q \rho(u) + 1)\_l}{[l]\_q! 2^l} 3q^2 [l]\_q^2 \\ &=& \left( \frac{3 \rho^3(u)}{[m]\_q} + \frac{9 \rho^2(u)}{[m]\_q^2} + \frac{6 \rho(u)}{[m]\_q^3} \right) q^3 + \left( \frac{3 \rho^2(u)}{[m]\_q^2} + \frac{3 \rho(u)}{[m]\_q^3} \right) q^2. \end{aligned}$$

Also,

$$\begin{aligned} G &= \frac{2^{-(\lfloor m \rfloor\_q \rho(u) + 1)} \rho(u)}{[m]\_q^3} \sum\_{l=0}^\infty \frac{([m]\_q \rho(u) + 1)\_l}{[l]\_q! 2^l} 3q[l]\_q \\ &= \quad \left( \frac{3 \rho^2(u)}{[m]\_q^2} + \frac{3 \rho(u)}{[m]\_q^3} \right) q. \end{aligned}$$

On adding *D*, *E*, *F*, and *G* we have,

$$\begin{aligned} \mathcal{L}\_{m,q}^{\rho}(\rho^4;u) &= \begin{array}{c} \rho^4(u)q^6 + \frac{(6q^6+q^5+2q^4+3q^3)}{[m]\_q}\rho^3(u) \\ + \frac{(11q^6+3q^5+6q^4+10q^3+3q^2+3q)}{[m]\_q^2}\rho^2(u) \\ + \frac{(6q^6+2q^5+4q^4+7q^3+3q^2+3q+1)}{[m]\_q^3}\rho(u) \end{array} \end{aligned}$$

**Corollary 1.** *For n* = 1, 2, 3, 4 *the nth order central moments of* <sup>L</sup>*ρ<sup>m</sup>*,*<sup>q</sup> defined as <sup>μ</sup>ρ<sup>n</sup>*,*<sup>m</sup>*(*q*; *u*) = <sup>L</sup>*ρ<sup>m</sup>*,*<sup>q</sup>*((*ρ*(*t*) − *ρ*(*u*))*nq* ; *<sup>u</sup>*)*, by using linearity of operators* (4) *and by Lemma 1 we have*

*(i)* <sup>L</sup>*ρ<sup>m</sup>*,*<sup>q</sup>*(*ρ*(*t*) − *ρ*(*u*); *u*) = <sup>L</sup>*ρ<sup>m</sup>*,*<sup>q</sup>*(*ρ*(*t*); *u*) − *<sup>ρ</sup>*(*u*)L*ρ<sup>m</sup>*,*<sup>q</sup>*(1; *u*) = 0,

$$\begin{array}{rcl} \mathcal{L}^{\rho} \left( \mathrm{ii} \right) \mathcal{L}^{\rho}\_{m,\emptyset} \left( (\rho(\mathrm{t}) - \rho(\mathrm{u}))^{2}; \mathrm{u} \right) &= \mathcal{L}^{\rho}\_{m,\emptyset} \left( \rho^{2}(\mathrm{t}); \mathrm{u} \right) + \rho^{2}(\mathrm{u}) \mathcal{L}^{\rho}\_{m,\emptyset} \left( \mathrm{1}; \mathrm{u} \right) - 2\rho(\mathrm{u}) \mathcal{L}^{\rho}\_{m,\emptyset} \left( \rho(\mathrm{t}); \mathrm{u} \right) = \frac{(1+q)}{|m|\_{\emptyset}} \rho(\mathrm{u}),\\ \mathrm{iii} \quad \mathcal{L}^{\rho}\_{m,\emptyset} \left( (\rho(\mathrm{t}) - \rho(\mathrm{u}))^{3}; \mathrm{u} \right) &= \mathcal{L}^{\rho}\_{m,\emptyset} \left( \rho^{3}(\mathrm{t}); \mathrm{u} \right) - \rho^{3}(\mathrm{u}) \mathcal{L}^{\rho}\_{m,\emptyset} \left( \mathrm{1}; \mathrm{u} \right) - 3\rho(\mathrm{u}) \mathcal{L}^{\rho}\_{m,\emptyset} \left( \rho^{2}(\mathrm{t}); \mathrm{u} \right) - 3\rho^{2}(\mathrm{u}) \mathcal{L}^{\rho}\_{m,\emptyset} \left( \rho^{3}(\mathrm{t}); \mathrm{u} \right). \end{array}$$

$$\begin{array}{ccccccccc} & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ & (\rho(t); u) = (q^3 - 1)\rho^3(u) + \frac{(3q^3 + q^2 - q - 3)}{[m]\_q}\rho^2(u) + \frac{(2q^3 + q^2 + 2q + 1)}{[m]\_q}\rho(u), & & & & & & & \\ \rho(\dots) & \mathcal{A}^{\rho} & ((\rho(\dots) & \rho(\dots))\mathcal{A}\_{\sigma}) & \mathcal{A}^{\rho} & (\mathcal{A}(\rho(\dots)) & \mathcal{A}(\rho(\dots)) & \mathcal{A}(\rho(\dots)) & \mathcal{A}(\rho(\dots)) & \mathcal{A}(\rho(\dots)) & \mathcal{A}(\rho(\dots)) \\ \end{array}$$

$$\begin{array}{rcl} \langle \text{iv)} \quad \mathcal{L}\_{m,q}^{\rho}((\rho(t)-\rho(u))^{4};u) &=& \mathcal{L}\_{m,q}^{\rho}(\rho^{4}(t);u) + \rho^{4}(u)\mathcal{L}\_{m,q}^{\rho}(1;u) + 6\rho^{2}(u)\mathcal{L}\_{m,q}^{\rho}(\rho^{2}(t);u) - 4\rho^{3}(u)\mathcal{L}\_{m,q}^{\rho}(1;u) \\ & (\rho(t);u) - 4\rho(u)\mathcal{L}\_{m,q}^{\rho}(\rho^{3}(t);u) &=& (q^{6}-4q^{3}+3)\rho^{4}(u) + \frac{(6q^{6}+q^{5}+2q^{4}-9q^{3}-4q^{2}-2q+6)}{[m]\_{q}}\rho^{3}(u) + \frac{(11q^{6}+2q^{5}+4q^{4}+7q^{3}+3q^{2}+3q+1)}{[m]\_{q}}\rho(u). \end{array}$$

**Remark 1.** *We observe from Lemma 1 and Corollary 1, that for q* = 1*, we get the moments and central moments of generalized Lupa¸s operators [7].*

#### **2. Weighted Approximation**

We start by noting that *ρ* not only defines a Korovkin-type set {1, *ρ*, *ρ*<sup>2</sup>} but also characterizes growth of the functions that are approximated.

Let *φ*(*u*) = 1 + *ρ*<sup>2</sup>(*u*) be a weight function satisfying the conditions (*ρ*1) and and (*ρ*2) given above let <sup>B</sup>*φ*[0, ∞) be the weighted space defined by

$$\mathcal{B}\_{\Phi}[0,\infty) = \{ f : [0,\infty) \to \mathbb{R} \, | \, |f(\mu)| \le \mathcal{K}\_f \phi(\mu), \mu \ge 0 \},$$

where <sup>K</sup>*f* is a constant which depends only on *f* . <sup>B</sup>*φ*[0, ∞) is a normed linear space equipped with the norm

$$\|\|f\|\|\_{\Phi} = \sup\_{u \in [0,\infty)} \frac{|f(u)|}{\Phi(u)}.$$

Also, we define the following subspaces of <sup>B</sup>*φ*[0, ∞) as

$$\mathcal{C}\_{\Phi}[0,\infty) = \{ f \in \mathcal{B}\_{\Phi}[0,\infty) \, : \, \, f \text{ is continuous on } \, [0,\infty) \},$$

$$\mathcal{C}\_{\Phi}^{\*}[0,\infty) = \left\{ f \in \mathcal{C}\_{\Phi}[0,\infty) \, : \, \lim\_{u \to \infty} \frac{f(u)}{\phi(u)} = \mathcal{K}\_{f} \right\},$$

where <sup>K</sup>*f* is a constant depending on *f* and

$$\mathcal{U}\_{\Phi}[0,\infty) = \{ f \in \mathcal{C}\_{\Phi}[0,\infty) \, : \, \frac{f(u)}{\phi(u)} \text{ is uniformly continuous on } [0,\infty) \}.$$

Obviously,

$$\mathcal{C}^\*\_{\Phi}[0,\infty) \subset \mathcal{U}\_{\Phi}[0,\infty) \subset \mathcal{C}\_{\Phi}[0,\infty) \subset \mathcal{B}\_{\Phi}[0,\infty) \dots$$

For the weighted uniform approximation by linear positive operators acting from <sup>C</sup>*φ*[0, ∞) to <sup>B</sup>*φ*[0, <sup>∞</sup>), we state the following results due to Gadjiev in [9,10].

**Lemma 2** ([9])**.** *Let* (A*m*)*m*≥<sup>1</sup> *be a sequence of positive linear operators which acts from* <sup>C</sup>*φ*[0, ∞) *to* <sup>B</sup>*φ*[0, ∞) *if and only if the inequality*

$$|\mathcal{A}\_{\mathfrak{m}}(\phi;\mathfrak{u})| \le \mathcal{K}\_{\mathfrak{m}}\phi(\mathfrak{u}), \ \mathfrak{u} \ge 0,$$

*holds, where* K*m* > 0 *is a constant depending on m.*

**Theorem 1** ([10])**.** *Let* (A*m*)*m*≥<sup>1</sup> *be a sequence of positive linear operators, acting from* <sup>C</sup>*φ*[0, ∞) *to* <sup>B</sup>*φ*[0, ∞) *and satisfying*

$$\lim\_{m \to \infty} \parallel \mathcal{L}\_m \rho^i - \rho^i \parallel\_{\phi} = 0, \quad i = 0, 1, 2.$$

*Then we have*

$$\lim\_{m \to \infty} \parallel \mathcal{L}\_m(f) - f \parallel\_{\phi} = 0, \text{ for } \text{ any} \quad f \in \mathcal{C}^\*\_{\phi}[0, \infty).$$

**Remark 2.** *It is clear from Lemma 1 and Lemma 2 that the operators* <sup>L</sup>*ρ<sup>m</sup>*,*<sup>q</sup> act from* <sup>C</sup>*φ*[0, ∞) *to* <sup>B</sup>*φ*[0, <sup>∞</sup>)*. Also the convergence of these operators are applicable in studying switched linear systems, see: Subspace confinement for switched linear systems, also see in [24,25].*

**Theorem 2.** *Let qm be a sequence in* (0, <sup>1</sup>), *such that qm* → 1 *as m* → ∞. *Then for each function f* ∈ *<sup>C</sup>*<sup>∗</sup>*φ*[0, ∞) *we have*

$$\lim\_{m \to \infty} \parallel \mathcal{L}^{\rho}\_{m, \emptyset m}(f) - f \parallel\_{\phi} = 0.$$

**Proof.** By Lemma 1 (i) and (ii), it is clear that

$$\|\|\mathcal{L}\_{m,\rho\_{\rm{fw}}}^{\rho}(1;u) - 1\|\|\_{\Phi} = 0.$$

$$\|\|\mathcal{L}\_{m,\rho\_{\rm{fw}}}^{\rho}(\rho;u) - \rho\|\|\_{\Phi} = 0.$$

and by Lemma 1 (iii), we have

$$\|\|\mathcal{L}\_{m,q\_m}^{\rho}(\rho^2;u) - \rho^2\|\|\_{\phi} = \sup\_{u \in [0,\infty)} \frac{(1+q)\rho(u)}{[m]\_q(1+\rho^2(u))} \le \frac{1+q}{[m]\_q}.\tag{5}$$

Then from Lemma 1 and (5) we ge<sup>t</sup> lim*m*→∞ <sup>L</sup>*ρ<sup>m</sup>*,*qm* (*ρ<sup>i</sup>*) − *ρi φ*= 0, *i* = 0, 1, 2. Hence, the proof is completed.

#### **3. Rate of Convergence or Order of Approximation**

In this section, we determine the rate of convergence for <sup>L</sup>*ρ<sup>m</sup>*,*<sup>q</sup>* by weighted modulus of continuity *ωρ*(*f* ; *δ*) which was recently considered by Holho¸s [11] as follows:

$$\omega\_{\rho}(f;\delta) = \sup\_{u,z \in [0,\infty), |\rho(z)-\rho(u)| \le \delta} \frac{|f(z)-f(u)|}{\Phi(z)+\phi(u)}, \quad \delta > 0,\tag{6}$$

where *f* ∈ <sup>C</sup>*φ*[0, <sup>∞</sup>), with the following properties:

$$(\mathbf{i})\quad\omega\_{\boldsymbol{\rho}}(f;0)=0,$$

$$\text{(ii)}\quad \omega\_{\mathbb{P}}(f;\delta) \ge 0, \delta \ge 0 \text{ for } f \in \mathcal{C}\_{\Phi}[0,\infty),$$

(iii) lim*δ*→<sup>0</sup> *ωρ*(*f* ; *δ*) = 0, for each *f* ∈ *<sup>U</sup>φ*[0, <sup>∞</sup>).

**Theorem 3** ([11])**.** *Let* A*m* : <sup>C</sup>*φ*[0, ∞) → <sup>B</sup>*φ*[0, ∞) *be a sequence of positive linear operators with*

$$\left. \right| \left. \mathcal{A}\_{\mathfrak{m}}(\rho^0) - \rho^0 \left. \right|\_{\mathfrak{q}^0} = \left. a\_{\mathfrak{m}} \right|\_{\mathfrak{q}} \tag{7}$$

$$\left. \right| \left. \mathcal{A}\_{m}(\rho) - \rho \right|\_{\phi^{\frac{1}{2}}} = \left. \right|\_{m\_{\prime}} \qquad \left. \right|\_{m\_{\prime}} \tag{8}$$

$$\|\|\mathcal{A}\_{\mathfrak{m}}(\rho^2) - \rho^2\|\|\_{\Phi} = \|\mathfrak{c}\_{\mathfrak{m}\_{\prime}}\tag{9}$$

$$\left. \right| \left. \mathcal{A}\_{\mathfrak{m}}(\rho^{\mathfrak{3}}) - \rho^{\mathfrak{3}} \right|\_{\phi^{\mathfrak{2}}} = \left. \right|\_{\phi^{\mathfrak{2}}} = \left. d\_{\mathfrak{m}} \right|\_{\phi^{\mathfrak{2}}} \tag{10}$$

*where the sequences am, bm, cm, and dm converge to zero as m* → ∞*. Then*

$$\left\| \left| \mathcal{A}\_{m}(f) - f \right| \right\|\_{\phi^{\frac{3}{2}}} \le (\mathcal{T} + 4a\_{m} + 2\varepsilon\_{m})\omega\_{\rho}(f; \delta\_{m}) + \left\| \left| f \right| \right\|\_{\Phi} a\_{m} \tag{11}$$

*for all f* ∈ <sup>C</sup>*φ*[0, <sup>∞</sup>)*, where*

$$\mathcal{S}\_m = 2\sqrt{(a\_m + 2b\_m + c\_m)(1 + a\_m)} + a\_m + \mathfrak{A}b\_m + \mathfrak{A}c\_m + d\_m.$$

**Theorem 4.** *Let for each f* ∈ <sup>C</sup>*φ*[0, ∞) *with* 0 < *q* < 1. *Then we have*

$$\left\| \left| \mathcal{L}\_{m,q}^{\mathcal{P}}(f) - f \right| \right\|\_{\phi^{\frac{2}{2}}} \leq \left( \mathcal{T} + \frac{2(1+q)}{[m]\_{q}} \right) \omega\_{\rho}(f; \delta\_{m,q})\_{\*}$$

*where ωρ is the weighted modulus of continuity defined in (6) and*

$$\delta\_{m,q} = -2\sqrt{\frac{(1+q)}{[m]\_q}} + \frac{3(1+q)}{[m]\_q} + \left( (q^3 - 1) + \frac{(3q^3 + q^2 + 2q)}{[m]\_q} + \frac{(2q^3 + q^2 + 2q + 1)}{[m]\_q^2} \right).$$

**Proof.** By using Lemma 1, we have

$$\begin{aligned} \left| \| \mathcal{L}\_{m,q}^{\mathcal{P}}(\rho^0) - \rho^0 \| \right|\_{\phi^0} &= a\_{m,q} = 0, \\ \left| \| \mathcal{L}\_{m,q}^{\mathcal{P}}(\rho) - \rho \| \right|\_{\phi^{\frac{1}{2}}} &= b\_{m,q} = 0, \end{aligned}$$

and

$$\|\,\mathcal{L}\_{m,q}^{\rho}(\rho^2) - \rho^2\|\_{\phi} \le \frac{(1+q)}{[m]\_q} = c\_{m,q}.$$

Finally,

$$\left\| \left| \begin{array}{c} \mathcal{L}\_{m,q}^{\rho}(\rho^3) - \rho^3 \right| \right\|\_{\rho^{\frac{3}{2}}} \leq \ (q^3 - 1) + \frac{(3q^3 + q^2 + 2q)}{[m]\_{\rho}} + \frac{(2q^3 + q^2 + 2q + 1)}{[m]\_{\overline{q}}^2} = d\_{m,q}. \end{array} \right\| $$

Thus, the sequences *am*,*q*, *bm*,*q*, *cm*,*q*, and *dm*,*<sup>q</sup>* are calculated. The sequences *am*, *bm*, *cm*, and *dm* converge to zero as *m* → ∞. Then

$$\left\| \left| \left| \mathcal{L}\_{m,q}^{\rho}(f) - f \right| \right|\_{\rho \frac{3}{2}} \le (\mathcal{T} + 4a\_{m,q} + 2c\_{m,q})\omega\_{\rho}(f; \delta\_{m,q}) + \left\| \left| f \right| \right\|\_{\phi} a\_{m,q} \tag{12}$$

for all *f* ∈ <sup>C</sup>*φ*[0, <sup>∞</sup>), where

$$\delta\_{m,q} = 2\sqrt{(a\_{m,q} + 2b\_{m,q} + c\_{m,q})(1 + a\_{m,q})} + a\_{m,q} + 3b\_{m,q} + 3c\_{m,q} + d\_{m,q}.$$

Hence, by substituting the values of *am*,*q*, *bm*,*q*, *cm*,*<sup>q</sup>* and *dm*,*<sup>q</sup>* we obtain the desired result.

**Remark 3.** *For* lim *δ*→0*ωρ*(*f* ; *δ*) = 0 *in Theorem 4, we find*

$$\lim\_{m \to \infty} \parallel \mathcal{L}\_{m,q}^{\rho}(f) - f \parallel\_{\phi^{\frac{3}{2}}} = 0, \quad \text{for} \quad f \in \mathcal{U}\_{\Phi}[0, \infty).$$

#### **4. Voronovskaya-Type Theorem**

In this section, using a technique developed in [5] by Cardenas-Morales, Garrancho and Ra¸sa, we prove pointwise convergence of <sup>L</sup>*ρ<sup>m</sup>*,*<sup>q</sup>* by obtaining Voronovskaya-type theorems.

**Theorem 5.** *Let f* ∈ <sup>C</sup>*φ*[0, <sup>∞</sup>)*, u* ∈ [0, ∞) *with* 0 < *qm* < 1, *qm* → 1 *as m* → ∞*. Suppose that* - *f oρ*<sup>−</sup><sup>1</sup> *and* - *f oρ*<sup>−</sup><sup>1</sup> *exist at ρ*(*u*)*. If* - *f oρ*<sup>−</sup><sup>1</sup> *is bounded on* [0, <sup>∞</sup>)*, then we have*

$$\lim\_{m \to \infty} [m]\_{q\_m} \left[ \mathcal{L}\_{m, q\_m}^{\rho} (f; \mathfrak{u}) - f(\mathfrak{u}) \right] = \rho(\mathfrak{u}) \left( f o \rho^{-1} \right)^{\prime \prime} (\rho(\mathfrak{u})).$$

**Proof.** By using the *q*-Taylor expansion of - *f oρ*<sup>−</sup><sup>1</sup> at *ρ*(*u*) ∈ [0, <sup>∞</sup>), there exist a point *w* lying between *u* and *z*, then we have

$$f(w) = \left(f o \rho^{-1}\right)\left(\rho(w)\right) \\
= \left(f o \rho^{-1}\right)\left(\rho(u)\right) + \left(f o \rho^{-1}\right)'\left(\rho(u)\right)\left(\rho(w) - \rho(u)\right) \\
\tag{13}$$

$$+ \quad \frac{\left(f o \rho^{-1}\right)''\left(\rho(u)\right)\left(\rho(w) - \rho(u)\right)^2}{[2]\_q} + \lambda\_u^q(w)\left(\rho(w) - \rho(u)\right)^2,$$

*Mathematics* **2020**, *8*, 68

where

$$
\lambda\_u^q(w) = \frac{\left(fo\rho^{-1}\right)''\left(\rho(w)\right) - \left(fo\rho^{-1}\right)''\left(\rho(u)\right)}{[2]\_q}.\tag{14}
$$

Therefore, by (14) together with the assumption on *f* ensures that

> |*λqu*(*w*)|≤K, for all *w* ∈ [0, ∞)

and is convergen<sup>t</sup> to zero as *w* → *u*. Now applying the operators (4) to the equality (13), we obtain

$$\begin{split} \left[\mathcal{L}\_{\boldsymbol{m},\boldsymbol{\eta}\_{\rm m}}^{\boldsymbol{\rho}}(f;\boldsymbol{u}) - f(\boldsymbol{u})\right] &= \quad \left(f\boldsymbol{o}\boldsymbol{\rho}^{-1}\right)'(\boldsymbol{\rho}(\boldsymbol{u})) \mathcal{L}\_{\boldsymbol{m},\boldsymbol{\rho}\_{\rm m}}^{\boldsymbol{\rho}}((\boldsymbol{\rho}(\boldsymbol{w}) - \boldsymbol{\rho}(\boldsymbol{u}));\boldsymbol{u}) \\ &+ \quad \frac{\left(f\boldsymbol{o}\boldsymbol{\rho}^{-1}\right)''(\boldsymbol{\rho}(\boldsymbol{u})) \mathcal{L}\_{\boldsymbol{m},\boldsymbol{\rho}\_{\rm m}}^{\boldsymbol{\rho}}\left((\boldsymbol{\rho}(\boldsymbol{w}) - \boldsymbol{\rho}(\boldsymbol{y}))^{2};\boldsymbol{u}\right)}{[2]\_{\boldsymbol{q}}} \\ &+ \quad \mathcal{L}\_{\boldsymbol{m},\boldsymbol{\rho}\_{\rm m}}^{\boldsymbol{\rho}}\left(\lambda\_{\boldsymbol{u}}^{\boldsymbol{q}}(\boldsymbol{w}) \left((\boldsymbol{\rho}(\boldsymbol{w}) - \boldsymbol{\rho}(\boldsymbol{u}))^{2};\boldsymbol{u}\right)\right). \end{split} \tag{15}$$

By Lemma 1 and Corollary 1, we ge<sup>t</sup>

$$\lim\_{m \to \infty} [m]\_{q\_m} \mathcal{L}\_{m, q\_m}^{\rho} \left( (\rho(w) - \rho(u)); \mu \right) = 0,\tag{16}$$

and

$$\lim\_{m \to \infty} [m]\_{\mathfrak{q}\_{\mathfrak{q}}} \mathcal{L}\_{m, \mathfrak{q}\_{\mathfrak{q}}}^{\rho} \left( (\rho(w) - \rho(u))^2; u \right) = [2]\_{\mathfrak{q}} \rho(u). \tag{17}$$

By estimating the last term on the right hand side of equality (15), we will ge<sup>t</sup> the proof. Since from (14), for every > 0, lim*w*→*u λqmu* (*w*) = 0. Let *δ* > 0 such that |*λqmu* (*w*)| < for every *w* ≥ 0. Using a Cauchy-Schwartz inequality, we have

$$\begin{split} \lim\_{m \to \infty} [m]\_{\mathfrak{q}\_{m}} \mathcal{L}\_{m,\mathfrak{q}\_{m}}^{\rho} \left( |\lambda\_{\mathfrak{u}}^{\mathfrak{q}\_{m}}(w)| \left( \rho(w) - \rho(u) \right)^{2}; u \right) &\leq \mathop{\rm c\limits\_{m \to \infty}} [m]\_{\mathfrak{q}\_{m}} \mathcal{L}\_{m,\mathfrak{q}\_{m}}^{\rho} \left( (\rho(w) - \rho(u))\_{\mathfrak{q}\_{m}}^{2}; u \right) \\ &+ \mathop{\rm c\limits\_{\delta^{2}}} \mathop{\rm l\limits\_{m \to \infty}} [m]\_{\mathfrak{q}\_{m}} \mathcal{L}\_{m,\mathfrak{q}\_{m}}^{\rho} \left( (\rho(w) - \rho(u))\_{\mathfrak{q}\_{m}}^{4}; u \right). \end{split}$$

Since

$$\lim\_{m \to \infty} [m]\_{q\_m} \mathcal{L}\_{m, q\_m}^{\rho} \left( (\rho(w) - \rho(u))\_{q\_m}^4; u \right) = 0,\tag{18}$$

we obtain

$$\lim\_{m \to \infty} [m]\_{q\_m} \mathcal{L}\_{m, q\_m}^{\rho} \left( |\lambda\_u^{q\_m}(w)| \left( \rho(w) - \rho(y) \right)\_{q\_m}^2; y \right) = 0. \tag{19}$$

Thus, by using Equations (16), (17) and (19) to Equation (15) the proof is completed.

#### **5. Local Approximation**

In this section, we present local approximation theorems for the operators <sup>L</sup>*ρ<sup>m</sup>*,*q*. By C*B*[0, <sup>∞</sup>), we denote the space of real-valued continuous and bounded functions *f* defined on the interval [0, <sup>∞</sup>). The norm · on the space C*B*[0, ∞) is given by

$$\|\|f\|\| = \sup\_{0 \le u < \infty} |\|f(\mathbf{x})\|\,.$$

For *δ* > 0 and *W*<sup>2</sup> = {*s* ∈ C*B*[0, ∞) : *s* ,*s* ∈ C*B*[0, <sup>∞</sup>)}. The K-functional is defined as

$$\mathcal{K}\_2(f,\delta) = \inf\_{s \in W^2} \{ \|f - s\| \, + \delta \, \mid \, \mathcal{g}^{\prime\prime} \, \vert \, \} \prime$$

*Mathematics* **2020**, *8*, 68

By Devore and Lorentz ([26], p. 177, Theorem 2.4), there exists an absolute constant C > 0 such that

$$\mathcal{K}(f,\delta) \le \mathcal{C}\omega\_2(f,\sqrt{\delta}).\tag{20}$$

The second order modulus of smoothness is as follows,

$$\omega\_2(f,\sqrt{\delta}) = \sup\_{0 < h \le \sqrt{\mathcal{J}}} \sup\_{u \in [0,\infty)} |f(u+2h) - 2f(u+h) + f(u)|$$

where *f* ∈ *CB*[0, <sup>∞</sup>). The usual modulus of continuity of *f* ∈ *CB*[0, ∞) is defined by

$$\omega(f,\delta) = \sup\_{0 < h \le \delta} \sup\_{\mathfrak{u} \in [0,\infty)} |f(\mathfrak{u}+h) - f(\mathfrak{u})| \, . $$

**Theorem 6.** *Let f* ∈ C*B*[0, ∞) *with* 0 < *q* < 1. *Let ρ be a function satisfying the conditions (ρ*1*), (ρ*2*), and* ||*ρ*|| *is finite. Then, there exists an absolute constant* C > 0 *such that*

$$\left|\mathcal{L}\_{m,q}^{\rho}(f;\mu) - f(\mu)\right| \le \mathcal{CK}\left(f, \frac{(1+q)}{[m]\_{\emptyset}}\rho(\mu)\right) \dots$$

**Proof.** Let *s* ∈ *W*<sup>2</sup> and *u*, *z* ∈ [0, <sup>∞</sup>). Using Taylor's formula we have

$$s(z) = s(u) + \left(s\rho \rho^{-1}\right)'(\rho(u))(\rho(z) - \rho(u)) + \int\_{\rho(u)}^{\rho(z)} (\rho(z) - v) \left(s\rho \rho^{-1}\right)''(v) dv. \tag{21}$$

Using the equality

$$\left(s\rho^{-1}\right)''\left(\rho(u)\right) = \frac{s''(u)}{\left(\rho'(u)\right)^2} - s''(u)\frac{\rho''(u)}{\left(\rho'(u)\right)^3}.\tag{22}$$

Now, put *v* = *ρ*(*y*) in the last term in equality (21), we ge<sup>t</sup>

$$\begin{split} \int\_{\rho(u)}^{\rho(z)} (\rho(z) - v) \left( s \rho \rho^{-1} \right)^{\prime\prime} (v) dv &= \int\_{u}^{z} (\rho(z) - \rho(y)) \left[ \frac{s^{\prime\prime}(y)\rho^{\prime}(y) - s^{\prime}(y)\rho^{\prime\prime}(v)}{(\rho^{\prime}(y))^{2}} \right] dy \\ &= \int\_{\rho(u)}^{\rho(z)} (\rho(z) - v) \frac{s^{\prime\prime}(\rho^{-1}(v))}{(\rho^{\prime}(\rho^{-1}(v)))^{2}} dv \\ &\quad - \int\_{\rho(u)}^{\rho(z)} (\rho(z) - v) \frac{s^{\prime}(\rho^{-1}(v))\rho^{\prime\prime}(\rho^{-1}(v))}{(\rho^{\prime}(\rho^{-1}(v)))^{3}} dv. \end{split} \tag{23}$$

By using Lemma 1 and (23) and applying the operator (4) to the both sides of equality (21), we deduce

$$\begin{split} \mathcal{L}\_{m,\eta}^{\rho}(s;u) &= \quad s(u) + \mathcal{L}\_{m,\eta}^{\rho} \left( \int\_{\rho(u)}^{\rho(z)} (\rho(z) - v) \frac{s''(\rho^{-1}(v))}{(\rho'(\rho^{-1}(v))^2} dv; u \right) \\ &\quad - \quad \mathcal{L}\_{m,\eta}^{\rho} \left( \int\_{\rho(u)}^{\rho(z)} (\rho(z) - v) \frac{s'(\rho^{-1}(v))\rho''(\rho^{-1}(v))}{(\rho'(\rho^{-1}(v))^3} dv; u \right). \end{split}$$

As we know *ρ* is strictly increasing on [0, ∞) and with condition (*ρ*2), we have

$$\left| \mathcal{L}\_{m,\emptyset}^{\rho}(\mathbf{s};\boldsymbol{\mu}) - \mathbf{s}(\boldsymbol{\mu}) \right| \leq \mathcal{M}\_{m,2}^{\rho}(\boldsymbol{\mu}) \left( \left||\mathbf{s}^{\prime\prime}|| + ||\mathbf{s}^{\prime}|| \left||\boldsymbol{\rho}^{\prime\prime}|| \right|| \right) \cdot$$

where

$$\mathcal{M}\_{m,2}^{\rho}(\mathfrak{u}) = \mathcal{L}\_{m,q}^{\rho}((\rho(t) - \rho(\mathfrak{u}))^2; \mathfrak{u}) .$$

For *f* ∈ C*B*[0, <sup>∞</sup>), we have

$$\begin{split} \|\mathcal{L}\_{m,q}^{\rho}(s;u)\| &\quad \leq \quad \|f o \rho^{-1}\| 2^{-[m]\_{q}\rho(u)} \sum\_{l=0}^{\infty} \frac{([m]\_{q}\rho(u))\_{l}}{2^{l}[l]\_{q}!} \\ &\leq \quad \|f\| \|\mathcal{L}\_{m,q}^{\rho}(1;u) = \|f\|. \end{split} \tag{24}$$

Hence we have

$$\begin{split} \left| \mathcal{L}\_{m,q}^{\rho}(f;\mathsf{u}) - f(\mathsf{u}) \right| &\leq \left| \mathcal{L}\_{m,q}^{\rho}(f - \mathsf{s};\mathsf{u}) \right| + \left| \mathcal{L}\_{m,q}^{\rho}(\mathsf{s};\mathsf{u}) - \mathsf{s}(\mathsf{u}) \right| + \left| \mathsf{s}(\mathsf{u}) - f(\mathsf{u}) \right| \\ &\leq 2 \left| \|f - \mathsf{g}\|\| + \frac{(1+q)}{[m]\_{q}} \rho(\mathsf{u}) \left( \|\mathsf{s}''\| + \|\mathsf{s}'\| \|\boldsymbol{\rho}''\| \right). \end{split}$$

If we choose C = max{2, *ρ*}, then

$$\left| \mathcal{L}\_{m,q}^{\rho}(f;\mu) - f(\mu) \right| \leq \mathcal{C} \left( 2||f - g|| + \frac{(1+q)}{[m]\_q} \rho(\mu) ||s'^\prime||\_{\mathcal{W}^2} \right).$$

Taking infimum over all *s* ∈ *W*<sup>2</sup> we obtain

$$\left| \mathcal{L}\_{m,q}^{\rho}(f;\mu) - f(\mu) \right| \le \mathcal{CK}\left( f, \frac{(1+q)}{[m]\_q} \rho(\mu) \right) \dots$$

Now, we recall local approximation in terms of *α* order Lipschitz-type maximal functions given in [27]. Let *ρ* be a function satisfying the conditions (*ρ*1), (*ρ*2), 0 < *α* ≤ 1 and *Lip*M(*ρ*(*u*); *<sup>α</sup>*), M ≥ 0 is the set of functions *f* satisfying the inequality

$$\left| f(z) - f(\mu) \right| \le \mathcal{M} \left| \rho(z) - \rho(\mu) \right|^\alpha , \mu, z \ge 0.$$

Moreover, for a bounded subset E ⊂ [0, <sup>∞</sup>), we say that the function *f* ∈ C*B*[0, ∞) belongs to *Lip*M(*ρ*(*u*); *<sup>α</sup>*), 0 < *α* ≤ 1 on E if

$$|f(z) - f(\mu)| \le \mathcal{M}\_{\mathfrak{a},f} |\rho(z) - \rho(\mu)|^\alpha, \mu \in \mathcal{E} \quad \text{and} \quad z \ge 0,$$

where M*<sup>α</sup>*, *f* is a constant depending on *α* and *f* .

**Theorem 7.** *Let ρ be a function satisfying the conditions (ρ*1*), (ρ*2.*) Then for any f* ∈ *Lip*M(*ρ*(*u*); *<sup>α</sup>*)*,* 0 < *α* ≤ 1 *with* 0 < *q* < 1 *and for every u* ∈ (0, <sup>∞</sup>), *m* ∈ N*, we have*

$$\left| \left| \mathcal{L}\_{m,q}^{\rho}(f;u) - f(u) \right| \leq \mathcal{M} \left( \frac{(1+q)}{[m]\_{q}} \rho(u) \right)^{\frac{\theta}{2}},\tag{25}$$

**Proof.** Assume that *α* = 1. Then, for *f* ∈ *Lip*M(*<sup>α</sup>*; 1) and *u* ∈ (0, <sup>∞</sup>), we have

$$\begin{aligned} |\mathcal{L}\_{m,q}^{\rho}(f;u) - f(u)| &\leq \mathcal{L}\_{m,q}^{\rho}(|f(z) - f(u)|; u) \\ &\leq \mathcal{M} \mathcal{L}\_{m,q}^{\rho}(|\rho(z) - f(u)|; u). \end{aligned}$$

By applying the Cauchy–Schwartz inequality, we find

$$\begin{aligned} \left| \mathcal{L}\_{m,q}^{\rho}(f;u) - f(u) \right| &\leq \mathcal{M} \left[ \mathcal{L}\_{m,q}^{\rho}((\rho(t) - \rho(u))^2; u) \right]^{\frac{1}{2}} \\ &\leq \mathcal{M} \sqrt{\frac{(1+q)\rho(u)}{[m]\_{q}}}. \end{aligned}$$

Let us assume that *α* ∈ (0, <sup>1</sup>). Then, for *f* ∈ *Lip*M(*<sup>α</sup>*; 1) and *u* ∈ (0, <sup>∞</sup>), we have

$$\begin{aligned} |\mathcal{L}\_{m,q}^{\rho}(f;u) - f(u)| &\leq \mathcal{L}\_{m,q}^{\rho}(|f(z) - f(u)|; u) \\ &\leq \quad \mathcal{M} \mathcal{L}\_{m,q}^{\rho}(|\rho(z) - f(u)|^{a}; u). \end{aligned}$$

From Hölder's inequality with *p* = 1*α* and *q* = 1 1−*<sup>α</sup>* , for *f* ∈ *Lip*M(*ρ*(*u*); *<sup>α</sup>*), we have

$$\left| \left| \mathcal{L}\_{m,\eta}^{\rho}(f;u) - f(u) \right| \right| \leq \left[ \mathcal{M} \left[ \mathcal{L}\_{m,\eta}^{\rho}(\left| (\rho(t) - \rho(u)| ; u) \right| \right]^{\alpha} \right]$$

Finally by the Cauchy–Schwartz inequality, we ge<sup>t</sup>

$$\left|\mathcal{L}\_{m,q}^{\rho}(f;u) - f(u)\right| \leq \mathcal{M}\left(\frac{(1+q)}{[m]\_q}\rho(u)\right)^{\frac{\alpha}{2}}.$$

A relationship between local smoothness of functions and the local approximation was given by Agratini in [28]. Here we will prove the similar result for operators <sup>L</sup>*ρ<sup>m</sup>*,*<sup>q</sup>* (*m* ∈ N) for functions from *Lip*M(*ρ*(*u*)) on a bounded subset.

**Theorem 8.** *Let* E *be a bounded subset of* [0, ∞) *and ρ be a function satisfying the conditions (ρ*1*), (ρ*2*). Then for any f* ∈ *Lip*M(*ρ*(*u*); *<sup>α</sup>*)*,* 0 < *α* ≤ 1 *on* E *α* ∈ (0, 1]*, we have*

$$|\mathcal{L}\_{m,q}^{\rho}(f;u) - f(u)| \leq \mathcal{M}\_{a,f} \left\{ \left[\frac{(1+q)\rho(u)}{[m]\_{q}}\right]^{\frac{q}{2}} + 2[\rho'(u)]^{a}d^{\kappa}(u,\mathcal{E}) \right\}, u \in [0,\infty), m \in \mathbb{N}\_{+} $$

*where d*(*<sup>u</sup>*, E) = *inf* {*u* − *y* : *y* ∈ E} *and* M*<sup>α</sup>*, *f is a constant depending on α and f .*

**Proof.** Let E be the closure of E in [0, <sup>∞</sup>). Then, there exists a point *u*0 ∈ E such that *d*(*<sup>u</sup>*, E) = |*u* − *<sup>u</sup>*0|. Using the monotonicity of <sup>L</sup>*ρ<sup>m</sup>*,*<sup>q</sup>* and the hypothesis of *f* , we obtain

$$\begin{split} |\mathcal{L}\_{m,q}^{\rho}(f;u) - f(u)| &\leq \mathcal{L}\_{m,q}^{\rho} \left( |f(z) - f(u\_0)|; u \right) + \mathcal{L}\_{m,q}^{\rho} \left( |f(u) - f(u\_0)|; u \right) \\ &\leq \mathcal{M}\_{a,f} \left\{ \mathcal{L}\_{m,q}^{\rho} \left( |\rho(z) - \rho(u\_0)|^a; u \right) + |\rho(u) - \rho(u\_0)|^a \right\} \\ &\leq \mathcal{M}\_{a,f} \left\{ \mathcal{L}\_{m,q}^{\rho} \left( |\rho(z) - \rho(u)|^a; u \right) + 2|\rho(u) - \rho(u\_0)|^a \right\}. \end{split}$$

By choosing *p* = 2*α* and *q* = 2 2−*α* , as well as the fact |*ρ*(*u*) − *ρ*(*<sup>u</sup>*0)| = *ρ*(*u*)|*ρ*(*u*) − *ρ*(*<sup>u</sup>*0)| in the last inequality. Then by using Hölder's inequality we easily conclude

$$\mathbb{E}\left|\mathcal{L}\_{m,\boldsymbol{\eta}}^{\boldsymbol{\rho}}(f;\boldsymbol{u})-f(\boldsymbol{u})\right|\leq \mathcal{M}\_{\boldsymbol{u},\boldsymbol{f}}\left\{\left[\mathcal{L}\_{m,\boldsymbol{\eta}}^{\boldsymbol{\rho}}((\boldsymbol{\rho}(\boldsymbol{z})-\boldsymbol{\rho}(\boldsymbol{u}))^{2};\boldsymbol{u})\right]^{\frac{1}{2}}+2\left[\boldsymbol{\rho}'(\boldsymbol{u})|\boldsymbol{\rho}(\boldsymbol{u})-\boldsymbol{\rho}(\boldsymbol{u}\_{0})|\right]^{\boldsymbol{\rho}}\right\}.$$

Hence, by Corollary 1 we ge<sup>t</sup> the proof.

Now, for *f* ∈ C*B*[0, <sup>∞</sup>), we recall local approximation in terms of *α* order generalized Lipschitztype maximal function given by Lenze [29] as

$$\widetilde{\omega}\_a^{\rho}(f;u) = \sup\_{z \neq u, z \in (0,\infty)} \frac{|f(z) - f(u)|}{|z - u|^a}, \ u \in [0,\infty) \text{ and } a \in (0,1]. \tag{26}$$

Then we ge<sup>t</sup> the next result

**Theorem 9.** *Let f* ∈ C*B*[0, ∞) *and α* ∈ (0, 1] *with* 0 < *q* < 1. *Then, for all u* ∈ [0, <sup>∞</sup>)*, we have*

$$\left|\mathcal{L}\_{\mathfrak{m},q}^{\rho}(f;u) - f(u)\right| \le \widehat{\omega}\_{\mathfrak{n}}^{\rho}(f;u) \left(\frac{(1+q)}{[m]\_{q}}\rho(u)\right)^{\frac{\rho}{2}}.$$

**Proof.** We know that

$$|\mathcal{L}\_{m,q}^{\rho}(f;\mu) - f(\mu)| \le \quad \mathcal{L}\_{m,q}^{\rho}(|f(t) - f(\mu)|; \mu).$$

From Equation (26), we have

$$|\mathcal{L}\_{m,q}^{\rho}(f;\mu) - f(\mu)| \le \varepsilon \quad \tilde{\omega}\_a^{\rho}(f;\mu)\mathcal{L}\_{m,q}^{\rho}(|\rho(z) - \rho(\mu)|^a;\mu).$$

From Hölder's inequality with *p* = 2*α* and *q* = 2 2−*α* , we have

$$\begin{aligned} \left| \mathcal{L}\_{m,q}^{\rho}(f;u) - f(u) \right| &\leq \left. \widetilde{\omega}\_{\mathfrak{a}}^{\rho}(f;u) \right[ \mathcal{L}\_{m,q}^{\rho}((\rho(t) - \rho(u))^2; u) ] \right|^{\frac{a}{2}} \\ &\leq \left. \widetilde{\omega}\_{\mathfrak{a}}^{\rho}(f;u) \left( \frac{(1+q)}{[m]\_{q}} \rho(u) \right)^{\frac{a}{2}} . \end{aligned}$$

which proves the desired result.
