1 Introduction

The significant influence of the Bernstein polynomials on modern mathematics—not only theoretical, but also applied and computational—brought about the emergence of its numerous versions and modifications. See, for example, [2, 3, 12]. While the Bernstein polynomials serve to approximate the continuous functions on [0, 1], the Kantorovich polynomials constructed with respect to the Bernstein basis are applicable for the approximation of integrable functions. Kantorovich’s breakthrough idea was further developed by Durrmeyer [7] and Derriennic [6]. The latter proved that the Bernstein–Durrmeyer polynomials approximate functions in \(L_1[0, 1]\), and also generate self-adjoint operators in \(L_2[0, 1]\).

With the increasing role of the q-Calculus (see, e.g. [1, 4, 5, 14]), the q-analogues of various Bernstein-type operators have come to the fore. The reader is referred to [3, 8, 15]. New versions of these operators, targeting a wide spectrum of various problems, are continuously coming out.

In 2008, Gupta [9] introduced a simple q-analogue of the Bernstein–Durrmeyer operators, denoted by \(D_{n,q}\), and studied its approximation properties. One of the properties that he proved was that \(\{D_{n,q}\}\) converges to the limit operator \(D_{\infty ,q}\) in the strong operator topology on C[0, 1]. More results on the q-Durrmeyer operator have been obtained in [10, 13].

In the present work, further investigation is carried out concerning the limit q-Bernstein–Durrmeyer operator. Distinct from the preceding studies on the subject, this paper is focused on the analytic properties that the image of \(f \in C[0,1]\) possesses under the operator \(D_{\infty ,q}\). Here, it is proved that, for each \( f \in C[0,1]\), the function \(D_{\infty ,q}f\) admits an analytic continuation from [0, 1] to the whole complex plane \(\mathbb {C}\). The growth estimates of the entire function \(D_{\infty ,q}f\) are provided, along with the interconnection between the growth of \(D_{\infty ,q}f\) and the behaviour of f. The sharpness of the obtained results is demonstrated.

To present the results, let us recall the necessary notation and definitions. The q-Pochhammer symbol denotes, for each \(a \in \mathbb {C}\),

$$\begin{aligned} (a;q)_0:=1, \qquad (a;q)_n=\prod _{j=0}^{n-1}(1-aq^j), \qquad (a;q)_\infty =\prod _{j=0}^{\infty }(1-aq^j). \end{aligned}$$

The Euler Identities

$$\begin{aligned} (z;q)_\infty =\sum _{k=0}^{\infty }\frac{ (-1)^k q^{k(k-1)/2}}{ (q;q)_k}z^k, \quad |q |< 1, \end{aligned}$$
(1.1)

and

$$\begin{aligned} \frac{1}{(z;q)_\infty }=\sum _{k=0}^{\infty } \frac{z^k}{(q;q)_k}, \quad |q|<1, \quad |z|<1, \end{aligned}$$
(1.2)

will be used throughout. See [1, Ch. 10, Cor. 10.2.2].

The q-integral over an interval [0, a], first introduced by Thomae [16] and later by Jackson [11], is defined as

$$\begin{aligned} \int _{0}^{a} f(t) \,d_qt:= (1-q)a \sum _{j=0}^{\infty } q^j f(aq^j). \end{aligned}$$
(1.3)

Definition 1.1

[9] Let \(q \in (0,1)\), \(f \in C[0,1]\). The limit q-Durrmeyer operator is defined by

$$\begin{aligned} (D_{\infty ,q}f)(x):=D_{\infty ,q}(f;x)= {\left\{ \begin{array}{ll} \sum _{k=0}^{\infty } A_{\infty k}(f) p_{\infty k}(q;x), &{} x \in [0,1), \\ f(1), &{} x= 1. \end{array}\right. } \end{aligned}$$

where

$$\begin{aligned} A_{\infty k}(f):= \frac{q^{-k}}{1-q} \int _{0}^{1} f(t) p_{\infty k}(q;qt) \,d_qt, \quad k=0,1,\ldots , \end{aligned}$$
(1.4)

and

$$\begin{aligned} p_{\infty k}(q;x)=\frac{(x;q)_\infty \, x^k}{(q;q)_k}, \quad k=0,\ldots . \end{aligned}$$
(1.5)

As coefficients (1.4) form a bounded sequence whenever \(f\in C[0, 1]\), the function \(D_{\infty , q}f\) admits an analytic continuation from [0, 1] to the open disc \(\{z:|z| < 1\}\). Taking into account (1.3), \(A_{\infty k}(f)\) can also be expressed as

$$\begin{aligned} A_{\infty k}(f)= \frac{(q;q)_\infty }{(q;q)_k} \sum _{j=0}^{\infty } \frac{f(q^j) q^{(k+1)j}}{(q;q)_j}. \end{aligned}$$
(1.6)

Throughout the paper, the letter C—with or without subscripts—denotes a positive constant whose specific value is of no importance. Subscripts, when used, indicate the dependence of C on certain parameters. It should be pointed out that the same letter may stand for different values. Moreover, if f is analytic in the closed disc \(\Delta _r:=\{z: |z| \leqslant r\}\), the notation

$$\begin{aligned} M(r;f):=\max _{z\in \Delta _r} |f(z)| \end{aligned}$$

will be employed.

The article is organized as follows: In Sect. 2, the main results are stated, while Sect. 3 contains the auxiliary technical lemmas. Finally, the proofs of the main results appear in Sect. 4.

2 Statement of Results

Theorem 2.1

For each \(f \in C[0,1]\), the function \((D_{\infty ,q}f)(x)\) admits an analytic continuation from [0, 1] as an entire function given by

$$\begin{aligned} (D_{\infty ,q}f)(z)=\sum _{j=0}^{\infty } \frac{f(q^j)q^j}{(q;q)_j} \sum _{n=0}^{\infty } \frac{(-1)^n q^{n(n+1)/2}}{(q;q)_n} (z;q)_{n+j}. \end{aligned}$$
(2.1)

The proof of Theorem 2.1 presented in Sect. 4 yields, apart from (2.1), the following corollary:

Corollary 2.2

The growth of \(D_{\infty ,q}f\), for each \(f \in C[0,1]\), enjoys the following estimate:

$$\begin{aligned} M(r;D_{\infty ,q}f)=O((-r;q)_{\infty }), \quad r \rightarrow \infty . \end{aligned}$$
(2.2)

It is worth pointing out that coefficients (1.6) can be viewed as the values of the function \(g(z):=(qz;q)_\infty \, \rho (z)\) at points \(z=q^k\), \(k=0,1, \ldots \), where

$$\begin{aligned} \rho (z)=\sum _{j=0}^{\infty } \frac{f(q^j) q^j}{(q;q)_j}z^j. \end{aligned}$$
(2.3)

Since \((qz;q)_\infty \) is entire and the series converges in the disc \(\{z: |z|<1/q \}\) for any \(f \in C[0,1]\), it follows that g is analytic in that disc. Clearly, the radius of convergence for \(\rho \) can be greater than 1/q. The representation below of \(D_{\infty ,q}\) with the help of divided differences of g is important.

Theorem 2.3

Given \(f \in C[0,1]\), let \(g(z)=(qz;q)_\infty \, \rho (z)\), where \(\rho \) is defined by (2.3). Then,

$$\begin{aligned} (D_{\infty ,q}f)(z)=\sum _{k=0}^{\infty } (-1)^k q^{k(k-1)/2} \, g[1;q; \ldots ; q^k]z^k, \quad z \in \mathbb {C}. \end{aligned}$$

Here, \(g[x_0;\ldots ; x_k]\) stands for the divided difference of g at the distinct nodes \(x_0,\ldots , x_k\).

This representation allows us to not only refine the estimate of Corollary 2.2, but also establish a connection between the behaviour of f and the growth of its image under \(D_{\infty ,q}\).

Theorem 2.4

Let \(R>1\) be such that \(\rho \) is analytic in \(\Delta _R\). Then,

$$\begin{aligned} M(r;D_{\infty ,q}f)=o \left( \frac{(-r;q)_{\infty }}{r^{\lambda }} \right) , \quad r \rightarrow \infty , \end{aligned}$$

for every \(\lambda <(\ln R)/\ln (1/q)\).

As a consequence of Theorem 2.4, the crude estimate (2.2) can be improved. Since \(\rho \) is analytic in \(\{z:|z|<1/q\}\), it is possible to assume \(\lambda =0\) in Theorem 2.4 and obtain the following result.

Corollary 2.5

For any \(f \in C[0,1]\),

$$\begin{aligned} M(r;D_{\infty ,q}f)=o((-r;q)_{\infty }), \quad r \rightarrow \infty . \end{aligned}$$

Corollary 2.6

If \(f(q^j)=O(q^{\alpha j})\), \(j \rightarrow \infty \), for some \(\alpha >0\), then

$$\begin{aligned} M(r;D_{\infty ,q}f)=o(r^{-\lambda }(-r;q)_{\infty }), \quad r \rightarrow \infty , \end{aligned}$$
(2.4)

for all \(\lambda < 1+ \alpha \).

Indeed, in this case, \(\rho \) is analytic in \(\{z:|z|<q^{-1-\alpha }\}\).

Corollary 2.7

If, for every \(\alpha >0\), the estimate \(f(q^j)=o(q^{\alpha j})\), \(j \rightarrow \infty \) holds, then, for every \(\lambda \geqslant 0\), (2.4) is true.

The estimate in Theorem 2.4 is sharp as demonstrated by the assertion below.

Theorem 2.8

For every \(\lambda > 1\), there exists \(f\in C[0, 1]\) such that

$$\begin{aligned} M(r;D_{\infty ,q}f)\geqslant C r^{-\lambda }(-r;q)_{\infty }, \quad r \rightarrow \infty . \end{aligned}$$

Theorem 2.4 and Corollaries 2.5–2.7 establish the connection between the radius of convergence for the series (2.3) and the rate of growth for \(D_{\infty ,q}f\). In a general sense, the greater the radius is, the slower the growth becomes. Approaching the problem from a different angle, the dependence of the growth on the differentiability of f at the origin is addressed in the next assertion. The statement makes it possible to obtain better estimates for \(M(r;D_{\infty ,q}f)\) than those guaranteed by Theorem 2.4 when f is differentiable at 0 even though the series (2.3) converges only in the smallest admissible disc.

Theorem 2.9

Let f be m times differentiable at 0 from the right. Then,

$$\begin{aligned} M(r;D_{\infty ,q}f)=o(r^{-\lambda }(-r;q)_{\infty }), \quad r \rightarrow \infty , \end{aligned}$$
(2.5)

for all \(\lambda <1+m\).

Corollary 2.10

If f is infinitely differentiable at 0 from the right, then (2.5) holds for all \(\lambda > 0\). In particular, (2.5) is valid whenever f is analytic in a neighbourhood of 0.

3 Auxiliary Results

In what comes next, the function \( \tau \) given by

$$\begin{aligned} \tau (z)= (z;q)_{\infty } \sum _{k=0}^{\infty } \frac{z^k}{(q;q)_k^{2}}, \quad |z|<1, \end{aligned}$$

plays a key role.

Lemma 3.1

The function \(\tau \) admits an analytic continuation from the open unit disc as an entire function.

Proof

Consider

$$\begin{aligned} \sum _{k=0}^{\infty } \frac{z^k}{(q;q)_k^2}=\sum _{k=0}^{\infty } \frac{z^k}{(q;q)_k} \frac{(q^{k+1};q)_\infty }{(q;q)_\infty }. \end{aligned}$$

By (1.1), with \(z=q^{k+1}\), one has

$$\begin{aligned} (q^{k+1}; q)_\infty =\sum _{n=0}^{\infty } \frac{(-1)^n q^{n(n-1)/2}}{(q;q)_n} (q^{k+1})^n, \end{aligned}$$

whence

$$\begin{aligned} \sum _{k=0}^{\infty } \frac{z^k}{(q;q)_k^2}&= \frac{1}{(q;q)_\infty } \sum _{k=0}^{\infty } \frac{z^k}{(q;q)_k} \sum _{n=0}^{\infty } \frac{(-1)^n q^{n(n-1)/2} q^{(k+1)n}}{(q;q)_n} \\&=\frac{1}{(q;q)_\infty } \sum _{n=0}^{\infty } \frac{(-1)^n q^{n(n+1)/2}}{(q;q)_n} \sum _{k=0}^{\infty } \frac{(q^n z)^k}{(q;q)_k}. \end{aligned}$$

By virtue of (1.2), it follows that

$$\begin{aligned} \sum _{k=0}^{\infty } \frac{z^k}{(q;q)_k^2}=\frac{1}{(q;q)_\infty } \sum _{n=0}^{\infty } \frac{(-1)^n q^{n(n+1)/2}}{(q;q)_n} \frac{1}{(q^n z;q)_\infty }, \quad |z|<1. \end{aligned}$$

Consequently, one obtains

$$\begin{aligned} \tau (z)&= \frac{1}{(q;q)_\infty }\sum _{n=0}^{\infty } \frac{(-1)^n q^{n(n+1)/2}}{(q;q)_n} \frac{(z;q)_\infty }{(q^n z;q)_\infty } \nonumber \\&=\frac{1}{(q;q)_\infty }\sum _{n=0}^{\infty } \frac{(-1)^n q^{n(n+1)/2}}{(q;q)_n} (z;q)_n, \quad |z|<1. \end{aligned}$$
(3.1)

Now, if \(z\in \Delta _R\), then

$$\begin{aligned} \sum _{n=0}^{\infty } \left| \frac{(-1)^n q^{n(n+1)/2}}{(q;q)_n}(z;q)_n\right| \leqslant \sum _{n=0}^{\infty } \frac{q^{n(n+1)/2}}{(q;q)_n} (1+R)^n < \infty . \end{aligned}$$

Hence, \(\tau (z)\) is analytic in \(\Delta _R\) for each \(R>0\) and (3.1) is valid for all \(z \in \mathbb {C}\). Therefore, \(\tau (z)\) is an entire function. \(\square \)

Lemma 3.2

Let \(R>1\) be such that \(\rho \) given by (2.3) is analytic in \(\{z: |z|\leqslant R \}\). Then,

$$\begin{aligned} \left| g[1;q; \ldots ; q^k]\right| \leqslant C q^{\lambda k} \end{aligned}$$

for every \( \lambda <(\ln R)/\ln (1/q)\).

Proof

It is known that (see for example, [12, Sect. 2.7., p.44, Eq. (4)])

$$\begin{aligned} g[a_0; \ldots ; a_k] =\frac{1}{2 \pi i} \oint _L \frac{g(\zeta )d\zeta }{(\zeta -a_0) \ldots (\zeta -a_k)}, \end{aligned}$$

where L is a positively-oriented, simple and closed curve encircling the distinct points \(a_0, \ldots , a_k\) and g is analytic everywhere on and inside L.

Therefore,

$$\begin{aligned} g[1;q; \ldots ; q^k] = \frac{1}{2 \pi i} \oint _{|\zeta |=R} \frac{g(\zeta )d\zeta }{(\zeta -1) (\zeta -q) \ldots (\zeta -q^k)}. \end{aligned}$$

Now, assume that \(0< \lambda _0 < (\ln R)/\ln (1/q)\), that is, \(1<q^{-\lambda _0}<R\). Two cases will be considered:

Case 1. If \(q^{-\lambda _0} \leqslant R-1\), then \(g[1;q;\ldots ;q^k]\) can be estimated as

$$\begin{aligned} \left| g[1;q; \ldots ; q^k]\right|&\leqslant \frac{1}{2 \pi } \cdot \frac{M(R;g)}{(R-1) (R-q) \ldots (R-q^k)} \cdot 2 \pi R\\&\leqslant \frac{M(R;g)R}{(R-1)^{k+1}} \leqslant 2M(R;g) q^{\lambda _0 k}. \end{aligned}$$

Case 2. If \(R-1<q^{-\lambda _0} \leqslant R\), then opt for \(m_0\in \mathbb {N}_0\) such that \(R-q^m>q^{-\lambda _0}\) whenever \(m \geqslant m_0\). Then, for \(k \geqslant m_0\), one has

$$\begin{aligned} \left| g[1;q; \ldots ; q^k]\right|&\leqslant \frac{M(R;g)R}{(R-1)\cdots (R-q^{m_0-1})(R-q^{m_0})\cdots (R-q^k)} \\&\leqslant \frac{M(R;g)R}{(R-1)\cdots (R-q^{m_0-1})} \cdot \frac{1}{(R-q^{m_0})^{k-m_0+1}} \\&\leqslant C_{R,q,g} \frac{1}{(R-q^{m_0})^{k}} < C q^{\lambda _0 k}, \quad k\geqslant m_0. \end{aligned}$$

As a result, \(\left| g[1;q; \ldots ; q^k]\right| \leqslant C q^{\lambda _0 k}\) for all k, possibly with a different C.

Combining the outcomes of the two cases yields \(\left| g[1;q; \ldots ; q^k]\right| \leqslant C q^{\lambda _0 k}\), and, in turn, \(\left| g[1;q; \ldots ; q^k]\right| \leqslant C q^{\lambda k}\) for all \(\lambda \leqslant \lambda _0\). Since \(\lambda _0\) has been chosen arbitrarily, it follows that the latter inequality holds for all \(\lambda < (\ln R)/\ln (1/q)\) as stated. \(\square \)

4 Proofs of Main Results

Proof of Theorem 2.1

Using (1.6), one obtains

$$\begin{aligned} (D_{\infty ,q}f)(z)&=\sum _{k=0}^{\infty } \left( \frac{(q,q)_\infty }{(q,q)_k} \sum _{j=0}^{\infty } \frac{f(q^j) q^{(k+1)j}}{(q,q)_j}\right) p_{\infty k}(q;z), \quad |z|<1. \end{aligned}$$

Recalling (1.5) leads to

$$\begin{aligned} (D_{\infty ,q}f)(z)&=\sum _{k=0}^{\infty } \frac{(q;q)_\infty }{(q;q)_k} \sum _{j=0}^{\infty } \frac{f(q^j)q^{(k+1)j}}{(q;q)_j} \frac{(z;q)_\infty z^k}{(q;q)_k} \\&=(q;q)_\infty (z;q)_\infty \sum _{j=0}^{\infty } \frac{f(q^j)q^{j}}{(q;q)_j} \sum _{k=0}^{\infty } \frac{(q^jz)^k}{(q;q)_k^2} \\&=(q;q)_\infty (z;q)_\infty \sum _{j=0}^{\infty } \frac{f(q^j)q^{j}}{(q;q)_j} \frac{\tau (q^jz)}{(q^jz;q)_\infty }, \\&=(q;q)_\infty \sum _{j=0}^{\infty } \frac{f(q^j)q^{j}}{(q;q)_j} (z;q)_j \tau (q^jz), \quad |z|<1. \end{aligned}$$

By (3.1),

$$\begin{aligned} \tau (q^jz)&= \frac{1}{(q;q)_\infty }\sum _{n=0}^{\infty } \frac{(-1)^n q^{n(n+1)/2}}{(q;q)_n} (q^j z;q)_n, \end{aligned}$$

and, hence,

$$\begin{aligned} (D_{\infty ,q}f)(z)&=\sum _{j=0}^{\infty } \frac{f(q^j)q^{j}}{(q;q)_j} (z;q)_j \sum _{n=0}^{\infty } \frac{(-1)^n q^{n(n+1)/2}}{(q;q)_n} (q^j z;q)_n \nonumber \\&=\sum _{j=0}^{\infty } \frac{f(q^j)q^{j}}{(q;q)_j} \sum _{n=0}^{\infty } \frac{(-1)^n q^{n(n+1)/2}}{(q;q)_n} (z;q)_{j+n}, \quad |z|<1. \end{aligned}$$
(4.1)

Since, for \(R>0\) and \(z\in \Delta _R\), one has \(|(z;q)_{j+n}| \leqslant (-R;q)_\infty \) for all \(j,n \in \mathbb {N}_0\), the series in (4.1) converges uniformly in any closed disc \(\Delta _R\). Therefore,

$$\begin{aligned} \left| \sum _{n=0}^{\infty } \frac{(-1)^n q^{n(n+1)/2}}{(q;q)_n} (z;q)_{j+n}\right| \leqslant (-R;q)_\infty \sum _{n=0}^{\infty } \frac{ q^{n(n+1)/2}}{(q;q)_n} =(-R;q)_\infty (-q;q)_\infty \end{aligned}$$

which implies that, when \(z\in \Delta _R\),

$$\begin{aligned} \left| (D_{\infty ,q}f)(z) \right|&\leqslant (-R;q)_\infty (-q;q)_\infty \sum _{j=0}^{\infty } \frac{ \left| f(q^j) \right| q^{j}}{(q;q)_j} \\ {}&\leqslant \Vert f\Vert _{C[0,1]} (-R;q)_\infty \frac{(-q;q)_\infty }{(q;q)_\infty }\\ {}&=:C_{f,q} (-R;q)_\infty . \end{aligned}$$

Consequently, \((D_{\infty ,q}f)(z)\) is analytic in any disc of radius \(R>0\). Thus, \((D_{\infty ,q}f)(z)\) is entire. This completes the proof. \(\square \)

Proof of Theorem 2.3

Starting from (1.6), one arrives at

$$\begin{aligned} A_{\infty k}(f)&=(q^{k+1};q)_\infty \sum _{j=0}^{\infty } \frac{f(q^j) q^{(k+1)j}}{(q;q)_j}=[(qz;q)_\infty \, \rho (z)]\Big |_{z=q^k}=g(q^k). \end{aligned}$$

Therefore,

$$\begin{aligned} (D_{\infty ,q}f)(z)=(z;q)_\infty \sum _{k=0}^{\infty } g(q^k) \frac{z^k}{(q;q)_k}, \quad |z|<1/q. \end{aligned}$$

Application of Euler’s identity (1.1) leads to

$$\begin{aligned} (D_{\infty ,q}f)(z)&=\sum _{k=0}^{\infty } \sum _{j=0}^{\infty } \frac{(-1)^k q^{k(k-1)/2} g(q^j) z^{k+j} }{(q;q)_k(q;q)_j} \\&=\sum _{k=0}^{\infty } \sum _{j=0}^{k} \frac{(-1)^{k-j} q^{(k-j)(k-j-1)/2} g(q^j) z^{k} }{(q;q)_{k-j}(q;q)_j} \\&=\sum _{k=0}^{\infty } (-1)^k q^{k(k-1)/2} \left( \sum _{j=0}^{k} \frac{(-1)^{-j} g(q^j)}{q^{j(j-1)/2} (q;q)_j q^{j(k-j)} (q;q)_{k-j}}\right) z^{k},\\&\qquad |z|<\frac{1}{q}. \end{aligned}$$

Employing [12, p. 44, Eq. (3)] with \(x_j=q^j\), one arrives at

$$\begin{aligned} g[1;q; \ldots ; q^k]=\sum _{j=0}^{k} \frac{(-1)^{-j} g(q^j)}{q^{j(j-1)/2} (q;q)_j q^{j(k-j)} (q;q)_{k-j}}. \end{aligned}$$

Therefore, formula

$$\begin{aligned} (D_{\infty ,q}f)(z)=\sum _{k=0}^{\infty } (-1)^k q^{k(k-1)/2} \, g[1;q; \ldots ; q^k]z^k \end{aligned}$$

holds for \(|z|<1/q\) and also in every disc where \(D_{\infty ,q}f\) possess an analytic continuation. Applying Theorem 2.1, one completes the proof. \(\square \)

Proof of Theorem 2.4

By Theorem 2.3,

$$\begin{aligned} (D_{\infty ,q}f)(z)=\sum _{k=0}^{\infty } (-1)^k q^{k(k-1)/2} \, g[1;q; \ldots ; q^k]z^k, \quad z \in \mathbb {C}. \end{aligned}$$

Select \(\lambda <(\ln R)/\ln (1/q)\) and take \(\mu \) such that \(\lambda<\mu <(\ln R)/\ln (1/q)\). Now, the growth of \(D_{\infty ,q}f\) may be estimated with the help of Lemma 3.2, which implies \(|g[1;q;\ldots ;q^k]| \leqslant C q^{\mu k}\). Therefore,

$$\begin{aligned} \left| (D_{\infty ,q}f)(z) \right|&\leqslant C \sum _{k=0}^{\infty } q^{k(k-1)/2} \left( q^{\mu } |z|\right) ^k \leqslant C \sum _{k=0}^{\infty } \frac{q^{k(k-1)/2}}{(q;q)_k} \left( q^{\mu } |z|\right) ^k, \end{aligned}$$

and, hence,

$$\begin{aligned} M(r;D_{\infty ,q}f)\leqslant C (-q^\mu r;q)_\infty . \end{aligned}$$

Recall [17, Eq.  (2.6)] that, for r large enough,

$$\begin{aligned} C_1 \exp \left\{ \frac{\ln ^2r}{2 \ln \frac{1}{q}}+\frac{\ln r}{2}\right\} \leqslant (-r;q)_\infty \leqslant C_2 \exp \left\{ \frac{\ln ^2r}{2 \ln \frac{1}{q}}+\frac{\ln r}{2}\right\} . \end{aligned}$$

Consequently,

$$\begin{aligned} C_1 \frac{(-r;q)_\infty }{r^\mu } \leqslant (-q^\mu r;q)_\infty \leqslant C_2 \frac{(-r;q)_\infty }{r^\mu } \end{aligned}$$
(4.2)

for r large enough.

As a result,

$$\begin{aligned} M(r;D_{\infty ,q}f)&=O\left( \frac{(-r;q)_{\infty }}{r^{\mu }} \right) , \quad r \rightarrow \infty , \\&=o \left( \frac{(-r;q)_{\infty }}{r^{\lambda }} \right) , \quad r \rightarrow \infty , \end{aligned}$$

as stated. \(\square \)

Proof of Theorem 2.8

For \(\lambda >1\), set \(\alpha =q^{\lambda -1} \in (0,1)\) and

$$\begin{aligned} s_j=\sum _{k=0}^j \frac{\alpha ^k}{(q;q)_{j-k}}, \quad j\in \mathbb {N}_0. \end{aligned}$$

Obviously, the sequence \(\{s_j\}\) is bounded. In addition, it is increasing because, for \(j\in \mathbb {N}_0\),

$$\begin{aligned} s_{j+1}-s_j=\sum _{k=0}^j \alpha ^k \left( \frac{1}{(q;q)_{j+1-k}}-\frac{1}{(q;q)_{j-k}}\right) + \alpha ^{j+1} > 0. \end{aligned}$$

Consequently, \(\{s_j\}\) converges. Now, let \(f\in C[0, 1]\) be such that \(f(q^j)=(q;q)_js_j\). This is possible due to the fact that \(\{(q;q)_js_j\}\) is convergent as a product of two convergent sequences. For this f, one has

$$\begin{aligned} \rho (z)=\sum _{j=0}^\infty s_j (qz)^j. \end{aligned}$$

Evidently, \(\rho \) is analytic in \(\{z:|z|<1/q\}\) and

$$\begin{aligned} \rho (z)&=\sum _{j=0}^\infty \left( \sum _{k=0}^j \frac{\alpha ^k}{(q;q)_{j-k}}\right) (qz)^j\\&=\sum _{j=0}^\infty \frac{(qz)^j}{(q;q)_{j}} \sum _{k=0}^\infty (\alpha qz)^k\\&= \frac{1}{(qz;q)_\infty } \cdot \frac{1}{1-\alpha q z}, \quad |z|<\frac{1}{q}. \end{aligned}$$

Hence, \(g(z)=\rho (z) (qz;q)_\infty = 1/(1-\alpha q z)\), whence g is analytic in \(\{z:|z|<1/(\alpha q)\}\). Simple calculations reveal:

$$\begin{aligned} g^{(k)}(z)= \frac{(\alpha q)^k k!}{(1-\alpha q z)^{k+1}}, \quad k \in \mathbb {N}_0. \end{aligned}$$

By the Intermediate Value Theorem,

$$\begin{aligned} g[1;q;\ldots ;q^k] = \frac{g^{(k)}(\xi )}{k!},\quad \xi \in (q^k, 1). \end{aligned}$$

Since all \(g^{(k)}(x)\) are increasing on [0, 1], there holds

$$\begin{aligned} g[1;q;\ldots ;q^k] \geqslant \frac{g^{(k)}(q^k)}{k!}=\frac{(\alpha q)^k}{(1-\alpha q^{k+1})^{k+1}} \geqslant (\alpha q)^k,\quad k \in \mathbb {N}_0. \end{aligned}$$

As a result,

$$\begin{aligned} M(r; D_{\infty , q}f)&= \sum _{k=0}^\infty q^{k(k-1)/2}g[1;q;\ldots ;q^k] r^k \\&\geqslant (q;q)_\infty \sum _{k=0}^\infty \frac{q^{k(k-1)/2}}{(q;q)_k} (\alpha q r)^k\\&=(q;q)_\infty (-\alpha q r;q)_\infty . \end{aligned}$$

Writing \(\alpha =q^{\lambda -1}\) and using (4.2), one obtains

$$\begin{aligned} M(r; D_{\infty , q}f) \geqslant C r^{-\lambda } (-r; q)_\infty , \quad r\rightarrow \infty , \end{aligned}$$

which completes the proof. \(\square \)

Proof of Theorem 2.9

By Taylor’s Theorem, one can write

$$\begin{aligned} f(x)=T_m(x)+S_m(x) \end{aligned}$$

where \(T_m(x)\) is a polynomial of degree at most m and \(S_m(x)=o(x^m)\) as \(x \rightarrow 0^+\). Since \(D_{\infty ,q}\) maps a polynomial to a polynomial of the same degree (see [9, Rem. 3]), there holds

$$\begin{aligned} (D_{\infty ,q}f)(z)=P_m(z)+(D_{\infty ,q}S_m)(z), \end{aligned}$$

where \(P_m(z)\) is a polynomial of degree at most m and, as such,

$$\begin{aligned} M(r;P_m)=o(r^{-\lambda }(-r;q)_{\infty }), \quad r \rightarrow \infty , \end{aligned}$$

for all \(\lambda >0\). As for \(M(r;D_{\infty ,q}S_m)\), it can be estimated by means of Corollary 2.6 with \(\alpha =m\). \(\square \)