\(^\ast \)Department of Mathematics and Computer Science, University of Oradea, Universităţii st. no. 1, 410087 Oradea, Romania, e-mail: galso@uoradea.ro

\(^\S \)School of Applied Sciences, Netaji Subhas Institute of Technology, Sector 3 Dwarka, New Delhi-110078, INDIA, e-mail: vijaygupta2001@hotmail.com

1 Introduction

If \(f:G\to \mathbb {C}\) is an analytic function in the open set \(G\subset \mathbb {C}\), with \(\overline{D}_1\subset G\) (where \(D_1=\{ z\in \mathbb {C} : |z|{\lt}1\} \)), then S. N. Bernstein proved that the complex Bernstein polynomials converges uniformly to \(f\) in \(\overline{D}_1\) (see e.g., Lorentz [ 13 ] , p. 88). Exact quantitative estimates and quantitative Voronovskaja-type results for these polynomials (see Gal [ 5 ] ), together with similar results for complex Bernstein-Stancu polynomials (see also the papers Gal [ 6 ] - [ 7 ] ), complex Kantorovich-Stancu polynomials (see also the paper Gal [ 8 ] ), complex Favard-Szász-Mirakjan operators, Butzer’s linear combinations of complex Bernstein polynomials, complex Baskakov operators and complex Balázs-Szabados operators were obtained by the first author in several recent papers collected by the recent book Gal [ 10 ] .

The approximation properties of certain complex Durrmeyer-type operators were studied in Gal [ 4 , 9 ] , Agarwal and Gupta [ 3 ] and Mahmudov [ 14 , 15 ] . Furthermore, the approximation properties of the complex Beta operators of fist kind was studied in Gal-Gupta [ 11 ] .

The aim of the present article is to obtain approximation results for the complex Stancu Beta operator of second kind, firstly introduced in the case of real variable in D. D. Stancu [ 17 ] . Then, Abel [ 1 ] , Abel-Gupta [ 2 ] and Gupta-Abel-Ivan [ 12 ] obtained various estimates of the rate of convergence in the real variable case.

The complex Stancu Beta operators of second kind will be defined for all \(n\in \mathbb {N}\) and \(z\in \mathbb {C}\) satisfying \(0{\lt}{\rm Re}(z)\), by

where \(B(\alpha , \beta )\) is the Euler’s Beta function, defined by

and \(f\) is supposed to be locally integrable and of polynomial growth on \((0, +\infty )\) as \(t\to \infty \). This last hypothesis on \(f\) assures the existence of \(K_{n}(f;z)\) for sufficiently large \(n\), that is there exists \(n_{0}\) depending on \(f\) such that \(K_{n}(f;z)\) exists for all \(n\ge n_{0}\) and \(z\in \mathbb {C}\) with \({\rm Re}(z){\gt}0\).

Note that because of the well-known formulas \(B(\alpha , \beta )=\tfrac {\Gamma (\alpha ) \cdot \Gamma (\beta )}{\Gamma (\alpha +\beta )}\) and \(\Gamma (\alpha +1)=\alpha \Gamma (\alpha )\), \({\rm Re}(\alpha ){\gt}0\), \({\rm Re}(\beta ) {\gt}0\), where \(\Gamma \) denotes the Euler’s Gamma function, for all \(z\in \mathbb {N}\) with \({\rm Re}(z){\gt}0\) and sufficiently large \(n\) we can easily deduce the form

The results in the present paper will show the overconvergence phenomenon for this complex Stancu Beta integral operator of second kind, that is the extensions of approximation properties with upper and exact quantitative estimates, from the real interval \((0, r]\) to semidisks of the right half-plane of the form

and to subsets of semidisks of the form

with \(r\ge 1\) and \(0{\lt}a{\lt}r\).

It is worth noting that due to the special form of the complex Stancu Beta operators of second kind, the methods of proof are different from those used in the cases of the other complex operators studied by the papers mentioned in References.

2 AUXILIARY RESULT

In the sequel, we shall need the following auxiliary results.

By the relationship (1.1) of the Stancu Beta operators of second kind, it is obvious that \(K_{n}(e_{0}, z)=1\) and \(K_{n}(e_{1})(z)=e_{1}(z)\) (see [ 1 ] , Proposition 2). Next

Since \(B(\alpha , \beta )\) is only defined for \({\rm Re}(\alpha ) {\gt}0\) and \({\rm Re}(\beta ){\gt}0\), it follows that the above recurrence is valid only for \(n-p{\gt}0\). This completes the proof of Lemma 2.1.

3 MAIN RESULTS

The first main result one refers to upper estimate.

In the definition of \(K_n(f, z)\) in (1.1), for \(z=x + {\rm i} y\) with \(x{\gt}0\), note that it follows \(t^{nz-1}={\rm e}^{(nz-1){\rm ln}(t)}={\rm e}^{(nx-1){\rm ln}(t)}\cdot {\rm e}^{{\rm i} n y {\rm ln}(t)}\) and \(|t^{nz-1}|=t^{nx-1}\), which implies

But it is well-known that because \(f(t)\) is of polynomial growth as \(t\to +\infty \), the last integral exists finite for sufficiently large \(n\).

Therefore, there exists \(n_{0}\) depending only on \(f\), such that \(K_{n}(f, z)\) is well-defined for sufficiently large \(n\) and for \(z\) with \(Re(z){\gt}0\).

It remains to prove that \(K_{n}(f, z)\) is in fact analytic for \({\rm Re}(z){\gt}0\) and \(n\) sufficiently large. For this purpose, from a standard result in the theory of improper integrals depending on a parameter, it suffices to prove that for any \(\delta {\gt}0\), the improper integral

is uniformly convergent for \({\rm Re}(z)\ge \delta {\gt}0\) and \(n\) sufficiently large.

But by simple calculation we obtain

and since \({\rm ln}(1+t)\le 1+t\) for all \(t\ge 0\), it easily follows that it remains to prove that the integral \(\int _{0}^{\infty }\tfrac {t^{nz-1}}{(1+t)^{nz+n+1}}\cdot {\rm ln}(t) f(t) d t\) is uniformly convergent for \({\rm Re}(z)\ge \delta {\gt} 0\) and \(n\) sufficiently large.

By \({\rm ln}(t){\lt}t\) for all \(t\ge 1\) and by

clearly that it remains to prove the uniform convergence, for all \({\rm Re}(z)\ge \delta {\gt}0\) and \(n\) sufficiently large, for the integral \(\int _{0}^{1}\tfrac {t^{nz-1}}{(1+t)^{nz+n+1}}\cdot {\rm ln}(t) f(t){\rm d}t\). But this follows immediately from the estimate

(see e.g. [ 16 ] , p. 19, Exercise 1.51), where \(|f(t)|\le C\) for all \(t\in [0, 1]\).

In what follows we deal with the approximation property. For this purpose, firstly let us define \(S_{n}(z)=\sum _{k=0}^{n}c_{k} z^{k}\) if \(|z|\le r\) and \(S_{n}(t)=f(t)\) if \(t\in (r, +\infty )\), where \(1\le r {\lt} \tfrac {1}{2 A}\). Evidently that for each \(n\in \mathbb {N}\), \(S_{n}\) is piecewise continuous on \([0, +\infty )\) (more exactly, has a discontinuity point of first kind at \(x=r\)), but locally integrable on \([0, +\infty )\) and of polynomial growth as \(t\to \infty \).

Clearly, \(f(z)-S_{n}(z)=\sum _{k=n+1}^{\infty }c_{k} z^{k}\) if \(|z|\le r\) and \(f(t)-S_{n}(t)=0\) if \(t\in (r, \infty )\). Also, it is immediate that \(K_{n}(S_{n})(z)\) is well-defined for all \(n\in \mathbb {N}\).

Therefore, for sufficiently large \(n\) and for \(z\in SD^{r}(0, r]\) we have

where \(e_{k}(z)=z^{k}\).

Firstly we will obtain an estimate for \(|S_{n}(z)-f(z)|\). Let \(1\le r {\lt} \tfrac {1}{2 A} {\lt} r_{1} {\lt} R\). By the hypothesis, we can make such of choice for \(r_{1}\).

Denoting \(M_{r_{1}}(f)=\max \{ |f(z)|: |z|\le r_{1}\} \) and \(\rho =\tfrac {r}{r_{1}}\), by \(0 {\lt}\rho =\tfrac {r}{r_{1}} {\lt} 2 A r {\lt}1\) and by the Cauchy’s estimate (see e.g. [ 18 ] , p. 184, Lemma 10.5) we get \(|c_{k}|=\tfrac {|f^{(k)}(0)|}{k !}\le \tfrac {1}{k !}\cdot \tfrac {M_{r_{1}(f)} k!}{r_{1}^{k}}=\tfrac {M_{r_{1}}(f)}{r_{1}^{k}}\), which implies

for all \(|z|\le r\) and \(n\in \mathbb {N}\).

By using now Lemma 2.1 and taking into account the inequalities

for all \(z\in SD^{r}(0, r]\) and \(n\ge p+1\) we get

for all \(p=0, 1, ..., n-1 \).

Therefore, denoting \(E_{p, n}(z)=|K_{n}(e_{p}, z)-e_{p}(z)|\), we have obtained

for all \(p=0, 1, ..., n-1 \).

Since \(E_{0, n}(z)=E_{1, n}(z)=0\), for \(p=1\) in the above inequality we get

In what follows we will use the obvious inequality \(p\le 2(p-1)+2\), valid for all \(p\ge 1\).

For \(p=2\) in the above recurrence inequality it follows

For \(p=3\) in the above recurrence inequality we get

By mathematical induction we easily obtain

for all \(n\ge p+1\) and \(z\in SD^{r}(0, r]\).

Therefore, we obtain

where the hypothesis on \(f\) obviously implies that \(\sum _{k=0}^{\infty } k\cdot (2 A r)^{k}{\lt}\infty \).

Now, let us estimate \(|K_{n}(f-S_{n}, z)|\). By the definition of \(S_{n}\) and by (1.2), we easily get

for all \(z\in SD^{r}[a, r]\), \(z=x+{\rm i}y\), and \(n\in \mathbb {N}\). Passing to the absolute value, it follows

Now, let us estimate the integral \(\int _0^r \tfrac {t^{n x-1}}{(1+t)^{n x+n+1}}{\rm d} t\). For sufficiently large \(n\) (such that \(na -1\ge 1\)) we have

which immediately implies the estimate for \(n\ge n_{0}\) (with \(n_{0}\) depending only on \(f\) and \(a\)) and \(z\in SD^{r}[a, r]\)

Collecting all the above estimates, for sufficiently large \(n\) and \(z\in S^{r}[a, r]\), we get

In (3.1) we need to choose \(n\ge 2/a\).

Now, denote

We can write

Note that because \(0{\lt}\rho {\lt}1\) and \(0{\lt}r/(r+1){\lt}1\), clearly that for sufficiently large \(n\) we have \(n \cdot \rho ^{n+1}\le \tfrac {c_{1}}{n}\) and \(n\big(\tfrac {r}{r+1}\big)^{na-1}\le \tfrac {c_{2}}{n}\), where \(c_{1}{\gt}0\) and \(c_{2}{\gt}0\) are independent of \(n\) and \(z\). On the other hand, by simple calculation we get

for all \(n\in \mathbb {N}\). We used here the inequality \(e{\lt}3\). Therefore, the sequence \((a_{n})_{n}\) is nonincreasing, which implies that it is bounded.

In conclusion, for sufficiently large \(n\) we have

which coupled with (3.1) immediately implies the order of approximation \({\mathcal O}(1/n)\) in the statement of Theorem 3.1.

The following Voronovskaja-type result with a quantitative estimate holds.

Keeping the notations in the proof of Theorem 3.1, we can write

We get

From the proof of Theorem 3.1, for all \(z\in SD^{r}[a, r]\) with \(a\in (0, r)\) and for sufficiently large \(n\), we have

where \(C_{1}{\gt}0\), \(C_{2}{\gt}0\) are independent of \(n\) and \(z\) but may depend on \(f\), \(r\) and \(a\) and \(0{\lt}\rho {\lt}1\).

In order to estimate \(A_{3}\), let \(0{\lt}a_{1}{\lt}a{\lt}r\), \(1\le r{\lt}r_{1}{\lt}\tfrac {1}{2 A}\) and denote by \(\Gamma =\Gamma _{a_{1}, r_{1}}=S_{1}\bigcup L_{1}\) the closed curve composed by the segment in \(\mathbb {C}\)

and by the arc

Clearly that \(\Gamma \) together with its interior is exactly \(SD^{r_{1}}[a_{1}, r_{1}]\) and that from \(r{\lt}r_{1}\) we have \(SD^{r}[a, r]\subset SD^{r_{1}}[a_{1}, r_{1}]\), the inclusion being strictly.

By the Cauchy’s integral formula for derivatives, we have for all
\(z\in SD^{r}[a, b]\) and \(n\in \mathbb {N}\) sufficiently large

which by the estimate of \(||f-S_{n}(\cdot )||_{SD^{r_{1}}[a_{1}, r_{1}]}\) in the proof of Theorem 3.1 and by the inequality \(|u-z|\geq d=\min \{ r_{1}-r, a-a_{1} \} \) valid for all \(z\in SD^{r}[a, r]\) and \(u\in \Gamma \), implies

with \(\rho =\tfrac {r}{r_{1}}\).

Note that here, by simple geometrical reasonings, for the length \(l(\Gamma )\) of \(\Gamma \), we get

where \(\arccos (\alpha )\) is considered expressed in radians.

Therefore, collecting all the above estimates we easily get \(A\le \tfrac {C}{n^{2}}\) for sufficiently large \(n\), with \(C{\gt}0\) independent of \(n\) and \(z\) (but depending on \(f\), \(r\) and \(a\)).

In the last part of the prof, we will obtain an estimate of the order \({\mathcal O}(1/n^{2})\) for \(B=\left|K_{n}(S_{n}, z)-S_{n}(z)-\tfrac {z(1+z)S_{n}^{\prime \prime }(z)}{2(n-1)}\right|\) too, which will implies the estimate in the statement.

Denoting \(\pi _{k, n}(z)=K_{n}(e_{k})(z)\) and

firstly it is clear that \(E_{0, n}(z)=E_{1, n}(z)=0\). Then, we can write

so it remains to estimate \(E_{k, n}(z)\) for \(2\le k\le n\), by using the recurrence in Lemma 2.1.

In this sense, simple calculation based on Lemma 2.1 too, leads us to the formula

Taking into account the inequalities valid for all \(2\le k\le n\) and \(r\ge 1\)

this immediately implies, for all \(2\le k\le n\) and \(|z|\le r\) with \(a\le {\rm Re}(z)\le r\)

Denoting \(A(k, r)=2(1+r)^{2}(k+1)k(k-1)(k-2)\), we have obtained

Obviously \(E_{0,n}(z)=E_{1,n}(z)=E_{2, n}=0\). Take in the last inequality, \(k=3, 4, ..., n\).

For \(k=3\) we obtain \(|E_{3, n}(z)|\le \tfrac {r^{2}}{n^{2}}\cdot A(3, r)\).

For \(k=4\) it follows

For \(k=5\) we analogously get

Reasoning by mathematical induction, finally we easily obtain

We conclude that

where since \(2 A r{\lt}1\) by hypothesis, we get that

Indeed, the fact that the last series is convergent follows form the uniform convergence of the series \(\sum _{k=0}^{\infty }z^{k}\) and its derivative of order 5, for \(|z|{\lt}1\). This finishes the proof of the theorem.

In what follows, we obtain the exact order in approximation by the complex Stancu Beta operators of second kind and by their derivatives. In this sense, we present the following three results.

For all \(|z|\le r\) and \(n{\gt}n_0\) (with \(n_{0}\) depending only on \(f\)), we have

Also, we have

It follows

Taking into account that by hypothesis \(f\) is not a polynomial of degree \(\le 1\) in \(D_R\), we get \(\left|\left|\tfrac {e_1(1+e_1)}{2}f^{\prime \prime }\right|\right|_{SD^{r}[a, r]}{\gt}0\).

Indeed, supposing the contrary it follows that \(\tfrac {z(1+z)}{2}f^{\prime \prime }(z)=0\) for all \(z\in \overline{D}_R\), which implies that \(f^{\prime \prime }(z)=0\), for all \(z\in \overline{D}_R \backslash \{ 0,-1\} \). Because \(f\) is analytic, by the uniqueness of analytic functions we get \(f^{\prime \prime }(z)=0\), for all \(z\in D_{R}\), that is \(f\) is a polynomial of degree \(\leq 1\), which contradicts the hypothesis.

Now by Theorem 3.2, for sufficiently large \(n\) we have

Therefore there exists an index \(n_1{\gt}n_0\) depending only on \(f\), \(a\) and \(r\), such that for any \(n\geq n_1\), we have

which immediately implies

This completes the proof.

As a consequence of Theorem 3.1 and Theorem 3.3, we immediately get the following:

Our last result is in simultaneous approximation and can be stated as follows.

Denote by \(\Gamma =\Gamma _{a_{1}, r_{1}}=S_{1}\bigcup L_{1}\) the closed curve composed by the segment in \(\mathbb {C}\)

and by the arc

Clearly that \(\Gamma \) together with its interior is exactly \(SD^{r_{1}}[a_{1}, r_{1}]\) and that from \(r{\lt}r_{1}\) we have \(SD^{r}[a, r]\subset SD^{r_{1}}[a_{1}, r_{1}]\), the inclusion being strictly.

By the Cauchy’s integral formula for derivatives, we have for all
\(z\in SD^{r}[a, b]\) and \(n\in \mathbb {N}\) sufficiently large

which by the estimate in Theorem 3.1 and by the inequality \(|u-z|\geq d=\min \{ r_{1}-r, a-a_{1} \} \) valid for all \(z\in SD^{r}[a, r]\) and \(u\in \Gamma \), implies

Note that here, by simple geometrical reasonings, for the length \(l(\Gamma )\) of \(\Gamma \), we get

where \(\arccos (\alpha )\) is considered expressed in radians.

It remains to prove the lower estimation for \(||K_n^{(p)}(f,\cdot )-f^{(p)}||_{SD^{r}[a, r]}.\)

By the proof of Theorem 3.3, for all \(u\in \Gamma \) and \(n\ge n_{1}\), we have

Substituting it in the above Cauchy’s integral formula, we get

Thus

Applying Theorem 3.2 too, it follows

But by the hypothesis on \(f\), we necessarily have

Indeed, supposing the contrary we get that \(e_1(1+e_1)f''\) is a polynomial of degree \(\le p-1\) in \(SD^{r}[a, r]\), which by the uniqueness of analytic functions implies that

where \(Q_{p-1}(z)\) is a polynomial of degree \(\le p-1\).

Now, if \(p=1\) and \(p=2\), then the analyticity of \(f\) in \(D_{R}\) easily implies that \(f\) necessarily is a polynomial of degree \(\le 1=\max \{ 1, p-1\} \). If \(p{\gt}2\), then the analyticity of \(f\) in \(D_{R}\) easily implies that \(f\) necessarily is a polynomial of degree \(\le p-1=\max \{ 1, p-1\} \). Therefore, in all the cases we get a contradiction with the hypothesis.

In conclusion, \(||[e_1(1+e_1)f''/2]^{(p)}||_{SD^{r}[a, r]}{\gt}0\) and furthermore, reasoning exactly as in the proof of Theorem 3.3, but for \(\| K_{n}^{(p)}(f, \cdot ) -f^{(p)}\| _{SD^{r}[a, r]}\) instead of \(\| K_{n}(f, \cdot ) -f\| _{SD^{r}[a, r]}\), we immediately get the desired conclusion.