1 Introduction
In 1912, Bernstein [1] proposed the famous polynomials, nowadays called Bernstein polynomials, to prove the Weierstrass approximation theorem as follows:
where \(x\in[0,1]\), \(n=1,2,\ldots \) , and Bernstein basis functions \(b_{n,k}(x)\) are defined as follows:
Based on this, there are many papers that mention Bernstein type operators, we illustrate some of them [2‐13]. In 2010, Ye et al. [14] defined the following new Bernstein bases with shape parameter λ:
where \(b_{n,i}(x)\) (\(i=0,1,\ldots,n\)) are defined in (2), \(x\in [0,1]\), \(\lambda\in[-1,1]\). They discussed some important properties of the basis functions and the corresponding curves and tensor product surfaces. It must be pointed out that we have more modeling flexibility when adding the shape parameter λ.
$$\begin{aligned} B_{n}(f;x)=\sum_{k=0}^{n}f \biggl(\frac{k}{n} \biggr)b_{n,k}(x), \end{aligned}$$
(1)
(2)
$$\begin{aligned} \textstyle\begin{cases} \widetilde{b}_{n,0}(\lambda;x)=b_{n,0}(x)-\frac{\lambda }{n+1}b_{n+1,1}(x),\\ \widetilde{b}_{n,i}(\lambda;x)=b_{n,i}(x)+\lambda (\frac {n-2i+1}{n^{2}-1}b_{n+1,i}(x)-\frac{n-2i-1}{n^{2}-1}b_{n+1,i+1}(x) )& (1\leq i\leq n-1),\\ \widetilde{b}_{n,n}(\lambda;x)=b_{n,n}(x)-\frac{\lambda}{n+1}b_{n+1,n}(x), \end{cases}\displaystyle \end{aligned}$$
(3)
Recently, Cai et al. [15] introduced the λ-Bernstein operators as follows:
where \(\widetilde{b}_{n,k}(\lambda;x)\) (\(k=0,1,\ldots,n\)) are defined in (3) and \(\lambda\in[-1,1]\).
$$\begin{aligned} B_{n,\lambda}(f;x)=\sum_{k=0}^{n} \widetilde{b}_{n,k}(\lambda;x)f \biggl(\frac{k}{n} \biggr), \end{aligned}$$
(4)
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In this paper, we propose the Kantorovich type λ-Bernstein operators
and the Bézier variant of Kantorovich type λ-Bernstein operators
where
\(\widetilde{b}_{n,k}(\lambda;x)\) (\(k=0,1,\ldots,n\)) are defined in (3), \(\alpha\geq1\), \(x\in[0,1]\), and \(\lambda\in[-1,1]\).
$$\begin{aligned} K_{n,\lambda}(f;x)=(n+1)\sum_{k=0}^{n} \widetilde{b}_{n,k}(\lambda;x) \int _{\frac{k}{n+1}}^{\frac{k+1}{n+1}}f(t)\,dt, \end{aligned}$$
(5)
$$\begin{aligned} L_{n,\lambda,\alpha}(f;x)=(n+1)\sum_{k=0}^{n}Q_{n,k}^{(\alpha)}( \lambda ;x) \int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}f(t)\,dt, \end{aligned}$$
(6)
$$\begin{aligned} Q_{n,k}^{(\alpha)}(\lambda;x)= \bigl[J_{n,k}(\lambda;x) \bigr]^{\alpha }- \bigl[J_{n,k+1}(\lambda;x) \bigr]^{\alpha}, \qquad J_{n,k}(\lambda;x)=\sum_{j=k}^{n} \widetilde{b}_{n,k}(\lambda;x), \end{aligned}$$
Obviously, when \(\alpha=1\), \(L_{n,\lambda,1}(f;x)\) reduce to Kantorovich type λ-Bernstein operators (5); when \(\lambda=0\), \(L_{n,0,\alpha}(f;x)\) reduce to Bernstein–Kantorovich–Bézier operators defined in [13]; when \(\lambda=0\), \(\alpha=1\), \(L_{n,0,1}(f;x)\) reduce to Bernstein–Kantorovich operators defined in [13].
Let
and
where \(\chi_{k}(t)\) is the characteristic function on the interval \([\frac{k}{n+1},\frac{k+1}{n+1} ]\) with respect to \([0,1]\). By the Lebesgue–Stieltjes integral representations, we have
$$\begin{aligned} P_{n,\lambda,\alpha}(x,t)=(n+1)\sum_{k=0}^{n}Q_{n,k}^{(\alpha)}( \lambda ;x)\chi_{k}(t) \end{aligned}$$
$$\begin{aligned} R_{n,\lambda,\alpha}(x,t)= \int_{0}^{t}P_{n,\lambda,\alpha}(x,s)\,ds, \end{aligned}$$
$$\begin{aligned} L_{n,\lambda,\alpha}(f;x)= \int_{0}^{1}f(t)P_{n,\lambda,\alpha }(x,t)\,dt= \int_{0}^{1}f(t)\,d_{t}R_{n,\lambda,\alpha}(x,t). \end{aligned}$$
(7)
The aims of this paper are to study the rate of convergence of operators \(L_{n,\lambda,\alpha}\) for \(f\in C_{[0,1]}\) and the asymptotic behavior of \(L_{n,\lambda,\alpha}\) for some absolutely continuous functions \(f\in\Phi_{\mathrm{DB}}\), where the class of functions of \(\Phi_{\mathrm{DB}}\) is defined by
For a bounded function f on \([0,1]\), the following metric forms were first introduced in [12]:
where \(x\in[0,1]\) is fixed, \(0\leq\delta_{1}\leq x\), \(0\leq\delta_{2}\leq 1-x\), and \(\mu\geq1\). For the basic properties of \(\Omega_{x-}(f;\delta _{1})\), \(\Omega_{x+}(f;\delta_{2})\), and \(\Omega_{x}(f;\mu)\), refer to [12].
$$\begin{aligned} \Phi_{\mathrm{DB}}= \biggl\{ f \Big\vert f(x)-f(0)= \int_{0}^{x}\phi(u)\,du; x\geq0; \phi\mbox{ is bounded on }[0,1] \biggr\} . \end{aligned}$$
(8)
$$\begin{aligned}& \Omega_{x-}(f;\delta_{1})=\sup_{t\in[x-\delta_{1},x]} \bigl\vert f(t)-f(x) \bigr\vert ; \qquad\Omega_{x+}(f; \delta_{2})=\sup_{t\in[x,x+\delta_{2}]} \bigl\vert f(t)-f(x) \bigr\vert ; \\& \Omega_{x}(f;\mu)=\sup_{t\in[x-x/\mu,x+(1-x)/\mu]} \bigl\vert f(t)-f(x) \bigr\vert , \end{aligned}$$
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2 Some lemmas
For proving the main results, we need the following lemmas.
Lemma 2.1
([15])
Let
\(e_{i}=t^{i}\), \(i=0,1,2\), and
\(n>1\). For the
λ-Bernstein operators
\(B_{n,\lambda}(f;x)\), we have
$$\begin{aligned}& B_{n,\lambda}(e_{0};x)=1; \\& B_{n,\lambda}(e_{1};x)=x+\frac{1-2x+x^{n+1}-(1-x)^{n+1}}{n(n-1)}\lambda; \\& B_{n,\lambda}(e_{2};x)=x^{2}+\frac{x(1-x)}{n}+\lambda \biggl[\frac {2x-4x^{2}+2x^{n+1}}{n(n-1)}+\frac{x^{n+1}+(1-x)^{n+1}-1}{n^{2}(n-1)} \biggr]. \end{aligned}$$
Lemma 2.2
Let
\(e_{i}=t^{i}\), \(i=0,1,2\), and
\(n>1\), for the Kantorovich type
λ-Bernstein operators
\(K_{n,\lambda}(f;x)\), we have the following equalities:
$$\begin{aligned}& K_{n,\lambda}(e_{0};x)=1; \end{aligned}$$
(9)
$$\begin{aligned}& K_{n,\lambda}(e_{1};x)=x+\frac{1-2x}{2(n+1)}+\frac {1-2x+x^{n+1}-(1-x)^{n+1}}{n^{2}-1} \lambda; \end{aligned}$$
(10)
$$\begin{aligned}& K_{n,\lambda}(e_{2};x)=x^{2}+\frac{3nx(2-3x)-3x^{2}+1}{3(n+1)^{2}}+2\lambda \biggl[\frac{ (x-2x^{2}+x^{n+1} )n+x^{n+1}-x}{(n-1)(n+1)^{2}} \biggr]. \end{aligned}$$
(11)
Proof
We can obtain (9) easily by the fact that \(\sum_{k=0}^{n}\widetilde{b}_{n,k}(\lambda;x)=1\). Next, by (5) and using Lemma 2.1, we have
Finally,
Lemma 2.2 is proved. □
$$\begin{aligned} K_{n,\lambda}(e_{1};x) =&(n+1)\sum_{k=0}^{n} \widetilde{b}_{n,k}(\lambda ;x) \int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}t\,dt \\ =&\sum_{k=0}^{n}\widetilde{b}_{n,k}( \lambda;x)\frac{2k+1}{2(n+1)} \\ =&\frac{n}{n+1}B_{n,\lambda}(e_{1};x)+\frac{1}{2(n+1)} \\ =&x+\frac{1-2x}{2(n+1)}+\frac{1-2x+x^{n+1}-(1-x)^{n+1}}{n^{2}-1}\lambda. \end{aligned}$$
$$\begin{aligned} K_{n,\lambda}(e_{2};x) =&(n+1)\sum_{k=0}^{n} \widetilde{b}_{n,k}(\lambda ;x) \int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}t^{2}\,dt \\ =&\sum_{k=0}^{n}\widetilde{b}_{n,k}( \lambda;x)\frac {3k^{2}+3k+1}{3(n+1)^{2}} \\ =&\frac{n^{2}}{(n+1)^{2}}B_{n,\lambda} (e_{2};x )+\frac {n}{(n+1)^{2}}B_{n,\lambda}(e_{1};x)+ \frac{1}{3(n+1)^{2}} \\ =&x^{2}+\frac{3nx(2-3x)-3x^{2}+1}{3(n+1)^{2}}+2\lambda \biggl[\frac{ (x-2x^{2}+x^{n+1} )n+x^{n+1}-x}{(n-1)(n+1)^{2}} \biggr]. \end{aligned}$$
Lemma 2.3
For the Kantorovich type
λ-Bernstein operators
\(K_{n,\lambda}(f;x)\)
and
\(n>1\), using Lemma 2.2, we have
$$ \begin{gathered} K_{n,\lambda}(t-x;x)=\frac{1-2x}{2(n+1)}+\frac {1-2x+x^{n+1}-(1-x)^{n+1}}{n^{2}-1}\lambda, \\ K_{n,\lambda} \bigl((t-x)^{2};x \bigr)=\frac{nx(1-x)}{(n+1)^{2}}+ \frac {1-3x(1-x)}{3(n+1)^{2}}+\frac{2\lambda [x^{n+1}(1-x)+x(1-x)^{n+1} ]}{n^{2}-1} \\ \hphantom{K_{n,\lambda} \bigl((t-x)^{2};x \bigr)=}{}-\frac{4x(1-x)\lambda}{(n+1)^{2}(n-1)} \\ \hphantom{K_{n,\lambda} \bigl((t-x)^{2};x \bigr)}\leq\frac{4}{n+1}. \end{gathered} $$
(12)
Lemma 2.4
For the Bézier variant of Kantorovich type
λ-Bernstein operators
\(L_{n,\lambda,\alpha}(f;x)\)
and
\(f\in C_{[0,1]}\)
with the sup-norm
\(\Vert f \Vert:=\sup_{x\in[0,1]}|f(x)|\), we have
$$\begin{aligned} \bigl\Vert L_{n,\lambda,\alpha}(f) \bigr\Vert \leq\alpha \Vert f \Vert . \end{aligned}$$
Proof
Since, for \(\alpha\geq1\), we have
Then, from (9) and the definition of \(L_{n,\lambda,\alpha }(f;x)\), we have
□
$$\begin{aligned} 0< \bigl[J_{n,k}(\lambda;x)\bigr]^{\alpha}-\bigl[J_{n,k+1}( \lambda;x)\bigr]^{\alpha}\leq \alpha\bigl[J_{n,k}( \lambda;x)-J_{n,k+1}(\lambda;x)\bigr]=\alpha\widetilde {b}_{n,k}( \lambda;x). \end{aligned}$$
$$\begin{aligned} \bigl\Vert L_{n,\lambda,\alpha}(f) \bigr\Vert \leq\alpha \bigl\Vert K_{n,\lambda}(f) \bigr\Vert \leq\alpha \Vert f \Vert . \end{aligned}$$
Lemma 2.5
(i)
For
\(0\leq y\leq x<1\), we have
$$\begin{aligned} R_{n,\lambda,\alpha}(x,y)= \int_{0}^{y}P_{n,\lambda,\alpha}(x,t)\,dt\leq \frac{4\alpha}{(n+1)(x-y)^{2}}. \end{aligned}$$
(13)
(ii)
For
\(0< x< z\leq1\), we have
$$\begin{aligned} 1-R_{n,\lambda,\alpha}(x,z)= \int_{z}^{1}P_{n,\lambda,\alpha}(x,t)\,dt\leq \frac{4\alpha}{(n+1)(z-x)^{2}}. \end{aligned}$$
(14)
Proof
(i) Using (7) and (12), we have
Similarly, (ii) is proved. □
$$\begin{aligned} R_{n,\lambda,\alpha}(x,y) =& \int_{0}^{y}P_{n,\lambda,\alpha}(x,t)\,dt \\ \leq& \int_{0}^{y} \biggl(\frac{x-t}{x-y} \biggr)^{2}P_{n,\lambda,\alpha }(x,t)\,dt \\ \leq&\frac{1}{(x-y)^{2}} \int_{0}^{1}(t-x)^{2}P_{n,\lambda,\alpha}(x,t)\,dt \\ =&\frac{1}{(x-y)^{2}}L_{n,\lambda,\alpha} \bigl((t-x)^{2};x \bigr) \\ \leq&\frac{\alpha}{(x-y)^{2}}K_{n,\lambda} \bigl((t-x)^{2};x \bigr) \\ \leq&\frac{4\alpha}{(n+1)(x-y)^{2}}. \end{aligned}$$
3 Main results
As we know, the space \(C_{[0,1]}\) of all continuous functions on \([0,1]\) is a Banach space with sup-norm \(\Vert f \Vert:=\sup_{x\in [0,1]}|f(x)|\). Let \(f\in C[0,1]\), the Peetre’s K-functional is defined by \(K_{2}(f;t):=\inf_{g\in C_{[0,1]}^{2}}\{ \Vert f-g \Vert+t \Vert g' \Vert+{t}^{2} \Vert g'' \Vert\}\), where \(t>0\) and \(C_{[0,1]}^{2}:=\{g\in C_{[0,1]}: g', g''\in C_{[0,1]}\}\). By [16], there exists an absolute constant \(C>0\) such that
where \(\omega_{2}(f;t):=\sup_{0< h\leq t}\sup_{x,x+h,x+2h\in [0,1]}|f(x+2h)-2f(x+h)+f(x)|\) is the second order modulus of smoothness of \(f\in C_{[0,1]}\). We also denote the usual modulus of continuity of \(f\in C_{[0,1]}\) by \(\omega(f;t):=\sup_{0< h\leq t}\sup_{x,x+h\in [0,1]}|f(x+h)-f(x)|\).
$$ K_{2}(f;t)\leq C\omega_{2} (f;\sqrt{t} ), $$
(15)
Theorem 3.1
For
\(f\in C_{[0,1]}\), \(\lambda\in[-1,1]\), we have
where
C
is a positive constant.
$$\begin{aligned} \bigl\vert L_{n,\lambda,\alpha}(f;x)-f(x) \bigr\vert \leq C\omega _{2} \biggl(f;\sqrt{\frac{\alpha}{n+1}} \biggr), \end{aligned}$$
(16)
Proof
Let \(g\in C_{[0,1]}^{2}\), by Taylor’s expansion
As we know, \(L_{n,\lambda,\alpha}(1;x)=1\). Applying \(L_{n,\lambda,\alpha }(\cdot;x)\) to both sides of the above equation, we get
By the Cauchy–Schwarz inequality, (12) and Lemma 2.4, we have
Then, using the above inequality, we have
Hence, taking infimum on the right-hand side over all \(g\in C_{[0,1]}^{2}\), we get
By (15), we obtain
This completes the proof of Theorem 3.1. □
$$\begin{aligned} g(t)=g(x)+g'(x) (t-x)+ \int_{x}^{t}(t-u)g''(u)\,du. \end{aligned}$$
$$\begin{aligned} L_{n,\lambda,\alpha}(g;x)=g(x)+g'(x)L_{n,\lambda,\alpha }(t-x;x)+L_{n,\lambda,\alpha} \biggl( \int_{x}^{t}(t-u)g''(u)\,du;x \biggr). \end{aligned}$$
$$ \begin{aligned} \bigl\vert L_{n,\lambda,\alpha}(g;x)-g(x) \bigr\vert &\leq \bigl\vert g'(x) \bigr\vert \bigl\vert L_{n,\lambda,\alpha}\bigl( \vert t-x \vert ;x\bigr) \bigr\vert + \biggl\vert L_{n,\lambda,\alpha} \biggl( \int _{x}^{t}(t-u)g''(u)\,du;x \biggr) \biggr\vert \\ &\leq \bigl\Vert g' \bigr\Vert L_{n,\lambda,\alpha}\bigl( \vert t-x \vert ;x\bigr)+\frac{ \Vert g'' \Vert }{2}L_{n,\lambda ,\alpha} \bigl((t-x)^{2};x \bigr) \\ &\leq \bigl\Vert g' \bigr\Vert \sqrt{L_{n,\lambda,\alpha} \bigl((t-x)^{2};x \bigr)}+\frac{ \Vert g'' \Vert }{2}L_{n,\lambda ,\alpha} \bigl((t-x)^{2};x \bigr) \\ &\leq\sqrt{\alpha} \bigl\Vert g' \bigr\Vert \sqrt{K_{n,\lambda} \bigl((t-x)^{2};x \bigr)}+\frac{\alpha \Vert g'' \Vert }{2}K_{n,\lambda} \bigl((t-x)^{2};x \bigr) \\ &\leq\frac{2\sqrt{\alpha} \Vert g' \Vert }{\sqrt{n+1}}+\frac {2\alpha \Vert g'' \Vert }{n+1}. \end{aligned} $$
$$\begin{aligned} \bigl\vert L_{n,\lambda,\alpha}(f;x)-f(x) \bigr\vert \leq& \bigl\vert L_{n,\lambda,\alpha}(f-g;x) \bigr\vert + \bigl\vert (f-g) (x) \bigr\vert + \bigl\vert L_{n,\lambda,\alpha}(g;x)-g(x) \bigr\vert \\ \leq&2 \biggl( \Vert f-g \Vert +\sqrt{\frac{\alpha}{n+1}} \bigl\Vert g' \bigr\Vert +\frac{\alpha}{n+1} \bigl\Vert g' \bigr\Vert \biggr). \end{aligned}$$
$$\begin{aligned} \bigl\vert L_{n,\lambda,\alpha}(f;x)-f(x) \bigr\vert \leq2K_{2} \biggl(f;\frac{\alpha}{n+1} \biggr). \end{aligned}$$
$$\begin{aligned} \bigl\vert L_{n,\lambda,\alpha}(f;x)-f(x) \bigr\vert \leq C\omega _{2} \biggl(f;\sqrt{\frac{\alpha}{n+1}} \biggr). \end{aligned}$$
Next, we recall some definitions of the Ditzian–Totik first order modulus of smoothness and K-functional, which can be found in [17]. Let \(f\in C_{[0,1]}\), and \(\varphi(x):=\sqrt{x(1-x)}\), the first order modulus of smoothness is given by
The K-functional \(K_{\varphi}(f;t)\) is defined by \(K_{\varphi}(f;t):=\inf_{g\in C^{\varphi}_{[0,1]}} \{ \Vert f-g \Vert+t \Vert\varphi g' \Vert \}\), where \(t>0\), \(C^{\varphi}_{[0,1]}:= \{g:g\in AC_{[0,1]}, \Vert \varphi g' \Vert<\infty \}\), \(AC_{[0,1]}\) is the class of all absolutely continuous functions on \([0,1]\). Besides, from [17], there exists a constant \(C>0\) such that
$$\begin{aligned} \omega_{\varphi}(f;t):=\sup_{0< h\leq t, x\pm\frac{h\varphi(x)}{2}\in [0,1]} \biggl\vert f \biggl(x+\frac{h\varphi(x)}{2} \biggr)-f \biggl(x-\frac {h\varphi(x)}{2} \biggr) \biggr\vert . \end{aligned}$$
$$\begin{aligned} K_{\varphi}(f;t)\leq C\omega_{\varphi}(f;t). \end{aligned}$$
(17)
Theorem 3.2
For
\(f\in C_{[0,1]}\), \(\lambda\in[-1,1]\), and
\(\varphi (x)=\sqrt{x(1-x)}\), we have
where
C
is a positive constant.
$$\begin{aligned} \bigl\vert L_{n,\lambda,\alpha}(f;x)-f(x) \bigr\vert \leq C\omega _{\varphi} \biggl(f;\frac{2\sqrt{2\alpha}}{\sqrt{n+1}\varphi(x)} \biggr), \end{aligned}$$
Proof
Since
applying \(L_{n,\lambda,\alpha}(f;x)\) to the above equality, we have
We will estimate \(\int_{x}^{t}g'(u)\,du\): For any \(x,t\in(0,1)\), we have
From (18), using the Cauchy–Schwarz inequality, we obtain
Hence, using the above inequality, we have
Taking infimum on the right-hand side over all \(g\in C_{[0,1]}^{\varphi }\), we get
By (17), we obtain
Theorem 3.2 is proved. □
$$\begin{aligned} g(t)=g(x)+ \int_{x}^{t}g'(u)\,du, \end{aligned}$$
$$\begin{aligned} L_{n,\lambda,\alpha}(g;x)=g(x)+L_{n,\lambda,\alpha} \biggl( \int _{x}^{t}g'(u)\,du;x \biggr). \end{aligned}$$
(18)
$$\begin{aligned} \biggl\vert \int_{x}^{t}g'(u)\,du \biggr\vert \leq& \bigl\Vert \varphi g' \bigr\Vert \biggl\vert \int_{x}^{t}\frac{1}{\varphi(u)}\,du \biggr\vert \\ =& \bigl\Vert \varphi g' \bigr\Vert \biggl\vert \int_{x}^{t}\frac {1}{\sqrt{u(1-u)}}\,du \biggr\vert \\ \leq& \bigl\Vert \varphi g' \bigr\Vert \biggl\vert \int_{x}^{t} \biggl(\frac{1}{\sqrt{u}}+ \frac{1}{\sqrt{1-u}} \biggr)\,du \biggr\vert \\ \leq&2 \bigl\Vert \varphi g' \bigr\Vert \bigl( \vert \sqrt {t}-\sqrt{x} \vert + \vert \sqrt{1-t}-\sqrt{1-x} \vert \bigr) \\ =&2 \bigl\Vert \varphi g' \bigr\Vert \vert t-x \vert \biggl(\frac{1}{\sqrt{t}+\sqrt{x}}+\frac{1}{\sqrt{1-t}+\sqrt{1-x}} \biggr) \\ \leq&2 \bigl\Vert \varphi g' \bigr\Vert \vert t-x \vert \biggl(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{1-x}} \biggr) \\ \leq&2\sqrt{2} \bigl\Vert \varphi g' \bigr\Vert \frac{ \vert t-x \vert }{\varphi(x)}. \end{aligned}$$
$$\begin{aligned} \bigl\vert L_{n,\lambda,\alpha}(g;x)-g(x) \bigr\vert \leq&2\sqrt {2} \frac{ \Vert \varphi g' \Vert }{\varphi(x)}L_{n,\lambda ,\alpha}\bigl( \vert t-x \vert ;x\bigr) \\ \leq&2\sqrt{2}\frac{ \Vert \varphi g' \Vert }{\varphi (x)}\sqrt{L_{n,\lambda,\alpha} \bigl((t-x)^{2};x \bigr)} \\ \leq&2\sqrt{2\alpha}\frac{ \Vert \varphi g' \Vert }{\varphi (x)}\sqrt{K_{n,\lambda} \bigl((t-x)^{2};x \bigr)} \\ \leq&\frac{4\sqrt{2\alpha} \Vert \varphi g' \Vert }{\sqrt {n+1}\varphi(x)}. \end{aligned}$$
$$\begin{aligned} \bigl\vert L_{n,\lambda,\alpha}(f;x)-f(x) \bigr\vert \leq& \bigl\vert L_{n,\lambda,\alpha}(f-g;x) \bigr\vert + \bigl\vert (f-g) (x) \bigr\vert + \bigl\vert L_{n,\lambda,\alpha}(f;x)-g(x) \bigr\vert \\ \leq&2 \biggl( \Vert f-g \Vert +\frac{2\sqrt{2\alpha}}{\sqrt {n+1}\varphi(x)} \bigl\Vert \varphi g' \bigr\Vert \biggr). \end{aligned}$$
$$\begin{aligned} \bigl\vert L_{n,\lambda,\alpha}(f;x)-f(x) \bigr\vert \leq&2K_{\varphi } \biggl(f;\frac{2\sqrt{2\alpha}}{\sqrt{n+1}\varphi(x)} \biggr). \end{aligned}$$
$$\begin{aligned} \bigl\vert L_{n,\lambda,\alpha}(f;x)-f(x) \bigr\vert \leq&C\omega _{\varphi} \biggl(f;\frac{2\sqrt{2\alpha}}{\sqrt{n+1}\varphi(x)} \biggr). \end{aligned}$$
Finally, we study the approximation properties of \(L_{n,\lambda,\alpha }(f;x)\) for some absolutely continuous functions \(f\in\Phi_{\mathrm{DB}}\).
Theorem 3.3
Let
f
be a function in
\(\Phi_{\mathrm{DB}}\). If
\(\phi(x+)\)
and
\(\phi (x-)\)
exist at a fixed point
\(x\in(0,1)\), then we have
where
\([n]\)
denotes the greatest integer not exceeding
n, and
$$\begin{aligned} \bigl\vert L_{n,\lambda,\alpha}(f;x)-f(x) \bigr\vert \leq\frac{2\alpha ( \vert \phi(x+) \vert + \vert \phi(x-) \vert )}{\sqrt{n+1}}+ \frac{8\alpha+2x(1-x)}{nx(1-x)}\sum_{k=1}^{[\sqrt{n}]} \Omega_{x} \biggl(\phi_{x};\frac{1}{k} \biggr), \end{aligned}$$
$$\begin{aligned} \phi_{x}(u)= \textstyle\begin{cases} \phi(u)-\phi(x+),&x< u\leq1;\\ 0,&u=x;\\ \phi(u)-\phi(x-),&0\leq u< x. \end{cases}\displaystyle \end{aligned}$$
(19)
Proof
By the fact that \(L_{n,\lambda,\alpha}(1;x)=1\), using (7) and (8), we have
By the Bojanic–Cheng decomposition [18], we have
where \(\phi_{x}(u)\) is defined in (19), \(\operatorname{sgn}(u)\) is a sign function and \(\delta_{x}(u)= \bigl\{ \scriptsize{ \begin{array}{l@{\quad}l} 1,&u=x;\\ 0,&u\neq x. \end{array} } \) By direct integrations, we find that
where
Integration by parts derives
Note that \(R_{n,\lambda,\alpha}(x,t)\leq1\) and \(\phi_{x}(x)=0\), it follows that
From Lemma 2.5 (i) and change of variable \(t=x-x/u\), we have
Thus, it follows that
From Lemma 2.5(ii), using a similar method, we also obtain
By the Cauchy–Schwarz inequality, (12), and Lemma 2.4, we have
Hence, by (22), (23), (24), and (21), we have
Theorem 3.3 is proved. □
$$\begin{aligned} L_{n,\lambda,\alpha}(f;x)-f(x) =& \int_{0}^{1} \bigl[f(t)-f(x) \bigr]\,d_{t}R_{n,\lambda,\alpha}(x,t) \\ =& \int_{0}^{1} \biggl( \int_{x}^{t}\phi(u)\,du \biggr)\,d_{t}R_{n,\lambda,\alpha}(x,t). \end{aligned}$$
$$\begin{aligned} \phi(u) =&\frac{\phi(x+)+\phi(x-)}{2}+\phi_{x}(u)+\frac{\phi(x+)-\phi (x-)}{2}\operatorname{sgn}(u-x) \\ &{}+\delta_{x}(u) \biggl(\phi(x)-\frac{\phi(x+)+\phi(x-)}{2} \biggr), \end{aligned}$$
(20)
$$\begin{aligned} L_{n,\lambda,\alpha}(f;x)-f(x) =&\frac{\phi(x+)-\phi(x-)}{2}L_{n,\lambda ,\alpha}\bigl( \vert t-x \vert ;x\bigr)-U_{n,\lambda,\alpha}(\phi _{x};x)+T_{n,\lambda,\alpha}( \phi_{x};x) \\ &{}+\frac{\phi(x+)+\phi(x-)}{2}L_{n,\lambda,\alpha}(t-x;x), \end{aligned}$$
(21)
$$\begin{gathered} U_{n,\lambda,\alpha}(\phi_{x};x)= \int_{0}^{x} \biggl( \int_{t}^{x}\phi _{x}(u)\,du \biggr)\,d_{t}R_{n,\lambda,\alpha}(x,t), \\ T_{n,\lambda,\alpha}(\phi_{x};x)= \int_{x}^{1} \biggl( \int_{x}^{t}\phi _{x}(u)\,du \biggr)\,d_{t}R_{n,\lambda,\alpha}(x,t). \end{gathered} $$
$$\begin{aligned} U_{n,\lambda,\alpha}(\phi_{x};x) =& \int_{0}^{x} \biggl( \int_{t}^{x}\phi _{x}(u)\,du \biggr)\,d_{t}R_{n,\lambda,\alpha}(x,t) \\ =& \int_{t}^{x}\phi_{x}(u)\,duR_{n,\lambda,\alpha}(x,t) \bigg\vert _{0}^{x}+ \int_{0}^{x}R_{n,\lambda,\alpha}(x,t) \phi_{x}(t)\,dt \\ =& \int_{0}^{x}R_{n,\lambda,\alpha}(x,t) \phi_{x}(t)\,dt \\ =& \biggl( \int_{0}^{x-x/\sqrt{n}}+ \int_{x-x/\sqrt{n}}^{x} \biggr)R_{n,\lambda,\alpha}(x,t) \phi_{x}(t)\,dt. \end{aligned}$$
$$\begin{aligned} \biggl\vert \int_{x-x/\sqrt{n}}^{x}R_{n,\lambda,\alpha}(x,t)\phi _{x}(t)\,dt \biggr\vert \leq\frac{x}{\sqrt{n}}\Omega_{x} \biggl(\phi_{x};\frac {x}{\sqrt{n}} \biggr)\leq\frac{2x}{n}\sum _{k=1}^{[\sqrt{n}]}\Omega _{x} \biggl( \phi_{x};\frac{x}{k} \biggr). \end{aligned}$$
$$\begin{aligned} \biggl\vert \int_{0}^{x-x/\sqrt{n}}R_{n,\lambda,\alpha}(x,t)\phi _{x}(t)\,dt \biggr\vert \leq&\frac{4\alpha}{n+1} \int_{0}^{x-x/\sqrt{n}}\frac {\Omega_{x}(\phi_{x},x-t)}{(x-t)^{2}}\,dt \\ =&\frac{4\alpha}{(n+1)x} \int_{1}^{\sqrt{n}}\Omega_{x} \biggl( \phi_{x};\frac {x}{u} \biggr)\,du \\ \leq&\frac{8\alpha}{(n+1)x}\sum_{k=1}^{[\sqrt{n}]} \Omega_{x} \biggl(\phi _{x};\frac{x}{k} \biggr). \end{aligned}$$
$$\begin{aligned} \bigl\vert U_{n,\lambda,\alpha}(\phi_{x};x) \bigr\vert \leq& \frac{8\alpha }{(n+1)x}\sum_{k=1}^{[\sqrt{n}]} \Omega_{x} \biggl(\phi_{x};\frac{x}{k} \biggr)+ \frac{2x}{n}\sum_{k=1}^{[\sqrt{n}]} \Omega_{x} \biggl(\phi_{x};\frac {x}{k} \biggr) \\ \leq&\frac{8\alpha+2x^{2}}{nx}\sum_{k=1}^{[\sqrt{n}]} \Omega_{x} \biggl(\phi _{x};\frac{1}{k} \biggr). \end{aligned}$$
(22)
$$\begin{aligned} \bigl\vert T_{n,\lambda,\alpha}(\phi_{x};x) \bigr\vert \leq \frac{8\alpha +2(1-x)^{2}}{n(1-x)}\sum_{k=1}^{[\sqrt{n}]} \Omega_{x} \biggl(\phi_{x};\frac {1}{k} \biggr). \end{aligned}$$
(23)
$$\begin{aligned} L_{n,\lambda,\alpha}\bigl( \vert t-x \vert ;x\bigr)\leq\alpha K_{n,\lambda} \bigl( \vert t-x \vert ;x\bigr)\leq\alpha\sqrt {K_{n,\lambda} \bigl((t-x)^{2};x \bigr)}\leq\frac{2\alpha}{\sqrt {n+1}}. \end{aligned}$$
(24)
$$\begin{aligned} \bigl\vert L_{n,\lambda,\alpha}(f;x)-f(x) \bigr\vert \leq\frac{2\alpha ( \vert \phi(x+) \vert + \vert \phi(x-) \vert )}{\sqrt{n+1}}+ \frac{8\alpha+2x(1-x)}{nx(1-x)}\sum_{k=1}^{[\sqrt{n}]} \Omega_{x} \biggl(\phi_{x};\frac{1}{k} \biggr). \end{aligned}$$
4 Conclusion
In this paper, we have presented a Bézier variant of Kantorovich type λ-Bernstein operators \(L_{n,\lambda,\alpha}(f;x)\), and established approximation theorems by using the usual second order modulus of smoothness and the Ditzian–Totik modulus of smoothness. From Theorem 3.3 of Sect. 3, we know that the rate of convergence of operators \(L_{n,\lambda,\alpha}(f;x)\) for \(f\in\Phi_{\mathrm{DB}}\) is \(\frac {1}{\sqrt{n+1}}\). Furthermore, we might consider the approximation of these operators \(L_{n,\lambda,\alpha}(f;x)\) for locally bounded functions.
Acknowledgements
This work is supported by the National Natural Science Foundation of China (Grant No. 11601266), the Natural Science Foundation of Fujian Province of China (Grant No. 2016J05017), and the Program for New Century Excellent Talents in Fujian Province University. We also thank Fujian Provincial Key Laboratory of Data Intensive Computing and Key Laboratory of Intelligent Computing and Information Processing of Fujian Province University.
Competing interests
The author declares that there are no competing interests.
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