1 Introduction

Let X be a Banach space, endowed with norm \(\Vert \cdot \Vert\). Denote by L(X) the space of bounded linear operators \(T:X\rightarrow X\), endowed with norm \(\Vert L\Vert =\sup \{\Vert Lx\Vert ,\; x\in X, \Vert x\Vert =1\}.\) A \(C_0\) semigroup of operators on the space X is a family of operators \(\{T(t)\}_{t\ge 0}\), \(T(t)\in L(X)\), with the properties

  1. a)

    \(T(t+s)=T(t)T(s)\), for \(t,s\ge 0\);

  2. b)

    \(\lim _{t \rightarrow 0+}T(t)x=x\), for any \(x\in X\), in the sense of norm of X.

As a general bibliography of the subject we mention [1,2,3, 5, 10, 17, 18]. A basic result concerning \(C_0\) semigroups of operators is given by Trotter’s approximation theorem.

Theorem A

[21] Let \((L_n)_{n\in {\mathbb {N}}}\) be a sequence of bounded linear operators on a Banach space X and let \((\rho _n)_{n\in {\mathbb {N}}}\) be a decreasing sequence of positive real numbers tending to 0. Suppose that there exist \(M \ge 0\) and \(\omega \in {\mathbb {R}}\) such that

$$\begin{aligned}\Vert L_n^k\Vert \le M e^{\omega \rho _n k},\;(k,n\in {\mathbb {N}}). \end{aligned}$$

Moreover, assume that D is a dense subspace of X and for every \(f\in D\) the following Voronovskaja-type formula holds

$$\begin{aligned}Af := \lim _{n\rightarrow \infty }\frac{L_n(f)-f}{\rho _n} \end{aligned}$$

If \((\lambda I-A)(D)\) is dense in X for some \(\lambda >\omega\), then there exists a \(C_0\)-semigroup \((T(t))_{t\ge 0}\) such that for every \(f \in X\) and every sequence \((k(n))_n\in {\mathbb {N}}\) of positive integers satisfying \(\lim _{n\rightarrow +\infty } k(n)\cdot {\rho _n} = t\), we have

$$\begin{aligned} T(t)f = \lim _{n\rightarrow \infty }L_n^{k(n)}(f). \end{aligned}$$

A version of Trotter’s approximation theorem is the following

Theorem B

([3], a part of Corollary 2.2.11) Let \((L_n)_{n\ge 1}\) be a sequence of linear operators on the Banach space E, with \(\Vert L_n\Vert \le 1\) and let \((\rho _n)_{n\ge 1}\) be a sequence of positive real numbers such that \(\lim _{n\rightarrow \infty } \rho (n)=0\). Let \(A_0:D_0\rightarrow E\) be a linear operator defined on a subspace \(D_0\) of E and assume that (i) there is a family \((E_i)_{i\in I}\) of finite dimensional subspaces of \(D_0\) which are invariant under each \(L_n\) and whose union \(\bigcup _{i\in I} E_i\) is dense in E; (ii) \(\lim \limits _{n\rightarrow \infty }\frac{L_n(u)-u}{\rho (n)}=A(u)\) for every \(u\in D_0\).

Then \(A_0\) is closable and its closure \(A:D(A)\rightarrow E\) is the generator of a contraction \(C_0\)-semigroup \((T(t))_{t\ge 0}\) on E satisfying the following condition: if \((k(n))_{n\ge 1}\) is a sequence of positive integers with \(\lim \limits _{n\rightarrow \infty }k(n)/\rho (n)=t\), then, for every \(f\in E\),

$$\begin{aligned} T(t)(f)=\lim _{n\rightarrow \infty }L_n^{k(n)}(f). \end{aligned}$$
(1)

The iterates and the limiting semigroup generated by Bernstein operators were studied in [6,7,8,9, 12, 14, 16] among others. The semigroup generated by multidimensional Bernstein operators was considered in [6, 7, 15]. For the limiting groups generated by other positive linear operators we cite [4, 11, 13, 15, 19, 20].

2 Additional results for multidimensional Bernstein operators

We fix the following notation. Let \({\mathbb {N}}=\{1,2,\ldots \}\) and \({\mathbb {N}}_0={\mathbb {N}}\cup \{0\}\). Let \(d\in {\mathbb {N}}\) be fixed. For a multi-index \({\overline{k}}\in {\mathbb {N}}_0^d\), \({\overline{k}}=(k_1,\ldots ,k_d)\), denote \(|{\overline{k}}|=k_1+\ldots +k_d\) and \({\overline{k}}!=k_1!\ldots k_d!\). For \(n\in {\mathbb {N}}\), if \({\overline{k}}\in {\mathbb {N}}_0^d\), \(|{\overline{k}}|\le n\), define \(\left( {\begin{array}{c}n\\ {\overline{k}}\end{array}}\right) =\frac{n!}{{\overline{k}}!(n-|{\overline{k}}|)!}\).

Define the d-simplex

$$\begin{aligned} \varDelta _d:=\{{\overline{x}}=(x_1,\ldots ,x_d)\mid x_i\ge 0,\;(1\le i\le d),\;x_1+\ldots +x_d\le 1\}. \end{aligned}$$
(2)

Vectors \({\overline{e}}_i=(0,\ldots ,0,1,0,\ldots ,0)\), \(1\le i\le d\), form the standard base of the space \({\mathbb {R}}^d\). If \({\overline{x}}=(x_1,\ldots ,x_d)\in \varDelta _d\) put \(|{\overline{x}}|=x_1+\ldots +x_d\). Hence \(|{\overline{x}}|\le 1\). If, in addition we take \({\overline{k}}=(k_1,\ldots ,k_d)\in \varLambda _d^n\), then define \({\overline{x}}^{{\overline{k}}}=x_1^{k_1}\ldots x_d^{k_d}\). With this notation we define now

$$\begin{aligned} p_{n,{\overline{k}}}({\overline{x}}):=\left( {\begin{array}{c}n\\ {\overline{k}}\end{array}}\right) {\overline{x}}^{{\overline{k}}}(1-|{\overline{x}}|)^{n-|{\overline{k}}|}. \end{aligned}$$
(3)

We extend the definition of \(p_{n,{\overline{k}}}({\overline{x}})\), for \({\overline{k}}\in {\mathbb {Z}}^d\), putting

$$\begin{aligned} p_{n,{\overline{k}}}({\overline{x}}):=0\quad \text{ if } \ \exists i\ \text{ such } \text{ that } \ k_i<0\ \text{ or } \ |{\overline{k}}|>n. \end{aligned}$$
(4)

In the case \(d=1\) and \({\overline{k}}=k\), \({\overline{x}}=x\) we write simply \(p_{n,k}(x)\) instead of \(p_{n,{\overline{k}}}({\overline{x}})\).

With these preparations we can define the Bernstein operator on the simplex \(\varDelta _d\):

$$\begin{aligned} B_n(f,{\overline{x}}):=\sum _{|{\overline{k}}|\in \varLambda _d^n}p_{n,{\overline{k}}}({\overline{x}})f\left( \frac{{\overline{k}}}{n}\right) , \end{aligned}$$
(5)

where \(\frac{{\overline{k}}}{n}=\Big (\frac{k_1}{n},\ldots \frac{k_d}{n}\Big )\), \(n\in {\mathbb {N}}\), \(f:\varDelta _d\rightarrow {\mathbb {R}}\), \({\overline{x}}\in \varDelta _d\).

Let \(\alpha =(\alpha _1,\ldots ,\alpha _d)\in {\mathbb {N}}_0^d\). Suppose \(|\alpha |\ge 1\), where \(|\alpha |=\alpha _1+\ldots +\alpha _d\). If \(f\in C^{|\alpha |}(\varDelta _d)\) define

$$\begin{aligned} \frac{\partial ^{\alpha }f}{\partial {\overline{x}}^{\alpha }}:=\frac{\partial ^{|\alpha |}f}{\partial x_1^{\alpha _1}\ldots \partial x_d^{\alpha _d}}. \end{aligned}$$
(6)

If \(|\alpha |=0\), define \(\frac{\partial ^{\alpha }f}{\partial {\overline{x}}^{\alpha }}:=f\).

For \(\alpha \in {\mathbb {N}}_0^d\) denote by \(C^{\alpha }(\varDelta _d)\) the space of functions \(f:\varDelta _d\rightarrow {\mathbb {R}}\) which admits the partial derivative \(\frac{\partial ^{\alpha }f}{\partial {\overline{x}}^{\alpha }}\) continuous on \(\varDelta _d\). For \(1\le i\le d\) consider functions \(\pi _i:\varDelta _d\rightarrow {\mathbb {R}}\), \(\pi _i({\overline{x}})=x_i\).

The next lemma is easy to obtain and in great part well known, see for instance [2, Section 6.2].

Lemma 1

For \({\overline{x}}=(x_1,\ldots ,x_d)\in \varDelta _d\) we have

  1. i)

    \(B_n(\pi _i-x_i,{\overline{x}})=0\), \((1\le i\le d)\);

  2. ii)

    \(B_n((\pi _i-x_i)(\pi _j-x_j),{\overline{x}})=-\frac{x_i x_j}{n}\), \((1\le i,j\le d,\;i\not =j)\);

  3. iii)

    \(B_n((\pi _i-x_i)^2,{\overline{x}})=\frac{x_i(1-x_i)}{n}\), \((1\le i \le d)\);

  4. iv)

    \(B_n((\pi _i-x_i)^3,{\overline{x}})=\frac{x_i(1-x_i)(1-2x_i)}{n^2},\;(1\le i\le d)\);

  5. v)

    \(B_n((\pi _i-x_i)^2(\pi _j-x_j),{\overline{x}})=\frac{x_ix_j(2x_i-1)}{n^2};\;(1\le i,j\le d,\;i\not =j)\);

  6. vi)

    \(B_n((\pi _i-x_i)(\pi _j-x_j)(\pi _m-x_m),{\overline{x}})=\frac{2x_i x_jx_m}{n^2},\;(1\le i<j<m\le d)\);

  7. vii)

    \(B_n((\pi _i-x_i)^4,{\overline{x}})=\frac{1}{n^2}\left( 3-\frac{6}{n}\right) x_i^2(1-x_i)^2+\frac{x_i(1-x_i)}{n^3},\) \((1\le i\le d)\);

  8. viii)

    \(B_n((\pi _i-x_i)^2(\pi _j-x_j)^2,{\overline{x}})=\frac{1}{n^2}\left( 3-\frac{6}{n}\right) x_i^2x_j^3+\left( -\frac{1}{n^2}+\frac{2}{n^3}\right) (x_i^2x_j+x_ix_j^2)+\frac{n-1}{n^3}x_ix_j,\) \((1\le i,j\le d,\;i\not =j)\).

Theorem 1

Let \(\alpha \in {\mathbb {N}}_0^d\), \(|\alpha |\ge 1\). Then for any \(f\in C^{|\alpha |}(\varDelta _d)\), \(n\in {\mathbb {N}}\), \(n\ge |\alpha |\) and \({\overline{x}}\in \varDelta _d\) we have

$$\begin{aligned}&\frac{\partial ^{\alpha }}{\partial {\overline{x}}^{\alpha }}B_n(f,{\overline{x}})=\frac{n!}{(n-|\alpha |)!}\sum _{|{\overline{k}}|\le n-|\alpha |}p_{n-|\alpha |,{\overline{k}}}({\overline{x}})\times \nonumber \\&\quad \times \int \!\!\!\int \ldots \int _{\left[ 0,\frac{1}{n}\right] ^{|\alpha |}}\frac{\partial ^{\alpha }}{\partial {\overline{t}}^{\alpha }}f\Big (\frac{{\overline{k}}}{n}+\sum _{i\in I_{\alpha }}\Big (\sum _{j=1}^{\alpha _i}t_{i,j}\Big ){\overline{e}}_i\Big )d{\overline{t}}_{\alpha }, \end{aligned}$$
(7)

where \(I_{\alpha }=\{i\in \{1,\ldots ,d\}\mid \alpha _i\ge 1\}\) and

$$\begin{aligned}d{\overline{t}}_{\alpha }=\prod _{i\in I_{\alpha }}\prod _{j=1}^{\alpha _i}dt_{i,j}. \end{aligned}$$

In the case \(|\alpha |=0\), the term \(\int \!\!\!\int \ldots \int _{\left[ 0,\frac{1}{n}\right] ^{|\alpha |}}\frac{\partial ^{\alpha }}{\partial {\overline{t}}^{\alpha }}f\Big (\frac{{\overline{k}}}{n}+\sum _{i\in I_{\alpha }}\Big (\sum _{j=1}^{\alpha _i}t_{i,j}\Big ){\overline{e}}_i\Big )d{\overline{t}}_{\alpha }\) is reduced to \(f\left( \frac{{\overline{k}}}{n}\right) .\)

Proof

We consider only the case \(d\ge 2\), since the proof the case \(d=1\) can be easily deduced from the case \(d\ge 2\).

The following formula is well-known.

$$\begin{aligned} (p_{s,k}(x))'= s(p_{s-1,k-1}(x)-p_{s-1,k}(x)),\;s\in {\mathbb {N}},\;k\in {\mathbb {Z}},\;x\in [0,1]. \end{aligned}$$
(8)

We induct on \(r:=|\alpha |\). For \(r=0\) relation (7) is obvious. Suppose that relation (7) holds for any \(d\ge 1\) and any \(\alpha\) with \(|\alpha |=r\) and let show that it is a true for a multi-index \(\beta =(\beta _1,\ldots ,\beta _d)\) with \(|\beta |=r+1\). Then there are a multi-index \(\alpha =(\alpha _1,\ldots ,\alpha _d)\) with \(|\alpha |=r\) and an index \(1\le i\le d\) such that \(\beta _i=\alpha _i+1\) and \(\beta _j=\alpha _j\), for \(1\le j\le d\), \(j\not =i\). To simplify the notation, we can suppose that \(i=d\). In other cases we make a renumbering of the variables.

Let \({\overline{x}}\in \varDelta _d\), \({\overline{x}}=(x_1,\ldots ,x_d)\). Denote \(|{\overline{x}}|=x_1+\ldots + x_d\). Suppose \(x_d>0\). Define \({\overline{z}}=(x_1,\ldots ,x_{d-1})\) and \(|{\overline{z}}|=x_1+\ldots x_{d-1}\). Then \(|{\overline{z}}|<1\). Denote also \(y:=\frac{x_d}{1-|{\overline{z}}|}\in [0,1]\) and \(m:=n-|\alpha |=n-r\).

Let \({\overline{k}}\in {\mathbb {N}}_0^d\), with \(|{\overline{k}}|=m\). Denote \({\overline{\ell }}:=(k_1,\ldots k_{d-1})\). Then \(|{\overline{k}}|=|{\overline{\ell }}|+k_d\). We can write

$$\begin{aligned} p_{m,{\overline{k}}}({\overline{x}}) & = \frac{m!}{k_1!\ldots k_d!(m-|{\overline{k}}|)!}x_1^{k_1}\ldots x_d^{k_d}(1-|{\overline{x}}|)^{m-|{\overline{k}}|}\nonumber \\& = \frac{m!}{k_1!\ldots k_{d-1}!(m-|{\overline{\ell }}|)!}x_1^{k_1}\ldots x_{d-1}^{k_{d-1}}(1-|{\overline{z}}|)^{m-|{\overline{\ell }}|} \nonumber \\&\quad \times \frac{(m-|{\overline{\ell }}|)!}{k_d!(m-|{\overline{k}}|)!}\frac{x_d^{k_d}(1-|{\overline{x}}|)^{m-|{\overline{k}}|}}{(1-|{\overline{z}}|)^{m-|{\overline{\ell }}|}}\nonumber \\& = p_{m,{\overline{\ell }}}({\overline{z}})\cdot p_{m-|{\overline{\ell }}|,k_d}(y). \end{aligned}$$
(9)

For \({\overline{k}}\in {\mathbb {N}}_0^d\), \(|{\overline{k}}|\le m\), denote

$$\begin{aligned}T_{{\overline{k}}}=\int \!\!\!\int \ldots \int _{\left[ 0,\frac{1}{n}\right] ^{|\alpha |}}\frac{\partial ^{\alpha }}{\partial {\overline{t}}^{\alpha }}f\Big (\frac{{\overline{k}}}{n}+\sum _{i\in I_{\alpha }}\Big (\sum _{j=1}^{\alpha _i}t_{i,j}\Big ){\overline{e}}_i\Big )d{\overline{t}}_{\alpha }.\end{aligned}$$

By the hypothesis of induction we have

$$\begin{aligned}\frac{\partial ^{\alpha }}{\partial {\overline{x}}^{\alpha }}B_n(f,{\overline{x}})=\frac{n!}{m!}\sum _{|{\overline{k}}|\le m}p_{m,{\overline{k}}}({\overline{x}})T_{{\overline{k}}}.\end{aligned}$$

Using relation (9) and the decomposition of the sum

$$\begin{aligned} \sum _{|{\overline{k}}|\le m}=\sum _{|{\overline{\ell }}|\le m}\sum _{k_d=0}^{m-|{\overline{\ell }}|}, \end{aligned}$$
(10)

we can write

$$\begin{aligned} \frac{\partial ^{\alpha }}{\partial {\overline{x}}^{\alpha }}B_n(f,{\overline{x}})=\frac{n!}{m!}\sum _{|{\overline{\ell }}|\le m}p_{m,{\overline{\ell }}}({\overline{z}})\sum _{k_d=0}^{m-|{\overline{\ell }}|}p_{m-|{\overline{\ell }}|,k_d}(y)T_{{\overline{k}}}. \end{aligned}$$
(11)

By relation (11) it follows

$$\begin{aligned}\frac{\partial ^{\beta }}{\partial {\overline{x}}^{\beta }}B_n(f,{\overline{x}})&=\frac{\partial }{\partial x_d}\frac{\partial ^{\alpha }}{\partial {\overline{x}}^{\alpha }}B_n(f,{\overline{x}})\\&=\frac{n!}{m!}\sum _{|{\overline{\ell }}|\le m}p_{m,{\overline{\ell }}}({\overline{z}})\sum _{k_d=0}^{m-|{\overline{\ell }}|}\frac{\partial }{\partial x_d}p_{m-|{\overline{\ell }}|,k_d}(y)T_{{\overline{k}}}, \end{aligned}$$

where the first \(d-1\) components of \({\overline{k}}\) are fixed and form vector \({\overline{\ell }}\). Then

$$\begin{aligned}&\frac{\partial ^{\beta }}{\partial {\overline{x}}^{\beta }}B_n(f,{\overline{x}})\\&\quad =\frac{n!}{m!}\sum _{|{\overline{\ell }}|\le m}p_{m,{\overline{\ell }}}({\overline{z}})\sum _{k_d=0}^{m-|{\overline{\ell }}|}\frac{\mathrm{d}}{\mathrm{d}y}p_{m-|{\overline{\ell }}|,k_d}(y)\frac{1}{1-|{\overline{z}}|}\cdot T_{{\overline{k}}}\\&\quad =\frac{n!}{m!}\sum _{|{\overline{\ell }}|\le m}p_{m,{\overline{\ell }}}({\overline{z}})\sum _{k_d=0}^{m-|{\overline{\ell }}|}\frac{m-|{\overline{\ell }}|}{1-|{\overline{z}}|}\Big (p_{m-|{\overline{\ell }}|-1,k_d-1}(y)-p_{m-|{\overline{\ell }}|-1,k_d}(y)\Big )\cdot T_{{\overline{k}}}\\&\quad =\frac{n!}{m!}\sum _{|{\overline{\ell }}|\le m-1}\frac{m-|{\overline{\ell }}|}{1-|{\overline{z}}|}p_{m,{\overline{\ell }}}({\overline{z}})\sum _{k_d=0}^{m-|{\overline{\ell }}|}\Big (p_{m-|{\overline{\ell }}|-1,k_d-1}(y)-p_{m-|{\overline{\ell }}|-1,k_d}(y)\Big )\cdot T_{{\overline{k}}}\\&\quad =\frac{n!}{(m-1)!}\sum _{|{\overline{\ell }}|\le m-1}p_{m-1,{\overline{\ell }}}({\overline{z}})\sum _{k_d=0}^{m-|{\overline{\ell }}|}\Big (p_{m-|{\overline{\ell }}|-1,k_d-1}(y)-p_{m-|{\overline{\ell }}|-1,k_d}(y)\Big )\cdot T_{{\overline{k}}}\\&\quad =\frac{n!}{(m-1)!}\sum _{|{\overline{\ell }}|\le m-1}p_{m-1,{\overline{\ell }}}({\overline{z}})\sum _{k_d=0}^{m-|{\overline{\ell }}|-1}p_{m-|{\overline{\ell }}|-1,k_d}(y)\left[ T_{{\overline{k}}+{\overline{e}}_d}-T_{{\overline{k}}}\right] . \end{aligned}$$

Now, using similar relations to (9) and (10), but with \(m-1\) instead of m we obtain

$$\begin{aligned} \frac{\partial ^{\beta }}{\partial {\overline{x}}^{\beta }}B_n(f,{\overline{x}})=\frac{n!}{(m-1)!}\sum _{|{\overline{k}}|\le m-1}p_{m-1,{\overline{k}}}({\overline{x}})\left[ T_{{\overline{k}}+{\overline{e}}_d}-T_{{\overline{k}}}\right] . \end{aligned}$$
(12)

Finally, we have

$$\begin{aligned}T_{{\overline{k}}+{\overline{e}}_d}-T_{{\overline{k}}}&= \int \!\!\!\int \ldots \int _{\left[ 0,\frac{1}{n}\right] ^{|\alpha |}}\Big \{\frac{\partial ^{\alpha }}{\partial {\overline{t}}^{\alpha }}f\Big (\frac{{\overline{k}}}{n} +\frac{1}{n}{\overline{e}}_d+\sum _{i\in I_{\alpha }}\Big (\sum _{j=1}^{\alpha _i}t_{i,j}\Big ){\overline{e}}_i\Big )\\&\quad -\frac{\partial ^{\alpha }}{\partial {\overline{t}}^{\alpha }}f\Big (\frac{{\overline{k}}}{n} +\sum _{i\in I_{\alpha }}\Big (\sum _{j=1}^{\alpha _i}t_{i,j}\Big ){\overline{e}}_i\Big )\Big \}d{\overline{t}}_{\alpha }\\&=\int \!\!\!\int \ldots \int _{\left[ 0,\frac{1}{n}\right] ^{|\alpha |}}\Big [\int _0^{\frac{1}{n}}\frac{\partial }{\partial x_d}\frac{\partial ^{\alpha }}{\partial {\overline{t}}^{\alpha }}f\Big (\frac{{\overline{k}}}{n}+ s{\overline{e}}_d+\sum _{i\in I_{\alpha }}\Big (\sum _{j=1}^{\alpha _i}t_{i,j}\Big ){\overline{e}}_i\Big )ds\Big ]d{\overline{t}}_{\alpha }. \end{aligned}$$

Because \(\beta _d=\alpha _d+1\ge 1\) it follows that \(d\in I_{\beta }\). Then we can denote s by \(t_{d,\beta _d}\). Let us use the notation \(d{\overline{t}}_{\beta } =\prod _{i\in I_{\beta }}\prod _{j=1}^{\beta _i}dt_{i,j}\). Then

$$\begin{aligned}s{\overline{e}}_d+\sum _{i\in I_{\alpha }}\Big (\sum _{j=1}^{\alpha _i}t_{i,j}\Big ){\overline{e}}_i=\sum _{i\in I_{\beta }}\Big (\sum _{j=1}^{\beta _i}t_{i,j}\Big ){\overline{e}}_i.\end{aligned}$$

Also, \(\left[ 0,\frac{1}{n}\right] ^{|\alpha |}\times \left[ 0,\frac{1}{n}\right] =\left[ 0,\frac{1}{n}\right] ^{|\beta |}\), \(dt_{d,\beta _d}d{\overline{t}}_{\alpha }=d{\overline{t}}_{\beta }\) and \(\frac{\partial }{\partial x_d}\frac{\partial ^{\alpha }}{\partial {\overline{t}}^{\alpha }}=\frac{\partial ^{\beta }}{\partial {\overline{t}}^{\beta }}\). Thus,

$$\begin{aligned} T_{{\overline{k}}+{\overline{e}}_d}-T_{{\overline{k}}}=\int \!\!\!\int \ldots \int _{\left[ 0,\frac{1}{n}\right] ^{|\beta |}}\frac{\partial ^{\beta }}{\partial {\overline{t}}^{\beta }}f\Big (\frac{{\overline{k}}}{n}+\sum _{i\in I_{\beta }}\Big (\sum _{j=1}^{\beta _i}t_{i,j}\Big ){\overline{e}}_i\Big )d{\overline{t}}_{\beta }. \end{aligned}$$
(13)

From relations (12) and (13) and since \(m-1=n-|\beta |\) one obtains

$$\begin{aligned}\frac{\partial ^{\beta }}{\partial {\overline{x}}^{\beta }}B_n(f,{\overline{x}})&=\frac{n!}{(n-|\beta |)!}\sum _{|{\overline{k}}|\le m-1}p_{n-|\beta |,{\overline{k}}}({\overline{x}}) \nonumber \\&\quad \times \int \!\!\!\int \ldots \int _{\left[ 0,\frac{1}{n}\right] ^{|\beta |}}\frac{\partial ^{\beta }}{\partial {\overline{t}}^{\beta }}f\Big (\frac{{\overline{k}}}{n}+\sum _{i\in I_{\beta }}\Big (\sum _{j=1}^{\beta _i}t_{i,j}\Big ){\overline{e}}_i\Big )d{\overline{t}}_{\beta }. \end{aligned}$$
(14)

The relation above can be also extended by continuity at a point \({\overline{x}}\) with \(x_d=0\). The induction step is proved. \(\square\)

Let \(\alpha \in {\mathbb {N}}_0^d\). Denote

$$\begin{aligned} K^{\alpha }(\varDelta _d)=\left\{ f\in C^{\alpha }(\varDelta _d)\mid \frac{\partial ^{\alpha }f}{\partial {\overline{x}}^{\alpha }}({\overline{x}})\ge 0,\;({\overline{x}}\in \varDelta _d)\right\} . \end{aligned}$$
(15)

The following corollaries are immediate.

Corollary 1

For any \(n\in {\mathbb {N}}\) we have

$$\begin{aligned} B_n(K^{\alpha }(\varDelta ))\subset K^{\alpha }(\varDelta _d). \end{aligned}$$
(16)

Let \(\alpha \in {\mathbb {N}}_0^d\). If \(f\in C^{\alpha }(\varDelta _d)\) denote \(\left\| \frac{\partial ^{\alpha }f}{\partial {\overline{x}}^{\alpha }}\right\| =\max _{{\overline{x}}\in \varDelta _d}\left| \frac{\partial ^{\alpha }f}{\partial {\overline{x}}^{\alpha }}({\overline{x}})\right|\).

Corollary 2

For any \(n\in {\mathbb {N}}\), any \(\alpha \in {\mathbb {N}}_0^d\) and any \(f\in C^{\alpha }(\varDelta _d)\) we have

$$\begin{aligned} \left\| \frac{\partial ^{\alpha }}{\partial {\overline{x}}^{\alpha }}B_n(f)\right\| \le \frac{n!}{(n-|\alpha |)!\,n^{|\alpha |}} \left\| \frac{\partial ^{\alpha }f}{\partial {\overline{x}}^{\alpha }}\right\| . \end{aligned}$$
(17)

By induction one obtains

Corollary 3

For any \(n\in {\mathbb {N}}\), any \(\alpha \in {\mathbb {N}}_0^d\), any \(j\in {\mathbb {N}}_0\) and any \(f\in C^{\alpha }(\varDelta _d)\) we have

$$\begin{aligned} \left\| \frac{\partial ^{\alpha }}{\partial {\overline{x}}^{\alpha }}(B_n)^j(f)\right\| \le \left( \frac{n!}{(n-|\alpha |)!\,n^{|\alpha |}} \right) ^j\left\| \frac{\partial ^{\alpha }f}{\partial {\overline{x}}^{\alpha }}\right\| . \end{aligned}$$
(18)

Remark 1

For \(|\alpha |\ge 2\) it follows

$$\begin{aligned}\frac{n!}{(n-|\alpha |)!\,n^{|\alpha |}}\le \frac{n!}{(n-2)!\,n^2}=\frac{n-1}{n}.\end{aligned}$$

For \(k\in {\mathbb {N}}\), \(f\in C^k(\varDelta _d)\) define

$$\begin{aligned} \mu _k(f):=\sup _{\alpha \in {\mathbb {N}}_0^d,\;|\alpha |=k}\Big \Vert \frac{\partial ^{\alpha }f}{\partial {\overline{x}}^{\alpha }}\Big \Vert . \end{aligned}$$
(19)

Corollary 4

For any \(n\in {\mathbb {N}}\), any \(j\in {\mathbb {N}}_0\), any \(k\in {\mathbb {N}}\), \(k\ge 2\), and any \(f\in C^{k}(\varDelta )\) we have

$$\begin{aligned} \mu _k((B_n)^j(f))\le \Big (\frac{n-1}{n}\Big )^j\mu _k(f). \end{aligned}$$
(20)

Proof

Let \(j\ge 0\). There exists \(\alpha _0\in {\mathbb {N}}_0^d\) with \(|\alpha _0|=k\) such that \(\mu _k((B_n)^{j+1}(f))\) \(=\Big \Vert \frac{\partial ^{\alpha _0}}{\partial {\overline{x}}^{\alpha _0}}(B_n)^{j+1}(f)\Big \Vert\). Then using relation (17) and Remark 1 we obtain

$$\begin{aligned} \mu _k((B_n)^{j+1}(f))& = \Big \Vert \frac{\partial ^{\alpha _0}}{\partial {\overline{x}}^{\alpha _0}}B_n((B_n)^{j+1}(f))\Big \Vert \le \frac{n-1}{n} \Big \Vert \frac{\partial ^{\alpha _0}}{\partial {\overline{x}}^{\alpha _0}}(B_n)^{j}(f)\Big \Vert \\&\le \frac{n-1}{n}\mu _k((B_n)^{j}(f)). \end{aligned}$$

So that we can apply the induction. \(\square\)

Corollary 5

We have \(B_n(\varPi _m)\subset \varPi _m\), \(m\ge 0\), where \(\varPi _m\) is the set of polynomials with d variables with total degree at most m.

Proof

Take a monomial function \(f({\overline{x}})={\overline{x}}^{\gamma }\), \(\gamma =(\gamma _1,\ldots ,\gamma _d)\), with \(|\gamma |\le m\). Then \(\frac{\partial ^{\gamma _j+1}}{\partial x_j^{\gamma _j+1}}f=0\) on \({\mathbb {R}}^d\), for \(1\le j\le d\). From Theorem 1 we deduce that \(\frac{\partial ^{\gamma _j+1}}{\partial x_j^{\gamma _j+1}}B_n(f)=0\) on \(\varDelta _d\), for every \(1\le j\le d\). It is easy to see that \(B_n(f)\) is a polynomial of the form \(\sum _{s\in I}a_s{\overline{x}}^{\beta _s}\), where I is finite, \(\beta _s=(\beta _{s,1},\ldots ,\beta _{s,d})\in {\mathbb {N}}^d\), \(\beta _{s,j}\le \gamma _j\), for \(1\le j\le d\), \(s\in I\) and \(a_s\in {\mathbb {R}}\), for \(s\in I\). Therefore \(B_n(f)\in \varPi _m\). It follows \(B_n(\varPi _m)\subset \varPi _m\). \(\square\)

3 A quantitative estimate for Trotter’s theorem

Consider operator

$$\begin{aligned} Af({\overline{x}})=\frac{1}{2}\sum _{i=1}^d \frac{\partial ^2 f({\overline{x}})}{\partial x_i^2}x_i(1-x_i)-\sum _{1\le i<j\le d}\frac{\partial ^2 f({\overline{x}})}{\partial x_i\partial x_j}x_ix_j,\;f\in C^2(\varDelta _d). \end{aligned}$$
(21)

In the following lemma we give a Voronovskaja type theorem for operators \(B_n\).

Lemma 2

$$\begin{aligned} \lim _{n\rightarrow \infty }n(B_n(f,{\overline{x}})-f({\overline{x}}))=Af({\overline{x}}),\;f\in C^2(\varDelta _d). \end{aligned}$$

Remark 2

There exists a semigroup of bounded linear operators \(\{T(t)\}_{t\ge 0}\), \(T(t):C(\varDelta _d)\rightarrow C(\varDelta _d)\), such that

$$\begin{aligned}\lim _{n\rightarrow \infty }B_n^{m_n}(f)=T(t),\;t\ge 0,\end{aligned}$$

for any sequences of integers \((m_n)_n\) such that \(\frac{m_n}{n}=t\). This fact follows, for instance, from Theorem B, with the choices: \(L_n=B_n\), \(E=C(\varDelta _d)\), \(D_0=C^2(\varDelta _d)\), \(A_0=A\) and \(E_i=\varPi _i\), \(i\ge 0\), where \(\varPi _i\) is the space of polynomials with total degree i, see Corollary 5.

Lemma 3

For \(g\in C^4(\varDelta _d)\) we have

$$\begin{aligned} \Big \Vert B_n(g)-g-\frac{1}{n} Ag\Big \Vert \le \frac{C^1_d}{n^2}\mu _3(g), \end{aligned}$$
(22)

where

$$\begin{aligned} C_d^1=\frac{1}{3}d^3-\frac{1}{2}d^2+\frac{1}{3}d. \end{aligned}$$
(23)

and \(\mu _3(g)\) is defined in (19).

Proof

For \(g\in C^4(\varDelta _d)\), \({\overline{x}},{\overline{t}}\in \varDelta _d\) we get

$$\begin{aligned} g({\overline{t}})& = g({\overline{x}})+\sum _{i=1}^d\frac{\partial g_i({\overline{x}})}{\partial x_i}(t_i-x_i)\\&\quad +\frac{1}{2}\Big [\sum _{i=1}^d \frac{\partial ^2 g_i({\overline{x}})}{\partial x_i^2}(t_i-x_i)^2+2\sum _{1\le i<j\le d}\frac{\partial ^2 g_i({\overline{x}})}{\partial x_i\partial x_j} (t_i-x_i)(t_j-x_j)\Big ]\\&\quad +\frac{1}{6}\Big [\sum _{i=1}^d \frac{\partial ^3 g_i(\xi )}{\partial x_i^3}(t_i-x_i)^3+3\sum _{1\le i,j\le d,\;i\not =j}\frac{\partial ^3 g_i(\xi )}{\partial x_i^2\partial x_j}(t_i-x_i)^2(t_j-x_j)\\&\quad +6\sum _{1\le i<j<k\le d}\frac{\partial ^3 g_i(\xi )}{\partial x_i\partial x_j\partial x_k} (t_i-x_i)(t_j-x_j)(t_k-x_k)\Big ], \end{aligned}$$

where \(\xi\) belongs to the interval \([{\overline{x}},{\overline{t}}]\subset \varDelta _d\). Then, using relation (21) and Lemma 1 we obtain \(B_n(g,{\overline{x}})=g({\overline{x}})+\frac{1}{n}A(g,{\overline{x}})+R_3({\overline{x}})\), and

$$\begin{aligned} |R_3({\overline{x}})|&\le \frac{\mu _3(g)}{6}\Big |\sum _{i=1}^d\frac{x_i(1-x_i)(1-2x_i)}{n^2}+3\sum _{1\le i,j\le d,\;i\not =j}\frac{x_ix_j(2x_i-1)}{n^2}\\&+6\sum _{1\le i<j<k\le d}\frac{2x_i x_jx_k}{n^2}\Big |\\& = \frac{\mu _3(g)}{6n^2}[d+3d(d-1)+2d(d-1)(d-2)]\\& = \frac{\mu _3(g)}{n^2}\Big [\frac{1}{3}d^3-\frac{1}{2}d^2+\frac{1}{3}d\Big ]. \end{aligned}$$

\(\square\)

Lemma 4

For any \(g\in C^4(\varDelta _d)\) and \(t\ge 0\) we have

$$\begin{aligned} \Vert T(t)g-g-tAg\Vert\le & \frac{t^2}{2}\sum _{k=2}^4C_d^k\mu _k(g), \end{aligned}$$

where

$$\begin{aligned} C_d^2 =\frac{1}{2}d^2,\; C_d^3=d^3-d^2+\frac{1}{2}d,\; C_d^4=\frac{1}{4}d^4 \end{aligned}$$
(24)

and \(\mu _k(g)\), \(k=2,3,4\) are defined in (19).

Proof

First we use the known inequality

$$\begin{aligned}\Vert T(t)g-g-tAg\Vert \le \frac{t^2}{2}\Vert A^2g\Vert .\end{aligned}$$

In the sequel we use abbreviated notations for sums of the form \(\sum _i a_i\), \(\sum _{i,j} a_{i,j}\), \(\sum _{i,j,k} a_{i,j,k}\), \(\sum _{i,j,k,\ell } a_{i,j,k,\ell }\). We suppose that all the indices are in the set \(\{1,2,\ldots , d\}\) and are different from each other in the case of these sums. The terms are unique taken as indicated in the generic form described by the sum. For instance, \(\sum _{i,j}x_ix_j\) is the abbreviation of \(\sum _{1\le i<j\le d}x_ix_j\) and \(\sum _{i,j}x_i^2x_j\) is the abbreviation of \(\sum _{1\le i,j\le d,\;i\not =j}x_i^2x_j\). We also use the convention that if the number of indices is strictly greater than d, then the corresponding sum is null.

From (21) one obtains, after certain calculations, for \(g\in C^4(\varDelta _d)\) and \({\overline{x}}\in \varDelta _d\):

$$\begin{aligned} A^2(g,{\overline{x}})& = \frac{1}{4}\sum _i \frac{\partial ^4 g({\overline{x}})}{\partial x_i^4}x_i^2(1-x_i)^2\\&\quad +\frac{1}{2} \sum _{i}\frac{\partial ^3g({\overline{x}})}{\partial x_i^3}x_i(1-x_i)(1-2x_i)\\&\quad -\frac{1}{2} \sum _{i}\frac{\partial ^2g({\overline{x}})}{\partial x_i^2}x_i(1-x_i)\\&\quad +\frac{1}{2}\sum _{i,j}\frac{\partial ^4 g({\overline{x}})}{\partial x_i^2\partial x_j^2}x_i(1-x_i)x_j(1-x_j)\\&\quad -\frac{1}{2}\sum _{i,j} \frac{\partial ^4 g({\overline{x}})}{\partial x_i^3\partial x_j}x_i^2(1-x_i)x_j\\&\quad -\frac{1}{2}\sum _{i,j,k}\frac{\partial ^4 g({\overline{x}})}{\partial x_i\partial x_j\partial x_k^2}x_ix_jx_k(1-x_k)\\&\quad -\frac{1}{2}\sum _{i,j,k}\frac{\partial ^4 g({\overline{x}})}{\partial x_i^2\partial x_j\partial x_k}x_i(1-x_i)x_jx_k -\frac{1}{2}\sum _{i,j} \frac{\partial ^3 g({\overline{x}})}{\partial x_i^2\partial x_j}x_i(1-x_i)x_j\\&\quad -\frac{1}{2} \sum _{i,j} \frac{\partial ^4 g({\overline{x}})}{\partial x_i^3\partial x_j}x_i^2(1-x_i)x_j -\frac{1}{2} \sum _{i,j}\frac{\partial ^3 g({\overline{x}})}{\partial x_i^2\partial x_j}x_i(1-2x_i)x_j\\&\quad +\sum _{i,j}\frac{\partial ^4 g({\overline{x}})}{\partial x_i^2\partial x_j^2}x_i^2x_j^2 +\sum _{i,j}\frac{\partial ^3 g({\overline{x}})}{\partial x_i^2\partial x_j}x_i^2x_j\\&\quad +\sum _{i,j}\frac{\partial ^2 g({\overline{x}})}{\partial x_i\partial x_j}x_ix_j +6\sum _{i,j,k,\ell }\frac{\partial ^4g({\overline{x}})}{\partial x_i\partial x_j\partial x_k\partial x_{\ell }}x_ix_jx_kx_{\ell }\\&\quad +2\sum _{i,j,k}\frac{\partial ^4 g({\overline{x}})}{\partial x_i^2\partial x_j\partial x_k}x_i^2x_jx_k +6\sum _{i,j,k}\frac{\partial ^3 g({\overline{x}})}{\partial x_i\partial x_j\partial x_k}x_ix_jx_k. \end{aligned}$$

Therefore

$$\begin{aligned} \Vert A^2g\Vert&\le \frac{1}{4}d\mu _4(g)+\frac{1}{2}d\mu _3(g)\\&\quad +\frac{1}{2}d\mu _2(g)+\frac{1}{2}\cdot \frac{d(d-1)}{2}\mu _4(g)+\frac{1}{2}d(d-1)\mu _4(g)\\&\quad +\frac{1}{2}d(d-1)\mu _3(g)+\frac{1}{2}\frac{d(d-1)(d-2)}{2}\mu _4(g)\\&\quad +\frac{1}{2}\frac{d(d-1)(d-2)}{2}\mu _4(g)\\&\quad +\frac{1}{2}d(d-1)\mu _4(g)+\frac{1}{2}d(d-1)\mu _3(g)\\&\quad +\frac{d(d-1)}{2}\mu _4(g)+d(d-1)\mu _3(g)\\&\quad +\frac{d(d-1)}{2}\mu _2(g)\\&\quad +6\frac{d(d-1)(d-2)(d-3)}{24}\mu _4(d)\\&\quad +2\frac{d(d-1)(d-2)}{2}\mu _4(g)\\&\quad +6\frac{d(d-1)(d-2)}{6}\mu _3(g)\\& = \frac{1}{4}d^4\mu _4(g)+\Big (d^3-d^2+\frac{1}{2}d\Big )\mu _3(g)\\&+\frac{1}{2}d^2\mu _2(g). \end{aligned}$$

\(\square\)

The main result is the following.

Theorem 2

For \(f\in C^4(\varDelta _d)\), \(m\in {\mathbb {N}}\), \(n\in {\mathbb {N}}\), \(t\ge 0\) we have

$$\begin{aligned} \Vert (B_n)^mf-T(t)f\Vert\le & \left| \frac{m}{n}-t\right| \nonumber \\&\frac{d^2}{2}\mu _2(f)+\frac{1}{n}\Big [C^1_d\mu _3(f)+\frac{1}{2}\sum _{k=2}^4C_d^k\mu _k(f)\Big ] \end{aligned}$$
(25)

where \(C^k_d\), \(k=1,2,3,4\) are given in (23) and (24).

Proof

The method of proof is a modification of the method used in [12] and consists in a modification of a telescopic sum argument.

Since \(\Vert (B_n)^m\Vert =1\), for \(m\ge 1\) it follows \(\Vert T(t)\Vert =1\), \(t>0\).

Consider the decomposition

$$\begin{aligned} \Vert (B_n)^mf-T(t)f\Vert \le \Big \Vert T\Big (\frac{m}{n}\Big )f-T(t)f\Big \Vert +\Big \Vert (B_n)^mf-T\Big (\frac{m}{n}\Big )f\Big \Vert . \end{aligned}$$
(26)

From relation (21) we deduce \(\Vert Af\Vert \le \Big (\frac{1}{2}d+\frac{d(d-1)}{2}\Big )\mu _2(f)=\frac{d^2}{2}\mu _2(f)\). We obtain successively:

$$\begin{aligned} \Big \Vert T\Big (\frac{m}{n}\Big )f-T(t)f\Big \Vert& = \left\| \int _t^{\frac{m}{n}}T(u)Afdu\right\| \nonumber \\&\le \left| \frac{m}{n}-t\right| \sup _{u\in \left[ \frac{m}{n},t\right] }\Vert T(u)Af\Vert \nonumber \\&\le \left| \frac{m}{n}-t\right| \cdot \Vert Af\Vert \nonumber \\&\le \left| \frac{m}{n}-t\right| \cdot \frac{d^2}{2}\mu _2(f). \end{aligned}$$
(27)

For the second term one can use a telescopic sum:

$$\begin{aligned} \Big \Vert (B_n)^mf-T\Big (\frac{m}{n}\Big )f\Big \Vert= & \Big \Vert \sum _{j=0}^{m-1}T\Big (\frac{m-1-j}{n}\Big )\Big (B_n-T\Big (\frac{1}{n}\Big )\Big )(B_n)^jf\Big \Vert \nonumber \\\le & \sum _{j=0}^{m-1}\Big \Vert \Big (B_n-T\Big (\frac{1}{n}\Big )\Big )(B_n)^jf\Big \Vert . \end{aligned}$$
(28)

We can write

$$\begin{aligned}&\Big \Vert \Big (B_n-T\Big (\frac{1}{n}\Big )\Big )(B_n)^jf\Big \Vert \nonumber \\&\qquad \le \Big \Vert \Big (B_n-I-\frac{1}{n}A\Big )(B_n)^jf\Big \Vert +\Big \Vert \Big (T\Big (\frac{1}{n}\Big )-I-\frac{1}{n}A\Big )(B_n)^jf\Big \Vert \end{aligned}$$
(29)

From Lemmas 3 and 4 it results for any j

$$\begin{aligned} \Big \Vert \Big (B_n-I-\frac{1}{n}A\Big )(B_n)^jf\Big \Vert\le & \frac{C^1_d}{n^2}\mu _3((B_n)^jf) \end{aligned}$$
(30)
$$\begin{aligned} \Big \Vert \Big (T\Big (\frac{1}{n}\Big )-I-\frac{1}{n}A\Big )(B_n)^jf\Big \Vert\le & \frac{1}{2n^2}\sum _{k=2}^4C_d^k\mu _k((B_n)^jf). \end{aligned}$$
(31)

But using Corollary 4 we have \(\mu _k((B_n)^jf)\le \Big (\frac{n-1}{n}\Big )^j\mu _k(f)\), for \(j\ge 0\), \(k=2,3,4\). Then by using (28), (29), (30) and (31) one obtains

$$\begin{aligned} \Big \Vert (B_n)^mf-T\Big (\frac{m}{n}\Big )f\Big \Vert\le & \sum _{j=0}^{m-1} \Big (\frac{n-1}{n}\Big )^j\frac{1}{n^2} \Big [C^1_d\mu _3(f)+\frac{1}{2}\sum _{k=2}^4C_d^k\mu _k(f)\Big ]\nonumber \\= & \frac{1}{n}\Big [C^1_d\mu _3(f)+\frac{1}{2}\sum _{k=2}^4C_d^k\mu _k(f)\Big ]. \end{aligned}$$
(32)

From (27) and (32) it results (25). \(\square\)

Finally, we compare our result with others, obtained previously.

Remark 3

A quantitative version of Trotter’s theorem for the semigroup generated by Bernstein operators defined on the simplex \(\varDelta _d\) was obtained by Campiti and Tacelli [6, 7] for functions belonging to the space \(C^{2,\alpha }(\varDelta _d)\), with \(0<\alpha <1\). The space \(C^{2,\alpha }(\varDelta _d)\) consists of real functions f defined on \(\varDelta _d\), which admit second derivatives on \(\varDelta _d\) and for which the following condition

$$\begin{aligned} \sup _{{\mathop {x\not =y}\limits ^{x,y\in \varDelta _d}}}\frac{1}{\Vert x-y\Vert ^{\alpha }}\sum _{i,j=1}^d\left| \frac{\partial ^2 f}{\partial x_i\partial x_j}(x)-\frac{\partial ^2 f}{\partial x_i\partial x_j}(y)\right| <\infty \end{aligned}$$

is satisfied. In [6, Theorem 2.3], completed in [7], the following estimate is obtained:

$$\begin{aligned}\Vert T(t)f-(B_n)^{k(n)}f\Vert &\le \frac{t\psi (f)}{n^{\alpha /(4+\alpha )}}+ \left( \left| \frac{k(n)}{n}-t\right| \right. \nonumber \\&\quad \left. +\frac{\sqrt{k(n)}}{n}\right) \left( \Vert Af\Vert +\frac{\psi (f)}{n^{\alpha /(4+\alpha )}}\right) , \end{aligned}$$
(33)

for every \(t\ge 0\), \(f\in C^{2,\alpha }(\varDelta _d)\) and sequence \((k(n))_{n\ge 1}\) of positive integers, where \(\psi (f)\) depends only on f. On other hand, relation (25) with m replaced by k(n) is of the form

$$\begin{aligned}\Vert T(t)f-(B_n)^{k(n)}f\Vert &\le C_1(f)\left| \frac{k(n)}{n}-t\right| \nonumber \\&\quad + C_2(f)\frac{1}{n},\;t\ge 0,\;f\in C^4(\varDelta _d). \end{aligned}$$
(34)

The first remark is that in the hypothesis \(k(n)/n\rightarrow t\), \((n\rightarrow \infty )\), relation (33) is generally stronger, because it is valid for the greater space \(C^{2,\alpha }(\varDelta _d)\), instead of space \(C^4(\varDelta _d)\).

In the case when \(f\in C^4(\varDelta _d)\), in order to make an asymptotic comparison, let fix f and t and denote \(\beta =\frac{\alpha }{4(\alpha +1)}\in (0,1/8)\). We can make this comparison in two cases.

If \(\liminf _{n\rightarrow \infty }\left| \frac{k(n)}{n}-t\right| n^{\beta }\in (0,\infty )\cup \{\infty \}\), then the two estimates have the same order of convergence to 0, namely \(\mathrm{O}\left( \left| \frac{k(n)}{n}-t\right| \right)\),

In the case when \(\left| \frac{k(n)}{n}-t\right| =\mathrm{o}\left( n^{-\beta }\right) \;(n\rightarrow \infty )\) relation (34), i.e., (25) is stronger than relation (33).

Remark 4

Another estimate for approximation of the semigroup generated by the Bernstein operators on a simplex was given by Mangino and Raşa [15] in the form:

$$\begin{aligned}&\Vert (B_n)^{k_n}f-T(t)f\Vert _{\infty }\nonumber \\ &\quad \le \left( tC_n+\left( \left| t-\frac{k(n)}{n}\right| +\frac{\sqrt{k(n)}}{n}\right) (C_n+1)\right) \Vert f\Vert _3, \end{aligned}$$
(35)

where \(C_n=\frac{1}{n}+\frac{1}{6}d^3\sqrt{n}\sup _{{\overline{x}}\in \varDelta _d}\sqrt{B_n(\Vert {\overline{x}}-\bullet \Vert ^4,{\overline{x}})}\), \(f\in C^3(\varDelta _d)\), \(\Vert f\Vert _3=\sum \limits _{|\alpha |\le 3}\Vert D^{\alpha }f\Vert\). This estimate has also a larger domain of applicability: \(C^3(\varDelta )\). It remains to compare (35) with (25) for \(f\in C^4(\varDelta )\). Fix d and \(t>0\). Consider a sequence \((k(n))_n\), such that \(\lim _{n\rightarrow \infty }\frac{k(n)}{n}=t\). We can make the comparison in two cases.

If \(\liminf _{n\rightarrow \infty }\left| \frac{k(n)}{n}-t\right| \sqrt{n}\in (0,\infty )\cup \{\infty \}\), then, by taking into account that \(C_n=\mathrm{O}\left( \frac{1}{\sqrt{n}}\right)\), it follows that the two estimates have the same order, namely \(\mathrm{O}\left( \left| \frac{k(n)}{n}-t\right| \right)\).

In the case when \(\left| \frac{k_n}{n}-t\right| =\mathrm{o}\left( n^{-\frac{1}{2}}\right)\), estimate (25) is stronger than relation (35).