Relations Between the Bernstein Polynomials and q-monotone Functions

Abstract

We show that certain inequalities involving differences of the Bernstein basis polynomials and values of a function $f\in C[0,1]$, imply that the function is q-monotone. In view of previous results of the authors (see the list of references), the current results provide, among others, a characterization of q-monotone functions in C[0, 1].

1 Introduction and the Main Results

Given a function f defined on [0, 1], the classical Bernstein polynomials associated with it are defined by [8],

$$\begin{aligned} (B_nf)(x)=\sum _{j=0}^{n}p_{n,j}(x){f}\left( \frac{j}{n}\right) ,\qquad x\in [0,1], \end{aligned}$$

where

$$\begin{aligned} p_{n,j}(x)=\left( {\begin{array}{c}n\\ j\end{array}}\right) x^j(1-x)^{n-j},\qquad 0\le j\le n. \end{aligned}$$

The following theorem was proved by J. Mrowiec, T. Rajba and S. Wąsowicz [9]. It has given an affirmative answer to a conjecture by the third author.

Theorem A

If $f\in C[0,1]$ is convex, then for all $n\in \mathbb {N}$,

$$\begin{aligned} \sum _{i=0}^n\sum _{j=0}^n\bigl [ p_{n,i}(x)p_{n,j}(x)+p_{n,i}(y)p_{n,j}(y)-2p_{n,i}(x)p_{n,j}(y)\bigr ]f\left( \frac{i+j}{2n}\right) \ge 0. \nonumber \\ \end{aligned}$$

(1.1)

The proof given by J. Mrowiec, T. Rajba and S. Wąsowicz [9] makes heavy use of probability theory. As a tool they applied stochastic convex orderings (which they proved for binomial distributions) as well as the so-called concentration inequality. This probabilistic approach has been applied in several other papers (see [7] and references therein) connecting (1.1) with the theory of stochastic ordering.

Recently [1], the first author gave an elementary, analytic, proof of Theorem A, and this approach has been followed by several authors (see [5, 6] and references therein). Here we continue the analytic line.

Since the Bernstein polynomials preserve convexity, if $f\in C[0,1]$ is convex, then for all pairs $x,y\in [0,1]$,

$$\begin{aligned} (B_nf)(x)+(B_nf)(y)\ge 2(B_nf)\left( \frac{x+y}{2}\right) . \end{aligned}$$

(1.2)

It follows from Vandermonde’s identity

$$\begin{aligned} \sum _{\begin{array}{c} i,j=0\\ i+j=k \end{array}}^n\left( {\begin{array}{c}n\\ i\end{array}}\right) \left( {\begin{array}{c}n\\ j\end{array}}\right) =\left( {\begin{array}{c}2n\\ k\end{array}}\right) , \end{aligned}$$

that,

$$\begin{aligned} (B_{2n}f)(u)=\sum _{i=0}^n\sum _{j=0}^n p_{n,i}(u)p_{n,j}(u)f\left( \frac{i+j}{2n}\right) ,\quad u\in [0,1]. \end{aligned}$$

(1.3)

So, the first and third authors [3] asked whether one may actually prove a stronger inequality than (1.1), namely, for convex f, prove that

$$\begin{aligned} (B_{2n}f)\left( \frac{x+y}{2}\right) \ge \sum _{i=0}^n\sum _{j=0}^n p_{n,i}(x)p_{n,j}(y)f\left( \frac{i+j}{2n}\right) , \end{aligned}$$

(1.4)

and indeed, they proved it in [3].

A function $f\in C[0,1]$ is called q-monotone, $q\in {\mathbb {N}}$, if $\Delta ^q_h(f,x)\ge 0$, for all $x\in [0,1]$ and $h>0$, where $\Delta ^q_h(f,x)$ is the qth forward difference with step $h>0$. In particular, f is 1-monotone or 2-monotone, if it is nondecreasing, respectively, convex in [0, 1]. It is well known that $f\in C[0,1]$ is q-monotone, $q\ge 2$, if and only if $f\in C^{q-2}(0,1)$, and $f^{(q-2)}$ is convex in (0, 1).

Recently, Raşa’s conjecture has been extended to q-monotone functions, $q>2$, by the first two authors who proved in [2] that,

Theorem B

Let $q,n\in \mathbb {N}$. If $f\in C\left[ 0,1\right]$ is a q-monotone function, then for all $x,y\in \left[ 0,1\right]$,

$$\begin{aligned}&\mathrm {sgn}(x-y)^q\sum _{\nu _1,\dots ,\nu _q=0}^{n}\sum _{j=0}^q(-1)^{q-j} \left( {\begin{array}{c}q\\ j\end{array}}\right) \left( \prod \limits _{i=1}^{j}p_{n,\nu _{i}}(x)\right) \nonumber \\&\quad \times \left( \prod \limits _{i=j+1}^{q}p_{n,\nu _{i}}(y)\right) f\left( \frac{\nu _{1}+\cdots +\nu _{q}}{qn}\right) \ge 0. \end{aligned}$$

(1.5)

It is well known (see, e.g., [8, 1.5(2)]) that for $q\ge 1$,

$$\begin{aligned} \frac{d^q}{dx^q}(B_nf)(x)=q!\left( {\begin{array}{c}n\\ q\end{array}}\right) \sum _{i=0}^{n-q}p_{n-q,i}(x)\Delta ^q_{1/n}(f,i/n). \end{aligned}$$

Hence, the Bernstein polynomials preserve q-monotonicity of all orders $q\ge 1$, so that if $f\in C[0,1]$ is q-monotone, then for any pair $x,y\in [0,1]$,

$$\begin{aligned} \mathrm {sgn}(x-y)^q\sum _{j=0}^q(-1)^{q-j}\left( {\begin{array}{c}q\\ j\end{array}}\right) (B_nf)\left( \frac{jx+(q-j)y}{q}\right) \ge 0. \end{aligned}$$

(1.6)

The identity analogous to (1.3) for a general q is easily proved, applying the Vandermonde’s identity

$$\begin{aligned} \sum _{\begin{array}{c} \nu _1,\dots ,\nu _q=0\\ \nu _1+\cdots +\nu _q=k \end{array}}^n\left( \prod _{i=1}^q\left( {\begin{array}{c}n\\ \nu _i\end{array}}\right) \right) =\left( {\begin{array}{c}qn\\ k\end{array}}\right) . \end{aligned}$$

Namely,

$$\begin{aligned} (B_{qn}f)(u)=\sum _{\nu _1,\dots ,\nu _q=0}^n\left( \prod _{i=1}^q p_{n,\nu _i}(u)\right) f\left( \frac{\nu _1+\cdots +\nu _q}{qn}\right) ,\quad u\in [0,1]. \nonumber \\ \end{aligned}$$

(1.7)

Hence, one may ask whether, for $q>2$, a stronger inequality (analogous to (1.4)) is valid. That is, is it true that for a q-monotone f,

$$\begin{aligned}&\mathrm {sgn}(x-y)^q\sum _{j=1}^{q-1}(-1)^{q-j+1}\left( {\begin{array}{c}q\\ j\end{array}}\right) (B_{qn}f)\left( \frac{jx+(q-j)y}{q}\right) \nonumber \\&\quad \ge \mathrm {sgn}(x-y)^q\sum _{\nu _1,\dots ,\nu _q=0}^{n}\sum _{j=1}^{q-1}(-1)^{q-j+1} \left( {\begin{array}{c}q\\ j\end{array}}\right) \nonumber \\&\qquad \times \left( \prod \limits _{i=1}^jp_{n,\nu _{i}}(x)\right) \left( \prod \limits _{i=j+1}^{q}p_{n,\nu _{i}}(y)\right) f\left( \frac{\nu _{1}+\cdots +\nu _{q}}{qn}\right) ? \end{aligned}$$

(1.8)

In Sect. 2, we give an affirmative answer to this question. Namely, we prove

Theorem 1.1

If $f\in C[0,1]$ is q-monotone, $q\ge 3$, then (1.8) holds for all $n\in {\mathbb {N}}$.

Since, for $f\in C[0,1]$, $\lim _{n\rightarrow \infty }(B_nf)(x)=f(x)$, uniformly in $x\in [0,1]$, evidently, if (1.6) holds for any pair $x,y\in [0,1]$, for some subsequence $\{n_k\}_{k=1}^\infty$, then f is q-monotone in [0, 1]. Thus, one may ask whether this statement is true also for the weaker inequality (1.5). In Sect. 3, we answer this question affirmatively. Namely, in Sect. 3, we prove

Theorem 1.2

Let $q\ge 1$ and $f\in C[0,1]$. If for any pair $x,y\in [0,1]$, inequality (1.5) holds for a subsequence $\{n_k\}_{k=1}^\infty$, then f is q-monotone in [0, 1].

In view of the above, we propose the following open question. Prove or disprove that if $f\in C[0,1]$ and (1.8) holds for all $n\ge 1$, then f is q-monotone. We note that the answer is unknown even for $q=2$.

Remark 1.3

The above notwithstanding, if (1.8) with $q=2$, that is, (1.4) holds for $f\in C[0,1]$, which is twice differentiable in (0, 1), then, we nevertheless, can prove that f is convex. In fact, it suffices that for each pair $x,y\in [0,1]$, (1.4) holds for some subsequence $\{n_k\}$.

Proof

To this end, we substitute $x=0$, $y=2t$, $0<t\le \frac{1}{2}$, into (1.4), to obtain for $n=n_k$,

$$\begin{aligned} (B_{2n}f)(t)\ge \sum _{j=0}^np_{n,j}(2t){f}\left( \frac{j}{2n}\right) =\left( B_nf\left( \frac{\cdot }{2}\right) \right) (2t). \end{aligned}$$

Hence,

$$\begin{aligned} 2n\left[ (B_{2n}f)(t)-f(t)\right] \ge 2n\left[ \left( B_nf\left( \frac{\cdot }{2}\right) \right) ( 2t)-f\left( \frac{\cdot }{2}\right) (2t)\right] . \end{aligned}$$

By Voronovskaja’s theorem for the classical Bernstein polynomials, passing to the limit on $\{n_k\}$, yields

$$\begin{aligned} \frac{t(1-t)}{2}f''(t)\ge 2\frac{2t(1-2t)}{2}\frac{1}{4}f''(t), \end{aligned}$$

which in turn implies $f''(t)\ge 0$, $0<t\le \frac{1}{2}$.

Repeating the same with $g(s):=f(1-s)$, yields that $f''(t)\ge 0$, $\frac{1}{2}\le ~t<1$.

Hence, we conclude that f is convex in [0, 1]. $\square$

Final comment, note that, in our proof, we only used (1.4) with $x=0$ and $x=1$.

2 Proof of Theorem 1.1

We begin with some preparatory lemmas.

Lemma 2.1

If

$$\begin{aligned} P(u,v):=\frac{1}{(u-v)^q}\sum _{k=0}^q(-1)^{q-k}\left( {\begin{array}{c}q\\ k\end{array}}\right) \left( \frac{ku+(q-k)v}{q}\right) ^{qn},\quad u\ne v, \end{aligned}$$

(2.1)

then it is a homogeneous polynomial of total degree $q(n-1)$, in the variables u and v, with non-negative coefficients.

Proof

Denote $s:=qn$ and rewrite

$$\begin{aligned} P(u,v)=\frac{1}{(u-v)^q}\sum _w(-1)^{\sigma (w)}\left( \frac{w_1+\cdots +w_q}{q}\right) ^s, \end{aligned}$$

where the sum is over $2^q$ possible $w=\{w_1,\dots ,w_q\}$, with the variables $w_1,\dots ,w_q$ taking the values u and v, and for each w, $\sigma (w)$ is the number of $w_i$’s that take the value v.

Expanding the s power, applying the multinomial expansion, yields,

$$\begin{aligned} P(u,v)&=\frac{1}{(u-v)^q}\frac{1}{q^s}\sum _{m_1+\cdots +m_q=s}\left( {\begin{array}{c}s\\ m_1,\dots ,m_q\end{array}}\right) \sum _w(-1)^{\sigma (w)}w_1^{m_1}\cdots w_q^{m_q}\nonumber \\&=\frac{1}{q^s}\sum _{m_1+\cdots +m_q=s}\left( {\begin{array}{c}s\\ m_1,\dots ,m_q\end{array}}\right) \prod _{i=1}^q\frac{u^{m_i}-v^{m_i}}{u-v}\,. \end{aligned}$$

(2.2)

Since $\frac{u^{m_i}-v^{m_i}}{u-v}$ is a homogeneous polynomial in u and v of total degree $m_i-1$, the lemma follows. $\square$

Thus, we may write

$$\begin{aligned} P(u,v)=:\sum _{i=0}^{q(n-1)}a_iu^iv^{q(n-1)-i}, \end{aligned}$$

(2.3)

where all $a_i\ge 0$, $0\le i\le q(n-1)$.

We will compare the coefficients of P to the coefficients of the homogeneous polynomial of total degree $q(n-1)$, in the variables u and v,

$$\begin{aligned} Q(u,v):=\left( \frac{u^n-v^n}{u-v}\right) ^q=:\sum _{i=0}^{q(n-1)}b_iu^iv^{q(n-1)-i}. \end{aligned}$$

(2.4)

We will prove that

$$\begin{aligned} 0\le a_i\le b_i\quad \text {for all}\quad i=0,\dots ,q(n-1). \end{aligned}$$

(2.5)

To this end, we observe that, by (2.2), P(u, v) is a weighted average of the polynomials $\prod _{i=1}^q\frac{u^{m_i}-v^{m_i}}{u-v}$. Hence, it suffices to prove that, coefficient-by-coefficient, they maximize when $m_1=\cdots =m_q=n$.

Without loss of generality we may assume that $v=1$ and $u\ne 1$.

Lemma 2.2

Let $1\le m<l-1$. Then the coefficients of the polynomial $\frac{(u^m-1)(u^l-1)}{(u-1)^2}$ are no bigger than those of $\frac{(u^{m+1}-1)(u^{l-1}-1)}{(u-1)^2}$.

Proof

For any $1\le m\le l$,

$$\begin{aligned} \frac{(u^m-1)(u^l-1)}{(u-1)^2}&=\left( \sum _{j=0}^{l-1}u^j\right) \left( \sum _{i=0}^{m-1}u^i\right) =\sum _{j=0}^{l-1}\sum _{i=0}^{m-1}u^{i+j}\\&=\sum _{j=0}^{l-1}\sum _{k=j}^{j+m-1}u^k =\sum _{k=0}^{m+l-2}u^k\left( \sum _{j=\max \{0,k-m+1\}}^{\min \{k,l-1\}}1\right) \\&=\sum _{k=0}^{m-1}(k+1)u^k+\sum _{k=m}^{l-1}u^k\left( \sum _{j=k-m+1}^k1\right) \\&\quad +\sum _{k=l}^{m+l-2}u^k\left( \sum _{j=k-m+1}^{l-1}1\right) \\&=\sum _{k=0}^{m-1}(k+1)u^k+m\sum _{k=m}^{l-1}u^k+\sum _{k=l}^{m+l-2}(m+l-k-1)u^k. \end{aligned}$$

If $m<l-1$, then $m+1\le l-1$, so we may replace in the above m by $m+1$ and l by $l-1$, and obtain

$$\begin{aligned} \frac{(u^{m+1}-1)(u^{l-1}-1)}{(u-1)^2}&=\sum _{k=0}^m(k+1)u^k+(m+1)\sum _{k=m+1}^{l-2}u^k\\&\quad +\sum _{k=l-1}^{m+l-2}(m+l-k-1)u^k\\&=\sum _{k=0}^{m-1}(k+1)u^k+(m+1)\sum _{k=m}^{l-2}u^k+mu^{l-1}\\&\quad +\sum _{k=l}^{m+l-2}(m+l-k-1)u^k. \end{aligned}$$

Comparing the two equations proves the lemma. $\square$

Lemma 2.3

Let $s=qn$ and $m_1+\cdots +m_q=s$. Then the biggest coefficients in the expansion of $P(u):=\frac{(u^{m_1}-1)\cdots (u^{m_q}-1)}{(u-1)^q}$ are obtained for $m_1=\cdots =m_q=n$.

Proof

Suppose that for some k, $m_k\ne n$. Then, there exist two indices such that $m_i<n<m_j$. By Lemma 2.2, all coefficients of the polynomial $\frac{(u^{m_1}-1)\cdots (u^{m_i+1}-1)\cdots (u^{m_j-1}-1)\cdots (u^{m_q}-1)}{(u-1)^q}$ are no smaller than those of P(u). As long as there is an $m_k\ne n$, we continue. This completes the proof. $\square$

Corollary 2.4

The difference

$$\begin{aligned} D:&=\sum _{k=1}^{q-1}(-1)^{q-k+1}\left( {\begin{array}{c}q\\ k\end{array}}\right) \left[ \left( \frac{ku+(q-k)v}{q}\right) ^{qn}-u^{kn}v^{(q-k)n}\right] , \end{aligned}$$

is divisible by $(u-v)^q$, and the resulting polynomial is a homogeneous polynomial of total degree $q(n-1)$ in the variables u and v, with nonnegative coefficients.

Proof

Note that

$$\begin{aligned} D&=\sum _{k=0}^q(-1)^{q-k}\left( {\begin{array}{c}q\\ k\end{array}}\right) \left[ u^{kn}v^{(q-k)n}-\left( \frac{ku+(q-k)v}{q}\right) ^{qn}\right] \\&=(u-v)^q\left[ Q(u,v)-P(u,v)\right] . \end{aligned}$$

Hence, D is divisible by $(u-v)^q$, and the assertion that the coefficients are nonnegative follows by (2.5). $\square$

Proof of Theorem 1.1

We are ready to prove (1.8).

Using the straightforward representation

$$\begin{aligned} p_{n,i}(u)=\left. \frac{1}{i!}\left( \frac{\partial }{\partial z}\right) ^i(1+uz)^n\right| _{z=-1}, \end{aligned}$$

by [3, Sect. 2], we have the following equations.

$$\begin{aligned}&(B_{qn}f)\left( \frac{jx+(q-j)y}{q}\right) \\&=\sum _{k=0}^{qn}\frac{1}{k!}\left. \left[ \left( \frac{\partial }{\partial z}\right) ^k \left( \frac{j(1+xz)+(q-j)(1+yz)}{q}\right) ^{qn}\right] \right| _{z=-1}f\left( \frac{k}{qn}\right) \\&=:J_{1,j},\quad 1\le j\le q-1, \end{aligned}$$

and

$$\begin{aligned}&\sum _{\nu _1,\dots ,\nu _q=0}^n\left( \prod \limits _{i=1}^jp_{n,\nu _{i}}(x)\right) \left( \prod _{i=j+1}^qp_{n,\nu _{i}}(y)\right) f\left( \frac{\nu _1+\cdots +\nu _q}{qn}\right) \\&\quad =\sum _{k=0}^{qn}\frac{1}{k!}\left. \left[ \left( \frac{\partial }{\partial z}\right) ^k\left( (1+xz)^j(1+yz)^{q-j}\right) ^n\right] \right| _{z=-1}f\left( \frac{k}{qn}\right) \\&\quad =:J_{2,j},\quad 1\le j\le q-1. \end{aligned}$$

Thus,

$$\begin{aligned}&\sum _{j=1}^{q-1}(-1)^{q-j+1}\left( {\begin{array}{c}q\\ j\end{array}}\right) (J_{1,j}-J_{2,j})=\sum _{k=0}^{qn}\frac{1}{k!}f\left( \frac{k}{qn}\right) \nonumber \\&\quad \left( \frac{\partial }{\partial z}\right) ^k\left\{ \sum _{j=1}^{q-1}(-1)^{q-j+1}\left( {\begin{array}{c}q\\ j\end{array}}\right) \biggl [\left( \frac{j(1+xz)+(q-j)(1+yz)}{q}\right) ^{qn}\right. \nonumber \\&\qquad \left. \left. -\left( (1+xz)^j(1+yz)^{q-j}\right) ^n\biggr ]\right\} \right| _{z=-1}\nonumber \\&\quad =:\sum _{k=0}^{qn}\frac{1}{k!}\left. \frac{d^kh_{x,y}(z)}{dz^k}\right| _{z=-1}f\left( \frac{k}{qn}\right) =:I. \end{aligned}$$

(2.6)

Denoting $1+xz:=u$ and $1+yz:=v$, we conclude by Corollary 2.4 that,

$$\begin{aligned} h_{x,y}(z)=:(x-y)^qz^q g_{x,y}(z), \end{aligned}$$

where

$$\begin{aligned} g_{x,y}(z)=\sum _{j=0}^{q(n-1)}c_j(1+xz)^j(1+yz)^{q(n-1)-j}, \end{aligned}$$

is a polynomial in z of degree $q(n-1)$, and $c_j\ge 0$, $0\le j\le q(n-1)$. Note that all derivatives of $g_{x,y}(z)$, are nonnegative at $z=-1$.

Indeed, the kth derivative (with respect to z), $k\ge 0$, of the jth term in the above sum is,

$$\begin{aligned} c_jk!\sum _{m=0}^k\left( {\begin{array}{c}j\\ m\end{array}}\right) \left( {\begin{array}{c}q(n-1)-j\\ k-m\end{array}}\right) x^my^{k-m}(1+xz)^{j-m}(1+yz)^{q(n-1)-j-(k-m)}. \end{aligned}$$

Substituting $z=-1$, we have

$$\begin{aligned} c_jk!\sum _{m=0}^k\left( {\begin{array}{c}j\\ m\end{array}}\right) \left( {\begin{array}{c}q(n-1)-j\\ k-m\end{array}}\right) x^my^{k-m}(1-x)^{j-m}(1-y)^{q(n-1)-j-(k-m)}\ge 0, \end{aligned}$$

since $c_j\ge 0$ and $0\le x,y\le 1$.

Now,

$$\begin{aligned}&\left. \frac{d^kh_{x,y}(z)}{dz^k}\right| _{z=-1}\\&\,=(x-y)^q\sum _{j=0}^q(-1)^{q-j}\left( {\begin{array}{c}k\\ j\end{array}}\right) \frac{q!}{(q-j)!}g_{x,y}^{(k-j)}(-1). \end{aligned}$$

Substituting into (2.6), we obtain for f which is q-monotone,

$$\begin{aligned} (x-y)^qI=(x-y)^{2q}\sum _{k=0}^{q(n-1)}\frac{1}{k!}g_{x,y}^{(k)}(-1)\Delta ^qf\left( \frac{k}{qn}\right) \ge 0, \end{aligned}$$

where $\Delta$ is the forward difference, namely, for the sequence $\{a_k\}$, $\Delta ^1a_k:=\Delta a_k:=a_{k+1}-a_k$, and $\Delta ^{m+1}a_k:=\Delta (\Delta ^m a_k)$.

Therefore, (1.8) is proved, and the proof of Theorem 1.1 is complete. $\square$

3 Proof of Theorem 1.2

We have,

$$\begin{aligned} I_n&:=\sum _{\nu _{1},\ldots ,\nu _{q}=0}^{n}\sum _{j=0}^{q}\left( -1\right) ^{q-j}\left( {\begin{array}{c}q\\ j\end{array}}\right) \prod _{i=1}^{j}p_{n,\nu _{i}}(x) \prod _{i=j+1}^{q}p_{n,\nu _{i}}(y)f\left( \frac{\nu _{1}+\cdots +\nu _{q}}{qn}\right) \\&=\sum _{j=0}^{q}\left( -1\right) ^{q-j}\left( {\begin{array}{c}q\\ j\end{array}}\right) \sum _{k=0}^{qn}\sum _{\begin{array}{c} r,s\ge 0 \\ r+s=k \end{array}} \left( \sum _{\begin{array}{c} \nu _{1}+\cdots +\nu _{j}=r \\ \nu _{j+1}+\cdots +\nu _{q}=s \end{array}} \left( \prod _{i=1}^{q}\left( {\begin{array}{c}n\\ \nu _{i}\end{array}}\right) \right) \right) \\&\quad \times x^r(1-x)^{jn-r}y^s(1-y)^{(q-j)n-s}f\left( \frac{r+s}{qn}\right) . \end{aligned}$$

Applying the generalized Vandermonde identity we obtain

$$\begin{aligned} \sum _{\begin{array}{c} \nu _{1}+\cdots +\nu _{j}=r \\ \nu _{j+1}+\cdots +\nu _{q}=s \end{array}}\prod _{i=1}^{q}\left( {\begin{array}{c}n\\ \nu _{i}\end{array}}\right) =\left( {\begin{array}{c}jn\\ r\end{array}}\right) \left( {\begin{array}{c}\left( q-j\right) n\\ s\end{array}}\right) . \end{aligned}$$

Hence,

$$\begin{aligned} I_n=\sum _{j=0}^{q}\left( -1\right) ^{q-j}\left( {\begin{array}{c}q\\ j\end{array}}\right) \sum _{r=0}^{jn}\sum _{s=0}^{\left( q-j\right) n}p_{jn,r}\left( x\right) p_{\left( q-j\right) n,s}\left( y\right) f\left( \frac{r+s}{qn}\right) . \end{aligned}$$

Define

$$\begin{aligned} g_{j}\left( x,y\right) =f\left( \frac{jx+\left( q-j\right) y}{q}\right) ,\quad 0\le j\le q. \end{aligned}$$

Then,

$$\begin{aligned} g_0(x,y)=f(y)\quad \text {and}\quad g_q(x,y)=f(x), \end{aligned}$$

and for $1\le j\le q-1$,

$$\begin{aligned} g_{j}\left( \frac{r}{jn},\frac{s}{\left( q-j\right) n}\right) =f\left( \frac{j\frac{r}{jn}+\left( q-j\right) \frac{s}{\left( q-j\right) n}}{q}\right) =f\left( \frac{r+s}{qn}\right) . \end{aligned}$$

Hence,

$$\begin{aligned} I_n&=(B_{qn}f)(x)+(-1)^q(B_{qn}f)(y)\\&\quad +\sum _{j=1}^{q-1}\left( -1\right) ^{q-j}\left( {\begin{array}{c}q\\ j\end{array}}\right) \sum _{r=0}^{jn}\sum _{s=0}^{\left( q-j\right) n}p_{jn,r}\left( x\right) p_{\left( q-j\right) n,s}\left( y\right) g_{j}\left( \frac{r}{jn},\frac{s}{\left( q-j\right) n}\right) \\&=(B_{qn}f)(x)+(-1)^q(B_{qn}f)(y)+\sum _{j=1}^{q-1}\left( -1\right) ^{q-j}\left( {\begin{array}{c}q\\ j\end{array}}\right) (B_{jn,(q-j)n}g_j)(x,y), \end{aligned}$$

where, for a function g(x, y) of two variables, $(B_{\mu ,\kappa }g)(x,y)$ is the 2-dimensional tensor Bernstein polynomial.

It is well known, see, e.g., [4, p. 122], and actually goes back to Bernstein in Soc. Math. Charkow 13 (1912-13), that if g is continuous in $[0,1]\times [0,1]$, then

$$\begin{aligned} \lim _{\min \{\mu ,\kappa \}\rightarrow \infty }(B_{\mu ,\kappa }g)(x,y)=g(x,y)\quad \text {uniformly in}\quad [0,1]\times [0,1]. \end{aligned}$$

Since, by the assumptions of Theorem 1.2, inequality (1.5) is valid for a subsequence $\{n_k\}_{k=1}^\infty$, i.e., $(x-y)^q I_{n_k}\ge 0$, $k\ge 1$, we get

$$\begin{aligned} \lim _{k\rightarrow \infty }(x-y)^qI_{n_k}&=(x-y)^q\sum _{j=0}^{q}\left( -1\right) ^{q-j}\left( {\begin{array}{c}q\\ j\end{array}}\right) g_{j}\left( x,y\right) \\&=(x-y)^q\sum _{j=0}^{q}\left( -1\right) ^{q-j}\left( {\begin{array}{c}q\\ j\end{array}}\right) f\left( \frac{jx+\left( q-j\right) y}{q}\right) \ge 0, \end{aligned}$$

so that f is q-monotone on [0, 1]. This completes the proof of Theorem 1.2.