1 Introduction and the Main Result

Given a function f defined on [0, 1], the classical Bernstein polynomials associated with it are defined by [5],

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

where

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

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).

The first two authors proved in [2] that if \(f\in C[0,1]\) is a q-monotone function, then for \(q,n\in {\mathbb {N}}\) and all \(x,y\in [0,1]\),

$$\begin{aligned}&\textrm{sgn}(x-y)^q\sum _{\nu _1,\dots ,\nu _q=0}^{n}\sum _{j=0}^q(-1)^{q-j} \left( {\begin{array}{c}q\\ j\end{array}}\right) \left( \prod \limits _{i=1}^{j}p_{n,\nu _{i}}(x)\right) \nonumber \\&\quad \times \left( \prod \limits _{i=j+1}^{q}p_{n,\nu _{i}}(y)\right) f\left( \frac{\nu _{1}+\cdots +\nu _{q}}{qn}\right) \ge 0. \end{aligned}$$
(1.1)

This result extended what had been originally a conjecture by Raşa concerning convex functions in C[0, 1], that was first proved using probabilistic methods in [6] and then in [1] using analytic approach.

Recently, the authors [4] proved that if \(f\in C[0,1]\) satisfies (1.1) for some \(q\in {\mathbb {N}}\) and a subsequence \(\{n_k\}_{k=1}^\infty \), then f is q-monotone in [0, 1].

It is well known (see, e.g., [5]*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 \(x,y\in [0,1]\),

$$\begin{aligned} \textrm{sgn}(x-y)^q\sum _{j=0}^q(-1)^{q-j}\left( {\begin{array}{c}q\\ j\end{array}}\right) (B_nf)\left( \frac{jx+(q-j)y}{q}\right) \ge 0. \end{aligned}$$
(1.2)

Applying the Vandermonde’s identity,

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

it readily follows that,

$$\begin{aligned} (B_{qn}f)(u)=\sum _{\nu _1,\dots ,\nu _q=0}^n\left( \prod _{i=1}^q p_{n,\nu _i}(u)\right) f\left( \frac{\nu _1+\cdots +\nu _q}{qn}\right) ,\quad u\in [0,1]. \end{aligned}$$
(1.3)

In [4] we proved the stronger assertion that if \(f\in C[0,1]\) is q-monotone, for some \(q\ge 3\) (for \(q=2\) see [3]), then

$$\begin{aligned}&\textrm{sgn}(x-y)^q\sum _{j=1}^{q-1}(-1)^{q-j+1}\left( {\begin{array}{c}q\\ j\end{array}}\right) (B_{qn}f)\left( \frac{jx+(q-j)y}{q}\right) \nonumber \\&\quad \ge \textrm{sgn}(x-y)^q\sum _{\nu _1,\dots ,\nu _q=0}^{n}\sum _{j=1}^{q-1}(-1)^{q-j+1} \left( {\begin{array}{c}q\\ j\end{array}}\right) \nonumber \\&\qquad \times \left( \prod \limits _{i=1}^jp_{n,\nu _{i}}(x)\right) \left( \prod \limits _{i=j+1}^{q}p_{n,\nu _{i}}(y)\right) f\left( \frac{\nu _{1}+\cdots +\nu _{q}}{qn}\right) . \end{aligned}$$
(1.4)

We also asked whether for \(f\in C[0,1]\), (1.4) holds for all \(n\ge 1\), implies that f is q-monotone. We noted that the answer is unknown even for \(q=2\), but we remarked (see [4]*Remark 1.3) that if (1.4) with \(q=2\), holds for \(f\in C[0,1]\), which is twice differentiable in (0, 1), then we can prove that f is convex. In fact, it suffices that for every pair \(x,y\in (0,1)\), (1.4) holds for a subsequence \(\{n_k\}_{k=1}^\infty \).

It is the purpose of this note to give, necessary and sufficient conditions involving (1.4) for \(f\in C[0,1]\) to be q-monotone, when \(q\ge 3\).

We first prove

Theorem 1.1

Given \(q\ge 4\). If \(f\in C[0,1]\) is twice differentiable in (0, 1) and for any pair \(x,y\in (0,1)\), (1.4) holds for some subsequence \(\{n_k\}_{k=1}^\infty \), then f is q-monotone in [0, 1]. Conversely, if \(f\in C[0,1]\) is q-monotone in [0, 1], then f is twice differentiable in (0, 1) and (1.4) holds for all \(n\in {\mathbb {N}}\). In fact, \(f''\) is absolutely continuous in every closed subinterval \([a,b]\subset (0,1)\).

Second, we prove

Theorem 1.2

Let \(q=3\). Necessary and sufficient conditions for \(f\in C[0,1]\) to be 3-monotone are that \(f'\) is absolutely continuous in every closed subinterval \([a,b]\subset (0,1)\), and for any pair \(x,y\in (0,1)\), (1.4) with \(q=3\), holds for some subsequence \(\{n_k\}_{k=1}^\infty \).

Proof of Theorem 1.1

We begin with the proof of the necessity, because it is straightforward. We observe that if f is q-monotone in [0, 1], then \(f^{(q-2)}\) is convex in (0, 1) and \(q-2\ge 2\), so that \(f''\) is absolutely continuous in every subinterval \([a,b]\subset (0,1)\). The validity of (1.4) for all \(n\in {\mathbb {N}}\), follows by [4]*Theorem 1.1.

Conversely, assume that \(q\ge 4\) and \(x,y\in (0,1)\), and denote

$$\begin{aligned} g(x_1,\ldots ,x_q)=f\left( \frac{x_1+\cdots +x_q}{q}\right) . \end{aligned}$$

Then,

$$\begin{aligned}&\sum _{\nu _1,\dots ,\nu _q=0}^n\left( \prod \limits _{i=1}^jp_{n,\nu _i}(x)\right) \left( \prod \limits _{i=j+1}^qp_{n,\nu _i}(y)\right) f\left( \frac{\nu _1+\cdots +\nu _q}{qn}\right) \nonumber \\&\quad =({\mathcal {B}}_ng)(\underset{j}{\underbrace{x,\ldots ,x}},\underset{q-j}{\underbrace{y,\ldots ,y}}), \end{aligned}$$
(1.5)

where \({\mathcal {B}}_n\) denotes the q-dimensional tensor product Bernstein polynomial.

By Voronovskaya formula for tensor Bernstein polynomials,

$$\begin{aligned}&\lim _{n\rightarrow \infty }n\left( \left( {\mathcal {B}}_{n}g\right) \left( x_{1},\ldots ,x_{q}\right) -g\left( x_{1},\ldots ,x_{q}\right) \right) \nonumber \\&\quad =\sum _{\ell =1}^{q}\frac{x_{\ell }\left( 1-x_{\ell }\right) }{2}\frac{ \partial ^2g}{\partial ^2x_\ell }(x_1,\ldots ,x_q), \end{aligned}$$
(1.6)

provided the derivatives on the right-hand side exist.

Since \(\frac{\partial ^2g}{\partial ^2x_\ell }(x_1,\ldots ,x_q)=\left( \frac{1}{q}\right) ^2f'' \left( \frac{x_1+\cdots +x_q}{q}\right) \), for \(0<\frac{x_1+\cdots +x_q}{q}<1\), we obtain

$$\begin{aligned}&\lim _{n\rightarrow \infty }n\bigl (({\mathcal {B}}_ng)( \underset{j}{\underbrace{x,\ldots ,x}},\underset{q-j}{\underbrace{y,\ldots ,y}})-g(\underset{j}{\underbrace{x,\ldots ,x}}, \underset{q-j}{\underbrace{y,\ldots ,y}})\bigr )\\&\quad =\left( j\frac{x\left( 1-x\right) }{2}+\left( q-j\right) \frac{y\left( 1-y\right) }{2}\right) \left( \frac{1}{q}\right) ^2f^{\prime \prime }\left( \frac{jx+(q-j)y}{q}\right) . \end{aligned}$$

Hence, by (1.5) and (1.6), we have

$$\begin{aligned}&\lim _{n\rightarrow \infty }qn\biggl [\sum _{\nu _1,\dots ,\nu _q=0}^{n}\left( \prod \limits _{i=1}^{j}p_{n,\nu _i}(x)\right) \left( \prod \limits _{i=j+1}^q p_{n,\nu _i}(y)\right) f\left( \frac{\nu _1+\cdots +\nu _q}{qn}\right) \\&\qquad -f\left( \frac{jx+(q-j)y}{q}\right) \biggr ]\\&\quad =q\left( j\frac{x(1-x)}{2}+(q-j)\frac{y(1-y)}{2}\right) \left( \frac{1}{q}\right) ^2 f''\left( \frac{jx+(q-j)y}{q}\right) . \end{aligned}$$

At the same time, by the ordinary Voronovskaya formula for the one dimensional Bernstein polynomials, for each \(0\le j\le q\), we have

$$\begin{aligned}&\lim _{n\rightarrow \infty }qn\left[ (B_{qn}f)\left( \frac{jx+(q-j)y}{q}\right) -f\left( \frac{jx+(q-j)y}{q}\right) \right] \\&\quad =\frac{jx+(q-j)y}{2q}\left( 1-\frac{jx+(q-j)y}{q}\right) f''\left( \frac{jx+(q-j)y}{q}\right) . \end{aligned}$$

Substituting the last two formulae into (1.4), we conclude that

$$\begin{aligned}&\textrm{sgn}(x-y)^q\sum _{j=0}^{q}(-1)^{q-j+1}\left( {\begin{array}{c}q\\ j\end{array}}\right) \left[ \frac{jx+(q-j)y}{q}\left( 1-\frac{jx+(q-j)y}{q}\right) \right. \nonumber \\&\quad \left. -\left( \frac{j}{q}x(1-x)+\left( 1-\frac{j}{q}\right) y(1-y)\right) \right] f''\left( \frac{jx+(q-j)y}{q}\right) \ge 0. \end{aligned}$$
(1.7)

Now,

$$\begin{aligned}&\frac{jx+(q-j)y}{q}\left( 1-\frac{jx+(q-j)y}{q}\right) -\left( \frac{j}{q} x(1-x)+\left( 1-\frac{j}{q}\right) y(1-y)\right) \\&\quad =\frac{(x-y)^2}{q^2}j(q-j), \end{aligned}$$

and

$$\begin{aligned} \left( {\begin{array}{c}q\\ j\end{array}}\right) j(q-j)=q(q-1)\left( {\begin{array}{c}q-2\\ j-1\end{array}}\right) ,\quad 1\le j\le q-1. \end{aligned}$$

Hence, (1.7) becomes,

$$\begin{aligned} \textrm{sgn}(x-y)^q(x-y)^2\sum _{j=1}^{q-1}(-1)^{q-j+1} \frac{q-1}{q}\left( {\begin{array}{c}q-2\\ j-1\end{array}}\right) f''\left( \frac{jx+(q-j)y}{q}\right) \ge 0, \end{aligned}$$

so that,

$$\begin{aligned} \textrm{sgn}(x-y)^q\sum _{j=0}^{q-2}(-1)^{q-j}\left( {\begin{array}{c}q-2\\ j\end{array}}\right) f''\left( \frac{(j+1)x+(q-1-j)y}{q}\right) \ge 0. \end{aligned}$$
(1.8)

For any \(z\in (0,1)\), let \(h>0\) be such that \(h<\min \{z,\frac{1-z}{q-1}\}\). Let \(x:=z+(q-1)h\) and \(y:=z-h\). Then it follows that

$$\begin{aligned} \Delta ^{q-2}_hf''(z)=\sum _{j=0}^{q-2}(-1)^{q-2-j}\left( {\begin{array}{c}q-2\\ j\end{array}}\right) f''(z+jh)\ge 0. \end{aligned}$$
(1.9)

This shows that \(f''\) is \((q-2)\)-monotone in (0, 1), which implies that f is q-monotone in [0, 1].

Indeed, evidently,

$$\begin{aligned} \Delta ^q_hf(z)=\Delta ^{q-2}_h(\Delta ^2_hf)(z)&=\Delta ^{q-2}_h\left( \int _z^{z+h}\int _t^{t+h}f''(u)dudt\right) \\&=\int _z^{z+h}\int _t^{t+h}\Delta ^{q-2}_hf''(u)dudt\ge 0. \end{aligned}$$

This completes the proof. \(\square \)

Proof of Theorem 1.2

Again, we begin with the necessity. If f is 3-monotone, then \(f'\) is convex in (0, 1), whence, it is absolutely continuous in every subinterval \([a,b]\subset (0,1)\). The validity of (1.4) with \(q=3\), for all \(n\in {\mathbb {N}}\), follows by [4]*Theorem 1.1.

Conversely, take \(x,y\in (0,1)\) such that \(f''\left( \frac{x+2y}{3}\right) \) and \(f''\left( \frac{2x+y}{3}\right) \) exist, that is, almost all x, y. Since Voronovskaya’s formula is valid whenever \(f''\) exists, following the proof of Theorem 1.1, we obtain (1.8) with \(q=3\), for almost all \(x,y\in (0,1)\).

Hence, for \(z\in (0,1)\), and \(0<h<\min \{z,\frac{1-z}{2}\}\) such that x, y are as above, that is, for almost all h, we get by (1.9)

$$\begin{aligned} \Delta _hf''(z)=f''(z+h)-f''(z)\ge 0. \end{aligned}$$

Thus, \(f''\) (defined almost everywhere) is monotone nondecreasing in (0, 1), which implies that \(f'\) is convex in (0, 1), that is, f is 3-monotone in (0, 1), and by continuity in [0, 1]. \(\square \)