Notes on Higher-Order Convex Functions (III)

1.
- —
  
  Either $f(x)$ a non-concave function (of order 1) in the interval ( $a,b$ ). This function is characterized by the inequality

\left[x_{1},x_{2},x_{3};f\right]\geq 0,\left({}^{1}\right)

(1)

verified by any group of three distinct points $x_{1},x_{2},x_{3}$ of $(a,b)$ .
From (1) follow other well-known inequalities. Such is Jensen's inequality ( ² )

f\left(\frac{\Sigma p_{i}x_{i}}{\Sigma p_{i}}\right)\leqq\frac{\Sigma p_{i}f\left(x_{i}\right)}{\Sigma p_{i}},\quad p_{i}\geqq 0,\quad\Sigma p_{i}>0

(2)

mM G. Hardy, JE Littlewood and G. Polya determined all inequalities of the form

\sum_{i=1}^{m}f\left(x_{i}\right)\geqq\sum_{i=1}^{m}f\left(y_{i}\right)\left({}^{3}\right)

(3)

$\left({}^{1}\right)$ For notations see my earlier works.
( ² ) J.L.W.V. JENSEN, »On convex functions and inequalities between mean values. Acta Math., 30, 175-193 (1906).
( ³ ) G.H. Hardy, J.E. LITTLEWOOD, G. POLYA, »Some simple inequalities satisfied by convex functions. Mess. of Math. 58, 145-152 (1929). See also J. Karamata, »On an inequality relating to convex functions. Publ. Math. Univ. Belgrade, 1, 145-148 (1932).

ON HIGHER ORDER CONVEX FUNCTIONS

In particular, MK Toda demonstrated that

\frac{1}{m}\sum_{i=1}^{m}f\left(x_{i}\right)\geqq\frac{1}{m-1}\sum_{=1}^{m-1}f\left(y_{i}\right)

(4)

if $y_{i}$ are the zeros of the derivative of the polynomial $\prod_{i=1}^{m}\left(x-x_{i}\right)\left({}^{4}\right)$ . This result has already been established for the particular function $f(x)=x^{p}$ , in the case of $p$ positive integer by MH Bray ( ⁵ , and in the case of $p\geqq 1$ any by MS Kakeya (').
2. - It is easy to find the necessary and sufficient conditions that non-zero constants must fulfill $p_{i}$ and the points $x_{1}<x_{2}<\cdots<x_{m}$ of ( $a,b$ ) so that we have ( $m\geqq 3$ )

\sum_{i=1}^{m}p_{i}f\left(x_{i}\right)\geqq 0,

(5)

whatever the non-concave function $f(x)$ .
We can write

\sum_{i=1}^{m}p_{i}f\left(x_{i}\right)=\mathrm{A}_{0}f\left(x_{1}\right)+\mathrm{A}_{1}\left[x_{1},x_{2};f\right]+\sum_{r=1}^{n-2}\mathrm{C}_{r}\Delta_{2}^{r},

(6)

Or

\Delta_{2}^{r}=\left[x_{r},x_{r+1},x_{r+2};f\right],\quad r=1,2,\ldots,m-2,

the constants $\mathrm{A}_{0},\mathrm{\penalty 10000\ A}_{1},\mathrm{C}_{r}$ are completely determined and are independent of the function $f(x)$ . The necessary and sufficient conditions are

\mathrm{A}_{0}=0,\quad\mathrm{\penalty 10000\ A}_{1}=0,\quad\mathrm{C}_{r}\geqq 0;\quad r\simeq 1,2,\ldots,m-2.

Indeed, we can construct a non-concave function $f(x)$ such as $f\left(x_{1}\right),\left[x_{1},x_{2};f\right]$ have any values and $\Delta_{2}^{r}$ any non-negative values.

To determine the coefficients $\mathbf{A}_{0},\mathbf{A}_{1},\mathbf{C}_{r}$ just choose ( $x$ ) properly. If we first take $f(x)=\alpha x+\beta,\alpha,\beta$

being any constants, we find

\mathrm{A}_{0}=\sum_{i=1}^{m}p_{i},\quad\mathrm{\penalty 10000\ A}_{1}=\sum_{i=1}^{m}p_{i}x_{i}-x_{1}\sum_{i=1}^{m}p_{i}=\sum_{i=2}^{m}p_{i}\left(x_{i}-x_{1}\right).

Let us then take $f(x)=x-x_{r+1}+\left|x-x_{r+1}\right|,\quad 1\leqq r\leqq m-2$ , We have

\left[x_{i},x_{i+1},x_{i+2};f\right]=\begin{cases}0,&i\neq r\\ \frac{2}{x_{r+2}-x_{r}},&i=r\end{cases}

and equality (6) gives us

\mathrm{C}_{r}=\left(x_{r+2}-x_{r}\right)\sum_{i=r+2}^{m}p_{i}\left(x_{i}-x_{r+1}\right),\quad r=1,2,\ldots,m-2.

So we have the following property:
For inequality (5) to be verified for any nonconcave function (of order 1) $f(x)$ , it is necessary and sufficient that we have
(7) $\quad\sum_{i=1}^{m}p_{i}=\sum_{i=1}^{m}p_{i}x_{i}=0,\sum_{i=r+2}^{m}p_{i}\left(x_{i}-x_{r+1}\right)\geqq 0,r=1,2,\ldots,m-2$ .

If these relationships are verified and if moreover $f(x)$ is convex, we have

\sum_{i=1}^{m}p_{i}f\left(x_{i}\right)>0.

The last part of the statement results from the fact that one cannot have $\sum_{i=r+2}^{m}p_{i}\left(x_{i}-x_{r+1}\right)=0,r=1,2,\ldots,m-2$ , THE $p_{i}$ being $\neq 0$ by hypothesis.
3. - Under the term (7) the preceding conditions are not very convenient for applications.

Let us first consider inequality (3). Suppose that the $x_{i}$ are given and let us say that:

THE $y_{1},y_{2},\ldots,y_{m}$ verify condition (C) if we can find $m^{2}$ non-negative numbers $p_{ij}$ such as

1.

$\sum_{i=1}^{m}p_{ij}=\sum_{j=1}^{m}p_{ij}=1,\quad i,j=1,2,\ldots,m$ .
$2^{0}.y_{i}=p_{i1}x_{1}+p_{i2}x_{2}+\cdots+p_{in}x_{m},\quad i=1,2,\ldots,m$ .
Applying Jensen's inequality (2), we see that:
If $y_{1},y_{2},\ldots,y_{m}$ verify condition (C), inequality (3) is verified regardless of the non-concareous function $f(x)$ .

Consider, in ordinary space at $m$ dimensions, the point $\mathrm{P}\left(y_{1},y_{2},\ldots,y_{m}\right)$ of coordinates $y_{1},y_{2},\ldots,y_{m}$ . We will say that this point verifies condition (C) if its coordinates $y_{1}^{\prime},y_{2},\ldots,y_{m}$ verify this condition.

Let's suppose that $x_{1}\leqq x_{2}\leqq\ldots\leqq x_{m},y_{1}\leqq y_{2}\leqq\ldots\leqq y_{m}$ and then let's say that:

THE $y_{1},y_{2},\ldots,y_{m}$ check the condition ( $\mathrm{C}^{\prime}$ ) if we have

\left\{\begin{array}[]{l}x_{1}+x_{2}+\cdots+x_{i}\leqq y_{1}+y_{2}+\cdots+y_{i},i=1,2,\ldots,m-1\\ x_{1}+x_{2}+\cdots+x_{m}=y_{1}+y_{2}+\cdots+y_{m}(\eta).\end{array}\right.

We then have that ⁽⁸⁾
The necessary and sufficient condition for inequality (3) to be verified whatever the non-concave function $f(x)$ is that the $y_{1},y_{2},\ldots,y_{m}$ check the condition ( $\mathrm{C}^{\prime}$ ).
4. - We now propose to demonstrate that the conditions ( C ) and ( $\mathrm{C}^{\prime}$ ) are equivalent ( ⁹ ). To do this it is sufficient to demonstrate that if $y_{1},y_{2},\ldots,y_{m}$ check the condition ( $C^{\prime}$ ) the point $\mathrm{P}\left(y_{1},y_{2},\ldots,y_{m}\right)$ verifies condition (C).

Note that if any number of points $\mathrm{P}_{1},\mathrm{P}_{2},\ldots$ verify condition (C) any point that belongs to the smallest convex domain that contains the points $P_{1},P_{2},\ldots$ also verifies the condition ( C ). We see, on the other hand, that the points P verifying the condition ( $C^{\prime}$ ) form a convex and bounded domain $D$ . Now, the domain D is formed by the hyperplanes

\left\{\begin{array}[]{c}y_{i}=y_{i+1},y_{1}+y_{2}+\cdots+y_{i}=x_{1}+x_{2}+\cdots+x_{i},i=1,2,\ldots,m-1\\ y_{1}+y_{2}+\cdots+y_{m}=x_{1}+x_{2}+\cdots+x_{m}.\end{array}\right.

⁽⁷⁾ The condition (^′ ) can be put in the symmetric form

\begin{gathered}\min\left(x_{i_{1}}+x_{i_{2}}+\ldots+x_{i_{k}}\right)\leqq y_{j_{1}}+y_{j_{2}}+\ldots+y_{j_{k}}\leqq\max\left(x_{i_{1}}+x_{i_{0}}+\ldots+x_{i_{k}}\right)\\ k=1,2,\ldots,m\end{gathered}

where the min and max are relative to all combinations $i_{1},i_{2},\ldots,i_{k}$ numbers $1,2,\ldots,m$ taken $k$ has $k$ And $j_{1},j_{2},\ldots,jk$ are $k$ any numbers in the sequence $1,2,\ldots,m$ .
$\left({}^{8}\right)$ see loc. cit. $\left({}^{9}\right)$ .
( ⁹ ) This property is due to Messrs. Hardy, JE Littlewood and G. Polya. See: „Inequalities" Cambridge Univ. Press, XII-314 pp. (1934). (Note after the correction).

This domain can also be considered as the smallest convex domain containing those intersection points of the hyperplanes (9) which verify the condition ( $C^{\prime}$ ). It is easy to see that there is in all $2^{m-1}$ such points. These points are

(\underbrace{\xi_{1},\xi_{1},\ldots,\xi_{1}}_{i_{1}},\underbrace{\xi_{2},\xi_{2},\ldots,\xi_{2}}_{i_{2}},\ldots,\underbrace{\xi_{k},\xi_{k},\ldots,\xi_{k}}_{i_{k}})

Or
$i_{1}+i_{2}+\cdots+i_{k}=m,\quad\xi_{r}=\frac{1}{i_{r}}\sum_{s=1}^{i_{r}}x_{i_{1}+i_{2}+\cdots+i_{r-1}+s}\quad,\quad r=1,2,\ldots,k$ .
But, all these points obviously verify condition (C). The equivalence of the two conditions ( $C$ ) And ( $C^{\prime}$ ) is therefore demonstrated.
5. - We can therefore state the following property:

The necessary and sufficient condition for inequality (3) to hold, regardless of the non-concareous function $f(x)$ , is that the $y_{1},y_{2},\ldots,y_{m}$ verify condition (C).

We can always put an inequality (b) in the form

\sum_{i=1}^{r}\alpha_{i}f\left(x_{i}\right)\geqq\sum_{i=1}^{s}\beta_{i}f\left(y_{i}\right)\quad(r+s\geqq 3)

(10)

Or $\alpha_{i}>0,\beta_{i}>0,\alpha_{1}+\alpha_{2}+\cdots+\alpha_{r}=\beta_{1}+\beta_{2}+\cdots+\beta_{s}=1,x_{1}<x_{2}<\cdots<x_{r}$ , $y_{1}<y_{2}<\cdots<y_{s}$ . We obtain this inequality of (3) by first assuming that some of the points $x_{i},y_{j}$ come to coincide and then by passages at the limit (if the $\alpha_{i},\beta_{i}$ are not all rational). It is then easy to deduce from the above the following general property:

The necessary and sufficient condition for inequality (10) to hold, regardless of the non-concave function $f(x)$ , is that we can find rs non-negative numbers $p_{ij}$ so that we have
$1^{0}.\sum_{j=1}^{r}p_{ij}=1,\quad i=1,2,\ldots,s$
(11) $2^{0}.\sum_{i=1}^{s}\beta_{i}p_{ij}=\alpha_{j},j=1,2,\ldots,r$
$3^{0}.y_{i}=p_{i1}x_{1}+p_{i2}x_{2}+\cdots+p_{ir}x_{r},\quad i=1,\ldots,s$ .
It is clear that if $f(x)$ is convex it is the sign $>$ which is still suitable in (10).
6. - As an application let us take up inequality (4) of MK Toda.

Consider the polynomial $\mathbf{P}(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\ldots\left(x-x_{m}\right)$ where we can assume that the (real) zeros $x_{1},x_{2},\ldots,x_{m}$ are
distinct and are $y_{1},y_{2},\ldots,y_{m-1}$ the zeros of the derivative $\mathbf{P}^{\prime}(x)$ . We will show that we can indeed satisfy conditions (11) where, in this case,

\alpha_{1}=\alpha_{2}=\ldots=\alpha_{m}=\frac{1}{m}\quad,\quad\beta_{1}=\beta_{2}=\ldots=\beta_{m-1}=\frac{1}{m-1}.

We have

\frac{\mathrm{P}^{\prime}(x)}{\mathrm{P}(x)}=\frac{1}{x-x_{1}}+\frac{1}{x-x_{2}}+\ldots+\frac{1}{x-x_{m}}

from where

\frac{1}{y_{i}-x_{1}}+\frac{1}{y_{i}-x_{2}}+\cdots+\frac{1}{y_{i}-x_{n}}=0,\quad i=1,2,\ldots,m-1

who gives us

y_{i}=\frac{\sum_{i=1}^{m}\frac{x_{i}}{\left(y_{i}-x_{j}\right)^{2}}}{\sum_{j=1}^{m}\frac{1}{\left(y_{i}-x_{j}\right)^{2}}},\quad i=1,2,\ldots,m-1.

It remains to be demonstrated that

	$\displaystyle\sum_{i=1}^{m-1}\frac{\frac{1}{\left(y_{i}-x_{i}\right)^{2}}}{\sum_{l=1}^{m}\frac{1}{\left(y_{i}-x_{l}\right)^{2}}}=-\sum_{i=-1}^{m-1}\frac{\left.\mathrm{P}y_{i}\right)}{\mathrm{P}^{\prime\prime}\left(y_{i}\right)}\cdot\frac{1}{\left(x_{j}-y_{i}\right)^{2}}=\frac{m-1}{m}$		(12)
	$\displaystyle j=1,2,\ldots,m$

Consider the polynomial

\mathrm{F}(x)=\mathrm{P}\left(x_{1}-\frac{1}{m}\left(x-\frac{x_{1}+x_{2}+\cdots x_{m}}{m}\right)\mathrm{P}^{\prime}(x)\right.

of degree $m-2$ . We have

\frac{\mathrm{F}(x)}{\mathrm{P}^{\prime}(x)}=\sum_{i=1}^{m-1}\frac{\mathrm{\penalty 10000\ F}\left(y_{i}\right)}{\mathrm{P}^{\prime\prime}\left(y_{i}\right)\left(x-y_{i}\right)}=\sum_{i=1}^{m-1}\frac{\mathrm{P}\left(y_{i}\right)}{\mathrm{P}^{\prime\prime}\left(y_{i}\left(x-y_{i}\right)\right.}.

The first member of (12) can therefore be written

\left[\frac{\mathrm{F}(x)}{\mathrm{P}^{\prime}(x)}\right]_{x=x_{j}}^{\prime}=\left[\frac{\mathrm{P}(x)}{\mathrm{P}^{\prime}(x)}-\frac{1}{m}\left(x-\frac{x_{1}+x_{2}+\cdots+x_{m}}{m}\right)\right]_{x=x_{j}}^{\prime}=\frac{m-1}{m}.

We see therefore that the inequality of MK Toda is in this way an immediate consequence of the inequality (2).
7. - To fix the ideas supposous that

x_{1}\leqq y_{1}\leqq x_{2}\leqq y_{2}\leqq\cdots\leqq a_{m-1}\leqq y_{m-1}\leqq x_{m}

(13)

we can then write the inequalities (7) which express the condition ( $\mathrm{C}^{\prime}$ ). The first two equalities (7) are obviously verified. There remain $2m-3$ inequalities which, written in a suitable form, are

$\displaystyle(m-1)\left(x_{1}+x_{2}\right.$	$\displaystyle\left.+\cdots+x_{i}\right)+ix_{i+1}\leq m\left(y_{1}+y_{2}+\cdots+y_{i}\right)$	(14)
$\displaystyle i$	$\displaystyle=1,2,\ldots,m-2.$
$\displaystyle(m-1)\left(x_{1}+x_{2}+\ldots+x_{i}\right)\leq m\left(y+y_{2}+\ldots+y_{i}\right)-iy_{i}$		( $\prime$ )
$\displaystyle i$	$\displaystyle=1,2,\ldots,m-1.$

We can easily see that the inequalities (14') result from (13) and (14) and from

\frac{x_{1}+x_{2}+\ldots+x_{m}}{m}=\frac{y_{1}+y_{2}+\ldots+y_{m-1}}{m-1}

(15)

so
if $x_{1}\leqq x_{2}\leqq\ldots\leqq x_{m}$ are the zeros of a polynomial of degree $m$ And $y_{1}\leqq y_{2}\leqq\ldots\leqq y_{m-1}$ the zeros of the derived polynomial, we have the inequalities (14).

The first of these inequalities is only an inequality of war. According to this inequality, zero $y_{i}$ is always included in the interval $\left(x_{i}+\frac{x_{i+1}-x_{i}}{m},x_{i+1}-\frac{x_{i+1}-x_{i}}{m}\right)$ . Mr. R. Godeau generalized this theorem of Laguerre by demonstrating an inequality which amounts to the following ( ¹⁰ )

	$\displaystyle\frac{(m-i)x_{i}+r_{i+1}}{m-i+1}\leqq y_{i}\leqq\frac{ix_{i+1}+x_{i}}{i+1}$		(16)
	$\displaystyle i=1,2,\ldots,m-1$

It should be noted that inequalities (14) do not generally result from (13), (15) and (16). To see this, it is sufficient to take $m=5,\left\{\begin{array}[]{c}x_{1}=0,x_{2}=60,x_{3}=120,x_{4}=180,x_{5}=240\\ y_{1}=15,y_{2}=80,y_{3}=163,y_{4}=222.\end{array}\right.$

Inequalities (14) are therefore new unequalities between the zeros of a polynomial and the zeros of the derived polynomial.

The general theorem of No. 4 allows us to establish other inequalities for convex functions. We do not want to dwell on this question further here.
8. - Let us now move on to non-concave functions of order $n>1$ . Such a function is characterized by the inequality

\left[x_{1},x_{2},\ldots,x_{n+2};f\right]\geqq 0

(17)

verified by any group of $n+2$ distinct points $x_{1},x_{2},\ldots,x_{n+2}$ of the set E on which the function is defined.

Let us again seek the necessary and sufficient conditions that non-zero constants must fulfill. $p_{i}$ and the points $x_{1}<x_{2}<\ldots<x_{m}$ of E so that we have the inequality (5 ( $m\geqq n+2$ ) whatever the function $f(x)$ , non-concave of order $n$ .

In this case we can write

	$\displaystyle\sum_{i=1}p_{i}f\left(x_{i}\right)=\mathrm{A}_{0}f\left(x_{1}\right)+\mathrm{A}_{1}\left[x_{1},x_{2};f\right]+\ldots+$		(18)
	$\displaystyle\ldots+\mathrm{A}_{n}\left[x_{1},x_{2},\ldots,x_{n+1};f\right]+\sum_{i=1}^{m-n-1}\mathrm{C}_{i}\Delta_{n+1}^{i}$

where we posed

\Delta_{n+1}^{i}=\left[x_{i},x_{i+1},\ldots,x_{i+n+1};f\right],i=1,2,\ldots,m-n-1

(19)

and where the $A_{i}$ and the $C_{i}$ do not depend on the function $f(r)$ .
We can determine the coefficients $A_{i},C_{i}$ by an easy calculation which we do not insist on here. We find

		$\displaystyle\mathrm{A}_{0}=\sum_{i=1}^{m}p_{i}$
		$\displaystyle\mathrm{\penalty 10000\ A}_{1}=\sum_{i=1}^{m}p_{i}x_{i}-x_{1}\sum_{i=1}^{m}p_{i}=\sum_{i=2}^{m}p_{i}\left(x_{i}-x_{1}\right)$
		$\displaystyle\mathrm{A}_{2}=\sum_{i=1}^{m}p_{i}x_{i}^{2}-\left(x_{1}+x_{2}\right)\sum_{i=1}^{m}p_{i}x_{i}+x_{1}x_{2}\sum_{i=1}^{m}p_{i}=\sum_{i=3}^{m}p_{i}\left(x_{i}-x_{1}\right)\left(x_{i}-x_{2}\right)$
		$\displaystyle\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot$
		$\displaystyle\mathrm{A}_{n}=\sum_{i=n+1}^{m}p_{i}\left(x_{i}-x_{1}\right)\left(x_{i}-x_{2}\right)\ldots\left(x_{i}-x_{n}\right)$
		$\displaystyle\mathrm{C}_{r}=\left(x_{r+n+1}-x_{r}\right)\sum_{i=r+n+1}^{m}p_{i}\left(x_{i}-x_{r+1}\right)\left(x_{i}-x_{r+2}\right)\ldots\left(x_{i}-x_{r+n}\right)$
		$\displaystyle r=2,\ldots;m-n-1.$

It is clear that we obtain sufficient conditions for inequality (5) by writing

\mathrm{A}_{0}=\mathrm{A}_{1}=\ldots=\mathrm{A}_{n}=0\quad,\quad\mathrm{C}_{r}\geqq 0,r=1,2,\ldots,m-n-1

We can easily see that these relationships are equivalent to the following

\left\{\begin{array}[]{c}\sum_{i=1}^{m}p_{i}=\sum_{i=1}^{m}p_{i}x_{i}=\ldots=\sum_{i=1}^{n}p_{i}x_{i}^{n}=0\\ \sum_{i=r+n+1}^{m}p_{i}\left(x_{i}-x_{r+1}\right)\left(x_{i}-x_{r+2}\right)\ldots\left(r_{i}-x_{r+n}\right)\geq 0\\ r=1,2,\ldots,m-n-1.\end{array}\right.

THE $m-n-1$ The latter inequalities can also be written differently. If we assume, in fact, that the $n+1$ first equalities (20) are verified, we have

\begin{gathered}\sum_{i=1}^{m}p_{i}\left(x_{i}-x_{r+1}\right)\left(x_{i}-x_{r+2}\right)\ldots\left(x_{i}-x_{r+n}\right)=0\\ r=1,2,\ldots,m-n-1\end{gathered}

which gives us, in this case,

	$\displaystyle\mathrm{C}_{r}=-\left(x_{r+n+1}-x_{r}\right)\sum_{i=1}^{r}p_{i}\left(x_{i}-x_{r+1}\right)\left(x_{i}-x_{r+2}\right)\ldots\left(x_{i}-x_{r+n}\right)$		(21)
	$\displaystyle r=1,2,\ldots,m-n-1$

We can therefore write the inequalities (20) in the form

\begin{gathered}\sum_{i=1}^{r}p_{i}\left(x_{i}-x_{r+1}\right)\left(x_{i}-x_{r+2}\right)\ldots\left(x_{i}-x_{r+n}\right)\leqq 0\\ r=1,2,\ldots,m-n-1\end{gathered}

If we now assume that the function is defined only on the $m$ points $x_{i}$ , we can obviously take for $f\left(x_{1}\right),\left[x_{1},x_{2};f\right],\ldots,\left[x_{1},x_{2},\ldots,x_{n+1};f\right]$ any values and for $\Delta_{n+1}^{r},r=1,2,\ldots,m-n-1$ any non-negative values. It therefore follows that:

The necessary and sufficient condition for inequality (5) to be verified for any non-concave function of order n defined on the points $x_{1},x_{2},\ldots,x_{m}$ is that relations (20) are checked.

If in addition the function is convex of order $n$ this is the sign $>$ which fits in (5).
9. - Now suppose that it is a function defined in an interval ( $a,b$ ) which contains the points $x_{i}$ . Conditions (20) are still sufficient but they are no longer necessary. This results from my work on the extension of convex functions ( ¹¹ ). Indeed, in this case the numbers $\Delta_{n+1}^{r}$ verify certain inequalities, apart from the obvious inequalities $\div_{n+1}^{r}\geq 0$ . Let us first recall these results. Consider the functions $\psi_{i}(x)$ defined in the interval $\left(x_{1},x_{m}\right)$ in the following manner

\psi_{i}(x)=\left\{\begin{array}[]{l}0,\text{ dans }\left(x_{1},x_{i}\right)\\ -\sum_{j=0}^{k}\frac{\left(x_{i+j}-x\right)^{n}}{Q_{i}^{\prime}\left(x_{i+j}\right)},\text{ dans }\left(x_{i+k},x_{i+k+1}\right),k=0,1,\ldots,n\\ 0,\text{ dans }\left(x_{i+n+1},x_{m}\right)\end{array}\right.

where we posed

\mathrm{Q}_{i}(x)=\left(x-x_{i}\right)\left(x-x_{i+1}\right)\ldots\left(x-x_{i+n+1}\right),i=1,2,\ldots,m-n-1

We then demonstrated the following result:
For there to exist a non-concave function of order n defined in the interval $\left(x_{1},x_{m}\right)$ and taking the values $f\left(x_{i}\right)$ to the points $x_{i}$ it is necessary and sufficient that one has

\sum_{r=1}^{m-n-1}\lambda_{r}\Delta_{n+1}^{r}\geqq 0

whatever the numbers $\lambda_{1},\lambda_{2},\ldots,\lambda_{m-n-1}$ choose so that the function

\psi(x)=\sum_{r=1}^{m-n-1}\lambda_{r}\psi_{r}(x)

be non-negative in the interval $\left(x_{i},x_{m}\right)\left({}^{12}\right)$
Furthermore we know that we can take, for example,

\Delta_{n+1}^{r}=\psi_{r}(x),r=1,2,\ldots,m-n-1

$x$ being a point of ( $x_{1},x_{m}$ ).
Now let's get back to our problem. The $n+1$ first equalities (20) are still necessary. To see this, simply re-
$\left({}^{11}\right)$ Tiberiu Popoviciu "On the extension of convex functions of higher order" Bull. Math. Soc. Romania of Sc. 36, 75-108 (1934).
$\left({}^{12}\right)$ We take as interval $(a,b)$ the interval $\left(x_{1},x_{m}\right)$ . This does not restrict generality since we know that we can extend a function of order $n$ outside the interval. See loc. cit. $\left({}^{11}\right)$ .
mark that can be chosen as a function $f(x)$ any polynomial of degree $\leqq n$ .

According to the previous remark the condition

\psi^{*}(x)=\sum_{r=1}^{m-n-1}\mathrm{C}_{r}\psi_{r}(x)\geqq 0,\quad\text{ pour }\quad x_{1}\leqq x\leqq x_{m}

(23)

is necessary and we immediately see that this condition is also sufficient.

For inequality (5) to be verified for any nonconcave function of order $nf(x)$ , defined in an interval $(a,b)$ containing the points $x_{i}$ it is necessary and sufficient that one has

		$\displaystyle\sum_{i=1}^{m}p_{i}=\sum_{i=1}^{m}p_{i}x_{i}=\ldots=\sum_{i=1}^{m}p_{i}x_{i}^{n}=0$
		$\displaystyle\psi^{*}(x)\geqslant 0,\text{ pour }x_{1}\leqq x\leqq x_{m}$

$\psi^{*}(x)$ being the function (23).
If the function is convex of order $n$ This is the sign $>$ which fits in (5).
10. - Inequality (23) can be put in another form. Taking into account (21) and (22) we find, by a calculation on which it is useless to insist,

\psi^{*}(x)=-\sum_{s=1}^{r}p^{s}\left(x^{s}-x\right)^{n},\text{ dans }\left(x_{r},x_{r+1}\right),r=1,2,\ldots,m-1

(24)

Taking into account the $n+1$ first equalities (20) we also have
therefore

\sum_{s=1}^{m}p_{s}\left(x_{s}-x\right)^{n}=0

$\left(24^{\prime}\right)\psi^{*}(x)=\sum_{s=r+1}^{m}\mu_{s}\left(x_{s}-x\right)^{n}$ , In $\left(x_{r},x_{r+1}\right),r=1,2,\ldots,m-1$ .
We can therefore state the property sought in the following form

For inequality (5) to hold for any nonconcave function $f(x)$ , defined "in an interval ( $a,b$ ) containing the points $x_{i}$ , it is necessary and sufficient that

1^{0}\sum_{i=1}^{m}p_{i}=\sum_{i=1}^{m}p_{i}r_{i}=\ldots=\sum_{i=1}^{m}p_{i}x_{i}^{n}=0

$2^{0}$ The polynomial - $\left.\sum_{s=1}^{m}p_{s}{}^{\prime}x_{s}-x\right)^{n}\left[\right.$ Or $\left.\sum_{s=r+1}^{m}p_{s}\left(x_{s}-x\right)^{n}\right]$ be non-negative in the interval ( $x_{r},x_{r+1}$ ) For $r=1,2,\ldots,m-1$ .

Let us note, in passing, that the necessary and sufficient condition of prolongability can also be stated in the following way

For the non-concave function of order $nf(x)$ , defined on the points $x_{1}<x_{2}<\ldots<x_{m}$ be extendable in the meantime $\left(x_{1},x_{m}\right)$ it is necessary and sufficient that whatever the numbers $\mu_{1},\mu_{2},\ldots,\mu_{m}$ chosen so that
$1^{0}\sum_{i=1}^{m^{\prime}}\mu_{i}=\sum_{i=1}^{m}\mu_{i}x_{i}=\ldots=\sum_{i=1}^{m}\mu_{i}x_{i}^{n}=0$ .
20. The polynomial $-\sum_{s=1}^{r}\mu_{s}\left(x_{s}-x\right)^{n}\left[\right.$ Or $\left.\sum_{s=r+1}^{m}\mu_{s}\left(x_{s}-x\right)^{n}\right]$ be non-negative in the interval ( $x_{r},x_{r+1}$ ) For $r=1,2,\ldots,m-1$ , we have the inequality

\sum_{i=1}^{m}\mu_{i}f\left(x_{i}\right)\geqq 0

11.
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  Let us return to inequality (5). Taking $r=m-1$ we see that the condition $p_{m}>0$ is necessary. Taking $r=1$ we find that it is necessary that $-p_{1}\left(x_{1}-x\right)^{n}>0$ In $\left(x_{1},x_{2}\right)$ , SO $(-1)^{n+1}p_{1}>0$ is also necessary. We can therefore say that

If inequality (5) is verified for any non-concave function of order $n$ defined in an interval ( $a,b$ ) containing the points $x_{1}<x_{2}<\ldots<x_{m}$ it is necessary that one has
(25) $\quad(-1)^{n+1}p_{1}>0,\quad p_{m}>0$ .

Consider the polynomial $\mathrm{P}(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\ldots\left(x-x_{m}\right)$ having as zeros the numbers $x_{1}\leqq x_{2}\leqq\ldots\leqq x_{m}$ distinct or not and let us pose

\mathrm{M}_{f}(\mathbf{P})=\frac{f\left(x_{1}\right)+f\left(x_{2}\right)+\cdots+f\left(x_{m}\right)}{m}.

We have analogous expressions $\mathrm{M}_{f}\left(\mathrm{P}^{\prime}\right),\mathrm{M}_{f}\left(\mathrm{P}^{\prime\prime}\right),\ldots$ for successive derivatives $\mathrm{P}^{\prime},\mathrm{P}^{\prime\prime},\ldots$ of the polynomial P .

Let us then form the following expression

\mathrm{F}_{n}f(\mathrm{P})=\binom{n}{0}\mathrm{M}_{f}(\mathrm{P})-\binom{n}{1}\mathrm{M}_{f}\left(\mathrm{P}^{\prime}\right)+\cdots+(-1)^{n}\binom{n}{n}\mathrm{M}_{f}\left(\mathrm{P}^{(n)}\right)

which is a kind of difference in order $n\left({}^{13}\right)$ .
( ¹³ ) If $\mathrm{P}(x)=\left(x-x_{1}\right)\left(x-x_{1}-mh\right)^{n-1}$ the expression $\Gamma_{n}f(\mathrm{P})$ differs only by a constant positive factor from the divided difference $\left[x_{i},x_{i}+h,x_{i}+2h,\ldots,x_{i}+nh,x_{i}+mh;f\right]$ .

A problem of MK Toda, examined in his cited works $\left({}^{14}\right)$ can be asked as follows:

The expression $\mathrm{F}_{n}f(\mathrm{P})$ does it remain non-negative whatever the function $f(x)$ non-concave order $n$ and whatever the polynomial $\mathrm{P}(x)$ having its zeros in the interval $(a,b)$ ?

The answer is that this is only possible if $n=1$ .
MK Toda has clearly demonstrated that $\mathrm{F}_{n}f(\mathrm{P})$ is zero when $f(x)$ is a polynomial of degree $\leqq n$ and this property constitutes very interesting relations between the zeros of a polynomial and those of its successive derivatives.

The expression $\mathrm{F}_{n}f(\mathrm{P})$ is an expression of the form of the first member of (5).

It is now clear that the property cannot be true for $n$ pair. It is enough, in fact, to take $\mathrm{P}(x)$ such as $x_{1}<x_{2}\leqq\ldots$ … $\leqq x_{m-1}<x_{m}$ and then the first and last coefficient $p_{i}$ are both equal to $\frac{1}{m}$ . The necessary conditions (25) are not verified.

In the case of $n$ odd $>1$ let's take the polynomial $\mathrm{P}(x)==\left(x-x_{1}\right)^{2}\left(x-x_{2}\right)^{m-2},x_{1}<x_{2},m\geqq n+3$ . The first coefficient $p_{1}$ is equal to

\frac{2}{m}-\frac{n}{m-1}=-\frac{(n-2)m+2}{m(m-1)}<0

and the last coefficient $p_{i}$ is equal to

\begin{gathered}\frac{m-2}{m}-\binom{n}{1}\frac{m-3}{m-1}+\binom{n}{2}\frac{m-4}{m-2}-\cdots+(-1)^{n}\frac{m-n-2}{m-n}=\\ =(-1)^{n+1}2\left[\frac{1}{m-n}-\binom{n}{1}\frac{1}{m-n+1}+\cdots+(-1)^{n-1}\binom{n}{n-1}\frac{1}{m-1}+(-1)^{n}\frac{1}{m}\right]=\\ =(-1)^{n+1}2\int_{0}^{1}x^{m-n-1}(1-x)^{n}dx>0\end{gathered}

It therefore follows that the property cannot be true for $n$ odd $>1$ .

It can easily be seen that, more generally, $F_{n}f(P)$ cannot be a constant sign for everything $P$ and for any non-concave function of order $n$ that if $n=1$ .

Notes on Higher-Order Convex Functions (III)

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NOTES ON HIGHER ORDER CONVEX FUNCTIONS (III)

On certain inequalities verified by convex functions

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