Notes on higher-order convex functions (X)

(Manthematics, Physics, Chemistry)

Volume XXVIII, Year 1942

Booklet 1.

Pages
BARBILIAN C. - v. BOGDAN H.
BEDREAG CG - Das Element $\frac{239}{93}$ as uranide in the natural system of elements

—

Structure and natural systematics of the elements 143-148
—

Nuclear Stability II

BOGDAN CP - On the asymptotic lines of the Steiner surface.

BOGDAN H. (M-elle) and BARBILIAN C. - Combinations of the acid ( $\mathrm{H}_{2}$ (SNC) ₃ ) H with some organic bases

CERNATESCU R. - v. PONI MP (Ms)
CLIMESCU AL. C. - On the class of analytical functions which keep the half-planes determined by the real axis

COZUBSCHI E. (Ms.) - The action of isothiocyanates on benzoinoximes

GHEORGHIU C. V, and STOICESCU L. (M-me) - Condensation products of hydrogen derivatives with thio-2-phenyl-3-ethoxy-4-tetra-hydro-1, 2, 3, 4quinazoline

PAPAFIL E. - The action of phenyl isothiocyanate on the oximes of cyclic ketones

PAPAFIL E. and PAPAFIL M. (M-me). - Mercury salts with isomeric phenylenediamines

PAPPHIL M. (M-m). - v. PAPAFIL E.
PONI MP (M-11th) and CERNATESCU R. - Acid neutral salts $\mathrm{PSO}_{7}\mathrm{H}_{3}$

POPOVICIU T. - Notes on higher order convex functions

—

On the approximation of continuous functions of a real variable by polynomials

SIADBEI V. - On the determination of the points of convergence of Kapteyn's star currents,

STOICESCU N. (M-me). - v. GEORGE CV
TRIANDAF L. (M-me). - Lithium meta-arsenite

NOTES ON HIGHER ORDER CONVEX FUNCTIONS (X)

On some properties of divided differences and Lagrange polynomials.

Consider a function $f=f(x)$ , real, of the real variable $x$ , defined on a linear set $E$ . To any group of $n+1$ points of $E$ we can attach the Lagrange polynomial for the function $f$ , therefore the polynomial of minimum degree which takes the same values as this function at the points considered.

Assuming $E$ finished, in § 3 we study the problem of Lagrange polynomials which are majorant (or minorant) for the function $f$ . A polynomial $P(x)$ is majorant for the function $f(x)$ are you him and

P(x)\geq f(x),\quad x\in E

We establish the necessary and sufficient conditions for a Lagrange polynomial, attached to a given group of points of $E$ , solt majorant for the function $f$ . We deduce from this, for any function $f$ , the existence of at least one upper bound Lagrange polynomial of given degree.

Always assuming $E$ finished, in § 4 we look for whether a non-concave function of order $k$ can admit a Lagrange polynomial of given degree $n$ which is also non-concave in order $k$ on $E$ . For $k=-1$ , so for the case of non-negativity, the answer, always affirmative, is a consequence of the results of § 3. This is no longer the case for $k\geq 0$ . In this case, if $k$ And $n$ are of different parities there
always exists at least one Lagrange polynomial of degree $n$ which is also non-concave in order $k$ on $E$ . But, if $k$ And $n$ are of the same parity, we demonstrate by an example that the property may not be true. In the latter case the problem of existence depends not only on the function $f$ but also the distribution of points of $E$ We conclude this § with a more detailed study of some simple particular cases.

In § 5 we say a few words about the problems dealt with in §§ 3 and 4 in the case where $E$ is a finite and closed interval. This case is quite different from the previous one, therefore from the one where $E$ is finished. We will only give a few indications on these problems.

In § 1 we recall some well-known definitions and formulas and in § 2 we give some new properties which are used later.

The problems we pose in this work raise many others. We hope that the simple cases, which are also the easiest, that we have presented, are sufficient to show the interest of these questions.

§ 1.

Some preliminary properties and formulas ¹ ).

1.
- —
  
  We will consider only real, uniform and finite functions of the real variable $x$ .

Either $f=f(x)$ such a function, defined on the $n+1$ distinct points

x_{1},x_{2},\ldots,x_{n+1}.

(1)

The divided difference of the function $f$ on points (1) is completely characterized by the following three properties
I. It is a linear functional of $f$ .
II. It is zero for functions $f=1,x,x^{2},\ldots,x^{n-1}$ .
III. It is equal to 1 for the function $f=x^{n}$ .

Property I means that the divided difference is of the form

⁰⁰footnotetext: 1. For further details and proofs, see my earlier works. In particular; "On some properties of functions of one or two real variables". Mathematica, 8, 1-85 (1934) and "Introduction to the theory of divided differences". Bull, Math, Soc. Roumaine des Sci,, 42, 65-78 [1941]

\sum_{i=1}^{n+1}\lambda_{i}f\left(x_{i}\right),

$\lambda_{i}$ being independent of the function $f$ . Properties II, III then completely determine these coefficients.

We will designate, as usual, by
(2)

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

or also by

\left[x_{1},x_{2},\ldots,x_{n+1};f(x)\right],

the divided difference thus defined.
We have the following formulas
(3)

\left[x_{1},x_{2},\ldots,x_{n+1};x^{r}\right]=\left\{\begin{array}[]{l}0,r=0,1,\ldots,n-1,\\ 1,r=n_{1}\end{array}\right.

(4)

\left[x_{1},x_{2},\ldots,x_{n+1};Cf\right]=C\left[x_{1},x_{2},\ldots,x_{n+1};f\right],

(5) $\left[x_{1},x_{2},\ldots,x_{n+1};f+g\right]=\left[x_{1},x_{9},\ldots,x_{n+1};f\right]+\left[x_{1},x_{2},\ldots,x_{n+1};g\right]$ , Or $C$ is a constant and $f,g$ two functions defined on (1).

The divided difference (2) can also be put in the form of a quotient of two determinants of order $n+1$ ,

\left[x_{1},x_{2},\ldots,x_{n+1};f\right]=\frac{U\left(x_{1},x_{2},\ldots,x_{n+1};f\right)}{V\left(x_{1},x_{2},\ldots,x_{n+1}\right)}

(6)

U\left(x_{1},x_{2},\ldots,x_{n+1};f\right)=\left|1x_{i}x_{i}^{2}\ldots x_{i}^{n-1}f\left(x_{i}\right)\right|

And

\begin{gathered}V\left(x_{1},x_{2},\ldots,x_{n+1}\right)=U\left(x_{1},x_{2},\ldots x_{n+1};x^{n}\right)=\\ =\prod_{\begin{subarray}{c}i,j=1\\ i>j\end{subarray}}^{n+1}\left(x_{i}-x_{j}\right)=\prod_{i=2}^{n+1}\left(x_{i}-x_{i-1}\right)\left(x_{i}-x_{i-2}\right)\ldots\left(x_{i}-x_{1}\right)\end{gathered}

is the Vandermonde determinant of numbers $x_{i}$ .
La formule (3) peut se compléter par les suivantes, correspondantes aux valeurs -1 et $n+1$ de $r$ ,

	$\displaystyle{\left[x_{1},x_{2},\ldots,x_{n+1};\frac{1}{x}\right]}$	$\displaystyle=\frac{(-1)^{n}}{x_{1}x_{2}\ldots x_{n+1}}$		(7)
	$\displaystyle{\left[x_{1},x_{2},\ldots,x_{n+1};x^{n+1}\right]}$	$\displaystyle=x_{1}+x_{2}+\cdots+x_{n+1}$		(8)

La différence divisée (2) est symétrique par rapport aux points (1).
2. - Considérons maintenant une fonction $f$ définie sur un ensemble linéaire quelconque $E$ . Sur tout groupe de $n+1$ points (1) de $E$ on peut définir la différence divisée. Nous disons qu’une différence divisée définie sur $n+1$ points est d’ordre $n$ . Si, en particulier, l’ensemble $E$ est fini et est formé par $m$ points

x_{1},x_{2},\ldots,x_{m}

(9)

la fonction $f$ a des différences divisées d’ordre $0,1\ldots,m-1$ . En tout, la fonction a $\binom{m}{n+1}$ différences divisées d’ordre $n(m\geq n+1)$ . Si l’ensemble $E$ est infini la fonction a des différences divisées de tout ordre. Par définition, la différence divisée d’ordre 0 sur le point $x_{i}$ est la valeur $f\left(x_{i}\right)$ de la fonction en ce point,

\left[x_{i};f\right]=f\left(x_{i}\right).

Si nous posons
(10)

\varphi(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\ldots\left(x-x_{n+1}\right)

nous avons

\left[x_{1},x_{2},\ldots,x_{n+1};f\right]=\sum_{i=1}^{n+1}\frac{f\left(x_{i}\right)}{\varphi^{\prime}\left(x_{i}\right)}

(11)

où $\varphi^{\prime}$ est la dérivée du polynome $\varphi^{1}$ ).
A l’aide de la formule (11) il est facile d’établir la formule de récurrence des différences divisées,

\left[x_{1},x_{2},\ldots,x_{n+1};f\right]=\frac{\left[x_{2},x_{3},\ldots,x_{n+1};f\right]-\left[x_{1},x_{2},\ldots,x_{n};f\right]}{x_{n+1}-x_{1}}

(12)

⁰⁰footnotetext: 1. La dérivation est ici une opération linéaire applicable aux polynomes et telle que

(xn)^{\prime}=nx^{n-1}.

On prend d’habitude cette formule comme définition des différences divisées de divers ordres ¹ ).

La formule (11) permet aussi d’établir la suivante

\left[x_{1},x_{2},\ldots,x_{n+1};f\right]=\left[x_{1},x_{2},\ldots,x_{r};\frac{f(x)}{\left(x-x_{r+1}\right)\left(x-x_{r+1}\right)\ldots\left(x-x_{n+1}\right)}\right]+

+\left[x_{r+1},x_{r+1},\ldots,x_{n+1};\frac{f(x)}{\left(x-x_{1}\right)\left(x-x_{2}\right)\ldots\left(x-x_{r}\right)}\right]

(§13))

Les formules (3), (4) et (5) nous montrent que la différence divisée d’ordre $n$ d’un polynome de degré $<n$ est constamment nulle et la différence divisée d’ordre $n$ du polynome $c_{0}x^{n}+c_{1}x^{n-1}+\cdots$ est constamment égale à $c_{0}{}^{2}$ ).

Remarquons aussi la formule suivante
(14) $\left[x_{1}+h,x_{2}+h,\ldots,x_{n+1}+h;f(x)\right]=\left[x_{1},x_{2},\ldots,x_{n+1};f(x\div h)\right]$ .
3. - La formule (5) donne la différence divisée d’une somme de deux fonctions. De même, nous avons la formule
(15) $\left[x_{1},x_{2},\ldots,x_{n+1};fg\right]=\sum_{i=1}^{n+1}\left[x_{1},x_{2},\ldots,x_{i};f\right]\left[x_{i},x_{i+1},\ldots,x_{n+1};g\right]$ , qui donne la différence divisée du produit de deux fonctions $f$ et g. C’est la formule de Leibniz des différences divisées.

Considérons une fonction $f$ définie sur l’ensemble fini (9). Pour simplifier nous posons
(16) $\Delta_{j}^{i}(f)=\left[x_{i},x_{i+1},\ldots,x_{i+j};f\right],j=0,\ldots,m-1,i=1,2,\ldots,m$ , et

	$\displaystyle\Psi_{i,j+1}={}_{i,j+1}(x)=\left(x-x_{i}\right)\left(x-x_{i+1}\right)\ldots\left(x-x_{i+j}\right),\varphi_{i,0}=1,$		(17)
	$\displaystyle j=0,\ldots,m-i,i=1,2\ldots,m.$

1.

C’est de cette façon que A. M. AMPÈRE a, pour la première foís, introduit les différences divisées. Voir N. E. Nörlund, „Differenzenrechnung", J. Springer, Berlin 1924.
2.

Toute fonction de la foxme $c_{0}x^{n}+c_{1}x^{n-1}+\cdots$ est un polynome de degré $n$ , Si c $*\neq 0$ , ce polynome est de degré effectif $n$ ; co est le premier coefficient du polynome. On voit que la notion de premier coefficient est relative au degré et non pas au degré effectif du polynome, La constante 0 est un polynome de degré - 1 ou de degré effectif - 1 .

Nous avons donc, compte tenant de (11),

\Delta_{j}^{i}(f)=\sum_{r=i}^{i+j}\frac{f\left(x_{r}\right)}{\varphi_{i,j+1}^{\prime}\left(x_{r}\right)}.

(18)

Avec cette notation, la formule (15), pour $m$ points, devient

\Delta_{m-1}^{1}(fg)=\sum_{i=1}^{m}\Delta_{m-i}^{i}(g)\Delta_{i-1}^{1}(f)

(19)

Remarquons que si la fonction $g$ est donnée, $\Delta_{m-1}^{1}(fg)$ est une fonctionnelle linéaire de $f$ . Réciproquement, toute fonctionnelle linéaire de $f$ ,

F[f]=\sum_{i=1}^{m}\lambda_{i}f\left(x_{i}\right),

(20)

définie pour les fonctions $f$ définies sur l’ensemble fini (9), peut s’écrire sous la forme d’une différence divisée d’ordre $m-1$ du produit $f$ . $g$ , en déterminant convenablement la fonction g. Prenons, en effet,

	$\displaystyle g\left(x_{i}\right)=\frac{\lambda_{i}}{\left(x_{i}-x_{1}\right)\left[x_{i}-x_{2}\right)\ldots\left(x_{i}-x_{m}\right)}=\frac{\lambda_{i}}{\psi_{1,m}^{\prime}\left(x_{i}\right)}$		(21)
	$\displaystyle i=1,2,\ldots,m$

et nous avons alors

F[f]=\Delta_{m-1}^{1}(fg)=\left[x_{1},x_{3},\ldots x_{m};fg\right].

(22)

4.
- —
  
  La fonctionnelle linéaire (20) peut s’écrire sous la forme

F[f]=\sum_{i=1}^{n}u_{i}^{(n)}\Delta_{i-1}^{1}(f)+\sum_{i=1}^{m-n}v_{i}^{(n)}\Delta_{n}^{l}(f),\quad(m\geq n+1),

(23)

où les coefficients $\mu_{i}^{(n)},v_{i}^{(n)}$ sont indépendants de la fonction $f$ . Ces coefficients sont linéaires, homogènes en $\lambda_{i}$ et sont complètement déterminés. Il est facile de les obtenir en tenant compte des formules (19), (21) et (22) et de la relation de récurrence (12). On a d’abord $\mu_{i}^{(n)}=\Delta_{n-i}^{i}(g)$ et, en exprimant $\Delta_{i-1}^{1}(f),i>n$ , en fonction de $\Delta_{n}^{2}(f),\Delta_{n}^{2}(f),\ldots,\Delta_{n}^{i-n}(f)$ à l’aide de (12), on a les coefficients $v_{i}^{(n)}$ .

Une autre manière d’obtenir les coefficients $\mu_{i}^{(n)},v_{i}^{(n)}$ est de particulariser convenablement la fonction $f$ .

Avec les notations (17), nous avons

\begin{gathered}\mu_{i}^{(n)}=F\left[\varphi_{1,i-1}\right]=\sum_{j=i}^{m}\lambda_{j}\left(x_{j}-x_{1}\right)\left(x_{j}-x_{2}\right)\cdots\left(x_{j}-x_{i-1}\right)\\ i=1,2,\ldots,n\end{gathered}

Pour calculer les coefficients $v_{i}^{(n)}$ nous introduisons les fonctions
(24) $\quad f_{n,l}^{*}=f_{n,i}^{*}(x)=\left\{\begin{array}[]{l}\left.0,\text{ pour }x=x_{r},1\leqq\boldsymbol{r}\leqq n+i-1\text{ (ou }1\leqq\boldsymbol{r}\leqq i\right),\\ \varphi_{i+1,n-1}(x),\text{ pour }x=x_{r},\boldsymbol{\pi}+i\leqq\boldsymbol{r}\leqq m\text{ (ou }i<\boldsymbol{r}\leqq m\text{ ), }\end{array}\right.$

i=1,2,\ldots,m-n

Nous avons alors

		$\displaystyle\Delta_{j-1}^{1}\left(f_{n,i}^{*}\right)=0,j=2,\ldots,n$
		$\displaystyle\Delta_{n}^{j}\left(f_{n,i}^{*}\right)=0,j=2,\ldots,i-1,i+1,\ldots,m-n$

Compte tenant de (18) et des relations de récurrence

\varphi_{i,j+1}=\left(x-x_{i+j}\right)\varphi_{i,j}=\left(x-x_{i}\right)\varphi_{i+1,j},

nous trouvons

\Delta_{n}^{i}\left(f_{n,l}^{*}\right)=\frac{\varphi_{i+1,n-1}\left(x_{i+n}\right)}{\varphi_{i,n+1}^{\prime}\left(x_{i+n}\right)}=\frac{1}{x_{i+n}-x_{i}}

et nous en déduisons donc

\begin{gathered}v_{i}^{(n)}=\left(x_{i+n}-x_{j}\right)F\left[f_{n,}^{*}\right]=\left(x_{i+n}-x_{i}\right)\sum_{j=i+n}^{m}\lambda_{j}\left(x_{j}-x_{i+1}\right)\left(x_{j}-x_{i+2}\right)\cdots\left(x_{j}-x_{i+n-j}\right)\\ i=1,2,\ldots,m-n.\end{gathered}

La formule (23) devient

F[f]=\sum_{i=1}^{n}F\left[\varphi_{1,i-1}\right]\Delta_{i-1}^{1}(f)+\sum_{i=1}^{m-n}\left(x_{i+n}-x_{i}\right)F\left[f_{n,i}^{*}\right]\Delta_{n}^{i}(f).

(25)

C’est la formule fondamentale de transformation des différences divisées.

Pour $n=1$ cette formule devient la formule, bien connue, d’Abel.

Pour $n=m-1$ ce n’est qu’une autre forme de la formule de Leibniz.

Lorsque la fonctionnelle lineaire $F[f]$ est nulle pour tout polynome de degré $r(<n)$ , nous avons

\mu_{1}^{(n)}=\mu_{2}^{(n)}=\cdots=\mu_{r+1}^{(n)}=0.

Si $F[f]$ est nulle pour tout polynome de degré $n-1$ , tous les coefficients $\mu_{i}^{(n)}$ sont nuls et la formule fondamentale (25) devient

F[f]=\sum_{i=1}^{m-n}\left(x_{i+n}-x_{i}\right)F\left[f_{n,i}^{*}\right]\Delta_{n}^{i}(f).

(26)

En particulier, considérons la différence divisée

\left[x_{i,},x_{i,},\ldots,x_{i_{n+1}};f\right],

sur $n+1$ points $x_{ij}j=1,2,\ldots,n+1$ extraits de la suite (9). La formule (26) nous donne alors
(27) $\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n+1}};f\right]=\sum_{i=1}^{m-i}\left(x_{i+n}-x_{i}\right)\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n+1}};f_{n,j}^{*}\right]\Delta_{n}^{i}(f)$ .
5.-Supposons maintenant que la suite (9) soit ordonnée, done que

x_{1}<x_{2}<\cdots<x_{m^{*}}

Alors les coefficients des $\Delta_{n}^{i}(f)$ dans la formule (27) sont tous non-négatifs et nous obtenons le

Théorème 1. Si la suite (9) est ordonnée, toute différence divisée $\left[x_{i,},x_{i_{2}},\ldots,x_{i_{l+1}};f\right]$ , prise sur $n+1$ de ces points, est une moyenne arithmétique (généralisée) des différences divisées $\Delta_{n}^{1}(f),\Delta_{n}^{2}(f),\ldots,\Delta_{n}^{m-n}(f)$ .

Nous avons donc

	$\displaystyle{\left[x_{i_{1}},x_{i_{1}},\ldots,x_{i_{n+1}};f\right]=\sum_{i=1}^{m-n}A_{i}\Delta_{n}^{i}\|f\|,}$		(28)
	$\displaystyle A_{i}\geqq 0,i=1,2,\ldots,m-n,\sum_{i=1}^{m-n}A_{i}=1,$

où les $A_{i}$ sont indépendants de la fonction $f$ .
C’est le théorème de la moyenne des différences divisées. Nous avons, d’ailleurs,
(29) $A_{i}=\left(x_{i+n}-x_{i}\right)\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n+1}};f_{n,i}{}^{*}\right],i=1,2,\ldots,m-n$ .

On peut démontrer cette propriété par induction, en partant de l’identité

\begin{gathered}\left(x_{n+2}-x_{1}\right)\left[x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+2};f\right]=\\ =\left(x_{i}-x_{1}\right)\left[x_{1},x_{2},\ldots,x_{n+1};f\right]+\left(x_{n+2}-x_{i}\right)\left[x_{3},x_{3},\ldots,x_{n+2};f\right].\end{gathered}

Dans le § suivant nous démontrerons directement la nonnégativité des différences divisées d’ordre $n$ des fonctions $f_{n,i}^{*}$ , ce qui fournira une autre démonstration du théorème de la moyenne.

Il en résultera, d’ailleurs, de plus, que, si nous prenons $i_{1}<i_{2}<\cdots<i_{n+1}$ , nous avons

A_{i}=0,i=1,2,\ldots,i_{1}-1,i_{n+1}-n+1,i_{n,1}-n+2,\ldots,m-n_{,}

(30)

A_{i}>0,i=i_{1},i_{1}+1,\ldots,i_{n+1}-n.

Du théorème 1 il résulte immédiatement la propriété suivante

Théurème 2. - Si la suite (9) est ordonnée, nous avons $\min_{i=1,2,\ldots,n-n}\left\{\Delta_{n}^{i}(f)\right\}\leq\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n+1}};f\right]\leq_{i=1,\ldots,m-n}\left\{\Delta_{n}^{i}(f)\right\}$ quelle que soit la suite partielle $x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n+1}}$ de $n+1$ iermes de (9).
6. - Rappelons maintenant la définition des fonctions d’ordre $n$ .

Definition. - La fonction $f(x)$ , définie sur l’ensemble linéate $E$ , est convexe, non-concave, polynomiale, non-convexe resp. concave d’ordre $n$ sur $E$ si

\left[x_{1},x_{2},\ldots,x_{n+2};f\right]>,\geq,=,\leq\text{ resp. }<0,

quels que soient les $n+2$ points $x_{1},x_{2},\ldots,x_{n+2}$ de $E$ .
Toutes ces fonctions sont des fonctions d’ordre n.
Une fonction d’ordre $n$ sur E est donc une fonction dont la différence divisée d’ordre $n+1$ ne change pas de signe sur E. La convexité et la polynomialité d’ordre $n$ sont des cas particuliers de la non-concavité d’ordre $n$ . Si $f$ est convexe non-concave, polynomiale, non-convexe resp. concave d’ordre, $n$ sur $E$ , la fonction $-f$ est concave, non-convexe, polynomiale, non-concave resp. convexe d’ordre $n$ sur $E$ et réciproquement. On peut prendre comme type de fonction d’ordre $n$ la fonction non-concave d’ordre $n$ . Une fonction polynomiale d’ordre $n$ se réduit aux valeurs sur $E$ d’un polynome de degré $n$ .

La définition précédente est valable pour tout entier
$n\geqq-1$ . Les fonctions d’ordre -1 sont les fonctions de signe ¹ invariable et les fonctions d’ordre 0 les fonctions monotones.

Le théorème 1 nous montre que nous avons le
Théorème 3 - Pour que la fonction $f$ , définie sur la suite ordonnée (9) de $m(\geqq n+2)$ points, soit convexe, non-concave, polynomiale, non-convexe resp, concave d’ordre $n$ sur ces points, il faut et il suffit que l’on ait

\Delta_{n+1}^{i}|f|>,\geq,=\text{ resp. }<0,i=1,2,\ldots,m-n-1

Les formules (4), (5) nous montrent aussi que
Théoreme 4,-Si $C$ est une constante positive et si les fonctions $f$ et $g$ , définies sur l’ensemble linéaire $E$ jouissent d’une même propriété de convexité sur $E$ , les fonctions $Cf$ et $f+g$ jouissent aussi de la même propriété de convexité sur $E$ .

Si la fonction est définie dans un intervalle et si elle a une dérivée d’ordre $n+1$ , la non-négativité de cette dérivée est nécessaire et suffisante pour que la fonction soit non-concave d’ordre $n$ .
7. - Disons maintenant quelques mots sur le polynome de Lagrange. Reprenons la fonction $f$ , définie sur les points (1). Nous avous la propriété, bien connue, exprimée par le

Théorème 5.-Il existe un polynome et un seul de degré effectif minimum qui prend les valeurs $f\left(x_{i}\right)$ aux points $x_{i},i=1$ , $2,\ldots,n+1$ . Ce polynome est de degré $n$ .

Le polynome unique ainsi déterminé est le polynome (d’ūnterpolation) de Lagrange de la fonction $f$ sur les points (1). Nous le désignons par

L\left(x_{1},x_{2},\ldots,x_{n+1};f\mid x\right)

(31)

On voit immédiatement que c’est aussi l’unique polynome de degré $n$ prenant les valeurs $f\left(x_{i}\right)$ aux points $x_{i}$ . D’ailleurs, la forme générale des polynomes prenant les valeurs $f\left(x_{i}\right)$ aux points $x_{i}$ est

L\left(x_{1},x_{2},\ldots,x_{n+1};f\mid x\right)+\varphi(x)Q(x),

où $\varphi$ est le polynome (10) et $Q$ un polynome quelconque.
Les points $x_{i}$ sont les noeuds du polynome (31).
Si l’on donne à $x$ une valeur fixe, le polynome (31) devient une fonctionnelle linéaire de $f$ . Le polynome (31) est,
d’ailleurs, évidemment symétrique par rapport aux points $x_{i}{}^{1}$ ).
On trouve facilement la formule
(32) $L\left(x_{1},x_{2},\ldots,x_{n+1};f\mid x\right)=-\varphi(x)\left[x_{1},x_{2},\ldots,x_{n+1},x;fg\right]$ , ou

g\left(x_{i}\right)=\left\{\begin{array}[]{l}1,i=1,2,\ldots,n+1\\ 0,x_{i}=x\end{array}\right.

On en déduit diverses expressions, bien connues, du polynome (31) et, en particulier, la formule
(33) $L\left(x_{1},x_{2},\ldots,x_{n+1};f\mid x\right)-f(x)=-\varphi(x)\left[x_{1},x_{2},\ldots,x_{n+1},x;f\right]$ .

On voit aussi que

	$\displaystyle L\left(x_{1},x_{2},\ldots,x_{n+1};f\mid x\right)=L\left(x_{1},x_{2},\ldots,x_{n};f\mid x\right)+$
	$\displaystyle\quad+\left(x-x_{1}\right)\left(x-x_{2}\right)\ldots\left(x-x_{n}\right)\left[x_{1},x_{2},\ldots,x_{n+1};f\right].$		(34)

Le premier coefficient du polynome (31) est donc précisément la différence divisée de la fonction $f$ sur les points(1).

Remarquons aussi que si $f$ est un polynome de degré $n$ , nous avons

L\left(x_{1},x_{2},\ldots,x_{n+1};f\mid x\right)=f(x)

Soit maintenant $f$ une fonction définie sur l’ensemble linéaire $E$ . A tout groupe de $n+1$ points de $E$ correspond, pour la fonction, un polynome de Lagrange ayant ces points comme noeuds. Dans la suite nous convenons qu’un polynome de Lagrange est de degré $n$ s’il a $n+1$ knots. Two Lagrange polynomials are considered different if they do not have the same knots, but two different Lagrange polynomials can be identical on $E$ . In this case, it is clear that the function reduces to this polynomial on all the nodes of the polynomials considered.

§ 2.

New properties of divided differences of functions defined on a finite set.

8.
- —
  
  We will first resume the study of functions: $f_{n,i}^{*}$ , given by formula (24). These functions are defined

⁰⁰footnotetext: 1. A. Cauchy deduces the symmetry of the divided difference. This amounts to noting that the Lagrange polynomial is unique, independent of the order of the nodes and that its first coefficient is the divided difference taken on the nodes. See: "On interpolar functions". CR Acad. Sci, Paris, 11, 775-789 (1840).

on the ordered sequence (9), where we can assume, of course, $m\geq n+1$ . We will always assume that any partial sequence $x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n+1}}$ extracted from (9) is also ordered, so that $1\leq i_{1}<i_{2}<\ldots<i_{n+1}\leqq m$ .

We have seen that the mean value theorem results in non-concavity of order $n-1$ functions $f_{n,i}^{*}$ . But, there is interest in demonstrating this non-concavity directly. In this way the mean value theorem will be a consequence of this non-concavity. It should be noted that we cannot apply Theorem 3 here without committing a vicious circle. We will demonstrate directly that all the divided differences of order $n$ of the function $f_{n,i}^{*}$ are non-negative, so the following property

Theorem 6.-Divided differences of order $n$ on $n+1$ points of the ordered sequence (9] and relating to the functions $f_{n,i}^{*}$ , given by formula (24), are all positive or zero ¹ ).

The demonstration is done by induction on the number $n$ .
For $n=1$ , we immediately check that
(35) $\left[x_{i_{1}},x_{i_{2}};f_{1,i}^{*}\right]=\left\{\begin{array}[]{c}0,\quad i=1,2,\ldots,i_{1}-1,i_{2},i_{2}+1,\ldots,m-1,\\ \frac{1}{x_{i_{2}}-x_{i_{1}}}>0,i_{1}=i_{1},i_{1}+1,\ldots,i_{2}-1.\end{array}\right.$

Now suppose that the property is true for functions $f_{n-1,i}^{*}$ and demonstrate it for functions $f_{n,i}^{*}\{n>1\}$ . Let us first note that, following the definition of the functions $f_{n,i}^{*}$ , In

\left[x_{i_{1}},x_{i_{q}},\ldots,x_{i_{n+1}};f_{n,i}^{*}\right]\geqq 0,

(36)

the sign $=$ is valid for

⁰⁰footnotetext: 1. We assume

n\geq 1

. The property remains true also for

n=0

, the functions

f_{0,i}^{*}

being defined by the formula

f_{0,i}^{*}\left(x_{r}\right)=\left\{\begin{aligned} &0,r=1,2,\ldots,i-1,i+1,\ldots,m\\ &1,r=i\\ &i=1,2\ldots,m\end{aligned}\right.

Theorem 7 is also true for

n=0

i=1,2,\ldots,i_{1}-1,i_{n+1}-n+1,i_{n+1}-n+2,\ldots,m-n.

So we just need to look at the values of $i$ for which

i_{1}\leq i\leq i_{n+1}-n.

(37)

Considering the recurrence formula

f_{n,i}^{*}=\left(x-x_{i+n-1}\right)f_{n-1,i}^{*}

and applying formula (15), we have
(38)

\begin{gathered}{\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n+1}};f_{n,i}^{*}\right]=\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n+1}};f_{n-1,i}^{*}\right]\left(x_{i_{n+1}}-x_{i+n-1}\right)+}\\ +\left[x_{i_{i}},x_{i_{2}},\ldots,x_{i_{n}};f_{n-1,i}^{*}\right]\end{gathered}

But, the recurrence formula (12) allows us to write

\left[x_{i_{1}},x_{i_{s}},\ldots,x_{i_{n+1}};f_{n,-1,i}^{*}\right]=\frac{\left.\left.\left[x_{i_{s}},x_{i_{s}},\ldots,x_{i_{n+1}};f_{n-1,i}^{*}\right]-\right]_{x_{i_{1}}},x_{i_{2}},\ldots,x_{i_{n}};f_{n-1,i}^{*}\right]}{x_{i_{n+1}}-x_{i_{1}}}

and we deduce the formula

	$\displaystyle{\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n+1}};f_{n,i}^{*}\right]=}$		(39)
	$\displaystyle=\frac{\left(x_{i_{n+1}}-x_{i+i+1}\right)\left[x_{i_{2}},x_{i_{9}},\ldots,x_{i_{n+1}};f_{n-1,i}^{}\right]+\left(x_{i+n-1}-x_{i_{1}}\right)\left[x_{i_{1}},\ldots,x_{i_{n}};t_{n-1,i}^{}\right]}{x_{i_{n+1}}-x_{i_{1}}}$

from which the property results, taking into account the fact that (37) gives us

i_{1}<i_{1}+n-1\leqq i+n-1\leqq i_{n+1}-1<i_{n+1}.

We can also notice that the sign $>$ is valid in (36) for $i=i_{1},i_{1}+1,\ldots,i_{n+1}-n$ . Indeed, this results from (35) for $n=1$ . Formula (39) shows us that if the property is true for $n-1$ it will be true also for $n$ .
9. - Functions $f_{n,i}^{*}$ enjoy more complete properties. We have the

Theorem 7.-Functions $f_{n,i}^{*}$ are non-concave of orders $-1,0,1,\ldots,n-1$ on the ordered sequence (9).

So we have

\left[x_{i_{1}},x_{i_{q}},\ldots,x_{i_{k+1}};f_{n,i}^{*}\right]\geq 0,

(

\{40\}

)

whatever solent $i=1,2,\ldots,m-n,k=0,1,\ldots,n$ .
The demonstration can be done by induction, as for Theorem 6. For $n=1$ ownership is easily verified. For $n>1$ we have the recurrence formula

\begin{gathered}\text{ 41) }\\ {\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k+1}};f_{n,i}^{*}\right]=}\\ =\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k+1}};f_{n-1,}^{*}\right]\left(x_{i_{k+1}}-x_{i+n-1}\right)+\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}};f_{n-1,i}^{*}\right]\end{gathered}

generalization of (38) and which gives us the demonstration.
But, it is possible here to simplify the demonstration for $k<n$ . Indeed, theorem 6 being demonstrated, we can apply theorem 3. It is therefore sufficient to demonstrate the inequalities

\Delta_{k}^{j}\left(f_{n,i}^{*}\right)\geq 0,j=1,2,\ldots,m-k.

(42)

Formula (41) gives us, in particular,

\Delta_{k}^{j}\left(f_{n,i}^{*}\right)=\left(x_{i+k}-x_{i+n-1}\right)\Delta_{k}^{j}\left(f_{n-1,i}^{*}\right)+\Delta_{k-1}^{i}\left(f_{n-1,i}^{*}\right).

(43)

The proof of Theorem 7 is then immediate.
From the recurrence formula (43) we also deduce that

\Delta t_{n,i}^{\prime}\left(f_{n}^{*}\right)\begin{cases}=0,&j=1,2,\ldots,i+n-k-1,\\ >0,&j=i+n-k,i+n-k+1,\ldots,m-k.\end{cases}

So, taking into account formulas (28) and (30), we see that in (40) the sign $=$ is valid for $i>i_{k+1}-n$ and the sign $>$ is valid for $i\leqq i_{n+1}-n(k<n)$ .
10. - We will now say a few words about the polynomial (17). The function

\varphi_{i+1,n-1}-f_{n,i}^{*}

(44)

is analogous to $f_{n,i}^{*}$ with respect to the sequence (9) whose points are taken in the opposite direction. More precisely, the convexity properties of the function (44) are deduced from those of $f_{n,i}^{*}$ by changing the variable $x$ in $-x$ and function $f$ in $(-1)^{n-1}f$ . Now, these changes have the effect of preserving
any convexity property whose order is of the same parity with $n$ and to change the direction of any convexity whose order is of different parity with $n$ . We deduce the

Theorem 8.- Functions $\varphi_{i+1,n-1}-f_{n,i}^{*}$ are non-concave of orders $n-2,n-4,\ldots,n-2\left[\frac{n+1}{2}\right]$ ) and non-convex of orders $n-1,n-3,\ldots;n-2\left[\frac{n}{2}\right]-1$ on the ordered sequence (9).

If we notice that

\psi_{i+1,n-1}=\left(\psi_{i+1,n-1}-f_{n,i}^{*}\right)+f_{n,i}^{*}

and if we take into account theorems 4 and 8, we deduce the

Theorem 9. - Polynomials $\varphi_{i+1,n-1}$ are non-concave of orders $n-2,n-4,\ldots,n-2\left[\frac{n+1}{2}\right]$ on the ordered sequence (9). Moreover, the polynomial $\varphi_{i+1,n-1}$ is non-convex of orders $n-1,n-3,\ldots$ . $n-2\left[\begin{array}[]{c}n\\ 2\end{array}\right]-1$ on the points $x_{1},x_{2},\ldots,x_{i+n-1}$ and is non-concave in orders $n-1,n-3,\ldots,n-2\left[\frac{n}{2}\right]-1$ on the points $x_{i+1},x_{i+2},\ldots$ , $x_{m}{}^{,2}$ ).
11. - Now consider a function $f$ defined on the ordered sequence (9) of $m\geq n+1$ points, $n$ being a natural number.

Let's consider the difference

\psi_{r,n-1}=\psi_{r,n-1}(x)=L\left(x_{r},x_{r+1},\ldots,x_{r+n-1};f(x)-f(x)\right.

(45)

between the Lagrange polynomial having as nodes the a consecutive points $x_{r},x_{r+1},\ldots,x_{r+n-1}$ and between the function $f$ .

We propose to study the divided differences

\Delta_{k}^{j}\left(\psi_{r,n-1}\right)=\left[x_{j,}x_{j+1},\ldots,x_{j+k};\psi_{r,n-1}\right]

(46)

functions (45). From (45), it is clear that if $k\geq n$ on a

\Delta_{k}^{j}\left(\psi_{r,n-1}\right)=-\Delta_{k}^{l}(f),

⁰⁰footnotetext: 1.

[x]

denotes the largest integer

\leq a

, 2. The polynomial is also obviously of order

n-1

on (9).

but we want to examine, especially the case $k<n$ .
We see that (46) is a linear functional of $f$ , in which, moreover, only the values of $f$ to the points $x_{j},x_{j+1}$ , $\ldots,x_{j+k^{r}}x_{r},x_{r+1},\ldots,x_{r+n-1}$ intervene. Note that this linear functional is zero for any polynomial of degree $<n$ so formula (26) is applicable to it.

Let us therefore write formula (26), now denoting by $F[f]$ the divided difference (46).

Note that

L\left(x_{r},x_{r+1},\ldots,x_{r+n-1};f_{n,i}^{*}\mid x\right)=\begin{cases}0,&\text{ si }i\geq r\\ \varphi_{i+1,n-1}(x),&\text{ si }i\leqq r-1.\end{cases}

We must now distinguish two cases.
I. $j\leq r+n-k-1$ . We then have

F\left[f_{n,i}^{*}\right]=0,\quad l=r,r\div 1,\ldots,m-n

and, for the function $f=f_{n,i}^{*}$ ,

\psi_{r,n-1}=\varphi_{i+1,n-1}-f_{n,i}^{*},\text{ si }i\leq r-1.

Taking into account Theorem 8, we see that

F\left[f_{n,i}^{*}\right]\left\{\begin{array}[]{l}\geq 0,\text{ pour }k=n-1,n-3,\ldots,n+1-2\left[\frac{n+1}{2}\right]\\ \leq 0,\text{ pour }k=n-2,n-4,\ldots,n-2\left[\frac{n}{2}\right]\\ i=1,2,\ldots,r-1\end{array}\right.

II. $j\geq r$ . We then have

F\left[f_{n,i}^{*}\right]=0,\quad i=1,2,\ldots,r-1

and, for the function $f=f_{n,i}^{*}$ ;

\psi_{r,n-1}=-f_{n,i}^{*},\text{ si }i\geq r.

Taking into account Theorem 7, we deduce that

\begin{gathered}F\left[f_{n,i}^{*}\right]\leq 0,\text{ pour }k=0,1,\ldots,n-1,\\ i=r,r+1,\ldots,n-1.\end{gathered}

In summary, we will retain the following result from the previous analysis.

Théoreme 10. - Si la suite (9) est ordonnée, la différence divisée (46), d’ordre $k\leqq n$ , de la fonction (45), peut s’exprimer sous la forme

\Delta_{k}^{i}\left(\psi_{r,n-1}\right)=\sum_{i=1}^{m-n}B_{i}\Delta_{n}^{i}(f),

(47)

où les coefficients $B_{i}$ sont indêpendants de la fonction $f$ et sont
$1^{0}$ non-négatifs si $j\leq r+n-k-1$ et $k=n-1,n-3,\ldots,n+1-2\left[\frac{n+1}{2}\right]$
$2^{0}$ non-positifs si $j\leq r+n-k-1$ et $k=n,n-2,n-4,\ldots,n-2\left[\frac{n}{2}\right]$ et si $j\geq r$ et $k=0,1,\ldots,n$ .
12. - Du théorème précédent nous allons déduire un résultat intéressant. Prenons toujours une foncticn $f$ définie sur la suite ordonnée (9). Considérons la fonction

\psi_{r,n}(x)=L\left(x_{r},x_{r+1},\ldots,x_{r+n};f\mid x\right)-f(x).

La formule (34) permet d’écrire

\psi_{r,n}(x)=\psi_{r,n-1}(x)+\varphi_{r,n}(x)\Delta_{n}^{r}(f).

Compte tenant de la formule (47), nous pouvons écrire

\Delta_{k}^{j}\left(\psi_{r,n}\right)=\Delta_{k}^{j}\left(\psi_{r,n-1}\right)\div\Delta_{k}^{j}\left(\xi_{r,n}\right)\Delta_{n}^{r}(f)=\sum_{i=1}^{m-n}B_{i}\Delta_{n}^{i}(f)+\Delta_{k}^{j}\left(\varphi_{r,n}\right)\Delta_{n}^{r}(f).

Mais, si nous prenons $f=x^{n}$ , le premier membre se réduit à 0 , donc

\sum_{i=1}^{m-n}B_{i}+\Delta_{k}^{j}\left(\varphi_{r,n}\right)=0

et finalement nous avons

\Delta_{k}^{i}\left(\psi_{r,n}\right)=\sum_{i=1}^{m-n}B_{i}\left\{\Delta_{n}^{i}(f)-\Delta_{n}^{r}(f)\right\}.

Déterminons maintenant l’indice $r$ tel que

\Delta_{n}^{r}(f)=\max_{t=1,2,\ldots,m-n}\left\{\Delta_{n}^{i}(f)\right\}

Compte tenant alors des théorèmes 3 et 10 , nous obtenons le

Théorème 11. - Etant donnée une fonction $f$ , définie sur la suite ordonnée (9), on peut toujours trouver un polynome de Lagrange de degré $n$ , ayant comme noeuds $n+1$ points consécutifs $x_{r},x_{r+1},\ldots,x_{r+n}$ , tel que la différence

\psi_{r,n}(x)=L\left(x_{r},x_{r+1},\ldots,x_{r+n};f\mid x\right)-f(x)

soit une fonction non-concave d’ordres $n-1,n-3,\ldots,n-2\left\lceil\frac{n}{2}\right\rceil-1$ .
On voit facilement qu’on peut aussi déterminer le nombre $r$ de manière que $\psi_{r,n}(x)$ soit non-convexe d’ordres $n-1n-3,\ldots,n-2\left[\frac{n}{2}\right]-1$ sur (9).

§ 3.

Les polynomes de Lagrange majorants et minorants.

13.
- —
  
  Nous dirons qu’un polynome $P(x)$ est majorant resp. minorant pour la fonction $f$ définie sur un ensemble linéaire $E$ si nous avons

P(x)\geq f(x),x\in E,

resp.

P(x)\leqq f(x),x\in E.

Dans ce § nous nous proposons d’examiner les polynomes de Lagrange d’un degré donné $n$ de la fonction $f$ qui sont majorants ou minorants pour cette fonction. Il est clair qu’il suffit d’étudier seulement les polynomes majorants. Les propriétés correspondantes des polynomes minorants en résulteront par symétrie.

Nous supposerons toujours que la fonction $f$ soit définie sur la suite ordonnée (9), Toute suite partielle telle que $x_{i_{1}}$ . $x_{i_{2}},\ldots,x_{i_{n+1}}$ sera supposée ordonnée, donc $i_{1}<i_{2}<\cdots<i_{n+1}$ .

Pour que le polynome de Lagrange de degré $n$ ,

L\left(x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n+1}};f\mid x\right)_{,}

(48)

soit majorant pour la fonction $f$ , il faut et il suffit, d’après la formule (33), que l’on ait

	$\displaystyle\left(x_{i}-x_{i_{1}}\right)\left(x_{i}-x_{i_{2}}\right)\ldots\left(x_{i}-x_{i_{n+1}}\right)\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n+1}},x_{i};f\right]\leqq 0$		(49)
	$\displaystyle i=1,2,\ldots,m_{1}$

en convenant de remplacer par 0 tout symbole de différence divisée prise sur des points non tous distincts.
14. - Tout polynome de degré $n$ qui prend les mêmes valeurs que la fonction $f$ aux $n$ points $x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n}}$ est de la forme
50) $L\left(x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n}};t\mid x\right)+A\left(x-x_{i_{1}}\right)\left(x-x_{i_{3}}\right)\cdots\left(x-x_{i_{n}}\right)$ ,

A étant une constante.
Si, en particulier,

A=\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n+1}};f\right],

le polynome (50) coîncide avec le polynome de Lagrange (48)
Pour que le polynome (50) soit majorant pour la fonction $f$ , il faut et il suffit que l’on ait
(51) $\left(x_{i}-x_{i_{1}}\right)\left(x_{i}-x_{i_{2}}\right)\cdots\left(x_{i}-x_{i_{n}}\right)\left\{A-\left[x_{i_{1}},x_{i_{3}},\ldots,x_{i_{n}},x_{i};f\right]\right\}\geq 0$ ,

i=1,2,\ldots,m

Ce système est équivalent au suivant
(52)

\begin{gathered}(-1)^{n-j}\left\{A-\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n}},x_{i};f\right]\right\}\geq 0,\\ i_{j}<i<i_{j+1},j=0,1,\ldots,n,\end{gathered}

en convenant de poser $i_{0}=0,i_{n+1}=m+1$ . Nous supposons qu’on n’écrit que les formules pour les valeurs de $j$ telles que $i_{j+1}-i_{j}>1$ , autrement, en effet, il n’existe pas de valeurs admissibles pour $i$ . Dans (52) nous avons donc au plus $n+1$ groupes d’inégalités, les inégalités de chaque groupe étant précédées toutes d’un même signe + ou-.

Proposons-nous de chercher les polynomes majorants (48) ayant les $n$ noeuds $x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n}}$ donnés. On peut voir facilement que

Theorem 12. - The compatibility of the system (52) is necessary and sufficient for there to exist at least one upper bound Lagrange polynomial of degree n and having the given n nodes $x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n}}$ .
15. - Let us seek, in particular, the condition for a
majorant Lagrange polynomial of degree $n$ and having $n$ knots $x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n}}$ data exists for any function $f$ defined on points (9).

For this we note that the divided differences

\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n}},x_{i};f\right]

Or $i_{1},i_{2},\ldots,i_{n}$ are given and $i=1,2,\ldots,m,i\neq i_{1},i_{2},\ldots,i_{n}$ , can be taken arbitrarily. It follows that the necessary and sufficient condition sought is that in (52) we only have inequalities all preceded by the same sign. We can easily see that for this it is necessary and sufficient that we have either

i_{1}-i_{0}=i_{3}-i_{2}=\cdots=i_{2p+1}-i_{2p}=1,p=\left[\frac{n}{2}\right]

(53)

or else

i_{2}-i_{1}=i_{1}-i_{3}=\cdots=i_{2p}-i_{2p-1}=1,p=\left\lceil\frac{n+1}{2}\right\rceil

(54)

always posing $i_{0}=0,i_{n+1}=m+1$ 1).
Finally, therefore,
Theorem 13.-For any function $f$ , defined on the $m(\geq n+1)$ ordered points (9), has at least one upper bound Lagrange polynomial of degree $n$ and having $n$ given nodes $x_{i_{1}}$ , $x_{i_{2}},\ldots,x_{i_{n}}$ , it is necessary and sufficient that we have either (53) or (54).

The conditions are, in particular, verified if

i_{1}=1,i_{2}=2,\ldots,i_{n}=n

and also if

i_{1}=m-n+1,i_{2}=m-n+2,\ldots,i_{n}=m

We therefore deduce
Theorem 14. - For any function $f$ , defined on the $m(\geq n+1)$ ordered points (9), there exists at least one upper bound Lagrange polynomial of degree n (>0) ² ). In particular, at least one of the polynomials

1.

This condition can also be written in the following form. The numbers $i_{j}$ completely determine an index $k$ such that we have $n\geq k\geq 0$ And $i_{k}=k,i_{k+1}>k+1$ . The condition is then: It is necessary and sufficient that one has, or else $k=n$ , or else

i_{k+2p}-i_{k+2p-1}=1,p=1,2,\ldots,\left[\frac{n-k+1}{2}\right],(k<n).

2.

For $n=0$ , see further Theorem 19,

L\left(x_{1},x_{2},\ldots,x_{n},x_{n+i};f\mid x\right),i=1,2,\ldots,m-n,

and also at least one of the polynomials

L\left(x_{m-n+1},x_{m-n+2},\ldots,x_{m},x_{i};f\mid x\right),i=1,2,\ldots,m-n,

is majorant for the function $f$ .
16. - The existence of at least one upper bound Lagrange polynomial of given degree $n$ can also be demonstrated by induction on the number $m$ points of the sequence (9). It is sufficient, in fact, to demonstrate the second part of theorem 14, hence the

Theorem 15. - At least one of the polynomials

L\left(x_{1},x_{2},\ldots,x_{n},x_{n+i};f\mid x\right),i=1,2\ldots,m-n

is majorant for the function $f$ .
The property is obvious to $m=n+1$ . Suppose it is true for $m-1$ points and demonstrate it for $m$ points ( $m>n+1$ ). Either

L\left(x_{1},x_{2},\ldots,x_{n},x_{n+k};f\mid x\right),(0<k<m-n),

(55)

an upper bound polynomial for the function $f$ on the $m-1$ premiers points $x_{1},x_{2},\ldots,x_{m-1}$ . Two cases can arise.

1.

The polynomial (55) is majorant on the $m$ points (9) and then the property is demonstrated.

II, The polynomial (55) is not majorant on the points (9). We then have

L\left(x_{1},x_{2},\ldots,x_{n},x_{n+i};f\mid x_{n}\right)<f\left(x_{m}\right).

Formula (33) shows us that

\left[x_{1},x_{2},\ldots,x_{n},x_{n+k^{\prime}}x_{m};f\right]>0.

I say that, in this case, the polynomial

L\left(x_{1},x_{2},\ldots,x_{n},x_{m};fx\right)

is majorant on (9). Indeed, it suffices to demonstrate that it is majorant on the $m-1$ first points (9). Taking into account (12) and (34), it is easy to obtain the formula

\begin{gathered}L\left(x_{1},x_{2},\ldots,x_{n},x_{m};f|x|-f(x)=\right.\\ =L\left(x_{1},x_{2},\ldots,x_{n},x_{n+k};f\mid x\right)-f(x)+\\ +\left(x_{m}-x_{n+k}\right)\left(x-x_{1}\right)\left(x-x_{2}\right)\ldots\left(x-x_{n}\right)\left[x_{1},x_{2},\ldots,x_{n},x_{n+k}x_{m};f\right]\end{gathered}

which demonstrates the property.
17. - Let us complete Theorem 14. We first demonstrate, as above, the

Throrfme. 15. - For any function f, defined on the points (9), where $m\geq n+1>2$ and for everything $k=1,2,\ldots,n-1,l^{\prime}un$ at least polynomials

\begin{gathered}L\left(x_{1},x_{2},\ldots x_{k},x_{k+i},x_{m-n+k+1},x_{m-n+k+2},\ldots,x_{m-1},x_{m};f\mid x\right)\\ i=1,2,\ldots,m-n\end{gathered}

is majorant.
Suppose $m>n+1$ And

i_{1}=r,i_{2}=r+1,\ldots,i_{n}=r+n-1,1<r<m-n+1.

In this case, if $n$ is even. condition (54) is verified. On the contrary, if $n$ est impair, aucune des conditions (53), (54) n’est vérifiée. Nous avons donc le

Théorime 17. - Pour toute fonction $f$ , définie sur les m $(>n+1)$ points $(9)$ , il existe toujours au moins un polynome de Lagrange majorant de degre pair $n(>0)$ et ayant $n$ points consécutifs quelconques de la suite $x_{2},x_{n},\ldots,x_{m-1}$ comme noeuds. La propriété n’est pas vraie pour n impair.

Prenons encore $n$ impair $>1,m>n+1$ et

i_{1}=1,i_{2}=r,i_{3}=r+1,\ldots,i_{n}=r+n-2,2<r<m-n+1.

La condition (53) est alors vérifiée et, compte tenant des résultats précédents, nous pouvons énoncer le

Théortime 18. - Pour toute fonction $f$ , définie sur les $m|>n+1|$ points $(9)$ , il existe au moins un polynome de Lagrange majorant de degré $n>1$ et ayant $n-1$ noeuds consécutifs $x_{r},x_{r+1},\ldots,x_{r+n-y}$ donnés quelconques. La propriété est vraie pour tout entier $r$ si on convient de poser $x_{i}=x_{j}$ pour $i=j(\bmod m)$ .
18.-Proposons-nous de chercher la condition pour que le polynome de Lagrange

L\left(x_{r},x_{r+1},\ldots,x_{r+n};f\mid x\right),\quad 1\leq r\leq m-n_{s}

(56)

ayant comme noeuds $n+1$ points consécutifs de la suite (9), soit majorant pour la fonction $f$ . D’après ( 49 ), pour cela il faut et il suffit que l’on ait

Ces coaditions peuvent encore s’ecrire, comple tenant de la formule de récurrence (12) et en faisant usage des notations (16),

\begin{gathered}(-1)^{n+1}\left\{\Delta_{n}^{r}(f)-\left[x_{r+1},x_{r+2},\ldots,x_{r+n},x_{i};f\right]\right\}\leq 0,\quad i=1,2,\ldots,r-1,\\ {\left[x_{r+1},x_{r+2},\ldots,x_{r+n},x_{i};f\right]-\Delta_{n}^{r}(f)\leqq 0,\quad i=r+1,r+2,\ldots,m.}\end{gathered}

Si nous supposons maintenant $n$ pair $\geq 0$ , et si nous détérminons $r$ de manière que

	$\displaystyle\Delta_{n}^{\nu}(f)$	$\displaystyle=\max\left\{\Delta_{n}^{i}f\right\}.$		(58)
		$\displaystyle=1,2,\ldots,m-n$

nous voyons que les inégalités (57) sont toujours vérifiées. Nous avons donc le

D’ailleurs, ca n’est qu’une partie du théorème 11.
19. - Cherchons maintenant la condition pour que tous les polynomes (56) soient majorants. Les inégalités (57) nous montrent que pour qu’il en soit ainsi, il faut que l’on ait, en particulier,
$1^{0}$ . Si $n$ est pair,

	$\displaystyle-\Delta_{n+1}^{r-1}(f)\leq 0,$	$\displaystyle r=3,\ldots,m-n$
	$\displaystyle\Delta_{n+1}^{r}(f)\leq 0,$	$\displaystyle r=2,\ldots,m-n-1$

donc la fonction doit être polynomiale d’ordre $n$ .
$2^{0}$ . Si $n$ est impair,

\Delta_{n+1}^{r}(f)\leq 0,\quad r=1,2,\ldots,m-n-1

et la fonction doit être non-convexe d’ordre $n$ .

Il est facile de voir que ces conditions sont aussi suffisantes et nous pouvons donc énoncer le

Théorème 20. - Pour que les polynomes de Lagrange (56) soient tous majorants pour la fonction $f$ , il faut et il suffit que. cette fonction soit
$1^{10}$ . Polynomiale d’ordre $n$ pour $n$ pair.
$2^{0}$ . Non-convexe d’ordre $n$ pour $n$ impair.
Réciproquement d’ailleurs, la propriété de majoration de tous les polynomes (56) est, sous une autre forme, la définitlon même de la non-convexité d’ordre $n$ si $n$ est impair et une forme de la définition de la non-concavité et de la nonconvexité simultanées, donc de la polynomialité d’ordre $n$ si $n$ est pair. Ceci résulte de l’interprétation à l’aide des poly.. nomes de Lagrange de la définition des fonctions d’ordre $n$ et du théorème 3.

On peut maintenant compléter le théorème 17, pour $n$ impair, de la manière suivante

Théortme 21. - Si $n$ est impair, pour qu’il existe des polynomes de Lagrange majorants de degré n et ayant comme noeuds $n$ points donnés consécutifs quelconques de la suite (9), il faut et il suffit que la fonclion $t$ soit non-convexe d’ordre n sur les points (9).

La condition est suffísante d’après le théorème 20. Pour voir qu’elle est aussi nécessaire, nous devons écrire les conditions de compatibilité des systèmes (52) correspondants. Ces conditions sont.

\begin{gathered}\max_{i=r+n,r+n+1,\ldots,m}\left\{\left[x_{r},x_{r+1},\ldots,x_{r+n-1},x_{i};f\right]\right\}\leq\\ \leq\min_{i=1,2,\ldots,r-1}\left\{\left[x_{r},x_{r+1},\ldots,x_{r+n-1},x_{i};f\right]\right\}\\ r=2,3\ldots,m-n\end{gathered}

En particulier, nous devons avoir

\begin{gathered}\Delta_{n}^{r}(f)-\Delta_{n}^{r-1}(f)=\left(x_{r+n}-x_{r-1}\right)\Delta_{n+1}^{r-1}(f)\leq 0,\\ r=2,3,\ldots,m-n,\end{gathered}

which demonstrates the property.
20. - Finally, let us examine a little the upper bound Lagrange polynomials of degree $n$ order functions $n$ .

Inequality (49) shows us that we have
Theorem 22. - For the polyomy (48) to be majorant, for any non-concave resp, non-convex function of order n ( $\geq 0$ ) on (9), it is necessary and sufficient that we have

i_{n-sp+2}-i_{n-2p+1}=1,\quad p=0,1,\ldots,\left[\frac{n+1}{2}\right]

(59)

resp.

i_{n-2p+1}-i_{n-2p}=1,\quad p=0,1,\ldots,\left[\frac{n}{2}\right]

(60)

assuming $i_{0}=0,i_{u+2}=m+1$ ,
If the function is convex resp. concave of order $n$ on (9) the polynomials thus determined are the only majorant Lagrange polynomials of degree $n$ .

Indeed, for a convex resp. concave function of order $\pi$ conditions (49) become

\begin{gathered}\left(i-i_{1}\right)\left(i-i_{2}\right)\ldots\left(i-i_{n+1}\right)<0,\text{ resp. }>0,\\ i_{j}<i<i_{j+1},j=0,1,\ldots,n+1\end{gathered}

and we find (59) and (60) respectively.
The minimum number of upper bound Lagrange polynomials of degree $n$ for a non-concave or non-convex function of order $n$ is equal to the number of solutions of the Diophantine system (59) or (60), under the hypothesis

0=i_{0}<i_{1}<\ldots<i_{n+1}<i_{n+2}=m+1

These numbers are easily calculated. Let us denote them by $N_{m}^{n},N_{m}^{\prime n}$ respectively. We easily find the recurrence relations

N_{m}^{\prime 2p+1}=N_{m-q}^{\prime 2p-1}+N_{m-s}^{\prime 2p-1}+\cdots+N_{p}^{\prime 2p-1},N_{m}^{\prime 1}=m-1.

from where

N_{m}^{\prime\prime\mu+1}=\binom{m-p-1}{p+1}

We then have

N_{m}^{v_{0}}=1,\quad N_{m}^{u}=1,\quad N_{m}^{2p}=N_{m}^{vp}=N_{m-1}^{2p-1},N_{m}^{2p+1}=N_{m-2}^{2p-1}

from this we deduce

N_{m}^{n}=\left\{\begin{array}[]{c}m-\left[\begin{array}[]{c}n+3\\ 2\end{array}\right]\\ {\left[\frac{n}{2}\right]}\end{array}\right],\quad N_{m}^{\prime n}=\left[\begin{array}[]{c}m-\left[\frac{n+2}{2}\right]\\ {\left[\frac{n+1}{2}\right]}\end{array}\right].

We can also ask ourselves the problem of determining the minimum number of upper bound Lagrange polynomials of degree $n$ that any function defined on the $m$ points (9). This number is $\leqq N_{m}^{n}$ , according to Theorem 22. For $n=1$ it is obviously equal to 1. For $n>1$ it follows from the previous results that this number increases indefinitely with $m$ , but its exact value remains to be found. The same problem can be posed for the minimum number of Lagrange polynomials of given degree $n$ which are majorants or minorants. From the above it follows that this number is equal to 3 for $n=1$ And $m>2$ .
21. - The geometric interpretation of the properties studied is very simple for $n=0$ , and for $n=1$ .

Let us figure out the representative points $M_{i}\left(x_{1},f\left(x_{i}\right)\right)$ relative to the coordinate axes.

And $n=0$ , any Lagrange polynomial of degree $n$ is a constant, therefore represented by a parallel to the axis $Ox$ , not* passing through one of the points $M_{i}$ . The existence of an upper bound Lagrange polynomial simply means that the function reaches its maximum on (9). There is, in general, only one upper bound Lagrange polynomial of degree 0. Theorem 20 applies, but becomes a trivial property.

It is also almost obvious that, for $n$ any, if all Lagrange polynomials of degree $n$ are majorants, the function is polynomial of order $n$ . Indeed, there surely exists a lower bounding Lagrange polynomial. This polynomial must also be upper bounding and the function therefore coincides with it on (9).

And $n=1$ , a Lagrange polynomial of degree $n$ is represented by the line joining two points $M_{i}$ . Existence Theorem 14 means that the point system $M_{8}$ has at least one support line passing through two points and leaving all other points non-above $M_{i}$ . More precisely, there exists at least one such support line passing through $M_{1}$ and at least one passing through $M_{m}$ . It is easy to see that all sides of the smallest convex polygon containing the points $M_{i}$ represent Lagrange polynomials of degree 1 which are upper or lower bounds for the function $f$ . Theorem 20 also has a simple interpretation, It means that the polygon $M_{1}M_{2}\ldots M_{m}$ must be convex, the points $M_{i}$ being not-above-right $M_{1}M_{m}{}^{*}$

$\S 4$ .

On the Lagrange polynomials of n-order functions

22.-From the results of the previous § we deduce the following property

THEOREM 23. - Every non-negative function on the sequence (9) has at least one Lagrange polynomial of degree n which is also non-negative on the sequence (9).

It is sufficient, in fact, to take an upper bound polynomial. We see that the property remains true if instead of a non-negative function we take a positive function and we require the existence of a positive Lagrange polynomial.

We now propose to generalize this property and to seek whether, given a function $f$ nonconcave order $k$ on the ordered sequence (9) of $m(\geq n+1)$ points, we can find a Lagrange polynomial of degree $n$ which is also non-concave in order $k$ on (9).

Theorem 23 shows us precisely that for $k=-1$ the answer is always affirmative. We will see, on the contrary, that for $k\geq 0$ it is not so.

We will mainly examine conditions under which we can assert that for any non-concave function of order k there exists at least one Lagrange polynomial of degree $n$ which is also non-concave in order $k$ on (9).
23. - Let done $f$ a non-concave function of order $k\geq 0$ on the $m$ points (9).

It is clear that if $n\leq k+1$ , any Lagrange polynomial of degree $n$ is non-concave (even polynomial if $n\leqq k$ ) of order $k$ on (9). We also see that it is useless to examine the case $m=n+1$ since then there is only one Lagrange polynomial of degree $n$ , which coincides on (9) with the function $f$ and which is therefore, obviously, non-concave in order $k$ on (9).

So let's suppose that $n>k+1,m>n+1$ .
Theorem 11 allows us to state the following property
Theorem 24. - If $n$ is an integer $>k+1$ and of different parity with $k$ and if the function $f$ is non-concave of order $k$ on the $m>n+1$ points (9), there exists at least one Lagrange polynomial of degree n which is also non-concave of order $k$ on points (9).

Indeed, if $r$ is determined by condition (58), the function

\psi_{r,n}=L\left(x_{r},x_{r+1},\ldots,x_{r+n};f(x)-f(x)\right.

is non-concave of order $k$ on (9). Applying Theorem 4 we see that

\psi_{r,n}+f=L\left(x_{r},x_{r+1},\ldots,x_{r+n};f\mid x\right)

is also non-concave of order $k$ on points (9).
The existence of at least one Lagrange polynomial of degree $n$ and non-concave of order $k$ is therefore demonstrated for $n=k+3,k+5,\ldots$
24. - We will now construct an example which will demonstrate to us that for $n=k+2,k+4,\ldots,(k\geq 0)$ there do not always exist Lagrange polynomials of degree n and non-concave of order k on the points (9).

So we will assume $n\geqq k+2,n$ of the same parity with $k$ And $m\geq n+2$ .

As a continuation (9) we will choose the following

	$\displaystyle x_{1}=-1,x_{t}=(i-2)\lambda,i=2,3,\ldots,n+1,$		(61)
	$\displaystyle x_{n+1}<x_{n+2}<\cdots<x_{m}=(n-1)\lambda+1,$

Or $\lambda$ is a positive number. Points of the form $x_{i}$ with $i\geq n+1$ are written $x_{i}=(n-1)\lambda+\theta_{i}$ , THE $\theta_{i}$ being independent of $\lambda$ . We have done $0=\theta_{n+1}<\theta_{n+2}<\cdots<\theta_{m}=1$ .

Consider the polynomial of degree $n,P(x)$ , which is zero at points $x_{3},x_{1},\ldots,x_{n+1}$ et qui prend la valeur $(-1)^{k+1}$ au point $x_{1}$ et la valeur positive $b$ au point $x_{m}$ . Nous avons
(62) $P(x)=\frac{(x-\lambda)(x-2\lambda)\cdots(x-\bar{n}-1\lambda)\{[b(\bar{n}-1\lambda+1)-1]x+(1+b)(\bar{n}-1\lambda+1)\}}{(\lambda+1)(2\lambda+1)\cdots(\bar{n}-1\lambda+1)(\bar{n}-1\lambda+2)}$

Ce polynome prend la valeur

P\left(x_{2}\right)=\frac{(-1)^{k+1}(n-1)!\lambda^{n-1}(1+b)}{(\lambda+1)(2\lambda+1)\cdots(n-2\lambda+1)(n-1\lambda+2)}

au point $x_{2}{}^{1}$ ) et nous avons

\lim_{x\rightarrow+\infty}P\left(x_{i}\right)=(-1)^{k+1}(1+b)

Nous allons démontrer que pour $\lambda$ assez grand le polynome $P(x)$ n’est certainement pas non-concave d’ordre $k$ sur

⁰⁰footnotetext: 1. Pour

n=2

, ce qui exige

k:=0

, il n’y a pas de facteurs de la forme

i\lambda+1

, Cette remarque s’applique aussi plus loin.

(61). Pour cela il suffit de démontrer que la différence divisée $\Delta_{k+1}^{1}(P)$ , où nous faisons usage de la notation (16), est négative si $\lambda$ est assez grand. Un calcul simple, basé sur la formule (11), nous montre que

\Delta_{k+1}^{1}(P)=\frac{(-1)^{k+1}P\left(x_{1}\right)}{(\lambda+1)(2\lambda+1)\cdots(k\lambda+1)}+\frac{(-1)^{k}P\left(x_{2}\right)}{k!\lambda^{k}}

et nous en déduisons

\lim_{\lambda\rightarrow+\infty}\lambda^{k}\Delta_{k+1}^{1}(P)=-\frac{b}{k!}<0

d’où résulte notre propriété.
Mettons encore en évidence une autre propríété du polynome $P(x)$ .

Soient $3\leqq i_{1}<i_{2}<\cdots<i_{r}\leqq n+1,n+2\leqq j_{1}<j_{2}\cdots<j_{k-r+2}\leqq m_{1}1\leq r\leq k+1$ et considérons le polynome
$Q(x)=(\lambda+1)(2\lambda+1)\ldots(\overline{n-1}\lambda+1)\overline{(n-1}\lambda+2)\frac{P(x)}{\left(x-x_{i_{1}}\right)\left(x-x_{i_{2}}\right)\ldots\left(x-x_{i_{r}}\right)}$ .
Nous nous proposons d’étudier la différence divisée, qui, d’après la formule (14), s’écrit
$\left[x_{j_{1}},x_{j_{2}},\ldots,x_{j_{k-r+2}};Q(x)\right]=\left[\theta_{j_{1}},\theta_{j_{2}},\ldots,\theta_{j_{k-r+2}};Q(\bar{n}-\overline{1}\lambda+1]\right.$ , pour $\lambda$ très grand. Nous avons

Q\overline{(n-1}\lambda+x)=\sum_{i=0}^{n-r+1}\lambda^{n-r-i+1}\left(bA_{i}+B_{i-1}\right)

où $A_{i},B_{i}$ sont des polynomes de degré $i$ en $x$ , indépendants de $\lambda$ et de $b_{0}$ . On a, d’ailleurs, $B_{-1}=0,B_{0}=0,A_{n-n+1}=0$ et aussi $A_{0}=0$ lorsque $i_{r}\leq n$ .

Nous en déduisons le lemme suivant
Lemme 1. - Si $3\leq i_{1}<i_{2}<\cdots<i_{r}\leq n+1,n\div 2\leqq j_{1}<j_{2}<<\cdots<j_{k-m+2}\leqq m,1\leqq r\leq k+1$ , la limite

\lim_{\lambda\rightarrow+\infty}\lambda^{k}\left[x_{j_{1}},x_{j_{2}},\ldots,x_{j_{k-r+2}};\frac{P(x)}{\left(x-x_{i_{1}}\right)\left(x-x_{i_{2}}\right)\ldots\left(x-x_{i_{r}}\right)}\right]

existe et est de la forme $b\alpha$ , où $\alpha$ est un nombre indépendant de $b$ .
25. - Définissons maintenant la fonction $f$ sur les points (61) de la manière suivante

f\left(x_{2}\right)=0,\quad f\left(x_{i}\right)=P\left(x_{i}\right),\quad i=1,3,4,\ldots,m_{4}

(63)

où $P(x)$ est le polynome (62).
Cette fonction est non-concave d’ordre $k$ sur les points (61). En effet, d’après les théorèmes 4 et 9 , elle est non-concave d’ordre $k$ sur les points $x_{3},x_{4},\ldots,x_{m}$ . Nous avons donc $\Delta_{k+3}^{i}(f)\geq 0$ , $i=3,4,\ldots,m-k-1$ . De plus, nous avons $\Delta_{k+1}^{s}(f)=0$ et
$\Delta_{k+1}^{1}(f)=\frac{(-1)^{k+1}f\left(x_{1}\right)}{(\lambda+1)(2\lambda+1)\ldots(k\lambda+1)}=\frac{1}{(\lambda+1)(2\lambda+1)\ldots(k\lambda+1)}>0$ ,
ce qui démontre la propriété.
Nous allons démontrer qu’on peut choisir le nombre positif $b$ tel que pour $\lambda$ assez grand, la fonction $f$ n’ait aucun polynome de Lagrange de degré $n$ qui soit aussi non-concave d’ordre $k$ sur (61).

Pour les polynomes de Lagrange qui n’ont pas le point $x_{2}$ comme noeud il est clair qu’ils ne peuvent être non-concaves d’ordre $k$ . Ces polynomes coïncident, en effet, avec le polynome $P(x)$ , par suite de la définition de la fonction $f$ .

Considérons maintenant un polynome de Lagrange qui a le point $x_{2}$ comme noeud mais qui n’a pas le point $x_{1}$ comme noeud. Un tel polynome est de la forme

	$\displaystyle L(x)=L\left(x_{2},x_{i_{1}},x_{i_{2}},\ldots,x_{i_{r}},x_{j_{1}},x_{j_{2}},\ldots,x_{j_{n-r}};f\mid x\right),$
	$\displaystyle 3\leqq i_{1}<i_{2}<\cdots<i_{r}\leqq n+1,n+2\leq j_{1}<j_{2}<\cdots<j_{n-r}\leqq m,$		(64)
	$\displaystyle r=0,1,\ldots,n-1\text{ ). }$

Nous trouvons facilement

L(x)=P(x)+

$+\frac{(-1)^{n-1}P\left(x_{2}\right)\left(x-x_{i_{1}}\right)\left(x-x_{i_{3}}\right)\ldots\left(x-x_{i_{r}}\right)\left(x-x_{i_{1}}\right)\left(x-x_{i_{2}}\right)\ldots\left(x-x_{j_{n-r}}\right)}{x_{i_{1}}x_{i_{2}}\ldots x_{i_{r}}x_{i_{1}}x_{i_{2}}\ldots x_{i_{n-r}}}$ .

⁰⁰footnotetext: 1. Pour

r=0

il n’y a aucun noeud

x_{i}

avec

3\leqq i\leqq n+1

. Une remarque analogue s’applique aussi plus loin.

Nous allons chercher à préciser le signe de la différence divisée $\Delta_{k+1}^{1}(L)$ . Nous trouvons d’abord
(65) $\left\{\begin{array}[]{l}\lim_{\lambda\rightarrow+\infty}L\left(x_{1}\right)=(-1)^{k}b,\\ L\left(x_{2}\right)=0,\\ \lim_{\lambda\rightarrow+\infty}L\left(x_{i}\right)=\frac{(i-n-1)^{n-r}\left(i-i_{1}\right)\left(i-i_{2}\right)\ldots\left(i-i_{r}\right)}{(n-1)^{n-r}\left(i_{1}-2\right)\left(i_{2}-2\right)\ldots\left(i_{r}-2\right)}(1+b),\end{array}\right.$

i=3,4,\ldots,k+2.

Nous avons maintenant
$\Delta_{k+1}^{1}(L)=\frac{(-1)^{k+1}L\left(x_{1}\right)}{(\lambda+1)(2\lambda+1)\ldots(k\lambda+1)}+\frac{1}{k!\lambda^{k}}\sum_{y=0}^{k}(-1)^{k-r}\binom{k}{v}\frac{L\left(x_{v+2}\right)}{x_{v+2}-x_{1}}$
qui, d’après (65), nous donne

\lim_{\lambda\rightarrow+\infty}\lambda^{k}\Delta_{k+1}^{1}(L)=-\frac{b}{k!}<0.

Il en résulte que pour $\lambda$ assez grand les polynomes (64) ne sont certainement pas non-concaves d’ordre $k$ sur les points (61).

Il nous reste à examiner les polynomes de Lagrange qui ont les deux points $x_{1},x_{2}$ comme noeuds. Un tel polynome est de la forme

L\left(x_{1},x_{2},x_{i_{1}}x_{i_{2}},\ldots,x_{i_{r}}x_{j_{1}},x_{j_{3}},\ldots,x_{i_{n-r-1}};f\mid x\right),

(66)

\begin{gathered}3\leq i_{1}<i_{2}<\cdots<i_{r}\leq n+1,n+2\leqq j_{1}<j_{2}<\cdots<j_{n-r-1}\leqq m,\\ r=0,1,\ldots,n-1\end{gathered}

et nous trouvons facilement

L(x)=P(x)+

$\frac{(-1)^{n}P\left(x_{2}\right)(x-x)\left(x-x_{i_{1}}\right)\left(x-x_{i_{2}}\right)\ldots\left(x-x_{i_{r}}\right)\left(x-x_{j_{1}}\right)\left(x-x_{j_{3}}\right)\ldots\left(x-x_{j_{n-r-1}}\right)}{x_{i_{1}}x_{i_{2}}\ldots x_{i_{r}}x_{j_{1}}x_{j_{2}}\ldots x_{j_{n-r-1}}}$
en désignant maintenant par $L(x)$ le polynome (66).
Pour l’étude des polynomes (66) nous allons distinguer - plusieurs cas.
I. Supposons $r\leqq n-2$ . Considérons un point $x_{i}$ avec $3\leqq i\leqq n+1$ et qui ne coïncide pas avec l’un des points $x_{i_{1}},x_{i_{2}},\ldots,x_{i_{r}}$ . Such a point certainly exists. Let then the difference be divided

\left[x_{i},x_{i_{1}},x_{i_{2}},\ldots,x_{i_{s}},x_{i_{1}},x_{i_{3}},\ldots,x_{i_{k-s+1}};L\right].

Or $s=\min(k,r)$ This divided difference is, according to formula (13), the sum of the divided differences
(67) $\left[x_{i},x_{i_{1}},x_{i_{2}},\ldots,x_{i_{s}};\frac{L\mid x)}{\left(x-x_{i_{1}}\right)\left(x-x_{j_{2}}\right)\ldots\left(x-x_{i_{k-s+1}}\right)}\right]$ ,
(68) $\left[x_{j_{1}},x_{j_{2}},\ldots,x_{j_{k-s+1}};\frac{L(x)}{\left(x-x_{i}\right)\left(x-x_{i_{1}}\right)\left(x-x_{i_{2}}\right)\ldots\left(x-x_{i_{s}}\right)}\right]$ .

The difference divided (67) is equal to

\frac{L\left(x_{i}\right)}{\left(x_{i}-x_{i_{1}}\right)\left(x_{i}-x_{i_{2}}\right)\ldots\left(x_{i}-x_{i_{r}}\right)\left(x_{i}-x_{i_{1}}\right)\left(x_{i}-x_{i_{2}}\right)\ldots\left(x_{i}-x_{i_{k-s+1}}\right)}

But, we have

\lim_{\lambda\rightarrow+\infty}\frac{L\left(x_{i}\right)}{\lambda}=-\frac{(i-n-1)^{n-r-1}(i-2)\left(i-i_{1}\right)\left(i-i_{2},\ldots\left(i-i_{r}\right)\right.}{(n-1)^{n-r-1}\left(i_{1}-2\right)\left(i_{2}-2\right)\ldots\left(i_{r}-2\right)}(1+b)

and we deduce from it

\begin{gathered}\left.\lim_{\lambda\rightarrow+\infty}\lambda^{k}\left[x_{i}x_{i_{1}},x_{i_{2}},\ldots,x_{i_{s}};\left(x-x_{j_{1}}\right)\left(x-x_{j_{2}}\right)\ldots(x)\ldots x_{i_{h-s+1}}\right)\right]=\\ =-(1+b)\beta\end{gathered}

\beta=\frac{(l-n-1)^{n-k-r+s-2}(i-2)\left(i-i_{1}\right)\left(i-i_{2}\right)\ldots\left(i-i_{r}\right)}{(n-1)^{n-r-1}\left(i_{1}-2\right)\left(i_{2}-2\right)\ldots\left(i_{r}-2\right)\left(i-i_{1}\right)\left(i-i_{2}\right)\ldots\left(i-i_{s}\right)}.

In the divided difference (68) $L(x)$ can be replaced by $P(x)$ and Lemma 1 gives us

\lim_{\lambda\rightarrow+\infty}\lambda^{k}\left[x_{j_{1}},x_{j_{2}},\ldots,x_{i_{k-s+1}};\frac{L(x)}{\left(x-x_{i}\right)\left(x-x_{i_{1}}\right)\left(x-x_{i_{2}}\right)\ldots\left(x-x_{i_{s}}\right)}\right]=bx

Or $\alpha$ is a number independent of $b$ .
So finally we have

\lim_{\lambda\rightarrow+\infty}\lambda^{k}\left[x_{i},x_{i_{1}},x_{i_{2}},\ldots,x_{i_{s}},x_{i_{1}},x_{j_{2}}\ldots,x_{j_{k-s+1}};L\right]=-(1+b)\beta+b\alpha

(70)

Now let us discuss the result obtained.
$I_{1}$ And $n>2,i_{1}>3$ , we can take $i=3$ and we see that the number (69) is positive. The formula ('70) then shows us that if $b>0$ is chosen sufficiently small and $\lambda$ large enough, the polynomial (66) is certainly not non-concave of order $k$ on points (61). This case contains the case $r=0,n>2$ .
$\mathrm{I}_{2}$ And $1\leq r\leq n-3,i_{1}=3$ (this case requires $n\geq 4$ ), we can still choose $i$ so that $i\leqq n$ . We still have (70). The number (69) is not zero, but can be positive or negative. In this case we establish, exactly as above, the formula

\lim_{\lambda\rightarrow+\infty}\lambda^{\beta}\left[x_{i},x_{i_{2}},x_{i_{3}},\ldots,x_{i_{s}},x_{i_{1}},x_{i_{2}},\ldots,x_{j_{k-s+2}};L\right]=-(1+b)\beta^{\prime}+b\alpha^{\prime}

(71)

\beta^{\prime}=\frac{(i-n-1)^{n-k-r+s-3}(i-2)\left(i-i_{1}\right)\left(i-i_{2}\right)\ldots\left(i-i_{r}\right)}{(n-i)^{n-r-1}\left(i_{1}-2\right)\left(i_{2}-2\right)\ldots\left(i_{r}-2\right)\left(i-i_{1}\right)\left(i-i_{2}\right)\ldots\left(i-i_{s}\right)}=\frac{i-i_{1}}{i-n-1}\beta_{.}

The names $\beta$ And $\beta^{\prime}$ are therefore non-zero and of opposite signs. From (70) and (71) we see again that we can choose $b$ quite small and $\lambda$ large enough so that the polynomial (66) considered is not non-concave of order $k$ on points (61).
$\mathrm{I}_{3}$ And $r=n-2,n>2,i_{n-2}=n+1$ . In this case formula (70) is still valid with $\beta\neq 0$ , given by formula (69). We have $s=k$ . Consider the divided difference

\left[x_{2},x_{i},x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}};L\right]=\frac{L\left(x_{i}\right)}{\left(x_{i}-x_{2}\right)\left(x_{i}-x_{i_{1}}\right)\left(x_{i}-x_{i_{2}}\right)\ldots\left(x_{i}-x_{i_{k}}\right)}

We deduce from this

\lim_{\lambda\rightarrow+\infty}\lambda^{k}\left[x_{2},x_{i},x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}};L\right]=-(1+b)\beta^{\prime\prime},

(72)

\beta^{\prime\prime}=\frac{i-n-1}{i-2}\beta_{i}

Formulas (70), (72) also show us that the polynomial (66) cannot be non-concave of order $k$ and $b$ is quite small and $\lambda$ large enough.
II. Let us now consider the case $r=n-2,i_{n-2}=n$ , so the polynomial

L(x)=L\left(x_{1},x_{2},x_{3},\ldots,x_{n},x_{j};f\mid x\right),n+2\leq j\leq m.

(73)

We find

\lim_{\lambda\rightarrow+\infty}L\left(x_{n+1}\right)=(1+b)\theta_{j^{\prime}}\lim_{\lambda\rightarrow+\infty}L\left(x_{j}\right)=b\theta_{j^{\prime}}

If we notice that

\begin{gathered}{\left[x_{n-k+1},x_{n-k+2},\ldots,x_{n+1},x_{j};L\right]=}\\ =\frac{L\left(x_{n+1}\right)}{\left(x_{n+1}-x_{n-k+1}\right)\left(x_{n+1}-x_{n-k+1}\right)\ldots\left(x_{n+1}-x_{n}\right)\left(x_{n+1}-x_{j}\right)}\div\\ +\frac{L\left(x_{j}\right)}{\left(x_{j}-x_{n-k+1}\right)\left(x_{j}-x_{n-k+1}\right)\ldots\left(x_{j}-x_{n+1}\right)},\end{gathered}

we find

\lim_{\lambda\rightarrow+\infty}\lambda^{k}\left[x_{n-k+1},x_{n-k+2},\ldots,x_{n+1},x_{j};L\right]=-\frac{1}{k!}<0

which shows that for $\lambda$ large enough the polynomial (73) is not non-concave of order $k$ on points (61).

This result is also valid for $n=2$ . We then have $k=0,r=0$ And

\lim_{\lambda\rightarrow+\infty}\left[x_{3},x_{j};L\right]=-1

III. Finally, consider the polynomial

L(x)=L\left(x_{1},x_{2},\ldots,x_{n+1};f\mid x\right)

(74)

We have

L\left(x_{m}\right)=-1

and we deduce
$\left[x_{n-k+1},x_{n-k+2},\ldots,x_{n+1},x_{m};L\right]=-\frac{1}{(\lambda+1)(2\lambda+1)\ldots(k\lambda+1)}<0$ .
The polynomial (74) is therefore not non-concave of order $k$ on (61).

We have studied in this way all the polynomials of

Lagrange degree $n$ of the function (63) and we can state the following property

Theorem 25. - If the positive number b is small enough and the positive number $\lambda$ large enough, the function (63), non-concave of order $k\geq 0$ on the $m\geq n+2$ points (61), has no Lagrange polynomial of degree $n\geq k+2$ which is also non-concave of order k on the points (61), provided that $n$ be of the same parity with $k$ .
26. - Yes $n$ And $k$ are of the same parity, the existence of at least one Lagrange polynomial of degree $n$ and non-concave of order $k(>0)$ depends not only on the function $f$ , assumed to be non-concave of order $k$ , but also the distribution of points (9).

We will completely solve the problem in the simplest case which is $n=k+2,m=n+2=k+4$ . There then exists $k+4$ Lagrange polynomials of degree $n$ ,
$L_{i}(x)=L\left(x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{k+4};f\mid x\right),i=1,2,\ldots,k+4$ ,
and we have to study the sign of the three divided differences of order $k+1$ ,

\Delta_{k+1}^{1}\left(L_{i}\right),\quad\Delta_{k+1}^{2}\left(L_{i}\right),\quad\Delta_{k+1}^{3}\left(L_{i}\right),

still using the notation (16).
Taking into account the formula (33), we find

		$\displaystyle\Delta_{k+1}^{1}\left(L_{i}\right)=\Delta_{k+1}^{1}(f)-\left(x_{i}-x_{k+3}\right)\left(x_{i}-x_{k+4}\right)\Delta_{k+3}^{1}(f)$
		$\displaystyle\Delta_{k+1}^{2}\left(L_{i}\right)=\Delta_{k+1}^{2}(f)-\left(x_{i}-x_{1}\right)\left(x_{i}-x_{k+4}\right)\Delta_{k+3}^{1}(f)$
		$\displaystyle\Delta_{k+1}^{3}\left(L_{i}\right)=\Delta_{k+1}^{3}(f)-\left(x_{i}-x_{1}\right)\left(x_{i}-x_{2}\right)\Delta_{k+3}^{1}(f).$

The names

\Delta_{k+1}^{1}(f)=a,\quad\Delta_{k+1}^{2}(f)=b,\quad\Delta_{k+1}^{3}(f)=c,

are, by hypothesis, non-negative.
Formula (12) allows us to write
$\Delta_{k+3}^{1}(f)=\frac{\left(x_{k+4}-x_{2}\right)a-\left(x_{k+4}+x_{k+3}-x_{2}-x_{1}\right)b+\left(x_{k+3}-x_{1}\right)c}{\left(x_{k+3}-x_{1}\right)\left(x_{k+4}-x_{2}\right)\left(x_{k+4}-x_{1}\right)}$
and we deduce

\left\{\begin{array}[]{l}\Delta_{k+1}^{1}\left(L_{i}\right)=\frac{\left(x_{k+4}-x_{2}\right)\left(x_{i}-x_{1}\right)\left(x_{k+4}+x_{k+3}-x_{i}-x_{1}\right)a-}{-\left(x_{k+4}-x_{i}\right)\left(x_{k+3}-x_{i}\right)\left(x_{k+3}-x_{1}\right)c+}\\ +\left(x_{k+4}-x_{i}\right)\left(x_{k+3}-x_{i}\right)\left(x_{k+4}+x_{k+3}-x_{2}-x_{1}\right)b\\ \left(x_{k+3}-x_{1}\right)\left(x_{k+4}-x_{2}\right)\left(x_{k+4}-x_{1}\right)\\ \Delta_{k+1}^{2}\left(L_{i}\right)=\frac{\left.x_{1}\right)\left(x_{k+4}-x_{i}\right)\left[\left(x_{k+4}-x_{2}\right)a+\left(x_{k+3}-x_{1}\right)c\right]+}{+\left[\left(x_{k+4}-x_{2}\right)\left(x_{k+4}-x_{i}\right)\left(x_{k+3}-x_{i}\right)+\right.}\\ \left.+\left(x_{k+3}-x_{1}\right)\left(x_{i}-x_{1}\right)\left(x_{i}-x_{2}\right)\right]b\\ \left(x_{k+3}-x_{1}\right)\left(x_{k+4}-x_{2}\right)\left(x_{k+4}-x_{1}\right)\\ \Delta_{k+1}^{3}\left(L_{i}\right)=\frac{\left.1-x_{1}\right)\left(x_{k+4}-x_{i}\right)\left(x_{k+1}+x_{i}-x_{2}-x_{1}\right)c-3}{-\left(x_{i}-x_{1}\right)\left(x_{i}-x_{2}\right)\left(x_{k+1}-x_{2}\right)a+}\\ \frac{1}{\left(x_{i}-x_{1}\right)\left(x_{i}-x_{2}\right)\left(x_{k+4}+x_{k+3}-x_{2}-x_{1}\right)b}\\ \left(x_{k+3}-x_{1}\right)\left(x_{k+4}-x_{2}\right)\left(x_{k+4}-x_{1}\right)\end{array}\right.

We first have

\Delta_{k+1}^{2}\left(L_{i}\right)\geq 0,\quad i=1,2,\ldots,k+4,

We also have

		$\displaystyle\Delta_{k+1}^{1}\left(L_{k+3}\right)=\Delta_{k+1}^{1}\left(L_{k+4}\right)=\Delta_{k+1}^{2}(f)\geq 0,$
		$\displaystyle\Delta_{k+1}^{3}\left(L_{1}\right)=\Delta_{k+1}^{3}\left(L_{2}\right)=\Delta_{k+1}^{3}(f)\geq 0.$

and we see that the consequences

	$\displaystyle\Delta_{k+1}^{1}\left(L_{1}\right),\Delta_{k+1}^{1}\left(L_{2}\right),\ldots,\Delta_{k\div 1}^{1}\left(L_{k+1}\right),\Delta_{k+1}^{1}\left(L_{k+3}\right),$		(76)
	$\displaystyle\Delta_{k+1}^{3}\left(L_{2}\right),\Delta_{k+1}^{3}\left(L_{3}\right),\ldots,\Delta_{k+1}^{3}\left(L_{k+3}\right),\Delta_{k+1}^{3}\left(L_{k+4}\right),$		(77)

are monotonous.
There is no need to examine the case $\Delta_{k+3}^{t}(f)=0$ , because then the function is polynomial of order $k+2$ on the sequence (9) and all polynomials $L_{i}(x)$ coincide with $f$ And $\Delta_{k+s}^{1}(f)\neq 0$ , we can easily see that the sequence (76) is increasing resp. decreasing and the sequence (77) is decreasing resp. increasing depending on whether $\Delta_{k+s}^{1}(f)$ is positive resp. negative.

Now let's see in which cases we can find a function $f$ having no Lagrange polynomial of degree $k+2$ . which is non-concave of order $k$ on the $k+4$ points considered?

Either first $k=0$ . It follows that the necessary and sufficient condition sought is the compatibility of the system

\Delta_{1}^{1}\left(L_{2}\right)<0\quad\Delta_{1}^{3}\left(L_{3}\right)<0,

in non-negative unknowns $a,b,c$ . Ultimately, therefore, the compatibility of the system

\begin{gathered}a\geq 0,\quad b\geq 0,\quad c\geq 0\\ -\left(x_{2}-x_{1}\right)\left(x_{4}+x_{3}-x_{2}-x_{1}\right)a+\left(x_{3}-x_{1}\right)\left(x_{3}-x_{2}\right)c>\\ >\left(x_{3}-x_{3}\right)\left(x_{4}+x_{3}-x_{2}-x_{1}\right)b\\ \left(x_{3}-x_{2}\right)\left(x_{1}-x_{2}\right)a-\left(x_{4}-x_{3}\right)\left(x_{4}+x_{3}-x_{2}-x_{1}\right)c>\\ >\left(x_{3}-x_{2}\right)\left(x_{4}+x_{3}-x_{3}-x_{1}\right)b\end{gathered}

The condition sought is $\left(x_{3}-x_{1}\right)\left(x_{1}-x_{2}\right)\left(x_{3}-x_{2}\right)^{2}-\left(x_{2}-x_{1}\right)\left(x_{1}-x_{3}\right)\left(x_{4}+x_{3}-x_{9}-x_{1}\right)^{2}>0$ .

The first member is divisible by $x_{1}-x_{1}$ and we find
$\left(x_{3}-x_{2}\right)^{3}-3\left(x_{3}-x_{2}\right)\left(x_{2}-x_{1}\right)\left(x_{4}-x_{3}\right)-\left(x_{2}-x_{1}\right)\left(x_{4}-x_{3}\right)\left(x_{4}-x_{3}+x_{2}-x_{1}\right)>0$ from where, finally,

x_{3}-x_{2}>\sqrt[3]{\left(x_{2}-x_{1}\right)^{2}\left(x_{4}-x_{3}\right)}+\sqrt[3]{\left(x_{2}-x_{1}\right)\left(x_{1}-x_{3}\right)^{2}}

(78)

We can therefore state the following properties
Theorem 26. - The necessary and sufficient condition for there to exist a non-decreasing function on the four ordered points $x_{1},x_{2},x_{3},x_{4}$ and having no Lagrange plynome of degree 2 which is also non-decreasing on these points, is that inequality (78) is verified.

Theorem 27. - If the ordered points $x_{1},x_{2},x_{3},x_{4}$ verify the inequality

x_{3}-x_{2}\leq\sqrt[3]{\left(x_{2}-x_{1}\right)^{2}\left(x_{1}-x_{3}\right)}+\sqrt[3]{\left(x_{3}-x_{1}\right)\left(x_{4}-x_{3}\right)^{2}}

(79)

any non-decreasing function on these points has at least one Lagrange polynomial of degree 2 which is also non-decreasing on these same points.

Inequality (79) is verified, in particular, if the points $x_{1},x_{2},x_{3},x_{4}$ are equidistant. If the distribution of points $x_{1}$ , $x_{2},x_{3},x_{4}$ is symmetrical we can take, without restricting the generality, $x_{1}=-1,x_{2}=0,x_{3}=\lambda,x_{4}=\lambda+1$ and inequality (78) becomes $\lambda>2$ .
27. - Let us now suppose $k>0$ . According to the results of the previous No., the necessary and sufficient condition for
there to exist a function having no Lagrange polynomial of degree $k+2$ and non-concave of order $k$ is that there is a clue $i$ such as $2\leq i\leq k+2$ and such that the system

\Delta_{k+1}^{1}\left(L_{i}\right)<0,\quad\Delta_{k+1}^{3}\left(L_{i+1}\right)<0

be compatible in non-negative unknowns $a,b,c$ .
This compatibility condition is written, after calculations have been made,

	$\displaystyle\left(x_{k+s}-x_{i}\right)\left(x_{i+1}-x_{2}\right)\left(x_{i+1}-x_{i}\right)-$		(80)
	$\displaystyle-\left(x_{i}-x_{1}\right)\left(x_{k+4}-x_{i+1}\right)\left(x_{k+1}+x_{k+s}+x_{i+1}-x_{i}-x_{2}-x_{1}\right)>0.$

We deduce
Theorem 28. - If the $k+4$ ordered points $x_{1},x_{2},\ldots,x_{k+4}$ check the inequalities

	$\displaystyle\left(x_{k+3}-x_{i}\right)\left(x_{i+1}-x_{2}\right)\left(x_{i+1}-x_{i}\right)-$		(81)
	$\displaystyle-\left(x_{i}-x_{1}\right)\left(x_{k+1}-x_{i+1}\right)\left(x_{k+4}+x_{k+3}+x_{i+1}-x_{i}-x_{2}-x_{1}\right)\leq 0$
	$\displaystyle i=2,3,\ldots,k+2$

any non-concave function of order $k$ on these points has at least one Lagrange polynomial of degree $k+2$ which is also nonconcave in order $k$ on these same points.

For example, consider the sequence of points (61) for $n=k+2,m=k+4$ , done the points ( $\lambda>0$ ),
(82) $x_{1}=-1,x_{i}=(i-2)\lambda,i=2,3,\ldots,k+3,x_{k+i}=(k+1)\lambda+1$ .

Condition (80) becomes
$\delta_{i}(\lambda)=(k+3-i)(i-1)\lambda^{3}-(\overline{i-2}\lambda+1)(k+2-i\lambda+1)(\overline{2k+3}\lambda+1)>0$ .
More,

\delta_{i}(\lambda)-\delta_{i+1}(\lambda)=2(k+3-2i)\lambda^{2}(k+1\lambda+1),\quad\delta_{l}(\lambda)=\lambda_{k+4-i}(\lambda)

and it therefore follows that, if $\lambda$ checks the inequality

\delta_{2}(\lambda)>0

(83)

we can find a function $f$ non-concave order $k$ on the points (82) and having no Lagrange polynomial of degree $k+2$ non-concave order $k$ sur ces points. Une discussion simple nous montre que (82) est équivalent à $\lambda>\lambda_{0}$ , où $\lambda_{0}$ est la racine positive de l’équation en $\lambda_{\text{, }}$

\delta_{2}(\lambda)=(k+1)\lambda^{3}-k(2k+3)\lambda^{2}-(4k+3)\lambda-2=0.

Pour $k>0$ , on a $2k+1<\lambda_{0}<2k+2$ .
On peut encore voir que les inégalités (81) sont vérifiées si les points $x_{i}$ sont équidistants.
28. - Dans le cas $n=k+2$ , et même pour $m\geq n+2$ quelconque, on peut traiter autrement notre problème.

Démontrons d’abord le
Lemme 2.- Pour que le polynome $P(x)$ de degré $h+2$ soit non-concave d’ordre $k$ sur les points ordonnés (9), il faut et il suffit que l’on ait

\Delta_{k+1}^{1}(P)\geq 0,\quad\Delta_{k+1}^{m-k-1}\{P\}\geqq 0.

(84)

La démonstration est très simple. Les inégalités sont évidemment nécessaires. Montrons qu’elles sont aussi suffisantes. Soit

P(x)=c_{0}x^{k+2}+c_{1}x^{k+1}+\cdots

et, d’après (3) et (8),

\Delta_{k+1}^{i},(P)=c_{0}\left(x_{i}+x_{i+1}+\cdots+x_{i+k+1}\right)+c_{1}.

Nous en déduisons
$\Delta_{k+1}^{i}(P)=\frac{\left(\sum_{j=1}^{k+2}x_{m-j+1}-\sum_{j=1}^{k+2}x_{i+j-1}\right)\Delta_{k+1}^{1}(P)+\left(\sum_{j=1}^{k+2}x_{i+j-1}-\sum_{j=1}^{k+3}x_{j}\right)\Delta_{k+1}^{m-k-1}(P)}{\sum_{j=1}^{k+2}x_{m\cdot j+1}-\sum_{j=1}^{k+2}x_{j}}$
donc $\Delta_{h+1}^{i}(P)\geq 0,i=2,3,\ldots,m-k-2$ , ce qui démontre la propriété ¹ ),

Si maintenant $f$ est une fonction non-concave d’ordre $k$ sur les $m(\geq k+4)$ points ordonnés (9), nous pouvons écrire, en appliquant le lemme 2 , les conditions nécessaires et suffisantes pour que le polynome

L\left(x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k+2}},x_{i};f\mid x\right),\quad\left(1\leqq i_{1}<i_{2}<\cdots<i_{k+2}\leqq m\right)

⁰⁰footnotetext: 1. Des inégalités (84) iI résulte de plus que nous avons

(k+2)c_{0}x+c_{1}\geq 0

pour

x

compris dans l’intervalle

\left[\frac{1}{k+2}\sum_{j=1}^{k+2}x_{j},\quad\frac{1}{k+2}\sum_{j=1}^{k+2}x_{m-j+1}\right].

On en déduit que le polynome

P(x)

est non-concave d’ordre

k

dans cet intervalle.

soit aussi non-concave d’ordre $k$ sur (9). Compte tenant des formules (34), (13) et (8), nous trouvons que les inégalités (84) deviennent

\left\{\begin{array}[]{l}{\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k+2}};f\right]+\left(\sum_{j=1}^{k+3}x_{j}-\sum_{j=1}^{k+2}x_{i_{j}}\right)\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k+2}},x_{i};f\right]\geq 0,}\\ {\left[x_{i_{1}}x_{i_{2}},\ldots,x_{i_{k+2}};f\right]+\left(\sum_{j=1}^{k+2}x_{m-\rho+1}-\sum_{j=1}^{k+y}x_{i_{j}}\right)\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k+2}},x_{i};f\right]\geq 0.}\end{array}\right.

29. - Nous allons faire une application des résultats précédents, dans le cas $k=0$ . Prenons alors

i_{1}=1,i_{k+2}=m\quad\text{ et }\quad 1<i<m,

Les inégalités (85) deviennent

\left\{\begin{array}[]{l}{\left[x_{1},x_{m};f\right]-\left(x_{m}-x_{2}\right)\left[x_{1},x_{m},x_{i};f\right]\geq 0}\\ {\left[x_{1},x_{m};f\right]+\left(x_{m-1}-x_{1}\right)\left[x_{1},x_{m},x_{i};f\right]\geq 0}\end{array}\right.

Si nous remarquons que

\left[x_{1},x_{m},x_{i};\right]=\frac{f\left(x_{i}\right)}{\left(x_{i}-x_{1}\right)\left(x_{i}-x_{m}\right)}-\frac{\left[x_{1},x_{m};f\right]}{x_{i}-x_{m}}-\frac{f\left(x_{1}\right)}{\left(x_{i}-x_{1}\right)\left(x_{i}-x_{m}\right)}

nous déduisons que les inégalités (86) expriment que $f\left(x_{i}\right)$ doit être compris dans l’intervalle fermé $\left[a_{i},b_{i}\right]$ , oú

		$\displaystyle a_{i}=\frac{\left(x_{i}-x_{1}\right)\left(x_{i}-x_{2}\right)}{x_{m}-x_{2}}\left[x_{1},x_{m};f\right]+f\left(x_{1}\right)$
		$\displaystyle b_{i}=\frac{\left(x_{i}-x_{i}\right)\left(x_{m}+x_{m-1}-x_{i}-x_{1}\right)}{x_{m-1}-x_{1}}\left[x_{1},x_{m};f\right]+f\left\{x_{1}\right\}$

Nous pouvons supposer $\left[x_{1},x_{m};f\right]>0,\operatorname{car}\mathrm{si}\left[x_{1},x_{m};f\right]=0$ , la fonction (supposée non-décroissante) se réduit nécessairement à une constante. On voit alors que

a_{i}<b_{i},a_{i}<a_{i+1},b_{i}<b_{i+1},i=2,3,\ldots,m-1

Démontrons maintenant la propriété suivante
Théoreme 29. - Si les inégalités

a_{i+1}\leqq b_{i},i=2,3,\ldots,m-2

(87)

sont vérifiées, toute fonction non-décroissante définie sur les $m\geq 4$ points ordonnés (9) a au moins un polynome de Lagrange de degré 2 qui est aussi non-décroissant sur les points (9).

Montrons, en effet, que l’un au moins des polynomes

L\left(x_{1},x_{i},x_{m};f\mid x\right),\quad i=2,3,\ldots,m-1,

est non-décroissant.
Dans le cas contraire, il faudrait que $f\left(x_{i}\right)$ n’appartienne pas à $\left[a_{i},b_{i}\right]$ et ceci pour $i=2,3,\ldots,m-1$ . Mais, compte tenant de la non-décroissance de $f$ et de $a_{2}=f\left(x_{1}\right),b_{m-1}=f\left(x_{m}\right)$ , on voit qu’il faudrait alors que

f\left(x_{2}\right)>b_{2},f\left(x_{m-1}\right)<a_{m-1}

Il existe donc deux indices consécutifs $i,i+1$ tels que

f\left(x_{i}\right)>b_{i},f\left(x_{i+1}\right)<a_{i+1}

qui, en vertu de (87), nous donnerait $f\left(x_{i+1}\right)<f\left(x_{i}\right)$ , ce qui est absurde. La propriété est donc démontrée.

Les inégalités (87) peuvent s’écrire

\begin{gathered}\left(x_{m}-x_{2}\right)\left(x_{i}-x_{1}\right)\left(x_{m}+x_{m-1}-x_{i}-x_{1}\right)-\\ -\left(x_{m-1}-x_{1}\right)\left(x_{i+1}-x_{1}\right)\left(x_{i+1}-x_{2}\right)\geq 0\\ i=2,3,\ldots,m-1\end{gathered}

Ces inégalités sont vérifiées, en particulier, si les points $x_{i}$ sont équidistants, donc

Théorème 30. - Toute fonction non-décroissante, définie sur les $m\geq 4$ points ordonnés et équidistants (9), a au moins un polynome de Lagrange de degré 2 qui est aussi non-décroissant sur les points (9).
30.- Pour finir ce § et en revenant au cas $k>0$ , remarquons que si on suppose les points $x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k+2}}$ donnés, les inégalités (85) déterminent, en général, un intervalle auquel doit appartenir $f\left(x_{i}\right)$ . Les formules (13), (15), (14) et (7) permettent d’écrire

		$\displaystyle{\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k+2}},x_{i};f\right]=\frac{f\left(x_{i}\right)}{\left(x_{i}-x_{i_{1}}\right)\left(x_{i}-x_{i_{2}}\right)\ldots\left(x_{i}-x_{i_{k+2}}\right)}-}$
		$\displaystyle-\sum_{i=1}^{k++2}\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{j}};f\right]\frac{1}{\left(x_{i}-x_{i_{j}}\right)\left(x_{i}-x_{i_{j+1}}\right)\ldots\left(x_{i}-x_{i_{k+2}}\right)}$

Si nous posons

		$\displaystyle a_{i}=\frac{\left(x_{i}-x_{i_{1}}\right)\left(x_{i}-x_{i_{2}}\right)\ldots\left(x_{i}-x_{i_{k+1}}\right)\left(x_{i}-x_{i_{k+1}}+\beta-\alpha\right)}{\beta-\alpha}\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k+2}};f\right]+$
		$\displaystyle\quad+\sum_{j=1}^{k+1}\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i};f\right]\left(x_{i}-x_{i_{1}}\right)\left(x_{i}-x_{i_{2}}\right)\ldots\left(x_{i}-x_{i_{j-1}}\right),$
		$\displaystyle b_{i}=\frac{\left(x_{i}-x_{i_{1}}\right)\left(x_{i}-x_{i_{2}}\right),\ldots,\left\{x_{i}-x_{i_{k+1}}\right)\left(\gamma-\beta-x_{i}+x_{i_{k+2}}\right)}{\gamma-\beta}\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k+2}};f\right]+$
		$\displaystyle\quad+\sum_{j=1}^{k+1}\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{j}};f\right]\left(x_{i}-x_{i_{1}}\right)\left(x_{i}-x_{i_{2}}\right)\ldots\left(x_{i}-x_{i_{j-1}},\right.$

where, to simplify,

\alpha=\sum_{j=1}^{k+2}x_{j},\quad\beta=\sum_{j=1}^{k+2}x_{i_{j}},\quad\gamma=\sum_{j=1}^{k+2}x_{m-i+1}

$f\left(x_{i}\right)$ must belong to the closed interval of endpoints $a_{i},b_{i}$ , If we assume $i_{1}<m-k-1,\quad i_{k+y}>k+2$ , We have $\beta-\alpha>0,\gamma-\beta>0$ And $a_{i},b_{i}$ are indeed finite numbers. These numbers are, moreover, determined for all $i=1,2,\ldots,m$ and where to

a_{i_{j}}=b_{i_{j}}=f\left(x_{i_{j}}\right),\quad j=1,2,\ldots,k+2

Let us also note that the function $a(x)$ which takes the values $a_{i}$ to the points $x_{i}$ and the function $b(x)$ which takes the values $b_{i}$ to the points $x_{i}$ are non-concave of order $k$ on points (9). Indeed,

		$\displaystyle{\left[x_{i_{1}},x_{j_{2}},\ldots,x_{i_{k+2}};a\right]=\frac{\sum_{\gamma=1}^{k+2}x_{j_{r}}-\alpha}{\beta-\alpha}\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k+2}};f\right]\geq 0,}$
		$\displaystyle{\left[x_{i_{1}},x_{j_{2}},\ldots,x_{i_{k+2}};b\right]=\frac{\gamma-\sum_{\gamma=1}^{k+2}x_{j_{r}}}{\gamma-\beta}\left[x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k+2}};f\right]\geq 0.}$

§ 5.

32.
- —
  
  We will study a little the problems addressed in $\S\S 3$ and 4 in the case where the function $f$ is no longer defined
  on a finite number of points, but in an interval. To clarify the ideas we will assume that $f$ , finite and uniform, is defined in the finite and closed interval $[a,b]$ .

In these cases it is necessary to also consider: divided differences taken on points that are not all distinct and Lagrange polynomials having nodes that are not all distinct.

The Divided Difference

[\underbrace{x_{1},x_{1},\ldots,x_{1}}_{r_{1}},\underbrace{x_{2},x_{2},\ldots,x_{2},\ldots,x_{r_{s}},x_{s^{\prime}}\ldots,x_{s};f}_{r_{2}}]

(88)

Or $r_{i}$ points are confused with $x_{i},i=1,2,\ldots,s$ , is of order $n=r_{1}+r_{2}+\cdots+r_{s}-1$ and is obtained by passing to the limit. It is also expressed in the form of a quotient of two determinants $U$ And $V$ of the form (6). ME Nördund ¹ ) notes moreover that the divided difference (88) is equal to
$\frac{1}{\left(r_{1}-1\right)!\left(r_{2}-1\right)!\ldots\left(r_{s}-1\right)!}\frac{dr_{1}+r_{2}+\cdots+r_{s}-s}{dx_{1}^{r_{1}-1}dx_{2}^{r_{2}-1}\ldots dx_{s}^{r_{s}-1}}\left[x_{1},x_{2},\ldots,x_{s};f\right]$ .
This definition therefore requires the existence for $f$ of the order derivative $r_{i}-1$ to the point $x_{i}$ .

In the same way we define the Lagrange polynomial ² )

L(\underbrace{x_{1},x_{1},\ldots,x_{1}}_{r_{1}},\underbrace{x_{2},x_{2},\ldots,x_{2}}_{r_{2}},\ldots,\underbrace{x_{s},x_{s},\ldots,x_{s}}_{r_{s}};f(x)

of degree $n=r_{1}+r_{2}+\cdots+r_{s}-1$ , having multiple nodes. The node $x_{i}$ is in order $r_{i}$ of multiplicity (double, triple, etc. if $r=2,3,\ldots$ , etc.).

The existence of a knot $r_{i}^{upu}$ to the point $x_{i}$ still requires that the function be ( $r_{i}-1$ ) times differentiable at this point.

It is demonstrated that all the formulas established for divided differences and Lagrange polynomials taken at distinct points extend to this more general case.

⁰⁰footnotetext: 1. See loc. cit. above. 2. This is the Lagrange-Hermite polynomial. For more details on this polynomial see: N. E., Nörlund "Lectures on interpolation series". Paris 1926.

32. - Let us now examine the problem of upper bound Lagrange polynomials. In general, there are no such polynomials, but

Theorem 31.-Any upper semi-continuous function in the interval $[a,b]$ has at least one upper bound Lagrange polynomial of degree 0.

This is a trivial property. It simply expresses the well-known fact that a semi-continuous function above a finite and closed interval reaches its maximum, which is necessarily finite.

Let us now show that the upper semicontinuity hypothesis is still sufficient to assert the existence of at least one upper bound Lagrange polynomial of degree 1, so let us prove the

Theorem 32. - Any upper semi-continuous function in the interval $[a,b]$ has at least one upper bound Lagrange polynomial of degree 1.

Let us consider, in fact, the function

\alpha x+\beta-f(x),

Or $\alpha,\beta$ under two constants. This function is semi-continuous below, so reaches its minimum which is finite. Let $\mu(u,\beta)$ this minimum. Two cases are to be considered:
I. There are values of $\alpha,\beta$ such as $\mu(\alpha,\beta)$ be reached in two points $x_{1},x_{2}\in[a,b]$ , at least. So the polynomial

L\left(x_{1},x_{2};f\mid x\right)=\alpha x+\beta-\mu(\alpha,\beta)

is majorant.
II. The minimum $\mu(\alpha,\beta)$ is always reached at a single point. We then know that the function $f$ must be concave of order 1 in $[a,b]$ ). But, such a function is continuous in $[a,b]$ and has a derivative in $[a,b]$ , except perhaps on a set that is at most countable. There therefore exists a point $x_{1}\in[a,b]$ where the derivative $f^{\prime}\left(x_{1}\right)$ exists. The polynomial

L\left(x_{1},x_{1};f\mid x\right)

is then majorant.

⁰⁰footnotetext: 1. Tiberiu Popoviciu "Two remarks on convex functions", Bull. Sci. Acad, Roumaine, 20, 45-49 (1938).

33. - We do not know whether upper semi-continuity is sufficient for the generalization of Theorem 32 to the degree $n$ any. We will demonstrate the following less general property

Theorem 33. - Any continuous function is once differentiable in the interval $[a,b]$ has at least one upper bound Lagrange polynomial of degree n.

Indeed, either $T_{n}(x)$ the best-approximation Tchebycheff polynomial of degree $n$ and corresponding to the function $f$ in the meantime $[a,b].T_{n}$ is therefore the unique polynomial of degree $n$ for which the maximum

\max_{[a,b]}|f(x)-P(x)|

(89)

is as small as possible, $P(x)$ being a polynomial of degree $n$ . Let us designate by $\mu$ the minimum of (89).

We know that the difference $T_{n}-f$ alternately takes the values $\pm\mu$ in at least $n+2$ consecutive points of the interval $\left.[a,b]^{1}\right)$ . The polynomial $T_{n}+\mu$ is majorant for the function $f$ . It is a Lagrange polynomial of $f$ . Indeed, the function $T_{n}+\mu-f$ reaches its minimum 0 in at least: $\left[\frac{n+2}{2}\right]$ points. Yes $n$ is even, at most one of these points coincides with $a$ or $b$ And $n$ is odd it may be that two of these points coincide with $a$ or $b$ And $n$ is odd it may be that two of these points coincide with $a$ or $b$ respectively, but then the minimum is reached in at least $\frac{n+3}{2}$ points. The desired property results if we notice that any interior point of $[a,b]$ Or $T_{n}+\mu-f$ cancels out, can be taken as a double knot.
34. - Now let $f$ a non-negative function in $[a,b]$ . In the cases studied above, the existence of at least one Lagrange polynomial of degree $n$ which is also non-negative is demonstrated. But it seems to us that we can obtain more complete results. We will limit ourselves here to demonstrating the

Theorem 34. - Any non-negative function in $[a,b]\penalty 10000\ a$ at least one Lagrange polynomial of degree 2 which is also non-negative.

For the demonstration we will distinguish three cases:
I. $f(a)>0,f(b)>0$ . Let us then consider the polynomials

L(a,b;f\mid x)+A(x-a)(x-b),

(90)

There is a value $A_{0}$ such as for $A\leq A_{0}$ the polynomial (90) remains non-negative in $[a,b]$ . For $A=A_{0}$ this polynomial (90) has a double zero which is an interior point $x_{0}$ of $[a,b]$ , If we determine $A$ so that (90) takes the value $f\left(x_{0}\right)$ to the point $x_{0}$ , on a $A\leq A_{0}$ . It follows that the polynomial $L\left(a,b,x_{0};f\mid x\right)$ is non-negative in $[a,b]$ .
II. $f(a)=f(b)=0$ . We can easily see that the polynomial $L\left(a,b,x_{0};f\mid x\right)$ is non-negative whatever $x_{0}$ inside $[a,b]$ .
III. $f(a)=0,f(b)>0$ . Let us then consider the polynomial

\frac{f(b)}{(b-a)^{2}}(x-a)^{2}

(91)

If there is a point $x_{0}$ interior to $[a,b]$ where the function takes a value greater than the 'polynomial (91), the polynomial $L\left(a,b,x_{0};f\mid x\right)$ is non-negative in $[a,b]$ . Si l'on a

f(x)\leq\frac{f(b)}{(b-a)^{2}}(x-a)^{2},\quad x\in[a,b]

we immediately see that the derivative $f^{\prime}(a)$ to the point $a$ exists and is zero. In this case the polynomial (91) can be written $L(a,a,b;f\mid x)$ and is obviously non-negative in $[a,b]$ .

The conclusions are similar if $f(a)>0,f(b)=0$ .
Theorem 34 is true, obviously, for degree 0. For degree 1 we see that the polynomial $L(a,b;f\mid x)$ is non-negative in $[a,b]$ .
35. - Finally we will give a property of the Lagrange polynomials of non-concave functions of order $k$ in a very particular case.

Let's suppose $f$ non-concave order $k$ and admitting a continuous derivative of order $n$ In $[a,b]$ . We assume $n>k+2$
and of different parity with $k$ . Either $x_{0}$ a point where $f^{(n)}(x)$ reaches its maximum and then consider the Lagrange polynomial

L(x)=L(\underbrace{x_{0},x_{0},\ldots,x_{0}}_{n+1};f\mid x)=\sum_{i=0}^{n}\frac{\left(x-x_{0}\right)^{i}}{i!}f(i)\left(x_{0}\right).

I say that the polynomial $L(x)$ is non-concave of order $k$ In $[a,b]$ . It is sufficient to demonstrate that its derivative of order $k+1$ is non-negative in $[a,b]$ . We have

L^{(k+1)}(x)=\sum_{i=0}^{n-k-1}\frac{\left(x-x_{0}\right)^{i}}{i!}f^{(i+k+i)}\left(x_{0}\right)

But, Taylor's formula gives us

f^{(k+1)}(x)=\sum_{i=0}^{n-k-2}\frac{\left(x-x_{0}\right)^{i}}{i!}f^{(i+k+1)}\left(x_{0}\right)+\frac{\left(x-x_{0}\right)^{n-k-1}}{(n-k-1)!}f^{(n)}(\xi)

5 was a number between $x$ And $x_{0}$ .
We have done

		$\displaystyle L^{(k+1)}(x)=\left[L^{(k+1)}(x)-f^{(k+1)}(x)\right]+f^{(k+1)}(x)=$
		$\displaystyle=\frac{\left(x-x_{0}\right)^{n-k-1}}{(n-k-1)!}\left[f^{(n)}\left(x_{0}\right)-f^{(n)}(\xi)\right]+f^{(k+1)}(x)$

from which results the property.
Finally therefore
Theorem 35. - If $n$ is of different parity with $k$ , any non-concave function of order $k$ and admitting a derivative $n^{\text{ime }}$ continues in $[a,b]$ , has at least one Lagrange polynomial of degree n which is also non-concave of order $k$ In $[a,b]$ .

Note, moreover, that for a function having a derivative $n^{\text{ime }}$ continuous any, the difference $L(x)-f(x)$ is a non-concave function of order $k$ In $[a,b]$ yes, of course, $k$ And $n$ are of the same parity.

Mars 1943

Notes on higher-order convex functions (X)

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Booklet 1.

TABLE OF CONTENTS

NOTES ON HIGHER ORDER CONVEX FUNCTIONS (X)

On some properties of divided differences and Lagrange polynomials.

§ 1.

Some preliminary properties and formulas ¹ ).

§ 2.

New properties of divided differences of functions defined on a finite set.

§ 3.

Les polynomes de Lagrange majorants et minorants.

$\S 4$ .

On the Lagrange polynomials of n-order functions

Related Posts

Notes on higher-order convex functions (X)

Abstract

Authors

Keywords

Paper coordinates

PDF

About this paper

Journal

Publisher Name

DOI

Print ISSN

Online ISSN

References

Paper (preprint) in HTML form

(Manthematics, Physics, Chemistry)

Booklet 1.

TABLE OF CONTENTS

NOTES ON HIGHER ORDER CONVEX FUNCTIONS (X)

On some properties of divided differences and Lagrange polynomials.

§ 1.

Some preliminary properties and formulas 1 ).

§ 2.

New properties of divided differences of functions defined on a finite set.

§ 3.

Les polynomes de Lagrange majorants et minorants.

§​4\S 4.

On the Lagrange polynomials of n-order functions

Related Posts

Some preliminary properties and formulas ¹ ).

$\S 4$ .