ON A GENERALIZATION OF CONVEX FUNCTIONS

by
ELENA MOLDOVAN
in Cluj

This work is divided into 4 sections. In the $\S 1$ We present some properties of sets of interpolating functions. Sections 2 and 3 contain the definition and an introduction to the study of convex functions with respect to a set of interpolating functions. In Section 4, as an application, we present a mean value theorem for continuous functionals defined on the set of continuous functions on a given (finite and closed) interval. The proof of this theorem is based on the properties of convex functions as defined in Section 2.

This work is linked to the idea of studying the relationship of a set of functions with one of its subsets. Such as, for example, the relationship of functions defined on a given interval, with respect to the set of polynomials of a given degree, or the relationship of continuous functions that are differentiable a sufficient number of times, with respect to the integrals of a given differential equation.

In this work, we seek to avoid restricting the linearity of the sets of functions involved. We emphasize this point because current computational capabilities using high-speed electronic machines mean that the non-linear aspect of certain mathematical theories is no longer an obstacle to their practical applications.

In this work we have left aside the study of the differential properties of convex functions with respect to an interpolating set. We will return to these questions in another work.

Some properties of interpolating function sets

1.
- —
  
  Definition 1. - The set $\mathscr{F}_{n}$ , formed by real functions and of a real variable, is called an interpolating set of order $n$ on the en. seems linear $E$ , or simply a set of the type $I_{n}\{E\}$ if:
  (A). The elements of $\mathscr{F}_{n}$ are continuous on $E$ (
  B). Whatever the $n$ distinct points of $E$ (
  1)

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

and whatever the $n$ Numbers

y_{1},y_{2},\ldots,y_{n}

(2)

there is one function and only one $\varphi(x)\in\mathscr{F}_{n}$ such that one has

\varphi\left(x_{i}\right)=y_{i},\quad i=1,2,\ldots,n.

(3)

The function $\varphi(x)\in\mathscr{F}_{n}$ who meets conditions (3) will be designated by $L\left(x_{1},x_{2},\ldots,x_{n};y_{1},y_{2},\ldots,y_{n}\mid x\right)$ and also by $L\left(x_{1},x_{2},\ldots,x_{n};f\mid x\right)$ , when $f(x)$ is a function that takes the values (2) at the corresponding points (1).

If, in particular, the whole $E$ reduces to the closed interval $[a,b$ . resp. to the open interval ( $a,b$ ), For $I_{n}\{E\}$ we will also use $h$ rating $I_{n}[a,b]$ resp. $I_{n}(a,b)$ We use analogous notation in the case of half-open intervals. $(a,b],[a,b)$ The whole $p_{n}$ polynomials of degree $n-1$ is a set of the type $I_{n}(-\infty,\infty)$ and the whole $T_{2n+1}$ trigonometric polynomials of order $n$ is of the type $I_{2n+1}[a,b$ . if $ba<2\pi$ In general, the set of integrals of a linear and homogeneous differential equation of order $n$ with continuous coefficients over a given interval (the coefficient of $y^{(n)}$ (being equal to 1) is an interpolator of order $n$ on any sufficiently small subinterval [13].

In the previous examples, the set $f_{n}$ is linear*) There are also non-linear interpolation sets. A simple example of such a set is the set ' $D_{n}(A)$ polynomials $Ax^{n}+a_{1}x^{n-1}+\ldots+a_{n}$ Or $HAS$ is fixed and the other coefficients are variable. This set is of type dt $I_{n}(-\infty,\infty)$ and is not linear if $A\neq 0$ .

Sets of the type $I_{n}[a,b]$ were introduced by L. TORNHEN [14] under the name of sets with $n$ parameters and by Mr. I. MOROZOI [6] under the name of classes of approximation functions.

In this work we will use the mean theorems relating to sets of the type $I_{n}[a,b]$ and which we have already considered in other works [3,4]
2. Lemma 1. - $\mathscr{F}_{n}$ being a set of the type $I_{n}[a,b],n>1$ , s $\varphi_{1}(x),\varphi_{2}(x)\in F_{n}$ are distinct but coincide in $n-1$ distinct points

⁰⁰footnotetext:*). So, with two of its elements

f_{1},f_{2}

contains any linear combination

af_{1}+f^{i-}

of these elements,

\alpha,\beta

being any real numbers.

$x_{2},\ldots,x_{n-1}$ the difference $\varphi_{1}(x)-\varphi_{2}(x)$ changes sign while passing through every point $x_{i}\in(a,b)$ .

We have given the demonstration elsewhere [3].

Theorem 1. - Consider convergent sequences of numbers $\left\{x_{j}^{(i)}\right\}_{i=1}^{\infty},j=1,2,\ldots,n$ Or $x_{j}^{(i)}\in[a,b],j=1,2,\ldots,n,i=1,2,\ldots,x_{l}^{(i)}\neq x_{k}^{(i)}$ if $l\neq k$ and whose respective limits $x_{j},j=1,2,\ldots,n$ are assumed to be distinct. Let us also consider convergent sequences of numbers $\left\{y_{j}^{(i)}\right\}_{i=1}^{\infty}$ , $j=1,2,\ldots,n$ having respectively the limits of the numbers $y_{j},j=1,2,\ldots,n$ .

Under these conditions, the following
$\left\{L\left(x_{1}^{(i)},x_{2}^{(i)},\ldots,x_{n}^{(t)};y_{1}^{(i)},y_{2}^{(i)},\ldots y_{n}^{(i)}\mid x\right)\right\}_{i=1}^{\infty}$ whose terms belong to the set $\mathcal{F}_{n}$ of the type $I_{n}[a,b]$ , converges uniformly on $[a,b]$ towards the junction $L\left(x_{1},x_{2},\ldots,x_{n};y_{1},y_{2},\ldots,y_{n}\mid x\right)\in\mathscr{F}_{n}$ .

This is the theorem of L. Tornheim [14]. We also find it in M. I. Morozov [6].
3. - Definition 2. - The function $f(x)$ defined on the interval $[a,b]$ is said to be n-valent with respect to the set $f_{n}$ of the type $I_{n}[a,b]$ yes, whatever the function $\varphi(x)\in\mathcal{F}_{n}$ the difference $f(x)-\varphi(x)$ becomes zero, on the interval $[a,b]$ , on at most $n$ points.

This notion of function $n$ -valente was introduced, in the particular case $\mathcal{F}_{n}=\mathcal{P}_{n}$ , leaves. popoviciu [12]. In this case for $n=1$ , we find the classical notion of univalence.
Theorem 2. – If $f(x)$ is continuous on the interval $[a,b]$ and is $n$ -valued in relation to the set $\mathscr{F}_{n}$ of the type $I_{n}[a,b]$ and if the points $x_{i}\in[a,b]$ , $i=1,2,\ldots,n$ are distinct, the difference $f(x)-L\left(x_{1},x_{2},\ldots,x_{n};f\mid x\right)$ changes sign while passing through every point $x_{i}\in(a,b)$ .

To demonstrate this theorem, suppose that

x_{1}<x_{2}<\ldots<x_{n}

(4)

and either $n\geqq 2$ Suppose that the difference $f(x)-L\left(x_{1},x_{2},\ldots,x_{n};f\mid x\right)$ keep the same sign on the intervals ( $x_{k-1},x_{k}$ ), ( $x_{k},x_{k+1}$ ), Or $x_{k}\in(a,b)$ and where we placed $x_{0}=a,x_{n+1}=b$ To clarify, let's assume that the difference in question is $>0$ For $x\in\left(x_{k-1},x_{k}\right)\cup\left(x_{k},x_{k+1}\right)$ The elements of $f_{n}$ which coincide with $f(x)$ on the $n-1$ points $x_{1},x_{2},\ldots,x_{k-1},x_{k+1},\ldots,x_{n}$ form a set of the type $I_{1}\left(x_{k-1},x_{k+1}\right)$ If the sequence of numbers $\left\{\eta_{i}\right\}_{i=0}^{\infty}$ with positive terms tends towards zero, by virtue of Theorem 1, the sequence of functions

	$\displaystyle\left\{L\left(x_{1},x_{2},\ldots,\mathrm{x}_{n};f_{1},f_{2},\ldots,f_{k-1},\eta_{i}+f_{k},f_{k+1},\ldots,f_{n}\mid x\right)\right\}_{i=1}^{\infty}$		(5)
	$\displaystyle f_{i}=f\left(x_{i}\right),\quad i=1,2,\ldots,n$

tends uniformly towards $L\left(x_{1},x_{2},\ldots,x_{n};f\mid x\right)$ on any closed interval belonging to $\left(x_{k-1},x_{k+1}\right)$ From a certain rank onwards, the terms of the sequence (5) coincide with the function $f(x)$ on at least two distinct points of ( $x_{k-1},x_{k+1}$ This contradicts the assumption that the function $f(x)$ East $n$ -valent in relation to $\mathscr{f}_{n}$ The theorem is therefore proven for $n\geqq 2$ . If $n=1$ The demonstration is analogous, following on $\left\{L\left(x_{1};\eta_{i}+f_{1}\mid x\right)\right\}_{i=1}^{\infty}$ instead of (5).

The same procedure is followed if the difference $f(x)-L\left(x_{1},x_{2},\ldots,x_{n;1}\|_{x}\right.$ remains negative on $\left(x_{k-1},x_{k}\right)\cup\left(x_{k},x_{k+1}\right)$ .

In § 2 we will also give other properties of functions $n_{\text{va }}$ slow compared to a set of the type $I_{n}[a,b]$ 4.
- $F_{n}$ being always a set of the type $I_{n}[a,b]$ Let us consider a system of $n+1$ distinct points of the interval $[a,b]$ (
6)

x_{1}<x_{2}<\ldots<x_{n}<x_{n+1}

And $n+1$ any numbers
(7)

y_{1},y_{2},\ldots,y_{n},y_{n+1}

Definition 3. - The set of $n+1$ functions

	$\displaystyle L\left(x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1};y_{1},y_{2},\ldots,y_{i-1},y_{t+1},\ldots,y_{n+1}\mid x\right.$		(8)
	$\displaystyle i=1,2,\ldots,n+1$

is said to be an interpolatory system on the points (6) and with respect to the number (7).

For functions (8) we will also use the notation

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

If there is an element of $\mathscr{F}_{n}$ which at points (6) takes the corresponding values (i), the system (8) is formed by a single function. Otherwise, the system (8) is formed by $n+1$ distinct functions. Dan: in this case, any two of the functions (8) coincide on $n-1$ points of sequence (6). By virtue of Lemma 1, the difference

	$\displaystyle L\left(x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1};y\mid x\right)-$
	$\displaystyle-L\left(x_{1},x_{2},\ldots,x_{k-1},x_{k+1},\ldots,x_{n+1};y\mid x\right),\quad i\neq k$

changes sign by passing through the $n-1$ points where it cancels out. When $y_{n+1}>L\left(x_{1},x_{2},\ldots,x_{n};y\mid x_{n+1}\right)$ so we have

y_{i}>\text{ resp. }<L\left(x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1};y\mid x_{i}\right)

(9)

depending on $i+n+1$ is even or odd. When $y_{n+1}<L\left(x_{1},x_{2},\ldots,x_{n};y\mid x_{n+1}\right.$ the inequalities (9) occur depending on whether $i+n+1$ is odd resp even.

The functions (8) have important properties that follow from Definition 1 and Lemma 1. We will give some of these properties later.
5. - Lemma 2. - If $x_{n+1}<x_{0}\in[a,b]$ , we have
following that

	$\displaystyle\geqq L\left(x_{2},x_{3},\ldots,x_{n+1};y\mid x_{0}\right)\leqq\text{ resp. }\geqq$		(10)
	that

L\left(x_{1},x_{2},\ldots,x_{n};y\mid x_{n+1}\right)\geqq\text{ resp. }\leqq y_{n+1}

(11)

For the proof, let us first note that if in (11) we have equality, the set (8) is composed of a single function and Lemma 2 follows. Otherwise, taking into account that two of the functions (8) coincide in $n-1$ points (6), from lemma 1 it follows that (12) $L\left(x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1};y\mid x_{i}\right)>$ resp. $<y_{i},\quad i=2,3,\ldots,n$ depending on $i+n+1$ is even or odd, or depending on whether $i+n+1$ is odd resp. even if we have the first resp. the second inequality (11). In the first case, for $x_{n}<x<x_{n+1}$ , We have

		$\displaystyle L\left(x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1};y\mid x\right)<$
		$\displaystyle<L\left(x_{2},x_{3},\ldots,x_{n+1};y\mid x\right),\quad i=3,\ldots,n$

and done, for $x>x_{n+1}$ we will have

		$\displaystyle L\left(x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1};y\mid x\right)>$
		$\displaystyle>L\left(x_{1},x_{2},\ldots,x_{n+1};y\mid x\right),\quad i=3,\ldots,n$

In the second case we have for $x_{n}<x<x_{n+1}$ And $x>x_{n+1}$ respectively the opposing inequalities.

Lemma 3. - If $x_{n+1}<x_{0}\in[a,b]$ , We have

	$\displaystyle L\left(x_{1},x_{2},\ldots,x_{n};y\mid x_{0}\right)\geqq\text{ resp. }\leqq$		(13)
	$\displaystyle\leqq L\left(x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1};y\mid x_{0}\right),i=2,3,\ldots,n.$

following that we have (11).
The proof of this lemma is analogous to that of Lemma 2.
The inequalities indicated by Lemmas 2 and 3 give indications about the functions (8), which play a role in the study of sets of the type $I_{n}[a,b]$ 6.
- Let us consider the $m(\geqq n+1)$ points

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

(14)

and the numbers

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

(15)

On any group of $n$ distinct points $x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n}}$ extracts from sequence (14) we can construct the function $L\left(x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n}};y\mid x\right)$ which belongs to $\mathscr{F}_{n}$ of the type $I_{n}[a,b]$ We always assume that $1\leqq i_{1}<i_{2}<<\ldots<i_{n}\leqq m$ And $x_{m}<b$ We then have
theorem 3. - If $x_{i_{n}}<x_{0}\leqq b$ the number

L\left(x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n}};y\mid x_{0}\right)

(16)

is between the smallest and largest of the numbers

L\left(x_{j}x_{j+1},\ldots,x_{j+n-1};y\mid x_{0}\right),j=i_{1},i_{1}+1,\ldots,i_{n}-n+1

(17)

We can say that the value of the function $L\left(x_{i_{1}},\ldots,x_{i_{n}};y\mid x\right)_{\text{su }}$ the point $x_{0}$ is an average of the values on $x_{0}$ functions $L\left(x_{j},x_{j+1},\ldots,x_{j+n-1};y\mid x\right),j=i_{1},i_{1}+1,\ldots,i_{n}-n+1$ built on groups of $m$ consecutive points of the sequence (14). We have therefore

	$\displaystyle\min_{j=i_{1},i_{1}\div 1,\ldots,i_{n}-n\div 1}L\left(x_{j},x_{j\div 1},\ldots,x_{j\div\pi-1};y\mid x_{0}\right)\leqq$		(18)
	$\displaystyle\leqq L\left(x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n}};y\mid x_{0}\right)\leqq$

\leqq\max_{j=i_{1},i_{1}\div 1,\ldots,i_{n}-n\div 1}L\left(x_{j},x_{j\div 1},\ldots,x_{j\div n-1};y\mid x_{0}\right),\quad\left(x_{i_{n}}<x_{0}\leqq b\right)

The equalities hold if and only if the numbers (17) are all equal.
The proof of Theorem 3 is based on Lemmas 2 and 3. These two lemmas actually constitute the special case $1=i_{1}<i_{2}<\ldots<i_{n}==n+1$ of the theorem. We proceed by complete induction on the number $i_{n}-i_{1}$ points of the sequence (14) included between $x_{i_{1}}$ it $x_{i_{n}}$ We have $i_{n}-i_{1}\geqq n-1$ . If $i_{n}-i_{1}=n-1$ , the points $x_{i_{j}}$ are consecutive and the property is obviously true, with the sign $=$ in (18). If $i_{n}-i_{1}=n$ there is a clue $k$ , such as $i_{k}<i_{k}+1<i_{k+1}$ and the point $x_{i_{k+1}}$ is between $x_{i_{k}}$ And $x_{i_{k+1}}$ Let's consider the points

x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}},x_{i_{k}+1},x_{i_{k+1}},\ldots,x_{i_{n}}

(19)

If $x_{i_{n}}<x_{0}<b$ The first two, three show us that
when

	$\displaystyle L\left(x_{i_{2}},x_{i_{3}},\ldots,x_{i_{k}},x_{i_{k}+1},x_{i_{k+1}},\ldots,x_{i_{n}};y\left(x_{0}\right)<\right.$	(20)
	$\displaystyle<L\left(x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}},x_{i_{k+1}},\ldots,x_{i_{n}};y\left(x_{0}\right)<\right.$
$\displaystyle<$	$\displaystyle L\left(x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}},x_{i_{k+1}},x_{i_{k+1}},\ldots,x_{i_{n-1}};yx_{0}\right)$

(21)

L\left(x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}},x_{i_{k}+1},x_{i_{k+1}},\ldots,x_{i_{n-1}};y\mid x_{i_{n}}\right)>y_{i_{n}}

If instead of (21) we have the opposite inequality, then in (20) we must also take the opposite inequalities everywhere. The sign $=$ takes place simultaneously in (20), (21). It follows that Theorem 3 is true for $i_{n}-i_{1}=n$ .

Let us now suppose that the property takes place for $i_{2}-i_{1}\leqq v,v\geqq n$ and let's demonstrate that it remains true for $i_{n}-i_{1}=v+1$ . If $i_{n}-i_{1}=v+1$ there is a clue $p$ , for which $i_{p+1}-i_{p}>1$ Let us consider the points (22)

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

Applying the same reasoning to these points as we have done

	$\displaystyle\min_{j=1,2}L\left(x_{i_{j}},x_{i_{j+1}},\ldots,x_{i_{p}},x_{i_{p+1}},x_{i_{p+1}},\ldots,x_{i_{j+n-2}};y\mid x_{0}\right)\leqq$		(23)
	$\displaystyle\quad\leqq L\left(x_{i_{1}},x_{i_{2}},\ldots,x_{i_{p}},x_{i_{p+1}},\ldots,x_{i_{n}};y\mid x_{0}\right)\leqq$
	$\displaystyle\leqq\max_{j=1,2}L\left(x_{i_{j}},x_{i_{j+1}},\ldots,x_{i_{p}},x_{i_{p+1}},x_{i_{p+1}},\ldots,x_{i_{j+n-2}};y\mid x_{0}\right)$

Let
be the average value of the numbers

L\left(x_{j},x_{j+1},\ldots,x_{j+n-1};y\mid x_{0}\right),j=i_{1},i_{1}+1,\ldots,i_{n-1}-n+1

(24)

And $L\left(x_{i_{3}},x_{i_{3}},\ldots,x_{i_{p}},x_{i_{p+1}},x_{i_{p+1}},\ldots,x_{i_{n}};y\mid x_{0}\right)$ an average value of the numbers

L\left(x_{j},x_{j+1},\ldots,x_{j+n-1};y\mid x_{0}\right),j=i_{2},i_{2}+1,\ldots,i_{n}-n+1

(25)

From (23) it follows that (17) and Theorem 3 is proven.
7. - Consider the points (6) and the function $f(x)$ defined on these points.

		$\displaystyle D\left[x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1},x_{i};f\right]=$		(26)
	$\displaystyle=$	$\displaystyle f\left(x_{i}\right)-L\left(x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1};f\mid x_{i}\right)$		(6)

For each value of $i=1,2,\ldots,n+1$ (26) is a functional defined on the functions $f=f(x)$ defined on the points (6). We have
Theorem 4. - If the points (6) remain fixed and if the sequence $\left\{f_{k}(x)\right\}_{k=1}^{\infty}$ functions $l_{k}(x)$ , defined on the interval $[a,b]$ converges on $[a,b]$ towards the limit function $/(x)$ then for each $i$ , We have

	$\displaystyle\lim_{k\rightarrow\infty}D\left[x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1},x_{i};f_{k}\right]=$		(27)
	$\displaystyle\quad=D\left[x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1},x_{i};f\right]$

The proof follows from Theorem 1. Indeed, under the hypotheses of Theorem 4, the sequence of functions
$\left\{L\left(x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1};f_{k}\mid x\right)\right\}_{k=1}^{\infty}$ , converges uniformly, on $[a,b]$ , towards the function $L\left(x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1};f\mid x\right)$ It follows that the sequence of numbers
$\left\{L\left(x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1};f_{k}\mid x_{0}\right)\right\}_{k=1}^{\infty}$ tends, for $k\rightarrow\infty$ towards the number $L\left(x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1};f\mid x_{0}\right)$ . Of $f_{k}\left(x_{i}\right)\rightarrow f\left(x_{i}\right)$ , For $k\rightarrow\infty$ relation (27) results.

If we fix the function $f(x)$ and if we vary in a continuous way, the points $x_{1},x_{2},\ldots,x_{n+1}$ , so that they remain distinct, for each $i,D\left[x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1},x_{i};f\right]$ becomes a function $\Phi_{i}\left(x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1},x_{i}\right)$ of $n+1$ variables. We have
theorem 5. - If $f(x)$ is a continuous function on the interval $[a,b]$ , the functions $\Phi_{i}\left(x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1},x_{i}\right)i=1,2,\ldots,n+1$ are continuous at every point where they have been defined.

The proof follows easily from Theorem 1. It is unnecessary to reproduce it here.
Theorem 6. - If $f(x)$ is a continuous function on the interval $[a,b]$ and if for two groups of $n+1$ points $x_{1}<x_{2}<\ldots<x_{n+1},x_{1}^{\prime}<<x_{2}^{\prime}<\ldots<x_{n+1}^{\prime}de[a,b]$ We have
$D\left[x_{1},x_{2},\ldots,x_{n},x_{n+1};t\right]=A,D\left[x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{n}^{\prime},x_{n+1}^{\prime};t\right]\approx B$ Or $A\neq B$ , regardless of the number $C$ between $A,B$ (in the strict sense), i] exists, in the smallest interval containing the points $x_{i},x_{i}^{\prime},i=1,2$ , , $n+1$ , a system of $n+1$ points $\xi_{1}<\xi_{2}<\ldots<\xi_{n+1}$ such as one would $D\left[\xi_{1},\xi_{2},\ldots,\xi_{n},\xi_{n+1};f\right]=C$ .

To demonstrate*) the theorem let $t_{t}(\lambda)=\lambda x_{i}+(1-\lambda)x_{i}^{\prime}i=1,2,\ldots,n+1,\lambda$ being a real number, $0\leqq\lambda\leqq 1$ By virtue of the theorem $5,\psi(\lambda)=D\left[t_{1}(\lambda),t_{2}(\lambda),\ldots,t_{n}(\lambda),t_{n+1}(\lambda);f\right]$ is a continuous function of $i$ . on the interval $[0,1]$ . We have $\psi(0)=A,\psi(1)=B$ and by virtue of a well-known property, there exists a $\lambda_{0}\in(0,1)$ , such as $\psi\left(\lambda_{0}\right)=C$ The points $\xi_{i}=t_{i}\left(\lambda_{0}\right),i=1,2,\ldots,n+1$
verify the conclusion of Theorem 6. $6^{\prime}$ .- If $f(x)$ is a continuous function on the interval $[a,b]$ and if on the points $x_{1}<x_{2}<\ldots<x_{n+1},x_{1}^{\prime}<x_{2}^{\prime}<\ldots<x_{n+1}^{\prime}$ of $[a,b]$ We have $D\left[x_{1},x_{2},\ldots,x_{n+1};f\right]<0,D\left[x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{n}^{\prime},x_{n+1}^{\prime};f\right]>0$ , there exists, in the smallest interval that contains the points $x_{i},x_{i}^{\prime},i=1,2,\ldots,n+1$ , a system of $n+1$ points $\xi_{1}<\xi_{2}<\ldots<\xi_{n+1}$ such as one might have

D\left[\xi_{1},\xi_{2},\ldots,\xi_{n},\xi_{n+1};f\right]=0

This is a special case of Theorem 6. We have stated it since it will appear in this form later.
Theorem 7. - If $f(x)$ is a continuous function on the interval $[a,b]$ and if for the $n+1$ points $x_{1}<x_{2}<\ldots<x_{n}<x_{n+1}$ of $[a,b]$ We have $D\left[x_{1},x_{2},\ldots,x_{n},x_{n+1};t\right]=0$ , there is a point $\bar{\zeta}\in\left(x_{1},x_{n+1}\right)$ who enjoys the property that in each of his neighborhoods one can find $n+1$ points $\xi_{1}<\xi_{2}<\ldots<\xi_{n}<\xi_{n+1}$ such as one has $D\left[\xi_{1},\xi_{2},\ldots,\xi_{n},\xi_{n+1};t\right]=0$ .

The proof is based on the following lemmas:
Lemma 4. - If $f(x)$ is a continuous function on the interval $[a,b]$ and if:
$1^{\circ}.\varphi(x)\in F_{n}$ and the difference $f(x)-\varphi(x)$ cancels out on the points $x_{1}<<x_{2}<\ldots<x_{n+1}de[a,b]$ .
$2^{\circ}.f(x)-\varphi(x)$ does not vanish over the intervals $\left(x_{i},x_{i+1}\right)i=1,2,\ldots,n$ .
$3^{\circ}.f(x)-\varphi(x)$ does not change sign in $k$ points $x_{i},i=2,3,\ldots,n$ , $1\leqq k\leqq n-1(n\geqq 2)$ ,
then there is a $\varphi_{1}(x)\in F_{n}$ such as the difference $f(x)-\varphi_{1}(x)$ cancels out on $n+k+1$ points of $\left[x_{1},x_{n+1}\right]$ by changing sign on $n+k$ of these points located within this interval.
$f_{n}$ is always a set of the type $I_{n}[a,b]$ For the proof of the lemma ,
let us first assume that $k=n-1$ The subset of $\mathscr{F}_{n}$ whose elements take the value $f\left(x_{1}\right)$ to the point $x_{1}$ , is a set of the type $I_{n-1}\left(x_{1},b\right]$ Suppose that $\varphi(x)-f(x)\geqq 0$ For $x\in\left(x_{1},x_{n+1}\right)$ So if $\varepsilon>0$ is small enough, the difference $L\left(x_{1},x_{2},\ldots,x_{n};f\left(x_{1}\right),f\left(x_{2}\right)-\varepsilon,\ldots,f\left(x_{n}\right)-\varepsilon(x)-f(x)\right.$ cancels out, changing sign in at least $2k+1=2n-1$ points of ( $x_{1},x_{n+1}$ ). The same procedure is followed if $\varphi(x)-f(x)\leqq 0$ .

⁰⁰footnotetext: *). The demonstration is analogous to that given in the mean theorem on page 8.

If $k=n-j$ Or $2\leqq j\leqq n-1$ , either $x_{i_{1}}<x_{i_{2}}<\ldots<x_{i_{n-1-k}}$ those of the points $x_{i},i=2,3,\ldots,n$ on which $f(x)-\varphi(x)$ cancels out by changing sign. The elements of $\mathscr{F}_{n}$ which take the same values as $f(x)$ on the points $x_{1},x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n-1-k}}$ form a set of the type $I_{k}\{M\}$ Or $\mathcal{M}$ is the union of the intervals $\left[x_{1}+\eta,x_{i_{1}}-\eta\right],\left[x_{i_{1}}+\eta,x_{i_{2}}-\eta\right],\ldots$ , $\left[x_{i_{n-2-k}}+\eta,x_{i_{n-1-k}}-\eta\right],\left[x_{i_{n-1-k}}+\eta,b\right]$ Or $\eta>0$ is quite small. Let $x_{1}^{*},x_{2}^{*},\ldots,x_{k}^{*}$ those of the points $x_{i},i=2,3,\ldots,n$ Or $f(x)-\varphi(x)$ cancels out without changing sign. Let's consider the difference

	$\displaystyle L\left(x_{1},x_{i_{1}},x_{i_{2}}\ldots,x_{i_{n-1-k}},x_{1}^{},x_{2}^{},\ldots,x_{k}^{*};f_{1},f_{i_{1}},\ldots,f_{i_{n-1-k}},y_{1}\right.$		(28)
	$\displaystyle,y_{2},\ldots,y_{k}(x)-f(x),\left(f_{i}=f\left(x_{i}\right)\right)$

y_{i}=\left\{\begin{array}[]{l}f\left(x_{i}^{*}\right)+\varepsilon\operatorname{si}f(x)-\varphi(x)\geqq 0\text{ au voisinage de }x_{i}^{*}\\ f\left(x_{i}^{*}\right)-\varepsilon\operatorname{si}f(x)-\varphi(x)\leqq 0\text{ au voisinage de }x_{i}^{*}\end{array}\right.

If $s>0$ is quite small, the difference (28) becomes zero by changing sign over at least $n+k$ points of the interval ( $x_{1},x_{n+1}$ Lemma 4 is proven. The proof uses Theorem 1 and Lemma 1.

Lemma 5. - If the function $f(x)$ is continuous on the interval $[a,b]$ and if for the points $x_{1}<x_{2}<\ldots<x_{n+1}$ , We have $D\left[x_{1},x_{2},\ldots,x_{n+1};f\right]=0$ , there exists, in the open interval ( $x_{1},x_{n+1}$ ), $n+1$ points $x_{1}<x_{2}<\ldots<x_{n+1}$ , such as $D\left[x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{n}^{\prime},x_{n+1}^{\prime};j\right]=0$ .

For the demonstration, we distinguish between two cases:
$(\alpha)$ The difference $f(x)-L\left(x_{1},x_{2},\ldots,x_{n};f\mid x\right)$ does not change sign on intervals $\left(x_{i},x_{i+1}\right),i=1,2,\ldots,n$ .
( $\beta$ The difference $f(x)-L\left(x_{1},x_{2},\ldots,x_{n};f\mid x\right)$ changes sign on $m$ point of ( $x_{1},x_{n+1}$ ) different from the points $x_{i},i=1,2,\ldots,n$

In the case $(\alpha)$ We have $\varphi\left(x_{1}\right)=f\left(x_{i}\right),i=1,2,\ldots,n+1$ if $\varphi(x)==L\left(x_{1},x_{2},\ldots,x_{n};f\mid x\right)$ If the difference $\varphi(x)-f(x)$ changes sign on the points $x_{i},i=2,3,\ldots,n$ , we consider the interpolation function $\varphi_{1}(x)$ who on the points $x_{i}^{\prime}=1/2\left(x_{i}+x_{i+1}\right),i=1,2,\ldots,n$ takes the values of the function $f(x)$ The numbers $f\left(x_{i}^{\prime}\right)-\varphi\left(x_{i}^{\prime}\right),i=1,2,\ldots,n$ are alternately positive and negative, and as a result, the difference $\varphi(x)-\varphi_{1}(x)$ cancels out $n-1$ times, in particular on at least one point of each of the intervals $\left(x_{i}^{\prime},x_{i+1}^{\prime}\right),i=1,2,\ldots,n$ We distinguish two sub-cases:
$\left(\alpha^{\prime}\right)\varphi_{1}(x)-f(x)$ cancels out, without changing sign at least at one point $x_{i}$ .
$\left(\alpha^{\prime\prime}\right)\varphi_{1}(x)-f(x)$ changes sign at each of the points $x_{i}^{\prime}$ In
the case ( $\alpha^{\prime}$ ), in the same way as in the proof of Lemma 4, we can construct an element of $\mathscr{F}_{n}$ which coincides with $j(x)$ at least $n+1$ points of the interval ( $x_{1},x_{n+1}$ ).

In the case $\left(\alpha^{\prime\prime}\right)$ , if $\varphi_{1}(x)-f(x)$ does not change sign on any of the intervals $\left(x_{i}^{\prime},x_{i+1}^{\prime}\right),i=1,2,\ldots,n$ we have signed $\{\varphi(x)-f(x)\}==\operatorname{sgn}\left\{\varphi_{1}(x)-f(x)\right\}$ when $x$ is either in a left neighborhood or in a right neighborhood of each of the points $x_{i}^{\prime},i=1,2,\ldots,n$ But the functions.
$\varphi_{1}(x),\varphi(x)$ cannot coincide on more than $n-1$ points. The difference $\varphi_{1}(x)-f(x)$ must therefore cancel out on ( $x_{1},x_{n+1}$ ) on a point different from the points $x_{i},i=1,2,\ldots,n$ The result is that the difference $\varphi_{1}(x)-f(x)$ becomes null on at least $n-1$ points of the interval ( $x_{1},x_{n+1}$ ).

If $\varphi_{1}(x)-f(x)$ changes sign on one of the intervals $\left(x_{i},x_{i+1}^{\prime}\right)i=1,2,\ldots$ , $n$ , this difference cancels out on $n+1$ points of ( $x_{1},x_{n+1}$ ).

If $\varphi_{1}(x)-f(x)$ does not change sign on at least one of the points $x_{i}$ , $i=2,3,\ldots,n$ , we apply lemma 4 and thus return to the case ( $\alpha^{\prime}$ ), ( $\alpha^{\prime\prime}$ ) from the highest point.

Let's move on to the case study ( $\beta$ Suppose $m=1$ and either $x^{*}$ a point of $\left(x_{j},x_{j+1}\right)$ such as $f\left(x^{*}\right)-L\left(x_{1},\mathrm{x}_{2},\ldots,x_{n};f\mid x^{*}\right)=0$ . So we apply the reasoning done in the case ( $\alpha$ ) either for the points $x_{1},x_{2},\ldots$ , $x_{j},x^{*},x_{j+1},\ldots,x_{n}$ either for the points $x_{2},x_{3},\ldots,x_{j},x^{*},x_{j+1},\ldots,x_{n+1}$ .

When $m>1$ the existence of points $x_{1}<x_{2}<\ldots<x_{n+1}$ is obvious.
We will now return to the proof of Theorem 7. According to Lemma 5, we can construct an infinite sequence of intervals $\left[x_{1},x_{n+1}\right],\left[x_{1}^{\prime},x_{n+1}^{\prime}\right],\ldots,\left[x_{1}^{(k)},x_{n+1}^{(k)}\right],\ldots$ , such as
$1^{\circ}x_{1}<x_{1}^{\prime}<\ldots<x_{1}^{(k)}\ldots,x_{n+1}>x_{n+1}^{\prime}>\ldots>x_{n+1}^{(k)}\ldots$
$2^{\circ}$ In each of the intervals $\left[x_{1}^{(k)},x_{n+1}^{(k)}\right]$ they exist $n-1$ points $x_{2}^{(k)},x_{3}^{(k)},\ldots,x_{n}^{(k)}$ so that $D\left[x_{1}^{(k)},x_{2}^{(k)},\ldots,x_{n}^{(k)},x_{n+1}^{(k)};I\right]=0$
$3^{\circ}\lim_{k\rightarrow\infty}\left(x_{n+1}^{(k)}-x_{1}^{(k)}\right)=0$
To construct the sequence of intervals $\left[x_{1}^{(k)},x_{n+1}^{(k)}\right],k=1,2,\ldots$ Using Lemma 5, we start from the assumption that we can always highlight $n+1$ points, "consecutive" on which a function of $\mathscr{F}_{n}$ coincides with $f(x)$ This hypothesis is justified by the fact that if we cannot find $n+1$ consecutive roots of the difference between $f(x)$ and an element of $\mathscr{F}_{n}$ , the number of intervals contiguous to the closed set belonging to $\left[x_{1},x_{n+1}\right]$ and on which this difference is zero, is smaller than $n$ It is then easy to see that $f(x)$ coincides with the interpolation function over an entire interval. In this case, the existence of the point $\boldsymbol{\xi}$ The principle of theorem 7 is evident.

The property $3^{\circ}$ follows immediately from Theorem 1. Let $\lim x_{1}^{(k)}==\alpha,\lim_{k\rightarrow\infty}x_{n+1}^{(k)}=\beta$ We can assume that $\beta-\alpha.\quad$ is equal to the lower bound of the interval lengths $[\alpha,\beta]$ which can be constructed in the manner indicated. It suffices to demonstrate that this lower bound is equal to 0, therefore that $\beta-\alpha=0$ Or $\alpha=\beta$ Indeed, let us suppose the opposite, that we have $\alpha\neq\beta$ For everything $k$ there exists an interpolation function that coincides with $f(x)$ on $n+1$ points of $\left[x_{1}^{(k)},x_{n+1}^{(k)}\right]$ From all these functions we can extract a sequence $\left\{\varphi_{k}(x)\right\}^{\infty}=1,\varphi_{k}\left(x_{i}^{(k)}\right)=f\left(x_{i}^{(k)}\right),i=1,2,\ldots,n+1,k=1,2,\ldots$ which tends uniformly on $[a,b]$ towards a function $\varphi(x)$ of $\mathscr{F}_{n}$ We can always choose the points $x_{i}^{(k)}$ so that $\lim x^{(k)}-\lim x^{(k)}=$
$=\beta$ and that the consequences $\left\{x_{i}^{(k)}\right\}_{k=1}^{\infty},i=2,3,\ldots,n$ are convergent. The limits, for $k\rightarrow\infty$ points $x_{i}^{(k)},i=1,2,\ldots,n+1$ are not necessarily all distinct but the following $\left\{\varphi_{k}(x)\right\}_{k=1}^{\infty}$ can be chosen in such a way that $n$ at least these limits are distinct. This property follows from the following remark: If $n=1$ The property is clear. If $n=2$ And $D\left[x_{1},x_{2},x_{3};f\right]=0$ , whatever the point $x_{2}^{\prime}\in\left(x_{1},x_{3}\right)$ there is a point $x_{1}^{\prime}<x_{2}^{\prime}$ or a point $x_{3}^{\prime}>x_{2}^{\prime}$ such as $D\left[x_{1}^{\prime},x_{2}^{\prime},x_{3};f\right]=0$ Or $D\left[x_{1},x_{2}^{\prime},x_{3}^{\prime};f\right]==0$ In the case $n>2$ Note that the set of elements of $\mathscr{F}_{n}$ which coincide with $f(x)$ on the points $x_{1}^{(k)},x_{2}^{(k)},\ldots,x_{i-1}^{(k)},x_{i+2}^{(k)},\ldots$ , $x_{n+1}^{(k)}$ ( $k$ given) form a set of the type $I_{2}\left[x_{i-1}^{(k)},x_{i+2}^{(k)}\right]$ , regardless of $i=2,3,\ldots,n-1$ We can apply, at one of the points $x_{i}^{(4)},\quad x_{i+1}^{(k)}$ the reasoning we used in the case $n=2$ for the point $x_{2}$ Since the $n-2$ points $x_{3}^{(k)},x_{4}^{(k)},\ldots,x_{n}^{(k)}$ can be replaced by $n-2$ any other points of the interval ( $x_{1}^{(k)},x_{n+1}^{(k)}$ ), it follows that we can find the intervals $\left[\alpha_{i},\beta_{i}\right],i=1,2,\ldots,n-2$ so that the differences $\alpha_{i}-\beta_{i-1},i=2,3,\ldots,n-2$ are greater than a positive number $\eta$ and in such a way that $x_{i}^{(k)}\in\left(\alpha_{i},\beta_{i}\right),i=1,2,\ldots$ , $n-2$ It follows that the limit function $\varphi(x)$ coincides with $f(x)$ on $n+1$ points of $[\alpha,\beta]$ among which he $y$ a at most two that are coincident. It also follows that we can find $n+1$ distinct points of $[\alpha,\beta]$ on which $f(x)$ coincides with an element of $\mathscr{F}_{n}$ This contradicts the definition of the interval. $[\alpha,\beta]$ if $\alpha\neq\beta$ . We have $\alpha=\beta$ that's precisely what needed to be demonstrated.

It is easy to see that the point $\xi$ can always be chosen so that among the points $\xi_{i}$ there is at least one "no" vote on the left and at least one "no" vote on the right. $\xi$ This separation property can be further specified when the function is suitably particularized. We will return to this question in § 2.8.
We will now look at some applications. Theorems 6 and 7 generalize certain well-known mean theorems from classical analysis.

Let us consider the whole, $P_{n+1}$ of the type $I_{n+1}(-\infty,\infty)$ formed by all polynomials of degree $n$ If the function $f(x)$ is defined on the points $x_{1}$ , $x_{2},\ldots,x_{n+1}$ the interpolation function $L\left(x_{1},x_{2},\ldots,x_{n+1};f\mid x\right)$ reduces to the Lagrange polynomial

P\left(x_{1},x_{2},\ldots,x_{n+1};f\mid x\right)=\sum_{i=1}^{n+1}f\left(x_{i}\right)\quad\iota_{i}^{(n)}(x)

(29)

\ell_{i}^{(n)}(x)=\frac{l(x)}{\left(x-x_{i}\right)\ell^{\prime}\left(x_{i}\right)},\quad\ell(x)=\prod_{i=1}^{n+1}\left(x-x_{i}\right)

(30)

If $x_{0}\neq x_{i},i=1,2,\ldots,n+1$ , We have

	$\displaystyle D\left[x_{1},x_{2},\ldots,x_{n},x_{n+1},x_{0};f\right]=f\left(x_{0}\right)-$		(31)
	$\displaystyle-P\left(x_{1},x_{2},\ldots,x_{n+1};f\mid x_{0}\right)=l\left(x_{0}\right)\left[x_{1},x_{2},\ldots,x_{n+1},x_{0};f\right]$

Or $\left[x_{1},x_{2},\ldots,x_{n},x_{n+1},x_{0};f\right]$ is the difference divided by order $n+1$ on the nodes $x_{1},x_{2},\ldots,x_{n},x_{n+1},x_{0}$ , of the function $f(x)$ The coefficient $L\left(x_{0}\right)$ Since it is always different from zero, the following property of the set follows from Theorem 7: $\rho_{n+1}$ :

If $f(x)$ is a continuous function on the interval $I$ and if for the points $x_{1}^{\#}<x_{2}^{*}<\ldots<x_{n\div 2}^{*}$ of $I$ We have $\left[x_{1}^{*},x_{2}^{*},\ldots,x_{n+2}^{*};j\right]=0$ , there is a point $\xi\in\left(x_{1}^{*},x_{n+2}^{*}\right)$ such as in any neighborhood of $\xi$ they exist $n+2$ points $\xi_{1}^{*}<\xi_{2}^{*}<\ldots<\xi_{n+2}^{*}$ for which $\left[\xi_{1}^{*},\xi_{2}^{*},\ldots,\xi_{u+2}^{*};f\right]-0[9]$ .

The difference divided $\left\lfloor x_{1},x_{2},\ldots,x_{n+1};f\right\rfloor$ is the coefficient of $x^{n}$ in the interpolation polynomial (29). Now let the set $D_{n+1}(A)$ polynomials of degree $n$ in which the coefficient of $x^{n}$ is equal to $A$ and which is of the type $I_{n}(-\infty,\infty)$ If it exists in $D_{n+1}(A)$ a polynomial that takes the values $f\left(x_{i}\right)$ on the $n+2$ points $x_{i}$ respective, then $\left[x_{1},x_{2},\ldots,x_{n+1};f\right]=A$ In this case, Theorem 7 gives us the property valid for the set $\mathcal{P}_{n+1}(A)$ :

If $f(x)$ is a continuous function on the interval $I$ and if on the points $x_{1}^{*}<x_{2}^{*}<\ldots<x_{n+1}^{*}$ of I we have $\left[x_{1}^{*},x_{2}^{*}\ldots,x_{n+1}^{*};f\right]=A$ , there is a point of $\left(x_{1}^{*},x_{n+1}^{*}\right)$ such that each of its neighborhoods contains $n+1$ points $\xi_{1}^{*},\xi_{2}^{*},\ldots,\xi_{n+1}^{*}$ (distinct) for which $\left[\xi_{1}^{*},\xi_{2}^{*},\ldots,\xi_{n+1}^{*};f\right]=A$ .

If $n=1$ and if the function $f(x)$ is differentiable on the interval ( $x_{1}^{*},x_{2}^{*}$ ) we deduce the mean value formula. In this case $A=\frac{i\left(x_{2}\right)-f\left(x_{1}\right)}{x_{2}-x_{1}}$ .

Note that if instead of the set $\mathcal{D}_{n+1}$ , we consider the linear envelope of a Tschebyshev-Markoff system formed by the functions $\varphi_{1}(x),\varphi_{2}(x),\ldots,\varphi_{n+1}(x)$ , we recover the mean theorems for the respective generalized divided differences (see T. popoviciu [10], where this divided difference is defined)

Finally, let us note that in the case of the set $F_{n+1}$ , if we fix the last coefficient of the polynomial of the form $a_{0}x^{n}+a_{1}x^{n+1}+\ldots++a_{n}$ Therefore, if we consider the set of polynomials $a_{0}x^{n}+a_{1}x^{n-1}++\ldots+a_{n-1}x+A$ , Or, $a_{i},i=0,1,\ldots,n-1$ are variable and $A$ is fixed, we obtain an interpolated set of order $n$ on any interval not containing the origin. In this case, for $n=1$ From theorem 7, it follows that there is a discrete analogue of Pompeii's mean value theorem [7] and, under the hypothesis of differentiability, Pompeii's own theorem*.) The case $n>1$ This gives us an extension of this theorem. The mean theorems valid for the other coefficients of the Lagrange polynomial are also contained in Theorem 7.

of Pompeii. This formula, moreover, comes back to the f $x_{1}-x_{2}$ attached to the function $xf(1/x)$ .

We know that for divided differences there also exists a mean value theorem on a discrete set of points [9]. Its generalization for the case of a set of the type $I_{n}[a,b]$ results from Theorem 3. In § 3 we will give this generalization.
9. - Theorem 8. - If $f(x)$ is a continuous function on the interval $[a,b],n$ -valued in relation to the set $\mathscr{F}_{n}$ of the type $I_{n}[a,b]$ , $D\left[x_{1},x_{2},\ldots,x_{n},x_{n+1};f\right]$ keep the same sign for the dots $x_{1}<x_{2}<<\ldots<x_{n+1}$ of $[a,b]$ .

The proof of this theorem follows from Theorem 6'. Indeed, suppose that, under the hypotheses of Theorem 8, there exist two systems, each of $n+1$ points $x_{1}<x_{2}<\ldots<x_{n+1};x_{1}^{\prime}<x_{2}^{\prime}<\ldots<x_{n+1}^{\prime}$ such as $D\left[x_{1},x_{2},\ldots,x_{n},x_{n+1};f\right]>0,D\left[x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{n}^{\prime},x_{n+1}^{\prime};f\right]<0$ From the theorem $6^{\prime}$ This results in the existence of a system of $n+1$ points $x_{1}^{\prime\prime}<x_{2}^{\prime\prime}<<\ldots<x_{n+1}^{\prime\prime}$ such as $D\left[x_{1}^{\prime\prime},x_{2}^{\prime\prime},\ldots,x_{n}^{\prime\prime},x_{n+1}^{\prime\prime};f\right]=0$ which contradicts the $n$ -valence of the function $f(x)$ .

In the statement of Theorem 8, the hypothesis of the continuity of the function $f(x)$ is essential*).

The importance of studying the properties of functions $n$ -valent with respect to a set of the type $I_{n}[a,b]$ Its origin lies precisely in theorem 8. We will return to this question in § 2.

§ 2.

The notion of convex tonction with respect to a set of the type $I_{n}[a,b]$ .

1.
- —
  
  Let us consider the whole $\mathscr{F}_{n}$ of the type $I_{n}[a,b]$ , the points

x_{1}<x_{2}<\ldots<x_{n}<x_{n+1}

(32)

of the interval $[a,b]$ and the function $f(x)$ defined on the points (32)
Definition 4. - The function $f(x)$ is said to be convex, polynomial, or concave, with respect to the set $\mathscr{F}_{n}$ , on points (32), depending on

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

(33)

We can see immediately, taking into account the properties of the interpolation system (8), that if $f(x)$ is convex with respect to the set $\mathscr{F}_{n}$ , on points (32), we have the inequalities

\begin{gathered}(-1)^{n+1-i}\left\{f\left(x_{i}\right)-L\left(x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1};f\mid x_{i}\right)\right\}>0\\ i=1,2,\ldots,n+1\end{gathered}

and of course, if $f(x)$ is concave with respect to $f_{n}$ On points (32), we have the opposite inequalities (everywhere $<$ instead of $>$ ).

⁰⁰footnotetext: *). See in [12] an example in the case of the set

D_{1}

Let us now consider the subset $E$ of $[a,b]$ having at least $n+1$ points and suppose that the function $f(x)$ either defined on $E$ .

Definition 5. - The function $f(x)$ is said to be convex, non-concave, polynomial, non-convex, or concave with respect to $\mathscr{F}_{n}$ across the board $E_{\text{, }}$ depending on
(34) $D\left[x_{1},x_{2},\ldots,x_{n},x_{n+1};f\right]>,\geqq,=,\leqq$ resp. $<0$ for any system of $n+1$ point $x_{1}<x_{2}<\ldots<x_{n+1}$ of $E$ .

Any function that satisfies one of the properties in definition 5 will be called a function of order $n$ compared to the whole $\mathscr{F}_{n}$ and across the board $E$ To simplify the language, we will also use the terms $\mathscr{F}_{n}$ -convex, $\mathscr{F}_{n}$ -non-concave, $\mathscr{F}_{n}$ -polynomial, $\mathscr{F}_{n}$ -non-convex resp. of F _n -concave for functions belonging to the respective classes specified by definition 5.

Order functions $n$ compared to the whole $\varnothing_{n}$ were studied by T. Popoviciu [9]). The case $n=2,\mathscr{F}_{2}$ any, was considered by Beckenbach [2]. In this § we also use a definition from L. Tornheim [14].
2. - Theorem 9. - For the junction $f(x)$ either $f_{n}$ - convex at points
(35)

x_{1}<x_{2}<\ldots<x_{m},\quad m\geqq n+1

it is necessary and sufficient that it be $\mathscr{F}_{n}$ -convex on any system of $n+1$ consecutive points $x_{j},x_{j+1},\ldots,x_{j+n},\quad j=1,2,\ldots,m-n$ of the suite (35).

The proof of this theorem is based on Theorem 3. The necessity of the condition is obvious. The fact that the condition is also sufficient follows from the following. Let $n\geqq 2$ and the following inequalities are satisfied

D\left[x_{j},x_{j+1},\ldots,x_{j+n};f\right]>0,\quad j=1,2,\ldots,m-n

(36)

So we also have

\begin{array}[]{cl}L\left(x_{j},x_{j+1},\ldots,x_{j+n-1};\right.&\left.f\mid x_{j+n}\right)<\\ <L\left(x_{j+1},x_{j+2},\ldots,x_{j+n};f\mid x_{j+n}\right),&j=1,2,\ldots,m-n-1\end{array}

By virtue of Theorem 3, we also have

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

(38)

regardless of the system $n+1$ points $x_{i_{1}}<x_{i_{2}}<\ldots<x_{i_{n+1}}$ extracts from sequence (35). The property results from the inequalities (38) If $n=1$ we take into account that two distinct elements of $\mathscr{F}_{1}$ cannot coincide on any point of $[a,b]$ In this case of $D\left[x_{j},x_{j+1};f\right]>0$ , $j=1,2,\ldots,m-1$ it follows that $L\left(x_{1};f\mid x_{i}\right)<f\left(x_{i}\right),i=2,3,\ldots m_{\text{, }}L\left(x_{2};f\mid x_{i}\right)<f\left(x_{i}\right),i=3,4,\ldots,m,\ldots,L\left(x_{m-2};f\mid x_{i}\right)<f\left(x_{i}\right),i=\ldots--1,m$ .

A theorem analogous to Theorem 9 holds for other types of order functions $n$ compared to $f_{n}$ on the set (35).
3. - Theorem 10, - If the function $f(x)$ , defined on the interval $[a,b]$ is of order $n$ compared to $\mathscr{F}_{n}$ and if $n\geq 2$ , it is continuous on the open interval ( $a,b$ ).

For the demonstration, either $x_{0}$ a point of $(a,b)$ and suppose $f(x)\mathscr{F}_{n}$ - convex on the interval $[a,b]$ Let's also consider the $n+1$ points of $[a,b]$ ,

x_{1}<x_{2}<\ldots<x_{i}<x_{0}<x_{i+1}<\ldots<x_{n}.

(39)

The functions

	$\displaystyle L\left(x_{1},x_{2},\ldots,x_{i},x_{0},x_{i+1},\ldots,x_{n-1};f\mid x\right)$		(40)
	$\displaystyle L\left(x_{2},x_{3},\ldots,x_{i},x_{0},x_{i+1},\ldots,x_{n};f\mid x\right)$

coincide on $x_{0}$ with $f\left(x_{0}\right)$ and, for $h>0$ small enough, we have

	$\displaystyle L\left(x,x_{3},\ldots,x_{i},x_{0},x_{i+1},\ldots,x_{n};f\mid x_{0}+h\right)<f\left(x_{0}+h\right)<$		(41)
	$\displaystyle<L\left(x_{1},x_{2},\ldots,x_{i},x_{0},x_{i+1},\ldots,x_{n-1};f\mid x_{0}+h\right)$

And

	$\displaystyle L\left(x_{2},x_{3},\ldots,x_{i},x_{0},x_{i+1},\ldots,x_{n};f\mid x_{0}-h\right)>f\left(x_{0}-h\right)>$		(42)
	$\displaystyle\quad>L\left(x_{1},x_{2},\ldots,x_{i},x_{0},x_{i+1},\ldots,x_{n-1};f\mid x_{0}-h\right)$

or the opposite inequalities (without equality) depending on whether $n+i+1$ is even or odd. Since the functions (40) are continuous, by doing $h\rightarrow 0$ , the continuity of $f(x)$ on the point $x_{0}$ The result is...

The property remains true for other types of order functions as well. $n$ Theorem
11. - If the function $f(x)$ is of order 1 with respect to the set $\mathcal{F}_{1}$ on the interval $[a,b]$ it can only have discontinuities of the first kind.

The proof of this theorem is based on the fact that two distinct functions of $\mathscr{F}_{1}$ cannot coincide on any point of $[a,b]$ To clarify, let's suppose that $f(x)$ either $f_{1}$ -non-concave on $[a,b]$ So we have $D\left[x_{1},x_{2};f\right]\geqq 0$ , whatever $x_{1}<x_{2}$ Either $x_{0}\in[a,b]$ And $\left\{x_{v}\right\}_{v=1}^{\infty},x_{v}>x_{0},v=1,2,\ldots$ a series of points $[a,b]$ tending towards $x_{0}$ In light of the remark made above regarding the elements of $\mathscr{F}_{1}$ , we can assume that the following $\left\{x_{v}\right\}_{v=1}^{\infty}$ either decreasing. The sequence of functions $\left\{L\left(x_{v};f\mid x\right)\right\}_{v=1}^{\infty}$ is non-increasing and bounded below by the function 1a $L\left(x_{0};f\mid x\right)$ This sequence, by virtue of Theorem 1, is uniformly convergent on $[a,b]$ From this results the convergence of the sequence of numbers. $\left\{f\left(x_{\nu}\right)\right\}_{\nu=1}^{\infty}$ We have thus demonstrated the existence of the right-hand limit. $f\left(x_{0}+0\right)$ on the point $x_{0}$ The existence of the left-hand limit is demonstrated in the same way. $f\left(x_{0}-0\right)$ on the point $x_{0}\in[a,b]$ .

The demonstration is performed in the same way for the other categories of first-order functions.
Theorem 12. - If the function $f(x)$ is of order $n$ compared to the whole $\mathscr{F}_{n}$ , on the interval $[a,b]$ and if she is $\mathscr{F}_{n}$ -polynomial on the points (32), it is also $\mathscr{F}_{n}$ -polynomial on the interval ( $x_{1},x_{n+1}$ ).

To prove the theorem, it suffices to show that, under the stated hypotheses, on any point of ( $x_{1},x_{n+1}$ ) the value of the function $f(x)$ coincides with that of $L\left(x_{1},x_{2},\ldots,x_{n};j\mid x\right)$ Suppose the opposite, therefore there exists a point $x_{0}\epsilon\left(x_{1},x_{n+1}\right)$ for which

f\left(x_{0}\right)\neq L\left(x_{1},x_{2},\ldots,x_{n};f\left(x_{0}\right)\right.

(43)

Either $x_{k}<x_{0}<x_{k+1}$ and consider the functions

	$\displaystyle L\left(x_{1},\ldots,x_{k},x_{0},x_{k+1},\ldots x_{n-1};f\mid x\right)$		(44)
	$\displaystyle L\left(x_{2},x_{3},\ldots,x_{k},x_{0},x_{k+1},\ldots,x_{n};f\mid x\right)$

If $x_{0}<x_{n}$ the differences

		$\displaystyle L\left(x_{1},x_{2},\ldots,x_{n};f\mid x\right)-L\left(x_{1},x_{2},\ldots,x_{k},x_{0},x_{k+1},\ldots,x_{n-1};jx\right)$
		$\displaystyle L\left(x_{1},x_{2},\ldots,x_{n};f\mid x\right)-L\left(x_{2},x_{3},\ldots,x_{k},x_{0},x_{k+1},\ldots,x_{n};jx\right)$

have contrary signs for $x>x_{n}$ In this case, the number $f\left(x_{n+1}\right)$ is between the values of the functions (44) at the point $x_{n+1}$ This contradicts the hypothesis that $f(x)$ either of order $n$ compared to $\mathscr{F}_{n}$ on $[a,b]$ . If $x_{n}<x_{0}<x_{n+1}$ , as a result of (43), the numbers $\left.Dx_{1},x_{2},\ldots,x_{n},x_{0};f\right]$ , $D\left[x_{2},x_{3},\ldots,x_{n},x_{0},x_{n+1};f\right]$ are of opposite signs. This also contradicts the assumption made about the function $f(x)$ Theorem 12 is therefore proven. Note that the proof also extends to the case $n=1$ 4.
- Definition 6. - A subset & of $[a,b]$ having at least $n+1$ points, will be said to be a set of $\mathscr{F}_{n}$ -polynomiality of the function $f(x)$ defined on $[a,b]$ if for any system of $n+1$ points $x_{1}<x_{2}<\ldots<x_{n+1}$ of $\mathcal{E}$ we have $\mathrm{D}\left[x_{1},x_{2},\ldots,x_{n},x_{n+1};f\right]=0$ .

Based on Theorem 12, we can conduct a detailed study of the sets of $\mathscr{F}_{n}$ -polynomiality of a function $f(x)$ order $n$ compared to $\mathscr{F}_{n}$ on $[a,b]$ . If $n\geqq 2,f(x)$ is continuous on $(a,b)$ In this case, everything together $\mathscr{L}\in(a,b)$ of $\mathscr{F}_{n}$ -polynomiality of $f(x)$ contains all its accumulation points that do not coincide with $a$ with $b$ . If $\lim\mathscr{E}\neq b$ , film $\mathscr{E}\neq a$ , $\mathscr{E}$ is a closed set. If $a\in\mathscr{E}^{\prime},a\in\mathscr{L}$ , the function $f(x)$ must be continuous on $a$ The same observation applies to the right end. $b$ By virtue of Theorem 12, the set $\mathcal{E}$ is convex, therefore it contains, with two of its points, every point between those points. It follows that $\mathscr{C}$ is an interval (which is not reduced to a point).

Let $\mathscr{L}_{1},\mathscr{L}_{2}$ two sets of $\mathscr{f}_{n}$ -polynomiality. If the intersection $\mathscr{E}_{1}\cap\mathscr{E}_{2}$ On more than one point, the meeting $\mathscr{E}_{1}\cup\mathscr{E}_{2}$ is a set of $\mathscr{F}_{n}$ -polynomiality. Otherwise, $\mathscr{L}_{1}\cup\mathscr{E}_{2}$ may not be a set of $\mathscr{F}_{n}$ -polynomiality. By always combining two sets of $\mathscr{F}_{n}$ -polynomiality which have more than one point in common, the set $F$ of all the sets of $\mathscr{F}_{n}$ -polynomiality of the function $f(x)$ , of order $n$ compared to $\mathscr{F}_{n}$ on $[a,b]$ , is a set of intervals, any two having at most ${}^{\text{un }}$ common point and of which at most two are not closed. $F$ is at most countable.

If $n=1,f(x)$ may also have discontinuities within the interval $[a,b]$ Any set of $\mathscr{F}_{n}$ -polynomialities is an interval, but, unlike the case $n\geqq 2$ the whole $F$ can contain more than two open intervals.
5. - Theorem 13. - If $f(x)$ is a continuous function on the interval $[a,b]$ , the necessary and sufficient condition for it to be $\mathscr{F}_{n}$ -convex or $\mathscr{F}_{n}$ -concave on $[a,b]$ is that she $n$ -valent in relation to $\mathscr{F}_{n}$ on $[a,b]$ .

For the demonstration, we rely on Theorem 6. Let's assume that the hypotheses of this theorem are verified. Then $D\left[x_{1},x_{2},\ldots,x_{n},x_{n+1};f\right]$ keeps its sign for any point system (32) of $[a,b].f(x)$ is therefore $\mathscr{F}_{n}$ -convex or $\mathscr{F}_{n}$ -concave depending on whether this sign is + or -. The condition of the stated theorem is therefore sufficient. The necessity of the condition is evident from definition 4. For the sets, $D_{n}$ the theorem was given by t. popoviciu [12].

The importance of the concept, given in this work, of the $n$ -valence with respect to a set of the type $I_{n}[a,b]$ results precisely from Theorem 13. In his work, L. TORNHEIM [14] calls convex with respect to a set of the type $I_{n}[a,b]$ , a continuous function on the closed interval $[a,b]$ And $n$ -valued with respect to the interpolating set considered*). Our definition 5 is more general.

§ 3.

Functions of order n with respect to a set of the type $I_{n}[a,b]$ , which contains an interpolation chain of order $(1,2,\ldots,n-1)$ .

1.
- —
  
  Definition 7. If $\mathscr{F}_{n}$ is a set of the type $I_{n}[a,b],n\geqq 2$ we will say that it contains an interpolation chain of order ( $1,2,\ldots,n-1$ ), if they exist $n-1$ subsets $\mathscr{F}_{1}\subset\mathscr{F}_{2}\subset\ldots\subset\mathscr{F}_{n-1}$ of $\mathscr{F}_{n}$ such as $\mathcal{F}_{k}$ either of the type $I_{k}[a,b],k=1,2,\ldots,n-1$ To be more precise, we will say that the subsets $f_{k},k-1,2\ldots,n-1$ form an interpolation chain of order $(1,2,\ldots,n-1)$ on $[a,b]$ .

Definition 8. - The point $\xi\in(a,b)$ is called an order point $k$ , of the function $f(x)$ defined on the interval $[a,b]$ , in relation to the whole $F_{k}du$ kind $I_{n}[a,b]$ if they exist in each of its neighborhoods $k+1$ distinct points $x_{1},x_{2},\ldots,x_{k},x_{k+1}$ such as one might have $D\left[x_{1},x_{2},\ldots,x_{k},x_{k+1};f\right]=0$ .

⁰⁰footnotetext: *) L. TORNHEIM [14], uses other names.

Lemma 6. - Either $f(x)$ a function $\mathscr{F}_{n}$ -convex on $[a,b]$ and either $D\left[x_{1},x_{2},\ldots,x_{n};f\right]=0$ , Or $x_{1}<x_{2}<\ldots<x_{n}<b$ . If $x_{k}<x_{0}<x_{k+1}$ , $3\leqq k\leqq n-1,n\geqq 4$ , so for $x>x_{n}$ We have

		$\displaystyle f(x)-L\left(x_{2},x_{3},\ldots,x_{k-1},x_{0},x_{k+1},\ldots,x_{n};f\mid x\right)>0.\text{ Si }x_{2}<x_{0}<x_{2}$
		.

A similar property applies to the function $f(x),\mathscr{F}_{n}$ -concave on $[a,b]$ , by respectively changing the direction of inequalities.

The demonstration is immediate. If $x_{2}<x_{0}<x_{3}$ the functions

L\left(x_{2},x_{3},\ldots,x_{n};f\mid x\right),\quad L\left(x_{0},x_{3},\ldots,x_{n};f\mid x\right)

(45)

coincide on $n-2$ points, so their difference changes sign when passing through these points. If we had $f(x)-L\left(x_{0},x_{3},\ldots,x_{\mathrm{n}};f\mid x\right)<0$ For $x>x_{n}$ , the difference of the functions (45) should cancel out twice in ( $x_{2},x_{3}$ ) (otherwise the functions (45) would coincide in $n-1$ points, which is impossible) and we would run into a contradiction with the $\mathscr{F}_{n}$ -convexity of the function $f(x)$ For the same reason, one cannot have $f(x)--L\left(x_{0},x_{3},\ldots,x_{n};f\mid x\right)=0$ on one point $x>x_{n}$ .

If $3\leqq k\leqq n-1$ the functions $L\left(x_{2},x_{3},\ldots,x_{n};f\mid x\right)$ And $L\left(x_{2},x_{3},\ldots,x_{k-1},x_{0},x_{k+1},\ldots,x_{n};f\mid x\right)$ coincide on $n-2$ points. As above, we see that we cannot have $f(x)--L\left(x_{2},x_{3},\ldots,x_{k-1}.x_{0},x_{k+1},\ldots,x_{n};f\mid x\right)\leqq 0$ for a $x>x_{n}$ .

Lemma 7. - If $f(x)$ is a function $\mathcal{F}_{n}$ -convex (or $\mathcal{F}_{n}$ -concave) on $[a,b],n\geqq 2$ and if $\xi$ is an order point $n-1$ of this function in relation to the whole $F_{n-1}$ They exist. $n-1$ points $x_{1}^{\prime}<x_{2}^{\prime}<\ldots<x_{n-1}^{\prime}$ , Or $\xi\leqq x_{1}$ such as for $x_{0}>x_{n-1}^{\prime}$ we have $D\left[x_{1}^{\prime}x_{2}^{\prime},\ldots,x_{n-1}^{\prime},x_{0};t\right]>0(\mathrm{ou}<0)$ Under the same assumptions, they exist. $n-1$ points $x_{1}^{\prime\prime}<x_{2}^{\prime\prime}<\ldots<x_{n-1}^{\prime\prime}$ , Or $x_{n-1}^{\prime\prime}<\xi$ such as $D\left[x_{1}^{\prime\prime},x_{2}^{\prime\prime},\ldots,x_{n-1}^{\prime\prime},\xi;f\right]>0\quad(\mathrm{ou}<0)$ .

To prove Lemma 7, we rely on Lemma 6. We can always assume that the points $x_{i}$ from the neighborhood of $\xi$ on which $D\left[x_{1},x_{2},\ldots,x_{n};f\right]=0$ verify the inequalities $x_{1}<x_{2}<\ldots<x_{n}<b$ Suppose $x_{k}<\xi<x_{k+1},2\leqq k\leqq n-1$ (if $k=1$ (we have nothing to study). We construct, by virtue of Lemma 6, the function $L\left(x_{2},x_{3},\ldots,x_{k-1},\xi,x_{k+1},\ldots,x_{n};f\mid x\right)$ and the points $\xi,x_{k+1},\ldots,x_{n}$ will be $n-k+1$ points $x_{i}^{\prime}$ We construct the functions successively

		$\displaystyle L\left(x_{3},x_{4},\ldots,x_{k-1},\xi,x_{2}^{\prime},x_{k+1},\ldots,x_{n};f\mid x\right),\ldots$
		$\displaystyle L\left(x_{k-1},\xi,x_{2},x_{3}^{\prime},\ldots,x_{k-2}^{\prime},x_{k+1},\ldots,x_{n};f\mid x\right)$
		$\displaystyle L\left(\xi,x_{2}^{\prime},x_{3}^{\prime},\ldots,x_{k-1}^{\prime},x_{k+1},\ldots,x_{n};f\mid x\right)$

(the points $x_{2}^{\prime},x_{3}^{\prime},\ldots,x_{k-1}^{\prime}$ being arbitrarily chosen from $\left(\xi,x_{k+1}\right)$ ).
The points $x_{1}=\xi,x_{2},x_{3},\ldots,x_{k+1},x_{k}^{\prime},=x_{k+1},\ldots,x_{n-1}^{\prime}=x_{n}$ , verify the first part of the conclusion of lemma 7.

We can state a lemma analogous to lemma 6, in failing of the point $x_{1}$ instead of point $x_{n}$ And $x_{k}$ instead of $x_{k+1}$ The role of the point $x>x_{k}$
will be played by any point in the interval ( $x_{k+1},x_{k}$ ). We therefore do not dwell on the second part of the conclusion of Lemma 7.
2. - Theorem 14. - If $f(x)$ is a function $\mathscr{F}_{n}$ -convex or $\mathscr{F}_{n}$ concave on the interval $[a,b],n\geqq 2$ it has at most one order point $n-1$ compared to $\mathscr{F}_{n-1}$ on $(a,b)$ .

By virtue of Lemma 7, if $\xi_{1}<\xi_{2}$ was two points of order $n-1$ compared to $\mathscr{F}_{n-1}$ , of the function $f(x)$ , there would be two systems, each formed by $n$ points $x_{1}^{\prime}<x_{2}^{\prime}<\ldots<x_{n}^{\prime};x_{1}^{\prime\prime}<x_{2}^{\prime\prime}<\ldots<x_{n}^{\prime\prime}$ , so that $x_{1}^{\prime}\leqq\xi_{1},x_{n}^{\prime\prime}\leqq\xi_{2}$ and that $D\left[x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{n}^{\prime};f\right],D\left[x_{1}^{\prime\prime},x_{2}^{\prime\prime},\ldots,x_{n}^{\prime\prime};f\right]$ let (non-zero and) have opposite signs. Applying Theorem 5 to the set $\mathscr{F}_{n-1}$ We can see that they exist $n$ distinct points $x_{1},x_{2},\ldots x_{n}$ belonging to $\left(\zeta_{1},\zeta_{2}\right)$ such as $D\left[x_{1},x_{2},\ldots,x_{n};f\right]=0$ .

Either $M_{n-1}$ the set of all order points $n-1$ compared to $\mathscr{F}_{n-1}$ of the function $f(x)$ Let $\xi^{\prime}=\inf M_{n-1},\xi^{\prime\prime}=\sup M_{n-1}$ and suppose $\xi^{\prime}<\zeta^{\prime\prime}$ The whole $M_{n-1}$ is dense everywhere in the interval $\left[\xi^{\prime},\xi^{\prime\prime}\right]^{*}$ ), SO $f(x)$ coincides with an element of $\mathscr{F}_{n-1}$ over this interval, which contradicts the fact that $f(x)$ East $n$ -valent in relation to $\mathscr{F}_{n}$ on $[a,b].M_{n-1}$ Therefore, it can only contain at most one point.

From Theorem 14, we can deduce some interesting consequences. $f(x)$ has no order points $n-1$ compared to $\mathscr{F}_{n-1}$ , she is $(n-1)$ valente compared to $\mathscr{F}_{n-1}$ , so she is $\mathscr{F}_{n-1}$ -convex or $\mathscr{F}_{n-1}$ -concave. By virtue of Theorem 14, in this case $f(x)$ cannot have more $un$ order point $n-2$ compared to $f_{n-2}$ .

If $f(x)$ indeed has a point $\xi$ order $n-1$ compared to $\mathscr{F}_{n-1}$ this point divides the interval $[a,b]$ in the subintervals $[a,\xi],[\xi,b]$ , on each the function being ( $n-1$ )-valent with respect to $\mathscr{F}_{n-1}\mathrm{Si}f(x)$ East $\mathscr{F}_{n-1}$ -convex resp. $\mathscr{F}_{n-1}$ -concave on the interval $[a,\xi]$ , he is $\mathscr{F}_{n-1}$ concave resp. $\mathscr{F}_{n-1}$ -convex on the interval $[\xi,b]$ .

Definition 9. - Under the hypotheses of Theorem 14, the point $\xi$ order $n-1$ compared to $\mathscr{F}_{n-1}$ , of the function $f(x)$ , is said to be of the first class if the function to its left is $\mathfrak{f}_{n-1}$ -concave and is said to be of the second class otherwise.**).

From the demonstration of theorem 7 it follows that the point $\xi$ , which appears in the statement, separates the points $\xi_{1}<\xi_{2}<\ldots<\xi_{n+1}$ .

Either $f(x)$ a function $n$ -valent in relation to $\mathscr{F}_{n}$ And $\xi$ a point of order $n-1$ compared to $\mathscr{F}_{n-1}$ of this function. We then have
Theorem 15. - If $n\geqq 2$ they exist, in every neighborhood of $\xi$ , $n$ points $\xi_{1}<\xi_{2}<\ldots<\xi_{n}$ such as $\xi_{n-1}<\xi<\xi_{n}$ And $D\left[\xi_{1},\xi_{2},\ldots,\xi_{n};f\right]=0$ Similarly, there are n points $\xi_{1}^{\prime\prime}<\xi_{2}^{\prime\prime}<\ldots<\xi_{n}^{\prime}$ such as $\xi_{1}^{\prime}<\xi<\xi_{2}^{\prime}$ And $D\left[\xi_{1}^{\prime},\xi_{2}^{\prime},\ldots,\xi^{\prime};f\right]=0$ .

To demonstrate Theorem 15, it suffices to consider the case $n=3$ Suppose that $f(x)$ either $\mathscr{F}_{n}$ -convex on $[a,b]$ . Either $\xi_{1}<\xi<\xi_{2}<\xi_{3}$

⁰⁰footnotetext: *).

M_{n-1}

is reduced, in effect, to the interval [

\xi^{\prime},\xi^{\prime\prime}

].
has

f_{n-k}

And $D\!\left[\xi_{1},\xi_{2},\xi_{3};f\right]=0$ . We have

L\!\left(\xi_{2},\xi_{3};f\mid x\right)<f(x)\quad\text{pour }x>\xi_{3},

And

L\!\left(\xi_{1},\xi_{2};f\mid x\right)>f(x)\quad\text{pour }\xi_{2}<x<\xi_{3}.

Among the elements of the form $L\!\left(x_{0},\xi_{3};f\mid x\right)$ , Or $\xi_{1}<x_{0}<\xi$ belonging to $\mathscr{F}_{2}$ , there is one that coincides with $f(x)$ at one point $\xi^{\prime}$ between $x_{0}$ And $\xi$ Indeed, if such a function did not exist, then, for $x_{0}=\xi$ the difference

f(x)-L\!\left(x_{0},\xi_{3};f\mid x\right)

would be zero at a point between $\xi$ And $\xi_{2}$ , or would cancel each other out in $x_{0}$ without changing sign. In all cases, $f(x)$ should have an order point $n-1$ relative to $\mathscr{F}_{n-1}$ , distinct from $\xi$ , which, by virtue of Theorem 14, is impossible.

So we have

\xi_{1}<\xi_{2}^{\prime}<\xi<\xi_{3}\quad\text{et}\quad D\!\left[\xi_{1},\xi_{2}^{\prime},\xi_{3};f\right]=0.

If $n<3$ And

\xi_{1}<\xi_{2}<\ldots<\xi_{k}<\xi<\xi_{k+1}<\ldots<\xi_{n},\qquad 2\leq k\leq n-2,

we proceed in the same way with the elements of $\mathscr{F}_{n-1}$ which coincide with $f(x)$ at the points

\xi_{1},\xi_{2},\ldots,\xi_{k-1},\xi_{k+2},\ldots,\xi_{n}.

In the case $k=1$ we delete the period $\xi_{1}$ This set is then of the type $I_{2}\!\left[\xi_{k-1},\xi_{k+2}\right]$ .

We can therefore find a point $\xi_{k+1}^{\prime}$ such as

\xi_{1}<\xi_{2}<\ldots<\xi_{k}<\xi_{k+1}^{\prime}<\xi<\xi_{k+2}<\ldots<\xi_{n},

And

D\!\left[\xi_{1},\xi_{2},\ldots,\xi_{k},\xi_{k+1}^{\prime},\xi_{k+2},\ldots,\xi_{n};f\right]=0.

By repeating this construction, we obtain the first part of the conclusion of the theorem. The second part of the conclusion is proven in a similar way. Theorem 16. - If $f(x)$ is a function $\mathscr{F}_{n}$ -convex or $\mathscr{F}_{n}$ -concave on $[a,b],n\geqq 2$ the interval $[a,b]$ can be broken down into at most $k+1$ consecutive subintervals on which $f(x)$ either alternately convex and concave with respect to $\mathscr{F}_{n-k}$ (we assume) $\left.1\leqq k\leqq n-1\right)^{*}$ ).

For the demonstration, let's first consider the case $k=1$ and suppose that $f(x)$ either $\mathscr{F}_{n}$ -convex on $[a,b]$ Two cases must be distinguished: either it exists in $\mathscr{F}_{n-1}$ an element that coincides with $f(x)$ on $n$ points of $[a,b]$ , or such a function does not exist in $\mathscr{F}_{n-1}$ In the second case $f(x)$ East $(n-1)$ -valent in relation to $\mathscr{F}_{n-1}$ on $[a,b]$ In the first case $f(x)$ at a point $\xi^{(n-1)}$ order $n-1$ compared to $\mathscr{F}_{n-1}$ The point $\xi^{(n-1)}$ divides the interval $[a,b]$ in two consecutive sub-intervals $\left[a,\xi^{(n-1)}\right]$ And $\left[\xi^{(n-1)},b\right]$ on each $f(x)$ being ( $n-1$ )-valent with respect to $\mathscr{f}_{n-1}$ It follows, taking into account Lemma 6, that $f(x)$ East $\mathscr{F}_{n-1}$ -concave on the first and $\mathscr{F}_{n-1}$ -convex on the second of these subintervals. We proceed in the same way if $f(x)$ East $\mathscr{F}_{n}$ -concave on $[a,b]$ .

For $k=1$ The property expressed by Theorem 16 therefore occurs as a result of the uniqueness of the decomposition that results from the definition of the point $\xi^{(n-1)}$ .

Under the assumption of the existence of the point $\xi^{(n-1)}$ , let us designate by $f_{1}^{(n-1)}(x)$ And $\operatorname{par}f_{2}^{(n-1)}(x)$ the restrictions of $f(x)$ respectively on the intervals $\left[a,\xi^{n-1)}\right],\left[\xi^{(n-1)},b\right]$ The function $f_{1}^{(n-1)}(x)$ can have at most one order point $n-2$ compared to $\mathscr{F}_{n-2}$ and the same remark also applies to the function $f_{2}^{(n-1)}(x)$ As a result, $f(x)$ has at most two points $\xi_{1}^{(n-2)}<\xi_{2}^{(n-2)}$ order $n-2$ compared to $\mathscr{F}_{n-2}$ on $[a,b]$ Indeed, if the point $\xi^{(n-1)}$ was

⁰⁰footnotetext: *). The subintervals

\left[\alpha_{1},\beta_{1}\right],\left[\alpha_{2},\beta_{2}\right],\ldots,\left[\alpha_{k},\beta_{k}\right]

[a,b]

are consecutive if

a\leqq\alpha_{1},\beta_{1}=\alpha_{2},\ldots,\beta_{k-1}=\alpha_{k},\beta_{k}\leqq b

at the same time a point of order $n-2$ compared to $\mathscr{F}_{n-2}$ , then one of the order points $n-2$ compared to $\mathscr{F}_{n-2}$ , located in $\left(a,\xi^{(n-1)}\right)$ And $\left(\xi^{(n-1)},b\right)$ should be missing. Otherwise they would be from different classes and $\xi^{(n-1)}$ would be of the same class as one of them, which is impossible, since then there would exist an interval on which $f(x)$ would coincide with an element of $\mathscr{F}_{n}$ .

So be it $\xi_{1}^{(n-2)}<\xi_{2}^{(n-2)}$ order points $n-2$ compared to $\mathscr{F}_{n-2}$ of the function $f(x)$ The property expressed by Theorem 16 remains valid for $k=2$ on the intervals $\left[a,\xi_{1}^{(n-2)}\right],\left[\xi_{1}^{(n-2)},\xi_{2}^{(n-2)}\right],\left[\xi_{2}^{(n-2)},b\right]$ . If $\xi^{(n-1)}$ does not exist, $f(x)$ being $\mathscr{f}_{n-1}$ -convex or $\mathscr{f}_{n-1}$ -concave, there is no on $[a,b]$ at most one order point $n-2$ compared to $\mathscr{F}_{n-2}$ .

For $k>2$ It is easy to see that the number of order points $n-k$ compared to $\mathscr{f}_{n-k}$ is finite. Let's assume this number $m\geqslant k+1$ Note that between two consecutive points $\xi_{i}^{(n-k)}<\xi_{i+1}^{(n-k)}$ order $n-k$ compared to $\mathscr{F}_{n-k}$ , there exists $un$ order point $n-k+1$ compared to $\mathscr{F}_{n-k+1}$ Otherwise, indeed, the points $\xi_{i}^{(n-k)},\xi_{i+1}^{(n-k)}$ would belong to an interval on which $f(x)$ would be convex or concave with respect to $\mathscr{F}_{n-k+1}$ , therefore there could not be two points of order on this interval $n-k$ The hypothesis $m\geqq k+1$ leads us to the existence of at least two order points $n-1$ compared to $f_{n-1}$ of the function $f(x)$ , which is impossible. Theorem 16 is therefore true, since on the subintervals determined by the points of various orders, the alternation of convexity with concavity is equally ensured.

This property can also be stated in the following form:
THEOREM 17. - A Junction $\mathscr{F}_{n}$ -convex or $\mathscr{F}_{n}$ -concave on $[a,b]$ at most $k$ order points $n-k$ compared to $\mathcal{F}_{n-k},k=1,2,\ldots,n-1$ Points of the same order belong alternately to different classes.

The first part is clear. The second part follows immediately if we note that between two points of the same order, which belong to the same class, there always exists at least one more point of the same order.

If they exist exactly $k$ order points $n-k$ compared to $\mathscr{F}_{n-k}$ Under the previous assumptions, we also have
Theorem 18. - Points of order $n-k$ compared to $\mathscr{F}_{n-k}$ and those of order $n-k+1$ compared to $\mathscr{f}_{n-k+1}$ separate ( $2\leqq k\leqq n-1$ 3.
- By specifying the whole $\mathscr{F}_{n}$ We deduce some conclusions from the previous theorems.

A function that is convex or concave with respect to the set $\mathscr{F}_{n}$ at most $n-1$ points of order 1, with respect to the set $\mathscr{F}_{1}$ These are the points where $f(x)$ has relative maxima and minima.

Theorems 15 and 16 can be extended to functions that are non-concave or non-convex with respect to the set $\mathscr{F}_{n}$ In this case, the decomposition of the interval $[a,b]$ , as required in the statements, may not be unique.
4. - To conclude this section, we will give a theorem relating to an interesting property of the elements of $\mathscr{F}_{n}$ , under the assumption
of the existence of an interpolation chain $\mathscr{F}_{1}\subset\mathscr{F}_{2}\subset\ldots\subset\mathscr{F}_{n-1}$ order $(1,2,\ldots,n-1)$ .

In accordance with the definition of the set $\mathscr{F}_{n}$ , the difference between two distinct elements of $\mathscr{F}_{n}$ cancels out at most $n-1$ points of $[a,b]$ It follows that any function $\varphi(x)\in\mathscr{F}_{n}$ which does not belong to $\mathscr{F}_{n-1}$ East ( $n-1$ ). valente compared to $\mathscr{F}_{n-1}$ so being continuous on $[a,b]$ , by definition is convex or concave with respect to $\mathscr{F}_{n-1}$ on the interval $[a,b]$ In any case, such a function $\varphi(x)$ at most $n-2$ points of order 1 with respect to $\mathscr{F}_{1}$ on ( $a,b$ ), in accordance with Theorem 17.

Lemma 8. - If $n\geqq 3$ , if $\varphi(x)\in\mathscr{F}_{n}$ does not belong to $\mathscr{F}_{n-1}$ and if $g(x)$ is any element of $\mathscr{F}_{1}$ , such as $g(x)-\varphi(x)$ cancels out on the points $x_{1}<x_{2}<\ldots<x_{n-1}$ of $[a,b]$ , SO $\varphi(x)$ has $n-2$ points of order 1 with respect to $\mathscr{F}_{1}$ .

The proof of this lemma follows from Theorem 7. On each of the intervals $\left(x_{i},x_{i+1}\right),i=1,2,\ldots,n-2,\varphi(x)$ has a point of order 1 with respect to $\mathscr{F}_{1}$ .

Lemma 9. — Under the assumptions of Lemma 8, if we denote by $\xi_{i}$ the order point $1$ compared to $\mathcal{F}_{1}$ located in the interval $(x_{i},x_{i+1})$ , For $i=1,2,\ldots,n-2$ So, the differences

\varphi(x)-L\!\left(\xi_{i};\varphi\mid x\right)

cancel each other out at the points $\xi_{i}$ , without changing sign.

The demonstration follows from the fact that the difference $L\left(\xi_{i};\varphi(x)-g(x)\right.$ , $i=1,2,\ldots,n-2$ cannot be canceled on $[a,b]$ .

The same 10. - If the elements of $\mathscr{F}_{n}$ are differentiable, under the assumptions of Lemma 9, we have

\left[\frac{dL\left(\xi_{i};\varphi\mid x\right)}{dx}\right]_{x=\xi_{i}}=\left[\frac{dg(x)}{dx}\right]_{x=\xi_{i}},i=1,2,\ldots,n-2

The proof follows from Lemma 9.5
. - Definition 10. - Let $g(x)\in\mathscr{F}_{n}$ And $x_{1}<x_{2}<\ldots<x_{n-1},n-1$ points of $(a,b)$ The elements of $\mathscr{F}_{n}$ , which on the points $x_{i},i=1,2,\ldots,n-1$ coincide with $g(x)$ form a group that we will call an ear of corn of order $n-1$ and we will refer to it as $\mathrm{S}\left(x_{1},x_{2},\ldots,x_{n-1};g(x)\right)^{*}$ )

It is clear that every ear of corn in order $n-1,n>1$ , contains an infinite number of functions. Every element, distinct from $g(x)$ , of the ear $S\left(x_{1},x_{2},\ldots,x_{n-1};g(x)\right)$ a, in each of the intervals ( $x_{i},x_{i+1}$ ), a point of order 1 with respect to $F_{1}$

Definition 11. - The ear of corn $S\left(x_{1},x_{2},\ldots,x_{n-1};g(x)\right)$ is said to be normal if all its elements, different from $g(x)$ have the same points of order 1 with respect to $\stackrel{{\scriptstyle a}}\mathscr{F}_{1}$ .

In what follows we will assume that the elements of $\mathscr{F}_{n}$ are derivable on $(a,b)$ . Either $g(x)\in\mathscr{F}_{1}$ And $\varphi_{1}(x),\varphi_{2}(x)$ two distinct elements of $\mathscr{F}_{n}$ , which do not belong to $\mathscr{F}_{n-1}$ and have the same convexity character with respect to $f_{n-1}$ on $[a,b]$ We then have the

THEOREM 19. - Suppose that $\varphi_{1}(x)-g(x)$ cancels out on the points $x_{1}<x_{2}<\ldots<x_{n-1}$ and that $\varphi_{2}(x)-g(x)$ cancels out on the points

⁰⁰footnotetext: *). It is unnecessary to emphasize that the definition is relative to the set

\mathscr{F}_{n}

because this set remains unchanged.

$x_{1}^{\prime}<x_{2}^{\prime}<\ldots<x_{n-1}^{\prime}$ Furthermore, the points $x_{i}$ separate the points $x_{i}^{\prime}{}^{*}$ If the ears $S\left(x_{1},x_{2},\ldots,x_{n-1};g(x)\right),S\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{-1}^{\prime};g(x)\right)$ are normal, the $n-2$ points $\xi_{1}<\xi_{2}<\ldots<\xi_{n-2}$ of order 1 with respect to $\mathcal{G}_{1}$ of $\varphi_{1}(x)$ separate the $n-2$ points $\xi_{1}^{\prime}<\xi_{2}^{\prime}<\ldots<\xi_{n-2}^{\prime}$ of order 1 with respect to $\mathscr{F}_{1}$ of the function $\varphi_{2}(x)$ .

A special case of this theorem is constituted by the well-known Markoff theorem [2] relating to the mutual separation of the extrema of two polynomials of the same degree.

The proof of the theorem is based on the lemmas. $8,9,10$ Suppose that

x_{1}<x_{1}^{\prime}<x_{2}<x_{2}^{\prime}<\ldots<x_{n-1}<x_{n-1}^{\prime}

(46)

Let's demonstrate that between two consecutive terms of the sequence

\xi_{1},\xi_{2},\ldots,\xi_{n-2}

(47)

there is always one and only one point in the sequence

\xi_{1}^{\prime},\xi_{2}^{\prime},\ldots,\xi_{n-2}^{\prime}

Indeed, the difference, $\varphi_{1}(x)-\varphi_{2}(x)$ cancels out on each of the intervals $\left(x_{i},x_{i+1}\right),i=1,2,\ldots,n-2$ . Either $x_{i}^{\prime\prime}$ the point of $\left(x_{i},x_{i+1}\right)$ on which this difference becomes zero and, to clarify, either $i$ an indication of which $\varphi_{1}\left(x_{i}^{\prime\prime}\right)=\varphi_{2}\left(x_{i}^{\prime\prime}\right)>g\left(x_{i}^{\prime\prime}\right)$ Let's demonstrate that $\xi_{i}^{\prime}>\xi_{i}$ We will first demonstrate that we cannot have $\xi_{i}^{\prime}<\xi_{i}$ Suppose that $\xi_{i}^{\prime}<\xi_{i}$ We can first note that we cannot have $\zeta_{i}^{\prime}<x_{i}^{\prime\prime}<\xi$ because otherwise we would have $\varphi_{2}\left(\xi_{i}^{\prime}\right)<\varphi_{1}\left(\xi_{i}^{\prime}\right),\varphi_{1}\left(\xi_{i}\right)<<\varphi_{2}\left(\xi_{i}\right)$ , from which it would follow that the difference $\varphi_{1}(x)-L\left(\xi_{i};\varphi_{2}\mid x\right)$ cancels out at two points in the interval ( $x_{i},x_{i}^{\prime\prime}$ ), SO $\varphi_{1}(x)$ would have in the meantime $\left(x_{1},x_{2}\right)$ two points of order 1 with respect to $\mathscr{F}_{1}$ , which contradicts Lemma 8. Therefore, we cannot have $\xi_{i}^{\prime}<x_{i}^{\prime\prime}<\xi_{i}$ .

Suppose $\xi_{i}^{\prime}<\xi_{i}<x_{i}^{\prime\prime}$ The ear of corn $S\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{n-1}^{\prime};g(x)\right)$ contains a function which at the point $\xi_{i}$ takes the value $\varphi_{1}\left(\xi_{i}\right)$ The point $\xi_{i}^{\prime}$ is a point of order 1 with respect to $\mathscr{F}_{1}$ of this function. We see that of the hypothesis $\xi_{i}^{\prime}<\xi_{i}$ it would result in $\varphi_{1}(x)$ a point of order 1 with respect to $G_{1}$ , located to the left of $\xi_{1}$ which is impossible. So we have $\xi_{i}\leqq\xi_{i}^{\prime}$ By using the differentiability of functions $\mathscr{F}_{n}$ the case $\xi_{i}=\xi_{i}^{\prime}$ is excluded since this equality would imply the existence of an element of the ear $S\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{n}^{\prime};g(x)\right.$ ) whose difference with $\varphi_{1}(x)$ would cancel out, without changing sign on $\xi_{i}^{\prime}$ We are thus in contradiction with Lemma 9. We have therefore demonstrated that $\xi_{i}<\xi_{i}^{\prime}$ Similarly, it can be demonstrated that $\xi_{i}^{\prime}<\xi_{i+1}$ So we finally have

\xi_{1}<\xi_{1}^{\prime}<\xi_{2}<\ldots<\xi_{n-2}<\xi_{n-2}^{\prime}

(49)

what needed to be taken apart.

⁰⁰footnotetext: ^*) . therefore

x_{i}<x_{i}^{\prime},i=1,2,\ldots,n-1

x_{i}^{\prime}<x_{i},i=1,2,\ldots,n-1

Theorem 19 has been proven, by various methods, in the particular case where $\mathscr{F}_{n}$ reduces to $\mathcal{D}_{n}$ This theorem plays an important role in the study of best approximation problems. Definition 11 leads us to a classification of interpolating sets, which we will not elaborate on here.
6. – In this section, the essential fact arose that the set $\mathcal{F}_{n}$ contains an interpolation chain of order $(1,2,\ldots,n-1)$ A classic example in this sense is the set ' $p_{n}$ which contains the interpolation chain $p_{n}C\subset\mathcal{D}_{2}\subset\ldots\subset\mathcal{F}_{n-1}$ Every system of $n$ continuous functions $\varphi_{1}(x),\varphi_{2}(x),\ldots,\varphi_{n}(x)$ on the interval $[a,b]$ , which satisfies the condition that any linear combination $\sum_{i=1}^{n+1}\alpha_{i}\varphi_{i}(x)$ cancels out at most $k-1$ points of $[a,b]$ and this is for $k=1,2,\ldots,n$ .

Let the set $\mathscr{F}_{n}$ formed by the functions $F\left(x;\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)$ defined on $[a,b]$ and dependent on $n$ actual parameters $\left.\alpha_{i},i=1,2,\ldots,n\right)^{*}$ .

Under certain additional assumptions, there still exists an interpolation chain of order $(1,2,\ldots,n-1)$ These assumptions can be deduced from the properties that the parameters then possess. $\alpha_{i},i=1,2,\ldots,n$ Suppose that one of these parameters is $\alpha_{l}$ , has the property that for each of its values $\alpha_{l}=\tilde{\alpha}_{l}$ the set of functions $F\left(x;\alpha_{1},\alpha_{2},\ldots,\alpha_{l-1},\tilde{\alpha}_{l}\right.$ , $\left.\alpha_{l+1},\ldots,\alpha_{n}\right)$ either of the type $I_{n-1}[a,b]$ , which means that for any system of $n-1$ distinct points $x_{1},x_{2},\ldots,x_{n-1}$ of $[a,b]$ the system of equations

F\left(x_{i};\alpha_{1},\alpha_{2},\ldots,\alpha_{l-1},\tilde{\alpha}_{l},\alpha_{l+1},\ldots,\alpha_{n}\right)=y_{i},i=1,2,\ldots,n-1

in relation to the unknowns $\alpha_{1},\alpha_{2},\ldots,\alpha_{l-1},\alpha_{l+1},\ldots,\alpha_{n}$ has one and only one solution, regardless of the numbers. $y_{i},i=1,2,\ldots,n-1$

We then have
Lemma 12. - Whatever the system of $n-1$ distinct points $x_{1}$ , $x_{2},\ldots,x_{n+1}$ and whatever the numbers $y_{1},y_{2},\ldots,y_{n-1}$ , the paramèri $\alpha_{l}$ is a montone function, with respect to the values at a point $x_{0}\neq x_{i}$ , $i=1,2,\ldots,n-1$ elements of the ear of corn $\mathrm{S}\left(x_{1},x_{2},\ldots,x_{n-1};y\right)^{***}$ ).

Indeed, in accordance with the hypothesis made on $\alpha_{l}$ in every ear $S\left(x_{1}\right.$ . $x_{2},\ldots,x_{n-1};y$ ), there is only one function of $\mathscr{F}_{n}$ with $\alpha_{l}=\tilde{\alpha}_{l}$ It follows that the value on $x_{0}$ functions of the ear $S\left(x_{1},x_{2},\ldots,x_{n-1};y\right)$ is a continuous function $y_{0}=\Phi\left(\alpha_{l}\right)$ compared to $\alpha_{l}$ Its inverse $\alpha_{l}=\Phi^{-1}\left(y_{0}\right)$ is univaled, therefore continuous, therefore a monotonic function in $y_{0}$ .

We also have the converse of Lemma 12,
Lemma 13. - If the monotonicity property of Lemma 12 holds, for any value $\hat{\alpha}_{l}$ parameter $\alpha_{l}$ the whole range of functions $F\left(x;\alpha_{1},\alpha_{2},\ldots\right)\left.\alpha_{l-1},\tilde{\alpha}_{l},\alpha_{l+1},\ldots,\alpha_{n}\right)$ is of the type $I_{n-1}[a,b]$ .

⁰⁰footnotetext: *). We always assume that

F\left(x;\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)

is continuous with respect to the set of

\operatorname{ses}n+1

variables.
**) To simplify

S\left(x_{1},x_{2},\ldots,x_{n-1}\right.x_{i}

and to the values

y_{i},i=1.2,\ldots,n-1

If $A[f]=0$ We are dealing with the case of Lemma 14. Let us therefore suppose that $A[f]=C\neq 0$ and consider the function $g(x)=f(x)-\frac{A[f]}{A\left[\varphi_{0}\right]}\varphi_{0}$ , Or $\varphi_{0}$ is an element of $\mathscr{F}_{n+1}$ which does not belong to $\mathscr{F}_{n}$ We have done $A\left[\varphi_{0}\right]\neq 0$ And, we'll check it right away, $A[g]=0$ As a result, $g(x)$ cannot be nor $\mathscr{F}_{n}$ -convex nor $\mathscr{F}_{n}$ -concave. $g(x)$ being continuous on $[a,b]$ , we can apply lemma 14 and we have $A\left[L\left(\xi_{1},\xi_{2},\ldots,\xi_{n+1};g\mid x\right)\right]=0$ for a certain points system $\xi_{1},\xi_{2},\ldots,\xi_{n+1}$ of $[a,b]$ Taking into account that $L\left(\xi_{1},\xi_{2},\ldots,\xi_{n+1};\varphi_{0}\mid x\right)$ coincides with $\varphi_{0}$ we deduce (50), by calculating $A\left[L\left(\xi_{1},\xi_{2},\ldots,\xi_{n+1};g\mid x\right)\right]$ .

The conclusion of Lemma 15 leads us to a few remarks. If $f_{n+1}$ , $\mathscr{f}_{n}$ are linear, we know that in $\mathscr{f}_{n+1}$ they exist $n+1$ elements $f_{1}(x),f_{2}(x),\ldots,f_{n+1}(x)$ , which form an interpolating basis [5]. Any function $\varphi(x)\in\mathscr{F}_{n+1}$ is of the form $\varphi(x)=\sum_{i=1}^{n+1}\alpha_{i}i_{i}(x)$ If for $\alpha_{n+1}==0$ we get the whole set $\mathscr{F}_{n}$ We are dealing with the classic case of a Chebyshev system $f_{1}(x),f_{2}(x),\ldots,f_{n}(x),f_{n+1}(x)$ where the $n$ The first functions also form a Chebyshev system. We have $\mathrm{L}\left(x_{1},x_{2},\ldots,x_{n},x_{n+1};f\mid x\right)=\sum_{i=1}^{n+1}d_{i}[f]f_{i}(x)$ , Or $d_{n+1}[f]$ is the generalized divided difference of the function $f(x)$ in relation to the system of functions $f_{1}(x),f_{2}(x),\ldots,f_{n+1}(x)$ (
51)

d_{n+1}[f]=\frac{\left.\left|\begin{array}[]{ccccccc}f_{1}\left(x_{1}\right)&f_{2}\left(x_{1}\right)&\cdots&\cdots&f_{n}\left(x_{1}\right)&f\left(x_{1}\right)\\ f_{1}\left(x_{2}\right)&f_{2}\left(x_{2}\right)&\cdots&\cdot&\cdot&f_{n}\left(x_{2}\right)&f\left(x_{2}\right)\\ \cdots&\cdots&\cdot&\cdot&\ddots&\cdot&\cdot\\ f_{1}\left(x_{n+1}\right)&f_{2}\left(x_{n+1}\right)&\cdot&\cdot&\cdot&\cdot&f_{n}\left(x_{n+1}\right)\end{array}\right|\begin{array}[]{c}f\left(x_{n+1}\right)\\ \cdot\end{array}\right\rvert\,}{\left|\begin{array}[]{ccccccc}f_{1}\left(x_{1}\right)&f_{2}\left(x_{1}\right)&\cdots&\cdot&\cdot&f_{n}\left(x_{1}\right)&f_{n+1}\left(x_{1}\right)\\ f_{1}\left(x_{2}\right)&f_{2}\left(x_{2}\right)&\cdots&\cdot&\cdot&f_{n}\left(x_{2}\right)&f_{n+1}\left(x_{2}\right)\\ \cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\\ f_{1}\left(x_{n+1}\right)&f_{2}\left(x_{n+1}\right)&\cdots&\cdot&f_{n}\left(x_{n+1}\right)&f_{n+1}\left(x_{n+1}\right)\end{array}\right|}

Under the assumptions of Lemma 15, formula (50) becomes
(52)

A\left[L\left(\xi_{1},\xi_{2},\ldots,\xi_{n+1};f\mid x\right)\right]=d_{n+1}[f]\mathrm{A}\left[f_{n+1}\right]=A[f]

The statement of Lemma 14 therefore contains, as a special case, the mean value theorem for linear functionals, given by T. Popoviciu [113. - In what follows we will use the properties of an ear of order " of the set $f_{n+1}$ First, let us note that every ear of corn of order $n$ of the whole $\mathscr{F}_{n+1}$ contains one and only one function belonging to $\mathscr{F}_{n}$ Let's consider the points $M_{i}\left(x_{i},y_{i}\right),i=1,2,\ldots,n$ Or $x_{1},x_{2},\ldots,x_{n}$ are $n$ distinct points
of the interval $[a,b]$ The function $\mathrm{L}\left(x_{1},x_{2},\ldots,x_{n};y\mid x\right)$ determines in the considered ear of corn two non-empty subsets: that of the functions of $\mathscr{F}_{n+1}$ who are $\mathscr{F}_{n}$ -convex and that of the functions of $\mathscr{F}_{n+1}$ who are $\mathscr{F}_{n}$ -concave.

We have
Lemma 16. - If:
$1^{\circ}A[f]$ is continuous on its domain of definition *)
$2^{\circ}A[f]=0$ , For $f\in\mathscr{F}_{n}$ .
$3^{\circ}A[f]\neq 0$ For $f\mathscr{F}_{n}$ -convex or $\mathscr{F}_{n}$ -concave on $[a,b]$ ,
SO $A[f]$ maintains a constant sign for all functions $\mathscr{F}_{n}$ convex ( $f_{n}$ -concave) belonging to an ear of corn of order $n$ of the whole $f_{n+1}$ .

Let's consider the points $M_{i}\left(x_{i},y_{i}\right)$ From higher up. We have $A\left[L\left(x_{1},x_{2},\ldots,x_{n};y\mid x\right)\right]=0$ Suppose that among the functions $\mathscr{F}_{n}$ -convex ( $\mathscr{f}_{n}$ -concave) belonging to the corresponding lep, there are two of them, $g_{1}(x),g_{2}(x)$ such as $A\left[g_{1}\right]<0,A\left[g_{2}\right]>0$ Suppose that for $x>x_{n}$ we have $\mathrm{g}_{1}(x)<g_{2}(x)$ . If $x_{n}=b$ we choose a point $x$ of the interval $\left(x_{n-1},x_{n}\right)$ on which we assume $g_{2}(x)>g_{1}(x)$ By virtue of Theorem 1, the set of elements of the epi, whose values on $x$ are between $g_{1}(x)$ And $g_{2}(x)$ is compact. It follows that $A[f]$ must cancel out on an element of this set. This contradicts the hypothesis $3^{\circ}$ of the lemma.

Suppose that $\mathscr{F}_{n+1}$ that is, the set generated by the Tschebycheff system $\varphi_{1}(x),\varphi_{2}(x),\ldots,\varphi_{n+1}(x)$ and that $\mathscr{F}_{n}$ and the subset of $f_{n+1}$ generated by the functions $\varphi_{1}(x),\varphi_{2}(x),\ldots,\varphi_{n}(x)$ , which is also assumed to form a Chebyshev system.

Lemma 17. - Under the assumptions of Lemma 15, $A[f]$ is monotonic with respect to the value, on a fixed point different from the points $x_{i},i=1,2,\ldots,n$ , functions of the ear of corn relative to a system of points $\left.Mi\left(x_{i},y_{i}\right),i=1,2,\ldots,n^{**}\right)$ .

The proof follows from Lemma 15. By virtue of formula (52), $A\left[f_{n+1}\right],A[f]$ coincide, apart from a constant factor, with the parameter $\alpha_{n+1}$ of the function $F\left(x;\alpha_{1},\alpha_{2},\ldots,\alpha_{n+1}\right)=\sum_{k=1}^{n+1}\alpha_{k}\varphi_{k}(x)$ which generates the whole $\mathscr{F}_{n+1}$ Lemma 17 then follows from lemma 13 of the preceding paragraph.

Note that, under the assumptions of Lemma 16, the set of elements of $\mathscr{F}_{n+1}$ , for which $A[f]$ takes the same value $C$ is of the type $I_{n}[a,b]$ So we have the

Lemma 18. - If $\mathscr{F}_{n},\mathscr{F}_{n+1},\mathscr{F}_{n}\subset\mathscr{F}_{n+1}$ are arbitrary interpolating sets (not necessarily linear) and if the hypotheses of Lemma 15 are verified, of the monotonicity of the functional $A[f]$ on any ear of corn (in the sense of Lemma 16), it follows that the set of elements of $\mathscr{F}_{n+1}$ on which $A[t]$ takes the same value $C$ is of the type $I_{n}[a,b]$ *
) If the sequence of functions $\left\{f_{k}(x)\right\}_{k=1}^{\infty}$ , tends uniformly towards $f(x)$ on $[a,b]$ , We have $A\left[f_{k}\right]\rightarrow A[f]$ .
^∗∗ ). This monotony will be called a montony on the ear of corn $S\left(x_{1},x_{2},\ldots,x_{n};y\right)$ .

We also deduce
theorem 20, - If the conditions of lemma 16 are verified and if.
$1^{\circ}$ . HAS $[f]$ is monotonous on every ear of the whole $f_{n+1}$ ,
$2^{\circ}.C\neq 0$ being arbitrary, we have $A[\varphi]\neq C$ for any function $\varphi(x)$ which is convex or concave with respect to the set of functions $f$ of $\mathcal{F}_{n+1}$ for which we have $A[f]=C$ ,
then for any function $f(x)$ continue on $[a,b]$ , there is a system of $n+1$ points $\xi_{1},\xi_{2},\ldots,\xi_{n+1}$ of $[a,b]$ such as $A[f]=A\left[L\left(\xi_{1},\xi_{2},\ldots,\xi_{n};f\mid x\right)\right]$ .

The proof amounts to applying Lemma 14 to the subset formed by the functions $f$ of $\mathscr{F}_{n+1}$ for which $A[f]=A[\rho]=C_{\neq 0}$ If $C=0$ We return to Lemma 14.
4. - We will give an application of Theorem 20, in particular concerning sets $\mathscr{F}_{n}$ And $\mathscr{F}_{n+1}$ Consider the set of polynomials of degree (at most equal to) $n$ and the subset formed by the polynomials of degree $n-1,n\geqq 1$ . Either $f(x)$ a continuous function on the interval $[a,b]$ , And $T_{n-1}^{*}(x)$ , the polynomial of best approximation of the degie $n-1$ , of the function $f(x)$ , on the interval $[a,b]$ In accordance with the definition of the polynomial $T_{n-1}^{*}(x)$ ,

\begin{array}[]{r}\mu_{n-1}[f]=\max_{x\in[a.b]}\left|f(x)-T_{n-1}^{*}(x)\right|,\text{ où }\mu_{n-1}[f]=\\ =\min_{T_{n-1}\in\mathcal{D}_{n}}\left|\max_{x\in[a,b]}\right|f(x)-T_{n-1}(x)\mid\end{array}

We know that the polynomial $T_{n-1}^{*}(x)$ exists and is unique.
Designs by $\mathrm{V}\left(x_{1},x_{2},\ldots,x_{n+1}\right)$ the Vandermonde determinant of numbers $x_{1},x_{2},\ldots,x_{n+1}$ and by $\left[x_{1},x_{2},\ldots,x_{n+1};j\right]$ the difference divided by order $n$ of the function $f(x)$ on the (supposedly distinct) points $x_{1},x_{2},\ldots$ . $x_{n+1}$ We know that

=\frac{\rho_{n-1}\left(x_{1},x_{2},\ldots,x_{n+1};f\right)=}{\sum_{i=1}^{n+1}V\left(x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1}\right)}\left|\left[x_{1},x_{2},\ldots,x_{n+1};f\right]\right|

(53)

is the best approximation of the function $f(x)$ , by polynomials of degree $n-1$ on the points $x_{1}<x_{2}<\ldots<x_{n+1}$ According to the well-known de la Vallée Poussin theorem, we have

\mu_{n-1}[f]=\max\rho_{n-1}\left(x_{1},x_{2},\ldots,x_{n+1};f\right)

(54)

the maximum of the second member being relative to all systems of $n+1$ points $x_{1}<x_{2}<\ldots<x_{n+1}$ of the interval $[a,b]$ We can note $q^{\text{1è }}$
the maximum of

\left|\sum_{i=1}^{n+1}V\left(x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1}\right)\right|

under the same conditions, is the best approximation of the function $f(x)==x^{n}$ by polynomials of degree $n-1$ , on the interval $[a,b]$ , therefore is equal to $\frac{(b-a)^{n}}{2^{2n}}$ It follows that if $p(x)\in\nabla_{n+1}$ And $p(x)\bar{\epsilon},\nabla_{n}$ We have

\mu_{n-1}[p]=\frac{(b-a)^{n}}{2^{2n}}\left|\left[x_{1},x_{2},\ldots,x_{n+1};p\right]\right|

(55)

the postman $\left|\left[x_{1},x_{2},\ldots,x_{n+1};p\right]\right|$ being constant and equal to the absolute value of the coefficient of $x^{n}$ of the polynomial $p(x)$ .
name $p\left(x_{n}\right)\in p_{n+1}(\alpha)$ such as $p\left(x_{i}\right),i=1,2,\ldots,n$ . But $L\left(x_{1},x_{2},\ldots,x_{n},x_{n+1;q_{1}}\right.$ coincides on $n+1$ points with the function $f(x)$ which is concave with respect to $p_{n+1}(\alpha)$ , SO $p\left(x_{n+1}\right)-L\left(x_{1},x_{2},\ldots,x_{n},x_{n+1};f\left(x_{n+1}\right)>0\right.$ which is equivalent to $\left[x_{1},x_{2},\ldots,x_{n+1};f\right]<\alpha$ In the demonstration of this inequality, we took into account the fact that for all points $M_{1}\left(x_{1},y_{1}\right),M_{2}\left(x_{2},y_{2}\right),\ldots,M_{n}\left(x_{n},y_{n}\right.$ . $x_{1},<x_{2}<\ldots<x_{n}$ if $x_{n+1}>x_{n}$ and if $p_{1}(x),p_{2}(x)$ , are two polynomials of the ear of corn relative to these points, then of $p_{1}\left(x_{n+1}\right)>p_{2}\left(x_{n+1}\right)$ it results $q\left(x_{k}\right.\left[x_{1},x_{2},\ldots,x_{n+1};p_{1}\right]>\left[x_{1},x_{2},\ldots,x_{n+1};p_{2}\right]$ .

From the above, we can deduce some details about the best approximation of the function $f(x)$ by polynomials of the set $\bar{p}_{3}$ We always maintain the assumption of the convexity of the function $f(x)$ , which we also assume to be continuous on the interval $[a,b]$ So then $\mathscr{E}_{\text{ni }}$ all the points of $[a,b]$ for which we have $\left|f(x)-T_{n-1}^{*}(x)\right|==\mu_{n-1}[f]$ , Or $T_{n-1}^{*}(x)$ is the polynomial of best approximation: of the degree $n-1$ of the function $f(x)$ on the interval $[a,b]$ The whole $\mathscr{E}_{n-1}$ and closed and then $\xi=\inf\mathscr{E}_{n-1}$ . We have $\left|f(\xi)-T_{n-1}^{+}(\xi)\right|=\mu_{n-1}[j$ Let's ask.

\delta_{n-1}[f]=(-1)^{n}\left[f(\xi)-T_{n-1}^{*}(\xi)\right].

(57)

We can note that if $f(x)$ coincides with an element $p(x)$ of $P_{n-1}$ which is not reducible to a polynomial of degree $n-1$ , the functional defined by (57) coincides with $\delta_{n-1}$ [ $p$ ] given by (56).

Under the assumptions made about the function $f(x)$ , $\partial_{\,n-1}[f]$ can never coincide with $\delta_{\,n-1}[p]$ , if $p\in\mathcal{P}_{n+1}(\alpha)$ Indeed, according to the remark made on the divided difference

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

where the points $x_{i}$ , $i=1,2,\ldots,n+1$ are arbitrary in $[a,b]$ It follows that $f(x)$ is a function with bounded divided differences, in the sense of the definition given by T. Popoviciu [9] .

This result is easily obtained. From the properties of higher-order convex functions, it follows first that the set of elements of $\mathcal{P}_{n+1}^{(2)}$ which coincide in $n$ distinct points with $f(x)$ is also bounded. It also follows that the set of elements of $\mathcal{P}_{n+1}$ which coincide in $n+1$ points with $f(x)$ and which are all concave with respect to $\mathcal{P}_{n+1}(\alpha)$ is also limited.

The fact that the differences divided

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

that they are bounded then follows from the well-known upper bound of the coefficients of an equally bounded set of polynomials of the same degree.

So we have…

\mu_{n-1}[f]=\frac{(b-a)^{n}}{2^{2n}}\left\{\max_{x_{1},x_{1},\ldots,x_{n+1}}\left|\left[x_{1},x_{2},\ldots,x_{n+1};f\right]\right|\right\}

(58)

Because we have always $\left[x_{1},x_{2},\ldots,x_{n+1};f\right]\neq\alpha$ , it follows $q^{\text{qu }}$ we can never have $\delta_{n-1}[f]=\delta_{n-1}[p]$ if $p\in\mathcal{O}_{n+1}(\alpha)$ .

Let's move on to the application of Theorem 20. Consider the function ull $A[f]=\delta_{n-1}[f]$ Based on the remarks made above, it satisfies all the conditions of the statement of Theorem 20. Regarding ${}^{\text{l }}$ continuity of this functional, with respect to. As for the colement of the space of continuous functions on the interval $[a,b]$ and with respect to the 1st
uniform metric*), it is well known. We are here in the case of the application of theorem 20 to a nonlinear functional.

If $f(x)$ is a continuous function on the interval $[a,b]$ , we have theorem 21. - If $\mu_{n-1}[f]$ is the best approximation of the function $f(x)$ by polynomials of degree $n-1$ on the interval $[a,b]$ they exist in $[a,b]n+1$ points $\xi_{1},\xi_{2},\ldots,\xi_{n+1}$ , such as the Lagrange interpolation polynomial $L\left(\xi_{1},\xi_{2},\ldots,\xi_{n+1};f\mid x\right)$ on the nodes $\xi_{i},i=1,2,\ldots,n+1$ , has the property**)

\mu_{n-1}\left[L\left(\xi_{1},\xi_{2},\ldots,\xi_{n+1};f\mid x\right)\right]=\mu_{n-1}[f].

(59)

This theorem leads us to an interesting observation about numbers. $\mu_{k}[f],k=0,1,\ldots$ We know that the sequence of these numbers is non-increasing. By virtue of Weierstrass's theorem, we have $\lim_{k\rightarrow\infty}\mu_{k}[f]=0$ Theorem 21 shows us that we have $\mu_{k}[f]=\frac{(b-a)^{k\rightarrow+1}}{2^{2k+2}}M_{k}[f]$ The study of the structure of numbers $M_{k}[f],k=0,1,\ldots$ , under the assumption of the continuity of the function $f(x)$ This may lead to a proof of Weierstrass's theorem, using best-approximate polynomials. The problem of such a proof has also been noted elsewhere [8].

Theorem 20 also has other applications in the study of the structure of certain functionals involved in numerical analysis. A particularly interesting application is presented for a given functional. $A[f]$ , the search for an interpolated set of order $n+1\geqq 2$ , such that the hypotheses of Theorem 20 are verified.

BIBLIOGRAPHY

1.

EF Beckenbach. Generalized convex functions. Bull. Bitter. Math. Soc. Vol. 43, 363-371, 1937.
2.

W. Markoff. Uber Polynomy, die in einem gegebenen Intervalle moglichts wenig von Null abweichen. Math. Ann. 77, 213-258, 1916.
3.

E. Moldovan. Asupra generalizări a noțiunii de convexitate. Studii şi Cercetăr științifice, Cluj, VI, 65-73, 1955.
4.

E. Moldovan. Asupra unor teorem de media. Comunicările Academiei RPR t. VI, no. 1, 7-12, 1956.
5.

E. Moldovan. Properties with multiple interpolation functions. Bul. Univ. ,,V. Babeş" si ,,Bolyai". Seria st. naturii I, 1-2, 1957.

⁰⁰footnotetext: *). The metric

\rho\left(f_{1},f_{2}\right)=\underset{x\in[a,b]}{\max}\left|f_{1}(x)-f_{2}(x)\right|

**). Of

\delta_{n-1}[f]=A

It obviously follows that

\mu_{n-1}[f]=|A|

7. D. Pompeiu. On a proposition analogous to the mean value theorem. Mathematica T. 22, 143-146, 1946.
8. T. Popoviciu. Cea mai bună aproximatic a functiilor continu prin polinoame. Cluj, 19ji
9. T. Popoviciu. On some properties of functions of one or two real variables. Mathematica t. VIII, 1-86, 1934.
10. T. Popoviciu. On a generalization of the notion of higher-order convexity. Mathematica t. 12, 227-333, 1936.
11. T. Popoviciu. Notes on lower-order convex functions ( $IX$ ). Bull. Math. of ? Soc. Romanian of Sc: t. 43, 85-141, 1941.
12. T. Popoviciu. Notes on higher order convex functions (1). Mathematicat 12 81-92, 1936.
13. G. Póly a. On the mean-value theorem corresponding to a given linear homogeneous different equation. Trans. Am. Math. Soc. 24, 312-324, 1922.
14. L. Tornheim. On n-parameter families of functions and associated convex function: Trans. Bitter. Mast. Soc. t. 69, 457-467, 1950.

Received on November 7, 1958.

On a generalization of convex functions

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ON A GENERALIZATION OF CONVEX FUNCTIONS

Some properties of interpolating function sets

We have given the demonstration elsewhere [3].

§ 2.

The notion of convex tonction with respect to a set of the type $I_{n}[a,b]$ .

§ 3.

BIBLIOGRAPHY

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On a generalization of convex functions

Abstract

Authors

Keywords

Paper coordinates

PDF

About this paper

Journal

Publisher Name

DOI

Print ISSN

Online ISSN

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Paper (preprint) in HTML form

ON A GENERALIZATION OF CONVEX FUNCTIONS

Some properties of interpolating function sets

We have given the demonstration elsewhere [3].

§ 2.

The notion of convex tonction with respect to a set of the typeIn​[has,b]I_{n}[a,b].

§ 3.

BIBLIOGRAPHY

Related Posts

The notion of convex tonction with respect to a set of the type $I_{n}[a,b]$ .