ON THE EXTENSION OF HIGHER-ORDER CONVEX FUNCTIONS

BY
TIBERIU POPOVICIU
(Cluj)

The present work aims to supplement, on certain points, the theory of higher-order convex functions that was presented in our Thesis ¹ ).

We consider functions $f(x)$ defined, uniform and real of the real variable $x$ on a linear and bounded set E.

Let us designate by $\mathrm{V}\left(\alpha_{1},\alpha_{2},\ldots,\alpha_{k+1}\right)$ the Van Der determinant of the world of quantities $\alpha_{i}$ and by $U\left(\alpha_{1},\alpha_{2},\ldots,\alpha_{k+1};f\right)$ the determinant that we deduce from $\mathrm{V}\left(\alpha_{1},\alpha_{2},\ldots,\alpha_{k+1}\right)$ when the elements of the last column are replaced by

f\left(\alpha_{1}\right),f\left(\alpha_{2}\right),\ldots,f\left(\alpha_{k+1}\right)

respectively.
The quotient

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

is the difference divided by order $k$ of the function $f(x)$ for distinct points $\alpha_{1},\alpha_{2},\ldots,x_{k+1}$ .

The divided differences are linked by the recurrence relation.

\begin{gathered}{\left[\alpha_{1},\alpha_{2},\ldots,\alpha_{k+1};f\right]=\frac{\left[\alpha_{2},\alpha_{3},\ldots,\alpha_{k+1};f\right]-\left[\alpha_{1},\alpha_{2},\ldots,\alpha_{k};f\right]}{\alpha_{k+1}-\alpha_{1}}}\\ {[\alpha;f]=f(\alpha)}\end{gathered}

The function $f(x)$ will be called convex, non-concave, polynomial, non-convex or concave of order $n$ on the set E following the differences divided by order $n+1$ across all groups of $n+2$ points of E $>0,\equiv 0,=0,\leq 0$ Or $<0$ .

These functions form the class of functions of order n.

⁰⁰footnotetext: ¹⁾ „On some properties of functions of one or two real variables" Paris 1933.

It can happen that a function possesses several convexity properties of different orders. We will say that it belongs to the class ( $a,b,c,\ldots$ ) if it possesses order properties $a,b,c,\ldots$ To highlight the nature of the function, we will assign numbers $a,b,c,\ldots$ clues in the following manner: $a,a^{*},\bar{a},a^{\prime},a^{**}$ depending on whether the function is non-concave, convex, polynomial, non-convex or concave of order $a$ It is useful to distinguish functions with an invariant sign. We will agree to call them functions of order -1, and we will assign this number of indices, as above, depending on whether the function remains $\geq 0,>0,=0,\leq 0$ Or $<0$ .

We will also refer to it as:

\mathrm{P}\left(\alpha_{1},\alpha_{2},\ldots,\alpha_{k};f\mid x\right)

the Lagrange polynomial for points $\alpha_{1},\alpha_{2},\ldots,\alpha_{k}$ relative to the function $f(x)$ , therefore the polynomial of degree $k-1$ who takes the values $f\left(\alpha_{i}\right)$ to the points $\alpha_{i}$ .

In what follows, we will constantly make use of the properties of functions of a given order or class that are demonstrated in our Thesis. We ask the reader to refer to it.

I. Statement of the extension problem

1.

We will say that the function $f(x)$ of a given class on $\mathrm{E}se$ extends onto another set $\mathrm{E}_{1}$ , if a function exists $f_{1}(x)$ of the same class defined on $\mathrm{E}+\mathrm{E}_{1}$ and which coincides with $f(x)$ on E. We will say that the function is extended in the broad sense if we consider convexity and polynomiality as special cases of non-concaveness. Otherwise, we will say that the function is extended in the strict sense. This extension is more restrictive and implies the extension in the broad sense.

To simplify the language, we call the maximum order of $f(x)$ the highest order that appears in its class.
2. We will study in particular the extension of functions defined on a finite set of $m$ points

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

(1)

The following properties are immediate.
Any function of order 0 or 1 defined on (1) can be strictly extended to any set $E_{1}$ .

Any function of maximum order 1 defined on (1) can be extended in a wide sense to any set $E_{1}$ .

It is obviously sufficient to demonstrate that the function can be extended in the interval ( $x_{1},x_{m}$ ), since it is on a point to the left of $x_{1}$ or to the right of $x_{m}$ Let's construct a differentiable function that performs the extension.

Simply looking at the geometric representation of the function shows that we can take

f_{1}^{\prime}\left(x_{1}\right),f_{1}^{\prime}\left(x_{2}\right),f_{1}^{\prime}\left(x_{3}\right)

such that this sequence precisely satisfies the convexity properties of the derivative of a function of the form considered, and such that we have

\begin{gathered}f_{1}^{\prime}\left(x_{1}\right)\leq\left[x_{1},x_{2};f\right],f_{1}^{\prime}\left(x_{m}\right)\geq\left[x_{m-1},x_{m};f\right]\\ {\left[x_{i-1},x_{i};f\right]\leq f_{1}^{\prime}\left(x_{i}\right)\leq\left[x_{i},x_{i+1};f\right]}\\ i=2,3,\ldots,m-1\end{gathered}

without equality if the function is convex. We can now construct $f_{1}(x)$ In $\left(x_{i},x_{i+1}\right)$ such that it has a continuous derivative reducing to $f_{1}^{\prime}\left(x_{i}\right),f_{1}^{\prime}\left(x_{i+1}\right)$ to the points $x_{i},x_{i+1}$ and fulfilling the conditions required for the extension. (For example, one could take $f_{1}(x)$ represented by an arc of an ellipse satisfying the desired conditions).

It should be noted that the strict extension of a function of maximum order 1 is not always possible. For example, if the function is of the class $(0,1*)$ and if $f\left(x_{i}\right)=f\left(x_{i+1}\right)$ its extension is necessarily of the class $(0,1)$ the function must reduce to a constant in the interval ( $x_{i},x_{i+1}$ On the contrary, a function of the class ( $0*,1*$ ) is always strictly extendable.
3. Let $f(x)$ of order $n>1$ on (1). We can assume $m>n+2$ because if $m=n+2$ the problem is entirely solved by the polynomial

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

We will examine the extension on one point $x$ distinct from points (1). This point lies between $x_{i},x_{i+1}$ (to the left of $x_{1}$ if $i=0$ , to the right of $x_{m}$ if $i=m$ The extension depends only on the $n+1$ first points to the left of $x$ and $n+1$ first points to the right of $x^{2}$ ). Therefore, we simply need to write that the function is of the desired class on the set

x_{i-n},x_{i-n+1},\ldots x_{i},x,x_{i+1},x_{i+2},\ldots,x_{i+n+1}

where we agree to consider the points $x_{1},x_{2},\ldots,x_{i}$ to the left of $x$ if $i<n+1$ (none if $i=0$ ) and the points $x_{i+1},x_{i+2},\ldots,x_{m}$ to the right of $x$ if $i>m-n-1$ (none if $i=m$ ) and we will keep this convention in the following without explicitly stating it.

⁰⁰footnotetext: ² ) See loc. cit. 1) p. 19.

The conditions for extension are therefore

	$\displaystyle y_{j}=\left[x_{j},x_{j+1},\ldots,x_{j+n},x;f\right]\geq 0\text{ ou }>0$		(2)
	$\displaystyle j=i-n,i-n+1,\ldots,i+1$

depending on whether the function is non-concave or convex and depending on whether it is a wide or strict extension.

For the function to extend to the point $x$ It is necessary and sufficient that the inequalities (2) be compatible with $f(x)$ viewed as a parameter. The function is then extended by any value of $f(x)$ verifying these inequalities.

Inequality (2) means that $f(x)$ has a precise position relative to the polynomials

		$\displaystyle\mathrm{P}_{j}=\mathrm{P}\left(x_{j},x_{j+1},\ldots,x_{j+n};f\mid x\right)$
		$\displaystyle j=i-n,i-n+1,\ldots,i+1$

The following geometric interpretation is easily found:
Polynomials $\mathrm{P}_{i-n},\mathrm{P}_{i-n+2},\mathrm{P}_{i-n+4}$ have an upper limit function $\bar{g}_{i}(x)$ and a lower limit function $g_{i}(x)$ Likewise, let them $\bar{h}_{i}(x),h_{i}(x)$ the two corresponding limit functions of the polynomials $\mathrm{P}_{i-n+1},\mathrm{P}_{i-n+3},\mathrm{P}_{i-n+5},\ldots$

We then have
$\bar{g}_{i}(x)\leq f(x)\leq\underline{h}_{i}(x)$ Or $\bar{g}_{i}(x)<f(x)<\underline{h}_{i}(x)$ if $\frac{|n-i+1|+n-i+1}{2}$ peer
$\underline{g}_{i}(x)\leq f(x)\leq\bar{h}_{i}(x)$ Or $\underline{g}_{i}(x)>f(x)>\bar{h}_{i}(x)$ if $\frac{|n-i+1|+n-i+1}{2}$ The necessary and sufficient conditions for extendability are therefore
expressed as follows:
$\bar{g}_{i}(x)\leq\underline{h}_{i}(x)$ Or $\bar{g}_{i}(x)<\underline{h}_{i}(x)$ if $\frac{|n-i+1|+n-i+1}{2}$ is even
$\underline{g}_{i}(x)\geq\bar{h}_{i}(x)$ Or $\underline{g}_{i}(x)>\bar{h}_{i}(x)$ if $\frac{|n-i+1|+n-i+1}{2}$ is odd
4. Let's introduce the following notation

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

Let's ask

\delta_{j,j+1}^{i}=y_{j}\left(x-x_{j}\right)+y_{j+1}\left(x_{j+n+1}-x\right)

we then

\delta_{j,j+1}^{i}=\left(x_{j+n+1}-x_{j}\right)\Delta_{n+1}^{j}

(3)

The conditions $\partial_{j,j+1}^{l}\geq 0$ Or $>0$ are therefore indeed verified, in other words the inequalities $y_{j}\geq 0\mathrm{ou}>0,y_{j+1}\geq 0\mathrm{ou}>0$ are still compatible.

Let us generally assume
(4)

\begin{gathered}\partial_{j,j+2p+1}^{i}=\delta_{j,j+2p+1}^{i}(x)=y_{j}\left(x-x_{j}\right)\left(x-x_{j+1}\right)\ldots\left(x-x_{j+2p}\right)+\\ +y_{j+2p+1}\left(x_{j+n+1}-x\right)\left(x_{j+n+2}-x\right)\ldots\left(x_{j+n+2p+1}-x\right).\end{gathered}

These quantities do not contain the parameter $f(x)$ and inequality

\begin{gathered}\delta_{j,j+2p+1}^{i}\geq 0\text{ ou }>0\\ i-n\leq i\leq i+1,p>0,j+2p+1\leq i+1,0<j\leq m-n-1\end{gathered}

precisely expresses the compatibility of inequalities $y_{j}\geq 0$ Or $>0,y_{j+2p+1\geq 0}$ Or $>0$ .

Quantities (4) can also be easily expressed using (3). These are polynomials of degree $2p$ in $x$ dependent on the given function only through its divided differences of order $n+1$ which was to be expected.

The inequalities (5) therefore express the conditions of prolongation.

When there are at most three quantities $y_{j}$ The number of conditions is equal to zero, therefore:

Any function of order n on (1) can be extended in the strict sense to any point to the left of $\mathrm{x}_{3}$ and any point to the right of $\mathrm{x}_{\mathrm{m}-2}$ .

Any function of order n defined on $\mathrm{n}+3$ points extend in the strict sense to every point.

The following property results from the fact that (4) are polynomials:
If a convex function of order $n$ on points (1) extends in the strict sense on a point $x$ , it also extends in the strict sense to every point in the vicinity of $x$ .

The points on which a convex function extends in the strict sense form a finite number of open intervals. The endpoints of these intervals ( $\pm\infty$ excluded) are points where we can extend in the broad sense; we will call them the singular points of the convex function.

II. A study of some simple extension cases

5.

We will now fully study second-order functions defined on (1). In this case, the quantities (4) can be written

	$\displaystyle\delta_{i-2,i+1}^{i}$	$\displaystyle=\left(x-x_{i-1}\right)\left(x-x_{i}\right)\left(x_{i+1}-x_{i-2}\right)\Delta_{3}^{i-2}+$
		$\displaystyle+\left(x-x_{i}\right)\left(x-x_{i+1}\right)\left(x_{i+2}-x_{i-1}\right)\Delta_{3}^{i-1}+$
		$\displaystyle+\left(x-x_{i+1}\right)\left(x-x_{i+2}\right)\left(x_{i+3}-x_{i}\right)\Delta_{3}^{i}$
		$\displaystyle i=4,\ldots,m-3.$

It is easy to see that if the function is convex, it extends in the neighborhood of every point $\mathrm{x}_{\mathrm{i}}$ .

In the meantime ( $x_{i},x_{i+1}$ ) can exist 2, 1, or 0 singular points depending on the equation $\delta_{i-2,i+1}^{i}(x)=0$ has two unequal real roots, a double root, or no real roots in the interval ( $x_{i},x_{i+1}$ ). Let us designate by $\alpha_{i},\beta_{i+1}$ these points

x_{i}<\alpha_{i}\leq\beta_{i+1}<x_{i+1}

Let's take $\alpha_{i},\beta_{i+1}$ arbitrarily in ( $x_{i},x_{i+1}$ ) and let's write the identity in $x$

\Sigma_{i-2,i+1}^{i}=\mathrm{A}\left(x-x_{i}\right)\left(x-\beta_{i+1}\right).

We deduce from this

\frac{\left(x_{i+1}-x_{i-2}\right)\Delta_{3}^{i-2}}{\mathrm{\penalty 10000\ A}_{i}}=\frac{\left(x_{i+2}-x_{i-1}\right)\Delta_{3}^{i-1}}{\mathrm{\penalty 10000\ A}_{i}^{\prime}}=\frac{\left(x_{i+3}-x_{i}\right)\Delta_{3}^{i}}{\mathrm{\penalty 10000\ A}_{i}^{\prime\prime}}

(6)

\begin{gathered}\mathrm{A}_{i}=\left(x_{i+2}-x_{i}\right)\left(\alpha_{i}-x_{i+1}\right)\left(\beta_{i+1}-x_{i+1}\right)\\ \mathrm{A}_{i}^{\prime}=\left(x_{i}+\beta_{i+1}\right)\left(x_{i+1}x_{i+2}-x_{i}x_{i-1}\right)+x_{i}\beta_{i+1}\left(x_{i}+x_{i-1}-x_{i+2}-x_{i+1}\right)-\\ -x_{i-1}x_{i+1}\left(x_{i+2}-x_{i}\right)-x_{i}x_{i+2}\left(x_{i+1}-x_{i-1}\right)\\ \mathrm{A}_{i}^{\prime\prime}=\left(x_{i+1}-x_{i-1}\right)\left(x_{i}-x_{i}\right)\left(\beta_{i+1}-x_{i}\right).\end{gathered}

These relationships determine $\Delta_{3}^{i-2},\Delta_{3}^{i-1},\Delta_{3}^{i}$ When $\alpha_{i},\beta_{i+1}$ are given and we see that $\mathrm{A}_{i},\mathrm{\penalty 10000\ A}_{i}^{\prime},\mathrm{A}_{i}^{\prime\prime}$ are positive when the points $\alpha_{i},\beta_{i+1}$ are in ( $x_{i},x_{i+1}$ ).

It follows that we can arbitrarily choose the two singular points $x_{i},\beta_{i+1}$ .

More generally, we can determine all second-order functions having the points as singular points

\alpha_{i_{1}},\beta_{i_{1}+1},\alpha_{i_{2}},\beta_{i_{2}+1},\alpha_{i_{3}},\beta_{i_{3}+1},\ldots

(7)

taken arbitrarily from the intervals $\left(x_{i_{1}},x_{i_{1}+1}\right),\left(x_{i_{2}},x_{i_{2}+1}\right),\ldots$ provided that $i_{1}>i_{i}+1,\mathrm{i}_{3}>i_{2}+1,\ldots$
6. We will now demonstrate that in two consecutive intervals $\left(\mathrm{x}_{\mathrm{i}},\mathrm{x}_{\mathrm{i}+1}\right),\left(\mathrm{x}_{\mathrm{i}+1},\mathrm{x}_{\mathrm{i}+2}\right)$ Singular points cannot exist at the same time.

We should indeed have
(8)

A_{i}^{\prime\prime}A_{i+1}--A_{i}^{\prime}A_{i+1}^{\prime}=0.

Let us designate the first member by $\mathrm{F}\left(\alpha_{i},\beta_{i+1};\alpha_{i+1},\beta_{i+2}\right)$ and let's ask

		$\displaystyle\mathrm{F}_{1}\left(\alpha_{i},\beta_{i+1};\alpha_{i+1}\right)=\mathrm{F}\left(\alpha_{i},\beta_{i+1};\alpha_{i+1},\alpha_{i+1}\right)$
		$\displaystyle\mathrm{F}_{2}\left(\alpha_{i},\beta_{i+1};x_{i+1}\right)=\mathrm{F}\left(\alpha_{i},\beta_{i+1};\alpha_{i+1},x_{i+2}\right)$
		$\displaystyle\mathrm{F}_{1}\left(\alpha_{i},\beta_{i+1};x_{i+1}\right)=\left(x_{i+3}-x_{i+1}\right)\left(x_{i+2}-x_{i+1}\right)\mathrm{F}_{3}\left(\alpha_{i},\beta_{i+1}\right)$
		$\displaystyle\mathrm{F}_{4}\left(\alpha_{i}\right)=\mathrm{F}_{3}\left(\alpha_{i},\alpha_{i}\right),\mathrm{F}_{5}\left(\alpha_{i}\right)=\mathrm{F}_{3}\left(\alpha_{i},x_{i+1}\right)$

$\mathrm{F}_{4}\left(\alpha_{i}\right)$ is a second-degree polynomial in $\alpha_{i}$ and we have a coefficient of $\alpha_{i}^{2}$ In $\mathrm{F}_{4}$
SO

\mathrm{F}_{4}\left(x_{i}\right)=-\left(x_{i+1}-x_{i}\right)^{2}\left(x_{i+2}-x_{i}\right)\left(x_{i}-x_{i-1}\right),\mathrm{F}_{4}\left(x_{i+1}\right)=0

\mathrm{F}_{4}\left(\alpha_{i}\right)=\mathrm{F}_{3}\left(\alpha_{i},\alpha_{i}\right)<0\quad\left(x_{i}<\alpha_{i}<x_{i+1}\right)

We also have $\mathrm{F}_{5}\left(x_{i}\right)<0$ , SO

\mathrm{F}_{3}\left(x_{i},x_{i+1}\right)<0\quad\left(x_{i}<\alpha_{i}<x_{i+1}\right)

$\mathrm{F}_{3}\left(\alpha_{i},\beta_{i+1}\right)$ being linear with respect to $\alpha_{i}$ And $\beta_{i+1}$
Therefore, we deduce

\mathrm{F}_{3}\left(\alpha_{i},\beta_{i+1}\right)<0\quad\left(x_{i}<\alpha_{i}\leq\beta_{i+1}<x_{i+1}\right)

\mathrm{F}_{1}\left(\alpha_{i},\beta_{i+1};x_{i+1}\right)<0\quad\left(x_{i}<x_{i}\leq\beta_{i+1}<x_{i+1}\right)

and also

\begin{gathered}\mathrm{F}_{1}\left(\alpha_{i},\beta_{i+1};\alpha_{i+1}\right)=\mathrm{F}\left(\alpha_{i},\beta_{i+1};x_{i+1},\alpha_{i+1}\right)<0\\ \left(x_{i}<x_{i}\leq\beta_{i+1}<x_{i+2}<\alpha_{i+1}<x_{i+2}\right)\end{gathered}

But

\mathrm{F}_{2}\left(x_{i},\beta_{i+1};x_{i+1}\right)<0,\left(x_{i}<\alpha_{i}\leq\beta_{i+1}<x_{i+2}\right)

so also

\begin{gathered}\mathrm{F}_{2}\left(\alpha_{i},\beta_{i+1};\alpha_{i+1}\right)=\mathrm{F}\left(\alpha_{i},\beta_{i+1};x_{i+1},x_{i+2}\right)<0\\ \left(x_{i}<x_{i}\leq\beta_{i+1}<x_{i+1}<\alpha_{i+1}<x_{i+2}\right)\end{gathered}

Finally, we can deduce that

\begin{gathered}\mathrm{F}\left(\alpha_{i},\beta_{i+1};x_{i+1},\beta_{i+2}\right)<0\\ \left(x_{i}<\alpha_{i}\leq\beta_{i+1}<x_{i+1}<\alpha_{i+1}\leq\beta_{i+2}<x_{i+2}\right)\end{gathered}

This contradicts (8), so the property is proven.
It follows that the distribution (7) of singular points is the most general possible, and we can also find, using formulas (6), all convex functions of order 2 having these points as singular points. We can also determine the functions that have no other singular points.

These conclusions remain true for non-concave functions, only in this case the points $\alpha_{i},\beta_{i+1}$ may coincide with $x_{i},x_{i+1}$ respectively. We see, for example, that if the function cannot be extended at any point in the interval ( $x_{i},x_{i+1}$ ) it can be extended to any point belonging to one of the intervals ( $x_{i-1},x_{i}$ ), ( $x_{i+1},x_{i+2}$ ).

Taking

\Delta_{3}^{1}=\Delta_{3}^{3}=\Delta_{3}^{5}=\ldots=0,\quad\Delta_{3}^{2}>0,\Delta_{3}^{4}>0,\Delta_{3}^{6}>0,\ldots

We obtain all the second-order functions that do not extend to any point in the intervals

\left(x_{3},x_{4}\right),\left(x_{5},x_{6}\right),\left(x_{7},x_{8}\right),\ldots

The number of singular points of a convex function of order 2 is at most equal to $m-4$ Or $m-5$ following that $m$ is even or odd.
7. If the order of the function is greater than 2, much more complicated cases of non-extendability can arise. For example, functions of order $n>2$ which are such that

\Delta_{n+1}^{1}=\Delta_{n+1}^{3}=\ldots=0,\Delta_{n+1}^{2}>0,\Delta_{n+1}^{4}>0,\ldots

do not extend to any point in the interval ( $x_{3},x_{m-2}$ ) if $m--n-1$ is odd and on no point of ( $x_{3},x_{m-3}$ ) if $m-n-1$ if is even. In the latter case, they also extend to every point of ( $x_{m-3},x_{m-2}$ ).

If the function class contains multiple convexity conditions, more complicated circumstances may arise. Nevertheless, some interesting observations can be made. Suppose the function belongs to the class $(1,2)$ : for it to be extendable at a point, it is necessary and sufficient that it be extendable as a function of order 2. Indeed, let the polynomials

\begin{array}[]{ll}\mathrm{P}_{1}=\mathrm{P}\left(x_{i-2},x_{i-1},x_{i};f\mid x\right)&\\ \mathrm{P}_{2}=\mathrm{P}\left(x_{i-1},x_{i},x_{i+1};f\mid x\right)&\mathrm{Q}_{1}=\mathrm{P}\left(x_{i-1},x_{i};f\mid x\right)\\ \mathrm{P}_{3}=\mathrm{P}\left(x_{i},x_{i+1},x_{i+2};f\mid x\right)&\mathrm{Q}_{2}=\mathrm{P}\left(x_{i},x_{i+1};f\mid x\right)\\ \mathrm{P}_{4}=\mathrm{P}\left(x_{i+1},x_{i+2},x_{i+3};f\mid x\right)&\mathrm{Q}_{3}=\mathrm{P}\left(x_{i+1},x_{i+2};f\mid x\right)\end{array}

The conditions for extension are
$P_{2}\geqslant Q_{1},P_{2}\geqslant Q_{3},P_{4}\geqslant P_{1},P_{4}\geqslant Q_{1},P_{4}\geqslant Q_{3},Q_{2}\geqslant P_{1},Q_{2}\geqslant P_{3}$ However
, we have, by examining the representative figure of the function $\mathrm{P}_{2}\geq\mathrm{Q}_{1}$ In ( $x_{i},x_{i+1}$ ) the function being of order 1
$P_{2}\geq Q_{3}$
$P_{4}\geq P_{1}$ by hypothesis
$\mathrm{P}_{4}\geqslant\mathrm{Q}_{1}$ to the point $x$ because $\mathrm{P}_{1}\leftrightarrows\mathrm{Q}_{1}$ In ( $x_{i},x_{i+1}$ )
$\mathrm{P}_{4}\geqslant\mathrm{Q}_{3}$ In ( $x_{i},x_{i+1}$ ) the function being of order 1
$\mathrm{Q}_{2}\geqslant\mathrm{P}_{1}\quad,\quad,\quad$ of the class $(1,2)$
$\mathrm{Q}_{2}\geqslant\mathrm{P}_{3}$ In ( $x_{i},x_{i+1}$ ) the function being of order 1.
8. Consider a function defined on any set E. We can extend the function in the strict sense on the derivative set $E$ except perhaps the extremities a and $b$ When the maximum order is greater than 0, this extension occurs by continuity; therefore, it is only possible in one way. If the maximum order is 0, the property
is fairly obvious, but uniqueness does not generally hold. The extension occurs in the strict sense by virtue of a known property ( ³ ).

Any function of a given class, bounded in the neighborhood of an endpoint, extends in the strict sense to that endpoint.

We assume, of course, that this extremity, $b$ for example, does not belong to E. We can then immediately see that it suffices to take for $f(b)$ the limit of $f(x)$ when $x\rightarrow b$ This limit exists and is clearly defined.

If the end $b$ (Or $a$ ) belongs to E the function can be discontinuous there and it is then clear that in general it is impossible to extend the function beyond this point.

Let us now examine the possibility of an extension beyond one end, for example in $b$ . If $f(x)$ is of maximal order $n$ , so that it can be extended beyond $b$ it is necessary that it be at $n^{\text{ième}}$ The difference divided is bounded in the neighborhood of this point. It can easily be seen that, for a strict extension, this condition is not sufficient.

For example, if the function is of class $(0^{*},1^{*})$ and if the right-hand derivative at the point $b$ is zero, so the function only extends in the broad sense , remaining constant beyond $b$ .

In general, derivatives $f^{\prime}(b)$ , $f^{\prime\prime}(b)$ , $\ldots$ , $f^{(n)}(b)$ exist to the point $b$ - Or $f^{(n)}(b)$ is, by definition, the $n^{\text{ième}}$ left derivative — and some of them have a specific sign. For simplicity, we assume that $E$ is an interval $(a,b)$ If the extension is possible beyond the point $b$ , particularly in the interval $(b,c)$ by function $f_{1}(x)$ , it is necessary that

f_{1}(b)=f(b),f_{1}^{(i)}(b)=f^{(i)}(b)i=1,2,\ldots n-1

and that $f_{1\bar{d}}^{(n)}(b)$ either $\geq$ Or $\leq f^{(n)}(b)$ depending on whether the function is non-concave or non-convex of order $n$ .

We can now see that even in a broad sense, extension is not always possible ⁴ ).

If the polynomial

\sum_{i=0}^{n}\frac{f^{(i)}(b)}{i!}(x-b)^{i}

is of the same class as $f(x)$ In $(b,c)$ He performs the extension (au seus Iarge). The same applies to

\sum_{i=0}^{n}\frac{f^{(i)}(b)}{i!}(x-b)^{i}+\mathrm{A}(x-b)^{n+1}

Being $=$ Or $\leq 0$ following that $f(x)$ is non-concave or non-convex
³ ) See loc. cit. 1) p. 26.
⁴ ) For example, a function of the class ( $0,1^{\prime},2,3^{\prime}$ ) cannot be extended beyond $b$ if $f^{\prime}(b)=0,f^{\prime\prime}(b)\neq 0$ (It is easy to construct such functions).
of order $n$ In some cases this extension is in the strict sense, such as for example if $f(x)$ is of order $n$ simply or if she is of the class $(k,k+1,\ldots,n)$ Or $\left(k^{*},k+1^{*},\ldots,n^{*}\right)\ldots$ etc.

We can also state the following property:
The necessary and sufficient condition for a function $\mathrm{f}(\mathrm{x})$ enjoying ( $\mathrm{a},\mathrm{b}$ ) order properties $\mathrm{k},\mathrm{k}+1,\ldots,\mathrm{n}$ of the same meaning, can be extended beyond point b is that it is a divided nth difference bounded in the neighborhood of b.

The extension is in the strict sense and can go as far as desired.
A similar property applies to the extremity $a$ , it is then necessary to consider order properties $k,k+1,\ldots,n$ alternating directions, as it results from a change in the orientation of the axis of $x$ We also deduce that:

The necessary and sufficient condition for a function of order n in $(\mathrm{a},\mathrm{b})$ that it can be extended in the strict sense in any interval is that it is a bounded divided nth difference.

III

Extension of convex functions in an interval.

9.

Let us consider a function $f(x)$ non-concave of order $n$ in the meantime $(a,b)$ and either

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

a sequence of points in this interval.
We can assume, without restricting the generality, that $x_{1}=ax_{m}=b$ as will result from the method we are about to describe.

Let us add
(9) $\quad\Delta_{n+1}^{i}=\left[x_{i},x_{i+1},\ldots,x_{i+n+1};f\right],i=1,2,\ldots,m-n-1$

We will first assume that the points $x_{2},x_{3},\ldots,x_{m-1}$ rationally divide the interval ( $x_{1},\mathrm{x}_{m}$ We can then find a positive number $\delta$ and positive integers $\rho_{1},\rho_{2},\ldots,\rho_{m-1}$ such as

x_{i+1}-x_{i}=\rho_{i}\delta\quad i=1,2,\ldots,m-1

We will use the notations
$\rho_{i}+\rho_{2}+\ldots+\rho_{j}=\rho_{j}^{\prime},\quad j=1,2,\ldots,m-1,\quad\rho_{0}^{\prime}=0,\quad\rho_{m-1}^{\prime}=\rho$
Let's divide the interval ( $x_{1},x_{m}$ ) in $p\circ$ equal parts, $p$ being a sufficiently large positive integer and are

x_{i}^{\prime}=x_{1}+(i-1)h,\quad h=\frac{x_{m}-x_{1}}{p_{\rho}}

the division points. We then have

x_{p_{\varphi^{\prime}j-1}+1}^{\prime}=x_{j},\quad j=1,2,\ldots,m.

Let us consider a function $\Psi(x;p)$ defined in the interval $\left(x_{p},x_{m}\right)$ reducing to a polynomial of degree $n$ in each of the intervals $\left(x_{1},x_{2}\right),\left(x_{2},x_{3}\right),\ldots,\left(x_{n-1},x_{m}\right)$ By asking

\mathrm{P}_{j}(x)=\sum_{i=0}^{n}\lambda_{i}^{j}\left(x-x_{j}\right)^{n-i},j=1,2,\ldots,m-1

we will be able to write

\psi(x;p)=\sum_{j=1}^{k}\mathrm{P}_{j}(x)\text{ dans }\left(x_{k},x_{k+1}\right),\quad k=1,2,\ldots,m-1.

We assume that this function is continuous, therefore

\lambda_{n}^{j}=0,\quad j=2,3,\ldots,m-1

and that the other coefficients are generally functions of $p$
We will examine the sum in the following section .

\sum_{i=1}^{p\varrho-n}\Psi\left(x_{i}^{\prime};p\right)\left[x_{i}^{\prime},x_{i+1}^{\prime},..,x_{i+n+1}^{\prime};f\right]=\sum_{i=1}^{p\varphi+1}\gamma_{i}f\left(x_{i}^{\prime}\right).

(10)

10.

The first member of (10) can be written

\frac{1}{(n+1)!h^{n+1}}\sum_{i=1}^{p\varphi-n}\Psi^{\prime}\left(x_{i}^{\prime};p\right)\left[\sum_{r=0}^{n+1}(-1)^{n+1-r}\binom{n+1}{r}f\left(x_{i+r}^{\prime}\right)\right]

It immediately follows that

\frac{1}{(n+1)!h^{n+1}}\sum_{(j)}(-1)^{n+1-j}\binom{n+1}{j}\Psi\left(x_{i-j}^{\prime};p\right)

where the summons extends from

\begin{array}[]{cll}0&\text{ à }&i-1\text{ pour }i\leq n+1\\ i-p_{p}+n&\text{ à }&n+1\text{ pour }i\geq p_{p}-n+1\\ 0&\text{ à }&n+1\text{ pour }n+1<i<p_{p}-n+1.\end{array}

The form of the function shows us that the only coefficients $\gamma_{i}$ The following are not zero:
$\gamma_{i},\gamma_{p\rho-n+i^{*}}i=1,2,\ldots n+1;\gamma_{p\varphi_{j}^{\prime}}+i,j=1,2,\ldots,m-2,i=2,3,\ldots,n+1$ We therefore
see the following appear in the second member of (10) $m$ groups of terms. The extreme groups contain $n+1$ the others $n$ terms.

For the first group, a simple calculation gives us
(11) $\sum_{i=1}^{n+1}7_{i}f\left(x_{i}^{\prime}\right)=\sum_{i=0}^{n}\left[\sum_{r=0}^{n-i}\binom{r+j}{r}7_{i+r+1}\right]\left[\sum_{r=0}^{i}(-1)^{i-r}\binom{i}{r}f\left(x_{r+1}^{\prime}\right)\right]=$

=\sum_{i=0}^{n}h^{i}\left[\sum_{r=0}^{n-i}\binom{r+i}{r}y_{i+r+1}\right]i!\left[x_{i}^{\prime},x_{2}^{\prime},\ldots,x_{i+1}^{\prime};f\right]

In this case, we have

\gamma_{i}=\frac{1}{(n+1)!h^{n+1}}\sum_{r=0}^{n}\lambda_{n-r}^{1}h^{r}\Gamma_{i}^{r}

\Gamma_{i}^{\mathrm{r}}=\sum_{j=0}^{i-1}(-1)^{n+1-j}\binom{n+1}{j}(i-j-1)^{r}\left(\Gamma_{1}^{r}=0,r>0,\Gamma_{1}^{0}=(-1)^{n+1}\right)

We deduce from this

\sum_{r=0}^{n-i}\binom{r+i}{r}\gamma_{i+r+1}=\frac{1}{(n+1)!h^{n+1}}\sum_{s=0}^{n}\lambda_{n-s}^{1}h^{s}\mathrm{G}_{i}^{s}

with

\mathrm{G}_{i}^{s}=\sum_{r=0}^{n-i}\binom{r+i}{r}\mathrm{r}_{i+r+1}^{s}

The numbers $\mathrm{G}_{i}^{s}$ are easily calculated. Let's consider the polynomial

\mathrm{F}_{i}(z)=(-1)^{n+1}\sum_{r=0}^{i-1}(-1)^{r}\binom{n+1-i+r}{r}(z-1)^{i-r-1}

we then

		$\displaystyle\mathrm{r}_{i}^{r}=\left[\left(z\frac{d}{dz}\right)^{(r)}\mathrm{F}_{i}(z)\right]_{z=1}$
		$\displaystyle\mathrm{G}_{i}^{s}=\left[\left(z\frac{d}{dz}\right)^{(s)}\sum_{r=1}^{n-i}\binom{r+i}{r}\mathrm{\penalty 10000\ F}_{i+r+1}(z)\right]_{z=1}$

hence in particular

\mathrm{G}_{i}^{s}=0\text{ pour }s=0,1,2,\ldots,n-i-1

$\mathrm{G}_{i}^{n-i}=(-1)^{n+1-i}\left[\left(z\frac{d}{dz}\right)^{(n-i)}(z-1)^{n-i}\right]_{z=1}=(-1)^{n+1-i}(n-i)!$
Note now that we know nothing about the quantities
(12)

i!\left[x_{1}^{\prime},x_{2}^{\prime}\ldots,x_{i+1}^{\prime};f\right],\quad i=1,2,\ldots,n.

Let's take the coefficients $\lambda_{i}^{\perp}$ values such as $1^{0}$
$\lim_{p\rightarrow\infty}\lambda_{0}^{1}=\Lambda_{1}$ exists and is finished
$2^{0}$ equalities

\sum_{r=0}^{n-i}\binom{r+i}{r}\gamma_{i+r+1}=\frac{1}{n+1)!h^{n+1}},\sum_{s=n-i}^{n}\lambda_{n-s}^{1}h^{s}\mathrm{G}_{i}^{s}=0

are verified identically in $p$
The result is that :
$1^{0}\quad\lambda_{1}^{1},\lambda_{2}^{1},\ldots,7_{n}^{1}$ tend towards zero for $p\rightarrow\infty$ .
$2^{0}$ All quantities (12) disappear in the sum (11). We can therefore write:

		$\displaystyle\lim_{p\rightarrow\infty}\frac{1}{p}\sum_{i=1}^{n+1}\gamma_{i}f\left(x_{i}^{\prime}\right)=\lim\cdot\frac{1}{p}\left[\sum_{r=0}^{n}\binom{r+1}{r}\gamma_{r+1}\right]f\left(x_{1}\right)=$
	$\displaystyle=$	$\displaystyle\lim\cdot\frac{1}{p}\frac{1}{(n+1)h^{n+1}}h^{n}\lambda_{0}^{1}\mathrm{G}_{0}^{n}=\frac{(-1)^{n+1}\rho\Lambda_{1}}{(n+1)\left(x_{m}-x_{1}\right)}f\left(x_{1}\right).$

Let us now consider a group of intermediate terms:

\sum_{i=1}^{n}\gamma_{p\varrho_{j}^{\prime}+i+1}f\left(x_{p\ell_{j}+i+1}^{\prime}\right)\quad 1\leq j\leq m-2

and we know the quantities.

\left[x_{p\ell_{j}^{\prime}+2}^{\prime},x_{p\ell_{j}^{\prime}+3}^{\prime},\ldots,x_{p\ell_{j}^{\prime}+i+2}^{\prime};f\right],\quad i=0,1,2,\ldots,n-1

those involved in this sum all remain limited in their scope. $\mathrm{p}\rightarrow\infty$ Without going into the details of the calculation, let
's just say that if we take the coefficients $\lambda_{i}^{j+1}$ values such as:
$1^{0}.\quad\lim_{p\rightarrow\infty}\lambda_{0}^{j+1}=\Lambda_{j+1}$ exists and is finished.
$2^{\prime}.\quad\lim_{p\rightarrow\infty}.\lambda_{i}^{j+1}=0,i=1,2,\ldots,n-1$ ,
we find the relationship.

\lim_{p\rightarrow\infty}\frac{1}{p}\sum_{i=1}^{n}\gamma_{p\varrho_{j}^{\prime}+i+1}f\left(x_{p\varrho_{j}^{\prime}+i+1}\right)=\frac{(-1)^{n+1}p\Lambda_{j+1}}{\left(n+1\left(x_{m}-x_{1}\right).\right.}f\left(x_{j+1}\right).

We still need to see what happens with the last group of terms.

\sum_{i=1}^{n+1}\gamma_{p\rho-n+i}f\left(x_{p\rho-n+i}^{\prime}\right).

(13)

We will try to arrange things so that this sum multiplied by $\frac{1}{p}$ also has a finite limit for $p\rightarrow\infty$ To do this, simply repeat the previous demonstration, this time starting from point $x_{m}$ towards the point $x_{1}$ , therefore by reversing the order of the points $x_{1},x_{2},\ldots,x_{m}$ This process amounts to performing an operation on the function $\Psi(x;p)$ the transformation $x\mid x_{m}+x_{1}-x$ , SO.

\Psi_{1}(x;p)=\Psi\left(x_{m}+x_{1}-x;p\right)

and we have
$\Psi_{1}(x;p)=\sum_{j=1}^{k}\mathrm{Q}_{j}(x)\quad\begin{gathered}\text{ dans }\left(x_{m}+x_{1}-x_{m-k+1},x_{m}+x_{1}-x_{m-k}\right)\\ k=1,2,\ldots,m-1.\end{gathered}$
with

\mathrm{Q}_{j}(x)=\sum_{i=0}^{n}\mu_{i}^{j}\left(x-x_{1}-x_{m}+x_{m-j+1}\right)^{n-i}\quad j=1,2,\ldots,m-1

The following three properties must be verified:
$1^{0}$ .

\lim_{p\rightarrow\infty}\mu_{0}^{j}\text{ existe et est finie. }

$2^{0}$ Equality.

\sum_{s=n-i}^{n}\mu_{n-s}^{1}h^{s}\mathrm{G}_{i}^{s}=0\quad i=1,2,\ldots,n

(14)

are verified identically on p.
30. $\quad\lim_{p\rightarrow\infty}\mu_{i}^{j}=0,\quad i>0$ .

Here are the coefficients $\mathrm{G}_{i}^{\prime s}$ are still independent $p$ but are not identical to $\mathrm{G}_{i}^{s}$ This stems from the fact that while in formula (10) the values of $\Psi(x;p)$ In $\left(x_{1}^{\prime},x_{p,-n}^{\prime}\right)$ In the reversed problem, we use the values of $\Psi_{1}(x;p)$ In $\left(x_{n+2}^{\prime},x_{pq+1}^{\prime}\right)$ So we have

		$\displaystyle\mathrm{p}_{i}^{\prime s}=\sum_{j=0}^{i-1}(-1)^{n+1-j}\binom{n+1}{j}(n+1-j)^{s}$
		$\displaystyle\mathrm{G}_{i}^{\prime s}=\sum_{r=0}^{n-i}\binom{r+i}{r}\mathrm{I}_{i+r+1}^{\prime s}$

But :

	$\displaystyle\mathrm{r}_{i}^{\prime s}$	$\displaystyle=\sum_{t=0}^{s}\binom{s}{t}(n+1)^{s-t}\mathrm{r}_{i}^{t}$
	$\displaystyle\mathrm{G}_{i}^{\prime s}$	$\displaystyle=\sum_{t=0}^{s}\binom{s}{t}(n+1)^{s-t}\mathrm{G}_{i}^{t}$

donations in particular:

\mathrm{G}_{i}^{\prime s}=0,s=0,1,2\ldots,n-i+1,\mathrm{G}_{i}^{\prime n-i}=\mathrm{G}_{i}^{n-i}\neq 0

The property $1^{0}$ is obviously verified. Note that:

\mathrm{Q}_{j}(x)=-\mathrm{P}_{m+1-j}\left(x_{m}+x_{1}-x\right),\quad j=2,3,\ldots,m-1

and it immediately follows that $3^{0}$ " For $j>1$ is verified.
We also have:

\mathrm{Q}_{1}(x)=\sum_{j=1}^{m-1}\mathrm{P}_{j}\left(x_{m}+x_{1}-x\right)

hence the values of the coefficients $\mu_{i}^{1}$

\mu_{i}^{1}=\sum_{j=1}^{m-1}\left[\sum_{r=0}^{i}(-1)^{i-r}\binom{n-r}{i-r}\left(x_{m}-x_{j}\right)^{i-r}\lambda_{r}^{j}\right],\quad i=0,1,2,\ldots,n.

So that $3^{0}$ should also be checked for $j=1$ it is necessary that.

\sum_{j=1}^{m-1}\left(x_{m}-x_{j}\right)^{i}\Lambda_{j}=0,\quad i=1,2,\ldots,n.

(15)

Once these equalities are satisfied, we see that we can always arrange things so that $2^{0}$ that should also be checked.

We now know that (13) multiplied by $\frac{1}{p}$ has a limit when the preceding conditions are met and this limit depends only on $f\left(x_{m}\right)$ We can write

\lim_{p\rightarrow\infty}\frac{1}{p}\sum_{i=1}^{n+1}\gamma_{p\zeta-n+i}f\left(x_{p\rho-n+i}^{\prime}\right)=\frac{(-1)^{n+1}\rho\Lambda_{m}}{(n+1)\left(x_{m}-x_{1}\right)}f\left(x_{m}\right)

and the coefficient $\mathbf{\Lambda}_{m}$ is obviously determined by the condition that

\lim\cdot\frac{1}{p}\sum_{i=1}^{p\varphi+1}\gamma_{i}f\left(x_{i}^{\prime}\right)

(16)

depends only on divided differences of order $n+1$ In other words, it must be zero if $f(x)$ is an arbitrary polynomial of degree $n$ . SO

\sum_{j=1}^{m}x_{j}^{k}\Lambda_{j}=0,\quad k=0,1,2,\ldots,n

and (15) are precisely the compatibility conditions of this system.
11. When $p\rightarrow\infty$ the function $\Psi(x;p)$ converges uniformly ⁵ ) in ( $x_{1},x_{m}$ ) to the function

\begin{array}[]{r}\Psi(x)=\lim_{p\rightarrow\infty}\Psi(x;p)=\sum_{j=1}^{k}\Lambda_{j}\left(x-x_{j}\right)^{n}\operatorname{dans}\left(x_{k},x_{k+1}\right)\\ k=1,2,\ldots,m-1.\end{array}

This function can be expressed as a sum of $m-n-1$ analogous functions.

To simplify the notation, let us assume for the Van der Monde determinants

\begin{array}[]{r}\mathrm{V}_{i}=\mathrm{V}\left(x_{i},x_{i+1},\ldots,x_{i+n+1}\right)\quad i=1,2,\ldots,m-n-1\\ \mathrm{\penalty 10000\ V}_{i}^{(k)}=\mathrm{V}\left(x_{i},x_{i+1},\ldots,x_{k-1},x_{k+1},\ldots,x_{i+n+1}\right)\\ \quad k=i,i+1,\ldots,i+n+1\end{array}

and consider the functions

\Psi_{i}(x)=\left\{\begin{array}[]{l}0\quad\text{ dans }\quad\left(x_{1},x_{i}\right)\\ \sum_{r=0}^{k}(-1)^{r}\frac{\mathrm{\penalty 10000\ V}_{i}^{(i+r)}}{\mathrm{V}_{i}}\left(x-x_{i+r}\right)^{n}\quad\text{ dans }\quad\left(x_{i+k},x_{i+k+1}\right)\\ 0\quad\text{ dans }\quad\left(x_{i+n+1},x_{m}\right)\\ \quad i=1,2,\ldots,m-n-1.\end{array}\right.

⁰⁰footnotetext: ⁵ ) The uniformity of convergence is immediate if we notice that

\Psi^{5}(x;p)

is formed by polynomials of invariable degree.

Taking into account (15) we see that

\Psi(x)=\sum_{j=1}^{m-n-1}\lambda_{j}\Psi_{j}(x)

Expression (16) will then dilate, by removing a common positive factor:

\sum_{j=1}^{m-n-1}\lambda_{j}\Delta_{n+1}^{j}.

12.

The preceding calculations allow us to state the following property:

If the function

\Psi(x)=\sum_{j=1}^{m-n-1}\lambda_{j}\Psi_{j}(x)

is non-negative in the interval ( $\mathrm{x}_{1}\mathrm{x}_{\mathrm{m}}$ ), any function $\mathrm{f}(\mathrm{x})$ nonconcave of order n in ( $\mathrm{x}_{1},\mathrm{x}_{\mathrm{m}}$ ) verifies the inequality.

\sum_{j=1}^{m-n-1}\lambda_{j}\Delta_{n+1}^{j}\geq 0

(17)

I'll start by saying that the function $\Psi_{\mathrm{i}}(x)$ is positive in the open interval $\left(\mathrm{x}_{\mathrm{i}},\mathrm{x}_{1+\mathrm{n}+1}\right)$ It suffices to demonstrate this. $i=1$ Because of the symmetry, it is even sufficient to show that

\sum_{r=1}^{k}(-1)^{r-1}\mathrm{\penalty 10000\ V}_{1}^{(r)}\left(x-x_{r}\right)^{n}>0,\quad\text{ dans }\left(x_{k},x_{k+1}\right)

For $k=2,3,\ldots,\left[\frac{n+2}{2}\right]^{6}$ (For $k=1,n+1$ (Obviously), This property can be verified by induction. It is immediate for $n=1$ Suppose it is true up to $n-1$ and let us show that it is a fortiori true for $n$ By hypothesis, the property is true for sequences. $x_{1},x_{2},\ldots,x_{n+1};x_{2},x_{3},\ldots,x_{n+2}$ Let us designate by $\mathrm{V}^{\prime}(1),\mathrm{V}^{\prime}(2),\ldots,\mathrm{V}^{\prime\prime}(2)$ , $\mathrm{V}^{\prime\prime}(3),\ldots$ , the analogues of $\mathrm{V}_{1}^{(r)}$ for these sequels, then

		$\displaystyle\mathrm{V}^{\prime(r)}=\mathrm{V}\left(x_{1},x_{2}\ldots,x_{r-1},x_{r+1},\ldots,x_{n+1}\right)$
		$\displaystyle\mathrm{V}^{\prime\prime(r)}=\mathrm{V}\left(x_{2},x_{3},\ldots,x_{r-1},x_{r+1},\ldots,x_{n+2}\right)$

⁰⁰footnotetext: ⁶ ) [ø] is equal to the number of integers included in u.

The relationship (which is easily verifiable)

\begin{gathered}\mathrm{V}^{\prime(r)}\left(x_{n+2}-x_{2}\right)\left(x_{n+2}-x_{3}\right)\ldots\left(x_{n+2}-x_{n+1}\right)\left(x-x_{1}\right)-\\ -\mathrm{V}^{\prime\prime(r)}\left(x_{2}-x_{1}\right)\left(x_{3}-x_{1}\right)\cdot\left(x_{n+1}-x_{1}\right)\left(x_{n+2}-x\right)=\mathrm{V}_{1}^{(r)}\left(x-x_{1}\right)\end{gathered}

allows writing

\sum_{r=1}^{k}(-1)^{r-1}\mathrm{\penalty 10000\ V}_{1}^{(r)}\left(x-x_{r}\right)^{n}=\left(x_{n+2}-x_{2}\right)\left(x_{n+2}-x_{3}\right)\ldots

(18)

$\left(x_{n+2}-x_{n+1}\right)\left(x-x_{1}\right)\sum_{r=1}^{k}(-1)^{r-1}\mathrm{\penalty 10000\ V}^{(r)}\left(x-x_{r}\right)^{n-1}+$

+\left(x_{2}-x_{1}\right)\left(x_{3}-x_{1}\right)\ldots\left(x_{n+1}-x_{1}\right)\left(x_{n+2}-x\right)\sum_{r=2}^{k}(-1)^{r-2}\mathrm{\penalty 10000\ V}^{\prime\prime(r)}\left(x-x_{r}\right)^{n-1}

which proves property ⁷ by induction ).
We can now state that if $\Psi(x)$ is non-negative in the open interval ( $x_{1},x_{m}$ ) there exists another analogous function that can be as close as one wants to $\Psi(x)$ and which is positive within that interval.

Therefore, it suffices to demonstrate the stated property for $\Psi(x)$ positive in the open interval ( $x_{1},x_{m}$ ). Our analysis then shows us that we can find a sequence of functions $\Psi(x;p)p=1,2,\ldots$ in such a way that:
$1^{0}$ The expression (10) multiplied by $\frac{1}{p}$ tends, up to a constant positive factor, towards the first member of (17).
$2^{0}$ . The function $\Psi(x;p)$ converges uniformly towards $\Psi(x)$ In $\left(x_{1},x_{m}\right)$ .

It follows that, from a certain value of $p\Psi(x;p)$ stay positive in $\left(x_{1},x_{m}\right)$ This is not entirely certain for the neighborhood of the points. $x_{1},x_{m}$ , but the special form of the functions $\Psi(x;p)$ shows us that we can always assume that this circumstance is realized. Inequality (17) is now immediate since from this value of $p$ All terms on the left-hand side of (10) are non-negative.
⁷ ) The polynomial

\sum_{r=1}^{k}(-1)^{r-1}V_{1}^{(r)}\left(x-x_{r}\right)^{n}

is even "very positive" in the interval ( $x_{k},x_{k+1}$ If we put it in the form

\sum_{i=0}^{n}\mathrm{\penalty 10000\ A}_{i}\left(x-x_{k}\right)^{i}\left(x_{k+1}-x\right)^{n-i}

the coefficients $A_{i}$ are all positive. (Immediate proof by induction).

Let's not forget that we assumed that the points $x_{2},x_{3},\ldots x_{m-1}$ rationally divide the interval ( $x_{1},x_{m}$ Note that our statement is not affected by this restriction. We also know that we can find points $x_{2}^{*},x_{3}^{*},\ldots x_{m-1}^{*}$ as close as we want to the respective points $x_{2},x_{3},\ldots,x_{m-1}$ and such that they rationally divide the interval ( $x_{1},x_{m}$ ). By taking the limit, which is perfectly legitimate because of the continuity of the function ⁸ ), it follows that the property is general.

To simplify the language, we will call the property thus highlighted: restricted convexity property of order $n$ on the $m$ points considered ⁹ ).

Note that we obtain the restricted convexity conditions on a partial sequence of $x_{1},x_{2},\ldots,x_{m}$ by determining the coefficients $\lambda_{i}$ so that the left-hand side of (17) depends only on the divided differences taken at these points. We then see that restricted convexity of order n implies non-concavity of the same order. If $m=n+2$ For restricted convexity, it suffices that the function be non-concave. This property is still true for $\mathrm{m}=\mathrm{n}+3$ Indeed, in this case we have the function:

\Psi(x)=\lambda_{1}\Psi_{1}(x)+\lambda_{2}\Psi_{2}(x)

And

\Psi\left(x_{2}\right)=\lambda_{1}\Psi_{1}\left(x_{2}\right),\quad\Psi\left(x_{n+2}\right)=\lambda_{2}\Psi_{2}\left(x_{n+2}\right)

so that $\Psi(x)$ either non-negative, it is necessary that $\lambda_{1},\lambda_{2}$ are non-negative, hence the property. The property is no longer true if $m>n+3$ except in the case $n=1$ In this case, indeed:

\Psi\left(x_{i}\right)=\lambda_{i-1}\Psi_{i-1}\left(x_{i}\right)\quad i=2,3,\ldots,m-2

and the non-negativity of the coefficients is necessary for that of $\Psi(x)$ In ( $x_{1},x_{m}$ 13.
Now let us consider the function $f(x)$ defined on the sequence of points

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

(19)

Suppose we have

\Delta_{n+1}^{j}=\Psi_{j}(\xi)\quad j=1,2,\ldots,m-n-1\quad\left(x_{1}\leq\xi\leq x_{m}\right)

(20)

Let us show that in this case the function can be extended in any interval containing the points (19).
⁸ ) In fact, in this way we exclude functions of order 0, which are not necessarily continuous, but for this simple case the problem is of no interest since the function is always and everywhere extendable.
⁹ ) To avoid complicating matters, it is unnecessary to make a more precise distinction regarding the nature of convexity. It is actually a matter of restricted non-concavity.

Let's suppose that $m=2n+2,x_{n+1}\leq\xi<x_{n+2}$ The formulas established above give us:

		$\displaystyle\mathrm{P}\left(x_{n+2},x_{n+3},\ldots,x_{2n+2};f\mid x\right)-\mathrm{P}\left(x_{1},x_{2},\ldots,x_{n+1};f\mid x\right)=$
		$\displaystyle=\sum_{i=1}^{n+1}\left(x_{n+i+1}-x_{i}\right)\Delta_{n+1}^{i}\left(x-x_{i+1}\right)\left(x-x_{i+2}\right)\ldots\left(x-x_{i+n}\right)$
		$\displaystyle=\sum_{i=1}^{n+1}\left(x_{n+i+1}-x_{i}\right)\Psi_{i}(\xi)\left(x-x_{i+1}\right)\left(x-x_{i+2}\right)\ldots\left(x-x_{i+n}\right).$

To calculate this expression, we will proceed by induction again. Relations (18) can be written in condensed form:

\begin{gathered}\left(x_{n+j+1}-x_{j}\right)\Psi_{j}(x)=\left(x-x_{j}\right)\Psi_{j}^{\prime}(x)+\left(x_{n+j+1}-x\right)\Psi_{j+1}^{\prime}(x)\\ j=1,2,\ldots,n+1\end{gathered}

Or $\Psi_{1}^{\prime},\Psi_{2}^{\prime},\ldots,\Psi_{n+2}^{\prime}$ are functions analogous to $\Psi_{1},\Psi_{2},\ldots,\Psi_{n+1}$ on the sequel $x_{1},x_{2},\ldots,x_{2n+2}$ , but by lowering the value of by one unit $n$ .

We can therefore deduce:

		$\displaystyle\sum_{i=1}^{n+1}\left(x_{n+i+1}-x_{i}\right)\Psi_{i}(\xi)\left(x-x_{i+1}\right)\left(x-x_{i+2}\right)\ldots\left(x-x_{i+n}\right)=$
	$\displaystyle=$	$\displaystyle(x-\xi)\sum_{i=1}^{n}\left(x_{n+i+1}-x_{i+1}\right)\Psi_{i+1}^{\prime}(\xi)\left(x-x_{i+2}\right)\left(x-x_{i+3}\right)\ldots\left(x-x_{i+n}\right)$

so little by little

\mathrm{P}\left(x_{n+2},x_{n+3},\ldots x_{2n+2};f\mid x\right)-\mathrm{P}\left(x_{1},x_{2},\ldots x_{n-1};f\mid x\right)=(x-\xi)^{n}

and the extension is obviously achieved by the function equal to:

\begin{array}[]{ll}\mathrm{P}\left(x_{1},x_{2},\ldots,x_{n+1};f\mid x\right)&\text{ pour }x\leq\xi\\ \mathrm{P}\left(x_{n+2},x_{n+3},\ldots,x_{2n+2};f\mid x\right)&\text{ pour }x\geq\xi\end{array}

it being manifestly non-concave in order $n$
The general case of $m$ any and $\dot{\xi}$ anything can be reduced to this. If, generally speaking, one has $x_{i}\leqslant\xi\div x_{i+1}$ the quantities:

\begin{array}[]{ll}\Psi_{j}^{\circ}(\xi)\quad j=1,2,\ldots,i-n-1\quad&(\text{ aucun si }i\leq n+1)\\ \Psi_{j}^{\circ}(\xi)\quad j=i+1,i+2,\ldots,m-n-1&(\text{ aucun si }i\geq m-n-1)\end{array}

are zero by definition. The function is therefore necessarily polynomial in the intervals $\left(x_{1},x_{i}\right),\left(x_{i+1},x_{m}\right)$ The previous demonstration then shows us that it suffices to keep the $n+1$ first points to the left of $x_{i}$ This includes and the $n+1$ first points to the right of $x_{i+1}$ the latter included. If $i<n+1$ we complete the function on the left $x_{1}$ by introducing $n+1-i$ points to the left of $x_{1}$ and taking differences divided in such a way that on this new sequence they are
still of the form (20). If $i>m-n-1$ we perform the same operation on the right of $x_{m}$ and if $m<2n+2$ we make the complement both on the left of $x_{1}$ and to the right of $x_{m}$ .

We can now state the following property:
For the function $\mathrm{f}(\mathrm{x})$ defined and non-concave of order n on the points (19) is everywhere extendable it is necessary and sufficient that it satisfies on these points the property of restricted convexity.

Indeed, in ordinary space, $m-n-1$ dimensions the curve

	$\displaystyle y_{1}=\Psi_{1}(x),y_{2}=\Psi_{2}(x),\ldots,y_{m-n-1}=\Psi_{m-n-1}(x)$		(C)
	$\displaystyle x_{1}\leq x\leq x_{m}.$

We then see that the restricted convexity condition means that any hyperplane passing through the origin and leaving the curve (C) on the same side, leaves the point with coordinates

\left(\Delta_{n+1}^{1},\Delta_{n+1}^{2},\ldots,\Delta_{n+1}^{m-n-1}\right)

(21)

on this same side. Geometrically this means that point (21) is inside or on the boundary of the smallest convex cone containing the curve $(\mathcal{C})$ and having the origin as its apex.

It immediately follows that if the restricted convexity condition is satisfied, then we have

\Delta_{n+1}^{j}=\Sigma_{\mu_{k}}\Psi_{j}\left(\xi_{k}\right)\quad\hat{\imath}=1,2,\ldots,m-n-1

(22)

Or $\mu_{k}$ are a finite number of positive constants and $\varepsilon_{k}$ of points in the interval ( $x_{1},x_{m}$ ). The stated property then follows from the preliminary position demonstrated at the beginning of this No.
14. Let us call $(\Gamma)$ the convex domain defined previously. If there exists a function $\Psi(x)$ non-negative and not identically zero, vanishing at the points $\xi_{k}$ the point (21) given by formulas (22) is on the frontier of ( $\mathrm{I}^{\prime}$ ). In this case, we can indeed find a point as close as we want to (21) that does not satisfy the restricted convexity condition. Any boundary point can be obtained in this way, which follows from the fact that if the points $\xi_{k}$ If the previous condition is not met, point (21) is necessarily an interior point. Indeed, if $\Psi(x)=\Sigma\lambda_{j}\Psi_{j}(x)$ is a non-negative function the expression

\left|\frac{\Sigma\mu_{k}\Psi\left(\dot{s}_{x}\right)}{\Sigma\lambda^{j}}\right|

has a minimum which is not zero in our case, otherwise we could, by a process of passing to the limit, conclude the existence of a function $\Psi(x)$ not identically zero and vanishing at the points $\xi_{k}$ , which contradicts the hypothesis.

Of course we are not talking about the origin point, the apex of the cone (l'), which is of no interest.

Suppose that one of the differences is divided, for example $\Delta_{n+1}^{i}$ , is zero. Then all the points $\xi_{k}$ are outside the open interval ( $x_{i},x_{i+n+1}$ ). But in this case the function $\Psi(x)=\Psi_{i}(x)$ cancels itself out in all respects $\xi_{k}$ , SO :

The points (21) corresponding to non-concave (and not convex) functions satisfying the restricted convexity condition lie on the boundary of the domain (1'). It should be noted that convex functions do not always yield an interior point. For example $n=2,m=6x_{3}<\xi<x_{4}$ ,

\Delta_{3}^{1}=\Psi_{1}(\xi),\Delta_{3}^{2}=\Psi_{2}(\xi)\Delta_{3}^{3}=\Psi_{3}(\xi)

All these functions are convex, but point (21) is on the boundary of ( $1^{\prime}$ since the function
$\Psi(x)=\left(\xi-x_{2}\right)\left(\xi-x_{3}\right)\Psi_{1}(x)+\left(\xi-x_{3}\right)\left(\xi-x_{4}\right)\Psi_{2}(x)+\left(\xi-x_{4}\right)\left(\xi-x_{5}\right)\Psi_{3}(x)$ is non-negative and vanishes at the point $\underset{.}{.}$ 15.
Let us say a few more words about the functions that can be extended at each point. For this to be the case, it is obviously necessary that the point (21) be inside or on the boundary of a certain convex conic domain ( $\Gamma^{\prime}$ ) having the origin as its vertex. The domain ( $\Gamma^{\prime}$ ) always contains the domain ( $\Gamma$ ), but is generally more extensive than the latter. It can be stated that ( $1^{\prime}$ ) is greater than ( $1^{\prime}$ ) as soon as there exists on the boundary of the latter a point such that any point sufficiently close to it corresponds to non-concave functions that can be extended to any point. Thus, in the case $n>2,m\geq n+4$ the domains ( $\mathrm{l}^{\prime}$ ), ( $\mathrm{l}^{\prime}$ ) do not coincide ¹⁰ ). The same applies if $n=2,m\geq 8{}^{11}$ ).

		$\displaystyle\text{ 10) Il suffit de prendre }m=n+4\text{. Supposons, pour simplifier, que }$
		$\displaystyle\qquad x_{1}=0,x_{2}=1,x_{3}=2,\ldots,x_{i}=i-1,\ldots,x_{n+4}=n+3$

and consider a function such that

\begin{gathered}\triangle_{n+1}^{1}=5^{n}-(n+1)3^{n}+\frac{n(n+1)}{2},\triangle_{n+1}^{2}=3^{n}-(n+1),\triangle_{n+1}^{3}=1\\ \text{ On a }\\ (x-1)(x-2)\triangle_{n+1}^{1}+(x-2)(x-n-1)\triangle_{n+1}^{2}+(x-n-1)(x-n-2)\triangle_{n+1}^{3}>0\\ \quad 3\leq x\leq n+2,n>2\end{gathered}

therefore any point quite close to $\left(\triangle_{n+1}^{1},\triangle_{n+1}^{2},\triangle_{n+1}^{3}\right)$ gives convex functions of order $n$ extendable at every point. On the other hand, it can be shown that there exists a function $\Psi(x)\geq 0$ canceling out at the point $\frac{5}{2}$ and conclude that it is indeed a boundary point.
¹¹ ) It suffices to consider $m=8$ . Example

\begin{gathered}x_{1}=0,x_{2}=1,\ldots,x_{8}=7\\ \triangle_{3}^{1}=1,\triangle_{3}^{2}=6,\triangle_{3}^{3}=2,\triangle_{3}^{4}=6,\triangle_{3}^{5}=1\end{gathered}

We arrive at the property as in the previous example by noting that we have

3\psi_{1}-\psi_{2}+3\psi_{3}-\psi_{4}+3\psi_{5}\geq 0

equality being verified for $x=\frac{5}{2},\frac{9}{2}$ .

On the contrary, we will show that if $n=2,m=6,7$ the domains ( $\mathrm{l}^{\prime}$ ), ( $\mathrm{l}^{\prime}$ ) coincide. It suffices to consider $m=7$ the case $m=6$ which can always be reduced to this. We will show that in any neighborhood of any boundary point of (I) we can find a point corresponding to functions that are either not of order 2, or, while being second order, are not extendable to every point. This property is immediate for boundary points arising from non-concave functions (of which at least one of the divided differences is zero). This is true, moreover, regardless of $n$ And $m{}^{12}$ ). It remains to examine the boundary points arising from convex functions. It is then necessary that in each open interval ( $x_{1},x_{4}$ ), $\left(x_{2},x_{5}\right),\left(x_{3}x_{6}\right)\left(x_{4},x_{7}\right)$ at least one point $\xi_{k}$ It is easy to see that the only case that can provide a boundary point is that where one of the points is in the open interval ( $x_{3},x_{4}$ ) all the others being in ( $x_{6},x_{7}$ ) [or one in ( $x_{4},x_{5}$ ) and the others in ( $x_{1},x_{2}$ )].

These boundary points are therefore of the form

\begin{gathered}\Delta_{3}^{1}=\lambda\Psi_{1}(\xi),\Delta_{3}^{2}=\lambda\Psi_{2}(\xi),\Delta_{3}^{3}=\lambda\Psi_{3}(\xi),\Delta_{3}^{4}=\mu\\ \lambda>0,\mu>0x_{3}<\xi<x_{4}\end{gathered}

(similar form if $x_{4}<\xi<x_{5}$ The existence of a neighboring point corresponding to a non-extendable function follows from the fact that

\begin{gathered}\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x_{4}-x_{1}\right)\Delta_{3}^{1}+\left(x-x_{3}\right)\left(x-x_{4}\right)\left(x_{5}-x_{2}\right)\Delta_{3}^{2}+\\ +\left(x-x_{4}\right)\left(x-x_{5}\right)\left(x_{6}-x_{3}\right)\Delta_{3}^{3}=\lambda\left(x-\frac{5}{4}\right)^{2}\text{ dans }\left(x_{3},x_{4}\right)\end{gathered}

It is now clear that ( $\Gamma^{\prime}$ ), ( $\Gamma^{\prime}$ ) must coincide. Let A be a point on ( $\mathrm{l}^{\prime}$ ) outside of ( $\Gamma$ ). The entire cone with vertex A, circumscribed about ( $\Gamma^{\prime}$ ) must belong to ( $\Gamma^{\prime}$ ) which contradicts what was demonstrated earlier ⁽¹³ ).
16. Let us call an elementary function of degree n with m vertices any function whose $(n-1)^{\dot{e}me}$ derivative is a polygonal line at $m$ vertices without counting the endpoints of the interval where these functions are defined.

It follows that an elementary function of degree $n$ is of the form

f(x)=\left\{\begin{array}[]{r}\mathrm{P}(x)\text{ polynome de degré }n\text{ dans }\left(a,\xi_{1}\right)\\ \mathrm{P}(x)+\sum_{i=1}^{k}a_{i}\left(x-\xi_{i}\right)^{n}\quad\text{ dans }\left(\xi_{k},\xi_{k+1}\right)\\ k=1,2,\ldots,m\left(\xi_{m+1}=b\right).\end{array}\right.

⁰⁰footnotetext: ¹² ) From this property it immediately follows that (

\Gamma

), (

\Gamma

) have parts of a common border.
¹³ ) In this reasoning, it is of course assumed that (

\boldsymbol{\Gamma}

) is indeed a 5-dimensional manifold. Moreover, it can easily be verified that in the general case (

\Gamma

) is indeed at

m-n-1

dimensions.

For such a function to be non-concave of order $n$ it is necessary and sufficient that

a_{i}\geq 0,i=1,2,\ldots,m

An elementary function of degree $n$ having a single vertex is always of order $n$ .

If a function of order $n$ defined on the points (19) is extendable in any interval; we have seen that it is extendable by an elementary function of degree $n$ .

It follows that if $f(x)$ is non-concave of order $n$ within an interval ( $x_{1},x_{m}$ ) there exists an elementary function of degree $n$ and non-concave of order $n$ taking the values $f\left(x_{i}\right)$ to the points

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

Let's examine more closely the functions that provide the extension. Since these are non-concave functions in any interval, they will be at $n^{\text{ème }}$ differences divided bounded in ( $x_{1},x_{m}$ ), therefore their ( $n-1$ The second derivative will always be continuous in this closed interval, and the derivatives $f^{(n)}\left(x_{1}\right),f^{(n)}\left(x_{m}\right)$ exist $\left[\right.$ we ask $f^{(n)}\left(x_{1}\right)=\left(\frac{d}{dx}\right)_{d}f^{(n-1)}\left(x_{1}\right)$ , $\left.f^{(n)}\left(x_{m}\right)_{1}=\left(\frac{d}{dx}\right)_{g}f^{(n-1)}\left(x_{m}\right)\right]$ I say that:

Every non-concave function of order n is the limit of a sequence of uniformly converging convex functions of order n. The sequence of derivatives converges uniformly to those of the given function up to order n. $\mathrm{n}-1$ and also for the nth derivative at both ends $x_{1},x_{m}$ .

By successive derivations we see that it suffices to demonstrate the property for $n=1$ . In this case, it can be easily seen by considering the representative figure of the function and proceeding as in No. 2.

Now suppose that point (21) is inside ( $\Gamma$ ). Let us consider a hypertetrahedron $\mathrm{A}_{1}\mathrm{\penalty 10000\ A}_{2}\ldots\mathrm{\penalty 10000\ A}_{m-n-1}$ not degenerated completely internal to ( $\mathrm{l}^{\prime}$ ) and containing within it point (21). Let $f_{1}(x),f_{2}(x),\ldots,f_{m-n-1}(x)$ giving the extension and corresponding to the points $\mathrm{A}_{1},\mathrm{\penalty 10000\ A}_{2},\ldots,\mathrm{\penalty 10000\ A}_{m-n-1}$ By modifying the points as needed, however slightly, $\mathrm{A}_{i}$ We can see that we can assume that $f_{i}(x)$ let them be convex. Let them now be $\lambda_{1},\lambda_{2},\ldots,\lambda_{m-n-1}(\geq 0)$ The homogeneous tetrahedral coordinates of point (21), we then see that the function

\frac{\sum\lambda_{i}f_{i}(x)}{\sum\lambda_{i}}+\text{ polynome convenable de degré }n

is convex and takes the values given at the points $x_{1},x_{2},\ldots x_{m}$ . SO.

If (21) is an interior point, the extension can always be done by a convex function ¹⁴ ).

It also follows that if (21) is inside, any function extending it can be approximated as closely as desired by other convex functions. Indeed, $f(x)$ such a function and $f*(x)$ a convex function performing the extension, it suffices to consider

\frac{f(x)+\lambda f^{*}(x)}{1+\lambda},\lambda>0

We also intend to demonstrate that, if the point $(21)$ is on the border of $(\Gamma)$ The extension is not possible with a convex function.

In other words, if $f(x)$ is a convex function of order $n$ and if we construct the point $(21)$ from its divided differences, then this point lies within $(I^{\prime})$ .

Let us consider the interval $(x_{i},x_{i+1})$ From a property that will be demonstrated later (property A of no ^. 22), it follows that there exists a function $\varphi(x)$ non-concave, and even convex of order $n$ In $(x_{i},x_{i+1})$ , such as …

\begin{gathered}\mathrm{f}^{(k)}\left(x_{i}\right)=f^{(k)}\left(x_{i}\right),\quad\mathrm{p}^{(k)}\left(x_{i+1}\right)=f^{(k)}\left(x_{i+1}\right)\\ k=1,2,\ldots,n-1\\ \mathrm{P}_{d}^{(n)}\left(x_{i}\right)=f_{d}^{(n)}\left(x_{i}\right),\quad\varphi_{g}^{(n)}\left(x_{i+1}\right)=f_{g}^{(n)}\left(x_{i+1}\right)\end{gathered}

and the values $\varphi\left(x_{i}\right),\varphi\left(x_{i+1}\right)$ being subject to the sole condition

\varphi\left(x_{i}\right)-f\left(x_{i}\right)|<\varepsilon,\quad|\varphi\left(x_{i+1}\right)-f\left(x_{i+1}\right)\mid<\varepsilon

$\varepsilon$ being a sufficiently small positive number.
It follows that any function $f_{1}(x)$ such as

\left|f\left(x_{i}\right)-f_{1}\left(x_{i}\right)\right|<\varepsilon,i=1,2,\ldots,m

is still extendable provided that $\varepsilon>0$ is quite small, which proves that it is indeed an internal point.

Properties demonstrated a little later will further follow that any function performing the extension is the limit of a sequence of elementary functions of degree $n$ carrying out the extension and converging uniformly in any finite interval.

For an interior point of ( $\Gamma$ ) the extension is always possible by an infinite number of functions. The same is true for boundary points in general, except for some of them for which the extension is only possible in one way within the interval ( $x_{1},x_{m}$ ). The problem addressed in the following issues allows us to study this uniqueness.

Finally, we could propose to express restricted convexity by explicit inequalities between quantities $\Delta_{n+1}^{i}$ . We don't have

⁰⁰footnotetext: ¹⁴ ) As we can see, the property is demonstrated for the interval (

x_{i},x_{m}

), but, we can extend the function outwards while respecting the convexity.

The intention here is to look for these inequalities, which present themselves in a rather complicated form. Let us simply note that we can look at the $\Delta_{n+1}^{i}$ as generalizing the sequence of coefficients of certain quadratic forms.

IV

Case study of an extension

17.

We have assumed up to this point that all points (19) are distinct. We can study limit problems by assuming that several of these points coincide. We will choose one of these problems, the most interesting one in fact.

By taking $m=2n+2$ and assuming that

\begin{gathered}x_{2},x_{3},\ldots,x_{n+1}\rightarrow x_{1}=a\\ x_{n+2},x_{n+3},\ldots,x_{2n+2}\rightarrow x_{2n+2}=b>a\end{gathered}

We obtain the following problem:
Determine a non-concave function of order n in the interval (a, b) taking the given values for its first n derivatives

		$\displaystyle f(a),f^{\prime}(a),\ldots,f^{(n)}(a)$
		$\displaystyle f(b),f^{\prime}(b),\ldots,f^{(n)}(b)$

We can obtain the possibility condition of this problem by taking the limit in the restricted convexity condition and we can easily show that this method is perfectly justified.

Thus we obtain

	$\displaystyle\Delta_{n+1}^{i+1}=\frac{(-1)^{i}}{i!(b-a)^{n+1}}$	$\displaystyle{\left[\sum_{k=0}^{i}(-1)^{k}(b-a)^{k}\binom{i}{k}\frac{(n-k)!}{(n-i)!}f(k)(b)-\right.}$		(23)
		$\displaystyle\left.-\sum_{k=0}^{n-i}(b-a)^{k}\binom{i}{k}\frac{(n-k)!}{(n-i)!}f^{(k)}(a)\right].$

On the other hand, the limits of the functions $\psi_{i}(x)$ are polynomials

\lim.\Psi_{i}(x)=\frac{1}{(b-a)^{n+1}}\binom{n}{i-1}(b-x)^{n-i+1}(x-a)^{i-1}.

Therefore, for our problem to be possible, it is necessary that for every polynomial

\sum_{i=1}^{n+1}\lambda_{i}\binom{n}{i-1}(b-x)^{n-i+1}(x-a)^{i-1}

(24)

not negative in ( $a,b$ ) we have

\sum_{i=1}^{n+1}\lambda_{i}\Delta_{n+1}^{i}\supseteq 0

The polynomial (24) can also be written in the form

\sum_{k=0}^{n}\mu_{k}(b-\cdots x)^{k}

\mu_{k}=\binom{n}{k}(b-a)^{n-k}\sum_{i=n-k+1}^{n+1}(-1)^{i+k-n-1}\lambda_{i}\binom{k}{n-i+1}

We also have

\sum_{i=1}^{n+1}\lambda_{i}\Delta_{n+1}^{i}=\frac{1}{n!(b-a)^{n+1}}\sum_{i=0}^{n}\mu_{i}c_{i}

where by performing the calculations using formulas (23)

c_{i}=i!\left[f^{(n-i)}(b)-\sum_{k=0}^{i}\frac{(b-a)^{k}}{k!}f^{(n-i+k)}(a)\right],i=0,1,\ldots,n.

(25)

Therefore, for the problem to be possible, it is necessary that for every polynomial

\sum_{i=0}^{n}\mu_{i}(b-x)^{i}

non-negative in $(a,b)$ we have

\sum_{i=0}^{n}\mu_{i}c_{i}\geqslant 0

If we have $c_{0}=0$ we must also have $c_{1}=c_{2}=\ldots=c_{n}=0$ as can be easily verified, and therefore the only solution is a polynomial of completely determined degree. In what follows, we will always assume that $c_{0}>0$ .

The condition of possibility can be expressed in the following geometric form:

For the problem posed to be possible, the point must be

\mathrm{M}_{n}\left(\frac{c_{1}}{c_{0}},\frac{c_{2}}{c_{0}},\ldots,\frac{c_{n}}{c_{0}}\right)

either inside or on the boundary of the smallest convex domain $\left(\mathrm{W}_{n}\right)$ containing curve ¹⁵ )

y_{1}=(b-x),y_{2}=(b-x)^{2},\ldots,y_{n}=(b-x)^{n}a\leq x\leq b.

Let's say once and for all that, given the problem data, every function of order $n$ to consider is $n^{\text{ème }}$ difference divided is bounded, therefore

⁰⁰footnotetext: ¹⁵⁾ This property is due to MS Kakeya. „On some Integral Equations III" Tôhoku Math. Journ. t. 8 p. 14. The author studies a particular case of the problem.

has a $(n-1)^{\text{ème }}$ continuous derivative in the closed interval $(a,b)$ and the $n^{\text{ème }}$ The derivative exists at the extremities.
18. Every point of $\left(\mathrm{W}_{n}\right)$ or its border can be represented in the form

	$\displaystyle c_{i}=\sum_{k=1}^{m}\lambda_{k}\left(b-\xi_{k}\right)^{i}i=0,1,2,\ldots,n$		(26)
	$\displaystyle m\leq n,\lambda_{i}\geq 0,a\leq\xi_{1}<\xi_{2}<\ldots<\xi_{m}\leq b.$

For this point to be on the boundary, it is necessary and sufficient that one of the following three conditions be met.

		$\displaystyle\text{ (type I) }m\leq\left[\frac{n}{2}\right]\text{ si }a<\xi_{1},\xi_{m}<b$
		$\displaystyle\text{ (type II) }m\leq\left[\frac{n+2}{2}\right]\text{ si }a=\xi_{1},\xi_{m}=b$
		$\displaystyle\text{ (type III) }m\leq\left[\frac{n+1}{2}\right]\text{ si }a<\xi_{1},\xi_{m}=b$
		$\displaystyle\text{ (type IV) }m\leq\left[\frac{n+1}{2}\right]\text{ si }a=\xi_{1},\xi_{m}-b$

which immediately follows from the distribution of zeros of a non-negative polynomial of degree $n$ In ( $a,b$ ).

It can easily be shown, by comparing two systems of linear equations, that the representation (26) of a boundary point is unique.

Note now that the projection of the domain ( $\mathrm{W}_{n}$ ) on the hyperplane $\mathrm{O}y_{1}y_{2}\ldots y_{n-1}$ is precisely the domain $\left(\mathrm{W}_{n-1}\right)$ . Likewise $\left(\mathrm{W}_{n}\right)$ can be viewed as the projection of $\left(\mathrm{W}_{n+1}\right)\ldots$ etc. The projection of the point $M_{n}$ and the point

\mathrm{M}_{n-1}\left(\frac{c_{1}}{c_{0}},\frac{c_{2}}{c_{0}},\ldots,\frac{c_{n-1}}{c_{0}}\right)

and corresponds, as we can see, to our problem posed with the same data for the derivative. $f^{\prime}(x)$ of the function. The line $M_{n}M_{n-1}$ cuts the border $\left(\mathrm{W}_{n}\right)$ in two points: $\mathrm{M}_{n}^{\prime}$ whose ordinate is the smallest and $\mathrm{M}_{n}^{\prime\prime}$ whose ordinate is the largest. If $\mathrm{M}_{n-1}$ is a border point $\mathrm{M}_{n}\mathrm{M}_{n-1}$ cannot cut ( $\mathrm{W}_{n}$ ) that at a single border point; $\mathrm{M}_{n}^{\prime}\mathrm{M}_{n}^{\prime\prime},\mathrm{M}_{n}$ therefore coincide. It follows that every point of ( $\mathrm{W}_{n}$ ) is the projection of a boundary point of ( $\mathrm{W}_{n+1}$ ).

It also follows that any interior point $M_{n}$ has infinitely many representations of the form (26). It can be shown that there are infinitely many for any value of $m>\left\lceil\frac{n}{2}\right\rfloor+1$ , with $a<\xi_{1},\xi_{m}<b$ . If $n$ is even there are an infinite number of representations with $m=\left[\frac{n}{2}\right]+1$ Among these re-
presentations there is one for which $a=\xi_{1},\xi_{m}<b$ and one for which $a<\xi_{1},\dot{\xi}_{m}=b$ . If $n$ is odd there is only one representation with $m=\left[\frac{n}{2}\right]+1$ and then $a<\xi_{1},\xi_{m}<b$ Among the representations with $m=\left[\frac{n}{2}\right]+2$ there is one for which $a=\xi_{1},\xi_{m}=b$ .

If $M_{n}$ is inside ( $W_{n}$ ), $M_{n-1}$ is inside ( $W_{n-1}$ ). It follows that if $n$ is odd $\mathrm{M}_{n}^{\prime},\mathrm{M}_{n}^{\prime\prime}$ are of type (III) and (IV) and if $n$ is even, they are of type (I) and (II). It can easily be shown that $M_{n}^{\prime}$ is of type (I) or (III) and $M_{n}^{\prime\prime}$ of type (II) or (IV).

For $n=1$ the domain $\left(\mathrm{W}_{n}\right)$ reduces to the segment $(0,b-a)$ and we can easily verify that our problem is impossible for the right-hand side.
19. Let us now examine the sufficient conditions. Consider the polynomials:

\mathrm{P}_{1}(x),\mathrm{P}_{2}(x),\ldots,\mathrm{P}_{n+2}(x)

defined by the relationships:

\begin{array}[]{ll}\mathrm{P}_{i}^{(k)}(a)=f^{(k)}(a)&k=0,1,\ldots,n-i+1\\ \mathrm{P}_{i}^{(k)}(b)=f^{(k)}(b)&k=0,1,\ldots,i-2\end{array}

The initial conditions being removed for $\mathrm{P}_{n+2}$ and the last ones for $P_{1}$ A simple calculation reveals

\mathrm{P}_{n+2}(x)-\mathrm{P}_{1}(x)=\frac{1}{n!}\sum_{i=1}^{m}\lambda_{i}\left(x-\dot{\xi}_{i}\right)^{n}

(27)

using representation (26). If $M_{n}$ is an interior point; there are an infinite number of representations with $a<\xi_{1},\xi_{m}<b$ and then our problem is solved by the function

\Phi(x)=\left\{\begin{array}[]{l}\mathrm{P}_{1}(x)\quad\text{ dans }\left(a,\xi_{1}\right)\\ \mathrm{P}_{1}(x)+\frac{1}{n!}\sum_{i=1}^{k}\lambda_{i}\left(x-\xi_{i}\right)^{n}\text{ dans }\left(\xi_{k},\xi_{k+1}\right)\\ \quad k=1,2,\ldots,m\quad\xi_{m+1}=b.\end{array}\right.

Let us call an elementary solution any elementary function of degree $n$ solving the problem.

So, if $\mathrm{M}_{n}$ is an interior point it $y$ always has infinitely many solutions. There are always infinitely many elementary solutions. In particular, there is always an elementary solution to $\left\lfloor\frac{n}{2}\right\rfloor+1$ summits. If $n$ is odd the solution of this form is unique, but there are infinitely many if $n$ is even. The set of all solutions forms a
bounded family. Their derivatives up to order $n-1$ as well as their two derivatives of order $n$ also form bounded families. The solutions have an upper limit function $\boldsymbol{\Phi}_{1}(x)$ and a lower limit function $\Phi_{2}(x)$ We must have it in particular!

P_{2}(x)\geqslant\Phi_{1}(x)\geqslant f(x)\geqslant\Phi_{2}(x)\geqslant P_{1}(x)\quad(a\leq x\leq b)

for any solution $f(x)$
Note that :

\mathrm{P}_{2}(x)-\mathrm{P}_{1}(x)=\left[\frac{f(b)-f(a)}{(b-a)^{n}}+\sum_{i=1}^{n}\frac{1}{i!(b-a)^{n-i}}f^{(i)}(a)\right](x-a)^{n}

It follows that if the problem data remains smaller than a fixed number in modulo, then... $\varepsilon-0$ corresponds to a $\eta>0$ such that if:
we have

		$\displaystyle\|f(b)-f(a)\|<\eta\quad\|b-a\|<\eta$
		$\displaystyle\left\|\Phi_{1}(x)-\Phi_{2}(x)\right\|<\varepsilon\quad\text{ dans }(a,b)$

20.

The function (28) can be constructed with any representation (26). If we have $a=\xi_{1}$ of course, we remove the definition in the interval ( $a,\dot{\xi}_{1}$ ). Let us designate in particular by $\Phi_{1}^{*}(x)$ the function (28) constructed with the representation for which $a=\xi_{1},m=\left[\frac{n}{2}\right]+1$ if $n$ is even and $a=\dot{\xi}_{1},b=\dot{\xi}_{m},m=\left[\frac{n}{2}\right]+2$ if $n$ is odd and by $\Phi_{2}^{*}(x)$ the function (28) constructed with the representation for which $\dot{\boldsymbol{s}}_{m}=b$ , $m=\left[\frac{n}{2}\right]+1$ if $n$ is even and $m=\left[\frac{n}{2}\right]+1$ if $n$ is odd.

We intend to demonstrate that.

\Phi_{1}(x)\equiv\Phi_{1}^{*}(x),\quad\Phi_{2}(x)=\Phi_{2}^{*}(x).

(30)

If $n$ is even $\Psi_{1}^{*}(x),\Psi_{2}^{*}(x)$ These are not solutions. Let them be:

\begin{array}[]{ll}\left(a=\xi_{1}^{\prime}\right),\xi_{2}^{\prime},\ldots,\xi^{\prime}\left[\frac{n}{2}\right]+1&\text{ les sommets de }\Phi_{1}^{*}\\ \xi_{1}^{\prime\prime},\xi_{2}^{\prime\prime},\ldots,\left(\xi^{\prime\prime}\right]\left[\frac{n}{2}\right]+1&=b)\end{array}

and consider an elementary solution $f(x)$ having $\left[\frac{n}{2}\right]+1$ summits $\xi_{1},\xi_{2},\ldots,\xi_{\left[\frac{n}{2}\right]+1}$ . It is easily demonstrated that if $\xi_{1}\rightarrow\xi_{1}^{\prime}$ we also have $\dot{\xi}_{i}\rightarrow\dot{\xi}_{i}^{\prime},i>1$ And $f(x)$ tends uniformly towards $\Phi_{1}^{*}(x)$ In $(a,b)$ Similarly if $\xi_{1}\rightarrow\xi_{1}^{\prime\prime}$ we also have $\xi_{i}\rightarrow\xi_{i}^{\prime\prime},i>1$ And $f(x)$ tends uniformly towards $\Phi_{2}^{*}(x)$ We also have the property of separation:

a=\xi_{1}^{\prime}<\xi_{1}^{\prime\prime}<\xi_{2}^{\prime}<\xi_{2}^{\prime\prime}<\cdots<\xi_{\left[\frac{n}{2}\right]+1}^{\prime}<\xi_{\left[\begin{array}[]{l}n\\ 2\end{array}\right]+1}^{\prime\prime}=b

which can be easily verified by comparing the two corresponding representations (26).

If $n$ is odd $\Phi_{2}^{*}(x)$ is one solution, but $\Phi_{1}^{*}(x)$ That's not a solution. Let's

\left(a=-\xi_{1}^{\prime}\right),\xi_{2}^{\prime},\xi_{3}^{\prime},\ldots,\left(\xi_{\left[\frac{n}{2}\right]+2}^{\prime}=b\right)\text{ les sommets de }\Phi_{1}^{*}

If $\xi_{1},\xi_{2},\ldots,\xi_{\left[\frac{n}{2}\right]+2}$ are the vertices of an elementary solution $f(x)$ has $\left[\frac{n}{2}\right]+2$ summits, $f(x)$ converges uniformly towards $\Psi_{1}^{*}(x)$ when $\xi_{1}\rightarrow a,\xi_{\left[\frac{n}{2}\right]+2}\rightarrow b$ We still have the separation property

a=\dot{\xi}_{1}^{\prime}<\dot{\xi}_{1}^{\prime\prime}<\dot{\xi}_{2}^{\prime}<\dot{\xi}_{2}^{\prime\prime}<\cdots<\dot{\xi}_{\left[\frac{n}{2}\right]+1}^{\prime\prime}<\dot{\xi}_{\left[\frac{n}{2}\right]+2}^{\prime\prime}=b.

We can therefore finally say that the functions $\Phi_{1,}^{*}(x),\Phi_{2}^{*}(x)$ are the limits of uniformly convergent sequences of solutions. It follows that to prove relations (30) it suffices to show that if $f(x)$ is any solution we have

\Phi_{1}^{*}(x)\geqslant f(x)\geqslant\Phi_{2}^{*}(x)\quad\text{ dans }(a,b)

(31)

21.

Let us now turn to the proof of inequalities (31). Let a point be in the interval $(a,b)$ The restricted convexity condition must be expressed on $2n+3$ points which $n+1$ are confused in $a,n+1$ confused in $b$ and the $(2n+3)^{\text{ème }}$ is the point $\xi$ This condition is expressed by the fact that if we have

\sum_{i=0}^{n}\mu^{i}(b-x)^{i}+\mu\Psi(x)\geqslant 0\quad\text{ dans }(a,b)

we must also have

\sum_{i=0}^{n}\mu_{i}c_{i}+c^{*}\geqslant 0

\mathrm{T}(x)=\left\{\begin{array}[]{cc}(\xi-x)^{n}&\text{ dans }(a,\xi)\\ 0&\text{ dans }(\xi,b)\end{array}c^{*}=n!\left[f(\xi)-\mathrm{P}_{1}(\xi)\right]^{16}\right).

$\left.{}^{16}\right)$ This condition can easily be seen by assuming that $f^{(n)}(x)$ exists. We then have

\begin{gathered}f(x)=\frac{1}{n!}\int_{a}^{x}(x-t)^{n}df^{(n)}(t)+\mathrm{P}_{1}(x)\\ c_{i}=\int_{a}^{b}(b-t)^{i}df^{(n)}(t),i=0,1,\ldots,n,c^{*}=\int_{a}^{b}\Psi(t)df^{(n)}(t)\end{gathered}

And $f^{(n)}(x)$ is non-decreasing. We arrive at the same result, of course, by taking the limit in the general condition of restricted convexity.

For $n=1$ The inequalities (31) are clear. In this case indeed $\Phi_{1}^{*}(x)$ is none other than the polynomial $\mathrm{P}_{2}(x)$ the summit $\xi_{1}^{\prime\prime}$ of $\Phi_{2}^{*}(x)$ is the point where we have $\mathrm{P}_{1}(x)=\mathrm{P}_{3}(x)$ And

\Phi_{2}^{*}(x)=\begin{cases}\mathrm{P}_{1}(x)&\text{ dans }\left(a,\stackrel_{1}^{\prime\prime}\right)\\ \mathrm{P}_{3}(x)&\text{ dans }\left(\stackrel_{1}^{\prime\prime}b\right)\end{cases}

Let us take the general case and prove the second inequality (31) by assuming that $n$ is odd.

Either

\xi_{k}^{\prime\prime}\leq\xi\leq\xi_{k+1}^{\prime\prime}\quad\left(k=1,2,\ldots,\left\lfloor\frac{n}{2}\right\rfloor\right)

and let's construct the polynomial of degree $n,\mathrm{Q}(x)$ defined by the relationships:
$\mathrm{Q}\left(\xi_{i}^{\prime\prime}\right)=0,\quad\mathrm{Q}^{\prime}\left(\xi_{i}^{\prime\prime}\right)=0,\quad i=k+1,k+2,\ldots,\left[\frac{n}{2}\right]+1$
$\mathrm{Q}\left(\xi_{i}^{\prime\prime}\right)=-\left(\xi-\xi_{i}^{\prime\prime}\right)^{\prime\prime},\mathrm{Q}^{\prime}\left(\xi_{i}^{\prime\prime}\right)=n\left(\xi-\xi_{i}^{\prime\prime}\right)^{n-1},\quad i=1,2,\ldots,k$
I say that

\mathrm{Q}(x)+\Psi(x)\geqslant 0\quad\text{ dans }(a,b)

(32)

This property is immediate. Indeed $\Psi(x)$ is of order $n$ , SO $\mathrm{Q}(x)+\Psi(x)$ is also of order $n$ But this elementary function has a vertex and vanishes, as does its first derivative, at the points $\xi_{1}^{\prime\prime},\xi_{2}^{\prime\prime}\ldots,\xi_{[n}^{\prime\prime}$

I say that then $\mathrm{Q}(x)+\Psi(x)$ is of invariable sign. Indeed, otherwise it would necessarily have to change sign at some point $\xi^{*}$ at least and therefore be zero at this point ¹⁷ ). We then have:

\left.\left[\xi_{1}^{\prime\prime},\xi_{1}^{\prime},\xi_{2}^{\prime\prime},\xi_{2}^{\prime\prime},\ldots,\xi_{\left[\frac{n}{2}\right]+1,}^{\prime\prime}\xi_{2}^{\prime\prime}\right]+\xi_{1},\xi^{*};\mathrm{Q}+\Psi^{\prime}\right]=0

and the function should be zero identically in $\left.\left(\xi_{1}^{\prime\prime},\xi_{[}^{\prime\prime}\right]+1\right)$ at least, which is impossible. Just look at the graph representing the functions $\mathrm{Q}(x)$ And $\Psi(x)$ to see that it is indeed (32) which takes place.

Let us now consider the function $\Phi_{2}^{*}(x)$ and be $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\left[\frac{n}{2}\right]+1$ the coefficients of its representation (26). We have:

	$\displaystyle\mathrm{Q}(x)$	$\displaystyle=\sum_{i=0}^{n}\mu_{i}(b-x)^{n}$
	$\displaystyle{\left[\frac{n}{2}\right]+1}$	$\displaystyle\lambda_{i}\mathrm{Q}\left(\xi_{i}^{\prime\prime}\right)=\sum_{i=1}^{n}\mu_{i}c_{i}=\sum_{i=1}^{k}-\lambda_{i}\left(\xi-\xi_{i}^{\prime\prime}\right)^{n}$

17.

$\xi^{*}$ may coincide with a point $\xi_{i}^{\prime\prime}$ The second derivative of $\mathrm{Q}(x)+\Psi(x)$ must also be cancelled at this point and the conclusions remain the same.

Therefore, we must have:

-\sum_{i=1}^{k}\lambda_{i}\left(\xi-\xi_{i}^{\prime\prime}\right)^{n}+c^{*}\geq 0

which is none other than

f(\xi)\geq\Phi_{2}^{*}(\xi)\quad\xi_{k}^{\prime\prime}\leq\vdots\leq\xi_{k+1}^{\prime\prime}\quad\left(k=1,2,\ldots,\left[\frac{n}{2}\right]\right)

In the intervals $\left(a,\stackrel{{\scriptstyle*}}_{1}^{\prime\prime}\right),\left(\stackrel{{\scriptstyle*}}^{\prime\prime}\left[\frac{n}{2}\right]+1,b\right)$ this inequality is verified, reducing itself to

[\xi,\underbrace{a,a,\ldots,a}_{n+1};f]\geq 0,\quad[\xi,b,\underbrace{b,\ldots,b}_{n+1};f]\geq 0

respectively.
If $n$ is even, the second inequality (31) is proven in the same way. To obtain inequality (32), we will take the polynomial

		$\displaystyle\mathrm{Q}\left(\xi_{\left[\frac{n}{2}\right]+1}^{\prime\prime}\right)=0$
		$\displaystyle\mathrm{Q}\left(\xi_{i}^{\prime\prime}\right)=0,\quad\mathrm{Q}^{\prime}\left(\xi_{i}^{\prime\prime}\right)=0\quad i=k+1,k+2,\ldots,\left[\frac{n}{2}\right]$
		$\displaystyle\left(\xi_{k}^{\prime\prime}\leq\xi\leq\xi_{k+1}^{\prime\prime}\right)$

The proof is similar for the first inequality (31). We will look for the polynomial $\mathrm{Q}(x)$ such as

\mathrm{Q}(x)-\Psi(x)\geq 0\quad\text{ dans }(a,b)

We will take: If $n$ is odd

		$\displaystyle\mathrm{Q}\left(\xi_{\left[\frac{n}{2}\right]+2}^{\prime}\right)=0$
		$\displaystyle\mathrm{Q}\left(\xi_{i}^{\prime}\right)=0,\mathrm{Q}^{\prime}\left(\xi_{i}^{\prime}\right)=0,i=k+1,k+2,\ldots,\left[\frac{n}{2}\right]+1$
		$\displaystyle\mathrm{Q}\left(\xi_{i}^{\prime}\right)=\left(\xi-\xi_{i}^{\prime}\right)^{n},\mathrm{Q}^{\prime}\left(\xi_{i}^{\prime}\right)=-n\left(\xi-\xi_{i}^{\prime}\right)^{n-1}\quad\left(\xi_{k}^{\prime}\leq\xi_{k}\leq\xi_{k+1}^{\prime\prime}\right)$
		$\displaystyle\mathrm{Q}\left(\xi_{i}^{\prime}\right)=\left(\xi-\xi_{i}^{\prime}\right)^{n}.$

		$\displaystyle\mathrm{Q}\left(\xi_{i}^{\prime}\right)=0,\quad\mathrm{Q}^{\prime}\left(\xi_{i}^{\prime}\right)=0,\quad i=k+1,k+2,\ldots,\left[\frac{n}{2}\right]+1$
		$\displaystyle\mathrm{Q}\left(\xi_{i}^{\prime}\right)=\left(\xi-\xi_{i}^{\prime}\right)^{n}\quad\mathrm{Q}^{\prime}\left(\xi_{i}^{\prime}\right)=-n\left(\xi-\xi_{i}^{\prime}\right)^{n-1}$
		$\displaystyle\quad 1=3,\ldots,k\quad\left(\xi_{k}^{\prime}\leq\xi\leq\xi_{k+1}^{\prime}\right)$
		$\displaystyle\mathrm{Q}\left(\xi_{1}^{\prime}\right)=\left(\xi-\xi_{1}^{\prime}\right)^{n}$

So ultimately, in all cases, identities (30) are proven. 22. We have the following two properties:
A. The point $\mathrm{M}_{n}$ whose coordinates are constructed using the quantities (25) corresponding to a convex function in (a, b) is always inside ( $\mathrm{W}_{n}$ B.
Every solution to our problem is the limit of a sequence of elementary solutions converging uniformly in $(\mathrm{a},\mathrm{b})$ .

To demonstrate property B, it suffices to consider convex solutions. Indeed, just as in No. 16, we demonstrate that every solution is the limit of a sequence of convex solutions converging uniformly in ( $a,b$ ).

I now say that property B is a consequence of property A. Property A is independent of the interval ( $a,b$ ) and we can then in any partial interval of ( $a,b$ ) replace the elementary function of degree $n$ non-concave of order $n$ concurring as well as its $n$ premières derivées avec celles de la fonction donnée aux extrémités. Divisons l’intervalle ( $a,b$ ) en $p$ parties égales et faisons la construction précédente dans chaque intervalle partiel. Comme la solution considérée est continue et est, ainsi que ses $n$ premières dérivées, bornée par les données du problème, de la propriété (29) résulte qu’à tout $\varepsilon>0$ correspond un $\eta$ tel que si $p>\eta$ la fonction élémentaire ainsi construite, qui est évidemment une solution, diffère de moin de s de la solution donnée, dans tout l’intervalle.

Considérons maintenant la projection $\mathrm{M}_{n-1}$ du point $\mathrm{M}_{n}$ . Soit $f(x)$ une solution et varions les valeurs $f(a),f(b)$ dans notre problème. L’expression

\int_{a}^{b}f^{\prime}(x)dx+\sum_{i=0}^{n}\frac{(b-a)^{i}}{i!}f^{(i)}(a)

a un maximum et un minimum. Il est clair que le maximum est égal à l’ $n^{\text{éme }}$ coordonnée du point $M_{n}^{\prime\prime}$ et le minimum à l’ $n^{\text{eme }}$ coordonnée de $M_{n}^{\prime}$ . Or le maximum est atteint pour la fonction $\Phi_{1}(x)$ et le minimum pour $\Phi_{2}(x)$ correspondant au problème relatif au point $M_{n-1}$ et par conséquence ces extrema ne sont atteints pour aucune autre solution correspondant à $\mathrm{M}_{n-1}$ .

Il en résulte que la propriété A est vraie et donc la propriété B est aussi vraie.

Nous avons ainsi démontré également que notre problème est impossible pour les points frontieres du type (II), (III) ou (IV) et admet une solution unique pour les points frontières du type (I). La solution est alors une fonction élémentaire de degré $n$ ayant $m\leq\left[\frac{n}{2}\right]$ sommets.

Nous avons considéré des points frontières tels que $M_{n}^{\prime}.M_{n}^{\prime\prime}$ se projetent sur un point intérieur de $\left(\mathrm{W}_{\mathrm{n}-1}\right)$ . Tous les autres se ramènent à cette forme par des projections succesives.

Remarquons encore que les conditions de possibilité du problème peuvent être exprimées par des inégalités explicites. Par une transformation facile on voit en effet que si on a

\sum_{l=0}^{n}\mu_{i}x^{i}\geqslant 0,\quad x\geqslant 0

on doit avoir

\sum_{i=0}^{n}\mu_{i}c_{i}^{\prime}\geqslant 0,\quad c_{i}^{\prime}=\sum_{r=0}^{i}(-1)^{r}\binom{i}{r}\frac{c_{n-i+r}}{\left(\begin{array}[]{ll}b&a\end{array}\right)^{n-i+r}}

On sait alors que les formes quadratiques

\sum_{i=0}^{\left[\frac{n}{2}\right]}\sum_{j=0}^{\left[\begin{array}[]{c}n\\ 2\end{array}\right]}c_{i+j}^{\prime}t_{i}t_{j},\quad\sum_{i=0}^{\left[\frac{n-1}{2}\right]}\sum_{j=0}^{\left[\frac{n-1}{2}\right]}c_{i+j+1}^{\prime}t_{i}t_{j}

doivent être positives. Le cas oû ces formes sont définies positives correspond aux points interieurs de $\left(\mathrm{W}_{n}\right)^{18}$ ).

Pour finir remarquons la liaisons étroite entre le problème du prolongement dans un intervalle et entre certains problèmes de moments. Le procédé par lequel nous établissons la condition de convexité restreinte rappelle d’ailleurs la définition d’une intégrale, mais tandis que dans les procédés d’intégration il s’agit toujours de certaines expressions ayant une limite, dans les problèmes des moments il suffit toujours que certaines sommes puisse ètre multipliées par des expressions de signe invariable telles que le produit ait une limite ¹⁹ ).

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Manuscrit reçu le 20 Avril, 1934.
¹⁸⁾ Voir Polya u. Szegõ „Aufgaben und Lehrsätze aus der Analysis" tome II p. 107. Cette propriété et la propriété projective des domaines $\left(\mathrm{W}_{n}\right)$ permettent de retrouver très simplement les formules donnant les coefficients de certaines formes quadratiques données par M. E. Fischer „Uber das Carathéodory’sche Problem Potenzreihen mit Positivem reellen Teil Bettreffend" Rendic. Circ t. 32 (1911) p. 240.
¹⁹ ) Les fonctions presentant des caractères de convexité de tout ordre et leur ilité pour le problème des moments ont été étudiées par M. S. Bernstein "Sur les

AL. PANTAZI. Sur les couples de congruences stratifiables par familles de surfaces réglées.
NICOLAS CIORĂNESCU. Sur une ralisant les sur une classe de polynomes à un paramètre généBARBILIAN. Zur Be he Legendre

Bewegungstheorie der Septuoren . . . . . . . . . . 29 superieur ${}^{\text{ICIU. Sur le prolongement des fonctions convexes d'ordre }}$ superieur fonctions absulument monotones" Acta Mathematica t. 52 (1928) p. 1.

On the extension of convex functions of higher order

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ON THE EXTENSION OF HIGHER-ORDER CONVEX FUNCTIONS

I.

Statement of the extension problem

II.

A study of some simple extension cases

III

Extension of convex functions in an interval.

IV

Case study of an extension

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