Notes on generalizations of higher-order convex functions (IV)

1.

In three notes on generalizations of higher-order convex functions ¹ ) we introduced, in particular, segment-monotone functions. It is first necessary to recall the main properties of these functions.

A function $f(x)$ , finite and uniform on the linear set $E$ is monotonic by segments on $E$ if we can decompose this set into a finite number of consecutive subsets

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

(1)

as on each $E_{i}$ , the function is monotone.
A segment-monotone function generally corresponds to an infinite number of decompositions (1). The number $m$ subsets of such a decomposition have a minimum $h$ . This number $h$ is the characteristic of the function $f(x)$ .

Among the decompositions (1) we have distinguished one that we have called the canonical decomposition of $E$ relative to the function $f(x)$ monotonic by segments. This decomposition

E_{1}^{*},E_{2}^{*},\ldots,E_{h}^{*},

(2)

exactly $h$ terms and is constructed as follows: $E_{1}^{*}$ is the subset of $E$ formed by all points $x$ s $E$ such as $f(x)$ be monotonic on the intersection of $E$ . with the closed interval ( $a,x$ ), Or $has$ is the left end of $E$ . The whole $E_{2}^{*}$ is deduced in the same way from $E-E_{1}^{*}$ and, in general, $E_{i+1}^{*}$ in the same way as $E-\left(E_{1}^{*}+E_{2}^{*}\right.+\ldots+E_{i}^{*}$ ).
2. Either $f(x)$ monotone by segments and (2) the corresponding canonical decomposition. Let us denote by $a_{i},b_{i}$ the ends (left and right) of $E_{i}^{\star}$ . We have $a_{i}<b_{i}$ unless $E_{i}^{\star}$ is formed by a single point and then, obviously, $a_{i}=b_{i}$ . Besides $a_{1}$ is the left end of $E$ And $b_{h}$ is the right end of $E$ .

We have defined a sequence of $2 hours$ proper or improper numbers

c_{1},c_{2},\ldots,c_{2h-1},c_{2h}

in the following manner: $c_{2i-1}=f\left(a_{i}\right)$ if $a_{i}\varepsilon E_{i}^{*}$ And $c_{2i-1}=\lim f(x)$ if $a_{i}$ does not belong to $E_{i}^{*}$ And $x$ tends towards $a_{i}$ on $E_{i}^{*}$ . Likewise $c_{2i}=f\left(b_{i}\right)$ if $b_{i}\varepsilon E_{i}^{*}$ And $c_{2i}=\lim f(x)$ if $b_{i}$ does not belong to $E_{i}^{*}$ And $x$ tends towards $b_{i}$ on $E_{i}^{*}$ .

Consider the sequence of differences
(3)

c_{2}-c_{1},c_{3}-c_{2},\ldots,c_{2h}-c_{2h-1}

and agree that

		$\displaystyle(+\infty)-u=u-(-\infty)=(+\infty)-(-\infty)>0,$
		$\displaystyle(-\infty)-u=u-(+\infty)=(-\infty)-(+\infty)<0,$
		$\displaystyle(+\infty)-(+\infty)=(-\infty)-(-\infty)=0,$

when $u$ is a finite number.
Then the terms of the sequence (3) are either zero or have a determined sign. The sequence (3) therefore presents a number $k$ of sign variations.
We say that the function $f(x)$ is of order ( $0\mid k$ ) on $E$ .
Consider a finite and ordered sequence $e=\left\{x_{1},x_{2},\ldots,x_{m}\right\}$ of $E$ , SO $x_{1},x_{2},\ldots,x_{m}\in E,x_{1}<x_{2}<\ldots<x_{m}$ . Let us pose, with the usual notations,

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

The sequel

\Delta_{1}^{1}(f),\Delta_{1}^{2}(f),\ldots,\Delta_{i}^{m-1}(f)

(4)

then presents at most $k$ variations of signs. More precisely the number $k$ enjoys the property that the maximum number of sign variations of (4), when considering the sequences (4) corresponding to all finite subsets $e$ of $E$ , is equal to $k$ .
3. A monotonic function by segments and of characteristic $h$ is at least of order ( $0\mid h-1$ ) and is at most of order ( $0\mid 2h-2$ ) ¹ ). There therefore exists $h$ types of characteristic functions $h$ . These are the functions of order respectively $(0\mid h-1),(0\mid h),(0\mid h+1),\ldots,(0\mid 2h-2)$ . It is easy to construct functions of each type.

Definition 1. We will say that a function $f(x)$ , monotonic by segments and of characteristic $h$ , is of the minimum type if it is of order ( $0\mid h-1$ ).

Let us study a little the functions of the minimum type. To do this, let us first demonstrate the

Lemma 1. For the function $f(x)$ is of the minimum type it is necessary. and it is sufficient that it is of the minimum type on each of the sets $E_{i}^{*}+E_{i+1}^{*}+\ldots+E_{j}^{*},j>i$ , which is equivalent to the

Lemma 1'. For the function $f(x)$ is of the minimum type it is necessary and sufficient that the sequence

c_{2i}-c_{2i-1},c_{2i+1}-c_{2i},\ldots,c_{2j}-c_{2j-1}

presents exactly $j-i$ variations of signs and this for $j=i+1$ , $i+2,\ldots,h,i=1,2,\ldots,h-1$ .

The condition is obviously necessary because we know that $E_{i}^{*},E_{i+1}^{*},\ldots E_{j}^{*}$ is the canonical decomposition of $E_{i}^{*}+E_{i+1}^{*}+\ldots+E_{i}^{*}$ . The condition is also sufficient. Indeed, the sequence (5) has at least $j-i$ variations. It cannot present more than $j-i$ variations because then the sequence (3) would present at least $h$ variations.

Now let us consider the following:

c_{2}-c_{1},c_{4}-c_{3},c_{6}-c_{5},\ldots,c_{2h}-c_{2h-1},

(6)

having $h$ terms. This sequence can have zero terms and terms of a given sign, in accordance with the conventions of No. 2. Two consecutive terms of the sequence (6) cannot be zero at the same time. Indeed, if $c_{2i}-c_{2i-1}=c_{2i+2}-c_{2i+1}=0$ , the function would be monotonic on $E_{i}^{*}+E_{i+1}^{*}$ which is impossible. Moreover

Lemma 2. If $f(x)$ is of the minimum type, a zero term (other than the first and the last) of the sequence (6) is always included between two terms of the same signs.

So, indeed, $c_{2i}-c_{2i-1}=0$ , SO
$\_\_\_\_$

c_{2i-2}-c_{2i-3}\neq 0,c_{2i+2}-c_{2i+1}\neq 0.

According to Lemma 1' the following

c_{2i-2}-c_{2i-3},c_{2i-1}-c_{2i-2},c_{2i}-c_{2i-1},c_{2i+1}-c_{2i},c_{2i+2}-c_{2i+1}

can only present one of the following arrangements of signs

+,-,0,-,+\quad-,+,0,+,-

and Lemma 2 is proved.
Note also that
Lemma 3. If $f(x)$ is of the minimum type two consecutive non-zero terms of the sequence (6) are always of opposite signs.

This is a consequence of Lemma 1'. More specifically, from the fact that the sequences

\begin{gathered}c_{2i}-c_{2i-1},c_{2i+1}-c_{2i},c_{2i+2}-c_{2i+1}\\ i=1,2,\ldots,h-1\end{gathered}

exhibit a variation of sign.
We can now state the following theorem
Theorem 1. For the function $f(x)$ , monotone by segments and of characteristic h, is of the minimum type it is necessary and sufficient that it is alternately non-decreasing and non-increasing on the sets $E_{1}^{*},E_{2}^{*},\ldots$ , $E_{h}^{*}$ of canonical decomposition.

The function is non-decreasing resp. non-increasing on $E_{i}^{*}$ following that $c_{2i}-c_{2i-1}$ is positive resp. negative. From this it immediately follows that the condition is sufficient. Lemmas 2 and 3 show us that it is also necessary. For the theorem to always be exact we assume, in accordance with the previous analysis, that a constant function (hence any function defined on a single point) is indifferently non-decreasing or non-increasing. If on $E_{i}^{*}$ the function is constant we consider it as non-decreasing or non-increasing depending on whether on $E_{i-1}^{*}$ And $E_{i+1}^{*}$ it is non-increasing or non-decreasing. In other words, we must replace the zero terms in the sequence (6) by signs such that this sequence presents the maximum possible number of sign variations.
4. We will now deal with the problem of the extension of monotonic functions by segments. We will examine, in particular, the extensions which preserve the order of the function.

Let us introduce the following definition
Definition 2. The function $f(x)$ of order ( $0\mid k$ ) on $E$ can be extended over the whole $E_{1}$ if we can find a function $f_{1}(x)$ of urdre ( $0\mid k$ ) on $E+E_{1}$ which coincides with $f(x)$ on $E$ .

An order function ( $0\mid k$ ) is not always extendable. For example, the function

f(x)=\begin{cases}-\frac{1}{x},&x\varepsilon(-\infty,0)\\ \frac{1}{x-2},&x\varepsilon(2,+\infty)\end{cases}

which is in order $(0\mid 1)$ , cannot be extended to any point in the interval [ 0,2 ].

Suppose that the function $f(x)$ be bounded on $E$ and let us examine a special extension of this function. Let us define the function $f_{1}(x)$ in the following manner:
$1^{\circ}.f_{1}(x)=f(x)$ on $E$ .
$2^{\circ}$ . At one point $x_{0}$ of $E^{\prime}$ which does not belong to $E$ we take for $f_{1}\left(x_{0}\right)$ one of the limit values (the left limit or the right limit) of $f(x)$ at this point.
$3^{\circ}$ . If $(\alpha,\beta)$ is a contiguous interval of the closure $\bar{E}$ of $E$ we take $f_{1}(x)$ linearly in the closed interval $[\alpha,\beta]$ . If $\alpha=-\infty$ Or $\beta=+\infty,f_{1}(x)$ is constantly equal to $f(\beta)$ Or $f(\alpha)$ In $(-\infty,\beta]$ Or $[\alpha,+\infty)$ .

I say that the function $f_{1}(x)$ same order as the function $f(x)$ . We can easily see that it is sufficient to demonstrate that the addition of an accumulation point to $E$ does not change the order and likewise the addition of an interval contiguous to $\bar{E}$ don't change the order.

If we add to $E$ the accumulation point $x_{0}$ ( $x_{0}$ not belonging to $E$ ), this point is added to a $E_{i}^{*}$ of canonical decomposition. The fact that $x_{0}=a_{i}$ means that $f_{1}\left(a_{i}\right)$ was defined as the right limit and the fact $x_{0}=b_{i}$ means that $f_{1}\left(b_{i}\right)$ was defined as the left limit. We then see that

E_{1}^{*},E_{2}^{*},\ldots,E_{i-1}^{*}\quad,E_{i}^{*}+x_{0},E_{i+1}^{*},\ldots,E_{h}^{*}

is a decomposition of $E+x_{0}$ . The sequence (3) corresponding to this decomposition is the same as the sequence (3) corresponding to the canonical decomposition of $E$ . The invariance of order results from this.

If we add to $\bar{E}$ a contiguous interval ( $\alpha,\beta$ ), two cases can arise. Or $\alpha,\beta$ both belong to the same subset $\bar{E}_{i}^{*}$ of the canonical decomposition of $\bar{E}$ and then the order obviously does not change. Or $\alpha$ belongs to a $\bar{E}_{i}^{\star}$ And $\beta$ has $\bar{E}_{i+1}^{*}$ . In this case $\alpha$ must be the right end of $\bar{E}_{i}^{*}$ And $\beta$ the left end of $\bar{E}_{i+1}^{*}$ . In this case, the canonical decomposition of $\bar{E}$ being ¹ ).
$\_\_\_\_$
(7)

\bar{E}_{2}^{*},\bar{E}_{2}^{*},\ldots,\bar{E}_{h}^{*},

$\left.{}^{1}\right)$ The characteristic of $f_{1}(x)$ on $\bar{E}$ is actually equal to that of $f(x)$ on $E$ .
For $\bar{E}+(\alpha,\beta)$ we have the decomposition ¹ )
(8) $\quad\bar{E}_{1}^{*},\bar{E}_{2}^{*},\ldots,\bar{E}_{i}^{*},(\alpha,\beta),\bar{E}_{i+1}^{*},\ldots,\bar{E}_{h}^{*}$ .

The sequence (3) corresponding to (8) differs from that corresponding to (7) only by the intercalation of two zero terms. The invariance of the order is still established.

We can therefore state
Theorem 2. Any function of order ( $0\mid k$ ) and bounded on $E$ is extendable (by conservation of order) on the entire real axis ( $-\infty,+\infty$ ).

This extension is carried out by the function $f_{1}(x)^{2}$ ).
The hypothesis that the function $f(x)$ is not necessary to be bounded. Let always $a,b$ the ends of $E$ and either $I_{E}$ the end interval $a,b$ and which is closed or open on the left or right depending on whether $a$ resp. $b$ belongs or not to $E$ . $\Pi$ It is easy to see that we then have the following property.

Theorem 3. Any function of order ( $0\mid k$ ) on $E$ and bounded on any completely interior subset. ${}^{\dot{a}}E^{3}$ ) is extendable (by preservation of order) throughout the entire interpalle $I_{E}$ .
5. Let's take a look at the functions defined in an interroom. $E$ .

Definition 3. We will say that the function $f(x)$ , monotonic by segments in the interval $E$ , is normal if the subsets $E_{i}^{*}$ of the canonical decomposition are all intervals of non-zero length.

For a normal function we therefore have
(9)

b_{1}<b_{2}<\ldots<b_{h-1}

THE $b_{i}$ always being the right ends of the sets $E_{i}^{*}$ .
The function is therefore monotonic in each of the intervals

\left(a,b_{1}\right),\left(b_{1},b_{2}\right),\ldots,\left(b_{h-1},b\right)

(10)

$a,b$ being the ends of $E$ .
Here we assume that:
$1^{\circ}$ . ( $a,b_{1}$ ) is closed or open on the left depending on whether $a$ belongs or not to $E$ .
$2^{\circ}$ . $\left(b_{h-1},b\right)$ is closed or open on the right depending on whether $b$ belongs or not to $E$ .

$3^{\circ}$ . Of two consecutive intervals $\left(b_{i-1},b_{i}\right)\left(b_{i},b_{i+1}\right)\left(b_{0}=a,b_{h}=b\right)$ if $b_{i}\varepsilon E_{i}^{*}$ the first is closed on the right and the second open on the left and if $b_{i}\varepsilon E_{i+1}^{*}$ the first is open on the right and the second closed on the left.

In this way (10) is precisely the canonical decomposition of $E$ .
If in addition the function is of the minimum type, this function is alternately non-decreasing and non-increasing in the intervals (10).

The points (9) can be called the nodes of the function $f(x)$ .
We still have
Theorem 4. Any function $f(x)$ , monotonous by segments and continuous in the interpalle $E$ , is normal and of the minimum type.

The demonstration is easy and there is no need to do it here. The nodes are points of relative maxima and minima. In particular, continuous functions with a finite number of relative extrema are segmentally monotone.

In the case of continuous functions the canonical decomposition is

\left(a,b_{1}\right],\left(b_{1},b_{2}\right],\left(b_{2},b_{3}\right],\ldots,\left(b_{h-1},b\right),

For $\left(a,b_{1}\right],\left(b_{h-1},b\right)$ the conventions $1^{\circ},2^{\circ}$ of the highest remaining valid and the other intervals being open on the left and closed on the right.
6. Let $f(x)$ a segmentally monotonic and normal function in the interval $E$ . Let (9) be the nodes of this function.

Consider an interval $I$ which is at the same time as $E$ open or closed at both ends and are

x_{1}<x_{2}<\ldots<x_{h-1},

(11)

$h-1$ points inside $I$ .
The interval $I$ can be, for example, the interval $E$ himself.
Either $\psi(x)$ a continuous and increasing function, defined in $\overline{\boldsymbol{E}}$ so that

\psi\left(b_{i}\right)=x_{i}\quad,\quad i=0,1,\ldots,h,

Or $b_{0},b_{h}$ are the ends of $E$ And $x_{0},x_{h}$ the ends

f^{*}(x)=f\left(\psi^{-1}(x)\right)\quad,\quad x\varepsilon I

is normal, of the same characteristic as $f(x)$ . The functions $f(x),f^{*}(x)$ are at the same time of the minimum type and at the same time continuous. We also have

f(x)=f^{*}(\psi(x))\quad;\quad x\varepsilon E.

7.

Let us return to any monotonic forictions by segments. Let us still denote by $a_{i},b_{i}$ the ends of $E_{i}^{*}$ .

Let us now establish a one-to-one correspondence $y=\chi(x)$ between $E(x\varepsilon E)$ and the subset $F(y\varepsilon F)$ of the interval [ $0,2h-1$ ] in the following manner:
$1^{\circ}$ . If $E_{i}^{*}$ is bounded and contains more than one point,

y=\frac{x+2(i-1)b_{i}-(2i-1)a_{i}}{b_{i}-a_{i}},x\varepsilon E_{i}^{*}

$2^{\circ}$ . If $E_{i}^{*}$ contains a single point, $y$ is equal to $2i-2$ for this value of $x$ .
$3^{\circ}$ . If $E_{1}^{*}$ is not limited $(a=-\infty)$ ,

y=\frac{1}{1+b_{1}-x},\quad x\in E_{1}^{*}

$4^{\circ}$ . If $E_{b}^{*}$ is not bounded ( $b=+\infty$ ),

y=2h-1-\frac{1}{x+1-a_{h}},\quad x\varepsilon E_{h}^{*}

The function $\chi(x)$ is uniform and increasing on $E$ and its inverse function $\chi^{-1}(x)$ is uniform and increasing on $F$ .

Let's define the function $g(x)$ on $F$ by the formula

g(x)=f\left(\chi^{-1}(x)\right),\quad x\varepsilon F

We then have

f(x)=g(\chi(x)),\quad x\varepsilon E

The correspondence between $E$ And $F$ establishes a correspondence between $E_{i}^{*}$ and a subset $F_{i}^{*}$ of $F$ .

We immediately see that $g(x)$ is segmentally monotone and the canonical decomposition of $F$ and precisely

F_{1}^{*},F_{2}^{*},\ldots,F_{h}^{*}

The numbers $c_{i}$ corresponding to the canonical decompositions of $E$ For $f(x)$ and of $F$ For $g(x)$ are the same. The functions $f(x),g(x)$ are of the same order $(0\mid k)$ and have the same characteristic $h$ .
8. Now suppose that the function $f(x)$ is limited, then $g(x)$ , which takes the same values, is also bounded. Consider the extended function $g_{1}(x)$ as we explained in No. 4. So $g_{1}(x)$ is, in the meantime $[0,2h-1]$ , of the same order $(0.\mid k)$ that the function $f(x)$ . Let us demonstrate more precisely that

Theorem 5. The function $g_{1}(x)$ is normal, of the minimum type and of order ( $0\mid k$ ) in the interroom $[0,2h-1]$ .

If we consider the canonical decomposition of $[0,2h-1]$ For $g_{1}(x)$ we see that each term of this decomposition must contain at least one of the intervals ( $i,i+1$ ), $i=0,1,\ldots,2h-2$ . The function $g_{1}(x)$ is therefore normal. Nodes can only be points $i=1,2,\ldots,2h-2$ . The point $i$ can only be a node if $\mathrm{g}_{1}(x)$ is of opposite monotony in the intervals $(i-1,i),(i,i+1)$ , as a result of the construction of the function $g_{1}(x)$ . Indeed, the function $g_{1}(x)$ is monotonic in the intervals $(i-1,i],[i,i+1)$ . The function $g_{1}(x)$ is therefore of the minimum type by virtue of Theorem 1.

Let's remember the formulas

	$\displaystyle g_{1}(x)$	$\displaystyle=f\left(\chi^{-1}(x)\right),$		$\displaystyle x\varepsilon F,$
	$\displaystyle f(x)$	$\displaystyle=g_{1}(\chi(x)),$		$\displaystyle x\varepsilon E.$

9.

A polynomial $P(x)$ is always segmentally monotone. It is therefore always normal, and of the minimum type. A polynomial of degree $h$ is at most of order ( $0\mid h-1$ ).

Theorem 6. If $P(x)$ is a polynomial of order $(0\mid k)$ And $\varphi(x)$ a non-decreasing function on $E$ , the function $P(\varphi(x))$ is monotone by segments and of order at most equal to ( $0\mid k$ ) on $E$ .

Indeed, otherwise, we can find $k+3$ points $x_{1}<x_{2}<\ldots<x_{k+3}$ of $E$ such as the following

\Delta_{1}^{1}(P(\varphi)),\Delta_{1}^{2}(P(\varphi)),\ldots,\Delta_{1}^{k+2}(P(\varphi))

presents $k+1$ variations. In this hypothesis $\Delta_{1}^{i}(P(\varphi))\neq 0$ , SO

\begin{gathered}\Delta_{1}^{i}(P(\varphi))=\frac{P\left(\varphi\left(x_{i+1}\right)\right)-P\left(\varphi\left(x_{i}\right)\right)}{\varphi\left(x_{i+1}\right)-\varphi\left(x_{i}\right)}\Delta_{1}^{i}(\varphi),\quad\Delta_{1}^{i}(\varphi)>0\\ i=1,2,\ldots,k+2\end{gathered}

This would result in $P(x)$ either of order ( $0\mid k+1$ ) at least, which is impossible.

The aim of this work is precisely to demonstrate the converse of this property.
10. We will first demonstrate the

Theorem 7. Any function $f(x)$ , segmentally monotonic, bounded; normal, of minimum type and characteristic $h$ in the meantime $E$ is of the form $Q(\varphi(x))$ , Or $Q(x)$ is a polynomial of degree h and $\varphi(x)$ a non-decreasing function in the interroom $E$ .

To demonstrate this property, let

b_{1}<b_{2}<\ldots<b_{h-1}

the nodes of the function $f(x)$ . The function is alternately non-decreasing and non-increasing in the intervals $E_{i}^{*}$ of the canonical decomposition. Moreover there exists a positive number $M$ such as

|f(x)|<M,\quad x\varepsilon E

We will now make a restrictive assumption. Suppose that we can find a polynomial $P(x)$ of degree $h$ having all its real zeros and included within the interval $E$ and having the nodes $b_{i}$ as zeros of its derivative $P^{\prime}(x)$ . The numbers

P\left(b_{1}\right),P\left(b_{2}\right),\ldots,P\left(b_{h-1}\right)

$2^{\circ}.\left|\lambda P\left(b_{i}\right)\right|>M,i=1,2,\ldots,h-1$ .
Let us pose $Q(x)=\lambda P(x)$ . The polynomial $Q(x)$ takes on $E_{i}^{*}(2\leqq i\leqq h-1)$ all values of $f(x)$ on $E_{i}^{*}$ and only once each of these values. If $x_{0}\varepsilon E_{i}^{*}$ we then take $x_{1}=\varphi\left(x_{0}\right),x_{1}\varepsilon E_{i}^{*}$ such as $Q\left(x_{1}\right)=f\left(x_{0}\right)$ . Likewise $Q(x)$ takes in $\left(-\infty,b_{1}\right)$ (closed or open on the right depending on whether $b_{1}$ belongs or not to $E_{i}^{\star}$ ) only once each value of $f(x)$ In $E_{i}^{*}$ . For a $x_{0}\varepsilon E_{1}^{*}$ we take again $x_{1}=\varphi\left(x_{0}\right),x_{1}\varepsilon\left(-\infty,b_{1}\right)$ such as $Q\left(x_{1}\right)=f\left(x_{0}\right)$ . We proceed in the same way in $E_{h}^{\star}$ .

We see that in this way if $x_{0}<x_{0}^{\prime}$ are two points of $E$ we have $\varphi\left(x_{1}\right)\leqq\varphi\left(x_{1}^{\prime}\right)$ . The function $\varphi(x)$ is therefore non-decreasing.

Let us now free ourselves from the restrictive hypothesis that the $b_{i}$ are the zeros of the derivative of the polynomial.

For this, let us consider a polynomial $P(x)$ of degree $h$ having all its real zeros included inside $E$ and be $x_{1}<x_{2}<\ldots<x_{h-1}$ the zeros of the derivative $P^{\prime}(x)$ . Now consider the function $\psi(x)$ defined in No. 6 ( $I\equiv E$ ) and either

f^{\star}(x)=f\left(\psi^{-1}(x)\right)\quad,\quad x\in E

We can then choose $Q(x)=\lambda P(x)$ and the non-decreasing function $\varphi_{1}(x)$ such as

We deduce from this

Q\left(\varphi_{1}(x)\right)=f^{*}(x)\quad,\quad x\varepsilon E

Q\left(\varphi_{1}(\psi(x))\right)=f^{\star}(\psi(x))=f(x),\quad x\varepsilon E

The function $\varphi(x)=\varphi_{1}(\psi(x))$ is still non-decreasing and Theorem 7 is completely proven.
11. We can now prove our fundamental theorem

Theorem 8. Any function $f(x)$ bounded and orderly ( $0\mid k$ ) on the whole $E$ is of the form $Q(\varphi(x))$ , Or $Q(x)$ is a polynomial of degree $k+1$ And $\varphi(x)$ a non-decreasing function on $E$ .

To demonstrate this property let us return to the functions $g(x),g_{1}(x)$ defined in No. 7. The function $g_{1}(x)$ satisfies the conditions of Theorem 7. This function being of characteristic $k+1$ we can find a polynomial $Q(x)$ of degree $k+1$ and a function $\varphi_{1}(x)$ non-decreasing in [ $0,2h-1$ ] such as

Q\left(\varphi_{1}(x)\right)=g_{1}(x)\quad,\quad x\varepsilon[0,2h-1]

But

g_{1}(x)=f\left(\chi^{-1}(x)\right)\quad,\quad x\in F

done

Q\left(\varphi_{1}(x)\right)=f\left(\chi^{-1}(x)\right)\quad,\quad x\in F

And

Q\left(\varphi_{1}(\chi(x))\right)=f(x)\quad,\quad x\varepsilon E

Just take $\varphi(x)=\varphi_{1}(\chi(x))$ and the theorem is proven.
12. The equation

Q(\varphi(x))=f(x)

is verified by an infinity of polynomials and an infinity of non-decreasing functions $\varphi(x)$ .

We can easily see that we can choose for $\varphi(x)$ a bounded function.

We could have demonstrated Theorem 8 directly without going through the transformations $y=\psi(x)$ And $y=\chi(x)$ and without using the extension, but we preferred to establish at the same time some properties of monotone functions by segments.

The assumption that the function is borried is not necessary. It is easy to see that Theorem 8 remains true for any function of order $(0\mid k)$ which is limited except perhaps in the vicinity of the ends of $E$ which do not belong to $E$ .

Thus the theorem applies, for example, to the function

f(x)=\left\{\begin{array}[]{cl}\frac{2x+1}{1+x}&,\quad x\in(-1,0)\\ 0&,\quad x=0\\ \frac{2x-1}{1-x}&,\quad x\in(0,1)\end{array}\right.

which is of order ( $0\mid 2$ ), but does not apply to the function

f(x)=\begin{cases}0&,\quad x=0\\ \frac{1}{x}&,\quad x\varepsilon(0,+\infty)\end{cases}

which is of order ( $0\mid 1$ ).
13. We will complete Theorem 8 a little, in the case where the function $f(x)$ is continuous. From our demonstration it does not follow that we can then choose the function $\varphi(x)$ continue on $E$ . We will demonstrate that

Theorem 9. Any function $f(x)$ continuous bounded and of order ( $0\mid k$ ) on $E$ is of the form $Q(p(x))$ , Or $Q(x)$ is a polynomial of degree $k+1$ And $\varphi(x)$ a continuous and non-decreasing function on $E^{1}$ ).

We can easily see that it is sufficient to demonstrate in the case where $E$ is reduced to an interval. Indeed, the extended function $g_{1}(x)$ is also continuous. In the case of an interval $f(x)$ is normal and of the minimum type.

Note, in passing, that Theorem 9 will remain true for certain unbounded functions, like Theorem 8 (see the previous No.).

Let us therefore prove Theorem 9 in the case where $E$ is an interval. If $b_{1}<b_{2}<\ldots<b_{k}$ are the nodes of the function $f(x)$ , by examining the proof of Theorem 7, we see that Theorem 9 is correct. if we can find a polynomial $Q(x)$ of degree $k+1$ such that we have

Q\left(b_{i}\right)=f\left(b_{i}\right),\quad Q^{\prime}\left(b_{i}\right)=0,\quad i=1,2,\ldots,k.

Now passing through the transformation of No. 6, we see that Theorem 9 will be completely proven if we prove the following theorem:

Theorem 10. Given $k$ Numbers $\lambda_{1},\lambda_{2},\ldots,\lambda_{k}$ , such as
or

\lambda_{1}>\lambda_{2},\lambda_{2}<\lambda_{3},\lambda_{3}>\lambda_{4},\ldots

—

$\quad\lambda_{1}<\lambda_{2},\lambda_{2}>\lambda_{3},\lambda_{3}<\lambda_{4},\ldots$
we can find a polynomial $P(x)$ of degree $k+1$ such that if $x_{1}<x_{2}<\ldots<x_{lk}$ are the zeros (assumed to be all real) of the derivation $P^{\prime}(x)$ , we have

P\left(x_{i}\right)=\lambda_{i},\quad i=1,2,\ldots,k.

14.

Let us first show that Theorem 10 follows from the following lemma: Lemma 4. Given $k-1$ positive numbers $p_{1},p_{2},\ldots,p_{k-1}$ we can always troop $k$ Numbers $x_{1}<x_{2}<\ldots<x_{k}$ such that we have

\int_{x_{i}}^{x_{i}+1}|R(x)|dx=\varrho p_{i}\quad,\quad i=1,2,\ldots,k-1,

Or $R(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\ldots\left(x-x_{k}\right)$ And $\varrho$ is a positive number.
¹ ) This is the answer to a question that was asked to me by Prof. S. Stoilo w. It is this problem that led me to undertake the study of functions of order $n$ by segments.

Indeed, let us take

p_{i}=\left|\lambda_{i}-\lambda_{i+1}\right|,\quad i=1,2,\ldots,k-1

and determine the polynomial $R(x)$ from Lemma 4. We have

\begin{gathered}\int_{x_{i}}^{x_{i+1}}|R(x)|dx=(-1)^{k-i}\int_{x_{i}}^{x_{i+1}}R(x)dx=e\mid\lambda_{i}-\lambda_{i+1}\\ i=1,2,\ldots,k-1\end{gathered}

Let us now take $\mu=\pm\frac{(-1)^{k-1}}{\varrho}$ following that $\lambda_{2}-\lambda_{1}$ is positive or negative. We have

\mu\int_{x_{i}}^{x_{i+1}}R(x)dx=\lambda_{i+1}-\lambda_{i},\quad i=1,2,\ldots,k-1

and the polynomial

P(x)=\mu\int_{x_{1}}^{x}R(x)dx+\lambda_{1}

verifies Theorem 10.
15. It remains to prove Lemma 4. To do this we will prove the following more general property, which is also of interest in itself,

Theorem 11. Given:
$1^{\circ}.k-1$ positive numbers $p_{1},p_{2},\ldots,p_{k-1}(k\geq 3)$ .
$2^{\circ}$ . A function $q(x)$ continuous and positive in the open interval ( $a,b$ ).
$3^{\circ}$ . Two points $x_{1},x_{k},x_{1}<x_{k}$ from the interroom $(a,b)$ .
We can always troop $k-2$ other points $x_{2},x_{3},\ldots,x_{k-1}$ such that we have:
I. $x_{1}<x_{2}<\ldots<x_{k}$ .
II.

A_{i}=\int_{x_{i}}^{x_{i+1}}q(x)|R(x)|dx=\varrho p_{i},\quad i=1,2,\ldots,k-1

Or $R(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\ldots\left(x-x_{l}\right)$ And $\varrho$ is a positive number.
Lemma 4 is obtained by taking $q(x)=1$ And $x_{1},x_{k}$ arbitrarily $\left(x_{1}<x_{k}\right)$ .

Let us now consider Theorem 11.
Theorem 11 is an existence theorem. We will attach a uniqueness theorem and a continuity theorem to it.

The uniqueness theorem is as follows:
Theorem 12. If the positive numbers $p_{1},p_{2},\ldots,p_{k-1}$ , the function $q(x)$ positive and continuous in the interval $(a,b)$ and the points $x_{1},x_{k},x_{1}<x_{k}$ of $(a,b)$ are given, there is only one system of $k-2$ Numbers $x_{2},x_{3},\ldots$ , $x_{k-1}$ such that conditions I and II of Theorem 11 are verified.

The continuity theorem is stated as follows:
Theorem 13. Let $p_{1},p_{2},\ldots,p_{k-1},k-1$ positive numbers and $q(x)$ a positive continuous function in the interpalle ( $a,b$ ). Let's take $k$ points $x_{1},x_{2},\ldots,x_{k}$ of the interval ( $a,b$ ) verifying conditions I and II of Theorem 11. Then if $x_{1}$ remains fixed, $x_{2},x_{3},\ldots,x_{k-1}$ are continuous functions of $x_{k}$ and have continuous derivatives with respect to $x_{k}$ in the interroom ( $x_{1},b$ ).

To demonstrate, we will proceed by complete induction by demonstrating the following lemmas:

Lemma 5. Continuity Theorem 13 is a consequence of Uniqueness Theorem 12.

Lemma 6. Theorems 11 and 12 are taken for $k=3$ .
Lemma 7. The existence theorem and the uniqueness theorem for $k+1$ are consequences of Theorems 11, 12 and 13 for $k$ .

The theorems $11,12,13$ are then completely demonstrated. Indeed, if we assume that they are true for $k$ we deduce, from Lemma 7, that Theorems 11 and 12 are true for $k+1$ . Theorem 13 for $k+1$ then results from Lemma 5.

For $k=3$ Theorem 13 follows from Lemmas 5 and 7.
It remains to prove Lemmas 5, 6 and 7.
16. First part of the proof of Lemma 5. In both Theorems 12 and 13 we assume, of course, that these are the same numbers $p_{i}$ and of the same function $q(x)$ .

To demonstrate continuity of functions $x_{2},x_{3},\ldots,x_{k-1}$ compared to $x_{k}$ it is necessary and sufficient to demonstrate that if $x_{1}<x_{2}<\ldots<x_{k}$ is a solution to the problem expressed by Theorem 11 and if

x_{1}=x_{1}^{(n)}<x_{2}^{(n)}<\ldots<x_{k}^{(n)},n=1,2,\ldots

(11)

is an infinite sequence of solutions to the problem such that

\lim_{n\rightarrow\infty}x_{k}^{(n)}=x_{k}

And

\lim_{n\rightarrow\infty}x_{1}^{(n)}=x_{i}^{\prime},i=1,2,\ldots,k-1,\left(x_{i}^{\prime}=x_{2}\right)

we have

x_{i}^{\prime}=x_{i},\quad i=2,3,\ldots,k-1

(12)

To do this, it is sufficient to demonstrate that

\text{ (13) }\quad x_{1}<x_{2}^{\prime}<\ldots<x_{k-1}^{\prime}<x

Indeed, if $\varrho^{(n)}$ are the numbers $\varrho$ corresponding to solutions (11), since $A_{1},A_{2},\ldots,A_{k-1}$ are continuous functions of $x_{2},x_{3},\ldots$ , $x_{k}$ , it follows that if we have (13) $\varrho^{(n)}$ tends towards a positive limit for $n\rightarrow\infty$ . So (13) is a solution to the problem and from Theorem 12 it follows that we have (12).

It remains to show that we have the inequalities (13), therefore that two consecutive points $x_{i}^{\prime}$ cannot coincide. This easily follows from the following lemma:

Lemma 8. If $x_{1}<x_{2}<\ldots<x_{k}$ is a solution to the problem expressed by conditions I and II of Theorem 11 and if $a<a_{1}\leqq x_{1}<x_{k}\leqq b_{1}<b$ , we adopt

x_{i+1}-x_{i}>\mu\left(x_{k}-x_{1}\right)^{k+1},i=1,2,\ldots,k-1

Or $\mu$ is a number independent of the points $x_{i}$ . ( $\mu$ depends only on $a_{1},b_{1}$ numbers $p_{i}$ and function $q(x)$ ).

Let us note, in fact, that

\begin{gathered}\frac{\min\left(p_{1},p_{2},\ldots,p_{k-1}\right)}{p_{1}+p_{2}+\ldots+p_{k-1}}=p>0\\ m_{1}=\max_{\left[a_{1},b_{1}\right]}q(x)>0,\quad m_{2}=\min_{\left[a_{1},b_{1}\right]}q(x)>0\\ |R(x)|<\left(b_{1}-a_{1}\right)^{k},\quad x\varepsilon\left(a_{1},b_{1}\right)\\ \int_{x_{1}}^{x_{k}}\left\lvert\,R\left(x\left\lvert\,dx\geqq\frac{\left(x_{k}-x_{1}\right)^{k+1}}{2^{2k}}{}^{1}\right.\right)\right.\end{gathered}

It follows that

	$\displaystyle p\leqq$	$\displaystyle\frac{p_{i}}{p_{1}+p_{2}+\ldots+p_{k-1}}=$
		$\displaystyle=\frac{\int_{x_{i}}^{x_{i+1}}q(x)\|R(x)\|dx}{\int_{x_{1}}^{x_{k}}q(x)\|R(x)\|dx}<\frac{2^{2k}m_{1}\left(b_{1}-a_{1}\right)^{k}\left(x_{i+1}-x_{i}\right)}{m_{2}\left(x_{k}-x_{1}\right)^{k+1}}$

¹⁾ This minimum is obtained by looking for the polynomial $P(x)=x^{k}+\ldots$ , of degree $k$ , minimizing the integral

\int_{x_{1}}^{x_{k}}|P(x)|dx

See: Mr. Fujiwara. Uber die Polynome von der kleinsten totalen Schwankung, The Tôhoku Math. Journal, 3, 129-136 (1913).

So just take

\mu=\frac{m_{2}p}{2^{2k}m_{1}\left(b_{1}-a_{1}\right)^{k}}

17.

Second part of the proof of Lemma 5. We will show that derivability is still a consequence of continuity and uniqueness, hence of uniqueness alone.

With our previous notations, the functions $x_{2},x_{3},\ldots,x_{k-1}$ are given by the system of equations
(14)

B_{i}=p_{1}A_{i}-p_{i}A_{1}=0,i=2,3,\ldots,k-1

THE $A_{i}$ , therefore also the first members of equations (14), are continuous and admit continuous partial derivatives with respect to the variables $x_{2},x_{3},\ldots,x_{k}$ , as is easily verified.

Let us first prove the following more general property:
Lemma 9. If:
$1^{\circ}.F_{i}\left(x,y_{1},y_{2},\ldots,y_{n}\right),i=1,2,\ldots,n$ are $n$ continuous functions and admitting continuous partial derivatives with respect to the pariables $x$ , $y_{1},y_{2},\ldots,y_{n}$ in the field

a<x<b\quad,\quad a<y_{i}<b\quad,\quad i=1,2,\ldots,n.

$2^{\circ}$ . At each $x_{0}\varepsilon(a,b)$ corresponds to a unique solution $y_{1}^{0},y_{2}^{0},\ldots,y_{n}^{0}$ of the system

F_{i}\left(x_{0},y_{1},y_{2},\ldots,y_{n}\right)=0,i=1,2,\ldots,n

(15)

$3^{\circ}$ . The functions $y_{i}(x),i=1,2,\ldots,n$ determined by the system (15) are continuous in ( $a,b$ ).
$4^{\circ}$ . We apons

\left[\frac{D\left(F_{1},F_{2},\ldots,F_{n}\right)}{D\left(y_{1},y_{2},\ldots,y_{n}\right)}\right]_{x=x_{0}}\neq 0

So the functions $y_{i}(x)$ are differentiable at the point $x_{0}$ and these drifts are continuous in any interval where the condition $4^{\circ}$ is perified.

The proof of this lemma follows easily from the fact that if we pose $x=x_{0}+\Delta x$ And

\Delta y_{i}=y_{i}\left(x_{0}+\Delta x\right)-y_{i}\left(x_{0}\right),i=1,2,\ldots,n,

as a result of the uniqueness, the solution of the system

F_{i}\left(x_{0}+\Delta x,y_{1},y_{2},\ldots,y_{n}\right)=0,i=1,2,\ldots,n

is precisely

y_{i}=y_{i}^{0}+\Delta y_{i},i=1,2,\ldots,n

We complete the demonstration, as usual, by applying, for example, the formula for finite increments and noting that, as a result of continuity, $\Delta y_{i}\rightarrow 0$ For $\Delta x\rightarrow 0,i=1,2,\ldots,n$ .

It follows that for Lemma 5 to be completely proven it is sufficient to show that the functional determinant

\frac{D\left(B_{2},B_{3},\ldots,B_{k-1}\right)}{D\left(x_{2},x_{3},\ldots,x_{k-1}\right)}

is different from zeros, $x_{1}<x_{2}<\ldots<x_{k}$ , being a solution to problem 11.
18. Third and last part of the proof of lemma 5. Let us set
(16) $\quad a_{i,j}=\left\{\begin{aligned} -\frac{\partial A_{i}}{\partial x_{j}}&,j=2,3,\ldots,i\text{ (aucun si }i=1\text{ ), }\\ \frac{\partial A_{i}}{\partial x_{j}}&,j=i+1,i+2,\ldots,k-1\text{ (aucun si }i=k-1\text{ ). }\end{aligned}\right.$

We then have

\frac{D\left(B_{2},B_{3},\ldots,B_{k-1}\right)}{D\left(x_{2},x_{3},\ldots,x_{k-1}\right)}=

(17)

$=p_{1}^{k-3}\left|\begin{array}[]{ccccccc}p_{1}&\alpha_{1,2}&\alpha_{1,3}&\alpha_{1,4}&\ldots&\alpha_{1,k-2}&\alpha_{1,k-1}\\ p_{2}-\alpha_{2,2}&\alpha_{2,3}&\alpha_{2,4}&\ldots&\alpha_{2,k-2}&\alpha_{2,k-1}\\ p_{3}-\alpha_{3,2}-\alpha_{3,3}&\alpha_{3,4}&\ldots&\alpha_{3,k-2}&\alpha_{3,k-1}\\ \ldots&.&.&.&.&.&.\\ p_{k-2}-\alpha_{k-2,2}-\alpha_{k-2,3}-\alpha_{k-2,4}&\ldots-\alpha_{k-2,k-2}&\alpha_{k-2,k-1}\\ p_{k-1}-\alpha_{k-1,2}-\alpha_{k-1,3}-\alpha_{k-1,4}&\ldots-\alpha_{k-1,k-2}-\alpha_{k-1,k-1}&.\end{array}\right|$ .
But, a simple calculation shows us that
(18) $\alpha_{i,j}=\int_{x_{i}}^{x_{i+1}}q(x)\left|\frac{R(x)}{x-x_{i}}\right|dx,j=2,3,\ldots,k-1,i=1,2,\ldots,k-1$ and we check, in passing, that the partial derivatives are continuous with respect to the variables $x_{2},x_{3},\ldots,x_{k}$ .

We see that we have the following inequalities:

\left\{\begin{array}[]{r}\alpha_{i,j}>0,j=2,3,\ldots,k-1,i=1,2,\ldots,k-1\\ \alpha_{i,i+1}>\alpha_{i,i+2}>\ldots>\alpha_{i,k-1},i=1,2,\ldots,k-3\\ \alpha_{i,2}<\alpha_{i,3}<\ldots<\alpha_{i,i},i=3,4,\ldots,k-1\end{array}\right.

The determinant of the second member of formula (17) can be written (with the sign $(-1)^{k-2}$ close),

\left|\begin{array}[]{cccccc}c_{1,1}-c_{1,2}-c_{1,3}&\ldots&&-c_{1,k-2}&p_{1}\\ -c_{2,1}c_{2,2}-c_{2,3}&\ldots&&-c_{2,k-2}&p_{2}\\ \ldots\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\ -c_{k-2,1}-c_{k-2,2}-c_{k-2,2}\ldots&c_{k-2,k-3}&c_{k-2,k-2}&p_{k-2}\\ -c_{1}-c_{2}-c_{3}\ldots&\ldots&-c_{k-2}&p_{k-1}\end{array}\right|

where the $p_{i},c_{i,j},c_{i}$ are positive and

	$\displaystyle c_{i,i}\geqq c_{i,1}+c_{i,2}+\ldots+c_{i,i-1}+c_{i,i+1}+\ldots+c_{i,k-2}$		(21)
	$\displaystyle i=1,2,\ldots,k-2$

(here, moreover, the signs $>$ are valid). We obtain this result by successively subtracting the elements of one column from those of the following column and taking into account the inequalities (19).

The property we are looking for is then a special case of the following more general property:

Lemma 10. If the numbers $p_{i},c_{j,j},c_{i}$ are positive and if the inequalities (21) are perified, the determinant (20) is positive ¹ ).

The demonstration of this property is very simple and can be done by induction. For $k=3$ the property is immediate. Let us assume it to be true for $k-1$ and let's demonstrate it for $k$ . Consider the determinant (20) as a function of $c_{1,1},c_{2,2},\ldots,c_{k-2,k-2}$ . The coefficient of $c_{i,i}$ is an analogous determinant of order $k-2$ , so is positive. Since the determinant is a linear function of $c_{i,i}$ it remains to be demonstrated that this determinant is positive if we take the sign $=$ in all formulas (21). But, in this case, it reduces to the product of $(-1)^{k-1}\sum_{i=i}^{k-2}c_{i}$ by the minor of the last element of the first column. This minor is, at the sign $(-1)^{k-3}$ near, an analogous determinant of order $k-2$ . Lemma 10 is therefore proven.

The property still remains true in one of the cases:
$1^{\circ}.c_{i}\geq 0,i=1,2,\ldots,k-2,\sum_{i=1}^{k-2}c_{i}>0$ .
¹⁾ This property is analogous to a well-known lemma of H. Minkowski. Moreover, the minor of the last element is precisely a Minkowski determinant (if the signs $>$ are valid in (21)). See also: G. Scorza, A proposito di un lemma di Minkowski, Boll. Un. Mat. It., 5, 229-231 (1926).
$2^{\circ}.c_{i}=0,i=1,2,\ldots,k-2$ , provided that in (21) at least one of the inequalities $>$ be valid.
$\left.3^{\circ}.c_{i,j}>0,c_{i}>0,p_{k-1}=0,p_{i}>0,i=1,2,\ldots,k-2^{1}\right)$ .
19. Before moving on to the proof of Lemmas 6 and 7, we will draw one more conclusion from Lemma 5.

Lemma 11. Expressions $A_{1},A_{2},\ldots,A_{k-1}$ regarded as functions of $x_{k}$ are continuous and increasing.

Indeed, let us substitute in $A_{1},A_{2},\ldots,A_{k-1}$ differentiable functions $x_{2},x_{3},\ldots,x_{k-1}$ by their values as a function of $x_{k}$ .

The definition system (1.4) shows us that

p_{1}\frac{dA_{i}}{dx_{k}}=p_{i}\frac{dA_{1}}{dx_{le}},i=2,3,\ldots,k-1

(22)

It follows that one of the derivatives $\frac{dA_{i}}{dx_{ic}}$ cannot be cancelled without all being zero. But this last case is impossible. Indeed, if

\begin{gathered}\frac{dA_{i}}{dx_{k}}=\frac{\partial A_{i}}{\partial x_{2}}\cdot\frac{dx_{2}}{dx_{k}}+\frac{\partial A_{i}}{\partial x_{3}}\cdot\frac{dx_{3}}{dx_{k}}+\ldots+\frac{\partial A_{i}}{\partial x_{k-1}}\cdot\frac{dx_{k-1}}{dx_{k}}+\frac{\partial A_{i}}{\partial x_{k}}=0\\ i=1,2,\ldots,k-1\end{gathered}

it would be necessary that

\left|\begin{array}[]{ccccc}\alpha_{1,2}&\alpha_{1,3}&\cdots&\alpha_{1,k-1}&\alpha_{1,k}\\ -\alpha_{2,2}&\alpha_{2,3}&\cdots&\alpha_{2,k-1}&\alpha_{2,k}\\ \cdot\cdot\cdot&\cdot&\cdots&\cdot&\cdot\\ -\alpha_{k-1,2}-\alpha_{k-1,3}&\cdots&-\alpha_{k-1,k-1}&\alpha_{k-1,k}\end{array}\right|=0

by extending notations (16) and (18) to $j=k$ .
But the inequalities (19) are completed by the following:
$a_{i,k}>0,i=1,2,\ldots,k-1,a_{i,k-1}>a_{i,k},i=1,2,\ldots,k-2$
Then the first member of (23) transforms, as above, into a positive determinant (Minkowski determinant).

From (22) it follows that the derivatives $\frac{dA_{i}}{dx_{k}}$ are always of the same sign. It follows that the functions $A_{i}$ are strictly monotone so, as is almost obvious, are increasing and Lemma 11 is proven.

20. Proof of Lemma 6. For $k=3$ just look at the report $\frac{A_{1}}{A_{2}}$ . We can easily verify that $A_{1}$ is a function of $x_{2}$ continuous, positive and growing in $\left(x_{1},x_{3}\right)$ And $A_{2}$ is a function of $x_{2}$ continuous, positive and decreasing in $\left(x_{1},x_{3}\right)$ . It follows that the report $\frac{A_{1}}{A_{2}}$ is continuous, positive and increasing. But this ratio tends towards 0 if $x_{2}\rightarrow x_{1}$ and tends towards $+\infty$ if $x_{2}\rightarrow x_{3}$ . Theorems 11 and 12 result from this.
21. First part of the proof of Lemma 7.

Suppose that Theorems 11, 12 and 13 are true for $k(k\geq 3)$ . Let us consider the points $x_{1},x_{k+1},a<x_{1}<x_{k+1}<b$ , be $p_{1},p_{2},\ldots$ , $p_{k}k$ positive numbers and $q(x)$ a continuous and positive function in $(a,b)$ . If we take a point $x_{k},x_{1}<x_{k}<x_{k+1}$ we can determine the points $x_{2}<x_{3}<\ldots<x_{k-1}$ between $x_{1}$ , $x_{k}$ such as

A_{i}=\int_{x_{i}}^{x_{i+1}}q_{1}(x)|R(x)|dx=\varrho p_{i},\quad i=1,2,\ldots,k-1

Or $\quad R(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\ldots\left(x-x_{k}\right)$ And $q_{1}(x)=q(x)\left(x_{k+1}-x\right)$ .
The sum $A_{1}+A_{2}+\ldots+A_{k-1}$ is a continuous positive and increasing function of $x_{k}$ in the meantime ( $x_{1},x_{k+1}$ ). This function tends to 0 when $x_{k}\rightarrow x_{1}$ .

Now consider the expression

A_{k}=\int_{x_{k}}^{x_{k+1}}q_{1}(x)|R(x)|dx=\int_{x_{k}}^{x_{k+1}}q(x)\left(x_{k+1}-x\right)|R(x)|dx

It is also a continuous and differentiable function of $x_{k}$ . Let's calculate its derivative $\frac{dA_{k}}{dx_{k}}$ compared to $x_{k}$ . We have

\frac{dA_{k}}{dx_{k}}=\frac{\partial A_{k}}{\partial x_{k}}+\frac{\partial A_{k}}{\partial x_{2}}\cdot\frac{dx_{2}}{dx_{k}}+\frac{\partial A_{k}}{\partial x_{3}}\cdot\frac{dx_{3}}{dx_{k}}+\ldots+\frac{\partial A_{k}}{\partial x_{k-1}}\cdot\frac{dx_{k-1}}{dx_{k}}

Using notation (14) we have (here we assume, of course, that $q(x)$ is replaced by $q_{1}(x)$ ),
(24) $\quad\frac{dA_{k}}{dx_{k}}\cdot\frac{D\left(B_{2},B_{3},\ldots,B_{k-1}\right)}{D\left(x_{2},x_{3},\ldots,x_{k-1}\right)}=\frac{D\left(B_{2},B_{3},\ldots,B_{k-1},A_{k}\right)}{D\left(x_{2},x_{3},\ldots,x_{k-1},x_{k}\right)}$ .

We know that

(-1)^{k-2}\frac{D\left(B_{2},B_{3},\ldots,B_{k-1}\right)}{D\left(x_{2},x_{3},\ldots,x_{k-1}\right)}>0

The determinant of the second member of formula (24) is written

p_{1}^{k-3}\left|\begin{array}[]{ccccc}p_{1}&\alpha_{1,2}&\alpha_{1,3}&\ldots&\alpha_{1;k}\\ p_{2}&-\alpha_{2,2}&\alpha_{2,3}&\ldots&\alpha_{2,k}\\ \ldots&\ldots&\ldots&\ldots&\ldots\\ p_{k-1}-\alpha_{k-1,2}-\alpha_{k-1,3}&\ldots&\alpha_{k-1,k}\\ 0&-\alpha_{k,2}&-\alpha_{k,3}&\ldots&-\alpha_{k,k}\end{array}\right|

still using the notations (16) for $j=2,3,\ldots,i$ And $j=i+1$ , $i+2,\ldots,k$ respectively and also $i=1,2,\ldots,k$ (taking $q_{1}(x)$ instead of $q(x)$ ). We still have

		$\displaystyle\alpha_{i,i+1}>\alpha_{i,i+2}>\ldots>\alpha_{i,k}\quad,\quad i=2,\ldots,k-2$
		$\displaystyle\alpha_{i,2}<\alpha_{i,3}<\ldots<\alpha_{i,i}\quad,\quad i=4,\ldots,k$

It follows, as above, that the determinant has the sign of $(-1)^{k-1}$ . From (24) it follows that

\frac{dA_{k}}{dx_{k}}<0

The function $A_{k}$ is therefore decreasing. It obviously tends towards zero for $x_{k}\rightarrow x_{k+1}$ .

The report

\frac{A_{1}+A_{2}+\ldots+A_{k-1}}{A_{k}}

(25)

is therefore a continuous positive and increasing function of $x_{k}$ In ( $x_{1},x_{k+1}$ ) and tends towards 0 for $x_{k}\rightarrow x_{1}$ and towards $+\infty$ For $x_{k}\rightarrow x_{k+1}$ . Existence theorem 11 follows for $k+1$ .
22. Second and last part of the proof of lemma 7. From the above it follows that the ratio (25) reaches the value

\frac{p_{1}+p_{2}+\ldots+p_{k-1}}{p_{k}}

for a single value of $x_{k}$ . So if

		$\displaystyle x_{1}<x_{2}<\ldots<x_{k-1}<x_{k}<x_{k+1}$
		$\displaystyle x_{1}<x_{2}^{\prime}<\ldots<x_{k-1}^{\prime}<x_{k}<x_{k+1}$

would be two solutions to the problem, the

		$\displaystyle x_{1}<x_{2}<\ldots<x_{k-1}<x_{k}$
		$\displaystyle x_{1}<x_{2}^{\prime}<\ldots<x_{k-1}^{\prime}<x_{k}$

would constitute two solutions to the problem for $k$ (taking $q_{1}(x)$ instead of $q(x)$ ). But, as a result of the uniqueness,

x_{i}^{\prime}=x_{i},\quad i=2,3,\ldots,k-1,

which demonstrates the property.
23. By the above, Theorem 9 is completely demonstrated.

Finally, let's make an application.
A function $f(x)$ is said to be of order $n$ on $E$ if its difference divided by order $n+1,\left[x_{1},x_{2},\ldots,x_{n+2};f\right]$ , does not change sign on $E$ . Such a function is always continuous on any subset completely interior to $E$ and can only be unbounded in the neighborhoods of the ends of $E$ which do not belong to $E$ . An order function $n$ is always segmentally monotone and of the minimum type. It is at most of order $(0\mid n)$ . So we have the following property:

Theorem 14. Any function of order n on $E$ is of the form $Q(\varphi(x))$ , Or $Q(x)$ is a polynomial of degree at most equal to $n+1$ And $\varphi(x)$ a non-decreasing function on $E$ , continues on any subset completely interior to $E$ .

If $f(x)$ is defined in an interval and is convex of order $n$ we can even choose an increasing function $\varphi(x)$ .

Bucharest, April 27, 1942.

Notes on generalizations of higher-order convex functions (IV)

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Notes on generalizations of higher-order convex functions (IV)

ON A REPRESENTATION OF MONOTONE FUNCTIONS BY SEGMENTS

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