Notes on higher order convex functions (VII)

Note presented by Mr. S. Stoilov, MC. AR, in the session of July 14, 1939.

I. A function$f=f(x)$, defined and uniform on any linear set E is said to be convex, non-concave, polynomial, non-convex, or concave of order$n$following that the inequality

is satisfied whatever$\left.x_{1},x_{2},\ldots,x_{n+2}\in E^{1}\right)$.
All these functions are order functions$n$.
We will say that the set E is decomposed into$m$consecutive subsets

if$\mathrm{E}_{i}\subset\mathrm{E},i=1,2,\ldots,m$, if every point of E belongs to an E and if every point of$\mathrm{E}_{i}$is to the left of any point of$\mathrm{E}_{i+1},i=1,2,\ldots,m-1$.

In our Thesis ² ) we have demonstrated the following property:
Theorem I. If$f$is in order$n$on E, we can decompose this set into at most$k+1$consecutive subsets on each the function being of order$nk$.

The property is true for$k=I,2,\ldots,n+I$.
This property follows from
Theorem II. If$f$is in order$n$and if
(I)

is any tini subset of E , the sequence
(2)

present at most$k$variations of signs. Here we have posed
(3)$\Delta_{k}^{i}=\left[x_{i},x_{i+1},\ldots,x_{i+k};f\right],i=\mathrm{I},2,\ldots,mk,k=0,\mathrm{I},\ldots,m-\mathrm{I}$

In this note we propose to give a converse of Theorem II.
2. Before going further let us make some remarks on the order functions$n$. Using notation (3), any divided difference of order$n+I$of$f$on$n+2$points, chosen from points (I), is an arithmetic mean of the divided differences$\Delta_{n+1}^{1},\Delta_{n+1}^{2},\ldots,\Delta_{n+1}^{mn-1}$, SO

THE$\mathrm{A}_{i}$being independent of the function$f$. We have surely
$\mathrm{A}_{i}>0,\mathrm{\penalty 10000\ A}_{i_{n+2}-n-\mathrm{I}}>0$if$i_{1}<i_{2}<\ldots<i_{n+2}$.
We immediately deduce
Lemma I. The necessary and sufficient condition for the function$f$, defined on the finite set (1), either convex, non-concave, polynomial, nonconvex or concave of order$n$is that we have

From this lemma immediately follows Theorem II.
Let us prove Lemma II again
. For the function$f$either of order$n$on E having at least$n+3$points it is necessary and sufficient that one has

whatever the points$x_{1}<x_{2}<\ldots<x_{n+3}$of E.
The condition is obviously necessary. Let us show that it is also sufficient. It is sufficient to show that the property is not true for a function which is not of order$n$, it is therefore sufficient to demonstrate the

Lemma III. If the function f is not of order$n$on E we can find$n+3$points$x_{1}<x_{2}<\ldots<x_{n+3}$of E such that we have

The fact that$f$is not in order$n$means that we can find$n+2$points$\alpha_{1},\alpha_{2},\ldots,\sigma_{n+2}$of E such that$\left[\alpha_{1},\alpha_{2},\ldots,\alpha_{n+2};f\right]>0$And$n+2$points$\beta_{1}\beta_{2},\ldots,\beta_{n+2}$of E such that$\left[\beta_{1},\beta_{2},\ldots,\beta_{n+2};f\right]<0$Let's put all the separate points in order.$o_{i},\beta_{i}$in an increasing sequence (I). We then have$n+3\leq m\leq 2n+2$. The continuation of the divided differences

has at least one sign variation, therefore contains at least two non-zero terms with opposite signs. Let$\Delta_{n+1}^{p}$the first non-zero term in (4) and$\Delta_{n+1}^{s}$the first non-zero term and of opposite sign with$\Delta_{n+1}^{p}$. Finally, be it$\Delta_{n+1}^{r}$the last non-zero term in the sequence$\Delta_{n+1}^{1},\Delta_{n+1}^{2}$, ,$\Delta_{n+1}^{s-1}$. We then have$\Delta_{n+1}^{r}.\Delta_{n+1}^{s}<0$And$r<s$. If$s=r+1$inequality$\Delta_{n+1}^{r}\Delta_{n+1}^{r+1}<\mathrm{o}$proves Lemma III. If$s>r+\mathrm{I}$we have$\Delta_{n+1}^{r+1}=\Delta_{n+1}^{r+2}=\ldots=\Delta_{n+1}^{s-1}=0$and inequality
$\left[x_{r},x_{r+1},\ldots,x_{r+n},x_{s+n};f\right]\left[x_{r+1},x_{r+2},\ldots,x_{r+n},x_{s+n},x_{s+n+1};f\right]<0$proves Lemma III.
3. Let us now establish the converse of Theorem II. To do this, let us note that$\left[x_{1},x_{2},\ldots,x_{n+2};\mathrm{P}\right]=\mathrm{o}$identically if P is a polynomial of degree$n^{1}$). From this results this important property that any function$f-\mathrm{P}$, where P is a polynomial of degree$n$, enjoys the same property as$f$with respect to any convexity character of order$\geq n$. We then have the

Theorem III. If, whatever the polynomial P of degree$n$and the finite subset ( I ) of E , the sequence (2) corresponding to ( I ) and to the function$f-\mathrm{P}$present at most$k$variations of signs, the function$f$is in order$n$on E.

It is enough to demonstrate that if the function is not of order$n$, we can find a sequence (I) of E and a polynomial P such that the sequence (2) has more than$k$variations. Let us take for this, as a continuation (I),$n+3$points$x_{1}<x_{2}<\ldots<x_{n+3}$of E such that$\Delta_{n+1}^{1}.\Delta_{n+1}^{2}<0$, which is possible according to Lemma III. Let us first determine the polynomial$\mathrm{P}=a_{0}x^{n}+a_{1}x^{n-1}+\ldots+a_{k-1}\lambda^{n-k+1}$so that for the function$f_{1}=f-\mathrm{P}$we have

The corresponding sequence (2) becomes, up to positive factors,

But (5), regarded as a system of$k$linear equations in the$k$unknowns$a_{0},a_{1},\ldots,a_{k-1}$, has its determinant different from zero ² ). We can therefore modify these coefficients so that the sequence (6) has all its terms non-zero and with alternating signs, therefore it presents$k+I$variations of signs. Theorem III is therefore demonstrated. We see that in the statement we can only consider the finite subsets of E having$n+3$points.

4. We have said that Theorem I follows from Theorem II. Theorem I also has a converse which we will study in another work. Let us make here only a remark on the case$n=I$. We demonstrated in a previous note ¹ ) that inequality
(7)$\quad f\left(x_{2}\right)\leq\max\left[f\left(x_{1}\right),f\left(x_{3}\right)\right],x_{1}<x_{2}<x_{3},x_{1},x_{2},x_{3}\in\mathrm{E}$
is necessary and sufficient so that we can decompose the set E into at most two consecutive subsets such that on each the function$f$Or -$f$is monotonic, the monotonicity being in opposite directions on the two subsets. In particular, the first-order functions verify this property and are therefore such that$f$Or -$f$verifies inequality (7). Here we have again a reciprocal and thus we can state the

Theorem IV. For the function$f$is of order I on E it is necessary and sufficient that, whatever the number a, we can decompose the set E into at most two consecutive subsets on each the function$f-\alpha x$being monotonic, the monotony being in opposite directions on the two subsets.

The condition is necessary since$f-\alpha x$is of order i if$f$is of order I. Let us show that it is also sufficient. It suffices to show that if$f$is not in order$I$we can find a number$\alpha$so that$f_{1}$And$-f_{1}$, with$f_{1}=f-\alpha x$, do not verify the property expressed by inequality (7). If$f$is not of order I we can find 4 points$x_{1}$of$<x_{3}<x_{4}$of E so that

Let's determine the number$\alpha$so that.
$\left[x_{2},x_{3};f\right]>a>\max\left(\left[x_{1},x_{2};f\right],\left[x_{3},x_{4};f\right]\right)$if$\left[x_{1},x_{2},x_{3};f\right]>0\left[x_{2},x_{3};f\right]<\alpha<\min\left(\left[x_{1},x_{2};f\right],\left[x_{3},x_{4};f\right]\right)\operatorname{si}\left[x_{1},x_{2},x_{3};f\right]<0$.

By posing then$f_{1}=f-\alpha x$, we have

in the first case and

in the second case. We immediately check that$f_{1}$And$-f_{1}$do not verify inequality (7).

We can choose the numbers$\alpha,\beta$so that if$f_{1}=f+\alpha x+\beta$we have$f_{1}\left(x_{2}\right)=\mathrm{A},f_{1}\left(x_{3}\right)=\mathrm{B},\mathrm{A},\mathrm{B}$being any two numbers. Taking A negative, B positive sufficiently small if$\left[x_{1},x_{2},x_{3};f\right]>\mathrm{o}$and A positive, B negative sufficiently small if$\left[x_{1},x_{2},x_{3};f\right]<0$, we have

in the first case and

in the second case. We deduce
Theorem V. For the function j to be of order I on E it is necessary and sufficient that, whatever the numbers$\alpha,\beta$, we can decompose the set E into at most three consecutive subsets on each the function$f+\alpha x+\beta$being of invariable sign.

We can now glimpse the converse of Theorem I in the general case but, as we have said, we will return to this question in another paper.

Cernăuți, July 5, 1939.