THEOREM 1. So that the partition (1) of$E$either the support for the function$f$constant across segments and of effective degree$n$on$E$It is necessary and sufficient that$f$reduce ($\mathrm{si}n>1$) with two different constants over any two consecutive terms$E_{\alpha},E_{\alpha+1}(\alpha=1,2,\ldots,n-1)$.

The partition (1) being a support of$f$This function must be a constant on each of the terms.$E_{\alpha},\alpha=1,2,\ldots,n$.

Let us now proceed to the proof of the theorem.
The condition is necessary, since otherwise, by a suitable contraction, one could obtain a support having less than$n$terms.

To demonstrate that the condition is also sufficient, let us assume that it is satisfied. Let (2) then be any support of the function$f$We will demonstrate that partition (2) is obtained from (1) by dilation. If$n=1$the property is immediate If i$F$is of 1a partition (2) I1 exic whatever ($\mu$)$\mathrm{indice}\alpha$such that (A denotes the empty set)

Otherwise, indeed, the definition of a partition would imply that$F_{\beta}\cap\left(\cup^{n}E_{\alpha}\right)=F_{\beta}\cap E=\Lambda$which is impossible since$F_{\beta}$is not empty. We have$F_{\beta}\subseteq E_{\alpha}$Indeed, otherwise we might find a clue$\alpha^{\prime}$, different from$\alpha$, such as

But from (3) and (4) it follows that one of the relations

certainly takes place. This is impossible since then the function$f$would not be constant on$F_{\beta}$, contrary to the definition of this set. It therefore follows that any segment$F_{\beta}$is a subset of one and only one (the uniqueness of$\alpha$in (3)) segment$E_{\alpha}$.

Let us also note that at all$\alpha$must correspond to at least one$\beta$such as$E_{\alpha}\cap F_{\beta}\neq\boldsymbol{\Lambda}$Otherwise, indeed, it would follow, as above, that$E_{\alpha}\cap E=\Lambda$which is impossible since$E_{\alpha}$is not empty.

This demonstrates that, under the conditions of the demonstration, partition (2) is obtained from (1) by dilation.

Theorem 1 is proven.
Note that we have proven a little more than what was stated in Theorem 1, and in particular that any support of a function constant under segments can be obtained by dilating its support with the minimum number of terms.
6. Consider two functions$f,g$constants per segment on$E$and be
  

the respective supports of these functions.
We will show that we can find a common support for the functions$f$And$g$.

The intersections$F_{\alpha}\cap G_{\beta},\alpha=1,2,\ldots,m,\beta=1,2,\ldots,n$are pairwise disjoint segments and their union is equal to$E$Among these intersections, some are certainly non-empty. These are the intersections$F_{1}\cap G_{1}$And$F_{m}\cap G_{n}$But also, among these intersections, there are some that are generally always empty. By arranging the non-empty intersections in a specific order, we obtain the partition

which is obtained from each of the partitions (6) by dilation and it therefore follows that it is a common support of the functions$f$And$g$.

The preceding property results from the following analysis. First, at any$\alpha(1\leqslant\alpha\leqslant m)$corresponds, a$\beta(1\leqslant\beta\leqslant n)$such as$F_{\alpha}\cap G_{\beta}$not be empty. Then if$F_{\alpha}\cap G_{\beta}$not all is empty$F_{\alpha^{\prime}}\cap G_{\beta^{\prime}}$with$\alpha^{\prime}>\alpha,\beta^{\prime}<\beta$is empty, because otherwise, every point of$F_{\alpha^{\prime}}\cap G_{\beta^{\prime}}$should be both to the left and to the right of any point of$F_{\alpha}\cap G_{\beta}$It follows that we can find the non-negative integers$k_{1},k_{2},\ldots,k_{m}$such as$k_{1}+k_{2}+\ldots+k_{m}=n$and such that all intersections$F_{\alpha}\cap G_{\beta}$different segments ($k_{0}=0$)

are empty.
The number of segments (8) is equal to$m+n-1$and so we have$p\leqslant m+n-1$for the number$p$terms of the partition (7).

Those of the sets (8) that are not empty, arranged in "lexicographical" order of their indices, form the following$E_{1},E_{2},\ldots,E_{p}$This means that if$E_{\gamma}=F_{\alpha}\cap G_{\beta},E_{\gamma^{\prime}}=F_{\alpha^{\prime}}\cap G_{\beta^{\prime}}$, of$\gamma<\gamma^{\prime}$it follows that$\alpha\leqslant\alpha^{\prime}$,$\beta\leqslant\beta^{\prime},\alpha+\beta<\alpha^{\prime}+\beta^{\prime}$This shows us, at the same time, that the partitions (6) are obtained from (7) by contraction.

It also follows that$p\geqslant m,p\geqslant n$Therefore, that$p\geqslant\max(m,n)$7.
From the previous results we can deduce some properties of the set of constant functions by segments.

The sum and product of two segment-constant functions are segment-constant functions. This property follows immediately if we perform the operations of addition and multiplication on the two functions considered with respect to a common support.

In particular, the constants are segment-constant functions. It follows that the set of segment-constant functions is a linear (vector) set with respect to the usual addition (value by value) of functions and the usual multiplication by a number (real or complex) of functions.

From previous results it also follows that if two functions constant on segments are respectively of degree$m$And$n$, their sum and their product are of degree$m+n-1$It should be noted that the functions considered can have the effective degrees$m$And$n$and, at the same time, their sum and their product of degree 1 (can be constants) without uuu be y that$n$duelconch of$E$and if the functions$f,g$are defined by the formulas

then the sum$f+g$reduces to the constant 1 and the product$fg$to the constant 0.8.
We will say that a segment$E^{\prime}$of$E$is an initial segment (of$E$) if there is no point of$E$to the left of all the points of$E^{\prime}$and similarly we will say that the segment$E^{\prime}$of$E$is a final segment (of$E$) if there is no point of$E$to the right of all points of$E^{\prime}$The empty set (or segment) is not considered an initial or final segment. If (1) is a partition of$E$the segment$E_{\text{est }}$in initial degree and$E_{n}$a final segment of$E$.

A function defined on$E$will be said to be initially constant and finally constant if there exists an initial segment and a final segment of$E$, on which this function is constant. Every constant, and in general, every segment-constant function, is initially constant and also ultimately constant, regardless of$E$.

But, in general, a function is neither initially constant nor ultimately constant. For example, function 1a$x$(and also any non-constant polynomial) is neither initially constant nor ultimately constant on an open interval. If$a=\inf E\in E$(so if$E$(at a minimum), any function defined on$E$is initially constant, since the set {$a$} formed by the single point$a$is an initial segment of$E$and every function is constant at a single point. Similarly, we see that if sup$E\in E$(so if$E$(a maximum) any function defined on$E$is ultimately constant.

THEOREM 2. A necessary and sufficient condition for the function$f$, defined on the set$E$(having at least$n+1$points), or constant
by segments and of degree$n$is that whatever the$n+1$points$x_{1}<<x_{2}<\ldots<x_{n+1}$of$E$one has

The condition is necessary. Indeed, if (1) is a support to$n$terms of$f$among the points$x_{\alpha},\alpha=1,2,\ldots,n+1$There are certainly at least two consecutive ones.$x_{\beta},x_{\beta+1}$which belong to the same term$E_{\alpha}$and equality (9) is verified.

Before going further, let's prove
Lemma 1. If the condition expressed by equality (9), from Theorem 2, is satisfied, the function$f$is ultimately constant.

Let us therefore assume that the condition of Theorem 2 is satisfied and assume that the function$f$is not ultimately constant. Therefore, in particular, this function is not constant, and thus there exist points$x_{1},x_{2}\in E$such as$x_{1}<x_{2}$And$f\left(x_{1}\right)\neq f\left(x_{2}\right)$On the final segment$E\cap\left[x_{2},\infty\right)$the function is not constant and therefore there is a point$x_{3}$of this segment, such as$x_{2}<x_{3}$and on which$f\left(x_{2}\right)\neq f\left(x_{3}\right)$We can continue in this way and form the (infinite) increasing sequence.$x_{1},x_{2},\ldots$, such as one might have$f\left(x_{\alpha}\right)\neq\neq f\left(x_{\alpha+1}\right)$for everything$\alpha$This contradicts equality (9) and this contradiction proves lemma 1.

It can be shown in the same way that, under the same conditions, the function is initially constant.

We will now proceed by complete induction to demonstrate the sufficiency of the condition of Theorem 2.

For$n=1$the property is true because if$f\left(x_{1}\right)=f\left(x_{2}\right)$for everything$x_{1},x_{2}\in E$, the function is constant and is therefore constant over segments of degree 1.

Let us now suppose that$n>1$and that the theorem is proven for constant functions by segments of degree$n-1$Suppose then that the condition expressed by equality (9) is satisfied. Let$E^{\prime}$the meeting of all the final segments on which the function$f$is constant. According to Lemma 1, the set$E^{\prime}$is not empty. So, either$E^{\prime}=E$the function is constant on$E$and the theorem is proven. Or else$E^{\prime}\subset E$, SO$E-E^{\prime}$is not empty. We will demonstrate that in this case the function$f$is constant by degree segments$n-1$on$E-E^{\prime}$Indeed, otherwise, one might find$n$points$x_{1}<x_{2}<<\ldots<x_{n}$of$E-E^{\prime}$such as one might have

But on the segment$E\cap\left[x_{n},\infty\right)$the function cannot be constant, by definition of$E^{\prime}$We can therefore find a point$x_{n+1}$of this
segment (therefore of$E$), different from$x_{n}$such as$f\left(x_{n}\right)\neq f\left(x_{n+1}\right)$The result would be$x_{n}<x_{n+1}$And

which contradicts equality (9).
This contradiction demonstrates Theorem 2 for segment constant functions of degree$n$.

Theorem 2 is therefore proven.
We see, by very slight modifications to the proof, that Theorem 2 also holds if$E$contains one or both improper points$-\infty,\infty$.

The property expressed by Theorem 2 can also be expressed in the form of

THEOREM 3. A necessary and sufficient condition for a function defined on the set$E$(having at least$n+1$points) is constant by segments of degree n is that this property is satisfied, by the function considered, on every subset of$E$formed by$n+1$points.
10. In this work we only consider segment-constant functions of finite degree. We can define partitions of$E$whose terms form an infinite sequence or, more generally, an ordered set of a given ordinal type. To these partitions correspond segment-constant functions of a corresponding type (a kind of degree). It is clear that such functions are only of interest if the ordinal type of the supports is different from that of the ordered set.$E$Thus, any function defined on a well-ordered set of ordinal numbers$\omega$(of an infinite sequence) is obviously constant over segments of degree$\omega$However, not every function defined on the real axis is constant over segments of degree$\omega$.

The notion of an ordered set is taken here in the sense of the ordering of real numbers according to their numerical magnitude. The ordinal type of the partitions, and therefore also the type (the degree) of the corresponding segment constant functions, can be specified as follows. Let$I$an ordered subset of the real (ordered) axis.$E=\cup E_{\alpha}$is a partition of the ordinal type of$I$if the$E_{\alpha}$are segments$f$of$E$, two by two disjoint. If$I=(1,2,\ldots,n)$we have constant functions of degree (finite) through segments$n$, studied in this work and if$I==(1,2,\ldots,n,\ldots)=1$Following the natural numbers, we have the constant functions by segments of degree$\omega$.