1. 1.

   We will consider functions$f(x)$, real, finite, uniform and defined on an arbitrary linear set$It is$We will denote by$M(f;E)$the upper edge and with$m(f;E)$the lower bound of this function. Finally we will use the notation

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The necessary and sufficient condition that$f(x)$to be monotonous on$It is$is to have$\left[x_{1},x_{2};f\right]\left[x_{2},x_{3};f\right]\geq 0$, for any group of 3 points$x_{1},x_{2},x_{3}$his/hers$It is$It follows that if the function is not monotone, the points can be found$x_{1}<x_{2}<x_{3}$, his$It is$, so that we have$\left[x_{1},x_{2};f\right]\left[x_{2},x_{3};f\right]<01$).

Let a be a real number and let us put ² ).

Let's note with$E_{a}^{*},E_{a}^{*}$crowds$E_{a},E_{a}^{\prime}$adding, if$a\in E$, the point a in any one of these sets. We will then have

$E_{a}^{*}=E_{a}\cup\{a\},E_{a}^{*}=E_{a}^{\prime}$or$E_{a}^{*}=E_{a},E_{a}^{*}=E_{a}^{\prime}\cup\{a\}$if$a\in E$.

We always have$E=E_{a}^{*}\cup E_{a}^{\prime*},E_{a}^{*}\cap E_{a}^{\prime*}=0$and we say that$It is$is decomposed into the union of two consecutive subsets. The point a is the separation point of this decomposition. There is only one separation point decomposition$A$, if this point does not belong to$It is$and two such decompositions if$a\in E$.

We will say that$f(x)$is a function ($m$) (on$It is$) if there is a decomposition$E=E_{a}^{*}\smile E_{a}^{*}$his/her$It is$in two consecutive subsets such that the function is monotone on each of the sets$E_{a}^{*},E_{a}^{\prime*}$, the monotonicity being of opposite sense on these two subsets of$It is$More precisely we will say that$f(x)$is then a function ($m^{+}$) respectively a function ($m^{-}$) as it is non-increasing and non-decreasing respectively on$E_{a}^{*}$(so non-decreasing respectively non-increasing on$E_{a}^{\prime*}$).

It is clear that any monotone function is at the same time a function$\left(m^{+}\right)$and a function ($m^{-}$). Conversely, any function that is at the same time a function ($m^{+}$) and a function ($m^{-}$) is a monotone function. If$f(x)$is a function ($m^{+}$) respectively a function ($m^{-}$), the function -$f(x)$is a function ($m^{-}$) respectively a function ($m^{+}$). Finally, any function ($m^{+}$) on$It is$is a function ($m^{+}$) on any subset of$It is$.
2. To simplify the exposition we will make some conventions. Linearity and increase (decrease) are particular cases of non-decrease (non-increase). We also agree that any function defined on at most one point is considered monotonic, that is, regardless of whether it is increasing or decreasing.

With these conventions one of the decomposition subsets$E_{a}^{*},E_{a}^{\prime*}$, corresponding to a function ($m$), can be reduced to the empty set. This happens if and only if the function is monotone. The separation point a can be one of the improper points -$\infty$or$+\infty$, which again occurs if and only if the function is monotonic.

Any function defined on at most 2 points is monotone and any function defined on at most 3 points is a function ($m$).

Calculations with proper and improper numbers are done according to the usual rules.
3. Because$f(x)$to be a function ($m^{+}$) on the points$x_{1}<x_{2}<x_{3}$it is necessary and sufficient for us to have

In the case of a crowd$It is$having at least 3 points, we have:
Theorem 1. Because$\mathrm{f}(\mathrm{x})$to be a function ($\mathrm{m}^{+}$) on E is necessary and sufficient for it to be a function ($\mathrm{m}^{+}$) on each of the three-point groups of E.

This property can also be stated in the following form:
Theorem 1'. Because$\mathrm{f}(\mathrm{x})$to be a function ($\mathrm{m}^{+}$) on E it is necessary and sufficient that inequality (1) is verified for any group of 3 points$\mathrm{x}_{1}<\mathrm{x}_{2}<\mathrm{x}_{3}$of E.

In the case of functions ($m^{-}$) we have analogous properties, inequality (1) being replaced by

It follows that if a function$f(x)$is not a function (m) we can find the points$x_{1}<x_{2}<x_{3},x_{1}^{\prime}<x_{2}^{\prime}<x_{3}^{\prime}$his/hers$It is$so that

A simple reasoning shows us that we can then find 4 points$x_{1}<x_{2}<x_{3}<x_{4}$his/hers$It is$so that the string

to have 2 sign variations ¹ ).
But on the points$x_{1}<x_{2}<x_{3}<x_{4}$the property of being$a$function ($m$) is expressed precisely by the fact that the string (2) has at most 1 sign variation. For functions defined on at least 4 points we therefore deduce:

Theorem 2. Because$\mathrm{f}(\mathrm{x})$to be a function (m) on E is necessary and sufficient for it to be a function (m) on any group of 4 points of E.

This property can also be stated in the form:
Theorem 2'. For f (x) to be a function (m) on E, it is necessary and sufficient that the sequence (2) has at least 1 sign variation, whatever the points.$\mathrm{x}_{1}<\mathrm{x}_{2}<\mathrm{x}_{3}<\mathrm{x}_{4}$of E.
4. The proof of theorem 1 has been given elsewhere ² ). The inequalities (1) are obviously necessary. We will give a new proof

of the sufficiency of these inequalities and then we will specify a little the structure of the functions ($m$).

We have first:
Lemma 1. If inequality (1) is verified for any group of 3 points$\mathrm{x}_{1}<\mathrm{x}_{2}<\mathrm{x}_{3}$of E and if a is any number, the function$\mathrm{f}(\mathrm{x})$It is monotonous on one of the smallest of the crowds.$\mathrm{E}_{a},\mathrm{E}_{a}^{\prime}$.

For the sake of demonstration, let's assume the opposite. We can then find the points

so that we have

and it is easy to verify that on at least one of the groups of 3, points$x_{1},x_{2},x_{3};x_{4},x_{5},x_{6};x_{2},x_{3},x_{4};x_{2},x_{4},x_{5}$inequality (1) is not verified.

One can still specify Lemma 1, but the above statement is sufficient for what follows. For example, it is easy to see that in the statement one can replace$E_{a},E_{a}^{\prime}$by$E_{a}^{*},E_{a}^{\prime*}$respectively.

We deduce from here:
Lemma 2. If inequality (1) is true for any group of 3 points$\mathrm{x}_{1}<\mathrm{x}_{2}<\mathrm{x}_{3}$of E , one can find a point c such that the function$\mathrm{f}(\mathrm{x})$to be unsightly on$\mathrm{E}_{c}$, and non-decreasing on$\mathrm{E}_{c}^{\prime}$.

It is enough to prove the property in the case when$f(x)$it is not monotonous on$E^{1}$).

There are points$a$so that$f(x)$to be monotonous on$\boldsymbol{E}_{a}$Indeed, if for the points$x_{1}<x_{2}<x_{3}$his/hers$E$HAVE$\left[x_{1},x_{2};f\right]\left[x_{2},x_{3};f\right]<0$then, by virtue of the lemma$\mathbf{1}$, anything$a<x_{1}$check the property. If then$a_{1}<a$and$f(x)$it is monotonous on$E_{a}$, it is also monotonic on$E_{a_{1}}$. Either$b$the upper edge of the points$a$The point$b$is the only point that enjoys the property that$f(x)$it is monotonous on$E_{a}$for anything$a\leqq b$and it is not monotonous on$E_{a}$if$a>b$We have$b>-\infty$and also$b<+\infty$because it is obvious that$b\leqq x_{3}$. It is also seen that there is a unique and proper point$b^{\prime}$who enjoys their own

father that$f(x)$it is monotonous on$E_{a}^{\prime}$for anything$a\geq b^{\prime}$and it is not monotonous on$E_{a}^{\prime}$if$a<b^{\prime}$We have the inequality$b^{\prime}\leqq b$because, otherwise, the function would not be monotone on any of the sets$E_{\frac{b+b^{\prime}}{2}},E_{\frac{b+b^{\prime}}{2}}^{\prime}$, which contradicts Lemma 1.

Lemma 2 will be satisfied if we take$c\in\left[b^{\prime},b\right]$The function is monotone on each of the sets$E_{c},E_{c}^{\prime}$and it remains to prove that it is effectively non-increasing on$E_{c}$and non-decreasing on$E_{c}^{\prime}$To consolidate the ideas, let us show that the function is non-increasing on$E_{c}$Otherwise we could find the points$x_{1},x_{2}\in E_{c}$,$x_{1}<x_{2}$so that we have$f\left(x_{1}\right)<f\left(x_{2}\right)$However, the function is not monotonic on$E$, the points can be found$x_{1}^{\prime},x_{2}^{\prime}\in E$,$x_{1}^{\prime}<x_{2}^{\prime}$so that we have$f\left(x_{1}^{\prime}\right)>f\left(x_{2}^{\prime}\right)$We should have then$x_{2}<x_{2}$If$f\left(x_{2}^{\prime}\right)<f\left(x_{2}\right)$inequality (1) is not verified on the points$x_{1},x_{2},x_{2}^{\prime}$If$f\left(x_{2}^{\prime}\right)\geq f\left(x_{2}\right)$we should also have$x_{2}<x_{1}^{\prime}$and inequality (1) would not be verified on the points$x_{1},x_{1}^{\prime},x_{2}^{\prime}$It is also proven that$f(x)$is non-decreasing on$E_{c}^{\prime}$.
5. Theorem 1 now follows easily. Let$c$a point that satisfies Lemma 2. If$c$does not belong to him$E$the decomposition sought is$E=E_{c}^{*}\cup E_{c}^{\prime*}$If$c\in E$, on at least one of the sets$E_{c}\cup\{c\}$,$E_{c}^{\prime}\cup\{c\}$the function is monotonic. Indeed, otherwise, we could find the points$x_{1}\in E_{c},x_{2}\in E_{c}^{\prime}$so that$f\left(x_{1}\right)<f(c),f\left(x_{2}\right)<<f(c)$and inequality (1) would not be verified on the points$x_{1},c,x_{2}$Theorem 1 still follows, adding the point$c$to one of the crowds$E_{c},E_{c}^{\prime}$and its decomposition$E$is still in shape$E=E_{c}^{\mathrm{k}}\cup E^{\prime*}{}_{c}$.

Analogous results exist for the functions ($m^{-}$).
If$f(x)$is a function ($m$) the points$c$separation of the decompositions$E=E_{c}^{*}\cup E_{c}^{\prime*}$in two consecutive subsets of opposite monotonicity form an interval$\left[b^{\prime},b\right]$, finite or infinite, being able to be reduced to a single point. This interval is always closed provided that we also consider improper numbers$-\infty,+\infty$If$f(x)$it is not monotonous,$b^{\prime},b$are precisely the finite numbers determined above. In this case$b^{\prime},b$must belong to any closed interval containing three points$x_{1}<x_{2}<x_{3}$his/hers$E$so that we have$\left[x_{1},x_{2};f\right]\left[x_{2},x_{3};f\right]<0$If$f(x)$is monotonous, the interval$\left[b^{\prime},b\right]$is infinite or reduces to one of the improper points$+\infty$,$-\infty$Indeed, if we look at
the function as a function ($m^{+}$) we have$b^{\prime}=-\infty$if it is non-decreasing and$b=+\infty$if it is non-increasing.

If the crowd$E\cap\left(b^{\prime},b\right)$is not empty, the function reduces to a constant on this set. Indeed, if$f(x)$is a function$\left(m^{+}\right)$and if$x_{0}\in E\cap\left(b^{\prime},b\right)$, he is non-increasing on$E_{x_{0}}\cup\left\{x_{0}\right\}$and unwaveringly on$E_{x_{0}}^{\prime}\cup\left\{x_{0}\right\}$It follows that$f\left(x_{0}\right)=m(f;E)$. It is also seen that if$f(x)$is a function$\left(m^{-}\right)$HAVE$f\left(x_{0}\right)=M(f;E)$.

From the above it follows that, for a function ($m$), its decomposition$E$into two consecutive subsets of opposite monotonicity can be done, in general, in several and even infinitely many ways. It is easy to see which are the cases in which this decomposition is unique. Assuming that$f(x)$is - function$\left(m^{+}\right)$, for uniqueness it is necessary that the set$E\cap\left(b^{\prime},b\right)$to be empty and furthermore it is necessary and sufficient that we have:$b^{\prime},b\in E$and$f(b)>m\left(f;E_{b^{\prime}}\right)$or$f\left(b^{\prime}\right)>m\left(f;E_{b}^{\prime}\right)$, if$b^{\prime}=b$ ; at most one of the points$b^{\prime},b$to belong to$E$and$f\left(b^{\prime}\right)>>m\left(f;E_{b}^{\prime}\right)$respectively$f(b)>m\left(f;E_{b^{\prime}}\right)$as$b^{\prime}\in E$respectively$b\in E$, if$b^{\prime}<b$.
6. Before going further, we will make an observation on the functions (m) that will be useful to us below.

Whether$f(x)$a function$\left(m^{+}\right),\left[b^{\prime},b\right]$the range of corresponding separation points and$I$an arbitrary interval (closed or not) that has no points in common with$E$(so$E\cap I=O$. Finally, either$E_{1}$some subset of$I$Under these conditions we have:

Lemma 3. If the intervals$\left[\mathrm{b}^{\prime},\mathrm{b}\right]$and I have at least one point in common and if$\gamma$is a constant such that$\gamma\leqq\mathrm{m}(\mathrm{f};\mathrm{E})$, function

is a function ($\mathrm{m}^{+}$) on$\mathrm{E}\cup\mathrm{E}_{1}$It is sufficient to prove the
property for$E_{1}=I$. Be it then$c\in\left[b^{\prime},b\right]\cap I$and let's put$F=E\cup I$Function$f_{1}(x)$is non-increasing on$F_{c}$, for it is easily seen that it is non-increasing on$E_{c}\cup I$and$F_{c}$is a subset of this set. It is also seen that$f_{1}(x)$is non-decreasing on$F_{c}^{\prime}$. Lemma 3 follows.

It can still be observed that the interval of the points of separation of the decompositions of$F$corresponding to the function$f_{1}(x)$contains the range$I$, so$\left[b^{\prime},b\right]$also contains the interval$I$.

An analogous property holds for the functions ($m^{-}$), the inequality from the hypothesis of lemma 3 is then$\gamma\geqq M(f;E)$.
7. We propose to generalize Theorem 2 by removing the restriction that the considered subsets of$E$to be consecutive.

We will say that$f(x)$is a function$(M)$(on$E$) if$E$can be decomposed into the union of two subsets (empty or not), on each of which the function is monotone and the monotonicity is of opposite sense on the two subsets. It is clear that any function ($m$) is a function$(M)$In particular, any function defined on at most 3 points is a function ($M$). Finally, it is obvious that any function$(M)$on$E$is a function ($M$) on any subset of$E$.

Let's first examine the functions defined on 4 distinct points$x_{1}<x_{2}<x_{3}<x_{4}$Let's consider the string
(3)

If the function$f(x)$is not univalent, it is obviously a function ($M$). Otherwise (i.e. if the function is univalent) two cases must be distinguished:
$1^{\circ}$The sequence (3) is not strictly monotone.
$2^{\circ}$. The sequence (3) is strictly monotone (increasing or decreasing).
In the case$1^{\circ}$we have:

and examining all possible cases it is seen that$f(x)$is a function$(M)$, the decomposition of the definition set being, for example: