Injectiveness and Discontinuity of Multiplicative Convex Functions

In the present work we study the set of multiplicative convex functions. In particular, we focus on the properties of injectiveness and discontinuity. We will show that a non constant multiplicative convex function is at most 2-injective, and construct multiplicative convex functions which are discontinuous at infinitely many points.


Introduction and Background Research on the Set of Multiplicative Convex Functions
The class of multiplicative convex functions was introduced as a way of extending the property of classical convexity from the arithmetic mean to the geometric mean. With this idea, Niculescu proposed that a multiplicative convex function would verify the relation for every choice of x, y > 0 and 0 < λ < 1 ( [1]). On the other hand, in [2] the authors proposed a different way to define multiplicative convexity were the aim was to upgrade the classical definition from addition to multiplication. In this direction, the following definition was introduced: . A function f : (0, ∞) → [0, ∞) is said to be multiplicative convex (or f is mc) if, for every µ > 0 and x, y ≥ 0 we have If, besides, f (1) = 1 then we will say that f is multiplicative convex 1 (or mc1, for short).
We will denote In [2] it was shown that the distinction between the sets MC and MC1 (namely, f (1) = 1) was crucial. In fact, the condition f (1) = 1 suffices to completely describe the set MC1: Theorem 1 ([2]). Let f : (0, ∞) → [0, ∞). Then, f is an mc1-function if and only if it can be written in the form where a, b, a and b satisfy the following conditions: 1. 0 < a < 1 and a > 1.
In particular, Theorem 1 implies that an mc1-function can show one (and only one) of the following behaviors: it is either monotone (increasing or decreasing) or it decreases over the interval (0, 1) and increases over the interval (1, ∞). Functions that behave like the latter ones (decrease over (0, b) and increase over (b, ∞) for certain number b > 0) will be called decreasing-increasing functions.
On the other hand, requiring f (1) > 1 (for an mc-function cannot have f (1) < 1) implied a huge difference with respect to the previous situation, where everything was under control: for one thing (see [3]), MC is closed under addition, product, multiplication by a scalar no smaller than 1 and composition (if the first function acting in the composition is not decreasing), but it was also proved the existence of discontinuous mc-functions: ). Let α > 1, α < β ≤ α 2 and f be an mc1-function. Define the function Then g is a discontinuous mc-function.
The following theorem can be used to prove the existence of an mc-function which is discontinuous at any given point. We include the constructive proof for the sake of completeness.

Proposition 1 ([3])
. Let x 0 > 0. Then, there exists an mc-function that is discontinuous at x 0 Proof. Assume first x 0 < 1 and let f be an mc1-function and g be an increasing discontinuous function from Theorem 2. Then, is an mc-function which is discontinuous at x 0 .
If now x 0 > 1, then we may just define Given the aim of this paper, we shall introduce some notation to denote the set of points of discontinuity of functions.
Definition 2. Let f : R → R be a function. We shall denote This set has some regular algebraic properties. For example, what follows is a standard calculus exercise: Proposition 2. Let f and g be two functions so that D( f ) ∩ D(g) = ∅. Then, In this paper we have two main objectives: the first one is to complete a description of the set of general mc-functions in a similar way as how the set of mc1-functions was described in [2] (where, before succeeding in giving a complete characterization of the set, the authors proved that mc1-functions were continuous and either monotone or non increasing-non decreasing). The results in Section 3 will lead to the conclusions that a general mc-function is of one of the mentioned two behaviors (monotone or decreasingincreasing), which in particular implies that the set MC is of cardinality c and that an mc-function is continuous everywhere except from possibly a countable set.
The second aim is to provide examples of mc-functions which are discontinuous at infinitely many points. The main result in Section 4 is held by some propositions and lemmas encompassed in Number Theory.
We will complete this paper by analyzing the set of discontinuous mc-functions from an algebraic point of view. Section 5 follows the line of action shown in [3] and focuses on the existence of certain algebraic structures whose non-zero elements fulfill some properties (see, e.g., [4][5][6][7][8]).

Other Fundamental Concepts and Classical Results
The following two theorems will be of great importance for the goals of this paper. Besides this, we will also make use of Steiner's problem, which we include bellow: Lemma 1 ([11]). The maximum of the function f (x) = x 1/x is attained at x = e. In fact, the function f is increasing on (0, e) and decreasing on (e, ∞).
With respect to the algebraic properties of the set MC, we remark that in general we cannot ensure that, given an mc-function f , α f is also an mc-function for α ≥ 0 but for α ≥ 1. Because of this, we have to consider the following definition.

Definition 3 ([3]
). Let X be a vector space and M ⊆ X. We say that M is an infinite dimensional truncated cone if M fulfills the following properties: 1.
M is closed under addition.

2.
M is closed under multiplication by scalars greater than or equal to 1.

3.
M contains a set of infinite cardinality of linearly independent elements.
The maximal possible cardinality of a set as in property (3) is the dimension of the truncated cone. If, furthermore, we can define a multiplication on X (satisfying the usual properties on itself and with respect to addition) and M is closed under multiplication, then we say that M is an algebraic truncated cone. In that case,

1.
the linear dimension of the truncated cone will be the maximal possible cardinality so that there exists a subset of such cardinality and consisting on linearly independent elements. 2.
The algebraic dimension will be the maximal possible cardinality so that there exists a subset of such cardinality and consisting on algebraic independent elements (that is, so that the only polynomial vanishing on them is the null polynomial).
Trivially, we obtain that the algebraic dimension is not greater than the linear dimension.

Study of the Injectiveness of an mc-Function
In order to conclude that an mc-function f is either monotone or decreasing-increasing, we will first show that we have that Theorem 5. Let f be an mc-function and define A = f −1 (0, 1). Then f |A is injective.
and therefore we can define which is a solution to θ This allows us to conlcude finding a contradiction to the condition f (θ 2 ) < 1 and proving our claim. Define next Then, , so by our previous claim, arriving at a contradiction. If 0 < θ 1 < 1 < θ 2 , then log θ 2 (θ 1 ) < 0 and therefore is always well-defined and positive, so we can again arrive at the contradiction f (θ 2 ) ≤ f (θ 2 ) µ+1/µ . Any other situation may be reduced to one of the previous two via the auxiliary function h(x) = f (1/x).

Theorem 6.
There is no mc-function, f , so that we can find Then, if we define the function we can apply Theorems 1 and 2 to conclude that g is a (discontinuous) mc-function. Hence, would also be an mc-function. Now, we can deduce the following chain of equivalent inequalities: On the other hand, applying the requisites over a, a , b, b and α from Equation (2), arriving at a contradiction.
Corollary 1. Let f be an mc-function and Proof. Assume otherwise. If θ 1 = 1, then this corollary is just Theorem 6. For 1 < θ 1 , x . In any case, g i would be an mc-function which contradicts Theorem 6.
Corollary 2. Let f be an mc-convex function which is not locally constant (that is, for every real number x and interval I with x in I we can find x 1 = x 2 also in I so that f (x 1 ) = f (x 2 )).
Then f is at most 2-injective (meaning that for every real number y # f −1 (y) ≤ 2).
Proof. Let y be a real number and x 1 , (2) ) and this is a contradiction with Theorem 6.
The following Corollary generalizes Theorem 2.1 from [2]: Corollary 3. Let f be an mc-function (continuous or not). Then, f is either monotone or decreasing-increasing.
Proof. This Corollary is easily proved if we consider the situation where f is not monotone and we apply Theorem 6 repeatedly.
Corollary 3, in combination with Theorem 4, implies that the set MC must be of cardinality less than c. On the other hand, since the function f (x) = b is an mc-function for b ≥ 1, we conclude the following Theorem, which answers several questions posted in [3]: The set MC has cardinality c.
In particular, this implies that the algebraic structures considered in [3] are of the greatest possible dimension.

On the Set of Points of Discontinuity of an mc-Function
It is obvious that, if A = {x 1 , x 2 , . . . , x n } is any finite set and, for 1 ≤ k ≤ n, f x k is an mc-function that is discontinuous at x k , then g(x) = f x 1 (x) + f x 2 (x) + . . . + f x n (x) is discontinuous over the set A. The question we will focus on now is whether it is posible to find an mc-function which is discontinuous over an infinite set. We remark that, taking Theorem 4 and Corollary 3 into account, the set A must be countable.
The main result of this section is as follows: be a decreasing sequence. Then, there exists an mcfunction which is discontinuous on X .
Before giving the proof, we will need some preliminary lemmas and definitions.
Proof. We will just show that β 1 is well-defined. Indeed, the product that defines β 1 converges if and only if ∑ ∞ k=1 log(a 2 k ) converges. Now, x m x m+1 x n x m x m x m x m+1 x m+1 x m+1 x n x m+1 x n x m+1 x n x m+1 x n x m+1 based on the fact that {x k } ∞ k=1 is a decreasing sequence.

Lemma 5.
Let {x n } ∞ n=1 ⊂ (0, 1) be a decreasing sequence. Then, for every n ≥ 1 and m ≥ n + 2, x n x n+1 x n+1 x n+1 x n+2 Proof. We will proceed by induction on m. If m = n + 2, then x n x n+1 x n+1 x n+1 x n+2 x n x n+1 x n+2 x n+1 x n+2 Assuming the result is true for m, then x n x n+1 x n+1 x n+1 x n+2 x n+2 , as desired.  Proof. Let l, n ∈ N. We notice that Therefore, sup (3) (3) cannot generally be attained: as a counterexample, we choose the functions
Proof of Theorem 8. Define, given m ∈ N, the following function: where x m , α m and β m are as in Definition 4.
The function f m is multiplicative convex and it is discontinuous at t = x m , since it has been defined following the contruction from Proposition 1.
We will show that the function g(t) = sup{ f n (t) : n ∈ N} is discontinuous on the set X = {x n } ∞ n=1 . First of all, for t > 0 the set { f n (t) : n ∈ N} is bounded since for every n ≥ 1 and x −x n n ≤ e 1/e because of Lemma 1.
Let n, m ∈ N. Then, x m x m x m+1 x m+1 x m+1 x m+2 x m+2 x m x m x m+1 x m+1 x m+1 x m+2 x m+2 by Lemma 4. If m ≥ n + 2, and using Corollary 4, we can prove that In conclusion, we can say that g(x n ) < β n . On the other hand, With a similar argument as before, sup lim Hence, by Corollary 5 As a consequence, g is not continuous at x n .
The following result complements Theorem 8 and shows that D(g) = X : Proposition 3. The function g considered in the proof of Theorem 8 is continuous on (0, ∞) \ X .
Proof. Let f m be the functions defined in the proof of Theorem 8, m 0 ≥ 1 and x m 0 < x < x m 0 −1 . Then, and therefore , leading us to conclude that . On the other hand, from x m+1 < x m we can deduce that , which allows us to deduce that This last inequality is equivalent to As a consequence, Hence, on (x m .x m−1 ), g can be expressed as the maximum of two continuous functions and in conclusion it is continuous over that interval.
If next 0 < x < x 0 = lim n→∞ x n (which in particular implies that x 0 = 0), then x m x m and, as above, we would have . Therefore, which again is a continuous function. For the case x = x 0 we notice that g(x 0 ) = 1. Assume {y n } ∞ n=1 is such that y n → x 0 as n → ∞. If y n < x 0 then g(y n ) = y n x 0 x 0 . On the other hand, if y n > x 0 then we can find {m n } ∞ n=1 ⊆ N so that x m n ≤ y n < x m n −1 . As before, g(y n ) = max α m n −1 y n x m n −1 we are able to conclude that lim n→∞ g(y n ) = g(x 0 ).
Finally, for the case x > x 1 we would obtain that f m (x) = β m x x m x m and hence (again, similarly as before) Remark 2. If X ⊆ (0, 1) is an increasing sequence, then there exists an mc-function so that D( f ) = X . Indeed, we only need to change the definition of the elements a n in Definition 4 as follows: a n = x n+1 x n x n . Corollary 6. If X is a monotone sequence, then we can find an mc-function which is discontinuous on X .
Proof. Let X 1 = X ∩ (0, 1) and X 2 = X ∩ [1, ∞). We can find two functions f 1 and f 2 so that f 1 is discontinuous only on X 1 and f 2 is discontinuous only on the set 1 x : x ∈ X 2 . The required function is then f (x) = f 1 (x) + f 2 ( 1 x ).

Algebraic
Structure on MC \ C(0, ∞) Theorem 10. There exists an algebraic truncated cone of algebraic dimension c every non-trivial algebraic combination of which is an mc-function which is discontinuous at infinitely many points.
In particular, taking a look at the proof of Theorem 8, the functionf ζ is defined as Because of Lemma 1 and since 1 is an increasing sequence with 1 From the proof of Lemma 3, β 1,ζ < e 4x 1,ζ = e and therefore 1 < 4 4 1/4 β 1,ζ , so g ζ is a (continuous) mc-function.
Define then the function Then f ζ is an mc-function (because it is the product of two mc-functions) which is discontinuous on the set X ζ .
Assume that Then, we notice that for 1 4 < x it must be Since the elements a ζ 1 , a ζ 2 , . . . , a ζ n are Q-linearly independent and the columns of N = {n i,j } n,m i,j=1 are pairwise different, we can conclude that the exponents are all different from each other. Therefore, f |(1/4,∞) is an identically null extended polynomial (with positive exponents). Hence, all its coefficients must be zero and, in conclusion, if 1 ≤ j ≤ m λ j 4 ∑ n i=1 n i,j = 0, which implies that λ j = 0 and in conclusion B is algebraically independent.
To finish with, let us choose an element f in the truncated cone generated by B. Then, we can find ζ 1 , ζ 2 , . . . , ζ n < c, λ 1 , . . . , λ m ≥ 1 and a matrix consisting of natural numbers as entries and without two equal columns N = {n i,j } n,m i,j=1 so that Without loss of generality we may assume that a ζ i < a ζ i+1 for every 1 ≤ i ≤ n − 1. Then, f ζ i is continuous on (0, a ζ 2 ) for every i ≥ 2.
For any function h : R → R we denote h a − = lim x→a − h(x) and h a + = lim x→a + h(x). We can then write where g j is an mc-function continuous on (0, a ζ 2 ).

Conclusions
In [2] the authors focused on those mc-functions that attained the value 1 at x = 1. It was proved that such functions were continuous and that in fact they could only be monotone, or decreasing until x = 1 and increasing from x = 1. In [3] some discontinuous mc-functions were introduced. This paper completes both of them in setting the cardinality of the set of all mc-functions at c (and therefore "there are not really many more mc-functions out of the cathegory of continuous functions") and showing that every of these functions show either monotonous or decreasing-increasing behavior.
It also continues the ideas shown in [3] when it constructs a trunkated cone consisting on mc-functions which this time have infinitely many points of discontinuity.
There are still some questions that remain open, namely whether the set of discontinuous mc-functions is a truncated cone itself, and what kind of properties can be concluded for the set of points of discontinuity of an mc-function.