Rest

On S-convexity and a few of our results I.M.R. Pinheiro∗ July 5, 2009

Abstract In this paper, we review a few of our results, those which have been published in the prestigious academic vehicles IJMA, NZJM, and DGDS. As a side dish, we discuss some fundamental issues on basics such as epigraph and possible connections between convex shapes, convex sets, and convex functions.

AMS (2000): 26A51 Key-words: Analysis, Convexity, Definition, S-convexity.

I. Introduction • Definitions & Symbols; • List of our published results and due remarks; • Epigraph; • Discussion on Jensen’s Inequality; • New result; • Updated results; • Conclusions; • References. ∗

Postal address: PO Box 12396, A’Beckett st, Melbourne, Victoria, Australia, 8006. Electronic address: [email protected].

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II. Definitions & Symbols Symbols ([1]) • Ks1 stands for the set of S-convex classes of type 1; • Ks2 stands for the set of S-convex classes of type 2; • 0 < s ≤ 1 is the index designating one of the classes for S-convexity; • K11 ≡ K12 and both classes are the same as the convex class of functions; • s1 stands for the s value designating a class in Ks1 ; • s2 stands for the s value designating a class in Ks2 . Definitions ([1]) Definition 1. A function f : X− > < is told to be S−convex in the first sense if 1 f (λx + (1 − λs ) s y) ≤ λs f (x) + (1 − λs )f (y) (1) holds ∀x, y ∈ X, ∀λ ∈ [0, 1], where X ⊆ <+ . Definition 2. A function f : X− > I is told to belong to Ks2 if the inequality f (λx + (1 − λ)(x + δ)) ≤ λs f (x) + (1 − λ)s f (x + δ)

(2)

holds ∀λ/λ ∈ [0, 1]; ∀x/x ∈ X; s = s2 /0 < s2 ≤ 1; X/X ⊆ < ∧ I/I ⊆ <+ ∧ X = [a, b]; ∀δ/0 < δ ≤ (b − x). Definition 3. A function f : X− > I is told to belong to Ks2 if the inequality f (λx + (1 − λ)(x + δ)) 1

1

≤ λ s f (x) + (1 − λ) s f (x + δ)

(3)

holds ∀λ/λ ∈ [0, 1]; ∀x/x ∈ X; s = s2 /0 < s2 ≤ 1; X/X ⊆ < ∧ I/I ⊆ <− ∧ X = [a, b]; ∀δ/0 < δ ≤ (b − x). Remark 1. If1 the inequality is obeyed in the reverse situation by f then f is said to be s2 −concave. 1

This remark applies to both definitions preceding it.

2

Definition 4. f is called s2 −convex, s2 6= 1, if the graph lies below the ‘bent chord’ between any two points of the curve for f , that is, for every compact interval J ⊂ I, with I being the domain of f , J having boundary ∂J, and a special function, of image curve L, L being of curvature Ψ, determined by s2 ∈ (0, 1), we have: Lx (s) ≥ sup(Lx (s) − f (x)) ≥ sup(Lx (s) − f (x)). J

∂J

III. List of results and due remarks 1. [4] Let V be a vector space over <. A subset X ⊂ V is called s1 -convex if every s1 −convex curve, defined by λs x1 + (1 − λs )x2 , ∀x1 , x2 ∈ X, intersects X in an interval, that is (λs x1 + (1 − λs )x2 ) ⊂ X when 0 < λ < 1 and x1 , x2 ∈ X. Pointed problems: None. 2. [4] Let V be a vector space over <. A subset X ⊂ V is called s2 -convex if every s2 −convex curve, defined by λs x1 + (1 − λ)s x2 , ∀x1 , x2 ∈ X, intersects X in an interval, that is (λs x1 + (1 − λ)s x2 ) ⊂ X when 0 < λ < 1 and x1 , x2 ∈ X. Pointed problems: We had a bit of difficulty deciding on whether we would simply extend the concept for convex sets or not because the concept for convex sets seems to match perfectly well the concept for convex shapes, but the concept of convex functions seem to have little in common with the other two. If the concept of convex functions were to be harmonized with that of convex sets and shapes then we would need to consider that a convex function would be only the straight line containing the convex combination of two initial function values, what seems weird, to the least. It must have been the case that the concepts of convex shape and convex set were developed in a time preceding that of the creation of the concept of convex functions and, by the time they create the concept for convex functions, they first 3

had the geometric need to then associate to the ‘closest’ geometric idea which existed in the scientific literature. This because we did not find enough coherence between all the other convex notions and that of convex functions. Notice that the graph of a convex function should have to do with the definition of convex shapes. Problem is that not mattering what you try there, you end up reaching the conclusion that the only thing which could make univocal sense with the shape concept would be the function with graphical representation being the straight line, otherwise any logical reason supporting a concave curve to be the representation for a convex function could also be supporting a convex curve to be such. Notwithstanding, we will simply accept the existing literature for not finding better alternative and keep our previously presented concept for S-convex sets in this paper, having already extended the concept for convex functions in the distant past respecting the existing definition for convex functions when doing so. 3. ([4]) Let f be a function from <m to <. Then f is s1 -convex iff f (λ1 x1 + ... + λm xm ) ≤ λs1 f (x1 ) + ... + λsm f (xm ) P s whenever λ1 ≥ 0, ..., λm ≥ 0, m p=1 λp = 1. Pointed problems: The same remarks, exposed for the second case of S-convexity, will apply here. We are having problems in accepting this definition because the function which is the basis of it departs from vectors and reaches real numbers. Of course it is very possible to create even a Cartesian correspondence between those vectors from the domain and the real numbers from the counter-domain (for example, locating the first coordinate of the vector in the ‘x’ axis and then ‘walking’ the distance corresponding to all the other coordinates together, like, for instance, the coordinates put one after the other in a decimal number). However, we do not know enough on the existence of such a system of representation to then come up with an extension which should, first of all, respect the geometric idea of convexity. 4. ([4]) Let f be a function from <m to <. Then f is s2 -convex iff f (λ1 x1 + ... + λm xm ) ≤ λs1 f (x1 ) + ... + λsm f (xm ) P whenever λ1 ≥ 0, ..., λm ≥ 0, m p=1 λp = 1.

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Pointed problems: The definition must be adapted to the evolution in our definitions. We will have to split this definition into two cases. To make it worse, this definition bears several mistakes of different orders. It has been inspired by other people’s work, published work, once more, also publishers of weight (see [5], p. 77, for instance). However, it is clearly the case that one cannot mix the dimensions as it is done in this definition, or in the original alike definition for convex functions, in which we have based ourselves. Notice that the line is the limit for the case in two dimensions (from which we only mention one in the definition, the graph being in two) and, in general, in Topology (a convex graphical object is that which does contain, in itself, every segment formed between any two points of it (see, for instance, p. 38 of [5])). According to [6], p.37, this definition should be about convex sets, rather than functions. However, in [7], for instance, we have a new variation of this same sort of definition, with a completely different concept, when compared to all definitions we have mentioned this far here, but more coherent with the initial idea for two dimensions. Basically, we feel as if there is not enough either coherent, or consistent, literature pieces providing similar references for the concept we should be ‘extending’, only, here. This way, we will refrain from working with several dimensions by now. We will have a similar definition using multiple domain points instead to replace this one (see Jensen’s inequality). 5. ([2]) For a real s1 −convex function Φ, numbers xi in its domain, and positive weights ai , we have  Pp Φ

1 s

i=1 ai xi Pp 1 ( i=1 ai ) s



Pp a Φ(xi ) Pp i ≤ i=1 . i=1 ai

Pointed problems: The proof presented by us for this theorem is wrong because we have based ourselves in the equivocated ‘extension’ of the definition of Convexity to several dimensions in the domain of the function (only there). We have recently noticed another major problem in the mathematical literature so far as well. This problem regards deciding on whether Jensen’s Inequality is about several dimensions or about several points in < and a usual bidimensional graph, with a one dimensional function. We will prove this result, however, as we prove Jensen’s extended inequality for S-convex functions, later on, in this very paper. 5

6. ([2]) For a real s2 -convex function Φ, numbers xi in its domain, and positive weights ai , we have:  Pp  Pp s ai xi i=1 i=1 ai Φ(xi ) Φ Pp ≤ P . p ( i=1 ai )s i=1 ai Pointed problems: The same remarks, made for the previous item, apply here. 7. ([3]) If f is an s1 −convex, non-negative, function and (x1 , x2 , ..., xn ) lies in its domain, then:   n X x1 + x2 + ... + xn f (xi ) − f 1 ns i=1        n−1 x1 + x2 xn−1 + xn xn + x1 ≥ f + ... + f +f . 1 1 1 n 2s 2s 2s Pointed problems: None so far. 8. ([3]) If f is an s2 −convex, non-negative, function and (x1 , x2 , ..., xn ) lies in its domain, then: n X



 x1 + x2 + ... + xn f (xi ) − f n i=1        x1 + x2 xn−1 + xn xn + x1 2s−1 (ns − 1) f + ... + f +f . ≥ ns 2 2 2 Pointed problems: None so far. At most, we could try to create a similar inequality for the second case of the definition of s2 −convex functions, that is, for when they are non-positive. 9. ([3]) If f is an s1 −convex, non-negative, function and (a1 , a2 , ..., an ) lies in its domain, then: (ns −1)[f (b1 )+f (b2 )+...+f (bn )] ≤ ns [f (a1 )+f (a2 )+...+f (an )]−nf (a), where a =

a1 +a2 +...+an 1 ns

1

and bi =

n s a−ai

Pointed problems: None so far. 6

1

(n−1) s

, i = 1, ..., n.

10. ([3]) If f is an s2 −convex, non-negative, function and (a1 , a2 , ..., an ) lies in its domain, then: (n−1)s [f (b1 )+f (b2 )+...+f (bn )] ≤ n[f (a1 )+f (a2 )+...+f (an )]−ns f (a), where a =

a1 +a2 +...+an n

and bi =

na−ai n−1 ,

i = 1, ..., n.

Pointed problems: None so far (the same remark, made above, applies). − − 11. [4] An s1 −convex combination of a set of vectors → v1 , → v2 , ..., − v→ n is a linear combination of those vectors in which the coefficients are all nonnegative, and the individual raise to a power ‘s’Presults in their n − − s sum being one, that is: λs1 → v1 + λs2 → v2 + ... + λsn − v→ n , with p=1 λp = 1 and λp ≥ 0, ∀p/p ∈ N, 1 ≤ p ≤ n, is an s1 -convex combination of vectors. Pointed problems: First of all, the wording in this definition is not the best, it may be improved substantially. Second, the notion does not seem to make much sense when compared to the concept of convex combinations. Reasons are plenty. Just for starters, with convexity both sides of the inequality contain the ‘same’ alignment elements (coefficients), so that there is no choice to be made, in terms of which combination we will write about. The same obviously does not happen with S-convexity. As a third remark, the algebraic meaning is the same, it is just the exponent which is inserted as a novelty. Geometrically, none of the definitions seem to make much sense with the concept of S-convexity (convexity included), unless we have vectors starting always from some fixed point, such as it happens with the Complex Numbers in the Argand Diagram, which apparently is also erroneously called Argand Diagram, deserving being called Wessel’s Diagram instead, as for the historical events surrounding its appearance in the mathematical literature ([12]). Basically, it would be best, for us all, from Mathematics, that this definition did not exist and we simply stuck to linear combinations instead (see [13], for instance). The major issue we would like to raise here, even though we will improve the wording of the concepts and re-list them in this paper, is that of how suitable the term ‘S-convex (convexity included)’ is for this specific situation of limitation of a linear combination to coefficients which are not only located between zero and one, as if in the ijk basis for vectors, but also sum one (interesting that the ijk basis also sums one for each 7

coordinate if we sum the basis vectors). Notice that convex combinations are directly related to convex sets and convex shapes because by making convex combinations of the elements of a set, or points in a space, we will reach convex sets or shapes. 12. [4] The set of all s1 −convex combinations of the vectors is their s1 −convex − − span denoted by SCS1 (→ v1 , → v2 , ..., − v→ n ). Pointed problems: None. − − 13. [4] An s2 −convex combination of a set of vectors → v1 , → v2 , ..., − v→ n is a ‘bent’ combination of those vectors in which the coefficients are all − − nonnegative, and hold sum one, that is: λs1 → v1 + λs2 → v2 + ... + λsn − v→ n , with Pn p=1 λp = 1 and λp ≥ 0, ∀p/p ∈ N, 1 ≤ p ≤ n, is an s2 -convex combination of vectors. Pointed problems: The same remarks, made for s1 -convex combinations, apply here. 14. [4] The set of all s2 −convex combinations of the vectors is their s2 − − convex span denoted by SCS2 (→ v1 , → v2 , ..., − v→ n ). Pointed problems: None. 15. [4] A finite set is s1 -convexly independent if none of the points is an s1 -convex combination of the others. A finite set, which is not s1 convexly independent, is s1 -convexly dependent. Pointed problems: We have reassessed the definition of the phenomenon S-convexity in Mathematics and have proved that continuity is a necessity, having therefore automatically excluded any cases of finite sets or discontinuous functions from our realm of possibilities. For this reason, we will be both adding a more suitable definition for convexly independent sets to our own produced scientific literature and withdrawing this one from our list of results in this paper. 16. [4] A finite set is s2 -convexly independent if none of the points is an s2 -convex combination of the others. A finite set, which is not s2 convexly independent, is s2 -convexly dependent.

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Pointed problems: We have reassessed the definition of the phenomenon S-convexity in Mathematics and have proved that continuity is a necessity, having therefore automatically excluded any cases of finite sets or discontinuous functions from our realm of possibilities. For this reason, we will be both adding a more suitable definition for convexly independent sets to our own produced scientific literature and withdrawing this one from our list of results in this paper. 17. [4] The closure of an s1 −convex set is s1 −convex. Pointed problems: None. 18. [4] The closure of an s2 −convex set is s2 −convex. Pointed problems: None. 19. [4] f : X ⊂ <+ − > < is considered s1 −convex iff the epigraph of f is s1 −convex, that is, iff epi(f ) = {(x, t)/x ∈ X, t ∈ <, t ≥ f (x)} is s1 −convex. Pointed problems: We were very unfortunate with the choice of this theorem and the literature we have researched in terms of epigraph proofs and definitions in the past. Basically, the definition we have used is incompatible with the generalized understanding of the concept, as for the majority of the scientific literature available. In this particular definition, we notice that epigraph would be a set of ‘pairs’, somehow ordered, once first you have the set element and next you hold a real constant. However, it is not an ordered pair, as in Cartesian notation, for when you read the proofs involving the concept, only one coordinate is used. It is also not a set because both the delimiters being inappropriate for a set and the proofs never dealing with the couple as a set, or subset. The generalized understanding of the concept, as for every mathematical proof we have found in the literature so far, returns what we have proposed as understanding in our proof as well. But the adequate statement of the definition would then have to be something else, which we shall introduce in this paper, along with a more adequate proof. 20. [4] f : X ⊂ <+ − > < is considered s2 −convex iff the epigraph of f is s2 −convex, that is, iff epi(f ) = {(x, t)/x ∈ X, t ∈ <, t ≥ f (x)} is 9

s2 −convex. Pointed problems: The same remarks, as for the previous item, apply here. 21. [4] The interior of an s1 −convex set is also an s1 −convex set. Pointed problems: None. 22. [4] The interior of an s2 −convex set is also an s2 −convex set. Pointed problems: None. 23. [4] For every E ⊂ V , where V is a vector space, the union, s1 ch(E), of all s1 -convex combinations of the elements of E, that is, N N X X s s1 ch(E) = { λj xj ; λj ≥ 0; λsj = 1; xj ∈ E; N = 1, 2, ...}, 0

0

is an s1 −convex set. Pointed problems: None. 24. [4] For every E ⊂ V , where V is a vector space, the union, s2 ch(E), of all s2 -convex combinations of the elements of E, that is, N N X X s2 ch(E) = { λsj xj ; λj ≥ 0; λj = 1; xj ∈ E; N = 1, 2, ...}, 0

0

is an s2 −convex set. Pointed problems: None.

IV. Epigraph From [14], p.74 (partial copy): Given f : < is the nonempty set

S

∞, the epigraph of f

epif = {(x, r) ∈
10

Pointed problems: What they probably meant was several elements from <, or multiple, rather than vectors, therefore they have mistakenly swapped the domain of f from < to < is the (n + 1)−dimensional set {(x, τ )/f (x) ≤ τ, x ∈ C, τ ∈ <}. Pointed problems: Here, things could have got better with the introduction of the Complex numbers in place of the multidimensional vectors, but then it is missing referring to the modulus for the Complex numbers somehow in the definition. Assuming you could be dealing with only the real part of the complex numbers also does not help because we still have the equivocated symbol for what is neither a set nor an ordered pair... From [16], p. 8 of Chapter 3, we get a proof which is similar to what we usually find for epigraphs regarding the theorem “f convex ⇐⇒ epif convex.” Proof. Given {(X, u), (Y, v)} ⊂ epif , we must show for all µ ∈ [0, 1] that µ(X, u) + (1 − µ)(Y, v) ∈ epif ; id est, we must show f (µX + (1 − µ)Y ) <M + µu+(1−µ)v (any constant, basically. As long as we get a constant to this side, such would be enough)2 . Yet, this holds by definition because f (µX + (1 − µ)Y )  µf (X) + (1 − µ)f (Y ) (and f (X) ≤ u, f (Y ) ≤ v, by assumption of the proof). The converse also holds. See that, when making the proof, all which counts is actually the fact that X and Y are obeying the rules for convexity. What they obviously mean with epigraph is what we will describe below, in accurate mathematical terms, instead of what they have been stating this far. 2

Notice that the original definition allows even for this constant to be infinity, so that there is no chance we will not get it!

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Definition 5. Given a function f : X ⊂ <− > <, the epigraph of f , denoted by epif , is the set of elements of X grouped by a limiting constant determining the sub-level set (see [14], p. 2, for instance) for the element, that is: epitn f = {x ∈ X/tn ≥ f (x)}. epif = {epit1 , epit2 , ..., epitn }, where n may be infinity and we consider our extended counter-domain sets, rather than the counter-domain sets, for consistency purposes. What they should have meant, for what is above is not compatible with all the proofs they have been using, is only the epitn notion above. And because no convex, or S-convex function, will have any point with image in infinity, but even infinity would have got a level set (special one, acquired via ‘extension’ of the functions, as it is in the literature), all domain points of such functions will be part of some level set. In this case, we would need to prove that it is possible to form convex, or S-convex, combinations with each couple of domain points and with them always get to obey the rule for classification of the level set as convex, or even S-convex, in order to prove that if a function is convex, or S-convex, then the epigraph of the function is a convex, or S-convex, set. Notice that if we disregard all traditional rules of Maths and accept a set of sets as a set of usual elements instead, where therefore the delimiters are ‘decoration’, it is obvious that this will be true. Notice, as well, that k ≥ f (x1 )∧k ≥ f (x2 ) ⇐⇒ λk ≥ λf (x1 )∧(1−λ)k ≥ (1−λ)f (x2 ). Once f ∈ K11 , f (λx1 +(1−λ)x2 ) ≤ λf (x1 )+(1−λ)f (x2 ) ≤ λk+(1−λ)k = k. Therefore, λx1 + (1 − λ)x2 ∈ epik as we wanted to show. Notice that the converse is not actually true. There is no direct implication between the epigraph of the function being convex and the function being convex. However, if they ‘confound’ the reader and state that the epigraph is epif instead, one may think that once the deductions would lead to λf (x1 ) + (1 − λ)f (x2 ) ∈ epi2k f while λx1 + (1 − λ)x2 ∈ epik f then it is all in epif (provided the initial mistake, gross mistake, on the foundations of Maths, is preserved), the mix, and then f is convex. It is just confusional thoughts, and all is deriving from the fact that someone would like to connect the notion of convex sets to that of convex functions, what we ALSO have felt the need of doing, having already been through this. Thus, it is actually the case that f convex =⇒ epif convex but not the converse and, even so, one must understand that epif we refer to should be the fixed epif , our old epitn f . Notice, as well, that we can only assert then that f ∈ Ks1 =⇒ epif is 12

s1 −convex, nothing else, once we would need the coefficients to sum one to have similar proof for Ks2 or, alternatively, less than one. Once more, in the ‘confusional’ epif , any condition will always be preserved, trivially, making the notion not only something irrelevant, trivial, in the mathematical literature, but something blatantly wrong in the foundations of the Maths (we may invent, but never disregarding what has already been built in Maths).

V. Discussion on Jensen’s Inequality The biggest drama in understanding Jensen’s Inequality regards a clear divergence, in the foundations of Mathematics, as to what Jensen has actually stated: While a few researchers seem to consider that Jensen has proposed an inequality for functions with domain set containing vectors, therefore of possibly several dimensions, other researchers seem to think that Jensen has proposed an inequality for functions with domain set containing real numbers, therefore one dimension. The dimensions are, by no means, like anything in Mathematics, something irrelevant, in fact they are one of the most relevant factors in mathematical concepts in general, with us having the trouble of dedicating a whole paper to fix the dimensions problem in the definition of convex functions in the recent past. We will here try to reproduce the different ways Jensen’s Inequality is found mentioned in the scientific literature. Hopefully, we will be able to provide our readers with the actual inequality intended by Jensen, which is probably only one of the versions presented so far by the literature items. Theorem 5.1. ([8]) Let f be a convex function on an open interval I in <. Then for any x1 , ..., xn ∈ I and λ1 , ..., λn ∈ [0, 1] such that λ1 + ... + λn = 1, we have f (λ1 x1 + ... + λn xn ) ≤ λ1 f (x1 ) + ... + λn f (xn ). Theorem 5.2. ([9]) If f : < is convex then X  X p p i f λi x ≤ λi f (xi ). i=1

i=1

Theorem 5.3. ([10]) For convex functions f , the inequality  X X n n f λ i xi ≤ λi f (xi ), 1

1

where the xi are arbitrary values in theP region on which f is convex and the λi are nonnegative numbers satisfying n1 λi = 1. 13

Given the poverty of the literature available on the life and works of Jensen (we actually have not been able to work out not even who Jensen was: According to some sources, he was a statistician, and according to others he was both an engineer and a mathematician), when compared to everyone else’s in Mathematics we know of, we will have to limit ourselves to ‘guessing’ who would be righter about what Jensen has actually created, or stated, regarding convex functions. Common sense tells us that sticking to the last version of the inequality should return more value in research and applications than not. If anything, it suffices that we change the name of the theorist in the future. Region obviously includes vectors of several dimensions as domain elements as well as lonely real values. Now that we have defined what Jensen’s Inequality actually is, we need to decide on a sound proof for it. Notice that the statements we have presented, collected from the scientific literature, bring Jensen’s Inequality as an inequality for convex functions. Therefore, it is not an actual definition, in case we can find a proof to it, but a property of the convex functions, or an implied result from the fact that we hold a convex function, what makes it impossible for us to think of simply ‘extending’ it to S-convexity, like it should be more complex than that. Proof of Jensen we have chosen (from [8], p. 320): Proof. Let the n points x1 , x2 , ..., xn in I be so numbered that x1 ≤ x2 ≤ Pn λ = 1. If we let ... ≤Pxn . Let λ1 , λ2 , ..., λn ∈ [0, 1] be such that i i=1 ξ = ni=1 λi xi then ξ ∈ [x1 , xn ] ⊂ I. Take an arbitrary real number m ∈ [(Dl f )(ξ), (Dr f )(ξ)]. By (b) of Theorem 14.8, we have f (x) ≥ m(x−ξ)+f (ξ) for x ∈ I. Then n X

λk f (xk ) ≥

k=1

n X

λk {m(xk − ξ) + f (ξ)}

k=1

=m

n X

λk (xk − ξ) + f (ξ)

k=1

= f (ξ) = f (

n X

λi xi ).

i=1

We will need to reproduce the theorem which is mentioned in this proof and the proof for it as well to make it complete, so there we go:

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Theorem 5.4. (Theorem 14.8) Let f be a convex function on an interval I in <. Then we have: ...(b) For every xo ∈ I and every real number m ∈ [(Dl f (xo ), Dr f (xo )], we have f (x) ≥ m(x − xo ) + f (xo ) for x ∈ I. Proof. Let xo ∈ I and let m ∈ [(Dl f )(xo ), (Dr f )(xo )]. Let x ∈ I. If x < xo , (xo ) then by (3) of Theorem 14.5, we have f (x)−f ≤ (Dl f )(xo ) ≤ m. Then x−xo since x − xo < 0, we have f (x) − f (xo ) ≥ m(x − xo ). On the other hand, if (xo ) x ≥ xo , then by (3) of Theorem 14.5, we have f (x)−f ≥ (Dr f )(xo ) ≥ m x−xo so that f (x) − f (xo ) ≥ m(x − xo ). Thus, we have shown that for x ∈ I such that x 6= xo , we have f (x) − f (xo ) ≥ m(x − xo ). This inequality holds trivially for x = xo . We then have f (x) ≥ m(x − xo ) + f (xo ) for x ∈ I. Theorem 5.5. (Theorem 14.5) Let f be a convex function on an interval I (x1 ) in <. Then we have: ...(3) (Dr f )(x1 ) ≤ f (xx22)−f ≤ (Dl f (x2 )). −x1 Proof. To prove the right differentiability of f at any xo ∈ I, let x1 , x2 ∈ I be such that xo < x1 < x2 . Identifying xo , x1 , x2 respectively with u, v, w (xo ) (xo ) in the first inequality in Proposition 14.4, we have f (xx11)−f ≤ f (xx22)−f . −xo −xo This shows that

f (x)−f (xo ) x−xo

↓ as x ↓ xo so that the right derivative of f at

(xo ) xo , that is, (Dr f )(xo ) = limx↓xo f (x)−f exists in < and moreover we have x−xo (xo ) (Dr f )(xo ) ≤ f (x)−f for any x > xo . This proves the first inequality in (3). x−xo Let u, x ∈ I be such that u < xo < x. Identifying u, xo , x respectively with (u) (xo ) u, v, w in the Proposition 14.4, we have f (xxoo)−f ≤ f (x)−f . Thus, for −u x−xo

x > xo ,

f (x)−f (xo ) x−xo

is bounded below by the constant

f (xo )−f (u) . xo −u

Therefore,

(xo ) limx↓xo f (x)−f x−xo

we have (Dr f )(xo ) = ∈ < and f is right-differentiable at xo . Similar proof does for left-differentiability. Theorem 5.6. (Proposition 14.4) Let f be a convex function on an interval I in <. Then, for any u, v, w ∈ I such that u < v < w, we have f (v) − f (u) f (w) − f (u) f (w) − f (v) ≤ ≤ . v−u w−u w−v Proof. Since v ∈ (u, w), ∃λ ∈ (0, 1) such that v = λu+(1−λ)w. Solving this w−v v−u equation for λ, we have λ = w−u and then 1 − λ = w−u . By the convexity condition satisfied by f , we have f (v) ≤ λf (u) + (1 − λ)f (w) =

w−v v−u f (u) + f (w) w−u w−u

and then 15

(1)

(w − u)f (v) ≤ (w − v)f (u) + (v − u)f (w).

To derive the first inequality of the Proposition, let us write (1) as (w − u)f (v) ≤ [(w − u) − (v − u)]f (u) + (v − u)f (w). Then we have (w − u)(f (v) − f (u)) ≤ (v − u)(f (w) − f (u)). Cross-multiplying, we have the first inequality of the Proposition. To derive the second inequality of the Proposition, rewrite (1) as (w − u)f (v) ≤ (w − v)f (u) + [(w − u) − (w − v)]f (w) and then (w − u)(f (v) − f (w)) ≤ (w − v)f (u) − f (w). Cross-multiplying and then multiplying by -1 we have the second inequality of the Proposition. The proof above may be adapted to fit s1 −convexity easily. See: Theorem 5.7. Let f be an s1 −convex function on an interval I in <. Then for any x1 , ..., xn ∈ I and λ1 , ..., λn ∈ [0, 1] such that λs1 + ... + λsn = 1, we have f (λ1 x1 + ... + λn xn ) ≤ λs1 f (x1 ) + ... + λsn f (xn ). Proof. We assume our s1 −convex function, f , is non-decreasing. Therefore, choosing xo , x1 , x2 from Df , such that x2 < x1 < xo , we get f (x2 ) ≤ f (x1 ) ≤ f (xo ). Thus: • x2 < x1 < xo =⇒ xo − x1 < xo − x2 =⇒ • f (x2 ) ≤ f (x1 ) ≤ f (xo ) =⇒

f (x2 )−f (xo ) xo −x2

≤

1 xo −x1

>

1 xo −x2 ;

f (x1 )−f (xo ) xo −x1

=⇒

f (xo )−f (x2 ) xo −x2

f (xo )−f (x1 ) ; xo −x1

• x1 < u < xo =⇒ xo − x1 > xo − u =⇒ • f (x1 ) ≤ f (u) ≤ f (xo ) =⇒

f (u)−f (xo ) xo −u

f (xo )−f (x1 ) ; xo −x1

16

≥

1 xo −x1

<

1 xo −u ;

f (x1 )−f (xo ) xo −x1

=⇒

f (xo )−f (u) xo −u

≤

≥

• Given the second item above, we hold a monotonic non-decreasing sequence, as for the ratio of the differences forming our limit for our derivative in the point xo . Given the fourth item above, the sequence is also limited from above, therefore it converges by the trivial consequences of the Bolzano-Weierstrass theorem ([11]). As the points considered approach xo from below, we do have a left differential and we (xo ) may assert that n ∈ N, n 6= o, xn < xo =⇒ f (xxnn)−f ≤ Dl f (xo ) = −xo M; • xo < x1 < x2 =⇒ x1 − x0 < x2 − xo =⇒ • f (xo ) ≤ f (x1 ) ≤ f (x2 ) =⇒

f (xo )−f (x2 ) x2 −xo

≤

1 x1 −xo

>

1 x2 −xo ;

f (xo )−f (x1 ) x1 −xo

=⇒

f (x2 )−f (xo ) x2 −xo

f (x1 )−f (xo ) ; x1 −xo

• xo < u < x1 =⇒ x1 − xo > u − xo =⇒ 1 xo −u ; • f (xo ) ≤ f (u) ≤ f (x1 ) =⇒

f (u)−f (xo ) xo −u

≤

1 x1 −xo

<

1 u−xo

f (x1 )−f (xo ) xo −x1

=⇒

1 xo −x1

>

f (xo )−f (u) xo −u

≥

=⇒

f (xo )−f (x1 ) ; xo −x1

• Given the second item above, we hold a monotonic non-decreasing sequence, as for the ratio of the differences forming our limit for our derivative in the point xo . Given the fourth item above, the sequence is also limited from above, therefore it converges by the trivial consequences of the Bolzano-Weierstrass theorem ([11]). As the points considered approach xo from above, we do have a right differential (xo ) and we may assert that n ∈ N, n 6= o, xn > xo =⇒ f (xxnn)−f ≥ −xo Dr f (xo ) = M 0 ; Having proved the same needed lemmas for Ks1 , we just need to extend the proof to this set of classes now. See: Let the n points x1 , x2 , ..., xn in I ⊂ <+ be so numbered P that x1 ≤ x2 ≤ ... ≤ xnP . Let λ1 , λ2 , ..., λn ∈ [0, 1] be such that ni=1 λsi = 1. If we let ξ = ni=1 λsi xi then ξ ∈ [x1 , xn ] ⊂ I. Take an arbitrary real number m ∈ [(Dl f )(ξ), (Dr f )(ξ)]. Notice that because our assumption is that f never decreases, we know that m ≥ 0, for if it is negative the function must have changed the slope in a negative direction, that is, the point which comes later went further down than the point before it, so that the subtraction of the numerator of the ratio determining the derivative would return a negative value. Also notice that if all domain 17

≥

points are equal the inequality is trivially verified. By the analogous for Ks1 of (b) of Theorem 14.8, we have f (x) ≥ m(x − ξ) + f (ξ) for x ∈ I. Then f (x) ≥ m(x − ξ) + f (ξ) for x ∈ I. Then n X

λsk f (xk )

≥

k=1

n X

λsk [m(xk − ξ) + f (ξ)]

k=1

=m

n X

λsk [(xk − ξ)] + f (ξ)

k=1

= f (ξ) = f (

n X

λsi xi )

i=1

Because f is non-decreasing, we actually reach our result: n X k=1

λsk f (xk )

≥ f(

n X i=1

λsi xi )

≥ f(

n X

λi xi ).

i=1

The same lemmas used for the proof of the extension of Jensen’s to Ks1 functions may easily be considered valid for the monotonic cases, all of them, as well as for any f which is differentiable, at most subtle changes being due in the proofs for the Ks2 functions. On the other hand, if we prove that no s2 -convex function will ever be short of lateral limits, we get the extension guaranteed for our second type of S−convex functions, the proof being then made below and the theorem also being stated below, with due adaptations. Lemma 1. Every non-negative s2 −convex function has got both lateral limits, as for differential calculus, in all its domain points. Proof. Suppose that a certain s2 −convex function does not hold either the lateral limit to the left or the lateral limit to the right of a certain point xo of its domain, as for differential calculus. We then hold at least one f (xk )−f (xo ) xk ∈ Df /|xk − xo | <  but xk −xo − L > δ(), ∀δ(). It happens that, by using the definition of s2 −convex function, we may get any point as close to the other as we wish in the domain and we still get an upper limit for the ratio. See: Consider xk . We then have xo < xk < xk+1 , or any variation of this triplet which preserves order. Such implies, by the definition of s2 −convexity majorized for the right bound, f (xk ) ≤ f (xo )+f (xk+1 ) =⇒ 18

f (x

f (xk )−f (xo ) xk −xo

)

k+1 ≤ xk −x . Subtracting L from both sides, and evaluating the o right member, we get our guaranteed δ(), with  interfering with the size of the element to the right of the inequality. Therefore contradiction and such is impossible.

Theorem 5.8. Let f be an s2 −convex function on an interval I in <+ . Then for any x1 , ..., xn ∈ I and λ1 , ..., λn ∈ [0, 1] such that λ1 + ... + λn = 1, we have f (λ1 x1 + ... + λn xn ) ≤ λs1 f (x1 ) + ... + λsn f (xn ). Proof. Let the n points x1 , x2 , ..., xn in I ⊂ <+ be so numbered that x1 ≤ Pn x2 ≤ ... P ≤ xn . Let λ1 , λ2 , ..., λn ∈ [0, 1] be such that i=1 λi = 1. If we n let ξ = i=1 λi xi then ξ ∈ [x1 , xn ] ⊂ I. Take an arbitrary real number m ∈ [(Dl f )(ξ), (Dr f )(ξ)]. By the analogous for Ks2 of (b) of Theorem 14.8, we have f (x) ≥ m(x − ξ) + f (ξ) for x ∈ I. Then f (x) ≥ m(x − ξ) + f (ξ) for x ∈ I. Then n n n X X X s λk {m(xk − ξ) + f (ξ)} λk f (xk ) ≥ λk f (xk ) ≥ k=1

k=1

k=1

=m

n X

λk (xk − ξ) + f (ξ)

k=1

= f (ξ) = f (

n X

λi xi ).

i=1

Theorem 5.9. Let f be an s2 −convex function on an interval I in <− . Then for any x1 , ..., xn ∈ I and λ1 , ..., λn ∈ [0, 1] such that λ1 + ... + λn = 1, we have 1 1 f (λ1 x1 + ... + λn xn ) ≤ λ1s f (x1 ) + ... + λns f (xn ). Proof.

Pk

1

1

1

s s s i=1 pi f (xi ) = pk f (xk )+(1−pk )

1

Pk−1 i=1

p00i f (xi ), where p00i =

By induction, suppose that it is true for the sum up to (k − 1) that X  k−1 k−1 1 X pis f (xi ) ≥ f pi xi . i=1

Then: Make p0i = 1

pi

pks f (xk ) + (1 − pk )f (

P

i=1

1

.

(1−pks )

i=1

to get

1 (1−pks )s 1 k−1 s

pis

Pk

1 s

i=1 pi

1

P p0i xi ) ≥ f ( ki=1 pi xi ), once f ∈ Ks2 . 19

1

f (xi ) = pks f (xk )+(1−pks )

Pk−1 i=1

p00i f (xi ) ≥

VI. New result Definition 6. Given a function f : X ⊂ <− > <, the epigraph of f , denoted by epif , is the set of elements of X grouped by a limiting constant determining the sub-level set for the elements, that is: epitn f = {x ∈ X, tn ∈
VII. Updated results 1. Let V be a vector space over <. A subset X ⊂ V is called s1 -convex if every s1 −convex curve, defined by λs x1 + (1 − λs )x2 , ∀x1 , x2 ∈ X, intersects X in an interval, that is (λs x1 + (1 − λs )x2 ) ⊂ X when 0 < λ < 1 and x1 , x2 ∈ X; 2. Let V be a vector space over <. A subset X ⊂ V is called s2 -convex if every s2 −convex curve, defined by λs x1 + (1 − λ)s x2 , ∀x1 , x2 ∈ X, intersects X in an interval, that is (λs x1 + (1 − λ)s x2 ) ⊂ X when 0 < λ < 1 and x1 , x2 ∈ X; 3. For a real s1 −convex function Φ, numbers xi in its domain, and positive weights ai , we have  Pp Φ

1 s

i=1 ai xi Pp 1 ( i=1 ai ) s



Pp a Φ(xi ) Pp i ≤ i=1 ; i=1 ai

4. For a real non-negative s2 -convex function Φ, numbers xi in its domain, and positive weights ai , we have: Φ

 Pp  Pp s ai xi i=1 ai Φ(xi ) Pi=1 ≤ P ; p p ( i=1 ai )s i=1 ai

5. Let f be an s1 −convex function on an interval I in <. Then for any x1 , ..., xn ∈ I and λ1 , ..., λn ∈ [0, 1] such that λs1 + ... + λsn = 1, we have f (λ1 x1 + ... + λn xn ) ≤ λs1 f (x1 ) + ... + λsn f (xn ); 20

6. Let f be an s2 −convex function on an interval I in <+ . Then for any x1 , ..., xn ∈ I and λ1 , ..., λn ∈ [0, 1] such that λ1 + ... + λn = 1, we have f (λ1 x1 + ... + λn xn ) ≤ λs1 f (x1 ) + ... + λsn f (xn ); 7. Let f be an s2 −convex function on an interval I in <− . Then for any x1 , ..., xn ∈ I and λ1 , ..., λn ∈ [0, 1] such that λ1 + ... + λn = 1, we have 1

1

f (λ1 x1 + ... + λn xn ) ≤ λ1s f (x1 ) + ... + λns f (xn ); − − 8. An s1 −convex combination of a set of vectors → v1 , → v2 , ..., − v→ n is a special linear combination of those vectors in which the coefficients, which appear with a fixed exponent ‘s’, 0 < s ≤ 1, are all nonnegative and, put − − together they give us one as result, that is: λs1 → v1 + λs2 → v2 + ... + Pnin a sum, → − s s λn v , p=1 λp = 1, λp ≥ 0, is an s1 -convex combination of vectors; 9. The set of all s1 −convex combinations of the vectors is their s1 −convex − − span denoted by SCS1 (→ v1 , → v2 , ..., − v→ n ); − − 10. An s −convex combination of a set of vectors → v ,→ v , ..., − v→ is a special 2

1

2

n

linear combination of those vectors in which the coefficients, which appear with a fixed exponent ‘s’, 0 < s ≤ 1, are all nonnegative and, put − − v1 + λs2 → v2 + ... + together in a sum, they give us one as result, that is: λs1 → P n → − s λn v , p=1 λp = 1, λp ≥ 0, is an s2 -convex combination of vectors3 ; 11. The set of all s2 -convex combinations of the vectors is their s2 −convex − − span denoted by SCS2 (→ v1 , → v2 , ..., − v→ n ); 12. If f is an s1 −convex, non-negative, function and {x1 , x2 , ..., xn } ⊂ Df then:   n X x1 + x2 + ... + xn f (xi ) − f 1 ns i=1        n−1 x1 + x2 xn−1 + xn xn + x1 ≥ f + ... + f +f ; 1 1 1 n 2s 2s 2s 13. If f is an s2 −convex, non-negative, function and {x1 , x2 , ..., xn } ⊂ Df then: n X i=1

 f (xi ) − f

x1 + x2 + ... + xn n

3



The remark from p. 259 of [4] and the definitions for S-convexly independent sets are to simply be kept as they appear there, so that we will not repeat them here

21

       x1 + x2 xn−1 + xn xn + x1 2s−1 (ns − 1) f + ... + f +f ; ≥ ns 2 2 2 14. If f is an s1 −convex, non-negative, function and {a1 , a2 , ..., an } ⊂ Df then: (ns −1)[f (b1 )+f (b2 )+...+f (bn )] ≤ ns [f (a1 )+f (a2 )+...+f (an )]−nf (a), where a =

a1 +a2 +...+an 1 ns

1

and bi =

n s a−ai 1

(n−1) s

, i = 1, ..., n;

15. If f is an s2 −convex, non-negative, function and {a1 , a2 , ..., an } ⊂ Df then: (n−1)s [f (b1 )+f (b2 )+...+f (bn )] ≤ n[f (a1 )+f (a2 )+...+f (an )]−ns f (a), where a =

a1 +a2 +...+an n

and bi =

na−ai n−1 ,

i = 1, ..., n;

16. The closure of an s1 −convex set is s1 -convex; 17. The closure of an s2 −convex set is s2 -convex; 18. The interior of an s1 −convex set is also an s1 −convex set; 19. The interior of an s2 −convex set is also an s2 −convex set; 20. For every E ⊂ V , where V is a vector space, the union, s1 ch(E), of all s1 -convex combinations of the elements of E, that is, n n X X s1 ch(E) = { λsj xj ; λj ≥ 0; λsj = 1; xj ∈ E; n = 1, 2, ...}, 0

j=0

is an s1 −convex set; 21. For every E ⊂ V , where V is a vector space, the union, s2 ch(E), of all s2 -convex combinations of the elements of E, that is, n n X X s2 ch(E) = { λsj xj ; λj ≥ 0; λj = 1; xj ∈ E; n = 1, 2, ...}, 0

0

is an s2 −convex set; 22. ([17], p. 117-118, re-worded) Let E be a vector space over <. Denote by conv(.) the operation of taking the convex hull of a subset of E. In other words, for any set X ⊂ E, conv(X) denotes the smallest convex set in E containing X. Let Z ⊂ E. We shall say that Z is convexly independent (in E) if, for each point z ∈ Z, we have z 6∈ conv(Z\{z}); 22

23. The operation s1 ch(.), the equivalent to the previous notion for s1 −convex sets, is redundant because the results are the same as in conv(.), that is, for any set X ⊂ E, where E is a vector space over <, s1 ch(X) ≡ conv(X); 24. The operation s2 ch(.), the equivalent to the previous notion for s2 −convex sets, is trivial, for it will always give the closure of < as a result (every time one considers all possible S-convex combinations between the elements of a set, one ends up with an expansion, which would become the smallest S-convex set were it not the case that the expansion will generate a larger necessary smallest set again, the smallest set itself becoming a delusional concept).

VIII. Conclusions In this paper, we have been able to review, and update, our past research results, communicated through three of the most prestigious mathematical journals of nowadays. Besides, we also have discussed a few ideas of meaning and weight, providing inspirational writing to researchers in the concerned fields.

23

IX. References [1] M. R. Pinheiro. On S-convexity and our paper with Aequationes Mathematicae. Submitted. 2009. [2] M.R. Pinheiro. Jensen’s Inequality in Detail and S-Convex Functions. International Journal of Mathematical Analysis, V. 3, no. 1 − 4, (2009), pp. 95 − 98. [3] M.R. Pinheiro. Lazhar’s Inequalities and the S-convex Phenomenon. New Zealand Journal of Mathematics, V. 38, (2008), pp. 57 − 62. [4] M.R. Pinheiro. S-convexity: Foundations for Analysis. Differential Geometry - Dynamical Systems, V. 10, (2008), pp. 257 − 262. [5] S. Boyd and L. Vandenberghe. Convex Optimization. Cambridge University Press. ISBN-10: 0521833787. ISBN-13: 978 − 0521833783, (2004). [6] L. Hormander. Notions of Convexity. Springer. ISBN: 9780817645847, (2007). [7] H. Blumberg. On Convex Functions. Transactions of the American Mathematical Society, Vol. 20, No. 1 (Jan., 1919), pp. 40-44. Accessed online in June of 2009 at http : //www.jstor.org/pss/1988982. [8] J. Yeh. Real analysis : theory of measure and integration. World Scientific, 2nd ed., ISBN: 9812566538 (hbk.), 9812566546 (pbk.), (2006). [9] S. Zlobec. Jensen’s inequality for nonconvex functions. Mathematical Communications, V. 9, pp. 119-124, (2004). [10] R. C. James. Mathematics dictionary. Chapman & Hall, 5th ed., ISBN: 0412990415 (pbk.), (1992). [11] T. Rowland and E. W. Weisstein, Eric W. MathWorld–A Wolfram Web Resource. http : //mathworld.wolf ram.com/Bolzano − W eierstrassT heo rem.html, accessed online on the 27th of June of 2009. [12] E. W. Weisstein. “Argand Diagram.” From MathWorld–A Wolfram Web Resource. http : //mathworld.wolf ram.com/ArgandDiagram.html. 24

Accessed on the 29th of June of 2009, (1999). [13] J. O. Smith III. DSPRelated.com. http : //www.dsprelated.com/dspbooks/ mdf t/LinearC ombinationV ectors.html. Accessed on the 29th of June of 2009, (1999). [14] J.-B. Hiriart-Urruty and C. Lemarchal. Fundamentals of convex analysis. Springer, ISBN: 3540422056, (2001). [15] E. de Klerk. Interior Point Methods: Slides week 2, http : //stuwww.uvt.nl / edeklerk/ipms lidesw eek2.pdf . Accessed on the 2nd July 2009. [16] J. Dattorro. Convex Optimization & Euclidean Distance Geometry. Meboo Publishing USA, ISBN 0-9764013-0-4 (English), ISBN 978-1-84728064-0 (International), (2006). [17] A. B. Kharazishvili. Nonmeasurable sets and functions. 1st ed, Elsevier, ISBN: 0444516263, (2004).

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