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First Note on the Definition ofs1convexity

I. M. R. Pinheiro

September 24, 2012

Abstract

In this note, we analyze a few major claims found in the scientific literatureabout K1

sand some of its characteristics/properties. As a consequence, we

nullify one major theorem and a couple of remarks of relevance.

MSC (2000): 26A51Key-words: Analysis, Convexity, Definition, S-convexity, geometry, shape.

1 Introduction

K1s

is a very interesting component of Sconvexity, not to say exotic: It differs,substantially, from K2

s, yet, in a certain sense, it seems to supplement it.

1.1 Notation

We use the symbols from [1] here:

K1s

for the class sconvex functions in the first sense, where s /0 < s 1;

K2s

for the class sconvex functions in the second sense, where s /0 < s 1;

K0 for the class convex functions;

s1 for the variable s, 0 < s1 1, used for the first type of sconvexity;

s2 for the variable s, 0 < s2 1, used for the second type of sconvexity.

Remark 1. The class 1convex functions is simply a subclass of the class convex

functions. If we make the domain of the convex functions be inside of the set of thenon-negative real numbers, we then have the class 1convex functions: K11 K

21

K0.

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1.2 Definition

We use the definition from [1] here:

Definition 2. A function f : X > is said to be s1convex if the inequality

f(1s x + (1 )

1s y)

f(x) + (1 )f(y)

holds / [0, 1]; x, y X; X +.

Remark 2. If the inequality is obeyed in the reverse1 situation by f, then f is told tobe s1concave.

2.1 Theorems that we discuss here

Dragomir and Pearce (in [2]) state that Hudzik and Maligranda, in [3], told us that:Theorem 2.1. Let 0 < s < 1. If f K1

s, then f is nondecreasing on (0, ) and

limu>0+ f(u) f(0).

We can infer, from the above theorem, that:(1) (Claim X) Any function in K1

s, with domain contained in (0, ), s specified,

s = 1, is nondecreasing;(2) (Claim Y) limu>0+ f(u) f(0) for f K

1s

, s = 1.In this paper, we prove that (1) is not true and (2) is incomplete, controversial, orunnecessary.

3 Analyzing Claim X

[2] presents the following sequence of implications as proof of the claim X:

Proof. We have, for u > 0 and [0, 1],

(PROBLEM 1) f[(1s + (1 )

1s )u] f(u) + (1 )f(u) = f(u).

The functionh() =

1s + (1 )

1s

is continuous on [0, 1], decreasing on [0, 12

], increasing on [ 12

, 1] and

(PROBLEM 2) h([0, 1]) = [h( 12

), h(1)] = [211s , 1].

This yields that(5.147) f(tu) f(u) for all u > 0, t [21

1s , 1].

If now t [211s , 1], then t

12 [21

1s , 1], and therefore, by the fact that (5.147) holds

for all u > 0, we getf(tu) = f(t 12 (t 12 u)) f(t 12 ) f(u) for2 all u > 0. By induction, we therefore obtainthat(5.148)(PROBLEM 3) f(tu) f(u) for all u > 0, t (0, 1].Hence, by taking 0 < u v and applying (5.148), we get

f(u) = f

(u

v.v

) f(v), which means that f is non-decreasing on (0, ).

1Reverse here means >, not .2Notice that the second member of the inequality should have been f(t

12 u), not f(t

12 ) as well

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We prove that PROBLEM 1, PROBLEM 2, and PROBLEM 3 will make the

proof not be a mathematical proof. There are more problems with the proof, how-ever.From [4], we learn that we cannot have x = y in the definition of Sconvexity (it isall tied to the geometric definition of convexity).This way, PROBLEM 1 should, per se, nullify the proof that we have just pre-sented.Notwithstanding, notice that 21

1s goes as close as we wish (PROBLEM 2) to 0

(limit when s > 0) and actually assumes the value 1 (when s = 1), what then makes

the interval [211s , 1] be a degenerated interval, or not be an interval, but just a point

instead.We do notice that s < 1 instead of s 1, as usual, in the theorem, however.Even if the result were true, and we are obviously entitled to seek it appealing tothe reasoning contained in the above proof, we cannot use the just-exposed lines as

proof.As for PROBLEM 3: Notice that, when s = 0.5, t [0.5, 1]. This means thatt [0.5, 1] instead of what is written there at least when s = 0.5.With that, we must have 0 < v 2u 2v in the next line, not 0 < u v, first of all,at least when s = 0.5.In this situation, just like in the original situation, in the proof, v and u cannot bevariables for f because whatever be a variable for f spans (0, ), with no discrimi-nation. We select only the values that obey the rule that we have created, which is,after due fixing, 0 < v 2u 2v. This way, we are using only a few selected valuesof the original variable of f. Because of that, either we create a new variable, whichwe call w, lets say, which is then going to be equal to u

v, with u and v assuming only

the values that we have selected, what means that the simplification is not valid in

the proof or we say that u and v are appearing as constants, one by one, in termsof values, whilst u, in f(u), is appearing as a variable, what then means that thesimplification is not valid and therefore the inference is not valid.We have then just proven that there is no proof of f being non-decreasing on (0, ).It is then the case that we either have to find a proper proof for the claim or a suitablecounter-example/proof of the contrary.Notice that any convex function is s1convex (K1s is supposed to extend K

11 , so that

this should be valid for any s1 we choose, provided that 0 < s1 1) if its domain is in+, in principle. Besides, when s = 1, we have precisely the class convex functions.This way, it suffices that we find a convex function that be decreasing to prove theclaim to be false.Easy enough, the quadratic function f(x) = x2 8x + 16, in the piece of domain(0, 4], is one of those.This claim is, therefore, FALSE.

4 Analyzing Claim Y

f(0) must exist because it appears in the theorem. Therefore, 0 is part of the domainof the function and we can replace the domain interval (0, ) with [0, ) at leastwhen stating the second part of the theorem.

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Because every sconvex function is continuous, we know that limu>c+ f(u) =

limu

>c f(u) = f(c) for each c in the domain of f, therefore limu

>0+ f(u) =limu>0

f(u) = f(0).As a consequence, limu>0+ f(u) f(0) and limu>0+ f(u) f(0) are both true.Since it is never true that limu>0+ f(u) < f(0) or limu>0+ f(u) > f(0), we shouldat most write that limu>0 f(u) = (0).The definition of K1

simplies any f such that f K1

sis continuous (simply imagine

that it be not. Imagine a discontinuity of first type, right at the vertex of a parabolathat has a point of minimum value, which we know is the image of a convex, thereforeSconvex in both senses, function. Now let (x1, f(x1)) be a point that is sufficientlydistant from the vertex, which is fully detached from the rest of the graph and liesmiles below it. Now make f(y), from the inequality of definition for K1

s, be the im-

age of the point that is a vertex. Notice that you will unavoidably find points in thegraph that are above the limiting line, what makes the function be both not convex

and not Sconvex, what is absurd).We will, however, try to find the mistake in the proof presented in [2].The proof in [2] is:

Proof. For u > 0, we have(PROBLEM 3) f(u) = f(u + 0) sf(u) + sf(0)and making u > 0+, we obtain(PROBLEM 4) limu>0+ f(u) limu>0+ f(u)

s limu>0+ f(u)+sf(0).

Hence,lim

u>0+

f(u) f(0).

PROBLEM 3 is that the assertion is only true if and satisfy the conditions ofthe definition ofK1

s, that is, if s + s = 1, what, as we know, implies 0 1 and

0 1.Even if such information is not relevant to the proof, we can only accept the proof asa proof if such constraints are mentioned.PROBLEM 4 is that it is missing explaining where the information f(u) f(u)came from. Since u u and their assumption was that the function, in this situation,does not decrease, what implies that f(u) f(u), not the opposite, things areunacceptable from this point onwards in the proof.Notice that this frontally opposes (5.147), for instance, and (5.147) is one of theirclaims.

5 Supplementary Remarks

Still in [2], we find a remark that is told to be in [3]:

Remark 3. If 0 < s < 1, then the function f K1s

is nondecreasing on (0, ) but notnecessarily on [0, ).

From Real Analysis, we know that this remark is absurd. It is not possible that onepoint, in a continuous function, change the nature of the function from nondecreasing

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to decreasing. That can only happen to a function with a discontinuity on x = 0.

Suppose that f : X + > is continuous and nondecreasing in (0, ) but notin [0, ).Then, for 0 < x1 < x2 < x3, we have f(x1) f(x2) f(x3) and f(0) > f(x1)( (f(0) f(x1))).We can then find (, x1), as close to zero as we wish but different from it, such thatf() f(x1).That would mean that limx>0+ f(x) = f(0) because f() = f(0), is as close as wewish to zero, and there is no other real number between them.Therefore, f is not continuous on x = 0.If we then say that K1

sfunctions do not have to be continuous, sustaining that they

extend the class convex functions will be a likely-to-be-impossible task.Besides, the definition of K1

sdoes not allow us to do that.

6 Conclusion

In this paper, we have completely nullified the following theorem:

Theorem 6.1. Let 0 < s < 1. If f K1s

, then f is nondecreasing on (0, ) andlimu>0+ f(u) f(0).

This theorem appears in [2] and is there reported to have originated in [3]. We provedthat there are plenty of decreasing functions that are in K1

sand if limu>0+ f(u)

f(0), then limu>0+ f(u) f(0) as well since limu>0 f(u) = f(0) for any sconvexfunction that be defined on (0, ) and also on zero. It does not make sense making atheorem to tell the just-written information because we aim objectivity and clarity inMathematics and the only actual piece of information that we need to know regarding

this is contained in the definition in an almost explicit way (continuity).

Remark 4. If 0 < s < 1, then the function f K1s

is nondecreasing on (0, ) but notnecessarily on [0, )

has also been nullified in this paper in what regards continuous functions and, there-fore, unless we change the definitions of convex and sconvex functions to now notbe about continuous functions, when one could then think of reassessing this remark,this remark has been nullified in full.

References

[1] M. R. Pinheiro. Convexity Secrets. Trafford Canada, 2008. ISBN 1-4251-3821-7.

[2] C. E. M. Pearce; S. S. Dragomir. Selected Topics on Hermite-Hadamard Inequalities andApplications. RGMIA monographs, http : //www.rgmia.vu.edu.au, 2002. Accessed inDecember of 2002.

[3] H. Hudzik; L. Maligranda. Some remarks on sconvex functions. Aequationes Mathe-maticae, 1994. Vol. 48, pp. 100 111.

[4] M. R. Pinheiro. Minima Domain Intervals and theSconvexity, as well as the Convexity,Phenomenon. Advances in Pure Mathematics, 2013. Vol. 3, no. 1.

I. M. R. Pinheiro, P.O. Box 12396, ABeckett St, Melbourne, Victoria, Australia, 8006, illmrpinheiro@gmail.com.

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