I presume the last term of the right-hand side of the defining inequality should be F(y)/2 rather than F(y/2).

I'll sketch a proof for F(x+y) >= F(x) + F(y), where x and y are integer multiples of the same factor, that is, x=ma and y=na for some integers m,n and nonzero real a.  Without loss of generality, we may assume that a=1 (define G(z)=F(az) and work with G).  For integer i, set g(i)=F(i+1)-F(i).  It follows directly from midpoint convexity that g(i)

yes that is correct, it should be F(y)/2 should be F(y/2);

Hmm, that is interesting as I noticed that under doubling as an equality or halving as an inequality F(y/2)=F(y)/2 or rather F(2x)=2F(x) that one achieves subadditivity

ie;s 1.F(x/2+y/2)

, I presume that generalizes to  F(3x)>=3f(x)

and thus if the super-addivity proof holds; So I presume that this in bold holds holds

F(3x)=F(2x+x)=>F(2x)+F(x)>=2F(x)+F(x))>=2F(x)+F(x), i via super-additivity again or but not to F(3x)>=3/2F(2x) or as to whether 2.F(2x +3y)>=F(2x)+F(3y)>=2F(x)+3F(y)

I presume standard convexity, and standard super additivity can express the former;F(2x +3y)>=F(2x)+F(3y)>=2F(x)+3F(y) even when F(0) is not set to zero, but maybe not F(3x)>=3/2F(2x) even when F(0)=1; I presume it expresses smaller then, for exponenents less then,rational exponents ie F(2x)1 although its unclear that convexity can always do so

,

Dear Taneli,

I presume that this weakened super-additivity then generalizes to G(x+x+x)? I will have to check the proof, it appears to make sense=1-1/2[F(1-x)/2+F(1-y)]/2 = 1/2[1-F(1-x)]+1/2[1-F(1-y)],

but due to symmetry again F(x)+F(1-x)=1, F(1-y)+F(y)=1

(1.6)1-F(1-y)=F(y), and 1-F(1-x)=F(x)

(Lhs of( 1.5) reduces to the equations below given (1.6)

1-1/2[F(1-x)/2+F(1-y)]/2 = 1/2[1-F(1-x)]+1/2[1-F(1-y)]= 1/2(F(x)+F(y)]

(1.7)so that F(x/2+y/2)>=1/2(F(x)+F(y)] but due to midpoint convexity;

midpoint concavity

(1) given midpoint convexity though (1)

(1.8)F(x/2+y/2)=1/2(F(x)+F(y)] which appear to reduce to

(1.9); F(x/2+y/2)=F(x)/2+F(y)/2 and jensens equatiion

F(x)/2+F(y)/2; F(x/2+y/2)>=F(x)+F(y)/2 but as I haveF(x+y)/2

 I presume that one either need convexity to express the rational fractions,' its not clear whether sub-addivity in itself always generalizes, as the inequalities may not be general enough. Once can express everything as an equality in the form of an integer ie

F(nx)=nF(x) one could easily express rational fractions of the sub-additive form. by finding the common terms. I don't know if sub-additivity or super-addivity in, and of itself can always do this; it can express F(sigma b (p)+ sigma2 y)>=F(sigmab) p)+F(sigma q); but I don't think that super-additivity with F(0)=0, in and of itself, can express that i.sF(sigma b (p)+ sigma2 y)>=sigma F(a)+sigma F(b), due to the sign changes in certain cases,  ie that where if sigma\times b= m/p. If one has integer equalities though one might be able to

F(sigma b)=m/qF(1/p)

F(b) =q F(1|p)

where m/q = sigma

F(sigma b)/F(b)=m/q so (simgab)=m/qF(b)=sigmaF(b) s; generally the equalities

For integer c F(cb)=cF(b)

I think all of the The integers equalities F(nx)=nF(x) are often to required nough to derive the rational equalities F(p/q x)=p/q F(x) for full addivtity,

but it would appear however that integer inequalities may not be enough to derive that result for rational inequalities in super-additivity; ie having F(cb)>=cF(b) wont be sufficient for F(t/q x)>=t/qF(x) ; it depends on the signs. If they are equalities though for the integers however, one should be able to derive the irrational inequalitiesF(t/q x)>=t/qF(x, if not come close to the rational equalitiesF(t/q x)=t/qF(x

(they would have to often equalities, if they are in integer form)

@Taneli Huuskonen  I note this post on maths stack exchange relating to super-addivity and midpoint convexity; particularly related to the inequalites F(x/3)

By Sierpinski theorem (see, i.g. [Hirsmann and Widder, Th. 9.1b]) every midpoint convex function is continuous. So if this function is rationally super-additive, it should be super-additive. On the other hand, there are midpoint convex functions that are not super-additive (see the Hille-Phillips monograf, Chapter 7, especially Th. 7.2.5).

Best regards,

Adolf.

A thanks for this; I was convinced of this. Ironically, if one has F(2x)=2F(x) and F(0)=0 the a midpoint convex function F:[0,1] to [0,1] comes out sub-additive. I wonder what A continuous/ (properly) convex function with this property, would have (ie that is also strictly monotonic increasing with F(1)=1, and F(0.5)=0.5). It would be nearly linear one would think (being both super-additive and sub-additive of a form) 

I am wondering if (without continuity) one can derive even rational super-additivity from midpoint convex F, F:[0,1] to [0,1] and F(1)=1 and F(0)=0, with F injective (increasing, or strictly increasing). I dont think one can unless monotone strict increasing-ness and bounded-ness (non-negativity) imply its continuity. 

I

@Adolf Mirotin

I made a post on mathstack exchange on a question by someone else, about examples of non-continuous or midpoint convex functions that are not convex or continuous. I suggested that if one could find one with F(0)=0 ; of just F(0)

@Adolf Mirotin

I note that F((1-x )+ (1-y))>=F(1-x) +F(1-y); I wonder whether letting y=t+1, and x =m+1, is an allowable substitution to infer super-additivity

I guess the real question is whether a midpoint convex function with f(0)=0 and f(1), F:[0,1] to [0,1] . f injective and increasing ,is sufficient in and of itself, for rational/dyadic super-additivity (inequality rational homogeneious) F(sigmax)>sigmaF(x) where sigma>1 and sigma rational, or just an integer

Generally if F midconvex and continuous and f(0)