https://math.stackexchange.com/questions/2265396/can-this-strictly-increasing-convex-function-f-meet-its-linear-line-segment-in three places. See my post
Is this correct that this result or intuition would follow even if the convex function neither of the two differentiability constraints.
But only F:[0,1] to [0,1]
1F is Convex and strictly monotonically increasing
2. F(0.5)=0.5 F(0)=0. F(1)=1.
I presume that bijectivity will follow via surjectivity resulting the continuity of convexity, given the min and max points, F(0)=0 F(1)=1 and F strictly increasing, INjectivity being directly entailed by strict monotonic Increasing ness. But is really enough to specify three trivial points (the three points that one can usually derive even in a two outcome system) to convert a convex function into convave and convex and thus linear in this case F(x)=x.
I know that strictly convex functions (or at least strongly convex functions) often cannot meet at more then two points with their linear line segment . But these functions are incompatible with linearity to begin with.
In my case I used symmetry to induce concavity, But i noticed that even without it, and one posited convexity rather then midpoint convexity that it may be just incompatible for a convex function to be merely convex if it has to agree at three points; and it cant have local minima or maxima.
Given that it has to stay on or below the curve. Perhaps the conditions for continuity are much easier to get to, if i can get to jensens equations and rational homogeineity and essentially cauchy equation first, as a result of my symmetry conditions.
I can only presume that the conditons that imply convexity or continuity given midpoint convexity are tougher to meet then as the corresponding conditions relating [jensen equation with F(0)=0 and or cauchy equation] to linearity continuous function ;
In cauchy ;s case, if its really true, that non-negativity of the domain and range, in the sense specified below is sufficient then it borders on being trivial
Essentially it somes only requires F:[0,1] to [0,1] and F(1)=1, that is non negative domain and interval; where cauchys equation holds for all x and rational homogeineity. If what I have been informed about is correct
I hope its not true.
It would render my symmetry equations (almost) obsolete, and I DO NOT want to get rid of them, or have them rendered redundant . They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)
But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F:[0,1] to [0,1] where F(0)=0 and F(0.5)=0.5 and F(1)=1 even without symmetry.
Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" In my case, or in case of others due to some other property (in my case the symmetry equations). I am also worried that even star convex functions at 0 over [0,1] might effectively linear at least for x>0.5 F(x)=x.... I think, given three pts (even without symmetry).
I don't like that either.
insofar that F(x)=x if
I assumed(or continuity) or merely continuity at 1
where F strict monotonic increasing and midpoint point convex; and f (merely continuous at 1 or just continuous, so at to render it bijective convex and continuous) with
F, F::[0,1] to [0,1] ,
F(0)=0, F(1)=1 and F(0.5)=0.5
that F would be F(x)=x . I want F(x)=x, but only because of my symmetry equations- which result in jensens equation.
And not because, if i did not use them (my symmetry equations), i could have gotten the result, anyway due to this artefact of mathematics, concerning three fixed points. I want to ensure that the result F(x)=x, holds only due to my symmetry identities. These entail concavity given convexity or jensens equation, in the non-continuous case, given midpoint convexity. And not because, it would have still held without them given continuity due to this.
So I need to weaken midpoint convexity (and in way am i getting rid of three fixed pts or symmetry) to a form so that its continuous version does not have this result given the three fixed pts without symmetry. So that F(x)=x only holds even when F is presumed continuous, once symmetry is added,
Of course symmetry helps one get to continuity to. I want to keep strict monotonicity as well, because it has to represent the ordering.
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