 #101
julian
Gold Member
 584
 105
Increasing because of claim 3 above.Or decreasing
There don't have to be only finitely many intersections.
Why can't there be mixed concavity on each interval? The points where ##f(x)## changes concavity won't be the same points where ##f(x)=x## in general.
I don't see how you are using concavity for this.
@archaic has an almost complete solution.
Convex up implies ##r \equiv f(f(f(r))) > r## for ##p< r < q## from attached figure. Seems the only way to not have a contradiction is if ##f(x) = x##.
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