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Let $f:(-\infty,0] \to \mathbb [0,\infty)$ be a $C^1$ strictly decreasing function satisfying $f(0)=0$.

Given $c \in (-\infty,0]$, we say that $f$ is midpoint-convex at the point $c$ if

$$ f((x+y)/2) \le (f(x) + f(y))/2, $$ whenever $(x+y)/2=c$, $x,y \in (-\infty,0]$.

Question: Let $r<s<0$, and suppose that $f$ is midpoint-convex at $r$. Is $f$ midpoint-convex at $s$?

Asaf Shachar
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1 Answers1

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No, not necessarily. Take, for example: $$f(x) = 1 - (x + 1)^3.$$ Then $f$ is $C^1$, strictly decreasing and midpoint convex at $x = -1$ (since $x^3$ is odd). But, it is not midpoint convex at any point in $(-1, 0)$, as the function is strictly concave on $(-1, 0)$.

user804886
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  • Thank you very much for your nice answer. I also thought of $-x^3$ but somehow got confused about where it is concave and where it is convex. Perhaps you will have an idea about this follow-up question as well? https://math.stackexchange.com/questions/3741024/does-midpoint-convexity-at-a-point-imply-midpoint-convexity-at-other-points – Asaf Shachar Jul 01 '20 at 08:37