Does midpoint-convexity at a point imply midpoint-convexity at a larger point?

Question

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 $s$?

Answer 1

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 $(-1, 0)$.