Let $f:[a,b] \to \mathbb [0,\infty)$ be a continuous function, and let $c \in (a,b)$ be a fixed point.
Suppose that $f$ is midpoint-convex at the point $c$, i.e.
$$ f((x+y)/2) \le (f(x) + f(y))/2, $$ whenever $(x+y)/2=c$, $x,y \in [a,b]$.
Is it true that $f$ is convex at $c$? i.e. does $$ f\left(\alpha x + (1- \alpha)y \right) \leq \alpha f(x) + (1-\alpha)f(y) $$ hold whenever $ \alpha \in [0,1]$ and $x,y \in [a,b]$ satisfy $\alpha x + (1- \alpha)y =c$?
Does the answer change if we assume $f$ is strictly decreasing?
The classic proofs do not seem to adapt to this case.
Pick $c=0$ and $f: [-1,1] \rightarrow [0,\infty), f(x)=-x^3+1$. Then we have for $c=0$ that $x=-y$.
$$ f(0)=1 = (-x^3 +1 +-(-x)^3+1)/2 = (f(x)+f(y))/2. $$ So $f$ is midpoint-convex. However, for $\alpha=2/3$ and $x=-1/2, y=1$ we have $$ \alpha x + (1-\alpha) y= \frac{2}{3}\cdot \frac{-1}{2} + \frac{1}{3} \cdot 1 = 0 = c $$ and $$ \alpha f(x) + (1-\alpha) f(y) = \frac{2}{3} \cdot \frac{9}{8} + \frac{1}{3} \cdot 0 = \frac{3}{4} < 1 = f(0). $$ So our function is not convex at $c=0$ (and $f$ is even strictly decreasing).