Let $f:(a,b) \to \mathbb{R}$ be a convex decreasing function that is constant in a nbh of the point $x$. Is it true that $f$ must be constant?
attempt: Let $c$ be this constant value and $z$ be a point with $f(z)\neq c$. I try to deduce a contradiction from here.
It seems obvious on a drawing because all the lines connecting the constant part and $z$ must lie above the graph which forces a discontinuity but I can't seem to make this formal.
If the hypothesis is for a fixed point $x$ the answer is NO. Take $f(x)=x^{2}$ for $-1<x<0$ and $0$ for $0 \leq x <1$.
If $f$ is constant in a neighborhood of each point $x$ the $f$ is a constant. This is easy consequence of compactness of any closed interval $[\alpha, \beta] \subset (a,b)$.