Let $f: [a,b] \rightarrow R$ be convex. If f is not constant, then the supremum of $f$ is not inside of [a,b].

Question

I could use some help with this proof.

Let $a,b \in \mathbb{R}, \ a<b \ \ f: [a,b] \rightarrow \mathbb{R}$ be convex.

Prove: If f is not constant, then the point, where the supremum of $f$ is, is not element of (a,b).

Answer 1

A non-constant convex function has no maxima on any open interval $(a,b)$:

Assume that $f \colon (a,b) \to \mathbb{R}$ is convex. Assume that $f$ is not a constant and has a maximum at $c \in (a,b)$. Then there exists some point $d \in (a,b)$ such that $f(c) > f(d)$. We can assume $d < c$, the other case goes similarly. Pick a point $e \in (c,b)$ and consider the line segment $$((1-t)d + te, (1-t)f(d) + t f(e)), \quad 0 \le t \le 1$$from $(d,f(d))$ to $(e,f(e))$. By convexity this lies above the graph of $f$, so we have $$f((1-t) d + t e)) \le (1-t) f(d) + t f(e).$$ Now choose $t'$ so that $(1-t')d + t' e = c$, then $$f(c) \le (1-t')f(d) + t'f(e) < (1-t')f(c) + t'f(c) = f(c),$$ which is a contradiction.