If f is a convex function defined over the set $[a,b]$ then it achieves the maximum at one of the boundry points.

Question

The function f is convex, then $\forall x,y \in [a,b], t\in [0,1] \hspace{10mm} f(tx + (1-t)y) \leq tf(x)+(1-t)f(y)$.

I would like to prove that $f(x) \leq max{f(a), f(b)} \forall x \in [a,b]$ without using a calculus argument.

I understand that since f is convex, then $f(tx_1+(1−t)x_2) \leq tf(x_1)+(1−t)f(x_2) ≤ max({f(x_1),f(x_2)})$ So the convex combination always lies below something. However, we know that a boundary point can not be expressed as a linear combination of two elements in a set, unless they are the boundary point.

However, I do not know how the proof would follow.

Answer 1

The inequality you have written shows that $f(t) \leq \max \{f(a),f(b)\}$ for all $t \in (a,b)$ because any point $t$ in $(a,b)$ can be written as $sa+(1-s)b$ for some $s \in (0,1)$. [$s=\frac {b-t} {b-a}$]. If $f(a) \geq f(b)$ we can conclude that $f(t) \leq f(a)$ for all $t \in [a,b]$ so $f(a)$ is the maximum. Similarly if $f(a) \leq f(b)$ we can conclude that $f(t) \leq f(b)$ for all $t \in [a,b]$ so $f(b)$ is the maximum

Answer 2

For all $x\in [a,b]$ there is a unique $t\in [0,1]$ such that $x=at+b(1-t)$. Indeed, $$ t=\frac{x-b}{a-b} $$ and this fraction is seen to always lie in $[0,1]$. Now apply your same argument to conclude.