strict convexity attains minimum

exercise:

Let $f$ be a strictly convex function defined in an interval $A$. Suppose that there exist distinct points $a$ and $b$ in $A$ such that $f(a)=f(b)$ Show that $f$ attains a minimum.

If $f$ continuous in the closed interval $[a,b]$ then by extreme value theorem it attains it's minimum. It is a trivial case.

By the previous theorems we know that strict convex functions is continuous in an open interval $(a,b)$ and by definition of convexity $f(t a+(1-t) b)<t f(a)+(1-t) f(b)$ we have $f(a)=f(b)>f(x)$ for all $x$ in $(a,b)$

I thought a lot considering theese facts but I couldn't figure out. Could anyone help me, please.

Solution 1:

It is not true, as stated. Take, for example, the function $$f : [0, 1] \to \Bbb{R} : x \mapsto \begin{cases} x^2 & \text{if } 0 < x \le 1 \\1 & \text{if }x = 0.\end{cases}$$ Then $f$ is a strictly convex function, satisfying $f(0) = f(1)$, with an infimum of $0$ that is not achieved.

I don't believe this can be done without some assumption of continuity at the endpoints, which would put you back in the "trivial" case.