Prove: If $f(x)$ is a local maximum, then $f(x)$ is a global maximum.

Question

Let $C \subseteq \mathbb{R}^n$ be a convex set. A function $f:C \to \mathbb{R}$ is convex if for all $x,y\in C$ and all $t \in [0,1]$, $$f(tx+(1-t)y) \le tf(x)+(1-t)f(y).$$ Prove or disprove the following. Assume that $C \subseteq \mathbb{R}^n$ is a convex set and $f: C \to \mathbb{R}$ is a convex function.

1. If $f(x)$ is a local maximum, then $f(x)$ is a global maximum.

2. If $f(x)$ is a local minimum, then $f(x)$ is a global minimum.

My instincts tell me that 1 is false and 2 is true. I am not sure how to find a counter-example for number 1 but I think I can prove the second one using a contradiction.

Answer 1

I would agree with your instincts.

To show 1, you just need an example. The function $f(x) = x^2$ on $[-1,2]$ will do.

Hint: what are the local maxima? And don't forget to prove $f$ is convex.