Convex function then local minimizer is also a global minimizer opposite

Question

It has been shown countless times on this forum that a local minimizer of a convex function is also a global minimizer.

However, I'm wondering if this can be an iif. Basically, can we say the sufficient condition: if all local minimizers of function are also global minimizers then f is convex. I have a feeling this is not true (e.g. sin(x), all local minimizers are also global) but I'm not sure if my intuition is right.

Also, how to formalize that is is false?

Answer 1

Your intuition is right, and your $\sin x$ example is good enough as a case that contradicts the claim. To formally prove, you need to show that

1. Your counter example satisfies the first part of the claim; i.e. prove that all local minimizers of $\sin x$ are also global minimizers
2. Your counter example does not satisfy the second part of the claim; i.e. prove that $\sin x$ is not convex. For this single-variable function, you can prove this by showing that the second derivative is sometimes negative.

If you want a 'simpler' example, you could use $$\int x(x-1)(x+1)dx = \frac{x^4}{4} - \frac{x^2}{2} + c$$