If every $x$ with $f(x) = f(\bar x)$ is local minimum, then $\bar x$ is global minimum.

Question

I am writing to ask for your help guys. The question I am having trouble with is:

Let $f: \mathbb{R}^n \to \mathbb{R}$ be a contionuous function and $\bar x \in \mathbb{R}$. Show that the following are equivalent:

1. $\bar x$ is a global minimum.
2. Every $x \in \mathbb{R}^n$ with $f(x) = f(\bar x)$, is a local minimum of $f$.

The proof that $(1)$ implies $(2)$ is quite direct, and for the other direction, from $(2)$ to $(1)$, I really have a really strong hunch that it is by contradiction but so far I am not successful. Any type of hint or comment or full solution is quite appreciated. Thanks to you all.

Answer 1

Indeed proof by contradiction is a good way to go about it. Suppose $\bar{x}$ is not a global minimum. That means there is some $y$ such that $f(y)<f(\bar{x})$. Now consider the function $$ g\colon t\in[0,1]\mapsto f(\bar{x}+t(y-\bar{x})). $$ Then let $t_0=\sup\{t\in[0,1]:g(t)=f(\bar{x})\}$ and $x_0=\bar{x}+t_0(y-\bar{x})$ satisfies ...

Answer 2

For all $x*\in\mathbb{R}^n$, if $f(x*)$ is local minimun for a open ball $B(x*)_r$ then there is $h>0$ such that $x\in B(x*)_{r+h}$ with $f(x)=f(\overline{x})$, then $f(\overline{x})\leq f(x*)$ Therefore $f(\overline{x})$ is a global minimun.