Prove that if $x$ is a local strict optima then $x$ is an unique global optima

Question

I'm thinking of the proof of the following Theorem:

Let's be $S\subset\mathbb{R}^n$ a non-empty and convex set and let $f:S\to\mathbb{R}$ a convex function in $S$. If $\bar{x}$ is a local optima of the problem "minimize $f(x)$ with $x\in S$" , then:

1. $\bar{x}$ is a local optima
2. If $\bar{x}$ is a strict optima or $f$ is strictly convex, then $\bar{x}$ is the unique global optima.

So far I have proved point 1 and point 2 when $f$ is strictly convex but I am having so trouble findig an argument to prove point 2 when $\bar{x}$ is a strict optima but $f$ is just convex.

I guess it is easy and that the argument has to be similar to that used when assuming $f$ strictly convex but I'm just not having "the happy idea" so any help would be appreciated.

What I've tried so far

Let $\bar{x}$ be a local strict optima that is not an unique global optima. Then,

• As $\bar{x}$ is a local strict optima there exists $\epsilon>0$ such as, for all $x\in B(x,\epsilon)\cap S, f(\bar{x})<f(x)$.
• And as it is not an unique global optima it exists $y\in S$ such that $f(y)<f(\bar{x})$.

$S$ is convex, so $\forall \lambda \in [0,1], z_\lambda=\lambda \bar{x} + (1-\lambda)y\in S$. Moreover, $\forall \lambda \in [0,1)$,

$f(z_\lambda)=f(\lambda\bar{x}+(1-\lambda)y)\leq\lambda f(\bar{x})+(1-\lambda)f(y)<\lambda f(\bar{x})+(1-\lambda)f(\bar{x})=f(\bar{x})$.

Then, if we take $\lambda$ close enough to $1$ we'll have $z_\lambda \in B(x,\epsilon)\cap S$ and that $f(z_\lambda)<f(\bar{x})$ which contradicts the fact of $\bar{x}$ being a local strict optima.

Answer 1

In the what I've tried so far section if we are supposing $\bar{x}$ not an unique global optima we have then that there exists $y\in S$ such that $f(y)=f(\bar{x})$. So we would have:

$S$ is convex, so $\forall \lambda \in [0,1], z_\lambda=\lambda \bar{x} + (1-\lambda)y\in S$. Moreover, $\forall \lambda \in [0,1)$,

$f(z_\lambda)=f(\lambda\bar{x}+(1-\lambda)y)\leq\lambda f(\bar{x})+(1-\lambda)f(y)=f(\bar{x})$.

Then, if we take $\lambda$ close enough to $1$ we'll have $z_\lambda \in B(x,\epsilon)\cap S$ and that $f(z_\lambda)\leq f(\bar{x})$ which contradicts the fact of $\bar{x}$ being a local strict optima.