Prove that an optimal solution $x^*$ of the problem 1 $\min f(x)$ s.t $x\in \mathbb{R}^n$ and an optimal solution $(\bar{x},\bar{z})$ of the problem 2 $\min z $ s.t $z\ge f(x)\,, x\in \mathbb{R}^n$ are such that $\bar{z}=f(\bar{x})=f(x^*).$ Provide an example where $x^*\neq \bar{x}$.
The problem with proof structure has been resolved!
Now here is an idea:
Can we prove $x^*$ is global optimal of problem 1?
Hypothesis: There does not exist $(x,z)$ s.t $z\ge f(x),\, z
Claim: There does not exist $x'\in \mathbb{R}^n$ s.t $f(x')
Proof: Suppose there exists $x'\in \mathbb{R}^n$ s.t $f(x') . Then we could create $z=f(x)$ but then $z . For the example, where $x^*\neq \bar{x}$, here's what I think: \begin{align} \min \,f(x,y)=2x-3y\\ &s.t \,\,\, -2< x < 2\\ & \, \, \,\,\,\,\,\,\,\, 0< y < 4\\ &\,\,\, x,y\in \mathbb{R}\\ \end{align} \begin{align} \min \,z=2x-3y+1\\ &s.t \,\,\, -2< x <2\\ & \, \, \,\,\,\,\,\,\,\, 0< y < 4\\ &\,\,\, x,y\in \mathbb{R}\\ \end{align}
Any comments/suggestions/answers are appreciated.