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In problem 4.55 of Boyd & Vandenberghe's Convex Optimization, the authors ask the following.

Show that in a multicriterion optimization problem, a unique solution of the scalar optimization problem

$$ \min. \max._{i =1,2\cdots q}F_i (x) $$ $$\text{s.t. } f_i(x)\leq 0 $$ $$ h_i(x)=0 $$ is Pareto optimal.

I know that, in a multicriterion optimization problem a solution is pareto optimal point if we can not find a better point. I assume that $x^*$ is the solution of the above scalar optimization problem. Now I have to show that for every feasible $y\neq x^*$ we have $$[F_1(x^*),~F_2(x^*), \cdots F_q(x^*)]\preceq [F_1(y),~F_2(y), \cdots F_q(y)].$$ I think the only other information that I have is that one of the $F_i(x^*)$ is greater than all of the rest of $F_j(x^*)'s$ for $j\neq i$. How to solve this problem? Thanks in advance.

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    $\begingroup$ Assume such a solution is not pareto optimal. Show a contradiction. $\endgroup$ – Mark L. Stone Aug 3 '19 at 22:53
  • $\begingroup$ You don't know that one of the $F_i(x^*)$ values is greater than all $F_j(x^*)$'s for $j \neq i$. For example, we might have $x^*=(3,3,3,3)$. What "unique solution" means here is that there is only one $x^*$ that solves the constrained min/max problem. $\endgroup$ – Michael Aug 4 '19 at 2:46
  • $\begingroup$ Keep in mind that the vectors might not always be ordered with respect to $\succeq$. Your definition of Pareto optimal is off. $\endgroup$ – Brian Borchers Aug 4 '19 at 4:53
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Assume $x^*$ is not Pareto optimal then wlog there exists $y \neq x^*$ such that $F_1(y) < F_1(x^*)$ and $F_i(y) \leq F_i(x^*)$ for all $i =2\cdots q$. Now, this implies $$ \max_{i =1,2\cdots q}F_i (y) \leq \max_{i =1,2\cdots q}F_i (x^*) = \min_{x} \max_{i =1,2\cdots q}F_i (x),$$ by definition of $x^*$. Therefore, $y$ is also a minimum to the original scalar optimization problem. By assumption, this minimum is unique, therefore $y = x^*$, a contradiction. Therefore, $x^*$ is Pareto optimal.

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