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Let $P$ be the problem \begin{alignat}{2} &\text{min} \ \ && f(\mathbf{x})\\ &s.t && g(x) \leq 0\\ & &&x\in X \end{alignat} for convex functions $f$ and $g$ over $\mathbb{R}^n$ and $X\subseteq\mathbb{R}^n$ is a convex set. Then if $\mathbf{x}^*$ is an optimal solution to $P$ with $g(x)<0$, then $\mathbf{x}^*$ is also an optimal solution to the problem $Q$: \begin{alignat}{2} &\text{min} \ \ && f(\mathbf{x})\\ & &&x\in X. \end{alignat}

I'm having trouble seeing how exactly $x^*$ would be an optimal solution to $Q$. Is there a way to exploit $g$ and its convexity to show this?

Thanks

user722447
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1 Answers1

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Let $h_P(x) = f(x)$ over $g(x) \le 0$ and $x \in X$ and prolongate it with $+\infty$ otherwise, and $h_Q(x)=f(x)$ over $x \in X$. Then $P$ and $Q$ can be rewritten as $$\min_{x \in \mathbb{R}} h_P(x)$$ and $$\min_{x \in \mathbb{R}} h_Q(x).$$

By optimality condition on convex problems using subgradients, we know that $x^*$ is an optimal solution of $P$ iff the subgradient of $h_P$ at $x^*$ contains $0$.

The set $\{x \in X,g(x)<0\}$ is open in $X$ since $g$ is continuous. See here for details.

If $x^*$ satifies $g(x^*) <0$, either $x^*$ lies in the interior of $X$ and $h_P=h_Q$ on an open set around $x^*$ and the subgradient of $h_P$ at $x^*$ is the same as the subgradient of $h_Q$. Or $x^*$ lies in the boundary of $X$ and $h_P=f=h_Q$ also around it since $g$ does not interfere, and we can make the same conclusion on subgradients.

Since $x^*$ is an optimal solution to $P$, the subgradient of $h_P$ at $x^*$ contains $0$. Hence, the subgradient of $h_Q$ at $x^*$ also contains $0$ and by optimality condition aforementionned, $x^*$ is a solution to $Q$.

nicomezi
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  • Sorry, i forgot to write the added condition, that $g(x)<0$ – user722447 Nov 06 '19 at 05:46
  • It should be fine now. I messed up with some details. – nicomezi Nov 06 '19 at 07:04
  • @nicomezi Can we say that the set of optimal solutions of $H_P(x)$ is subset of optimal solutions of $H_Q(x)$ and because $x^*$ is in the first he must be in the second and we get the same result? – convxy Nov 23 '20 at 08:52
  • I think @user722447 ment that $g(x^*)<0$ and not $g(x)<0$ for all $x$ – convxy Nov 23 '20 at 09:06
  • They have the same set of solutions as $h_P$ and $h_Q$ are only different where the (equivalents) constraints are not satisfied. – nicomezi Nov 23 '20 at 11:07