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Let $U$ be an open subset of $\mathbb{R}^n$ and $f:U\rightarrow \mathbb{R}$.

Consider the problem constrained optimization problem of minimizing $f(x)$ subject to $x\in A$, where $$ A=\{ x\in U\ \mid g(x)\le 0,\quad h(x)=0 \} $$ and $g:\mathbb{R}^n\rightarrow \mathbb{R}^p, h:\mathbb{R}^n\rightarrow \mathbb{R}^q$. As usual, $g,h$ can be thought of as $g=(g_1,...,g_p)$ and $h=(h_1,...,h_q)$. Take $x\in A$. If for an $i\in \{1,2,...,p\}$, one has $g_i(x)<0$, then we say that the restriction $g_i\le 0$ is inactive at $x$ and these kind of restrictions should be eliminated from the discussion. In the opposite case, when $g_i(x)=0$, we call yhis active (inequality) restriction. My question is: why does the restrcition $g_i\le 0$ not effectively influence the solution of the problem?

stefano
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  • It does influence the solution? The solution $x^$ must satisfy the constraint $g(x^)\leq 0$, not sure what you mean? – Jürgen Sukumaran Aug 15 '20 at 15:14
  • If for an $i$, one has $g_i(x)<0$, with $x\in M$, then the continuity of $g$ ensures the existence of a neighborhood $D$ of $x$ s.t. $g_i(d)<0$, for all $d\in D$. Therefore, when one looks for a certificate that $x^{}$ is a local solution of the optimization problem, the restriction does not effectively influence the set of points $y$ where one should compare $f(x^{})$ and $f(y)$. For this reason, one says that the restriction $g_i\le$ is inactive at $x$ and this restriction should be eliminated from the discussion. Why is this? I do not understand why the restriction doesn't matter. – stefano Aug 16 '20 at 11:22
  • You have to say what $M$ actually is before you make an argument using it. – Jürgen Sukumaran Aug 16 '20 at 12:28

1 Answers1

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The restriction (constraint) $g_i\le 0$ may influence the solution of the problem even when it is inactive, i.e., $g_i < 0$ at the optimum.

If the constraint $g_i\le 0$ were not imposed, the optimal solution might have $g_i > 0$, with corresponding lower value of f(x) than achievable with the constraint imposed, even if the constraint is inactive at the optimum.

Mark L. Stone
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