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Suppose $ x^*$ is a local minimiser of $f$ over $\Omega$ , $\Omega \subset \Omega ' $, $x^*$ is interior point, how to prove that $x^*$ is also local minimiser over $\Omega '$. It looks obvious, i try to show that the neighbourhood of $ x^*$ is same over $\Omega$ as it is over $\Omega '$.

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$x^*$ is an interior point of $\Omega$? –  littleO Nov 3 '12 at 17:45

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$x^*$ is a local minimizer on $\Omega$ iff there exists an open set $U$ such that $f(x)\geq f(x^*)$ for all $x \in U\cap \Omega$.

Since $x^* \in \Omega^\circ$, the interior of $\Omega$, this shows that we can take the open set $U$ to be contained in $\Omega$, ie, $U \subset \Omega$. Since $\Omega\subset \Omega'$, we have $f(x)\geq f(x^*)$ for all $x \in U\cap \Omega'= U.$ Hence $x^*$ is a local minimizer of $f$ on $\Omega'$.

To see a trivial example of why $x^*$ being in the interior matters, take $f(x) = x^2$. Then if $\Omega = [1,\infty)$, you see that $x^* = 1$ is a local minimizer on $\Omega$, but not on $\Omega' = \mathbb{R}$.

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