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Let $F(x)=\min\limits_{y\in \mathbb R^n}\{f(y)+\|x-y\|^2\} ,$ where $f(y)$ is convex and bounded below. How to show that

  1. if $x^*\in \arg \min \{F(x)\}$, then $x^*$ is in the closure of the effective domain of $f$.

  2. if $x^*$ is in the relative interior of the effective domain of $f$ and $x^*\in \arg \min \{F(x)\}$, then $x^*\in \arg \min \{f(x)\}$.

1 Answers 1

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Here are a few hints and sketches of proofs. If the effective domain of $f$ is empty, then it is trivial, so assume it is nonempty.

For 1, if $x^*$ is not in the closure of the effective domain, then there exists an open ball $B$ containing $x^*$ such that $f(x)=\infty$ for all $x\in B$. Let $y$ be such that $F(x^*)=f(y) + \|x^*-y\|^2$. Note that $y\neq x^*$ as $y$ is in the effective domain of $f$. Then choose x'\in B closer to $y$ than $x^*$. Then

F(x') \leq f(y) + \|x'-y\|^2 < f(y) + \|x^*-y\|^2 = F(x^*)

which is a contradiction.

For 2, let $y$ be such that $F(x^*)=f(y)+\|x^*-y\|^2$. Since $F(x^*) \leq F(y) \leq f(y)$ it follows that $y=x^*$. So $F(x^*)=f(x^*)$. Since $\inf F \leq \inf f$, it follows that $x^* \in \arg\min\{f\}$.

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    Consider $f(x)=x$ for x>0 and $f(x)=\infty$ for x < 0. Then ${\rm dom}f = (0,\infty)$, but $x^*=0$ minimizes $F$ (if you take $\inf$ instead of $\min$ in your definition of $F$, which I think you should be doing).2011-10-25