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I would like to ask for help with the proof of the following proposition:

Let f be a real continuous function, defined on a closed set $X \subset \mathbb{R}^n$, which is coercive, i.e. for every sequence $\{x_n\}_{n=1}^\infty$ with $||x_n|| \to \infty$ we have $f(x_n) \to +\infty$. Then $f$ achieves global minimum on $X$.

My idea for a proof: If $X$ is bounded we are done by the Weierstrass extreme value theorem. But we only have closeness. So let $f^* = \inf\limits_{x\in X}f$ which is achieved at $x^*$ for which we don't know if $x^* \in X$. Let $Y$ be the closed ball of radius $||x^*||+1$. It surely contains $x^*$ and is bounded and so is $$ Z = X \cap Y \subset Y $$ Now we know that $Z$ is compact and $f$ achieves its minimum there. But are we sure (and if yes, why) that this minimum is actually $f^*$ and thus $x^* \in X$?

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  • $\begingroup$ You don't have $ x^* $ as easily as you are saying. Consider $ X = \mathbb{R} $, $ f(x) = e^x $. Then $ \inf f = 0 $, but $ f(x) $ is never 0 $. $\endgroup$ – user81327 Apr 18 '17 at 15:12
  • $\begingroup$ But with your definition, $e^x$ is not coercive on $\mathbb{R}$ because for $x\to -\infty$ we have $e^x \to 0$. $\endgroup$ – Veliko Apr 18 '17 at 15:20
  • $\begingroup$ There is a problem with the statement, $X=\mathbb R$ is closed and $f(x)=x$ is coercive according to your definition, yet does not have a global minimum. You need to remove the absolute value around $f(x)$. $\endgroup$ – zwim Apr 18 '17 at 15:23
  • $\begingroup$ Thank you, @zwim. it is corrected. $|f(x_n)| \to +\infty$ should be $f(x_n) \to +\infty$. With this definition, $f(x) = x$ is not coercive. $\endgroup$ – Veliko Apr 18 '17 at 15:27
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    $\begingroup$ Yes, something along those lines. I can only think of the unwieldy idea of considering the balls $ B_n $ of radius $ n $. By coercivity and continuity of f, and compactness of the unit sphere, there exists an $ N $ for which $ f $ is "large" on $ X \backslash B_N $, so that the minimum is in $ B_N \cap X $. $\endgroup$ – user81327 Apr 18 '17 at 15:58
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Let choose any point in $X$, call it $x_0$.

Since $f$ is coercive, then $\exists k>0\mid ||x||\ge k\implies f(x)\ge1+f(x_0)$.

Note: this is a simple way to guarantee that $f(x)>f(x_0)$. It is not a restriction since coercivity allows to find $k$ for any $A$, in particular $A=f(x_0)+1$.


Now $K=X\cap \overline{B(0,k)}$ is compact (since $X$ is closed and closed ball compact) so $f$ reaches a minimum in $x^*\in K\subset X$.

Also $x_0\in K$ else $||x_0||>k\implies f(x_0)\ge 1+f(x_0)$ which is a contradiction.

In particular $f(x^*)\le f(x_0)$.


Yet $\forall x\in X\setminus K$ we have $||x||>k$ so $f(x)\ge 1+f(x_0)>f(x_0)\ge f(x^*)$

So $x^*$ is a global minimum for $f$ on all $X$.

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  • $\begingroup$ I got it, thank you. Just a comment: by [-k, k] you probably mean the closed ball of radius $k$ with a center 0. $\endgroup$ – Veliko Apr 18 '17 at 16:30
  • $\begingroup$ Ah yes, I forgot you were working in $\mathbb R^n$, just replace absolute values by norms and interval by the ball and it's ok. $\endgroup$ – zwim Apr 18 '17 at 16:31
  • $\begingroup$ Yes, nothing changes in the idea. $\endgroup$ – Veliko Apr 18 '17 at 16:32

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