# Coercive continuous function on a closed subset has a global minimum proof

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 $x^* \in X$?

• 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 $. – user81327 Apr 18 '17 at 15:12 • But with your definition,$e^x$is not coercive on$\mathbb{R}$because for$x\to -\infty$we have$e^x \to 0$. – Veliko Apr 18 '17 at 15:20 • There is a problem with the statement,$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)$. – zwim Apr 18 '17 at 15:23 • Thank you, @zwim. it is corrected.$|f(x_n)| \to +\infty$should be$f(x_n) \to +\infty$. With this definition,$f(x) = x$is not coercive. – Veliko Apr 18 '17 at 15:27 • Yes, something along those lines. I can only think of the unwieldy idea of considering the balls$ 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 $. – user81327 Apr 18 '17 at 15:58 ## 1 Answer Let choose any point in$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$. • I got it, thank you. Just a comment: by [-k, k] you probably mean the closed ball of radius$k$with a center 0. – Veliko Apr 18 '17 at 16:30 • Ah yes, I forgot you were working in$\mathbb R^n\$, just replace absolute values by norms and interval by the ball and it's ok. – zwim Apr 18 '17 at 16:31
• Yes, nothing changes in the idea. – Veliko Apr 18 '17 at 16:32