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A function $f:\mathbb{R}\rightarrow\mathbb{R}$ is $coercive$ if $$ \lim_{||x||\rightarrow\infty} f(x) = \infty.$$ Explicity, this condition means that for any $M>0$ there is an $R>0$ such that $||x||>R$ implies $f(x)\geq M$. Prove that if $f:\mathbb{R}^n\rightarrow \mathbb{R}$ is lower semi-continuous and coercive then $f$ is bounded from below and attains its infimum.

The definition I am using for LSC is: A function $f$ is lower semi-continuous on $X$ if for all $x\in X$ and every sequence $x_n\rightarrow x$, we have $$\lim_{n\rightarrow\infty} \inf f(x_n) \geq f(x). $$

This is my solution: Let $M>0$. Then since $f$ is $coercive$ there is an $R>0$ such that $||x||>R$ implies that $f(x)\geq M$. Let $x_n$ be a sequence such that $x_n\rightarrow x$. Since $f$ is LSC we have that $$ f(x)\leq \lim_{n\rightarrow\infty} \inf f(x_n)$$ which implies $$ M\leq f(x) \leq \lim_{n\rightarrow\infty}\inf f(x_n). $$ Hence $f$ is bounded below and attains its infimum.

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  • $\begingroup$ No you did not show that $f(x)$ is bounded below inside $|x|<R$, you need to use compactness for that (and lsc) $\endgroup$ – Evan Sep 20 '13 at 6:02
  • $\begingroup$ Ohhhh yes you are right I need to show that $f$ is bounded inside the n-cube $[-R,R]^n$. Thank you. $\endgroup$ – RDizzl3 Sep 20 '13 at 6:08
  • $\begingroup$ I have been thinking about this problem more and as you suggested I know you have to use the concept of compactness to show this is bounded from below inside the n-cube. My next question is how do I know $|x| < R$ is compact? $\endgroup$ – RDizzl3 Sep 20 '13 at 18:07
  • $\begingroup$ Well $|x|\leq R$ at least would be compact since it is closed and bounded. $\endgroup$ – Evan Sep 20 '13 at 23:22
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Let $x_n$ be a sequence such that $x_n\rightarrow x$.

The beginning of the attempted proof was okay, but this sentence is an aimless shot in the dark. "Let's take a sequence that converges somewhere" is not going to help you find where the infimum is attained.

This is the goal-oriented approach:

Let $x_n$ be a sequence such that $f(x_n)\rightarrow \inf f$.

Such a sequence exists for any function with finite $\inf f$, because for every $n$ we can find $x_n$ such that $f(x_n)<\inf f+1/n$. The next step is to extract a convergent subsequence from $x_n$; this is possible because the sequence $x_n$ is bounded (thanks to the coercitivity of $f$). Say, $x_{n_k}\to x$. The final step is to use the LSC property:

$$f(x)\le \lim f(x_{n_k}) = \inf f$$ hence $f(x)=\inf f$.

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  1. Suppose M is unbounded but closed. Let the level set denoted by $L:=\{x \in X | f(x) \le v\}, v> inf f(x), x \in X$ . And you need prove that $X \cap L$ is bounded(could be proven by means of contradiction). $$$$ 2.By Weirestress Theorem, if a lowersemicontinuous function is defined on a compact set then it is lower bounded and attains its minimum.
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  • $\begingroup$ Correction: Suppose X is unbounded but closed.( not M) $\endgroup$ – Sue Mar 1 '16 at 12:39
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Proof. Let $x_0\in R^{n}$ such that $f(x_0)< +\infty$. Since $F$ is coercive, there exist $r>0$ such that $$f(x_{1})\geq f(x_{0}),\quad\quad \forall x_{1}\in \{x|\|x\|>r\}.$$

If $r\geq \|x_{0}\|$, then $c_{0}=\max \{r,\|x_{0}\|\}=r$. Let $C=\{x\in R^{n} | \|x\|\leq c_{0}\}$. Since $f$ is lsc and $C$ is a compact set, there exist $x_{2}\in C$ such that $$x_{2} = \arg min\{f(x)| x\in C\}.$$ Thus, we have that $$f(x_2) = \inf \{f(x)| x\in R^{n}\}.$$

If $r<\|x_{0}\|$, then $c_{0}=\max \{r,\|x_{0}\|\}=\|x_{0}\|$. Let $C=\{x\in R^{n} | \|x\|\leq c_{0}\}$. Since $f$ is lsc and $C$ is a compact set, there exist $x_{2}\in C$ such that $$x_{2} = \arg min\{f(x)| x\in C\}.$$ Thus, we have that $$\inf \{f(x)| x\in R^{n}\}=\min\{f(x_0), f(x_2)\}.$$

The above two cases show that $f$ is bounded below.

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