I need help with the proof of this corollary of the Weierstrass theorem:
Let X $\subset$ $\mathbb{R}^n$ be a bounded set and f : $\mathbb{R}^n$ $\rightarrow$ $\mathbb{R}$ an inferiorly continuous function that verifies $\lim_{|x|\to \infty}$ f(x) = + $\infty$. Hence it exists a $\hat{x}$ $\in$ X where f ($\hat{x}$) $\leq$ f (x) $\forall$ x $\in$ X.
Thank you so much!
This is not true. For example $f(x)=x$ is continuous but if $K=(0,1)$ then its infimum is not attained.
However this is true if $K$ is also closed. If $m=\inf_x f(x)$. There exists a sequence $x_n $ such that $f(x_n) \to m$. Since $x_n$ is bounded there is a convergent subsequence $(x_{n_k})$. Let its limit be $\hat {x}$. By lower semi-continuity we get $f(\hat x) \leq \lim f(x_{n_k})=m$. But $m \leq f(x)$ for all $x$ so we are done.
