I am having trouble with a question, just hope that someone out there is kind enough to help me with it. Thanks in advance!!
Suppose $f := \Bbb R^n \rightarrow \Bbb R$ is continuous, and $f$ has the following properties:
(i) $f(x) \ge 1$ for all $x$ such that $x \in S$
(ii) there exists an $x$ such that $x \notin S$
Show that $f$ has a global minimizer.
I think , that in (ii) it should read $f(x)<1.$
Let $K:=\{x \in \mathbb R^n: ||x||_{\infty} \le r\}.$ $K$ is compact and $f$ is continuous, hence there is $x_0 \in K$ such that
$$(*) \quad f(x_0) = \min f(K).$$
Now assume that $||x_0||_{\infty}=r$. By (i) we see that $f(x_0) \ge 1.$This leads to a contradiction to $(*)$, since, by (ii), there is $x \in K$ with $f(x) <1.$
Therefore $||x_0||_{\infty}<r$. By (ii), there is $x \in K$ with $f(x) <1.$ Hence
$f(x_0) \le f(x) <1.$
With (i) and $(*)$ we get
$$f(x_0) = \min f( \mathbb R^n).$$
