If $f:\mathbb{R}^N\rightarrow\mathbb{R}$ is a continuous function and $\lim_{|x|\rightarrow \infty}f(x)=-\infty$, so for definition for all $N>0$ exists a $M>0$ such that $|x|>M$ implies $f(x)<-N$. Since I want search a global maximum, I can search it in $A=\{x\in\mathbb{R}^N : |x|\leq M\}$. It is a compact and so I can say that $f$ attains its maximum in some point $x_{0}$. How can I extend this for a upper semicontinuous function? Thank you.
Real Analysis – Maximum of an Upper Semicontinuous Function
analysisreal-analysissemicontinuous-functions
Best Answer
Recall that a function $f$ is upper-semicontinuous at $x_0 \in \mathbf R^N$ iff $$ \limsup_{x\to x_0} f(x) \le f(x_0) $$ Now let $K \subseteq \mathbf R^N$ be a compact subset, $m := \sup_{x \in K} f(x)$. For every $n \in \mathbf N$, choose $x_n \in K$ such that $f(x_n) \ge m - \frac 1n$. As $K$ is compact, some subsequence $(x_{n_k})$ converges, say $x_{n_k} \to x_0$. Then, by semi-continuity, $$ m \ge f(x_0) \ge \limsup_{x \to x_0} f(x) \ge \limsup_{k \to \infty} f(x_{n_k}) \ge \lim_k m - \frac 1{n_k} = m $$ Hence $m = f(x_0)$ and $f$ attains its maximum on $K$. Now use the same argument as for continuous functions.