Ex4-7

Chapter 4, Question 7

Question: Let $$f $$be usc and less than $$+\infty$$on a compact set $$E$$. Show that $$f$$ is bounded above on $$E$$. Show also that $$f$$assumes its maximum on $$E$$, that is, that there exists $$x_0\in E $$ such that $$f(x_0)\geq f(x)$$ for all $$x\in E$$.

Proof: Assume the contrary that $$f$$ is unbounded above. Then there is a sequence $$\{x_n\}$$ in $$E$$ such that $$\{f(x_n)\}$$is strictly increasing and

$$\lim_{n\to\infty} f(x_n)=+\infty. $$

Since a compact set is sequentially compact, we can choose a convergent subsequence $$\{x_{n_k}\}$$ which converges to $$x\in E.$$By definition, one has

$$f(x)\geq \limsup_{u\in E; u\to x}f(u)\geq \lim_{k\to\infty}f(x_{n_k})=+\infty,$$

the last inequality is valid because $$x_{n_k}\to x$$. This means $$f(x)=+\infty$$, contradicts the assumption that $$f<+\infty$$ on $$E$$.

Now let $$M=\sup_{x\in E}f(x) $$, we prove the supremum is attainable. For each $$n\in\mathbb N$$ we choose $$x_n\in E$$ such that $$f(x_n)>M-1/n$$. There is a convergent subsequence $$\{x_{n_k}\}$$ of $$E$$ which converges to $$ x\in E$$. Therefore, we have $$M\geq f(x)\geq \limsup_{u\to x; u\in E}f(u)\geq \lim_{k\to\infty} f(x_{n_k})\geq M.$$ Therefore, we have $$f(x)=M$$.