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This problem might seem very elementary to someone. I am using the following arguments. I would like to make sure that it is right.

Let $f$ be a continuous function defined on $R$. If $\lim_{x\rightarrow\pm\infty}f(x)$ exists, then $f$ is a continuous function over the extended real line $R\cup\{\pm\infty\}$. Then we can say that there exists $x\in R\cup\{\pm\infty\}$ such that

$$ f(y) \le f(x), \qquad \forall y\in R. $$

The usual extreme value theorem is stated over a compact set $[a,b]$; see the wikipedia.

Thank you very much!


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The extended real line is compact. Open neighborhoods of $-\infty$ contain a set $(-\infty, x)$ for some $x \in \mathbb{R}$ and similar for $+\infty$. –  WimC Feb 26 '12 at 22:58
Thanks WimC. :-) –  Anand Feb 26 '12 at 23:02

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up vote 4 down vote accepted

The argument works, because the extended real line is compact: $\mathbb{R}\cup\{\infty,-\infty\}$ with the natural order topology is homeomorphic to $[0,1]$. An explicit homeomorphism is

$$h(x)=\begin{cases} \frac12+\frac1\pi\arctan x,&\text{if }x\in\mathbb{R}\\\\ 1,&\text{if }x=\infty\\ 0,&\text{if }x=-\infty\;. \end{cases}$$

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Thanks Brian M. Scott. :-) –  Anand Feb 26 '12 at 23:02

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