Suppose we have $f : \mathbb{R} \to \mathbb{R}$ continuous and $$\lim\limits_{x \to \pm \infty} f(x) = \infty.$$
Then must $f$ have a minimum on $\mathbb{R}$? Intuitively it seems so, and it's easy to prove for some arbitrary closed interval in $\mathbb{R}$ with the Extreme Value Theorem, but how can we apply that to an open interval like $(-\infty, \infty)$?
Best Answer
There exists a number $M$ with the property that $|x| > M$ implies $f(x) > f(0)$. This follows from the definition of the limit at $\pm \infty$.
$f$ attains a minimum on the interval $[-M,M]$.
If this minimum occurs at $x_0$ you have $f(x_0) \le f(x)$ for all $|x| \le M$ and $f(x_0) \le f(0) \le f(x)$ for all $|x| > M$. Thus $f$ has a global minimum at $x_0$.