Let the continuous function $f:\mathbb R→\mathbb R$ satisfying
$$\lim_{x\rightarrow -\infty}{f(x)}=\lim_{x\rightarrow +\infty}{f(x)}=0$$
Show that the function $f$ is:(i) bounded;
(ii) attain its minimum or maximum value on $\mathbb R$, that is, there exists a point $x_0\in \mathbb R$ such that $f(x_0)$ be the maximum or minimum value on $\mathbb R$.
For the part (i), let
$$S=\{f(x) : x \in \mathbb R\}$$
Since both $\inf S$ and $\sup S$ are finite, so the function $f$ is bounded.
For the part (ii), I would like to discuss the following four cases:
(a) when $x>M$, $f(x)<\frac{\sup S}{2} \neq 0$ and $f(x)>\frac{\inf S}{2} \neq 0$
(b) when $x<N$, $f(x)<\frac{\sup S}{2} \neq 0$ and $f(x)>\frac{\inf S}{2} \neq 0$
(c) Both $\inf S= \sup S=0$
(d) Either $\inf S=0$ or $\sup S=0$, then the function $f$ either attains its global maximum or global minimum.
How do I elaborate my idea for part (ii) (a) and (b) to show that any local extremum point on the closed interval $[N,M]$ is the global extremum point on $\mathbb R$?
Best Answer
In part (i), you should elaborate more on why $S$ is bounded. You shouldn't just claim it has a supremum and infimum; you'll need to appeal to continuity (specifically, the extreme value theorem).
In fact, let's show it, because I think it might give you a kickstart towards (ii). Since $\lim_{x \to \pm \infty} f(x) = 0$, we know that, choosing $\varepsilon = 1$ arbitrarily, there is some $N, M$ such that \begin{align*} x > N &\implies |f(x)| < 1 \\ x < M &\implies |f(x)| < 1 \end{align*} Consider now the restriction of $f$ to $[M, N]$. Now, $f$ is a continuous function over a compact interval, and thus must be bounded, by the extreme value theorem. That is, there's some $K$ (which is any number larger than the absolute values of the maximum and minimum $f$ over the interval) such that $$M \le x \le N \implies |f(x)| < K.$$ Now, putting it together, we know that $|f(x)| < \max \{1, K \}$ everywhere, thus showing $S$ is bounded (above and below).
Now, to do (ii), you're right about splitting into cases. First deal with the contant case. Otherwise $\sup f > 0$ or $\inf f < 0$. Deal with these cases separately. Try using the limit definition, this time with $\varepsilon = \frac{\sup f}{2}$ for the first case, and use this to construct an appropriate $M$ and $N$. You're pretty close, you just need a little more rigour.