# Is $f$ bounded on $[0,\infty)$

Prove/Disprove:

Let $f:[0,\infty )\to \Bbb R$ be a continuous function with $\lim_{x\to \infty }f(x)=0$.

Then $f$ has a maximum in $[0,\infty )$.

$\lim_{x\to \infty }f(x)=0\implies |f(x)|<1\forall x>G$ for $G$ large.

Now $f$ is continuous on $[0,G]$ hence is bounded therein i.e. $|f(x)|<M\forall x\in [0,G]$.

Take $A=\max\{M,1\}$;then $|f(x)|<A$.

Hence true.But the answer given is that the statement is false.

• You are wrong because there is no reason to believe $M \in [0,\infty)$ as your conclusion requires. All you've shown is that $f$ is bounded. – dannum Feb 27 '17 at 18:02

Since nowhere it states that $0<\sup\limits_{x\in[0,\infty)}f(x)$, consider $$f(x)=-\frac{1}{x+1}$$

In fact, you are proving only statements on $\lvert f\rvert$, not on $f$ itself.

• But if $|f|$ is bounded then so is $f$? Please clarify – Learnmore Feb 27 '17 at 17:41
• Yes. But not all bounded functions have a maximum. – user228113 Feb 27 '17 at 18:50
• Yes thank you very much – Learnmore Feb 28 '17 at 2:45

Another counter-example : $f(x)=-e^{-x}$.

We have $f([0,+\infty))=[-1,0)$.

Note that "$f$ has maximum in $[0,\infty)$" means that there exists a point $a$ in $[0,+\infty)$ such that $f(a) \ge f(x) \; \forall x \in [0,+\infty) \;$ (In this example). Which is not possible in this case.

• Well done +1 for you – Learnmore Feb 28 '17 at 2:45