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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.

Where am I wrong?Please help.

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  • $\begingroup$ 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. $\endgroup$ – dannum Feb 27 '17 at 18:02
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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.

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

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    $\begingroup$ Well done +1 for you $\endgroup$ – Learnmore Feb 28 '17 at 2:45

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