0
$\begingroup$

I have that $f(x) : \mathbb{R}_+ \rightarrow \mathbb{R}$ is a uniformly continuous function s.t. $f(0) = \alpha \in \mathbb{R}_+$ and $lim_{x \rightarrow \infty} f(x)=0.$

I believe that if I can show that $\mathbb{R}_n$ is bounded, then I can use the extreme value theorem to show that the maximum must exist and be in $\mathbb{R}_+$. My idea is that I can define my distance function $d(x,y) = |x-y|$ and create a complete metric space. Thus my uniformly continuous function maps from a complete metric space to $\mathbb{R}$, making my function bounded. Since my function is uniformly continuous and bounded on a closed set $\mathbb{R}_+$, there exists $c$ and $d$ in $[0, \infty)$ such that:

$f(c) \leq f(x) \leq f(d) \forall x \in [0, \infty)$ with $d$ being my maximum.

$\endgroup$
6
  • 1
    $\begingroup$ So you use the convention $0\in\mathbb{R}_+$? $\endgroup$ – Jeppe Stig Nielsen Sep 16 '18 at 22:40
  • $\begingroup$ @JeppeStigNielsen sorry, I'm not understanding your question $\endgroup$ – Corran Horn Sep 16 '18 at 22:46
  • 1
    $\begingroup$ Is $\mathbb{R}_+ = \left[ 0,\infty \right)$, or is $\mathbb{R}_+ = \left( 0,\infty \right)$? I think it is true that if $f:\left[ 0,\infty \right) \to\mathbb{R}$ is continuous (uniformly continuous is not needed) and attains just one strictly positive value, and satisfies $\lim_{x\to+\infty}f(x)=0$, then $f$ attains its maximum. It will eventually be smaller than half that positive value, so you can look at a closed and bounded interval. $\endgroup$ – Jeppe Stig Nielsen Sep 16 '18 at 22:50
  • 1
    $\begingroup$ Since $f$ is apparently defined at $0$, I think you want $[0,\infty)$ as your domain. The fact that $f$ vanishes at infinity seems useful since for this positive number $\alpha$ we may find $A>0$ so that $|f(x)|<\alpha$ whenever $x>A$. $\endgroup$ – Matt A Pelto Sep 16 '18 at 22:50
  • $\begingroup$ @CorranHorn There are people who use the notation $\mathbb{R}_+$ for the poisitive reals. Anyway, the title is false when $f(0)<0$ as witnessed by $f(x)=-\exp(-x)$. $\endgroup$ – user10354138 Sep 16 '18 at 22:51
1
$\begingroup$

Split into three cases.

Case 1. $\alpha>0$. As $\lim_{x\rightarrow \infty} f(x) =0, \exists A >0$ s.t. $\forall x > A, |f(x)| < \alpha$.

Thus, we have a closed and bounded interval $[0,A]$, and so by the extreme value theorem we know that there must exist $c,d \in [0,A]$ s.t. $f(c) \leq f(x) \leq f(d) \,\forall x \in [0,A]$.

Thus, $d$ is the maximum of $f(x)$ on $[0,\infty)$.

Case 2. $\alpha =0$. Let $\exists x' \in \mathbb{R}_+$ s.t. $f(x')>\alpha$. As $\lim_{x\rightarrow \infty} f(x) =0, \exists B >0$ s.t. $\forall x > B, |f(x)| < f(x')$.

Thus, we have a closed and bounded interval $[0,B]$, and and we can apply the extreme value theorem.

Case 3. $\alpha =0$. Let $x'$ s.t. $f(x')> \alpha$ not exist $\in \mathbb{R}_n$. Then $\alpha \geq x \forall x \in \mathbb{R}_+$ and thus is our maximum.

Is this correct?

$\endgroup$
5
  • $\begingroup$ basically but please notice that $\mathbb{R}_+:=\{x \in \mathbb{R} : x>0\}$ and $[0,\infty):=\{x \in \mathbb{R} : x \geq 0\}$. These sets are not equal since $0$ is in the latter but it is not in the former. $\endgroup$ – Matt A Pelto Sep 16 '18 at 23:58
  • $\begingroup$ @MattAPelto I believe that $\mathbb{R}_+$ is $[0, \infty)$ while $\mathbb{R}_{++}$ is $(0, \infty)$? $\endgroup$ – Corran Horn Sep 17 '18 at 0:00
  • 2
    $\begingroup$ but you write $f(0) \in \mathbb{R}_+$. Hopefully you notice how we used the fact that $f(0)>0$ in finding this number $A>0$. If $f(0)\leq 0$ then the argument does not follow as written. $\endgroup$ – Matt A Pelto Sep 17 '18 at 0:08
  • 1
    $\begingroup$ If $\alpha=f(0)=0$, @MattAPelto is right you need another proof. Maybe treat two cases, (1) when $0$ is actually the supremum of $f$, and (2) when $f$ attains strictly positive values somewhere. $\endgroup$ – Jeppe Stig Nielsen Sep 17 '18 at 7:15
  • 1
    $\begingroup$ @MattAPelto I added to my answer to take into account your comment. Thank you for the help! $\endgroup$ – Corran Horn Sep 18 '18 at 3:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.