0
$\begingroup$

$f$ is continuous at $[0,\infty)$

$\lim_{x\to\infty}f(x)=L$, $L \in R$

Let $g(x) = f(x) \cdot \sin x$

  1. I had to prove that $g$ is bounded in $[0,\infty)$ which I did.
  2. Now It is said that $L=0$ and I need to prove that $g$ gets a minimum in $[0,\infty)$. I know now that $\lim_{x\to\infty}g(x)=0$ (since $\sin x$ is bounded), $g$ is continuous in $[0,\infty]$. How can I use it to show that $g$ has a minimum?
  • 1
    In English, we usually say "bounded" instead of "blocked". – 2017-02-26
  • 0
    extreme value theorem? – 2017-02-26
  • 0
    @Omnomnomnom I'm always happy to improve my english here as well – 2017-02-26

1 Answers 1

1

Hint: It's tricky to come up with the setup, but the proof is easy once you have that.

Suppose that $f(x) \to 0$ as $x \to \infty$. Let $M = |\min \{g(x): x \in [0,1]\}|$. There exists an $R > 0$ such that $|f(x)| < M$ whenever $x > R$. Note that $g$ must attain a minimum over the interval $[0,R]$ (why?). Note that this minimum is necessarily a minimum of $g$ over $[0,\infty)$ (why?).

  • 0
    1. $g$ must attain a minimum in that interval since it is continuous in a bounded interval. 2. I can't figure out why that minimum has to be the minimum over $[0,\infty)$. In $(R, \infty)$, $|f(x)| < M$, therefore, $g(x) \le |f(x)| < M$, what other than that can I get? – 2017-02-26
  • 0
    See my latest edit. I think it makes more sense now. – 2017-02-26
  • 0
    The idea is that the minimum over $[0,R]$ is at most $-M$, but $g(x) > -M$ for all $x \in (R,\infty)$. – 2017-02-26
  • 0
    Sorry I got confused again, it could be that M is negative. therefore, $|f(x)| < M$ makes no sense – 2017-02-26
  • 0
    See my latest edit. – 2017-02-26