2

Let $f:ℝ → ℝ$, continous, differentiable with $f(a)>0$, $a ∈ ℝ$

$\lim\limits_{x \rightarrow \infty}{f(x)\leq 0} $ & $\lim\limits_{x \rightarrow -\infty}{f(x)\leq 0}$

I want to show, that there exists an Maximum, but my problem is that I don't have a closed interval here, so I can't use my theorems from school.

It's clear to me that there has to be a maximum, but I don't know how to show this. Can somebody give me some hints?

Thanks in advance.

Sheosha

Sheosha
  • 23
  • 4

1 Answers1

0

We know that $f(a)/2>0$. Using the definition of limit we find that there exists $R>0$ such that if $|x|\ge R$, then $f(x)\le f(a)/2$. Since $f$ is continuous, it attains its maximum on $[-R,R]$. Moreove the maximum will be al least $f(a)>f(a)/2$. It follows that the maximum on $[-R,R]$ is also the maximum on $\mathbb{R}$.

Julián Aguirre
  • 76,354
  • Thanks for your answer, that's a nice way to create a compact interval. But how can you be sure, that it's also the maximum on $\mathbb{R}$? Couldn't there exist a point outside the interval which is bigger then $f(a)$? – Sheosha Jan 18 '17 at 10:50
  • Using the very definition of limit, you can assume that for $|x|>R$ large enough (i.e. when going to infinity) you will be at most epsilon above zero (as the limit is smaller or equal), and hence most surely smaller than $f(a)$, so the minimum is achieved inside the compact interval $[-R,R]$. – b00n heT Jan 18 '17 at 11:33
  • Observe that $f(x)\le f(a)/2<f(a)$ for $|x|\ge R$. – Julián Aguirre Jan 18 '17 at 13:25