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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

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    Do you know $f $ is continuous? I believe you also need that condition. – 2017-01-18
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    yes! I'm sorry, edited it. – 2017-01-18
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    I think, that some assumption about differentiability would be good, for example $f(x)=e^{-|x|}-0.5$ does not have maximum. – 2017-01-18
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    Thanks, I know that not every continous function is differentiable. So let's assume that, too. But the maximum should exist even if the function is not differentiable? For your example this would be for $x=0$. – 2017-01-18
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    @JaroslawMatlak $e^{-|x|}-0.5$ has a maximum at $x=0$. – 2017-01-18

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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 $\mathbb{R}$.

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    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 $f(a)$? – 2017-01-18
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    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 $[-R,R]$. – 2017-01-18
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    Observe that $f(x)\le f(a)/2$|x|\ge R$. – 2017-01-18