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Suppose $f:\Bbb R\to\Bbb R$ is continuous such that $f(x)\to0$ as $|x|\to\infty$. Prove that $f$ attains either a maximum or a minimum.

My attempt at the question :

Given $\epsilon > 0 \ \ \exists \ \alpha > 0$ such that
$\ |f(x)| < \epsilon\ \ \ \forall \ \ \ \ |x| > \alpha$

Now, when $\ \ x \in [-\alpha,\ \alpha] $, by the Extreme value theorem, since $f(x)$ is a continuous function on a closed interval, it attains a maximum (say $ M$) and a minimum (say $m$).

I am stuck at this part. I think the way to move forward is by using the fact that we can make $ \epsilon $ as small as possible. But I don't know how to prove that it will surely attain a maximum or a minimum or both.

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  • $\begingroup$ Can it attain only a maxima but not a minima and vice-versa? Is there any such example? $\endgroup$ – Diya May 10 '15 at 15:49
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You are basically done. Here are 2 proofs:

WLOG assume f is not a constant. For every ε we can find an α such that for all x outside of (-α,α), |f(x)| is ε-small.

Since f is not constant it has a supremum S and infimum I over the whole space with S≠I. Since f is bounded, S and I are not infinite. Since S≠I, at least one is not zero. Choose ε small enough so that max(|S|,|I|)>ε. we then know that there is α such that |f(x)| < ε for all |x|>α, so we only have to maximise/minimise the function on the compact set [-α,α] which we can do by EVT. Either the max is S or the min is I; this proves the result.

Alternatively, choose $\alpha$ minimally (as an infimum) and assume without loss $f$ is not constant. Suppose for a contradiction that $f$ does not achieve its sup and also does not achieve its inf. If $M\geq\varepsilon$ then M is a global maximum and if $m\leq-\varepsilon$ then m is a global minimum. So the only way for this to happen is if $m$ and $M$ are in $(-ε,ε)$; this leads to a contradiction as ε→0. (see chat/comments for details)

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  • $\begingroup$ So, in this proof have you proved the existence of both max and min? Because if f(x) = 2exp(-x^2), then min does not exist. $\endgroup$ – skankhunt42 Oct 12 '14 at 16:28
  • $\begingroup$ no, I only proved it either has a minimum or a maximum. The opposite of this statement, which is 'has neither a min nor a max' leads to a contradiction. $\endgroup$ – Calvin Khor Oct 12 '14 at 16:32
  • $\begingroup$ Have edited the wording to make it clearer. $\endgroup$ – Calvin Khor Oct 12 '14 at 16:42
  • $\begingroup$ So, does choosing alpha minimally mean that if even if x is slightly less that alpha, |f(x)| will become >= epsilon? If yes, then I think I understand. Are you implying that if both m and M are in (-eps,+eps), then you can choose an alpha less than the old alpha which still satisfies the constraints (upto where the function achieves max or min, whichever |x| is larger) if I'm not wrong? Is that it? $\endgroup$ – skankhunt42 Oct 12 '14 at 16:49
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    $\begingroup$ Almost, it means 'if you use $\tilde{\alpha} < \alpha$ then you can find an $x ∈ [-\tilde{\alpha},\tilde{\alpha}]$ such that $|f(x)|\geq ε$'. This value $f(x)$ should then be either the minimum or maximum of $[-\alpha,\alpha]$, but it isn't. $\endgroup$ – Calvin Khor Oct 12 '14 at 17:00

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