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.

  • $\begingroup$ 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

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)

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.