2
$\begingroup$

Definition: A function is said to be periodic with period $p>0$ if for every $x\in\mathbb{R}:f(x+p)=f(x)$. Prove that if $f:\mathbb{R}\to\mathbb{R}$ is continuous and $p$-periodic, then it has a global maximum and minimum in $\mathbb{R}$. That is, there exist points $x_1$ and $x_2$ such that $\forall x\in\mathbb{R}:f(x_1)\leq f(x)\leq f(x_2)$.

I think I need to use the mean value theorem but i'm not really sure. It is easy to see that there is a Min and Max point here, the proof is what makes me stuck.

$\endgroup$
  • 1
    $\begingroup$ The function only needs to be continuous, not differentiable. So you cannot use the mean value theorem. Instead, look for a theorem involving continuous functions and extremal points. $\endgroup$ – Florian Dec 22 '15 at 14:46
  • $\begingroup$ @Florian Extreme Value Theorem? $\endgroup$ – frosh Dec 22 '15 at 20:02
10
$\begingroup$

Hints : First prove that the function $f$ admits a min and a max on $[0,p]$. Secondly, try to prove that these min and max are global.

$\endgroup$
0
$\begingroup$

If it is continuous, then it can't have vertical asymptotes. If it is periodic, then it has to be continually bounce between the same sets of values. None of these can be infinity, because that would give you vertical asymptotes.

$\endgroup$

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.