0
$\begingroup$

Let $f:[0,1] \to (0,\infty)$ a continuous function. Then $\frac{f(x)}{x}$ attains its minimum in $(0,1]$.

I know by Weierstrass theorem that $f$ attains its minimum in $[0,1]$. I also know that $x$ attains its maximum on $x=1$. So it would make sense that $\frac{f(x)}{x}$ attains its minimum at some point, but I dont know how to write a proper proof to this.

Thanks.

$\endgroup$
  • $\begingroup$ This question has just been asked a few minutes ago and closed. $\endgroup$ – hamam_Abdallah May 23 at 1:41
  • $\begingroup$ Not the same question. Here I am asking for a proof. There I was asking for examples... $\endgroup$ – Wybie May 23 at 1:42
4
$\begingroup$

$f$ is (inferior) bounded by some $M>0$ so for $\epsilon_0>0$ small enough $M/\epsilon > f(1), 0<\epsilon \le \epsilon_0$; this means that the infimum of $f(x)/x$ on $(0,1]$ happens on $[\epsilon_0,1]$ but there $f(x)/x$ is continuos so the infimum is a minimum and is attained. Done!

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ good point - added inferior bounded as that was clearly meant in the context, but now I get that one can understand superior if not explicitly mentioned (the whole point is that $f(0)>0$ and $f$ is continuous at $0$ so $f(\epsilon)/\epsilon >f(1)$ when $\epsilon$ small enough $\endgroup$ – Conrad May 23 at 15:26

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.