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.


  • $\begingroup$ 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

$f$ 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!

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  • $\begingroup$ 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

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