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

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.

• This question has just been asked a few minutes ago and closed. – hamam_Abdallah May 23 at 1:41
• Not the same question. Here I am asking for a proof. There I was asking for examples... – 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!
• 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 – Conrad May 23 at 15:26