Let $f(x)$ be a given continuous function on a closed interval $[a,b]$. Prove that for a given $r>0$ there exists $\theta > 0$ so that for all $a \leq x \leq b$, $\theta |f(x)| \leq r$.
Hello, can anyone help me with this question? From what I understand, we should use the extreme value theorem?
Let $r>0$ be arbitrary. By the extreme value theorem, there exist $\overline x,\underline x\in[a,b]$ such that $f(\overline x)\geq f(x)$ and $f(\underline x)\leq f(x)$ for all $x\in[a,b]$. Since $c:=\max\{\vert f(\overline x)\vert, \vert f(\underline x)\vert\}$ is a real number, there exists a $\theta>0$ such that $\theta c\leq r$. Pick e.g. $\theta = \frac{r}{c+1}$. Finally, the statement to be proven follows from the fact that $\vert f(x)\vert \leq c$ for all $x\in[a,b]$.