I'm stuck on this question : $Let~E=C(\left [0,1 \right ],\mathbb{R})~and~let~g\in E.~For~all~f\in E~we~set~N(f)=\left\|fg \right\|_{\infty} $
Where $\left\|\phi\right\|_{\infty}=\sup_{t\in \left [ 0,1 \right ]}\phi(t)$
I've shown that $N$ is a norm iff $g$ doesn't vanish on an intervall of $\left [0,1 \right ]$ and now I have to show that :
$$g(x)\neq 0~for~all~x\in \left [ 0,1 \right ]\iff N~and~\left\|. \right\|_{\infty}~are~equivalent~norms~on~E.$$ For the direct implication, it's not that hard, but for the reciprocal, I'm struggling to figure it out. I'm reasoning by contraposition, supposing that $\exists t_{0}\in \left [ 0,1 \right ]$ such that $g(t_{0}) = 0$ and I'm trying to exhibit a sequence $f_{n}$ such that $\displaystyle \lim_{ n\to \infty}\frac{N(f_{n})}{\left\| f_{n}\right\|_{\infty}}=0~or~\displaystyle \lim_{ n\to \infty}\frac{N(f_{n})}{\left\| f_{n}\right\|_{\infty}}=+\infty$. But all the examples I've tried to find don't work. I'll be grateful if you could help me on this one.
Hint:
Take $f_\epsilon(x)=0$ for $x\not\in (t_0-\epsilon, t_0+\epsilon)$ and $f_\epsilon(x) =1-\left|\frac{x-t_0}{\epsilon} \right|$ for $x\in(t_0-\epsilon, t_0+\epsilon)$.