Consider $g\in C[0,1]$ a strictly positive function and bounded away from zero, prove $\|f\|_g=(\int_0^1g(x)|f(x)|^2dx)^{\frac{1}{2}}$ is equivalent to $\|f\|=(\int_0^1f^2(x)dx)^{\frac{1}{2}}$.
I know that I have to show $a\|f\|\leq\|f\|_g\leq b\|f\|$ for some constants $a$, $b$. But, I have some problems of computing constants $a$ and $b$.
Your assumptions are there exists $c >0$ such that $c \leqslant g(x), \forall $x$, and that $g$ is continuous. Thus $g$ is also bounded above, by a constant $C$.
Thus, for every $f$ any $t \in [0,1]$ \begin{align} c |f(t)|^2 \leqslant g(t)|f(t)|^2 \leqslant C |f(t)|^2 \end{align} Integrating on $[0,1]$ and taking the square root answers your question.