The definition of the two comparable metrics is that:
Two metrics $d_1$ and $d_2$ on a set are said to be comparable if there exist constants $A,B>0$ such that$$d_1(x,y)\leqslant Ad_2(x,y),\ d_2(x,y)\leqslant Bd_1(x,y).\quad\forall x,y \in X$$
I try to do this by $A\geqslant d_1(x,y)/d_2(x,y)$ and to show the right hand side is converge or goes infinite, but this seems do no work.
I know when the $p$ goes to infinity, the $p$-norm is equal to sup-norm. But I don't know how to use it in this equation.
Any suggestions appreciated.
Obviously, if $|h(t)|\leq M$ for all $t\in[0,1]$, it is $\int_0^1|h(t)|dt\leq\int_0^1Mdt=M$. So applying for $M=\sup_{t\in[0,1]}|h(t)|$, we have that $\int_0^1|h(t)|dt\leq\sup_{t\in[0,1]}|h(t)|.$ Thus $d_1(f,g)\leq d_\infty(f,g)$ (apply the previous observation for $h=f-g$).
On the other hand, if $n\in\mathbb{N}$ we can define the function $f_n(t)$ by $f_n(0)=1$ $f_n(t)=0$ for $t\geq 1/n$ and $f_n(t)=-nt+1$ for $t\in[0,1/n]$. Then $f_n$ is a well defined function. Also, $d_\infty(f_n,0)=\sup_{t\in[0,1]}|f_n(t)|=1$, but $d_1(f_n,0)=\int_0^1|f_n(t)|dt=\frac{1}{2n}$. This indicates that there is no constant $B>0$ such that $d_\infty(f,g)\leq Bd_1(f,g)$. Why? (you can answer this!)