Let $(C[0, 1], d_1)$ and $(C[0, 1], d_2)$ be the metric spaces where
$$d_1(f, g) = \sup_{x∈[0,1]} |f(x) − g(x)|\\ d_2(f, g) =\int_{0}^{1}|f(x) − g(x)|dx \,$$
Is $id:(C[0, 1], d_1) \to (C[0, 1], d_2)$ homeomorphism?
My attempt : I think No. For homeomorphism $f^{-1}$ must be continious.Here $d_1$ is complete but $d_2$ is not complete. So i think $f^{-1}$ is not continious.
For example take $f(x)=x^n $ and $g=0$.Then $d_2(f,g)=0$ but $d_1(f,g)=1\neq 0$
Continuous functions preserve convergence, therefore, if $id$ is a homeomorphism, then a sequence $(f_n)_{n=1}^\infty$ converges to $f$ in $d_1$ iff it converges to $f$ in $d_2$.
However, there is a sequence that converges to 0 in $d_2$ but not in $d_1$. Let $f_n$ be the function that goes from 1 to 0 linearly on $[0, 1/n)$, and then is constant 0 on $[1/n, 1]$. For large $n$, $\int_{0}^1 f_n(x) dx$ is arbitrarily small, while $sup(|f_n(x) - 0|)$ is always 1.
Therefore, $f_n$ converges to $0$ in $d_2$ but not in $d_1$, so $id$ is not a homeomorphism.
