The full question is below:
Let $S _ { \infty }$ be the set of continuous functions $f : [ 0,1 ] \rightarrow [0,1]$, with the distance $d _ { \infty } ( f , g ) = \sup | f ( x ) - g ( x ) |$, and let $S_1$ be the same set of continuous functions on $[0,1]$ but with the distance $d _ { 1 } ( f , g ) = \int _ { 0 } ^ { 1 } | f ( x ) - g ( x ) | d x$ . Now let $F : S _ { \infty } \rightarrow S _ { 1 }$ be the identity map taking any function $f$ to itself. Are $F$ and $F^{-1}$ continuous in this given metric space?
I believe that $F^{-1}$ will not be continuous due to a counter example of $f_n(x) = x^n$. Unless I am mistaken, this is because $f_n$ will converge to $0$ in $ (S_1,d_1),$ but won't converge in $(S_{\infty},d_{\infty})$. This would be enough proof that it does not converge, right? I am not sure how to prove that $F$ does or does not converge though. I would assume that it does converge, but I am not certain.
By the way, this is from a class on Real Analysis so my knowledge of topology is fairly limited.
Your example is fine for $F^{-1}$; do show that $f_n \to 0$ though under $d_1$.
For $F$ it's easy to see that it's Lipschitz:
$$d_1(f,g) \le d_\infty(f,g)$$
(we bound $|f-g|$ in the integral above by $d_\infty(f,g)$ and the length of the interval is $1$) and thus $F$ is (uniformly) continuous.