Let $X = C^0([0,1]$, then $\operatorname{Id} : (X, L^\infty) \to (X, L^1)$ is continuous.
The proof I seen goes like this:
$$||f||_{L^1} = \int_0 ^1 |f(x)| dx \leq \int_0^1 \sup\limits_{x \in [0,1]}|f(x)| = ||f||_{L^\infty}.$$
So we have $|| f -g ||_{L^\infty} \to 0 \implies || f - g||_{L^1} \to 0$.
I fail to see why this implies continuity in the topological sense. The way I see it we would have continuity of $Id : (X, L^1) \to (X, L^\infty)$.
Let $B_{\epsilon}^{L^i}(g)$, $i = 1, \infty$ be the open ball of radius $\epsilon$ centered in $g$ with respect to the metric $L^1$ or $L^\infty$.
Given any $\epsilon > 0$ and any $f \in B_{\epsilon}^{L^\infty}(0)$, we can find $\delta > 0$ such that $f \in B_{\delta}^{L^\infty}(f) \subset B_{\epsilon}^{L^\infty}(0)$ so in particular we have $f \in B_{\delta}^{L^1}(f) \subset B_{\delta}^{L^\infty}(f) \subset B_{\epsilon}^{L^\infty}(0)$. Meaning that $B_{\epsilon}^{L^\infty}(0)$ is also open in $(X, L^1)$.
What am I doing wrong?
To prove continuity at a point $f_0$ you have to show that given $\epsilon >0$ there exists $\delta >0$ such that $\{f: ||f-f_0||_1 <\epsilon\}$ contains $\{f: ||f-f_0||_{\infty} <\delta\}$. Just take $\delta = \epsilon$ and note that $||f-f_0||_{\infty} <\epsilon$ implies $||f-f_0||_1 <\epsilon$ since $||f-f_0||_1 \leq ||f-f_0||_{\infty}$