Exercise: For [a,b] a nondegenerate closed, bounded interval, show that there is no continuous mapping from $ L^{1}[a,b]$ onto $ L^{\infty}[a,b] $.
My initial thought is to prove by contradiction. Assuming that there indeed does exists a continuous mapping, say $f $, and then to use the definition of continuity by taking an open set in $L^{\infty}[a,b]$ and then show that it does not map to an open set in $L^1[a,b]$. However, I'm a little stuck as to how to begin (or if there is a better approach to the problem).
Hint: let $X,Y$ be metric spaces. If $X$ is separable and $f:X\rightarrow Y$ a continuous and surjective mapping, then $Y$ is separable. Now use the fact that $L^1[a,b]$ is separable, but $L^{\infty}[a,b]$ is not.