There is no continuous mapping from $L^1([0,1])$ onto $L^\infty([0,1])$.
Proof: suppose $T:L^1 \rightarrow L^\infty$ continuous and onto.
$L^1$ is separable, let $\{f_n\}$ be a countable dense subset. If $T$ is onto, for each $g\in L^\infty$, $g = Tf$ for some $f\in L^1$, and we can approximate $g$ with $Tf_{n_k}$ since $T$ is continuous. We get the contradiction that $\{Tf_n\}$ is a countable dense subset of $L^\infty$.
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