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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 $L^\infty$.
Is this okay? thank you very much!
Fix $\varepsilon>0$ and $g\in L^{\infty}([0,1])$. If $T$ is surjective, then $g=T(f)$ for some $f\in L^1([0,1])$, indeed. By the continuity of $T$ at $f$, there exists some $\delta>0$ such that \begin{align*} \left.\begin{array}{ll}\bullet\,h\in L^1([0,1])\\\bullet\,\|h-f\|_1<\delta\end{array}\right\}\Longrightarrow\|T(h)-T(f)\|_{\infty}=\|T(h)-g\|_{\infty}<\varepsilon. \end{align*} NB: It is not assumed that $T$ is linear! (If $T$ is linear, one could, alternatively, use a similar argument based on its operator norm.)
By the denseness of $\{f_n\}_{n\in\mathbb N}$, there exists some $n\in\mathbb N$ such that $\|f_n-f\|_1<\delta$. It follows that $\|T(f_n)-g\|_{\infty}<\varepsilon$. Therefore, $\{T(f_n)\}_{n\in\mathbb N}$ is dense in $L^{\infty}([0,1])$, so $L^{\infty}([0,1])$ is separable—which is impossible.
Short answer: yes.
