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!

  • 3
    $\begingroup$ It's fine, but might need a bit more detail for "and we can approximate $g$ with $Tf_{n_k}$ since $T$ is continuous". (You could show more generally that the continuous image of a separable space is separable.) $\endgroup$ Aug 27, 2014 at 0:06

1 Answer 1


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.

  • $\begingroup$ My pleasure, @Xiao. $\endgroup$
    – triple_sec
    Aug 27, 2014 at 0:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.