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For $1 \le p < \infty$, is $\mathscr B(L^p([0,1]))$ i.e. the space of all bounded linear operators on $L^p([0,1])$ separable in Operator norm topology?

My attempt: As $[0,1]$ a finite measure space, we have the inclusions $L^p[0,1] \subset L^1[0,1], \forall p$. Now as $C[0,1] \subset L^p[0,1]$ is dense, by Hahn-Banach and unique extension of bounded operators from dense subspaces, we have $\mathscr B(C[0,1]) \subset \mathscr B(L^p([0,1])) \subset \mathscr B(L^1([0,1]))$. Now using $C[0,1]$ is separable, can we proceed somehow?

Thanks in advance for help!

Brozovic
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  • I think that it's not separable. You can embed $L^{\infty}$ isometrically into it. – Olivier Moschetta Oct 08 '20 at 17:01
  • @OlivierMoschetta in this way? $L^\infty \to \mathscr B(L^p[0,1])$ by $f \mapsto (g \mapsto fg)$ ? – Brozovic Oct 08 '20 at 17:06
  • Yes. It might not be an isometry actually but it's a continuous embedding which is all that matters. – Olivier Moschetta Oct 08 '20 at 17:08
  • @OlivierMoschetta And a silly question: How do we show that $L^\infty[0,1]$ is not separable? By looking at indicator functions on the singletons of the Cantor set? – Brozovic Oct 08 '20 at 17:15
  • The top answer in this thread is top-notch. You look at indicators of $[0,r]$ for $r>0$. https://math.stackexchange.com/questions/97648/why-is-l-infty-not-separable

    Also it's not at all a silly question. It's a bit of a surprise at first glance that $L^{\infty}$ is not separable while all the $L^p$, $1\leq p<\infty$ are.

     – Olivier Moschetta Oct 08 '20 at 17:20
  • @OlivierMoschetta Thanks for the ideas. Would you like to post a complete answer, I'd like to close this question by accepting your answer. – Brozovic Oct 08 '20 at 17:26

1 Answers1

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No it's not separable. Consider the map $$L^{\infty}\rightarrow\mathcal{B}(L^p)$$ which maps $f\in L^{\infty}$ to $\psi_f$ which is the multiplication operator by $f$ as an operator from $L^p$ to itself. Clearly it is one-to-one. Since $$\|fg\|_{L^p}^p=\int_0^1 |fg|^p\,dx\leq \|f\|_{\infty}^p \|g\|_{L^p}^p$$ (where $g\in L^p$), it follows that $\|\psi_f\|_{\mathcal{B}(L^p)}\leq 1$ so that $L^{\infty}$ is continuously embedded in $\mathcal{B}(L^p)$. It suffices to show that $L^{\infty}$ is not separable.

However, this is a standard result of functional analysis, and the top answer in the following thread is a very neat proof.

Why is $L^{\infty}$ not separable?

Olivier Moschetta
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