This post is actually a refined question of here.

Let $N$ be an abelian von-Neumann algebra. Define densely defined derivation to be derivation $\delta:D(\delta)\rightarrow N$ where the domain is dense subalgebra in norm topology. Does $N$ admit any densely defined derivations? I am most interested in $l^{\infty}(\mathbb{N})$ and $L^{\infty}[0,1]$. My gut feeling is that it doesn't since their real rank is zero, but I don't know how to show that.

This post is actually a refined question of here.

Let $N$ be an abelian von-Neumann algebra. Define densely defined derivation to be derivation $\delta:D(\delta)\rightarrow N$ where the domain is dense subalgebra in norm topology. Does $N$ admit any closed densely defined derivations? I am most interested in $l^{\infty}(\mathbb{N})$ and $L^{\infty}[0,1]$. My gut feeling is that it doesn't since their real rank is zero, but I don't know how to show that.

This post is actually a refined question of here. Let $N$ be an abelian von-Neumann algebra. Define densely defined derivation to be derivation $\delta:D(\delta)\rightarrow N$ where the domain is dense subalgebra in norm topology. Does $N$ admit any densely defined derivations? I am most interested in $l^{\infty}(\mathbb{N})$ and $L^{\infty}[0,1]$. My gut feeling is that it doesn't since their real rank is zero, but I don't know how to show that.

This post is actually a refined question of here. 

Let $N$ be an abelian von-Neumann algebra. Define densely defined derivation to be derivation $\delta:D(\delta)\rightarrow N$ where the domain is dense subalgebra in norm topology. Does $N$ admit any densely defined derivations? I am most interested in $l^{\infty}(\mathbb{N})$ and $L^{\infty}[0,1]$. My gut feeling is that it doesn't since their real rank is zero, but I don't know how to show that.