Derivations on Banach $*$-ideals in von Neumann algebras

It is known that any derivation $\delta: \mathcal M \to \mathcal M$ on the von Neumann algebra $\mathcal M$ is an inner, i. e. $\delta(x) := \delta_a(x) =[a, x] =ax -xa$, $x \in \mathcal M$, for some $a \in \mathcal M$. If $H$ is a separable infinite-dimensional complex Hilbert space and $\mathcal K(H)$ is a $C^*$-subalgebra of compact operators in $C^*$-algebra $\mathcal B(H)$ of all bounded linear operators acting in $H$, then any derivation $\delta: \mathcal K(H) \to \mathcal K(H)$ is a spatial derivation, i.e. there exists an operator $ a \in \mathcal B(H)$ such that $\delta(x) = [x, a]$ for all $x \in K(H)$. In addition, it has recently been established by Ber A. F., Chilin V. I., Levitina G. B. and Sukochev F. A. (JMAA, 2013) that any derivation $\delta: \mathcal{E}\to \mathcal{E}$ on Banach symmetric ideal of compact operators $\mathcal{E} \subseteq \mathcal K(H)$ is a spatial derivation. We show that the same result is also true for an arbitrary Banach $*$-ideal in every von Neumann algebra $\mathcal{M}$. More precisely: If $\mathcal{M}$ is an arbitrary von Neumann algebra, $\mathcal{E}$ be a Banach $*$-ideal in $\mathcal{M}$ and $\delta\colon \mathcal{E}\to \mathcal{E}$ is a derivation on $\mathcal{E}$, then there exists an element $ a \in \mathcal{M}$ such that $\delta(x) = [x, a]$ for all $x \in \mathcal{E}$, i. e. $\delta $ is a spatial derivation.