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given two $C^{*}$ algebras $A\subset B $, acting on the same Hilbert space $H$, and $\delta $ is a derivation from $A$ into $B$,(in this case, a derivation is a linear mapping such that $\delta(ab)=a\delta(b)+\delta(a)b,\forall a,b \in A$) and assume it is bounded, then, is there an element $h$ of $A^{-}$(the weak operator closure of $A$), such that $\delta(a)=ha-ah, \forall a\in A$? \

Especially, I want to consider the case when $A,B$ are all commucative $C^{*}$ algebras.In other words, is there no nontrivial bounded derivation?

edited Mar 3 2011 at 4:25

asked Mar 3 2011 at 2:32

Jiang
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In the case where $A=B$, doesn't this follows from Sakai's theorem that all inner derivations from a von Neumann algebra to itself are inner? – Yemon Choi Mar 3 2011 at 2:58

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Actually, it seems to me that if the answer to your question were positive, then by taking B=B(H) one would deduce that every derivation from A to B(H) is inner, thus answering Kadison's similarity problem... – Yemon Choi Mar 3 2011 at 3:01

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ougao, if you have a more specific question you want to ask, you should either ask in the comments to the answer given, or ask a new question. – S. Carnahan♦ Mar 3 2011 at 11:48

I remarked in the comments to Kate's answer to the old version of this question that the new question is answered by an old paper of Kamowitz (Trans. AMS, 1962): ams.org/mathscinet-getitem?mr=170219 This paper was one of the ingredients that motivated Johnson to define amenability for Banach algebras, and the result I refer to can be generalized to the case of amenable commutative Banach algebras (of which $C(X)$ is a key example). – Yemon Choi Mar 3 2011 at 19:31

However, Kamowitz' paper may not be the best place to look for a proof of this result. Perhaps I will write up a short argument if I cannot find a self-contained argument; but there is nothing new about it, and it forms part of the basic knowledge needed by people thinking about bounded derivations – Yemon Choi Mar 3 2011 at 19:42

Kate Juschenko · Accepted Answer · 2011-03-03 03:11:24Z UTC

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The answer is no. Let $A$ be commutative von Neumann algebra. Let $T\in B(H)$ be such that $Ta-aT\neq 0$ for some $a\in A$. Then $\delta(a)=Ta-aT$ is bounded derivation but there is no element $h\in A^{-}=A$, such that $\delta(a)=ha-ah$ for all $a\in A$.

answered Mar 3 2011 at 3:11

Kate Juschenko
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	Ah, I missed the fact that the implementing element was meant to lie in the WOT-closure of A. – Yemon Choi Mar 3 2011 at 3:33
	good!but what if $A,B$ are all commutative $C^{*}$ algebras? – Jiang Mar 3 2011 at 4:22
	ougao, every (continuous) derivation from a commutative C*-algebra $A$ into a symmetric $A$-bimodule $X$ is zero; I belive this is due to Guichardet or Kamowitz. Do you know about amenability of Banach algebras? – Yemon Choi Mar 3 2011 at 5:09
	@Kate, sorry for my misfeasance, but I relly want to consider the special case, which I failed to explain clearly in my translation from the original problem faced in differential algebra to this one. – Jiang Mar 3 2011 at 10:00
	@Yemon Choi,thanks for your tip, I think the article written by J.R.Ringrose <Automatic continuity of derivations of operator algebras> is what I want. – Jiang Mar 3 2011 at 10:40

derivation between two $C^{*}$ algebras

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