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Let $A$ be a unital $C^*$ algebra. Assume that $D:A\to A$ is a bounded derivation.

Can one say that $1$ can not be in the image of $D$?

If the answer is no:

What is a counter example? What kind of $C^*$ algebra admits outer bounded derivation but stil they satisfy the above prevent property?

Motivation: I had intention to consider a simillar process to generat a kind of Legendre polynomial(a similar but not identical to it). So I wondered if requirement $D(P_1)=P_0$ is feasible for some appropriate derivation. On the other hand the impossibility $[x,y]=1$ leads us to search for outer derivation

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2 Answers 2

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$1$ cannot belong to the image of $D$.

Assume that $A\subset B(H)$. According to Theorem 4 from "Derivations of operator algebras" by Kadison any derivation $D$ is spatial, i.e. there is an operator $T \in B(H)$ such that $D(x) = [T,x]$. If $D(x) = 1$, then we would have represented the identity as a commutator of two bounded operators, which is not possible, as is well known.

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  • $\begingroup$ Thanks for your answer. In this representation is is it obvious that $1_A=1_{B(H)}$? $\endgroup$ Commented May 21, 2021 at 9:05
  • $\begingroup$ You can choose the representation to satisfy this. The usual procedure involving the GNS construction works, for example. $\endgroup$ Commented May 21, 2021 at 9:39
  • $\begingroup$ Yes I see The irreducible representation works otherwise the range of idempotent $1_A$ would be $A$-invariant. $\endgroup$ Commented May 21, 2021 at 10:16
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By adapting Popa's short (and slick) proof in

Popa, Sorin, On commutators in properly infinite W*-algebras, Invariant subspaces and other topics, 6th int. Conf., Timisoara and Herculane/Rom. 1981, Operator Theory: Adv. Appl. 6, 195-207 (1982). ZBL0529.46043.

of the Wiener-Weilandt theorem that the identity is not the commutator of two bounded operators, one can directly answer the OP's question without the need to invoke the result of Kadison, as follows.

If for contradiction we had $Dx=1$ for some $x \in A$, then we have $D(x^n) = n x^{n-1}$ for all $n \geq 1$, hence $D^n(x^n)=n!$. But $\|D^n(x^n)\| \leq \|D\|_{op}^n \|x\|^n$ grows at most exponentially in $n$, giving the required contradiction.

This argument in fact gives the more quantitative lower bound $\|Dx-1\| \geq \exp( - \|D\|_{op} \|x\|)$ when worked out more carefully; see Popa's paper for further discussion. There are also constructions that give somewhat matching upper bounds; see this recent paper of myself (for the case $A=B(H)$) and of Krishna-Johnson (for more general C^* algebras).

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  • $\begingroup$ Thank you very much for your very interesting answer and your attention to my question. $\endgroup$ Commented May 21, 2021 at 17:47

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