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



On solvability of equation $D(x)=1$ where $D:A\to A$ is a bounded outer derivation on a $C^*$ algebra
$1$ 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.


