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