Let $\omega$ be a positive linear functional on a $*$-algebra $A$ with identity $1$, and let $\ker(\omega) = \{x\in A:\omega(x) = 0\}$ be the kernel of $\omega$.
Does $\ker(\omega)$ have a left identity, i.e., does there exist an element $e$ in $\ker(\omega)$ such that $xe = x$ for each $x$ in $\ker(\omega)$?
And if not in general, are there any mild conditions under which this would be the case?
So far, it seems to me that the Cauchy-Schwarz inequality for positive linear functionals, and the fact that $\{x\in A:\omega(x^*x)=0\}$ is a left ideal, is related to my question, but does not really get me further (or at least I did not manage ;)).
Consider $C[0,1]$ and the positive linear functional $f \to \int f(x)\, dx$. Its kernel does not contain the identity.