Assume that $H$ is an infinite dimensional Hilbert space.The space of all bounded operators on $H$ is denoted by $B(H)$.We consider the Lie algebra structure $[T,S]=TS-ST$ on $B(H)$.



> 
> Is there a bounded linear operator $\phi: B(H)\to H$ such that$ \ker \phi$
>  is a Lie subalgebra but $\phi$ is not in the form of $\phi(T)=Th$ for some $h\in H$?
> 
> 
> 


The following post is an implicit motivation for our question:


[A particular Lie algebra $L\_{n}$ and (various) lie groups whose Lie algebra is isomorphic to $L\_{n}$](https://mathoverflow.net/questions/225811/a-particular-lie-algebra-l-n-and-various-lie-groups-whose-lie-algebra-is-i/225815)



Bounded operators $T: B(H)\to H$ whose Kernel is a Lie algebra
The answer is yes :


Choose injective $A \in B(H)$ and $h \in H$ such that $h$ and $Ah$ are linear independent and define $\phi(T) = A T h$.


Then it is easy to see that there doesn't exist a $k \in H$ such that $A T h = T k$ for all $T \in B(H)$ .


