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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 $L_{n}$

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    $\begingroup$ Yes. Write $\phi_h(T)=Th$. Then $\mathrm{ker}\phi_h$ determines the line generated by $h$ (namely, this line is the intersection of all kernels of all $T\in\mathrm{ker}\phi_h$). So if $\phi$ has the same kernel as $\phi_h$ but is not of the form $\phi_{th}$ for any scalar $t$ then $\phi$ is not of the form $\phi_{h'}$ for any $h'$. With the help of the Hahn-Banach theorem, it's then easy to cook up examples . $\endgroup$
    – YCor
    Commented Jul 2, 2016 at 9:09
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    $\begingroup$ My previous comments provides example where the kernel is the same as that of some $\phi_h$. Here's a way to get other kernels. Let $u:H\times H\to H$ be a bounded isomorphism. Let $h_1,h_2$ be non-collinear vectors. Define $s:T\mapsto u(\phi_{h_1}(t),\phi_{h_2}(t))=u(Th_1,Th_2)$. Then $s$ is a bounded linear operator and its kernel is $\mathrm{ker}\phi_{h_1}\cap \mathrm{ker}\phi_{h_2}$, which is distinct from any $\mathrm{ker}\phi_{h}$ and still is a Lie subalgebra. $\endgroup$
    – YCor
    Commented Jul 2, 2016 at 11:07
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    $\begingroup$ A variant of the previous one: pick now an isometric isomorphism $v$ from the $\ell^2$-direct sum of infinitely many copies of $H$ onto $H$. Let's assume that $H$ is separable and let $(e_n)_{n\ge 1}$ be an orthonormal basis. Then $T\mapsto v(2^{-1}Te_1,2^{-2}Te_2,\dots)$ is a well-defined bounded operator and is injective. That is, the kernel is $\{0\}$, which is not the case for any $\phi_h$. $\endgroup$
    – YCor
    Commented Jul 2, 2016 at 11:11

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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 $T \in B(H)$ .

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