Let $\mathcal{H}$ be a Hilbert space, $A:\mathcal{H}\rightarrow\mathcal{H}$ a Hilbert-Schmidt operator on it, and $H_0:\mathcal{D}(H_0)\rightarrow\mathcal{H}$ an unbounded self-adjoint operator on it. Suppose that the operator $AH_0:\mathcal{D}(H_0)\rightarrow\mathcal{H}$ is bounded and that there exists an orthonormal basis $(e_n)_{n\in\mathbb{N}}\subset\mathcal{D}(H_0)$ such that \begin{equation} \sum_{n\in\mathbb{N}}\|AH_0e_n\|^2<\infty, \end{equation} so that it can be uniquely extended to a bounded operator on the whole $\mathcal{H}$ (for brevity, I will still denote the latter as $AH_0$) which is still Hilbert-Schmidt.
Now let $V:\mathcal{D}(H_0)\rightarrow\mathcal{H}$ be a symmetric operator that is relatively bounded with respect to $H_0$, that is, there exist $a,b\geq0$ such that $\|V\varphi\|\leq a\|H_0\varphi\|+b\|\varphi\|$, and consider $H_0+V:\mathcal{D}(H_0)\rightarrow\mathcal{H}$. We may assume $a<1$ so that $H_0+V$ is self-adjoint by the Rellich-Kato theorem.
My questions: is $A(H_0+V)$ still bounded, and still Hilbert-Schmidt? I would intuitively expect so, but at the moment I have not been able to prove that.