Let $H=(H,(\cdot, \cdot))$ be a Hilbert space (over $\mathbb{R}$) and $v_0 \in H$ be a fixed element. Consider $A_0 : D(A_0) \subset H \longrightarrow H$, with $\overline{D(A_0)}=H$, a linear and self-adjoint operator and $T: D(A_0) \subset H \longrightarrow \mathbb{R}$ a linear function. Define $A:D(A_0) \subset H \longrightarrow H$ by $$A(u) := A_0(u) +T(u)v_0,\; \forall \; u \in D(A_0).$$
Question. The operator $A$ is also self-adjoint?
I know that: if $S$ and $B$ are two operator densely defined in $H$ such that $S$ is self-adjoint and $B$ is bounded and symmetric, with $D(S) \subset D(A)$ then $T :=B + S$ is self-adjoint.
Thus, my question comes down to: the sum of a self-adjoint operator with a "linear term" is still self-adjoint?
Not true even for bounded operators. If $A$ is self adjoint we would have $ \langle (A_0u+T(u)v_0), v \rangle=\langle u, (A_0v+T(v)v_0) \rangle$ for all $u,v$. Since $A_0$ is self adjoint this reduces to $ \langle T(u)v_0, v \rangle=\langle u, T(v)v_0 \rangle$. Take $u=v_0$ and $v$ orthogonal to $v_0$ to get $0=T(v) \langle v_0, v_0 \rangle$ which means $T(v)=0$ (assuming $v_0 \neq 0$). But this need not be true for every vector orthogonal to $v_0$.
[Take $T(x)=\langle x, y_0 \rangle$ where $y_0$ and $v_0$ are linearly independent].