Let $H$ ba a complex Hilbert space and Let $A : D(A)\subset H \to H$ be linear operator such that $${\rm Re} \langle Ax,x\rangle \geq \alpha \|x\|^2\quad \forall x\in D(A).$$ Let now $n$ be an integer such that $n\geq2$ and denote $$\Gamma_{r,\delta}:= \{z\in \mathbb{C}| -\pi r < {\rm Arg}(z) \leq \pi r, {\rm Re}(z) \geq \delta\}$$
Assume further that there exist $B :D(B)\subset H\to H$ and $C: D(C)\subset H\to H$ having the following properties:
- $D(A)\subset D(B)\cap D(C)$
There exist $\beta >0$ and $\gamma >0$ such that $\langle Bx,x\rangle \in \Gamma_{\frac{1}{2n},\beta}$ for all $x\in D(B)$ with $\|x\|=1$ and $\langle Cx,x\rangle \in \Gamma_{\frac{1}{2n},\gamma}$ for all $x\in D(C)$ with $\|x\|=1.$
Both $B$ and $C$ are invertible with inverses $B^{-1}, C^{-1}\in \mathcal{B}(H).$
$C^n = B^n =A$ (in the sense that $D(C^n) =D(B^n) =D(A)$ and $C^n x=B^nx = Ax$ for all $x \in D(A).$)
$B^{-1}C^{-1}= C^{-1}B^{-1}.$
I wonder whether we could prove that $B =C.$
Denote $B_1 = B^{-1}$ and $C_1 = C^{-1}$. The case where $n=2$ is based on the observation that $$0= A^{-1}-A^{-1}= C_1^2 - B_1^2= (C_1+B_1)(C_1-B_1).$$ Indeed, in view of property 5., $C_1, B_1$ commute. Then using property 2. to conclude that $C_1 = B_1.$
*We might try to prove that if, in general, $D,E$ satisfy assumption 2. with constants $r_1, r_2 < \frac 14$ and $\beta, \gamma >0,$ then $DE$ satisfy the same assumption with constants $r_1+r_2$ and $\gamma \beta \cos((r_1+r_2)\pi)$ *
Thank you in advanced
It is not true. Let $H = \mathbb{C}^2$, $A = B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $C = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ and let $D(A) = D(B) = D(C) = \left\{ (z_1,z_2) \in \mathbb{C}^2 : z_j = x_j + i y_j \text{ such that } x_j \geq \frac{1}{4},\: y_j \geq \frac{1}{4} \right\}$.
$A$, $B$ and $C$ are surjective on their domain. They verify properties 1,3,4,5 for $n=2$.
The property 2 is verified (for example) by $\beta = \gamma = \frac{1}{4}$.
Note that since $A$,$B$ and $C$ are hermitian matrices, so $$ \langle Ax,x \rangle,\langle Bx,x \rangle,\langle Cx,x \rangle \in \mathbb{R}. $$