I have a Hilbert space $H$ with:
- An unbounded self-adjoint operator $T$.
- Another unbounded operator $S$ such that $S=PTP^{-1}$, where $P$ is a bounded invertible operator (not necessarily unitary).
I am interested in proving a bound of the following type - for a given $\lambda\in \mathbb R, v\in H$: $$\|(S-\lambda)v\|\leq C\|(T-\lambda)v\|$$ for some $C>0$, which might depend on $\lambda$ ($\lambda$ is assumed to be in the resolvent set of $T$ and $S$).
What I tried to do (denoting the norm of $P$ by $a$): $$\|(S-\lambda)v\|=\|(PTP^{-1}-\lambda)v\|\leq a\|(TP^{-1}-P^{-1}\lambda)v\|$$ $$=a\|(TP^{-1}-\lambda P^{-1})v\|=a\|(T-\lambda)P^{-1}v\|$$
This is almost what I want. If I could pull $P^{-1}$ out of the norm, this would exactly give a bound of the type I am looking for. But since $T$ and $P^{-1}$ do not commute, I cannot do this, and I am wondering if there is still a way to obtain a helpful inequality.
I can try to write $TP^{-1}=P^{-1}T+[T,P^{-1}]$, and then maybe try to find bounds on the commutator as well. But I do not know too much about the commutator, and it feels like there should be a simpler way I am missing.
Any advice? I can try to give more information about the given operators if necessary.
Thanks in advance.
This is false. First notice that $S-\lambda = P(T-\lambda)P^{-1}$, so up to replacing $T$ by $T-\lambda$ and $S$ by $S-\lambda$, one can assume that $\lambda = 0$ and $S$ and $P$ are invertible.
Now take as an example $H = L^2(\Bbb R^2)$ and $Tv(x) = (1+|x_1|^2)\,v(x)$ where $x=(x_1,x_2)$. Then taking the bounded operator $Pv(x) = v(x_2,x_1)$ gives $Sv(x) = (1+|x_2|^2)\,v(x)$ and your inequality would correspond to $$ \int_{\Bbb R^2} v(x)^2 \,(1+|x_2|^2)\,\mathrm d x \leq C \int_{\Bbb R^2} v(x)^2 \,(1+|x_1|^2)\,\mathrm d x, $$ which is obviously false in general: take for example $v(x) = \varphi(x_1)\,(1+|x_2|^2)^{-1/2}$ with $\varphi$ smooth and compactly supported.