Let $A$ and $B$ be two bounded self-adjoint operators on a separable Hilbert space such that $i[A,B]\equiv i(AB-BA)$ is strictly positive, that is, $i[A,B]\geq c $ for some real number $c>0$.
How to prove that the resolvent of $A$, $R(z)\equiv(A-z)^{-1}$ is bounded in operator norm for all $z\in\mathbb{C}$ (and hence the spectrum of $A$ is empty)?
The closest I could get is proving that $R(z)$ is cannot diverge worse than $|\Im\{z\}|^{-1}$.
Indeed, since $i[A,B]\geq0$, $\exists C:|C|^2=i[A,B]$ (with $|C|^2\equiv C^\ast C$). In fact since $c>0$ we know that $C$ is invertible with bounded inverse. Then using the fact that $||R(z)||\leq|\Im\{z\}|^{-1}$ we get \begin{align} ||CR(z)||^2 & = |||CR(z)|^2|| \\ &= ||R(\bar{z})|C|^2R(z)|| \\ &= ||R(\bar{z})i[A,B]R(z)|| \\ &= ||R(\bar{z})i[(A-z),B]R(z)|| \\ &\leq ||R(\bar{z})B||+||BR(z)||+2|\Im\{z\}|||R(\bar{z})BR(z)|| \\ &\leq 4|\Im\{z\}|^{-1}||B|| \end{align}
But if $C$ is invertible with a bounded inverse, then $||CR(z)||\geq||R(z)||(||C^{-1}||)^{-1}$ so that finally the result is $$ ||R(z)|| \leq 4|\Im\{z\}|^{-1}||C^{-1}||||B|| $$
Possible solution:
We know that \begin{align} ||C R(z)||^2 &\leq ||R(\bar{z})B|| + ||BR(z)||+2|\Im\{z\}|||R(\bar{z})BR(z)|| \end{align}
However now instead of estimating $||R(z)||\leq|\Im\{z\}|^{-1}$ we use submultiplicativity of the norm and then $||R(z)||\leq|\Im\{z\}|^{-1}$ only within the parenthesis to obtain: \begin{align} ||C R(z)||^2 &\leq 2||B||||R(z)|| +2|\Im\{z\}|||B||||R(z)||^2 \\ &=2||B||||R(z)||(1+|\Im\{z\}|||R(z)||)\\ &\leq 4||B||||R(z)|| \end{align}
Now because $||C^{-1}||$ is bounded we find $$||R(z)||\leq4||B||||C^{-1}||^2$$ uniformly in $z\in\mathbb{C}$ and so the spectrum of $A$ is empty indeed!