I'm working on Schrödinger operators, currently struggling with the proof of $\nabla(\Delta + i)^{-1/2}$ bounded. The domain is $H^2(\mathbb R^3)$.
So, of course $(\Delta + i)^{-1}$ is bounded, and so is its square root - but how do I handle the nabla? The fact is given as obvious, and I think I'm missing something big here.
After some offline help by Daniel and thanks to the last comment, I think the riddle is solved by the following:
\begin{equation} \begin{split} &\langle\nabla(\Delta+i)^{-1}\psi, (\Delta + i)^{-1}\psi\rangle = \langle(-\Delta)(\Delta+i)^{-1}\psi, (\Delta+i)^{-1}\psi\rangle = \\ &= \langle(\Delta+i)^{-1}(-\Delta)(\Delta+i)^{-1}\psi, \psi\rangle\leq ||(\Delta+i)^{-1}||\cdot||\Delta(\Delta+i)^{-1}||\cdot||\psi||^2 \end{split} \end{equation}
Also observing $||\Delta(\Delta+i)^{-1}||<+\infty$ by \begin{equation} \Delta(\Delta+i)^{-1} = \int_{\mathbb R_+}\frac{\lambda}{\lambda + i}dE_{\Delta}(\lambda) \end{equation}
I'd still appreciate some external feedback on the solution. Is this correct?
Edit: still working on this: for the sake of clarity, I will add some further computation. By the last equation \begin{equation} \begin{split} ||\Delta(\Delta+i)^{-1}(\psi)|| &= ||\int_{\mathbb R_+}\frac{\lambda}{\lambda + i}dE_{\Delta}(\lambda)(\psi)||=\\ &= (2\pi)^{-3/2}||\int_{\mathbb R_+}\int_{|k|=\sqrt\lambda}\frac{\lambda}{\lambda + i}\hat{\psi}(k)\exp(ik\cdot x)d^3kd\lambda||\leq \\ &\leq (2\pi)^{-3/2}\int_{\mathbb R_+}\big|\frac{\lambda}{\lambda + i}\big|\cdot ||\int_{|k|=\sqrt\lambda}\hat{\psi}(k)\exp(ik\cdot x)d^3k||d\lambda \leq \\ &\leq || (2\pi)^{-3/2}\int_{\mathbb R^3}\hat{\psi}(k)\exp(ik\cdot x)d^3k|| = ||\psi|| \end{split} \end{equation} This would really settle things down.