Norm of the resolvent and inequalities

Question

I am considering the equation

$$ \lambda q- \frac{\mathrm{d}^{2} q}{\mathrm{d} x^{2}}=g, $$ where $g\in L^p(X)$ and $q\in W^{2,p}(X)$. If, I know that $$ \mid \mid \lambda (\lambda I- \frac{\mathrm{d}^{2} }{\mathrm{d} x^{2}})^{-1} \mid \mid \leq M $$ How can I conclude that $$ \left\|\lambda q\right\|_{L^{p}(X)}+\left\|q\right\|_{W^{2,p}(X)} \leqslant C\|g\|_{L^{p}(X)} ? $$

Answer 1

Let $A=\dfrac{\mathrm{d}^{2}}{\mathrm{d} x^{2}}$. Then $q=R(\lambda,A)g$. It follows that $$\begin{aligned} \left\|\lambda q\right\|_{L^{p}(X)}+\left\|q\right\|_{W^{2,p}(X)} & \le\|\lambda R(\lambda,A)g\|_{L^{p}(X)} + \|R(\lambda,A)g\|_{D(A)}\\ & \le M \|g\|_{L^{p}(X)} + (\frac{M}{\lambda}\|g\|_{L^{p}(X)}+ \|AR(\lambda,A)g\|_{L^{p}(X)}) \\ &= M \|g\|_{L^{p}(X)} + (\frac{M}{\lambda}\|g\|_{L^{p}(X)}+ \|\lambda R(\lambda,A)g - g\|_{L^{p}(X)})\\ & \le (1+M \frac{2\lambda +1}{\lambda}) \|g\|_{L^{p}(X)}. \end{aligned}$$