Let $\Delta$ denote the Laplacian on $\mathbb{R}$. For $\operatorname{Im}{\lambda} > 0$ it is given by the spectral theorem that for the resolvent $$R_0(\lambda) := (-\Delta - \lambda^2)^{-1}$$ we have the bound $$||R_0(\lambda)||_{L^2 \to L^2} \leq \frac{1}{|\lambda| \operatorname{Im}\lambda} \quad \operatorname{Im}\lambda > 0$$ Why is it then the case that $$||R_0(\lambda)||_{L^2 \to H^2} \leq C\frac{1 + |\lambda|^2}{|\lambda| \operatorname{Im}\lambda} \quad \operatorname{Im} \lambda > 0$$
I know that from the definition we have
\begin{align} ||R_0(\lambda)||_{L^2 \to H^2} &\leq C \big(||\Delta R_0(\lambda)||_{L^2 \to L^2} + ||R_0(\lambda)||_{L^2 \to L^2}\big)\\ &\leq C\big(||Id||_{L^2 \to L^2} + (|\lambda|^2+1)||R_0(\lambda)||_{L^2 \to L^2} \big) \end{align}
Where the second line followed from the resolvent being an inverse for $\Delta - \lambda^2$. However, I'm not sure how to absorb the norm from the identity into the $$(1+ |\lambda|^2)||R_0(\lambda)||_{L^2 \to L^2}$$