Bound of a resolvent against a laplacian

Question

Consider a Hamiltonian of the form $H = -\Delta + V$ acting on $L^2(\mathbb{R}^d)$, where lets say for simplicity $V$ is a $C_0^\infty$ function. Lets say the spectrum of $H$ is bounded below by $\lambda-1 < 0$ so that the resolvent $(H - \lambda)^{-1}$ is well-defined, and is a bounded operator from $L^2 \to L^2$. I would like to show that the operator: $$A := (H - \lambda)^{-1}(-\Delta)$$ is a bounded operator. Clearly $A:H^2 \to L^2$ is bounded, but I expect that maybe $A:L^2 \to L^2$ is also bounded. For example, considering via Plancherel's theorem, $$\lVert{(-\Delta + 1)^{-1}(-\Delta) \psi}\rVert_{L^2} = \lVert{\frac{k^2}{k^2 + 1} \hat{\psi}}\rVert_{L^2} \leqslant \lVert{\psi}\rVert_{L^2}$$ if I am not mistaken. Any insight or advice would be much appreciated!