Spectrum of inverse perturbed Laplacian Let $\Omega = B^n(0,1/2)$ with $n\geq 2$ and $q:\Omega\to\mathbb{R}$ be such that $\|q\|_\infty \leq 2$. Let $\lambda_1$ be the first eigenvalue of $-\Delta$ (with vanishing Dirichlet boundary conditions); it is known that $\lambda_1 > 5$.  I wish to prove the following claim:
$$\|(-\Delta + q)^{-1}\|\leq (\lambda_1-2)^{-1},$$
where the norm is the operator norm from $L^2$ to $L^2$.
My attempt is as follows:
I want to say that $\lambda_1 - 2$ lower bounds any eigenvalue of $-\Delta+ q$. Then since the eigenvalues of the inverse are the inverses of the eigenvalues and since $\lambda_1-2>0$, then $(\lambda_1-2)^{-1}$ upper bounds any eigenvalue of $(-\Delta+q)^{-1}$. Since $-\Delta+q$ is self-adjoint (thanks to the vanishing Dirichlet boundary conditions), so is $(-\Delta+ q)^{-1}$, and so $\|(-\Delta+ q)^{-1}\|$ is an eigenvalue of $(-\Delta+ q)$, and so $(\lambda_1-2)^{-1}\geq \|(-\Delta+ q)^{-1}\|$.
(The general fact I am using here is that if $H$ is a Hilbert space and $S:H\to H$ is a self-adjoint bounded linear operator, then the operator norm upper bounds the spectrum and is itself included in the spectrum.)
Here is what I am unsure about:  does $\lambda_1 -2$ really lower bound the spectrum of $-\Delta + q$? This seems like it should be true, but I'm having trouble actually proving it. If $q$ were a constant function, then adding $q$ to $-\Lambda$ surely shifts the eigenvalues by $q$. The question now is, if I shift the operator by a function taking values in $[-2,2]$, my intuition tells me that the eigenvalues are perturbed by no more than $\pm 2$, but it's not clear how one would actually show this.
For reference, this is coming from the proof of Lemma 3 in the following paper.
 A: Yes what you are saying appears to be true via an application of the Rayleigh quotient. We have that the first eigenvalue of $-\Delta$ is,
\begin{align}
\lambda_{1}=\min_{u\in H^{1}_{0}(\Omega)\setminus\{0\}}\frac{a(u,u)}{\|u\|_{2}^{2}},
\end{align}
where $a(\cdot,\cdot)$ is the associated bilinear form of $-\Delta$. Now consider $q\in L^{\infty}(\Omega)$ such that $\|q\|_{\infty}=2$ then $-2\leq q(x)\leq 2$ for almost all $x\in\Omega$, so,
\begin{align}
\frac{a(u,u)+(qu,u)_{L^{2}}}{\|u\|_{2}^{2}}\geq\frac{a(u,u)-2\|u\|_{2}^{2}}{\|u\|_{2}^{2}}=\frac{a(u,u)}{\|u\|_{2}^{2}}-2=\lambda_{1}-2.
\end{align}
Notice that by minimising the LHS of the inequality above this gives the first eigenvalue of $-\Delta+q$, that is,
\begin{align}
\lambda_{1}^{q}=\min_{u\in H^{1}_{0}(\Omega)\setminus\{0\}}\frac{a(u,u)+(qu,u)_{L^{2}}}{\|u\|_{2}^{2}}\geq\lambda_{1}-2.
\end{align}
Also just an additional note, $-\Delta+q$ is self-adjoint regardless of boundary conditions taken. In the Neumann case we would have $\frac{\partial u}{\partial\nu}=0$ and so the boundary integral would be zero. In the Robin case we would have $\frac{\partial u}{\partial\nu}=\beta u$, where we can take $\beta\in L^{\infty}(\Omega)$, then,
\begin{align}
\int_{\partial\Omega}\frac{\partial u}{\partial\nu}v\,d\sigma=\int_{\partial\Omega}\beta u v\,d\sigma=\int_{\partial\Omega}\beta vu\,d\sigma.
\end{align}
EDIT: The last part regarding boundary conditions is only meant in the context of homogeneous boundary conditions.
