Asymptotic behaviour of eigenvalues If you look at $-\Delta + q$ on the sphere in $\mathbb{R}^3$ for example and $||q|| < \infty,$ is there a way to asymptotically describe the behaviour of the eigenvalues? Probably they behave similar to the ones for $q=0$ which are given by $l(l+1)$. Unfortunately, I only found asymptotic estimates for the free case and general Laplacians on different sets. I also found this, but I don't know if this gives me the scaling of the eigenvalues http://en.wikipedia.org/wiki/Weyl_law#Generalizations
 A: I would say that the eigenvalues of $-\Delta+q$ stay within $\|q\|_{\infty}$ of eigenvalues of $-\Delta$ (and, as Noam Elkies says, note the multiplicity: the main term of asymptotics will be $\lambda_n\sim n$, as $l(l+1)\sim l^2$ is an eigenvalue of $-\Delta$ of multiplicity $2l+1$.)
If I'm not mistaken, you can re-formulate the eigenvalues problem for $A$ being any of the (symmetric!) operators $-\Delta$, $-\Delta+q$ in the following way: 
Lemma.
$\lambda_n(A)\le \lambda$ if and only if there exists an $n$-dimensional subspace $V$, on which $\langle f, Af\rangle \le \lambda \langle f,f\rangle$.
Proof.
For the "only if" part, take the space generated by the eigenvectors corresponding to $\lambda_1,\dots,\lambda_n$. For the "if", note that in the space $W$ that is a closure of the span of eigenvectors corresponding to $\lambda_n,\lambda_{n+1},\dots$, we have $\langle f, Af\rangle \ge \lambda_n \langle f,f\rangle$, and $W$ is a codimension $n-1$ subspace. Hence, if $\lambda_n$ was greater than $\lambda$, we would have a contradiction for a nonzero vector $f\in V \cap W$:
$$
\lambda \langle f,f\rangle\ge \langle f, Af\rangle \ge \lambda_n \langle f,f\rangle.
$$
(This proof mimics the standard proof from linear algebra, so I'm absolutely sure all of this should be known, but I do not know any references...)
Now, when you add or subtract $q$, you change $\langle f, Af\rangle$ at most by $\|q\|_{\infty} \langle f,f \rangle$, and hence shift the eigenvalues at most by $\|q\|_{\infty}$.
