# Weyl asymptotics vs. form perturbations

Consider Hilbert spaces $V$,$H$; a closed quadratic form $a$ with domain $V$; and its associated operator $A$ on $H$. (If necessary, the form can be assumed to be coercive.) For the sake of simplicity, assume the embedding of $V$ into $H$ to be compact (even trace class, if necessary), so that $A$ has purely point (real) spectrum.

Assume the operator $A$ to have a Weyl-type spectral asymptotics.

Now, take a new quadratic form $b$ which is in some sense small: form bounded or form compact, for example, so that $a+b$ still is a closed quadratic form. Are any conditions on $b$ available that would ensure the operator associated with $a+b$ to have Weyl-type spectral asymptotics, too?

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You should be able to encode Weyl asymptotics into an appropriate norm on the resolvent. Then analyze stability of this norm under perturbations. –  Helge Sep 12 '12 at 22:57
I am not quite sure I understand your idea. Weyl asymptotics is an assertion on the eigenvalues (which lie on the right half of the real line), whereas it seems to me that form methods typically give you some (rough) estimate on the norm of the resolvent in the left halfplane only. –  Delio M. Sep 13 '12 at 13:40
Let me elaborate a little bit on my idea. Denote by $\lambda_j$ the eigenvalues of the operator $A$. Then Weyl-asymptotics means that $$N(E) = \\#(j: \lambda_j < E) < C \cdot E^\alpha.$$ Consider now the operator $(A)^{-1}$ with eigenvalues $\lambda^{-1}$ ... and consider its Schatten $p$ norm $$\|A^{-1} \|_p = \sum _{j} \lambda_j^{- p} = p \int_0^{\infty} \frac{N(E) dE}{E^{p-1}}.$$ This is finite for $p >\alpha + 1$, so it suffices to investigate stability of the $p$ norm of the resolvent, which is standard.
Thanks for your answer. The first part of it is a very nice idea, but I cannot yet make complete sense of its second part. When you write that estimating the $p$-norm of a perturbation is standard, I cannot see why. This is indeed clear if the operator associated with $a+b$ is of the form $A+B$, but this is not necessarily true in this setting - think of the case where $b$ is a boundary integral, for example, so that the operator associated with $a+b$ is a boundary perturbation of $A$ (but there are other relevant cases). –  Delio M. Sep 14 '12 at 11:04
In fact, what I would need is exactly a way to estimate the $p$-norm of the operator associated with $a+b$ just by properties of $A$ and $b$. (Observe that similar problems arise if one tries to find a sectoriality estimate of the operator associated with $a+b$). –  Delio M. Sep 14 '12 at 11:04