# On a limit for the resolvent norm

Let $$H$$ be a complex, infinite dimensional, separable Hilbert space. Fix any two nonzero operators $$A,B \in B(H)$$ such that $$B$$ is not a scalar multiple of $$A$$. It is well known that:

$$\| R_A (z) \| \rightarrow 0 \quad \text{when}\, |z| \rightarrow +\infty$$

This easily follows from:

$$\|R_A(z) \| \leq \frac{1}{|z| - \| A \|} \quad \forall z \in \mathbb{C} : |z| > \| A \|$$

Now consider the operator $$A+ cB$$ where $$c$$ is a positive real number, and fix some complex number $$z$$. Can we say that, for all $$\epsilon >0$$, there exists $$M>0$$ such that:

$$c > M \, \, \text{such that} \, \, z \not \in \sigma(A+cB) \Rightarrow \|R_{A+cB} (z)\| < \epsilon$$

Intuitively, for $$c \rightarrow +\infty$$, we would have:

$$R_{A+cB}(z)=(z-A-cB)^{-1}= c^{-1}(z/c - A/c -B)^{-1} "\rightarrow" 0$$

because $$|z/c| \rightarrow 0$$, $$\|A\| / c \rightarrow 0$$, $$1/c \rightarrow 0$$ and $$B$$ is fixed. However, I'm not aware of any reference for a result of this kind. Does anybody have a reference or a proof of this?

• You need B to be invertible for this to work. Sep 9, 2020 at 12:51

As Michael already observed: It is simple to construct counterexamples if $$B$$ has a nontrivial kernel $$N(B)$$ and $$A$$ maps $$N(B)$$ into itself, since on $$N(B)$$ the size of $$c$$ plays no role.
But even if $$B$$ is an isomorphism, even the identity $$B=I$$, the conjecture is false: $$z-A-cB=(z-c)-A$$. The maximum of the norm of the inverse (over all $$z$$ from the corresponding resolvent set) is clearly independent of $$c$$.
Anyway, if $$B$$ is an isomorphism, there are some estimates:
There holds $$(z-A-cB)^{-1}=B^{-1}((z-A)B^{-1}-c)^{-1}$$, and one can use the resolvent estimate mentioned in the question to obtain $$\|R_{A+cB}(z)\|\le\|B^{-1}\|\frac1{|c|-\|(z-A)B^{-1}\|}.$$ This gives an estimate if $$|c|$$ is large compared to $$|z|\|B^{-1}\|$$.
If $$|z|$$ is large compared to $$|c|\|B\|$$, the estimate from the question can be used.
If neither is the case, there need not be a good estimate as mentioned above ($$B=I$$).