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?

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    $\begingroup$ You need B to be invertible for this to work. $\endgroup$ Sep 9, 2020 at 12:51

1 Answer 1


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 ($B=I$).


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