Let $A:D(A)\to X$ be closed linear operator in a Banach space. For some complex number $\lambda$, the resolvent operator $R(\lambda,A)$ is the inverse of the operator $\lambda I-A$ if it exists.
In this book about the theory of semigroup of operators, page 243, 1.13 Spectral Mapping Theorem for the Resolvent, the authors gave a formula which links the spectrum of the operator $A$ with the spectrum of its resolvent operator $R(\lambda,A)$ for some $\lambda \in \mathbb{C}$.
What I don't get in this formula is why the authors need to exlude $0$ from the spectrum of the resolvent operator $R(\lambda,A)$ ? isn't $0$ already not in the spectrum of $R(\lambda,A)$ because it is an invertible operator ?
Best Answer
The resolvent set of an operator $A$ is defined as those $\lambda\in\mathbb{C}$ such that $A-\lambda$ is invertible and the inverse is a bounded operator. The second part of this definition is important.
You are right that $R(\lambda,A) = (A-\lambda)^{-1}$ is obviously invertible, but the inverse (i.e. $(A-\lambda)$) is unbounded if $A$ is unbounded. That means that $0$ is always in the spectrum of any resolvent of an unbounded operator.