Spectral mapping theorem

For a holomorphic function near the spectrum, the spectrum of its functional-calculus value is exactly the image of the original spectrum. If a scalar is absent from that image, the reciprocal of the shifted function is holomorphic on a smaller spectral neighborhood and supplies an inverse. If it is attained at $\lambda$, a divided difference factors $f(x)-f(\lambda )$ through $x-\lambda$. These factors commute, so invertibility of their product would contradict $\lambda \in \sigma (x)$.

For a closed linear operator and $z$ in its resolvent, the displayed identity follows by factoring $(A-z{)}^{-1}-wI$ for $w\ne 0$. Its inverse exists exactly when $A-(z+1/w)I$ has a bounded inverse. Products involving $A$ are bounded after composing with a resolvent, since $A(A-\lambda {)}^{-1}=I+\lambda (A-\lambda {)}^{-1}$. Zero may be in the bounded resolvent's spectrum when $A$ is unbounded.