Can the inverse operator be expressed as a series?

Question

I've seen the claim that a function of an operator can be defined as a series. For example, say $A: H_1 \mapsto H_2$ is an operator. Then

$$ e^A \equiv \sum_{n=0}^\infty \frac{A^n}{n!}. \tag{1}$$

In general, for a function $f$

$$ f(A) \overset{?}{\equiv} \sum_{n=0}^\infty \frac{f^{(n)} (0)}{n!} A^n \tag{2}$$

where $\frac{f^{(n)} (0)}{n!}$ are scalars. Now what about function $f(x) = 1/x$? I would expect that it gives a series expansion for the inverse operator $A^{-1}$. But $1/x$ is undefined for $x=0$ so this function doesn't have the Maclaurin series given by $(2)$. Do we just use a general Taylor series around point $a$

$$ f(A) \overset{?}{\equiv} \sum_{n=0}^\infty \frac{f^{(n)} (a)}{n!} (A - a)^n ? \tag{3}$$

If so, shouldn't we prove that this definition is unambiguous for different $a$?

Answer 1

For an analytic function $f$, $f(A)$ can be defined as the Maclaurin series, as described in the OP. However, when $A$ is an element of a C*-algebra rather than just a Banach algebra, one can also consider continuous functions over the spectrum of $A$, which can still be defined as limits of certain polynomials of $A$ (as a consequence of the Stone-Weierstrass theorem). For a normal element $AA^* = A^*A$, the spectral mapping theorem asserts that $f(\sigma(A))=\sigma(f(A))$, where $\sigma(A)$ denotes the spectrum of $A$. From this one can see that, in order for $1/x$ to be continuous on $\sigma(A)$, $0$ must not be an element of $\sigma(A)$. This is precisely the condition that makes $A$ invertible.