I am interested in the conditions required for functions of operators to be manipulated as if it were a real valued function with a real domain. In an applied maths text I am using the following is stated:

Let $F(\hat{A})$ be a function of an operator $\hat{A}$. If $\hat{A}$ is a linear operator, we can Taylor expand $F(\hat{A})$ in a power series of $\hat{A}$: $$F(\hat{A}) = \sum_{n = 0}^{\infty}a_n \hat{A}^n$$ where $a_n$ is just an expansion coefficient. As an illustration of an operator function, consider $e^{a \hat{A}}$, where $a$ is a scalar which can be complex or real. We can expand it as follows: $$e^{a \hat{A}} = \sum_{n=0}^{\infty} \frac{a^n}{n !}\hat{A}^n = \hat I + a \hat A + \frac{a^2}{2 !} \hat A^2 + \frac{a^3}{3!}\hat A^3+...$$

Another excerpt from the text defined an operator function $$ U (\hat G) := \lim\limits_{N \to \infty} \prod_{k=1}^{N}(1+i\frac{\alpha}{N}\hat G) = \lim\limits_{N \to \infty}(1 + i\frac{\alpha}{N}\hat G)^N = e^{i \alpha \hat G}$$ where $\alpha$ is a constant. This is again analogous to the limit definition: $$e^z = \lim\limits_{n \to \infty}\big(1 + \frac{z}{n}\big)^n.$$


  • What requirements are necessary which allows us to deal with function operators in this way? Also, does anyone have a reference for a set of notes or a book which explains this?

1 Answer 1


If $A$ is a normal operator (and every selfadjoint operator is normal,) then the Spectral Theorem for normal operators gives you a way to define $F(A)$ for any function $F$ that is continuous on the spectrum $\sigma(A)$ of $A$. And, this correspondence preserves algebra, meaning $$ (F+G)(A) = F(A)+G(A),\;\;\;(FG)(A)=F(A)G(A),\;\;\; 1(A)=I. $$ And $F(A)=A$ if $F(z)=z$. Therefore, $F(z)=z^{n}$ must give $F(A)=A^n$, which is the usual definition for powers of operators. The above also imply that, if $F_{a}(t)=e^{at}$, the definition $e^{aA}$ as $e^{aA}=F_{a}(A)$ satisfies $$ e^{aA}e^{bA}=e^{(a+b)A},\;\; e^{0A}=I. $$ And $\|F(A)\|=\sup_{\lambda\in\sigma(A)}|F(\lambda)|$ holds, which allows the series approximation: $$ \lim_{N\rightarrow\infty}\|e^{aA}-\sum_{n=0}^{N}\frac{A^n}{n!}\|=0. $$

If the sequences of functions converge uniformly on the spectrum, then the corresponding operators convergen in operator norm. This calculus can be extended to Borel functions on the spectrum, but convergence must be weakened a bit.

The is the holomorphic functional calculus works for more general operators, but you don't get the tight bounds that you want, and you can only take holomorphic functions of operators. This calculus is based on the Cauchy integral formula $$ F(A) = \frac{1}{2\pi}\oint_{C} \frac{1}{(\lambda I -A)}F(\lambda)d\lambda $$ (Here $\frac{1}{\lambda I-A}$ stands for $(\lambda I-A)^{-1}$.) This works for all functions that are holomorphic on a neighborhood of the spectrum $\sigma(A)$, and where the contour must be a single or finite set of positively oriented contours enclosing the spectrum in their interiors. In this case you also get $(FG)(A)=F(A)G(A)$, $1(A)=I$, etc.. Obviously this functional calculus works great for power series functions. And it works for general bounded operators.

There is a functional calculus devoted to the exponential operator all by itself. This is the study of $C^0$ semigroups. The expression $e^{tA}$ can potentially make sense for bounded and unbounded operators provided $t \ge 0$ and $\sigma(A)$ contained in the left half plane of $\mathbb{C}$. No all operators are candidates for this functional calculus because $e^{A}$ must have have all positive power roots. For example $(e^{\frac{1}{2}A})^2=e^{A}$. And not all operators have roots, even for matrices. But it turns out there is a condition you can put on the resolvent of the operator that will rule out the nilpotent parts that keep you from making this happen. There's a necessary and sufficient condition. Then you get the nice exponential formulas, with a little trick to rewrite in terms of the resolvent. Instead of $(I+t\frac{A}{n})^{n}$, you replace $n$ by $-n$ and use $(I-t\frac{A}{n})^{-n}$. This formulation is equivalent to solve the Cauchy vector problem $$ \frac{dx(t)}{dt} = Ax(t),\;\;\; x(0)=x_0. $$ You end up with a solution operator $x(t)=e^{tA}x_0$. Then you can form functions of $A$ using a calculus related to the Laplace transform: $$ \int_{0}^{\infty}F(t)e^{tA}x_0 dt $$ If the Laplace transform $\mathscr{L}\{F\}$ is $f$, then the above corresponds to $f(A)$.


  1. Spectral Theorem
    Kreyszig, Introductory Functional Analysis with Applications

  2. Pazy, $C^0$ Semigroups

  3. Holomorphic Functional Analysis
    Angus Taylor, Introduction to Functional Analysis


You must log in to answer this question.