Define a fractional powers of an unbounded operator

Question

Let $(A, D(A))$ be an unbounded operator. What are the hypotheses to add to $A$ to define the operator $A^{s}$ for $s\in [0,1[$?

Thank you in advance

Answer 1

It should be noted that $A^s$ may not exist for a $2\times 2$ matrix. For example, $A^{1/2}$ does not exist for $$ A=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix} $$ The reason for this is that, if $B^2=A$, then $B^4=0$ forces the minimal polynomial for $B$ to divide $x^4$, which means either $B^2=0$ or $B=0$, either of which leads to the contradiction $A=B^2=0$.

A class of operators that do have roots $A^s$ for $s > 0$ are the generators of $C_0$ semigroups. The simplest condition to have such a generator is to require that $(0,\infty)$ be in the resolvent set of $A$, and $$ \|(A-\lambda I)^{-1}\| \le \frac{1}{\lambda},\;\; \lambda > 0. $$ Such a condition is not satisfied by the $2\times 2$ matrix $A$ given above because of the $1/\lambda^2$ term in $$ (A-\lambda I)^{-1} = -\frac{1}{\lambda}I-\frac{1}{\lambda^2}A. $$ (You may verify the above by multiplying $(A-\lambda I)(-\frac{1}{\lambda}I-\frac{1}{\lambda^2}A)$ and using $A^2=0$.)

Answer 2

It may also be that there exist infinite number of solutions. Like really many infinite. In general if you have an invertible operator or at least non-nilpotent that is far more likely. Then it is likely that you will need to add extra constraints what you want your solution to satisfy such that regularity or to encourage other behaviors.

A set of solution for $$ (A^{1/k})^k = A, k \in \mathbb{Z}\\\lambda_m(A^{1/k}) = \lambda_m(A)\cdot {(c_m)}^{l_m}$$ for a set of integers $l_m$ and roots of unity $c_m\in \mathbb F$ over our scalar field of eigenvalues.