Advanced Analysis, Notes 15: C*-algebras (square root)

by Orr Shalit

This post contains some make–up material for the course Advanced Analysis. It is a theorem about the positive square root of a positive element in a C*-algebra which does not appear in the text book we are using. My improvisation for this in class came out kakha–kakha, so here is the clarification.

Definition 1: A normal element a in a unital C*–algebra is said to be positive, denoted a \geq 0, if \sigma (a) \subset [0,\infty)

Theorem 2: Let a be a normal element in a unital C*–algebra A. Then a \geq 0 if and only if there exists a positive element b \in A such that b^2 = a. When this occurs, the element b is unique, and is contained in the C*–algebra generated by a and 1

Remark: The element b in the above theorem is referred to as the positive square root (or sometimes simply as the square root) of a and is denoted b = \sqrt{a} or b = a^{1/2}.

Proof: Suppose that a\geq 0. Let f(t) = \sqrt{t} be the positive square root function defined on \sigma(a). f \in C(\sigma(a)) because \sigma(a) \subset [0,\infty). Let b = f(a) be given by the continuous functional calculus. Then b \in C^*(1,a) is normal (because the functional calculus is a *–isomorphism from C(\sigma(a)) onto C^*(1,a)) and it has the same spectrum as f does in C(\sigma(a)), which is \sigma(b) = \{\sqrt{t} : t \in \sigma(a)\}. Thus b \geq 0. By the functional calculus b^2 = a.

Conversely, if a = b^2 with b \geq 0, then a is normal and satisfies \sigma(a) = \{t^2 : t \in \sigma(b)\} \subset [0,\infty).

Suppose now that a \geq 0, and let c \geq 0 in A satisfy c^2 = a. Let b = f(a) as in the first paragraph. We will prove that c = b. Let p_n be a sequence of polynomials converging uniformly to f on \sigma(a), and define another sequence q_n(t) = p_n(t^2). Then because \sigma(a) = \{t^2 : t \in \sigma(c)\}, we have  \sigma(a) = \sigma(c)^2  or  \sigma(c) = \sigma(a)^{1/2}. It follows that

q_n(t) = p_n(t^2) \rightarrow f(t^2) = t

uniformly on \sigma(c). Therefore q_n(c) \rightarrow c. On the other hand

q_n(c) = p_n(c^2) = p_n(a) \rightarrow f(a) = b.

Thus b=c.