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

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 $b=c$.