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 $latex a$ in a unital C*–algebra is said to be positive, denoted $latex a geq 0$, if $latex sigma (a) subset [0,infty)$.
Theorem 2: Let $latex a$ be a normal element in a unital C*–algebra $latex A$. Then $latex a geq 0$ if and only if there exists a positive element $latex b in A$ such that $latex b^2 = a$. When this occurs, the element $latex b$ is unique, and is contained in the C*–algebra generated by $latex a$ and $latex 1$.
Remark: The element $latex b$ in the above theorem is referred to as the positive square root (or sometimes simply as the square root) of $latex a$ and is denoted $latex b = sqrt{a}$ or $latex b = a^{1/2}$.
Proof: Suppose that $latex ageq 0$. Let $latex f(t) = sqrt{t}$ be the positive square root function defined on $latex sigma(a)$. $latex f in C(sigma(a))$ because $latex sigma(a) subset [0,infty)$. Let $latex b = f(a)$ be given by the continuous functional calculus. Then $latex b in C^*(1,a)$ is normal (because the functional calculus is a *–isomorphism from $latex C(sigma(a))$ onto $latex C^*(1,a)$) and it has the same spectrum as $latex f$ does in $latex C(sigma(a))$, which is $latex sigma(b) = {sqrt{t} : t in sigma(a)}$. Thus $latex b geq 0$. By the functional calculus $latex b^2 = a$.
Conversely, if $latex a = b^2$ with $latex b geq 0$, then $latex a$ is normal and satisfies $latex sigma(a) = {t^2 : t in sigma(b)} subset [0,infty)$.
Suppose now that $latex a geq 0$, and let $latex c geq 0$ in $latex A$ satisfy $latex c^2 = a$. Let $latex b = f(a)$ as in the first paragraph. We will prove that $latex c = b$. Let $latex p_n$ be a sequence of polynomials converging uniformly to $latex f$ on $latex sigma(a)$, and define another sequence $latex q_n(t) = p_n(t^2)$. Then because $latex sigma(a) = {t^2 : t in sigma(c)}$, we have $latex sigma(a) = sigma(c)^2$ or $latex sigma(c) = sigma(a)^{1/2}$. It follows that
$latex q_n(t) = p_n(t^2) rightarrow f(t^2) = t$
uniformly on $latex sigma(c)$. Therefore $latex q_n(c) rightarrow c$. On the other hand
$latex q_n(c) = p_n(c^2) = p_n(a) rightarrow f(a) = b. $
Thus $latex b=c$.