Given a unital $\mathcal{C}^*$-algebra $A$ and a positive element $a \in A$, I am trying to prove the existence of a square root $a^{\frac{1}{2}}$ i.e. a positive element $b \in A$ such that $b^2 = a$.
My strategy goes like this :
Restrict to a maximal commutative closed subalgebra $B$ containing the unit, $a$ and $a^*$ to allow using the Gelfand-Naimark theorem (i.e. $B \cong \mathcal{C}(\Phi_B)$ where $\Phi_B$ stands for the set of characters on $B$);
Show that $\widehat{a} \restriction \Phi_B$ (that is the continuous function that evaluates to $a$ all characters on $B$) has its image in $[0, \infty)$;
Define $\widehat{b}$ to be the pointwise square root of $\widehat{a}$;
Show that $\widehat{b}$ is hermitian and positive (easy given previous work);
At some point I got the following :
$$\widehat{a}[\Phi_B] ~=~ \sigma_{\mathcal{C}(\Phi_B)}(\widehat{a}) ~=~ \sigma_B(a)~=~ \sigma_A(a) ~\subseteq~ [0, \infty).$$
The first equality is because the point-spectrum of a continuous function on a compact is it's image. The second equality is due to Gelfand-Naimark theorem. The third equality is true for all maximal commutative subalgebra and the inequality is due to the fact that $a$ is a positive element.
My concern is the following : I'm using the fact that $B$ is maximal commutative subalgebra and I'm also using the fact that $B$ is a closed subalgebra (in Gelfand-Naimark). But are maximal commutative subalgebras always closed ? Is my hypothesis too strong to hold ?
A maximal commutative subalgebra containing the unit, $a$ and $a^*$ surely exists. It's closure surely exists, but is the commuativity preserved ? Because if not, we might get $\sigma_B(a) \supseteq \sigma_A(a)$ in which case the whole proof is no longer valid.
Here is the argument as outlined in Murphy's book (very similar argument in Conway's Functional Analysis).
Assume first that $b$ is selfadjoint. Then one can use the fact that in a Banach algebra, if the spectrum has no holes then it does not depend on the algebra. It applies here because, since $b=b^*$, $\sigma(b)\subset\mathbb R$.
For the general case, we want to show that if $b-\lambda 1$ is not invertible in $B$ then it is not invertible in $A$. Or, what is the same, to show that if some element $b$ is invertible in $A$, then it is invertible in $B$. So there exists $a\in A$ with $ab=ba=1$. Taking adjoints, $a^*b^*=b^*a^*=1$. Then $$ aa^*b^*b=ab=1,\ \ b^*baa^*=b^*a^*=1 $$ making $b^*b$ invertible in $A$. By the first part of the proof, the spectrum of $b^*b$ is the same in $B$ as it is in $A$, so $b^*b$ is invertible in $B$; that is, there exists $c\in B$ with $cb^*b=b^*bc=1$. Now we look, in $A$, at the equations $$ (cb^*)b=1=ab $$ and we conclude that $a=cb^*\in B$. So $b$ is invertible in $B$.