Theorem: Existence and Uniqueness of Positive Nth Root of Positive Element of C*-Algebra

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $x \in A$ be positive. 
Let $n \ge 2$. 
Then there exists a unique positive element $y \in A$ such that $x = y^n$.

Proof:
We first take $A$ to be unital.
Let $B$ be a commutative unital $\text C^\ast$-subalgebra of $A$ with $x \in B$.
We show the existence and uniqueness of positive $y \in B$ (''not'' in the entirety of $A$ yet) such that $x = y^n$. 
From Spectrum of Element of Unital C*-Subalgebra of Unital C*-Algebra we have $\map {\sigma_B} x = \map {\sigma_A} x$. 
Let $\Phi_B$ be the spectrum of $B$.
Let $\struct {\map \CC {\Phi_B}, \overline \cdot, \norm {\, \cdot \,} }$ be the $\text C^\ast$-algebra of continuous complex-valued functions vanishing at infinity.
Define $G : B \to \map \CC {\Phi_B}$ by:
:$\map G a = \hat a$
where $\hat a$ is the Gelfand transform of $a$ in $B$.
From Gelfand-Naimark Theorem: Commutative Case: Unital, $G$ is a isometric unital $\ast$-algebra isomorphism.
By Spectrum of Image of Element of Unital Algebra under Unital Algebra Homomorphism, we have:
:$\map {\sigma_{\map \CC {\Phi_B} } } {\map G x} = \map {\sigma_B} x$
Hence:
:$\map {\sigma_{\map \CC {\Phi_B} } } {\map G x} \subseteq \hointr 0 \infty$
For brevity write $f = \map G x$. 
From Spectrum of Element of Space of Continuous Functions on Compact Hausdorff Space, we have:
:$\map {\sigma_{\map \CC {\Phi_B} } } f = \set {\map f \phi : \phi \in \Phi_B}$
Hence $\map f \phi \in \hointr 0 \infty$ for each $\phi \in \Phi_B$.
We can therefore consider define $g : \Phi_B \to \hointr 0 \infty$ by $\map g \phi = \paren {\map f \phi}^{1/n}$ for each $\phi \in \Phi_B$, where $g$ denotes the square root.
From Composite of Continuous Mappings is Continuous, $g : \Phi_B \to \hointr 0 \infty$ is continuous.
Hence we can set $y = \map {G^{-1} } g$. 
From Inverse of Algebra Isomorphism is Algebra Isomorphism, we have that $G^{-1}$ is an algebra isomorphism.
Hence we have:
:$y^n = \map {G^{-1} } {g^n} = \map {G^{-1} } f = x$
Applying Spectrum of Element of Space of Continuous Functions on Compact Hausdorff Space again we have:
:$\map {\sigma_{\map \CC {\Phi_B} } } g = \set {\map g \phi : \phi \in \Phi_B} \subseteq \hointr 0 \infty$
Hence by Spectrum of Image of Element of Unital Algebra under Unital Algebra Homomorphism we have:
:$\map {\sigma_B} y = \map {\sigma_{\map \CC {\Phi_B} } } g \subseteq \hointr 0 \infty$
Hence by Continuous Function applied to Normal Element of Unital C*-Algebra is Positive iff Function is Non-Negative, we have that $y$ is positive element.
We have therefore shown the existence of $y$ in every commutative unital $\text C^\ast$-subalgebra $B$ of $A$ with $x \in B$.
We show uniqueness in $B$. 
Suppose that $z^n = x$ with $z \in B$ positive.
Then there exists $h \in \map \CC {\Phi_B}$ such that $z = \map {G^{-1} } h$ and $\map h \phi \ge 0$ for each $\phi \in \Phi_B$. 
Hence we have:
:$z^n = \map {G^{-1} } {h^n} = x$
and:
:$y^n = \map {G^{-1} } {g^n} = x$
So $\map {G^{-1} } {h^n} = \map {G^{-1} } {g^n}$.
Since $G^{-1}$ is an algebra isomorphism we have $h^n = g^n$. 
Hence for each $\phi \in \Phi_B$ we have $\cmod {\map h \phi} = \cmod {\map g \phi}$. 
Since $\map h \phi \ge 0$ and $\map g \phi$ for each $\phi \in \Phi_B$ we obtain $h = g$. 
So we have uniqueness ''within'' every commutative unital $\text C^\ast$-subalgebra $B$ of $A$ with $x \in B$.
We quickly assure ourselves that what we have shown so far is not vacuous.
We can take $B$ to be the $\text C^\ast$-algebra generated by $\set { {\mathbf 1}_A, x}$. 
By C*-Algebra Generated by Commutative Self-Adjoint Set is Commutative, $B$ is commutative and hence we obtain the existence of such a unique $y \in B$.
We now prove uniqueness in the whole algebra $A$.
Let $y, z \in A$ be positive with $y^n = z^n = x$. 
Let $B_1$ be the $\text C^\ast$-algebra generated by $\set { {\mathbf 1}_A, x, y}$.
Let $B_2$ be the $\text C^\ast$-algebra generated by $\set { {\mathbf 1}_A, x, z}$.
Note now that:
{{begin-eqn}}
{{eqn	| l = y x
		| r = y \paren {y^n}
}}
{{eqn	| r = y^{n + 1}
}}
{{eqn	| r = y^n y
}}
{{eqn	| r = x y
}}
{{end-eqn}}
Swapping $y$ for $z$ we have $x z = z x$ as well.
Hence from C*-Algebra Generated by Commutative Self-Adjoint Set is Commutative, $B_1$ and $B_2$ are commutative.
Hence from Intersection of C*-Subalgebras is C*-Subalgebra, we have that $B_1 \cap B_2$ is a $\text C^\ast$-subalgebra of $A$.
From what we have shown already, there is a unique positive element $u \in B_1 \cap B_2$ with $u^n = x$. 
Hence $u = y = z$. 
We therefore have uniqueness over the whole of $A$.
Now suppose that $A$ is not unital.
Let $\struct {A_+, \ast, \norm {\, \cdot \,}_\ast}$ be the unitization of $A$ as a $\text C^\ast$-algebra. 
From the definition of the spectrum in a non-unital algebra, we have:
:$\map {\sigma_A} x = \map {\sigma_{A_+} } {\tuple {x, 0} }$
Since $\map {\sigma_A} x \subseteq \hointr 0 \infty$, we have $\map {\sigma_{A_+} } {\tuple {x, 0} } \subseteq \hointr 0 \infty$.
So $\tuple {x, 0}$ is positive in $A_+$.
By the unital case, there exists a unique positive $\tuple {y, \lambda} \in A_+$ such that:
:$\tuple {y, \lambda}^n = \tuple {x, 0}$
From the definition of the unitization, we have that:
:$\tuple {y, \lambda}^n = \tuple {\map p {y, \lambda}, \lambda^n}$
for some $\map p {y, \lambda} \in A$ with $\map p {y, 0} = y^n$.
Hence, comparing the second component, we have $\lambda^n = 0$.
Hence $\lambda = 0$.
So $\tuple {y, 0}^n = \tuple {y^n, 0} = \tuple {x, 0}$. 
Hence $y^n = x$.
Further:
:$\map {\sigma_A} y = \map {\sigma_{A_+} } {\tuple {y, 0} } \subseteq \hointr 0 \infty$ 
Note now that if $z^n = x$ in $A$ with $z$ positive, then we have:
:$\tuple {z, 0}^n = \tuple {x, 0}$
and:
:$\map {\sigma_{A_+} } {\tuple {z, 0} } = \map {\sigma_A} z \subseteq \hointr 0 \infty$
Hence $\tuple {z, 0} = \tuple {y, 0}$, giving $y = z$ and hence uniqueness.
{{qed}}