[proofplan] We construct the square root by continuous functional calculus. In the unital case, the function $t\mapsto t^{1/2}$ on $\sigma(a)\subset [0,\infty)$ gives a positive element $d$ with $d^2=a$. In the non-unital case, we perform the same construction in the unitization and use the condition $f(0)=0$ to show that the resulting element lies in the closed subalgebra generated by $a$, hence in $A$. For uniqueness, any other positive square root $b$ commutes with $a$ and therefore with the functional calculus element $d$; the commutative algebra generated by $b$ and $d$ reduces the equality $b^2=d^2$ to pointwise equality of non-negative scalar functions. [/proofplan]

[step:Construct the square root in the unital case by functional calculus]Assume first that $A$ is unital, with unit $1_A$. Let $\sigma_A(a)$ denote the spectrum of $a$ computed in the unital algebra $A$. Since $a\in A_+$, the element $a$ is self-adjoint and its spectrum satisfies $\sigma_A(a)\subset [0,\infty)$ by [citetheorem:8552] and the spectral characterization of positivity. Define \begin{align*} f:\sigma_A(a)\to \mathbb C,\qquad t\mapsto t^{1/2}. \end{align*} The map $f$ is continuous on the compact set $\sigma_A(a)$. Let $C^*(a,1_A)$ denote the smallest closed $*$-subalgebra of $A$ containing $a$ and $1_A$. By the continuous functional calculus for the self-adjoint element $a$, there exists an element \begin{align*} d:=f(a)\in C^*(a,1_A)\subset A. \end{align*} Since $f$ is real-valued and non-negative on $\sigma_A(a)$, the positivity-preserving property of the continuous functional calculus gives $d\in A_+$. Moreover, by multiplicativity of the continuous functional calculus [citetheorem:8553], applied to the functions $f$ and $f^2$, we have \begin{align*} d^2=f(a)^2=(f^2)(a). \end{align*} For every $t\in\sigma_A(a)$, $f(t)^2=t$, so $f^2$ is the identity function on $\sigma_A(a)$. Hence \begin{align*} d^2=a. \end{align*}[/step]

[guided]We first treat the case where $A$ has a unit, because then the standard continuous functional calculus applies directly to the element $a$. Let $\sigma_A(a)$ denote the spectrum of $a$ computed in the unital algebra $A$. Since $a\in A_+$, the element $a$ is self-adjoint and has non-negative spectrum: \begin{align*} \sigma_A(a)\subset [0,\infty). \end{align*} This containment is exactly what allows us to use the scalar square-root function without ambiguity. Define the [continuous function](/page/Continuous%20Function) \begin{align*} f:\sigma_A(a)\to \mathbb C,\qquad t\mapsto t^{1/2}. \end{align*} The domain $\sigma_A(a)$ is compact, and $f$ is continuous on it. Let $C^*(a,1_A)$ denote the smallest closed $*$-subalgebra of $A$ containing $a$ and $1_A$. The continuous functional calculus for the self-adjoint element $a$ therefore produces an element \begin{align*} d:=f(a)\in C^*(a,1_A)\subset A. \end{align*} Because $f(t)\ge 0$ for every $t\in\sigma_A(a)$, the positivity-preserving part of the functional calculus gives $d\in A_+$. It remains to check that $d$ squares to $a$. The functional calculus is an algebra homomorphism from $C(\sigma_A(a))$ into $A$, so multiplication of functions becomes multiplication of elements [citetheorem:8553]: \begin{align*} d^2=f(a)^2=(f^2)(a). \end{align*} But $f^2$ is the identity function on $\sigma_A(a)$, since for every $t\in\sigma_A(a)$, \begin{align*} f(t)^2=t. \end{align*} The identity function on $\sigma_A(a)$ is sent by the functional calculus to $a$. Hence \begin{align*} d^2=a. \end{align*} Thus the desired positive square root exists when $A$ is unital.[/guided]

[step:Pass from the unitization back to the non-unital algebra] It remains to treat the case where $A$ is non-unital. Let $A^+$ denote the unitization of $A$, with unit $1_{A^+}$. We regard $A$ as a closed two-sided ideal of $A^+$. Since $a\in A_+$, the same element $a$, viewed in $A^+$, is positive. Applying the unital construction in $A^+$ gives an element \begin{align*} d\in (A^+)_+ \end{align*} such that \begin{align*} d^2=a. \end{align*} It remains to prove that $d\in A$. Let \begin{align*} f:\sigma_{A^+}(a)\to \mathbb C,\qquad t\mapsto t^{1/2}. \end{align*} Since $a$ belongs to the closed ideal $A\subset A^+$ and $0\in\sigma_{A^+}(a)$ whenever $A$ is non-unital, we have $f(0)=0$. By the polynomial approximation form of the continuous functional calculus, choose polynomials $p_n\in \mathbb C[z]$ such that \begin{align*} \lim_{n\to\infty}\sup_{t\in\sigma_{A^+}(a)} |p_n(t)-f(t)|=0. \end{align*} For each $n$, define $q_n\in \mathbb C[z]$ by $q_n(z):=p_n(z)-p_n(0)$. Then $q_n(0)=0$. Since $f(0)=0$, we also have \begin{align*} \lim_{n\to\infty}\sup_{t\in\sigma_{A^+}(a)} |q_n(t)-f(t)|=0. \end{align*} For each $n$, the element $q_n(a)$ is a finite linear combination of positive powers of $a$, hence $q_n(a)\in A$. Functional calculus convergence gives \begin{align*} \lim_{n\to\infty}\|q_n(a)-d\|_{A^+}=0. \end{align*} Because $A$ is closed in $A^+$, it follows that $d\in A$. Thus $d\in A_+$ and $d^2=a$ inside $A$. [/step]

[step:Show that any positive square root commutes with the constructed one]Let $b\in A_+$ satisfy \begin{align*} b^2=a. \end{align*} Then $b$ commutes with $a$, because \begin{align*} ba=b b^2=b^3=b^2 b=ab. \end{align*} Viewing all elements in $A^+$ if necessary, the element $b$ therefore commutes with every polynomial in $a$ and $1_{A^+}$. Since $d=f(a)$ is the norm limit in $A^+$ of polynomials in $a$ and $1_{A^+}$ under the continuous functional calculus, continuity of multiplication in the Banach algebra $A^+$ gives \begin{align*} bd=db. \end{align*}[/step]

[guided]Let $b\in A_+$ be another positive element satisfying \begin{align*} b^2=a. \end{align*} To compare $b$ with the constructed square root $d$, we first prove that they commute. The element $b$ commutes with $a$ because $a=b^2$, and associativity gives \begin{align*} ba=b b^2=b^3=b^2 b=ab. \end{align*} Therefore $b$ commutes with every positive power of $a$, with every polynomial in $a$, and, after viewing the elements in $A^+$ if $A$ is non-unital, with every polynomial in $a$ and $1_{A^+}$. The constructed element $d=f(a)$ is obtained by continuous functional calculus, so it is a norm limit in $A^+$ of polynomials in $a$ and $1_{A^+}$. If $(p_n)$ is such a polynomial approximating net or sequence with $p_n(a)\to d$ in $A^+$, then $bp_n(a)=p_n(a)b$ for every $n$. Multiplication in the Banach algebra $A^+$ is continuous, so passing to the norm limit gives \begin{align*} bd=db. \end{align*} This commutation is the input needed to place $b$ and $d$ in one commutative C*-algebra in the uniqueness step.[/guided]

[step:Reduce uniqueness to pointwise scalar square roots in a commutative algebra]Let $B:=C^*(b,d,1_{A^+})\subset A^+$. Since $b$ and $d$ are self-adjoint and commute, the $C^*$-algebra $B$ is commutative. Let $\Delta(B)$ be its character space, and let \begin{align*} \Gamma:B\to C(\Delta(B)) \end{align*} denote the Gelfand transform [citetheorem:8549]. Define the functions $\widehat b,\widehat d:\Delta(B)\to\mathbb C$ by \begin{align*} \widehat b(\tau):=\Gamma(b)(\tau)=\tau(b),\qquad \widehat d(\tau):=\Gamma(d)(\tau)=\tau(d) \end{align*} for each character $\tau\in\Delta(B)$. Because $b,d\in B_+$ and characters on commutative $C^*$-algebras are $*$-homomorphisms [citetheorem:8550], the scalar values $\widehat b(\tau)$ and $\widehat d(\tau)$ are non-negative [real numbers](/page/Real%20Numbers). Since $b^2=a=d^2$, multiplicativity of $\tau$ gives \begin{align*} \widehat b(\tau)^2=\tau(b^2)=\tau(d^2)=\widehat d(\tau)^2. \end{align*} The non-negative real square root is unique, so \begin{align*} \widehat b(\tau)=\widehat d(\tau) \end{align*} for every $\tau\in\Delta(B)$. Hence $\Gamma(b)=\Gamma(d)$. The Gelfand transform is injective on the commutative $C^*$-algebra $B$, so $b=d$. Thus the positive square root constructed above is unique. We denote it by $a^{1/2}$.[/step]

[guided]Since the previous step proved $bd=db$, the elements $b$ and $d$ generate a commutative unital C*-subalgebra \begin{align*} B:=C^*(b,d,1_{A^+})\subset A^+, \end{align*} where $C^*(b,d,1_{A^+})$ denotes the smallest closed $*$-subalgebra of $A^+$ containing $b$, $d$, and $1_{A^+}$. Let $\Delta(B)$ be the character space of $B$, and let \begin{align*} \Gamma:B\to C(\Delta(B)) \end{align*} denote the Gelfand transform [citetheorem:8549]. Define the scalar-valued functions $\widehat b,\widehat d:\Delta(B)\to\mathbb C$ by \begin{align*} \widehat b(\tau):=\Gamma(b)(\tau)=\tau(b),\qquad \widehat d(\tau):=\Gamma(d)(\tau)=\tau(d) \end{align*} for each character $\tau\in\Delta(B)$. The point of passing to $B$ is that equality can now be tested by scalar characters. Because $b,d\in B_+$ and characters on commutative $C^*$-algebras are $*$-homomorphisms [citetheorem:8550], every character sends them to non-negative real numbers. Since both elements square to $a$, we have $b^2=d^2$ inside $B$, and multiplicativity of each character $\tau$ gives \begin{align*} \widehat b(\tau)^2=\tau(b^2)=\tau(d^2)=\widehat d(\tau)^2. \end{align*} The two numbers $\widehat b(\tau)$ and $\widehat d(\tau)$ are non-negative real numbers with the same square, so uniqueness of the ordinary non-negative real square root gives \begin{align*} \widehat b(\tau)=\widehat d(\tau). \end{align*} This holds for every $\tau\in\Delta(B)$, hence $\Gamma(b)=\Gamma(d)$ as functions on $\Delta(B)$. The Gelfand transform is injective on the commutative C*-algebra $B$, so equality of the transforms implies equality of the elements: \begin{align*} b=d. \end{align*} Thus every positive square root of $a$ is equal to the constructed one. The positive square root is therefore unique, and we denote it by $a^{1/2}$.[/guided]