= Solution
Put $K=\sigma(T)$ and use the <continuous functional calculus> $\pi:C(K)\to\mathcal B(H)$. It is a <unital> star-homomorphism taking the coordinate function to $T$. This follows from the commutative C-star theorem applied to $C^*(I,T,T^*)$, together with spectral permanence for <C-star subalgebras>, which is among the permitted C-star results.
Take the <Hilbert space> inner product linear in its first variable. For a <vector> $\xi$, let $H_\xi=\overline{\{\pi(f)\xi:f\in C(K)\}}$. This cyclic subspace is invariant under every $\pi(f)$ and its adjoint, so it is reducing. The functional $f\mapsto\langle\pi(f)\xi,\xi\rangle$ is positive because $f\ge0$ implies $\pi(f)=\pi(\sqrt f)^*\pi(\sqrt f)$. It has, by the <Riesz-Markov-Kakutani representation theorem>, a finite regular positive <Borel measure> $\mu_\xi$ on $K$. The identity
$$
\|\pi(f)\xi\|^2=\int_K|f|^2\,d\mu_\xi
$$
makes $f\mapsto\pi(f)\xi$ an <isometry> on its quotient by the functions of zero $L^2$ <norm>. Density of <continuous> functions in $L^2(\mu_\xi)$ extends it to a unitary $U_\xi:L^2(\mu_\xi)\to H_\xi$. Multiplication by a <continuous> function intertwines with $\pi$ on this subspace.
Choose a maximal family of mutually <orthogonal> nonzero cyclic <reducing subspaces>. Their <orthogonal> sum is all of $H$: otherwise its reducing <orthogonal complement> contains a nonzero <vector> generating another such subspace. This argument does not require separability. For a bounded <Borel measurable function> $g$ on $K$, multiplication $M_g$ on every $L^2(\mu_\xi)$ is bounded with <norm> at most $\|g\|_\infty$. Define on that <orthogonal> sum
$$
\boxed{\beta_T(g)=\bigoplus_\xi U_\xi M_gU_\xi^{-1}.}
$$
<Multiplication operators> satisfy $M_{fg}=M_fM_g$, $M_{\overline g}=M_g^*$ and $M_1=I$. The same identities hold for their <direct sums>, and the <norm> bound is uniform in $\xi$. Thus $\beta_T$ is a norm-decreasing <unital> star-homomorphism on the algebra of actual bounded <Borel measurable functions>, and $\beta_T(Z)=T$ because it extends $\pi$. No quotient by a single unspecified measure is being substituted for that Borel algebra.
Choose the bounded Borel measurable square root $s(0)=0$ and $s(z)=|z|^{1/2}e^{i\operatorname{Arg}(z)/2}$ for $z\ne0$, taking $\operatorname{Arg}\in[0,2\pi)$. Let $S=\beta_T(s)$. Then
$$
\boxed{S^2=T,\qquad S^*S=\beta_T(|s|^2)=SS^*.}
$$
Thus every bounded <normal operator> has a normal square root. The Borel branch is essential in general: on a circle winding about zero a <continuous> square-root branch need not exist.
Back to article page
