Positive square root

For every $x$ one has $⟨(T-T^{\ast })x,x⟩=⟨Tx,x⟩-\overline{⟨Tx,x⟩}=0$; applying the four-term polarization expansion to the sesquilinear form $(x,y)\mapsto ⟨(T-T^{\ast })x,y⟩$ gives $⟨(T-T^{\ast })x,y⟩=0$ for all $x,y$, hence $T=T^{\ast }$.

The spectrum of the self-adjoint operator $T$ is a nonempty compact subset of $[0,+\mathrm{\infty })$, and the function $g(\lambda )=\sqrt{\lambda }$ is continuous and nonnegative on it.

Define $S:=g(T)$ by the self-adjoint calculus; then $S\in C^{\ast }(I,T)$, $S\ge 0$ because $g\ge 0$, and $S^2=g(T)g(T)=(g\cdot g)(T)=z(T)=T$. If polynomials $p_n$ converge uniformly to $g$ on $\sigma (T)$, isometry gives $\mathrm{\parallel }{p}_n(T)-S\mathrm{\parallel }=\mathrm{\parallel }{p}_n-g{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$, so this square root is obtained by continuous polynomial approximation.

$S$ commutes with every bounded operator commuting with $T$, by the commutant clause of the calculus.

Uniqueness: if $R\ge 0$ satisfies $R^2=T$, then $R$ commutes with $T$ and hence with $S$ by step 3.1, so $T,S$ and $R$ lie in the commutative unital C*-algebra $C^{\ast }(R,S)$; both $R$ and $S$ are self-adjoint with $\sigma (R),\sigma (S)\subseteq [0,+\mathrm{\infty })$ by [A3], so applying the self-adjoint calculus [A4] with the continuous function $\lambda \mapsto \sqrt{\lambda }$ on those spectra exhibits both as products inside $C^{\ast }(R,S)$: $R=(\sqrt{R}{)}^{\ast }(\sqrt{R})$ and $S=(\sqrt{S}{)}^{\ast }(\sqrt{S})$, so both elements are positive. Hence every character value satisfies $\chi (R)\ge 0$, $\chi (S)\ge 0$ [A5] and $\chi (R{)}^2=\chi (R^2)=\chi (T)=\chi (S^2)=\chi (S{)}^2$, whence $\chi (R)=\chi (S)$; injectivity of the Gelfand transform gives $R=S$.

The operator $S=T^{1\mathrm{/}2}$ is therefore the unique bounded positive square root of $T$; it lies in $C^{\ast }(I,T)$, commutes with the commutant of $T$, and was constructed by continuous polynomial approximation of $\sqrt{\lambda }$ on $\sigma (T)$, with no Borel calculus.