How prove that the square root operator commutes with every operator $X$ that commutes with its square

Question

Let $U ∈ \mathcal{L}(H)$ be such that $U^* = U$ and $U ≥ 0$ (i.e $(Ux,x)\geq0$), exists $V ∈ \mathcal{L}(H )$ such that $V^* = V$, $V \geq 0$, and $V^2 = U$. The operator $V$ is called the square root of $U$. How prove that the operator $V$ commutes with every operator $X$ that commutes with $U$ (i.e., $X \circ U = U \circ X$ implies $X \circ V = V \circ X$)? I really have no idea what to do, can someone give a tip?

Answer 1

The continuous functional calculus gives a continuous homomorphism $\varphi:C(X)\to\mathcal{L}(H)$ where $X$ is the spectrum of $U$ and $\varphi(f)=U$ where $f(x)=x$ is the identity function. The operator $V$ is then just $\varphi(g)$ where $g(x)=\sqrt{x}$ is the square root function. By the Weierstrass approximation theorem, $g$ is a limit of polynomials in $f$, and so applying $\varphi$ we see that $V$ is a limit of polynomials in $U$. Any operator which commutes with $U$ commutes with any polynomial in $U$, and thus taking limits also commutes with $V$.