Showing that ker$(x)=$ker$(|x|)$

Question

Let $\varphi:C(X)\rightarrow B(\mathcal{H})$ be a representation of $C(X)$ on the Hilbert space $\mathcal{H}$, where $X$ is compact. For $A\in B(\mathcal{H})$, denote the range projection of $A$ by $R(A)$, i.e., $R(A)$ is the projection onto the closure of $A(\mathcal{H})$. If $f\in C(X)$ is a continuous function, then $f=f_{+}-f_{-}$, where $f_{+},f_{-}\geq0$ and $f_{+}f_{-}=0$. My question is how can I show that $R(|\varphi(f)|)=R(\varphi(f))$ using the fact that $\varphi(f)=\varphi(f_{+})-\varphi(f_{-})$, $\varphi(|f|)=\varphi(f_{+})+\varphi(f_{-})$, and $f_{+}f_{-}=0$? I believe it is a general fact that $x$ and $(xx^{*})^{\frac{1}{2}}$ have the same kernel, but I want to know if this can be used to show that $R(|\varphi(f)|)=R(\varphi(f))$ directly using the relations above.

Answer 1

The algebra and the representation don't really play a role here.

You have $T\in B(H)$, selfadjoint (since you assumed $f$ real). You have $$ T=T_1-T_2,\qquad |T|=T_1+T_2,\qquad T_1,T_2\geq0,\quad T_1T_2=0. $$ The equality $T_1T_2=0$ gives you that $R(T_1)\perp R(T_2)$, since $$ \langle T_1x,T_2y\rangle=\langle T_2T_1x,y\rangle=0. $$ This forces $R(T)=R(T_1)+R(T_2)$. Indeed, the inclusion $R(T)\subset R(T_1)+R(T_2)$ is automatic. And if $z=T_1x+T_2y$, then (with $P$ the projection onto $R(T_1)$ and $Q$ the projeciton onto $R(T_2)$) $$ T\big[Px-Qy\big]=T_1x+T_2y=z. $$ Thus $R(T_1)+R(T_2)\subset R(T)$, and so $R(T)=R(T_1)+R(T_2)$.

As the analog computation can be made for $|T|=T_1+T_2$, $$ R(|T|)=R(T_1)+R(T_2)=R(T). $$

Answer 2

If $\varphi(f)x=0$, then $\varphi(f_{+})^{2}x=\varphi(f_{+})(\varphi(f_{+})-\varphi(f_{-}))x=0$ since $\varphi(f_{+})\varphi(f_{-})=0$. But $\varphi(f_{+})=\varphi(f_{+})^{*}$, so ker$(\varphi(f_{+})^{2})=$ker$(\varphi(f_{+}))$, and hence $\varphi(f_{+})x=0$. Similarly, $\varphi(f_{-})x=0$ and $\varphi(|f|)x=0$.

Answer 3

Why not use the definition directly?

\begin{align*}|x|\xi = 0 &\iff \||x|\xi\| = 0\\ &\iff \langle |x|^2 \xi, \xi\rangle = 0\\ &\iff \langle x^*x\xi, \xi\rangle = 0\\ &\iff \|x\xi\| = 0 \iff x\xi = 0\end{align*} so $\ker(x) = \ker(|x|)$.