I was going over some results from a class I finished, and I am stuck on the following in $\color{red}{\textrm{red}}$.
Theorem: Suppose that $A$ is a $C^\ast$-algebra and that $b\in A$. Then $b^\ast b$ is positive$^{[1]}$.
Proof. Start by defining \begin{align*} f(t)= \begin{cases} \sqrt{t} & \text{if } t\geq0\\ 0 & \text{otherwise} \end{cases} \quad\text{and}\quad g(t)= \begin{cases} 0 & \text{if } t\geq0\\ \sqrt{-t} & \text{otherwise}. \end{cases} \end{align*} Then for all $t\in\mathbb{R}$ we have $f(t)g(t)=g(t)f(t)=0$ and $f(t)^2-g(t^2)=t$. $\color{red}{\textrm{Since $f$ and $g$ both vanish at $0$, we get self-adjoint elements of $A$ via}}$ \begin{align*} \color{red}{\textrm{$u=f(b^\ast b)\quad\text{and}\quad v=g(b^\ast b)$}}. \end{align*} Then $uv=vu=0$ and $u^2-v^2=b^\ast b$. But then $vb^\ast bv=v(u^2-v^2)v=-v^4$. Thus $\sigma((bv)^\ast bv)=\sigma(-v^4)\subset(-\infty,0]$. Thus $-v^4=0$. Since $v$ is self-adjoint, this means $v=0$. But then $b^\ast b=u^2$, and $u^2$ is positive.
$^{[1]}$ Here an element $a$ is called positive if $a$ is self-adjoint and $\sigma(a)\subset[0,\infty)$.
I know that the element $b^\ast b$ is self-adjoint. Also, the maps are really the functional calculus, so that $f(b^\ast b)$ is not really $\sqrt{b^\ast b}$.