STATEMENT: This is a proof from one of Qiaochu's notes on $C^*$ algebras.
Proof: Let A be a $C^*$ algebra.We now want to show that for any $c\in A$ we have $c^*c\geq 0$. Suppose otherwise.We know that the $C^*$-subalgebra generated by $c^*c$ has the form $C(X)$ for some $X$. Choose $b\geq 0$ in this subalgebra such that $c^*cb\neq 0$ and $-c^*cb^2\geq 0$
QUESTION: I don't see why such a $b$ with those properties exist. Can someone clarify this for me.
Take a continuous function $f$ which is strictly positive on $(-\infty,0)\cap \operatorname{sp}(c^*c)$ and 0 on $(0,\infty)\cap \operatorname{sp}(c^*c)$. Take $b= f(c^*c)$. Then $c^*cb \neq 0$ since $xf(x) \neq 0$ for some $x$ (assuming that the spectrum of $c^*c$ contains a negative point). Also $xf(x) \leq 0$ for all $x$ since $f$ is zero on $(0,\infty)$ and strictly positive on $(-\infty,0)$.