Since $\text{}P(X)\text{}\mathrm{\perp}\text{}\text{}x-P(X)\text{}$ , we have $\text{}||x|{|}^{2}=||p(x)|{|}^{2}+||x-P(x)|{|}^{2}\text{}$ . thus $||x|{|}^{2}$=⟨x,x⟩ =$||P(x)|{|}^{2}$ +||x-$P(x)|{|}^{2}$

Therefore, $||P(x)|{|}^{2}\le ||x|{|}^{2}.since||P(x)||\ge 0$ and $||x||\ge 0,wehave||P(x)||\le ||x||$

We jump to the conclusion of $||P(x)|{|}^{2}\le ||x|{|}^{2}.$ because of $||P(x)||\ge 0$ and $||x||\ge 0.$

but where do we prove that $||P(x)||\ge 0$ and $||x||\ge 0$ in the first place?

Therefore, $||P(x)|{|}^{2}\le ||x|{|}^{2}.since||P(x)||\ge 0$ and $||x||\ge 0,wehave||P(x)||\le ||x||$

We jump to the conclusion of $||P(x)|{|}^{2}\le ||x|{|}^{2}.$ because of $||P(x)||\ge 0$ and $||x||\ge 0.$

but where do we prove that $||P(x)||\ge 0$ and $||x||\ge 0$ in the first place?