Two question on a lemma about C*-algebra

Question

I am reading Lin Hua xin's book "An introduction to the classification of amenable C*-algebras" and i am confused with the lemma 1.7.12 in this book.

Lemma 1.7.12 Let $A$ be a C*-algebra and $f\in A^{\ast}$ be self-adjoint. Then there are vectors $\xi$, $\eta\in H_{U}$ with $||\xi||^{2} ||\eta||^{2}\leq||f||$ such that $f(a)=\langle\pi_{U}(a)(\xi), \eta\rangle$ for all $a\in A$. (Here, $\pi_{U}(a):A\rightarrow B(H_{U})$ is the universal representation.)

Proof. To save notation, we may assume that $||f||=1$. By non-commutative Jordan decomposition theorem, there are positive linear functionals $f_{j}$, $j=1, 2$, such that $f=f_{1}-f_{2}$ and $||f||=||f_{1}||+||f_{2}||$. Each $f_{j}$ is a positive scalar multiple of state on $A$. Therefore there are mutually orthogonal vectors $\xi_{j}\in H_{U} (j=1,2)$ such that $||\xi_{j}||^{2}=||f_{j}||$ and $f_{j}(a)=\langle \pi_{U}(a)(\xi_{j}), \xi_{j}\rangle, j=1,2$. Set $\xi=\xi_{1}\oplus\xi_{2}$ and $\eta=\xi_{1}\oplus(-\xi_{2})$. Then $f(a)=\langle \pi_{U}(a)\xi, \eta\rangle$ for all $a\in A$.

My question are:

1. Why there exist mutually orthogonal vectors(why are they orthogonal?) $\xi_{j}\in H_{U} (j=1,2)$ such that $||\xi_{j}||^{2}=||f_{j}||$(why this normed equation hold?) and $f_{j}(a)=\langle \pi_{U}(a)(\xi_{j}), \xi_{j}\rangle, j=1,2$?

2. If we set $\xi=\xi_{1}\oplus\xi_{2}$ and $\eta=\xi_{1}\oplus(-\xi_{2})$, how to conclude that $f(a)=\langle \pi_{U}(a)\xi, \eta\rangle$ for all $a\in A$?

Answer 1

Let me denote by $(\pi_i,H_i,\xi_i)$ the GNS representation of a positive functional $f_i.$ That is $f_i(a)=\langle \pi_i(a)\xi_i,\xi_i\rangle$ for all $a\in A.$ The universal representation $\pi_U$ is the direct sum of $\pi_i$ where $f_i$'s run over all positive functionals on $A.$ That is $H_U=\oplus_i H_i.$

1. The vectors $\xi_i\in H_i$ are mutually orthogonal because $H_i\perp H_j.$ Equation $||\xi_i||^2=||f_i||$ follows from $|f_i(a)|=|\langle\pi_i(a)\xi_i,\xi_i\rangle|\leq||\pi_i(a)||\cdot ||\xi_i ||^2\leq||a||\cdot ||\xi_i ||^2$ and $f(1_A)=||\xi_i ||^2$

2. $f(a)=f_1(a)-f_2(a)=\langle\pi_{U}(a)\xi_1,\xi_1\rangle-\langle\pi_{U}(a)\xi_2,\xi_2\rangle=$

since $\xi_1$ and $\xi_2$ belong to mutually orthogonal invariant subspaces $H_1$ and $H_2$ in $H_U$ we have $\langle\pi_{U}(a)\xi_1,\xi_2\rangle=0=\langle\pi_{U}(a)\xi_2,\xi_1\rangle,$ hence we proceed

$ =\langle \pi_{U}(a)(\xi_1+\xi_2),(\xi_1-\xi_2)\rangle=\langle \pi_{U}(a)\xi, \eta\rangle$