For abelian C*-algebras we were able to completely determine the structure of the algebra in terms of the character space, that is, in terms of the one-dimensional representations. For the non-abelian case this is quite inadequate, and we have to look at representations of arbitrary dimension. There is a deep inter-relationship between the representations and the positive linear functionals of  a C*-algebra. Representations will be defined and some aspects of this inter-relationship investigated in the next section. In this section we establish the basis properties of positive linear functionals.

If \varphi:A\rightarrow B is a linear map between C*-algebras, it is said to be positive if \varphi(A^+)\subset B^+. In this case \varphi(A_{\text{sa}})\subset B_{\text{sa}}, and the restriction map \varphi:A_{\text{sa}}\rightarrow B_{\text{sa}} is increasing.

Every *-homomorphism is positive.

Example

Let A=C(\mathbb{T}) (continuous functions on the unit circle) and let m be the normalised arc length measure on \mathbb{T}, Then the linear functional

C(\mathbb{T})\rightarrow \mathbb{C}f\mapsto \int f\,dm,

is positive (and not a homomorphism — e.g. approximate to delta functions \delta_{\{-1\}} and \delta_{\{+1\}}. Now both integrals are both arbitrarily close to 1 but the integral of their product is 0).

Example

Let A=M_n(\mathbb{C}). The linear functional

\text{tr}:A\rightarrow\mathbb{C}[a_{ij}]\mapsto \sum_{i=1}^na_{ii},

is positive. It is called the trace. Observe that there are no non-zero *-homomorphisms from M_n(\mathbb{C}) to \mathbb{C} if n>1 (why is this — I can see that any such *-homomorphism \varphi must have \varphi(AB)=\varphi(BA) and a few calculations shows the result).

Let A be a C*-algebra and \rho a positive linear functional on A. Then the function

A^2\rightarrow \mathbb{C}(a,b)\mapsto \rho(b^*a),

is a positive sesquilinear form on A (postive because \rho(x,x)=\rho(x^*x) and x^*x\in A^+.). Hence \rho(b^*a)=\overline{\rho(a^*b)} and |\rho(b^*a)|\leq \sqrt{\rho(a^*a)}\sqrt{\rho(b^*b)} (by C-S). Morever the function a\mapsto \sqrt{\rho(a^*a)} is a semi-norm on A.

Suppose now that \rho is a linear functional on A and that M\in \mathbb{R}^+ such that |\rho(a)|\leq M for all a\in A^+\cap A^1. Then \rho is bounded with norm \|\rho\|\leq 4M. We show this: First suppose that a\in A_{\text{sa}}\cap A^1. Then a^+,a^- are positive elements of A^1, and therefore

|\rho(a)|=|\rho(a^+)-\rho(a^-)|\leq 2M.

Now suppose that a is an arbitrary element of A^1, so a=b+ic, where b,c are it’s real and imaginary (hermitian) parts, and \|b\|,\|c\|\leq 1. Then

|\rho(a)|=|\rho(b)+i\rho(c)|\leq 4M.

Theorem 3.3.1

If \rho is a positive linear functional on a C*-algebra A, then it is bounded.

Proof

If \rho is not bounded, then by the previous remarks

\sup_{a\in A^+\cap A^1}\rho(a)=+\infty.

Hence there is a sequence \{a_n\}\subset A^+\cap A^1 such that \rho(a_n)\geq 2^n for all n\in\mathbb{N}. Set

a=\sum_{n=0}^\infty \frac{a_n}{2^n},

so a\in A^+ (Theorem 5.7). Now \rho(a_n/2^n)\geq 1 and therefore

N\leq \sum_{n=0}^{N-1}\rho(a_n/2^n)=\rho\left(\sum_{n=0}^{N-1}\frac{a_n}{2^n}\right)\leq \rho(a).

Hence, \rho(a) is an upper bound for the set \mathbb{N}, which is impossible. Hence, by contradiction, \rho is bounded \bullet

Theorem 3.3.2

If \rho is a positive linear functional on a C*-algebra A, then \rho(a^*)=\overline{\rho(a)} and |\rho(a)|^2\leq \|\rho\|\rho(a^*a) for all a\in A.

Proof

Let \{u_\lambda\} be an approximate unit for A. Then (all bounded linear maps are continuous)

\rho(a^*)=\lim_\lambda \rho(a^*u_\lambda)=\overline{\lim_\lambda}(u_\lambda a)=\overline{\rho(a)}.

Also (the first inequality is C-S),

|\rho(a)|^2=\lim_\lambda |\rho(u_\lambda a)|^2\leq \|\rho\|\rho(a^*a) \bullet

Theorem 3.3.3

Let \rho be a bounded linear functional on a C*-algebra A. Then the following are equivalent:

  1. \rho is positive.
  2. For each approximate unit \{u_\lambda\} of A\|\rho\|=\lim_\lambda \rho(u_\lambda).
  3. For some approximate unit \{u_\lambda\} of A\|\rho\|=\lim_\lambda \rho(u_\lambda).

Proof

We may suppose that \|\rho\|=1. First we show that 1. \Rightarrow 2. Suppose that \rho is positive, and let \{u_\lambda\} be an approximate unit of A. Then \{\rho(u_\lambda)\} is an increasing net in \mathbb{R} (why increasing), so it converges to a supremum, which is obviously not greater than 1. Thus \lim_\lambda \rho(u_\lambda)\leq 1. Now suppose that a\in A^1. Then (why can we say \rho(u_\lambda^2)\leq \rho(u_\lambda) — perhaps by adjoining \Lambda to \Lambda^2? — but then \Lambda^4!! Unless we can use an ‘increasing’ argument to say

\rho(u_\lambda u_\lambda)\leq \lim_{\lambda'} \rho(u_{\lambda'} u_\lambda)=\rho(u_\lambda)\leq \lim_\lambda\rho(u_\lambda).

Would have no problem carrying this through if I could get a footing with ‘increasing’.)

|\rho(u_\lambda a)|^2\leq \rho(u^2_\lambda)\rho(a^*a)\leq \lim_\lambda (u_\lambda),

so |\rho(a)|^2\leq\lim_\lambda u_\lambda. Hence, 1\leq \lim_\lambda \rho(u_\lambda). Therefore, 1=\lim_\lambda \rho(u_\lambda), so 1. \Rightarrow 2.

That 2. \Rightarrow 3. is obvious.

Next we show that 3. \Rightarrow 1. Suppose that \{u_\lambda\} is an approximate unit such that \lim_\lambda u_\lambda=1. Let a\in A_{\text{sa}}\cap A^1 and write \rho(a)=\alpha+i\beta, where \alpha,\beta\in\mathbb{R}. To show that \rho(a)\in\mathbb{R}, we may suppose that \beta\leq 0. If n\in\mathbb{N}, then

\|a-inu_\lambda\|^2=\|(a+inu_\lambda)(a-inu_\lambda)\|

=\|a^2+n^2u_\lambda-in(au_\lambda-u_\lambda a)\|

\leq 1+n^2+n\|au_\lambda-u_\lambda a\|,

so

|\rho(a-inu_\lambda)|^2\leq 1+n^2+n\|au_\lambda-u_\lambda a\|.

However, \lim_\lambda \rho(a-inu_\lambda)=\rho(a)-in, and \lim_\lambda (au_\lambda-u_\lambda a)=0, so in the limit as \lambda increases, we get:

|\alpha+i\beta-in|\leq 1+n^2.

The left-hand side of this inequality is \alpha^2+\beta^2-2n\beta +n^2, so if we cancel and rearrange we get

-2n\beta\leq 1-\beta^2-\alpha^2.

Since  \beta\leq 0 and this inequality holds for all n\in\mathbb{N}\beta must be zero. Therefore, \rho(a) is real if a is hermitian.

Now suppose a\in A^+\cap A^1. Then u_\lambda -a is hermitian and \|u_\lambda-a\|\leq 1 (again problems with this inequality), so \rho(u_\lambda-a)\leq 1. But then

1-\rho(a)=\lim_\lambda \rho(u_\lambda-a)\leq 1,

and therefore \rho(a)\geq 0. Thus \rho is positive and we have shown 3. \Rightarrow\bullet

Corollary 3.3.4

If \rho is a bounded linear functional on a unital C*-algebra, then \rho is positive if and only if \rho(1)=1.

Proof

The constant sequence \{1\} is approximate unit for A. Apply the previous theorem \bullet

Corollary 3.3.5

If \rho_1\rho_2 are positive linear functionals on a unital C*-algebra, then \|\rho_1+\rho_2\|=\|\rho_1\|+\|\rho_2\|.

Proof

If \{u_\lambda\} is an approximate unit for the algebra, then

\|\rho_1+\rho_2\|=\lim_\lambda(\rho_1+\rho_2)(u_\lambda)=\lim_\lambda \rho_1(u_\lambda)+\lim_\lambda \rho_2(u_\lambda)=\|\rho_1\|+\|\rho_2\| \bullet

A state on a C*-algebra A is a positive linear functional on A of norm one. We denote by S(A) the set of states of A.

Theorem 3.3.6

If a is a normal element of a non-zero C*-algebra A, then there exists a state \rho\in S(A) such that |\rho(a)| =\|a\|.

Proof

We may assume that a\neq 0. Let B be the C*-algebra generated by 1 and a in \tilde{A}. Since B is abelian and \hat{a} is continuous on the compact space \Phi(B), there is a character \rho_2 on B such that \|a\|=\|\hat{a}\|_\infty=|\rho_2(a)|. By the Hahn-Banach theorem, there is a bounded linear functional \rho_1 on \tilde{A} extending \rho_2 and preserving the norm, so \|\rho_1\|=1. Since \rho_1(1)=\rho_2(1)=1\rho_1 is positive by Corollary 3.3.4. If \rho denotes the restriction of \rho_1 to A, then \rho is a positive linear functional on A such that \|a\|=|\rho(a)|. Hence,

\|a\|=|\rho(a)|\leq \|\rho\|\|a\|,

so \|\rho\|\geq 1, and the reverse inequality is obvious (because it’s a restriction). Therefore \rho\in S(A) \bullet

Theorem 3.3.7

Suppose that \rho is a positive linear functional on a C*-algebra A.

  1. For each a\in A\rho(a^*a)=0 if and only if \rho(ba)=0 for all b\in A.
  2. The inequality

\rho(b^*a^*ab)\leq \|a^*a\|\rho(b^*b)

holds for all a,b\in A.

Proof

Condition 1. follows from the Cauchy-Schwarz inequality (\Rightarrow is routine. To go the other way assume \rho(a^*a)\neq 0 and let b=a^* 😉).

To show Condition 2., we may suppose, using Condition 1. (if \rho(b*b)=0, then Condition 2. holds as \rho(b^*a^*ab)=0 also), that \rho(b^*b)>0. The function

\tau:A\rightarrow \mathbb{C}c\mapsto \frac{\rho(b^*cb)}{\rho(b^*b)},

is positive and linear (positive because b*b\in A^+, and if c\in A^+, then b^*cb is positive (positivity is preserved by pre- and post-multiplication by an element and its adjoint)), so if \{u_\lambda\} is any approximate unit for A, then

\|\tau\|=\lim_\lambda \tau(u_\lambda)=\lim_\lambda \frac{\rho(b^*u_\lambda b)}{\rho(b^*b)}=\frac{\rho(b^*b)}{\rho(b^*b)}=1.

Hence, \tau(a^*a)\leq \|a^*a\|, and therefore we have the result \bullet

We turn now to the problem of extending positive linear functionals.

Theorem 3.3.8

Let B be a C*-subalgebra of a C*-algebra A, and suppose that \rho is a positive linear functional on B. Then there is a positive linear functional \rho' on A extending \rho such that \|\rho'\|=\|\rho\|.

Proof

Suppose first that A=\tilde{B}. Define a linear functional \rho' on A by setting

\rho'(b+\lambda)=\rho(b)+\lambda \|\rho\|, (b\in B, \,\lambda\in \mathbb{C}).

Let \{u_\lambda\} be an approximate unit for B. By Theorem 3.3.3, \|\rho\|=\lim_\lambda \rho(u_\lambda). Now suppose that b\in B and \mu\in\mathbb{C}. Then

|\rho'(b+\mu)|=\left|\lim_\lambda \rho(bu_\lambda)+\mu\lim_\lambda \rho(u_\lambda)\right|=

\lim_\lambda|\rho((b+\mu)u_\lambda)|\leq \sup_{\lambda}\|\rho\|\|(b+\mu)u_\lambda\|\leq \|\rho\|\|b+\mu\|.

Hence, \|\rho'\|\leq \|\rho\|, and the reverse inequality is obvious (as it is an extension). Thus, \|\rho'\|=\|\rho\|=\rho'(1), so \rho is positive by Corollary 3.3.4. This proves the theorem in the case A=\tilde{B} (now an issue I have is why aren’t these kind of steps taken in general — in the next step we work in unitisations and this seems to be the usual framework).

Now suppose that A is an arbitrary C*-algebra containing B as a C*-subalgebra. Replacing B and A by \tilde{B} and \tilde{A} if necessary (can we do this because we can easily restrict to/ extend to maps between \tilde{A} and A/ (and vice versa)), we may suppose that A has a unit 1 which lies in B. By the Hahn-Banach Theorem, there is a functional \rho'\in A^* extending \rho and of the same norm (this theorem 3.3.8 is essentially Hahn-Banach for positive functionals). Since \rho'(1)=\rho(1)=\|\rho\|=\|\rho'\| (as 1\in B\cap A and \rho' is an extension \rho'(1)=\rho(1)), it follows as before from Corollary 3.3.4 that \rho' is positive \bullet

Let X be a compact Hausdorff space and denote by C(X,\mathbb{R}) the real Banach space of all real-valued continuous functions on X. The operations on C(X,\mathbb{R}) are the pointwise-defined ones and the norm is the supremum norm. The Riesz-Kakutani theorem (a bit of a nomenclature issue here — the Riesz-Kakutani theorem is a Riesz Representation Theorem — Murphy obviously prefers R-K and as Kakutani is the man who said one of my favourite maths quotes:

“A drunk man will find his way home, but a drunk bird may get lost forever”,

I’ll follow Murphy!) theorem asserts that if \rho:C(X,\mathbb{R})\rightarrow\mathbb{R} is a bounded real-linear functional, then there is a unique real measure \mu\in M(X) such that \rho(f)=\int f\,d\mu for all f\in C(X,\mathbb{R}). Moreover, \|\mu\|=\|\rho\| (I’m guessing here that M(X) is being normed by total variation), and \mu is positive iff \rho is; that is \rho(f)\geq 0 for all f\in C(X,\mathbb{R}) such that g\geq 0. The Jordan decomposition for a real measure \mu\in M(X) asserts that there are positive measures \mu^+,\mu^-\in M(X) such that \mu=\mu^+-\mu_- and \|\mu\|=\|\mu^+\|+\|\mu^-\|. We translate this via the Riesz-Kakutani theorem into a statement about linear functionals: If \rho:C(X,\mathbb{R})\rightarrow \mathbb{R} is  a bounded real-linear functional, then there exist positive bounded real-linear functionals \rho_+,\,\rho_-:C(X,\mathbb{R})\rightarrow \mathbb{R} such that \rho=\rho_+-\rho_- and \|\rho\|=\|\rho_+\|+\|\rho_-\|. We are now going to prove an analogue of this result for C*-algebras.

Let A be a C*-algebra. If \rho is a bounded linear functional on A, then

\|\rho\|=\sup_{a\in A^1}|\text{Re}(\rho(a))|   (*).

For if a\in A^1, then there is a number \lambda\in\mathbb{T} such that \lambda \rho(a)\in\mathbb{R}, so

|\text{Re}(\rho(a))||\rho(a)|=|\text{Re}(\rho(\lambda a))|\leq \|\rho\|,

which implies (*) (don’t see this — I didn’t need this \lambda to show that \|\rho\|\geq \sup |\text{Re}(\rho(a))|. Oh wait, let b\in A^1 such that \|\rho\|\approx |\rho(b)|. Suppose wlog that \text{Im}(\rho(b))\neq 0. Then \exists\,\lambda\in\mathbb{T} such that \lambda\rho(b)=\rho(\lambda b)\in\mathbb{R} and hence

|\rho(\lambda b)|=|\lambda \rho(b)|=|\lambda||\rho(b)|=|\rho(b)|,

that is the supremum is attained near a “\rho-real” element \lambda b so we can replace |\rho(a)| in \sup_{a\in A^1}|\rho(a)| by |\text{Re}(\rho(a))| as they will be equal for some “\rho-real element”.).

If \rho\in A^*, we define \rho^*\in A^* by setting \rho^*(a)=\overline{\rho(a^*)} for all a\in A. Note that \rho^{**}=\rho , \|\rho^*\|=\|\rho\|, and the map \rho\mapsto \rho^* is conjugate linear.

We say a functional \rho\in A^* is self-adjoint if \rho=\rho^*. For any bounded linear functional on A, there are unique self-adjoint bounded linear functionals \rho_1 and \rho_2 on A such that \rho=\rho_1+i\rho_2 (take \rho_1=(\rho+\rho^*)/2 and \rho_2=(\rho-\rho^*)/2i).

The condition \rho=\rho^* is equivalent to \rho(A_{\text{sa}})\subset\mathbb{R} (both sides are quick calculations), and therefore the restriction \rho':A_{\text{sa}}\rightarrow\mathbb{R} of \rho is a bounded real-linear functional. Moreover, \|\rho\|=\|\rho'\|; that is (you see I’m wondering why we don’t say, oh, let’s work in the unitisation and as the unit is self-adjoint, we have \|\rho\|=\rho(1)=\rho'(1)=\|\rho'\| — I assume that A_{\text{sa}} is a C*-algebra, then \rho' is a positive as required by this little calculation. Is A_{\text{sa}} complete? (http://www.springerlink.com/content/674642115688270q/fulltext.pdf says it is complete.) )

\|\rho\|=\sup_{a\in A_{\text{sa}}^1}|\rho(a)|.

For if a\in A, we have \text{Re}(\rho(a))=\rho(\text{Re}(a)) (a quick calculation with \rho self-adjoint), so

\|\rho\|=\sup_{a\in A^1}|\text{Re}(\rho(a))|=\sup_{a\in A^1}|\rho(\text{Re(a)})|\leq \sup_{b\in A_\text{sa}^1}|\rho(b)|\leq \|\rho\|.

(the inequality comes from the fact that the set \{\text{Re}(a):a\in A^1\}\subset A_{\text{sa}}).

We denote by A_{\text{sa}}^* the set of self-adjoint functionals in A^*, and by A^*_+ the set of positive functionals in A^*.

If X is a real-linear Banach space, we denote its dual (over \mathbb{R}) by X^{\mathbb{R}*}.

The space A_{\text{sa}} is a real-linear Banach space (there we go!) and it is an easy exercise to verify that A_{\text{sa}}^* is a real-linear vector space of A^* (\checkmark) and that the map A^*_{\text{sa}}\rightarrow A_{\text{sa}}^{\mathbb{R}*}\rho\mapsto \rho', is an isometric real-linear isomorphism (I had read this all wrong. A better notation for A_{\text{sa}}^* I feel is (A^*)_{\text{sa}}. Now let \varphi:(A^*)_{\text{sa}}\rightarrow A_{\text{sa}}^{\mathbb{R}*} be the relevant map \rho\mapsto\rho'. Now \varphi is certainly linear. It is surjective because we can extend a real functional \rho' on A_{\text{sa}} to the self-adjoint functional \rho' on the whole of  A by \rho(a)=\rho(b+ic)=\rho(b)+i\rho(c):

\rho^*(a)=\overline{\rho(b^*)-i\rho(c^*)}=\rho(b^*)+i\rho(c^*)=\rho(b)+i\rho(c)=\rho(a),

as required. It is injective because if \varphi(\rho_1)=\varphi(\rho_2) then \rho_1 and \rho_2 agree on hermitian elements. Let a=b+ic:

\rho_1(a)=\rho_1(b+ic)=\rho_1(b)+i\rho_1(c)=\rho_2(b)+i\rho_2(c)=\rho_2(a),

that is \rho_1=\rho_2. Finally,

\|\varphi(\rho)\|=\|\rho'\|=\sup_{b\in A_{\text{sa}}^1}|\rho(b)|=\sup_{b\in {A_{\text{sa}}}^1}|\rho(a)|=\|\rho\|,

by the above calculation.).

We shall use these observations these observations in the proof of the following result.

The Jordan Decomposition Theorem

Let \rho\in (A^*)_{\text{sa}} where A is a C*-algebra. Then there exists positive linear functionals \rho_+,\,\rho_- on A such that \rho=\rho_+-\rho_- and \|\rho\|=\|\rho_+\|+\|\rho_-\|.

Proof

Note that \Omega=(A^*_+)^1 is weak*-closed (I’m harking back to the question of whether or not A^+ is closed?), so by the Banach-Alaoglu Theorem ((A^*)^1 is weak* compact\Omega is a (Hausdorff — if \rho_1\neq \rho_2 then they differ at a point x. The continuous map \hat{x} separates \rho_1 and \rho_2) weak* compact space. If a\in A_{\text{sa}}, define \theta(a)\in C(\Omega,\mathbb{R}) by setting \theta(a)(\rho)=\rho(a). The map

\theta:A_{\text{sa}}\rightarrow C(\Omega,\mathbb{R})a\mapsto \theta(a),

is clearly real-linear, and also order-preserving; that is, if a is a positive element of A, then \theta(a)\geq 0 on \Omega. Moreover, \theta is isometric by Theorem 3.3.6 (we have that

\|\theta(a)\|=\sup_{\rho \in\Omega}|\rho(a)|\leq \|\rho\|\|a\|\leq\|a\|,

straightforwardly. However Theorem 3.3.6 says there exists \rho\in S(A)\subset \Omega such that |\rho(a)|=\|a\| so therefore \|\theta(a)\|=\|a\|.)

If \rho\in (A^*)_{\text{sa}}, then \rho'\in A_{\text{sa}}^{\mathbb{R}*}. By the Hahn-Banach Theorem, there exists a real-linear functional \tau\in C(\Omega,\mathbb{R})^{\mathbb{R}*} such that \tau\circ \theta=\rho' and \|\tau\|=\|\rho'\|. (O.K. I’m looking for a map \tau that go from C(\Omega,\mathbb{R})\rightarrow \mathbb{R} and make the following (poor attempt at a diagram(!)) commute:

\begin{array}{ccc}A_{\text{sa}}&\overset{\theta}{\rightarrow}&C(\Omega,\mathbb{R})\\ & \underset{\rho'}{\searrow} & \downarrow\\ & & \mathbb{R} \end{array}

I can’t see it. How about starting with \tau'=\rho\circ \theta^{-1}. This is well defined because \theta is injective. To see this let \theta(a)=\theta(b). Then for all \eta\in\Omega:

\theta(a)(\eta)=\theta(b)(\eta)\Rightarrow \eta(a)=\eta(b),

but because \Omega is Hausdorff we can find a separating functional. Hence a=b and \theta is injective. I need to show that \theta has continuous inverse — well I only need it to be continuous on \theta(A_{\text{sa}}).  I know that it is linear so all I need is boundedness. Let f\in \theta(A_{\text{sa}}) be equal to \theta(a) for some a\in A_{\text{sa}}. Now

\|\theta^{-1}(f)\|= \|\theta^{-1}(\theta(a))\|=\|a\|=\|\theta(a)\|\leq\|f\|,

(O.K. I presume that was a superfluous calculation ;)) Therefore \tau'=\rho\circ \theta^{-1} is continuous and bounded from \theta(A_{\text{sa}})\rightarrow \mathbb{R}. Maybe now I can use Hahn-Banach to extend \tau':\theta(A_{\text{sa}})\rightarrow \mathbb{R} to a bounded linear functional \tau:C(\Omega,\mathbb{R})\rightarrow\mathbb{R} such that \tau\in C(\Omega,\mathbb{R})^{\mathbb{R}*} and \|\tau\|=\|\tau'\|. Now let \theta(a)\in \theta(A_{\text{sa}}):

\|\tau'\|=\sup_{\theta(a)\in C(\Omega,\mathbb{R})^1}\|\tau'(\theta(a))\|=\sup_{a\in A_{\text{sa}}^1}|\rho(a)|=\|\rho\|=\|\rho'\|.

Hence we have a map \tau\in C(\Omega,\mathbb{R})^{\mathbb{R}*} such that \tau\circ \theta=\rho' and \|\tau\|=\|\rho'\|. Phew! I presume he uses a different form of the Hahn-Banach Theorem which encodes all of this stuff??!) By the remarks proceeding this theorem, there exist positive functionals \tau_+,\,\tau_-\in C(\Omega,\mathbb{R})^{\mathbb{R}*} such that \tau=\tau_+-\tau_- and \|\tau\|=\|\tau_+\|+\|\tau_-\|. Set \rho'_+=\rho_+\circ \theta and \rho'_-=\tau_-\circ \theta. Clearly, \rho'_+,\,\rho'_-\in (A_{\text{sa}})^{\mathbb{R}*}. We denote the corresponding self-adjoint functionals in (A_{\text{sa}})^* by \rho_+ and \rho_-. Then \rho=\rho_+-\rho_-, and since:

\|\rho\|=\|\rho'\|=\|\tau\|=\|\tau_+\|+\|\tau_-\|\geq \|\rho'_+\|+\|\rho'_-\|=\|\rho_+\|+\|\rho_-\|\geq \|\rho\|,

(the first inequality is a quick calculation) we have \|\rho\|=\|\rho_+\|+\|\rho_-\|. Clearly \rho_+,\,\rho_-\in A_+^* (this last part needs the fact that \theta is “positive”) \bullet

One can show that the functionals \rho_+ and \rho_- are unique, but we will have no need for this.

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