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.