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:

$\rho$ is positive.
For each approximate unit $\{u_\lambda\}$ of $A$ , $\|\rho\|=\lim_\lambda \rho(u_\lambda)$ .
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

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

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

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

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

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. $\Rightarrow$ 1 $\bullet$

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 $A$ .

For each $a\in A$ , $\rho(a^*a)=0$ if and only if $\rho(ba)=0$ for all $b\in A$ .
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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Ideals and Positive Functionals: 3.3 Positive Linear Functionals

Example

Example

Theorem 3.3.1

Proof

Theorem 3.3.2

Proof

Theorem 3.3.3

Proof

Corollary 3.3.4

Proof

Corollary 3.3.5

Proof

Theorem 3.3.6

Proof

Theorem 3.3.7

Proof

Theorem 3.3.8

Proof

The Jordan Decomposition Theorem

Proof

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Ideals and Positive Functionals: 3.3 Positive Linear Functionals

Example

Example

Theorem 3.3.1

Proof

Theorem 3.3.2

Proof

Theorem 3.3.3

Proof

Corollary 3.3.4

Proof

Corollary 3.3.5

Proof

Theorem 3.3.6

Proof

Theorem 3.3.7

Proof

Theorem 3.3.8

Proof

The Jordan Decomposition Theorem

Proof

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