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 is a linear map between C*-algebras, it is said to be *positive *if . In this case , and the restriction map is increasing.

Every *-homomorphism is positive.

## Example

Let (continuous functions on the unit circle) and let be the normalised arc length measure on , Then the linear functional

, ,

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

## Example

Let . The linear functional

, ,

is positive. It is called the *trace. *Observe that there are no non-zero *-homomorphisms from to if (why is this — I can see that any such *-homomorphism must have and a few calculations shows the result).

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

, ,

is a positive sesquilinear form on (postive because and .). Hence and (by C-S). Morever the function is a semi-norm on .

Suppose now that is a linear functional on and that such that for all . Then is bounded with norm . We show this: First suppose that . Then are positive elements of , and therefore

.

Now suppose that is an arbitrary element of , so , where are it’s real and imaginary (hermitian) parts, and . Then

.

# Theorem 3.3.1

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

## Proof

If is not bounded, then by the previous remarks

.

Hence there is a sequence such that for all . Set

,

so (Theorem 5.7). Now and therefore

.

Hence, is an upper bound for the set , which is impossible. Hence, by contradiction, is bounded

# Theorem 3.3.2

*If is a positive linear functional on a C*-algebra , then and for all .*

## Proof

Let be an approximate unit for . Then (all bounded linear maps are continuous)

.

Also (the first inequality is C-S),

# Theorem 3.3.3

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

*is positive.*

*For each approximate unit of , .**For*some*approximate unit of , .*

## Proof

We may suppose that . First we show that 1. 2. Suppose that is positive, and let be an approximate unit of . Then is an increasing net in (why increasing), so it converges to a supremum, which is obviously not greater than 1. Thus . Now suppose that . Then (why can we say — perhaps by adjoining to ? — but then !! Unless we can use an ‘increasing’ argument to say

.

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

,

so . Hence, . Therefore, , so 1. 2.

That 2. 3. is obvious.

Next we show that 3. 1. Suppose that is an approximate unit such that . Let and write , where . To show that , we may suppose that . If , then

,

so

.

However, , and , so in the limit as increases, we get:

.

The left-hand side of this inequality is , so if we cancel and rearrange we get

.

Since and this inequality holds for all , must be zero. Therefore, is real if is hermitian.

Now suppose . Then is hermitian and (again problems with this inequality), so . But then

,

and therefore . Thus is positive and we have shown 3. 1

# Corollary 3.3.4

*If is a bounded linear functional on a unital C*-algebra, then is positive if and only if .*

## Proof

The constant sequence is approximate unit for . Apply the previous theorem

# Corollary 3.3.5

*If , are positive linear functionals on a unital C*-algebra, then .*

## Proof

If is an approximate unit for the algebra, then

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

# Theorem 3.3.6

*If is a normal element of a non-zero C*-algebra , then there exists a state such that .*

## Proof

We may assume that . Let be the C*-algebra generated by and in . Since is abelian and is continuous on the compact space , there is a character on such that . By the Hahn-Banach theorem, there is a bounded linear functional on extending and preserving the norm, so . Since , is positive by Corollary 3.3.4. If denotes the restriction of to , then is a positive linear functional on such that . Hence,

,

so , and the reverse inequality is obvious (because it’s a restriction). Therefore

# Theorem 3.3.7

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

*For each , if and only if for all .**The inequality*

*holds for all .*

## Proof

Condition 1. follows from the Cauchy-Schwarz inequality ( is routine. To go the other way assume and let 😉).

To show Condition 2., we may suppose, using Condition 1. (if , then Condition 2. holds as also), that . The function

, ,

is positive and linear (positive because , and if , then is positive (positivity is preserved by pre- and post-multiplication by an element and its adjoint)), so if is any approximate unit for , then

.

Hence, , and therefore we have the result

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

# Theorem 3.3.8

*Let be a C*-subalgebra of a C*-algebra , and suppose that is a positive linear functional on . Then there is a positive linear functional on extending such that .*

## Proof

Suppose first that . Define a linear functional on by setting

, ().

Let be an approximate unit for . By Theorem 3.3.3, . Now suppose that and . Then

.

Hence, , and the reverse inequality is obvious (as it is an extension). Thus, , so is positive by Corollary 3.3.4. This proves the theorem in the case (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 is an arbitrary C*-algebra containing as a C*-subalgebra. Replacing and by and if necessary (can we do this because we can easily restrict to/ extend to maps between and / (and vice versa)), we may suppose that has a unit which lies in . By the Hahn-Banach Theorem, there is a functional extending and of the same norm (this theorem 3.3.8 is essentially Hahn-Banach for positive functionals). Since (as and is an extension ), it follows as before from Corollary 3.3.4 that is positive

Let be a compact Hausdorff space and denote by the real Banach space of all real-valued continuous functions on . The operations on 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 is a bounded real-linear functional, then there is a unique real measure such that for all . Moreover, (I’m guessing here that is being normed by total variation), and is positive iff is; that is for all such that . The Jordan decomposition for a real measure asserts that there are positive measures such that and . We translate this via the Riesz-Kakutani theorem into a statement about linear functionals: If is a bounded real-linear functional, then there exist positive bounded real-linear functionals such that and . We are now going to prove an analogue of this result for C*-algebras.

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

(*).

For if , then there is a number such that , so

,

which implies (*) (don’t see this — I didn’t need this to show that . Oh wait, let such that . Suppose wlog that . Then such that and hence

,

that is the supremum is attained near a “-real” element so we can replace in by as they will be equal for some “-real element”.).

If , we define by setting for all . Note that , , and the map is conjugate linear.

We say a functional is *self-adjoint *if . For any bounded linear functional on , there are unique self-adjoint bounded linear functionals and on such that (take and ).

The condition is equivalent to (both sides are quick calculations), and therefore the restriction of is a bounded real-linear functional. Moreover, ; 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 — I assume that is a C*-algebra, then is a positive as required by this little calculation. Is complete? (http://www.springerlink.com/content/674642115688270q/fulltext.pdf says it is complete.) )

.

For if , we have (a quick calculation with self-adjoint), so

.

(the inequality comes from the fact that the set ).

We denote by the set of self-adjoint functionals in , and by the set of positive functionals in .

If is a real-linear Banach space, we denote its dual (over ) by .

The space is a real-linear Banach space (there we go!) and it is an easy exercise to verify that is a real-linear vector space of () and that the map , , is an isometric real-linear isomorphism (I had read this all wrong. A better notation for I feel is . Now let be the relevant map . Now is certainly linear. It is surjective because we can extend a real functional on to the self-adjoint functional on the whole of by :

,

as required. It is injective because if then and agree on hermitian elements. Let :

,

that is . Finally,

,

by the above calculation.).

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

# The Jordan Decomposition Theorem

*Let where is a C*-algebra. Then there exists positive linear functionals on such that and .*

## Proof

Note that is weak*-closed (I’m harking back to the question of whether or not is closed?), so by the Banach-Alaoglu Theorem ( is weak* compact) is a (Hausdorff — if then they differ at a point . The continuous map separates and ) weak* compact space. If , define by setting . The map

, ,

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

,

straightforwardly. However Theorem 3.3.6 says there exists such that so therefore .)

If , then . By the Hahn-Banach Theorem, there exists a real-linear functional such that and . (O.K. I’m looking for a map that go from and make the following (poor attempt at a diagram(!)) commute:

I can’t see it. How about starting with . This is well defined because is injective. To see this let . Then for all :

,

but because is Hausdorff we can find a separating functional. Hence and is injective. I need to show that has continuous inverse — well I only need it to be continuous on . I know that it is linear so all I need is boundedness. Let be equal to for some . Now

,

(O.K. I presume that was a superfluous calculation ;)) Therefore is continuous and bounded from . Maybe now I can use Hahn-Banach to extend to a bounded linear functional such that and . Now let :

.

Hence we have a map such that and . 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 such that and . Set and . Clearly, . We denote the corresponding self-adjoint functionals in by and . Then , and since:

,

(the first inequality is a quick calculation) we have . Clearly (this last part needs the fact that is “positive”)

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

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