**This starts on P.58 rather than p.53. I underline my own extra explanations, calculations, etc. The notation varies on occasion from Murphy to notation which I prefer.**

is always a Hilbert space. is the set of unit vectors.

If is a finite-rank projection on , then the C*-algebra is finite dimensional. To see this, write , where . If , then

Hence, , () (*), and therefore is finite dimensional.

A closed vector subspace is *invariant for a subset * if . If is a C*-subalgebra of , it is said to be *irreducible*, or *act irreducibly *on , if the only closed vector subspaces of that are invariant for are and .

**Theorem 2.4.9**

*Let be a C*-algebra acting irreducibly on and having non-zero intersection with *(compact operators)*. Then .*

**Proof: ** is a non-empty self-adjoint set, so it contains a self-adjoint element, say. Now as self-adjoint operators are normal, , so contains non-zero elements. Hence (Th. 1.4.11: the non-zero elements of the spectrum of a compact operator are eigenvalues) admits a non-zero eigenvalue . The same theorem asserts that is isolated, so if is the (continuous) on and is a projection in (via the functional calculus). Moreover, is non-zero because is non-zero. If is the inclusion map , then , so , and therefore . The space is finite-dimensional (Th. 1.4.5: when is compact), so is therefore of finite rank.

Let be a non-zero projection in of minimal finite rank. Then the C*-algebra is finite dimensional; therefore, it is the linear span of its projections (by *). However, the minimal rank assumptions on implies that the only projectections in can be and , so . Now let be non-zero. If , then is a vector subpace of invariant for , and is non-zero as it contains . It follows from irreducubility, therefore that . Hence, if , then for some sequence . Therefore, (premultiplying by , , ), . But for some , because , so . This shows that , and therefore .

Now suppose . As before, there are operators such that so

.

Hence, . Therefore, all rank-one projections are in , so (Th. 2.4.6: is linearly spanned by rank-one projections), and therefore (Th. 2.4.5: is dense in )

Let be an operator on , and suppose that is an o.n.b. We define the *Hilbert-Schmidt norm of * to be

A quick calculation shows that this definition is independent of the choice of basis and also that . An operator is a *Hilbert-Schmidt operator *if . We denote the class of H-S operators by .

**Example 2.4.1: **Let be an o.n.b. for and let be diagonal with respect to , with diagonal sequence . Then is a H-S operator iff .

More generally, if and is its matrix with respect to , so that , then

**Example 2.4.2: **Let and denote the Lesbesgue spaces of and with the usual measures, normalised arc length , and the corresponding product measure . By elementary measure theory, and are -dense in and respectively. Define by , and by (). These sequences are orthonormal in the corresponding -spaces. By the Stone-Weiestrass theorem (if when is compact & Hausdorff,a subset seperates points , then the complex unital *-algebra generated by is dense in ), the sup-norm closed linear span of is itself. By a similar reasoning the sup-norm closed linear span of is . Thus, and have -dense linear span in, and therefore are orthonormal bases of, and , respectively.

Let . Then for almost all ,

since

Define the *integral operator * by

for almost all . That follows from another application of the Cauchy-Schwarz inequality,

Hence, is bounded with norm . Now we compute:

Thus and .

**Theorem 2.4.10**

*Let be operators on , and . *Then

- and ;
- ;
- . .

**Proof: **If is any finite set of orthonormal vectors of , then

It follows that . Positive Homogenity is trivial.

If , there is an orthonormal basis containing (“obvious” but hadn’t seen it before, it’s the General Basis Extension Theorem http://www.mathe2.uni-bayreuth.de/stoll/lecture-notes/LinearAlgebraI.pdf p.24). Hence,

,

so .

If is an o.n.b. of , then

.

Hence, . Therefore,

**Corollary 2.4.11**

*The set is a self-adjoint ideal of , and a normed *-algebra with norm .*

Note that if , then , so (a straightforward calculation). Hence .

**Lemma 2.4.12**

*Let be H-S operators on . If is an o.n.b of and , then the family is absolutely summable, and*

**Proof: **If is a finite non-empty subset of , then

This last inequality is Holder’s Inequality. Hence is absolutely summable. Also

by the polarisation identity, so

If is an operator on , we define its *trace-class norm *to be . If is an o.n.b. then

.

If , we call a *trace-class operator*. The connection between trace operators and H-S operators is given by the following result.

**Theorem 2.4.13**

*Let be an operator on . TFAE:*

*is trace-class.*

*is trace-class.*

*is a H-S operator.*

*There exist H-S operators such that .*

**Proof: **The implications 1. 2. 3. 4. are easy ( via the polar decomposition. A quick calculation shows that is H-S). We prove 4. 1. only.

Assume that , where . If , then . If is a o.n.b., then by Lemma 2.4.12 the result follows (after a quick calculation)

“It is clear from this this theorem that if is trace-class and is another bounded operator, that and are also of trace class.” – I’m guessing that this is a typo. Define operators on a separable Hilbert space :

Now is trace class but which certainly is not.

We define the *trace *of a trace-class operator to be

where is any o.n.b. of . This function is independent of the choice of basis.

**Theorem 2.4.14**

*Let and be operators on . Then*

*if either*

*and are both H-S operators*

*is trace-class*

**Proof: **If and are both H-S operators,

**Theorem 2.4.15**

*Let , be operators on and .*

*, and*

*and ,*

**Proof: **Beginning with 2. we have

.

This first equality holds as:

Now as is hermitian, taking adjoints:

The second equality is true because by the C*-equation ( is hermitian).

A justification of the third equality; on the one hand

by the C*-equation. However the LHS is also equal to:

If is the polar decomposition of , then , so and therefore (once we hit with , we are in and is the identity on this space). Hence .

Next, looking at 3. Let be the polar decomposition of and . Then . Hence (note that ; now pre- and post-multiply by the hermitian element ), hence Th. 2.2.6(the partial order on positive elements is preserved by root-taking). Consequently, if is an o.n.b. for ,

The inequality here is a consequence of the fact that is positive; hence the sesquilinear forms by Th. 2.3.5 (an operator-induced sesquilinear form is positive iff the operator is positive)

Also, .

Finally, 1. The equality is straightforward. Suppose that are trace-class, and let and , and . Then

.

If is an o.n.b.,

Now as is positive, it is equal to its own absolute value:

By Cauchy-Schwarz,

If is a Hilbert space we denote the set of trace-class operators on by . The previous theorem asserts that is a self-adjoint ideal of , and the function is a norm on making it a normed *-algebra.

**Theorem 2.4.16**

*Let be a Hilbert space. The function*

*is linear, and*

*, *

**Proof: **Linearity of trace is clear. To show the inequality let and let be an o.n.b. Then

By Th. 2.4.10 (3);

If , then and . The inclusions hold (Finite Rank implies finite sum).

**Theorem 2.4.17**

*Let be a Hilbert space. Then for each of , the ideal is contained in , and is dense in in the norm *

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July 6, 2011 at 3:33 pm

Von Neumann Algebras: The Weak and Ultraweak Topologies « J.P. McCarthy: Math Page[…] be a Hilbert space, and suppose that . It follows from Theorem 2.4.16 (http://irishjip.wordpress.com/2011/01/18/c-algebras-and-operator-theory-2-4-compact-hilbert-space-op…) that the […]

October 5, 2011 at 2:13 pm

Representations of C*-Algebras: Irreducible Representations and Pure States « J.P. McCarthy: Math Page[…] that the operator is positive. Since is a compact normal operator (), it is diagonalisable by Theorem 2.4.4; that is, there is an orthonormal basis and there is a family of scalars such that . Choose . […]