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.

H is always a Hilbert space. H^1 is the set of unit vectors.

If P is a finite-rank projection on H, then the C*-algebra A=PB(H)P is finite dimensional. To see this, write P=\sum_{j=1}^ne_j\otimes e_j, where e_1,\dots,e_n\in H. If T\in B(H), then

PTP=\sum_{j,k=1}^n(e_j\otimes e_j)T(e_k\otimes e_k)=\sum_{j,k=1}^n\langle Te_k,e_j\rangle (e_j\otimes e_k)

Hence, A\subset <e_j\otimes e_k>, (j,k=1,\dots,n) (*), and therefore is finite dimensional.

A closed vector subspace K\subset H is invariant for a subset A\subset B(H) if A(K)\subset K. If A is a C*-subalgebra of B(H), it is said to be irreducible, or act irreducibly on H, if the only closed vector subspaces of H that are invariant for A are \{0\} and H.

Theorem 2.4.9

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

Proof: A\cap K(H) is a non-empty self-adjoint set, so it contains a self-adjoint element, T say. Now as self-adjoint operators are normal, \nu(T)=\|T\|>0, so \sigma(T) contains non-zero elements. Hence (Th. 1.4.11: the non-zero elements of the spectrum of a compact operator are eigenvalues) T admits a non-zero eigenvalue \lambda. The same theorem asserts that \lambda is isolated, so if  f=\mathbf{1}_{\lambda} is the (continuous) on \sigma(T) and P=f(T) is a projection in A (via the functional calculus). Moreover, P is non-zero because f is non-zero. If z is the inclusion map \sigma(T)\rightarrow \mathbb{C}, then (z-\lambda)f=0, so (T-\lambda)P=0, and therefore P(H)\subset\ker(T-\lambda). The space \ker(T-\lambda) is finite-dimensional (Th. 1.4.5: when T is compact), so P is therefore of finite rank.

Let Q be a non-zero projection in A of minimal finite rank. Then the C*-algebra QAQ is finite dimensional; therefore, it is the linear span of its projections (by *). However, the minimal rank assumptions on Q implies that the only projectections in QAQ can be 0 and Q, so QAQ=<Q>. Now let y\in Q(H) be non-zero. If K=\overline{\{Ty:T\in A\}}, then K is a vector subpace of H invariant for A, and is non-zero as it contains y=QY. It follows from irreducubility, therefore that K=H. Hence, if x\in QH, then x=\lim T_ny for some sequence \{T_n\}\subset A. Therefore, (premultiplying by QQx=xQy=y), x=\lim QT_nQy. But QT_nQ=\lambda_nQ for some \lambda_n\in\mathbb{C}, because QAQ=<Q>, so x\in<y>. This shows that QH=<y>, and therefore q=y\otimes y.

Now suppose x\in H^1.  As before, there are operators T_n\in A such that x=\lim T_ny so

x\otimes x=\lim T_ny\otimes T_ny=\lim T_n(y\otimes y)T_n^*=\lim T_nQT_n^*.

Hence, x\otimes x\in A. Therefore, all rank-one projections are in A, so F(H)\subset A (Th. 2.4.6: F(H) is linearly spanned by rank-one projections), and therefore K(H)\subset A (Th. 2.4.5: F(H) is dense in K(H)) \bullet

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

\|T\|_2=\sqrt{\sum_{x\in E}\|Tx\|^2}

A quick calculation shows that this definition is independent of the choice of basis and also that \|T^*\|_2=\|T\|_2. An operator is a Hilbert-Schmidt operator if \|T\|_2<\infty. We denote the class of H-S operators by L^2(H).

Example 2.4.1: Let \{e_n\} be an o.n.b. for H and let T be diagonal with respect to \{e_n\}, with diagonal sequence \{\lambda_n\}. Then T is a H-S operator iff \sum |\lambda_n|^2<\infty.

More generally, if T\in B(H) and [A]_{ij} is its matrix with respect to \{e_n\}, so that a_{ij}=\langle Te_m,e_n\rangle, then

\|T\|_2=\sqrt{\sum_{m}\sum_n |a_{ij}|^2}

Example 2.4.2: Let L^2(\mathbb{T}) and L^2(\mathbb{T}^2) denote the Lesbesgue L^2 spaces of \mathbb{T} and \mathbb{T}^2 with the usual measures, normalised arc length m, and the corresponding product measure m\times m. By elementary measure theory, C(\mathbb{T}) and \mathbb{T}^2 are L^2-dense in L^2(\mathbb{T}) and L^2(\mathbb{T}^2) respectively. Define e_n\in C(\mathbb{T}) by e_n(\lambda)=\lambda_n, and e_{nm}\in C(\mathbb{T}^2) by e_{nm}(\lambda,\mu)=\lambda^n\mu^m (n,m\in\mathbb{Z}). These sequences are orthonormal in the corresponding L^2-spaces. By the Stone-Weiestrass theorem (if when X is compact & Hausdorff,a subset S\subset C(X) seperates points , then the complex unital *-algebra generated by S is dense in C(X)), the sup-norm closed linear span of \{e_n\}\subset C(\mathbb{T}) is C(\mathbb{T}) itself. By a similar reasoning the sup-norm closed linear span of \{e_{nm}\} is C(\mathbb{T}^2). Thus, \{e_n\} and \{e_{nm}\} have L^2-dense linear span in, and therefore are orthonormal bases of, L^2(\mathbb{T}) and L^2(\mathbb{T}^2), respectively.

Let k\in L^2(\mathbb{T}^2). Then for almost all \lambda \in\mathbb{T},

\int |k(\lambda,\mu)f(\mu)|dm\mu<\infty

since

\int\int |k(\lambda,\mu)f(\mu)|\,d(m\times m)(\lambda,\mu)

\leq\left(\int\int|k(\lambda,\mu)|^2\,d(m\times m)(\lambda,\mu)\right)^{1/2}\left(\int\int|f(\mu)|^2\,d(m\times m)(\lambda,\mu)\right)^{1/2}

=\|k\|_2\|f\|_2

Define the integral operator T=T_k\in L^(\mathbb{T}) by

Tf(\lambda)=\int k(\lambda,\mu)f(\mu)\,dm\mu

for almost all \lambda. That Tf\in L^2(\mathbb{T}) follows from another application of the Cauchy-Schwarz inequality,

\int|Tf(\lambda)|^2\,dm\lambda

=\int\left|\int k(\lambda,\mu)f(\mu)\,dm\mu\right|^2\,dm\lambda

\leq \int\left(\int|k(\lambda,\mu)|^2\,dm\mu\right)\left(\int|f(\mu)|^2\,dm\mu\right)\,dm\lambda

=\|k\|_2^2\|f\|_2^2

Hence, T is bounded with norm \|T\|\leq \|k\|_2. Now we compute:

\|T\|_2^2=\sum_{n\in\mathbb{Z}}\|Te_n\|^2

=\sum_{n,m\in\mathbb{Z}}|\langle Te_n,e_m\rangle|

=\sum_{n,m\in\mathbb{Z}}\left|\int Te_n(\lambda)\overline{e_m(\lambda)}\,dm\lambda\right|^2

=\sum_{n,m\in\mathbb{Z}}\left|\int\int k(\lambda,\mu)e_n(\mu)\overline{e_m(\lambda)}\,dm\mu\,dm\lambda\right|^2

=\sum_{n,m\in\mathbb{Z}}|\langle k,e_{m,-n}\rangle|^2

Thus \|T\|_2=\|k\|_2 and T\in L^2(\mathbb{T}).

Theorem 2.4.10

Let T,S be operators on H, and \lambda\in\mathbb{C}. Then

  1. \|T+S\|_2\leq\|T\|_2+\|S\|_2 and \|\lambda T\|_2=|\lambda|\|T\|_2;
  2. \|T\|\leq\|T\|_2;
  3. \|TS\|_2\leq \|T\|\|S\|_2\leq \|T\|_2\|S\|_2\|TS\|_2\leq \|T\|_2\|S\|.

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

\sqrt{\sum_{x\in F}\|Tx+Sx\|^2}\leq \sqrt{\sum_{x\in F}(\|Tx\|+\|Sx\|)^2}

\leq \sqrt{\sum_{x\in F}\|Tx\|^2}+\sqrt{\sum_{x\in F}\|Sx\|^2}

It follows that \|T+S\|_2\leq \|T\|_2+\|S\|_2. Positive Homogenity is trivial.

If x\in H^1, there is an orthonormal basis containing x (“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,

\|Tx\|^2\leq \sum_{y\in E}\|Ty\|^2=\|T\|^2_2,

so \|T\|\leq \|T\|_2.

If E is an o.n.b. of H, then

\|TS\|_2^2=\sum_{x\in E}\|TSx\|^2\leq \|T\|^2\sum_{x\in E}\|Sx\|^2=\|T\|^2\|S\|_2^2.

Hence, \|TS\|_2\leq \|T\|\|S\|_2. Therefore, \|TS\|_2=\|S^*T^*\|_2\leq \|S^*\|\|T^*\|_2=\|T\|_2\|S\| \bullet

Corollary 2.4.11

The set L^(H) is a self-adjoint ideal of B(H), and a normed *-algebra with norm \|\cdot\|_2.

Note that if x,y\in H, then \|x\otimes y\|_2=\|x\|\|y\|, so x\otimes y\in L^2(H) (a straightforward calculation). Hence F(H)\subset L^2(H).

Lemma 2.4.12

Let T_1,T_2 be H-S operators on H. If E is an o.n.b of H and S=T_1^*T_2, then the family \{\langle Sx,x\rangle\}_{x\in E} is absolutely summable, and

\sum_{x\in E}\langle Sx,x\rangle=\frac{1}{4}\sum_{k=0}^3i^k\|T_2+i^kT_1\|_2^2

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

\sum_{x\in F}|\langle Sx,x\rangle|=\sum_{x\in F}|\langle T_2x,T_1x\rangle|

\leq \sum_{x\in F}\|T_2x\|\|T_1x\|

\leq \sqrt{\sum_{x\in F}\|T_2x\|^2}\cdot\sqrt{\sum_{x\in F}\|T_1x\|^2}

This last inequality is Holder’s Inequality. Hence \{\langle Sx,x\rangle\}_{x\in E} is absolutely summable. Also

\langle Sx,x\rangle=\langle T_2x,T_1x\rangle=\frac{1}{4}\sum_{k=0}^3i^k\|T_2x+i^kT_1x\|^2

by the polarisation identity, so

\sum_{x\in E}\langle Sx,x\rangle=\frac{1}{4}\sum_{k=0}^3i^k\sum_{x\in E}\|(T_2+i^kT1)x\|^2

=\frac{1}{4}\sum_{k=0}^3 i^k\|T_2+i^kT_1\|_2^2 \bullet

If T is an operator on H, we define its trace-class norm to be \|T\|_1=\||T|^{1/2}\|_2^2. If E is an o.n.b. then

\|T\|_1=\sum_{x\in E}\langle |T|(x),x\rangle.

If \|T\|_1<\infty, we call T a trace-class operator. The connection between trace operators and H-S operators is given by the following result.

Theorem 2.4.13

Let T be an operator on H. TFAE:

  1. T is trace-class.
  2. |T| is trace-class.
  3. |T|^{1/2} is a H-S operator.
  4. There exist H-S operators S_1,\,S_2 such that T=S_1S_2.

Proof: The implications 1. \Rightarrow 2. \Rightarrow 3. \Rightarrow 4. are easy (T=U|T|^{1/2}|T|^{1/2} via the polar decomposition. A quick calculation shows that U|T|^{1/2} is H-S). We prove 4. \Rightarrow 1. only.

Assume that T=S_1S_2, where S_1,\,S_2\in L^2(H). If T=U|T|, then |T|=U^*T=U^*S_1S_2. If E is a o.n.b., then by Lemma 2.4.12 the result follows (after a quick calculation) \bullet

“It is clear from this this theorem that if T is trace-class and S is another bounded operator, that ST and TS are also of trace class.” – I’m guessing that this is a typo. Define operators T,\,S on a separable Hilbert space H:

S:e_n\mapsto 2^{-n}e_n

T:e_n\mapsto 2^ne_n

Now S is trace class but ST=I=TS which certainly is not.

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

\text{tr}(T)=\sum_{x\in E}\langle Tx,x\rangle

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

Theorem 2.4.14

Let T and S be operators on H. Then

\text{tr}(TS)=\text{tr}(ST)

if either

  1. T and S are both H-S operators
  2. S is trace-class

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

\text{tr}(TS)=\frac{1}{4}\sum_{k=0}^3i^k\|S+i^kT^*\|_2^2

=\frac{1}{4}\sum_{k=0}^3 i^k\|(S+i^kT^*)^*\|_2^2

=\frac{1}{4}\sum_{k=0}^3\|T+i^kS^*\|_2^2=\text{tr}(ST)

If S is trace-class, there exist T_1,T_2 such that S=T_1T_2, so
\text{tr}(TS)=\text{tr}((TT_1)T_2)=\text{tr}(T_2(TT_1))=\text{tr}(TT_1T_2)=\text{tr}(ST) \bullet
Theorem 2.4.15

Let TS be operators on H and \lambda\in\mathbb{C}.
  1. \|T+S\|_1\leq \|T\|_1+\|S\|_1, and \|\lambda T\|_1=|\lambda|\|T\|_1
  2. \|T\|\leq \|T\|_1=\|T^*\|_1
  3. \|TS\|_1\leq \|T\|\|S\|_1 and \|TS\|_1\leq \|T\|_1\|S\|,

Proof: Beginning with 2. we have

\|T\|_1=\||T|^{1/2}\|_2^2\geq \||T|^{1/2}\|^2=\||T|\|=\|T\|.

This first equality holds as:

\||T|^{1/2}\|_2^2=\sum_{x\in E}\langle |T|^{1/2}x,|T|^{1/2}x\rangle

Now as |T|^{1/2} is hermitian, taking adjoints:

\||T|^{1/2}\|_2^2=\sum_{x\in E}\langle |T|x,x\rangle=\||T|\|_1

The second equality is true because \||T|^{1/2}\|^2=\||T|\| by the C*-equation (|T|^{1/2} is hermitian).

A justification of the third equality; on the one hand

\|(T^*T)^{1/2}(T^*T)^{1/2}\|=\|(T^*T)^{1/2}\|^2=\||T|\|^2

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

\|T^*T\|=\|T\|^2\Rightarrow \||T|\|=\|T\|

If T=U|T| is the polar decomposition of T, then TT^*=U|T|^2U^*, so |T^*|^2=(U|T|U^*)^2 and therefore |T^*|=U|T|U^* (once we hit with |T|, we are in \text{ran } |T| and U^*U is the identity on this space). Hence \|T^*\|_1=\text{tr}(|T^*|)=\text{tr}(U|T|U^*)=\text{tr}(U^*T)=\text{tr}(|T|)=\|T\|_1.

Next, looking at 3. Let ST=W|ST| be the polar decomposition of ST and W'=W^*SU. Then |ST|=W^*ST=W^*SU|T|=W'|T|. Hence |ST|^2=|T|W'^*W'|T|\leq |T|^2\|W'\|^2\leq |T|\|S\|^2 (note that S^*S\leq \|S\|^2I; now pre- and post-multiply by the hermitian element |T|), hence |ST|\leq |T|\|S\| Th. 2.2.6(the partial order on positive elements is preserved by root-taking). Consequently, if E is an o.n.b. for H,

\|ST\|_1=\sum_{x\in E}\langle |ST|x,x\rangle\leq \sum_{x\in E}\langle |T|x,x\rangle\|S\|=\|T\|_1\|S\|

The inequality here is a consequence of the fact that |T|\|S\|-|TS| is positive; hence the sesquilinear forms \langle (|T|\|S\|-|TS|)x,x\rangle\geq 0 by Th. 2.3.5 (an operator-induced sesquilinear form is positive iff the operator is positive)

Also, \|TS\|_1=\|S^*T^*\|_1\leq |S|\|T\|_1.

Finally, 1. The equality \|\lambda T\|_1=|\lambda|\|T\|_1 is straightforward. Suppose that T,S are trace-class, and let T=U|T| and S=W|S|, and T+S=V|T+S|. Then

|T+S|=V^*(T+S)=V^*U|T|+V^*W|S|.

If E is an o.n.b.,

\|T+S\|_1=\sum_{x\in E}\langle |T+S|x,x\rangle

Now as \|T+S\| is positive, it is equal to its own absolute value:

=\left|\sum_{x\in E}\langle V^*U|T|x,x\rangle+\sum_{x\in E}\langle V^*W|S|x,x\rangle\right|

\leq \sum_{x\in E}\left|\langle |T|^{1/2}x,|T|^{1/2}U^*Vx\rangle\right|+\sum_{x\in E}\left|\langle |S|^{1/2}x,|S|^{1/2}W^*Vx\rangle\right|

By Cauchy-Schwarz,

\leq \left(\sum_{x\in E}\||T|^{1/2}x\|^2\right)^{1/2}\left(\sum_{x\in E}\||T|^{1/2}U^*Vx\|^2\right)^{1/2}

+\left(\sum_{x\in E}\||S|^{1/2}x\|^2\right)^{1/2}\left(\sum_{x\in E} \||S|^{1/2}W^*Vx\|^2\right)^{1/2}

=\|T\|_1^{1/2}\||T|U^*V\|_2+\|S\|_1^{1/2}\||S|W^*V\|_2

\leq \|T\|_1^{1/2}\|T\|_1^{1/2}+\|S\|_1^{1/2}\|S\|_1^{1/2}

=\|T\|_1+\|S\|_1 \bullet

If H  is a Hilbert space we denote the set of trace-class operators on H by L^1(H). The previous theorem asserts that L^1(H) is a self-adjoint ideal of B(H), and the function T\mapsto \|T\|_1 is a norm on L^1(H) making it a normed *-algebra.

Theorem 2.4.16

Let H be a Hilbert space. The function

\text{tr}:L^1(H)\rightarrow\mathbb{C},\,T\mapsto \text{tr}(T)

is linear, and

|\text{tr}(ST)|\leq \|S\|\|T\|_1\forall \,S\in B(H),\, T\in L^1(H)

Proof: Linearity of trace is clear. To show the inequality let T=U|T| and let E be an o.n.b. Then

|\text{tr}(ST)|=\left|\sum_{x\in E}\langle STx,x\rangle\right|

=\left|\sum_{x\in E}\langle |T|^{1/2}x,|T|^{1/2}U^* S^*x \rangle\right|

\leq \sum_{x\in E} \||T|^{1/2}x\| \||T|^{1/2}U^*S^*x\|

\leq \left(\sum_{x\in E}\||T|^{1/2}x\|^2 \right)^{1/2}\left(\sum_{x\in E}\||T|^{1/2}U^*S^*x\|^2 \right)^{1/2}

=\|T\|_1^{1/2}\||T|^{1/2}U^*S^*\|_2

\leq \|T\|^{1/2}_1\||T|^{1/2}\|_2\|S\|

By Th. 2.4.10 (3);

=\|T\|_1\|S\| \bullet

If x,y\in H, then \|x\otimes y\|_1=\|x\|\|y\| and \text{tr}(x\otimes y)=\langle x, y\rangle. The inclusions F(H)\subset L^1(H)\subset L^2(H) hold (Finite Rank implies finite sum).

Theorem 2.4.17

Let H be a Hilbert space. Then for each of i=1,2, the ideal L^{i}(H) is contained in K(H), and F(H) is dense in L^i(H) in the norm \|\cdot\|_i \bullet

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