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 $, ($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 $QAQ=$. 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 $Q$$Qx=x$$Qy=y$), $x=\lim QT_nQy$. But $QT_nQ=\lambda_nQ$ for some $\lambda_n\in\mathbb{C}$, because $QAQ=$, so $x\in$. This shows that $QH=$, 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 $T$$S$ 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 $\|\cdot\|_i$ $\bullet$