# Question 1

Let $A$ be a Banach algebra such that for all $a\in A$ the implication

$Aa=0$ or $aA=0$ $\Rightarrow a=0$

holds. Let $L$$R$ be linear mappings from $A$ to itself such that for all $a,b\in A$,

$L(ab)=L(a)b$$R(ab)=aR(b)$, and $R(a)b=aL(b)$.

Show that $L$ and $R$ are necessarily continuous.

# Question 2

Let $A$ be a unital C*-algebra.

## (a)

If $a,b$ are positive elements of $A$, show that $\sigma(ab)\subset \mathbb{R}^+$.

### Solution (Wills)

For elements $a,b$ of a unital algebra $A$:

$\sigma(ab)\cup\{0\}=\sigma(ba)\cup\{0\}$

If $a\in A^+$ then $a^{1/2}\in A^+$ so that

$\sigma(ab)\cup\{0\}=\sigma(a^{1/2}(a^{1/2}b))\cup\{0\}=\sigma(a^{1/2}ba^{1/2})\cup\{0\}$

Now if $b\in A^+$, for any $c\in A$$c^*bc\in A^+$. Hence $\sigma(a^{1/2}ba^{1/2})\subset \mathbb{R^+}$ and the result follows (note that $ab$ need not be hermitian) $\bullet$

## (b)

If $a$ is an invertible element of $A$, show that $a=u|a|$ for a unique unitary $u\in A$. Give an example of an element of $B(H)$ for some Hilbert space $H$ that cannot be written as the product of a unitary times a positive operator.

## (c)

Show that if $a\in G(A)$, then $\|a\|=\|a^{-1}\|=1$ iff $a$ is unitary.

### Solution

Suppose first that $a$ is unitary. Then $\|a\|=1$. Also

$a^*a=1$

$\Rightarrow (a^*)^{-1}a^{-1}=1=a^{-1}(a^*)^{-1}$

that is $a^{-1}$ is unitary also and hence $\|a^{-1}\|=1$

Suppose now that $a$ is not unitary… $\bullet$

# Question 3

Let $X$ be a locally compact Hausdorff space, and suppose that the C*-algebra $C_0(X)$ is generated by a sequence of projections $\{p_n\}\subset C_0(X)$. Show that the hermitian element

$h=\sum_{n=1}^\infty\frac{p_n}{3^n}$

generates $C_0(X)$.

## Solution

In the first instance, the only projection functions are the characteristic functions $\mathbf{1}_A$ for $A\subset X$. For now I am going to assume that $\{p_n\}=\{\mathbf{1}_{K_n}\}$ with $K_n$ compact. Then each of the $p_n\in C_0(X)$ is as required (eh — actually they’re not continuous!).

A polynomial in the $\{p_n\}$ is significantly simplified because $p_n^r=p_n$ for $r\geq 1$.  Also if $\Sigma\subset \{p_n\}$ is finite, then

$\prod_{K_i\in\Sigma}\mathbf{1}_{K_i}=\mathbf{1}_{\bigcap_i K_i}$

Hence, if these projections generate $C_0(X)$ then any  $f\in C_0(X)$ may be approximated in the supremum norm by:

$f\approx \sum_{i=1}^N\beta_i \mathbf{1}_{\Lambda_i}$

where the $\Lambda_i=\bigcap_{K_j\in\Sigma_i}K_j$ for finite $\Sigma_i\subset\{K_i\}$.

Hence we must now simultaneously approximate to $\{\beta_i\}$ on $\{\Lambda_i\}$ using polynomials in $h$ (not quite – sometimes the $\Lambda_i$ overlap). In the first instance

$h(x)=\sum_{i:x\in K_i}\frac{1}{3^i}$

# Question 4

We shall see in the next chapter that all closed ideals in C*-algebras are necessarily self-adjoint. Give an example of an ideal in the C*-algebra $C(\mathbb{D})$ that is not self-adjoint.

# Question 5

Let $\varphi:A\rightarrow B$ be an isometric linear map between unital *-algebras $A$ and $B$ such that $\varphi(a^*)=\varphi(a)^*$ (for all $a\in A$) and $\varphi(1)=1$. Show that $\varphi(A^+)\subset B^+$.

# Question 6

Let $A$ be a unital C*-algebra.

## If $\nu(a)<1$ and $b=\left(\sum_{n=0}^\infty a^{*n}a^n\right)^{1/2}$, show that $b\geq 1$ and $\|bab^{-1}\|<1$.

### Solution

Working under the slightly stronger hypothesis that $\|a\|<1$ (this guarantees convergence of $b$), we note that $A^+$ is closed (??). Now consider:

$b^2-1=\sum_{n=1}^\infty (a^n)^*a^n$

With the strengthening of the assumptions this series converges and $A^+$ closed, this sum of positive terms converges in $A^+$. Hence $b^2-1\geq 0$ $\Rightarrow b^2\geq 1$. Taking roots on both sides we get the result.

## (b)

For all $a\in A$, show that

$\nu(a)=\inf_{b\in G(A)}\|bab^{-1}\|=\inf_{b\in A_{\text{SA}}}\|e^ba e^{-b}\|$

# Question 7

Let $A$ be a unital C*-algebra.

## (a)

If $a,b\in A$, show that the map

$f:\mathbb{C}\rightarrow A$$\lambda\mapsto e^{i\lambda b}ae^{-i\lambda b}$

is differentiable and that $f'(0)=i(ba-ab)$

### Solution (Wills)

$e^{i\lambda b}=\sum_{n=0}^\infty \frac{(i\lambda b)^n}{n!}=\sum_{n=0}^\infty \frac{i^nb^n}{n!}\lambda^n$

$f'(\lambda)=ibe^{i\lambda b}ae^{-i\lambda b}+e^{i\lambda b}a(-i b)$.

If $\lambda=0$, then $f'(0)=i(ba-ab)$.

## (b)

Let $X$ be a closed vector subspace of $A$ which is unitarily invariant in the sense that $uXu^*\subset X$ for all unitaries $u\in A$. Show that $ab-ba\in X$ if $a\in X$ and $b\in A$.

### Solution (Wills)

If $X\subset A$ a closed subspace is unitarily invariant. Let $a,\,b\in X$ and make an additional assumption that $b^*=b$ ($\checkmark$). Then

$(e^{i\lambda b})^*=\sum_{n=0}^\infty\left[\frac{(i\lambda b)^n}{n!}\right]^*=\sum_{n=0}^\infty\frac{(-i\lambda b)^n}{n!}=e^{-i\lambda b}$.

Hence, $e^{-i\lambda b}=(e^{i\lambda b})^*$ and also $e^{-i\lambda b}e^{i\lambda b}=e^0=I_A$. Now if $f(\lambda)\in X$ then $f'(\lambda\in X)$ (since $X$ is closed???). Therefore $i(ba-ab)\in X$ and hence $ba-ab\in X$.

## (c)

Deduce that the closed linear span $X$ of the projections in $A$ has the property that $a\in X$ and $b\in A$ implies that $ba-ab\in X$.

### Solution

First we prove that if $p$ is a projection and if $u$ is a unitary then $upu^*$ is a projection.

$(upu^*)^2=upu^*upu^*=up^2u^*=upu^*$, also

$(upu^*)^*=(u^*)^{*}p^*u^*=upu^*$

Now it’s just a simple calculation to show:

$u\left(\sum_{i=1}^n\alpha_ip_i\right)u^*$

$=\sum_{i=1}^n\alpha_i (up_i u^*)\in X$.

Hence $X$ satisfies the hypotheses of the last exercise and hence we are done.

# Question 8 (Fuglede’s Theorem)

Let $a$ be a normal element of a C*-algebra $A$, and $b$ an element commuting with $a$. Show that $b^*$ also commutes with $a$.

## Solution (Hinted by Murphy)

Let $\tilde{A}$ be the unitisation of $A$ and let $f:\mathbb{C}\rightarrow \tilde{A}$ be defined by

$f(\lambda)=e^{i\lambda a^*}be^{-i\lambda a^*}$

From Exercise 2.7 this map is holmorphic with derivative at 0

$f'(0)=i(a^*b-ba^*)$

As $a$ commutes with $b$$b$ commutes with any continuous function of $a$ (or does this need to be polynomial – we don’t have compactness or a Hausdorff space either! Murphy suggests a different route certainly. However we do know that $a$ is normal). Now

$e^{i\lambda a}=e^{i\lambda (2 \text{Re}(a)-a)}$

Hence $f(\lambda)=b$ and hence constant. Therefore $f'(0)=0$ and we have $a^*b=ba^*$ as required $\bullet$

# Exercise 9

If  $I$ is an ideal of $B(H)$, show that it is self-adjoint.

## Solution

From Theorem 2.4.7 we know that $I$ contains $F(H)$. Suppose $B(H)$ has a basis $\{e_n\}$. Define $P_n$ as the projection onto $\langle e_1,e_2,\dots,e_n\rangle$. Then $P_n$ is of finite rank and hence in the ideal.

Now let $T\in B(H)$. Consider the operators $\{P_nT^*\}_{n\geq 1}$. All of these are in the ideal $I$ and furthermore as $P_nT^*\rightarrow T^*$ and $I$ is closed, we have that $T^*\in B(H)$. That is $I$ is self-adjoint $\bullet$

# Exercise 10

Let $T\in B(H)$.

## (a)

Show that $T$ is a left topological divisor in $B(H)$ iff it is not bounded below (cf. Exercise 1.11)

### Solution

From exercise 1.11 we know that:

1. An element $a$ in  unital Banach algebra is a left topological divisor if there is a sequence of unit vectors $\{a_n\}\in A^1$ such that $\lim_{n\rightarrow\infty}aa_n=0$. Equivalently $\zeta(a)=0$ where $\zeta(a)=\inf_{b\in A^1}\|ab\|$.
2. Left topological divisors are not invertible.

Hence assume that $T$ is a left-topological divisor. Then $T$ is not invertible and hence not bounded below.

Assume that $T$ is not bounded below. Then for any sequence $\{K_n\}\subset\mathbb{R}^+$ converging to zero, there exists a sequence $S=\{x_n\}\subset H$ such that:

$\|Tx_n\|

By homogeneity of the norm the sequence may be chosen to be unit vectors.

Let $S_n$ be the sequence of linear projections onto $\langle x_n\rangle$. Then

$\|TS_n\|=\inf_{x\in H^1}\|TS_nx\|$

$=\inf_{x_n\in S}\|T\lambda_n x_n\|=\inf_{x_n\in S}|\lambda_n|\|Tx_n\|.

Hence $\lim_{n\rightarrow \infty}\|TS_n\|=0$; hence $TS_n\rightarrow 0$ and so $T$ is a left-topological divisor.

## (b)

Define

$\sigma_{\text{ap}}(T)=\{\lambda\in\mathbb{C}:T-\lambda\text{ is not bounded below}\}$.

This set is called the approximate point spectrum of $T$ because $\lambda\in\sigma_{\text{ap}}(T)$ iff there is a sequence $\{x_n\}$ of unit vectors of $H$ such that $\lim_{n\rightarrow\infty}\|(T-\lambda)(x_n)\|=0$. Show that $\sigma_{\text{ap}}(T)$ is a closed subset of $\sigma(T)$ containing $\partial\sigma(T)$.

## (c)

Show that $T$ is bounded below iff it is left-invertible in $B(H)$.

## (d)

Show that $\sigma(T)=\sigma_{\text{ap}}(T)$ if $T$ is normal.

# Question 11

Let $T\in B(H)$ be a normal operator with spectral resolution of the identity $E$.

What the hell is a spectral resolution of the identity?

## (a)

Show that $T$ admits an invariant closed vector subspace other than $\{\mathbf{0}\}$ and $H$ if $\text{dim } H>1$.

## (b)

If $\lambda$ is an isolated point of $\sigma(T)$, show that $E(\lambda)=\text{ker}(T-\lambda)$ and that $\lambda$ is an eigenvalue of $T$.

# Question 12

An operator $T$ on $H$ is subnormal if there is a Hilbert space $K$ containing $H$ as a closed vector subspace and there exists a normal operator $S$ on $K$ such that $H$ is invariant for $S$, and $T$ is the restriction of $S$. We call $S$ a normal extension of $T$.

## (a)

Show that the unilateral shift is a non-normal subnormal operator.

### Solution

The unilateral shift is a non-normal subnormal operator vis the bilateral shift on a Hilbert Space $K$ with orthonormal basis $\{e_n:n\in\mathbb{Z}\}$ where $H$ is the subspace with orthonormal basis $\{e_n:n\in\mathbb{N}\}$.

## (b)

Show that if $T$ is subnormal, then $T^*T\geq TT^*$.

## (c)

A normal extension $S\in B(K)$ of a subnormal operator $T\in B(H)$ is a minimal normal extension if the only closed vector subspace of $K$ reducing $S$ and containing $H$ is $K$ itself. Show that $T$ admits a minimal normal extension. In the case that $S$ is a minimal normal extension, show that $K$ is the closed linear span of all $S^{*n}x$ ($n\in\mathbb{N}$$x\in H$).

## (d)

Show that if $S_1\in B(K_1)$ and $S_2\in B(K_2)$ are minimal normal extensions of $T$, then there exists a unitary operator $R:K_1\rightarrow K_2$ such that $S_2=RS_1R^*$ (so that there is obly one minimal normal extension).