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.

(a)

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\|<K_n\|x_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\|<K_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).

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