Functional Analysis/C*-algebras

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Functional Analysis
Chapter 5: C^*-algebras
0% developed  as of May 27, 2008 (May 27, 2008). Beware this chapter is less than a draft.

A Banach space \mathcal{A} over \mathbf{C} is called a Banach algebra if it is an algebra and satisfies

\|x y\| \le \|x\| \|y\|.

We shall assume that every Banach algebra has the unit 1 unless stated otherwise.

Since \|x_n y_n - xy\| \le \|x_n\| \|y_n - y\| + \|x_n  - x\| \|y\| \to 0 as x_n, y_n \to 0, the map

(x, y) \mapsto xy : \mathcal{A} \times \mathcal{A} \to \mathcal{A}

is continuous.

For x \in \mathcal{A}, let \sigma(x) be the set of all complex numbers \lambda such that x - \lambda 1 is not invertible.

5 Theorem For every x \in \mathcal{A}, \sigma(x) is nonempty and closed and

\sigma(x) \subset \{ s \in \mathbf{C} | |s| \le \|x\| \}.

Moreover,

r(x) \overset{\mathrm{def}}= \sup \{ |z| | z \in \sigma(x) \} = \lim_{n \to \infty} \|x^n\|^{1/n}

(r(x) is called the spectral radius of x)
Proof: Let G \subset \mathcal{A} be the group of units. Define f: \mathbf{C} \to A by f(\lambda) = \lambda 1 - x. (Throughout the proof x is fixed.) If \lambda \in \mathbf{C} \backslash \sigma(x), then, by definition, f(\lambda) \in G or \lambda \in f^{-1}(G). Similarly, we have: G \subset f(\sigma(x)). Thus, x \in f^{-1}(G) \subset \sigma(x). Since f is clearly continuous, \mathbf{C} \backslash \sigma(x) is open and so \sigma(x) is closed. Suppose that |s| > \|x\| for s \in \mathbf{C}. By the geometric series (which is valid by Theorem 2.something), we have:

\left(1 - {x \over s}\right)^{-1} = \sum_{n=0}^\infty \left({x \over s}\right)^n

Thus, 1 - {x \over s} is invertible, which is to say, s1 - x is invertible. Hence, s \not\in \sigma(x). This complete the proof of the first assertion and gives:

r(x) \le \|x\|

Since \sigma(x) is compact, there is a a \in \sigma(x) such that r(x) =  a. Since a^n \in \sigma(x^n) (use induction to see this),

r(x)^n \le \|x^n\|

Next, we claim that the sequence {x^n \over s^{n+1}} is bounded for |s| > r(x). In view of the uniform boundedness principle, it suffices to show that g \left( {x^n \over s^{n+1}} \right) is bounded for every g \in A^*. But since

\lim_{n \to \infty} g\left( {x^n \over s^{n+1}} \right) = 0,

this is in fact the case. Hence, there is a constant c such that \|x^n\| \le c|s|^{n+1} for every n. It follows:

r(x) \le \lim_{n \to \infty} \|x^n\|^{1/n} \le \lim_{n \to \infty} c^{1/n} |s| = |s|.

Taking inf over |s| > r(x) completes the proof of the spectral radius formula. Finally, suppose, on the contrary, that \sigma(x) is empty. Then for every g \in A^*, the map

s \mapsto g( (x - s)^{-1} )

is analytic in \mathbf{C}. Since \lim_{s \to \infty} g( (x - s)^{-1} ) = 0, by Liouville's theorem, we must have: g ( (x - s)^{-1} ) = 0. Hence, (x - s)^{-1} = 0 for every s \in \mathbf{C}, a contradiction. \square

5 Corollary (Gelfand-Mazur theorem) If every nonzero element of \mathcal{A} is invertible, then \mathcal{A} is isomorphic to \mathbf{C}.
Proof: Let x \in \mathcal{A} be a nonzero element. Since \sigma(x) is non-empty, we can then find \lambda \in \mathbf{C} such that \lambda 1 - x is not invertible. But, by hypothesis, \lambda 1- x is invertible, unless \lambda 1 = x.\square

Let \mathfrak{m} be a maximal ideal of a Banach algebra. (Such \mathfrak{m} exists by the usual argument involving Zorn's Lemma in abstract algebra). Since the complement of \mathfrak{m} consists of invertible elements, \mathfrak{m} is closed. In particular, \mathcal{A} / \mathfrak{m} is a Banach algebra with the usual quotient norm. By the above corollary, we thus have the isomorphism:

\mathcal{A} / \mathfrak{m} \simeq \mathbf{C}

Much more is true, actually. Let \Delta(\mathcal{A}) be the set of all nonzero homeomorphism \omega: \mathcal{A} \to \mathbf{C}. (The members of \Delta(\mathcal{A}) are called characters.)

5 Theorem \Delta(\mathcal{A}) is bijective to the set of all maximal ideals of \mathcal{A}.

5 Lemma Let x \in \mathcal{A}. Then x is invertible if and only if \omega(x) \ne 0 for every \omega \in \Delta(\mathcal{A})

5 Theorem \omega(x) = \omega(\hat x)

An involution is an anti-linear map x \to x^*: A \to A such that x^{**} = x. Prototypical examples are the complex conjugation of functions and the operation of taking the adjoint of a linear operator. These examples explain why we require an involution to be anti-linear.

Now, the interest of study in this chapter. A Banach algebra with an involution is called a C*-algebra if it satisfies

\|xx^*\| = \|x\|^2 (C*-identity)

From the C*-identity follows

\|x^*\| = \|x\|,

for \|x\|^2 \le \|x^*\| \|x\| and the same for x^* in place of x. In particular, \|1\| = 1 (if 1 exists). Furthermore, the C^*-identity is equivalent to the condition: \|x\|^2 \le \|x^*x\|, for this and

\|x^*x\| \le \|x^*\| \|x\| implies \|x\| = \|x^*\| and so \|x\|^2 \le \|x^* x\| \le \|x\|^2.

For each x \in \mathcal{A}, let C^*(x) be the linear span of \{ 1, y_1 y_2 ... y_n | y_j \in \{ x, x^* \} \} . In other words, C^*(x) is the smallest C*-algebra that contains x. The crucial fact is that C^*(x) is commutative. Moreover,

Theorem Let x \in \mathcal{A} be normal. Then \sigma_A (x) = \sigma_{C^*(x)} (x)

A state on C^*-algebra \mathcal{A} is a positive linear functional f such that \|f\| = 1 (or equivalently f(1) = 1). Since S is convex and closed, S is weak-* closed. (This is Theorem 4.something.) Since S is contained in the unit ball of the dual of \mathcal{A}, S is weak-* compact.


5 Theorem Every C^*-algebra \mathcal{A} is *-isomorphic to C_0(X) where X is the spectrum of \mathcal{A}.

5 Theorem If C_0(X) is isomorphic to C_0(Y), then it follows that X and Y are homeomorphic.

3 Lemma Let T be a continuous linear operator on a Hilbert space \mathcal{H}. Then TT^* = T^*T if and only if \|Tx\| = \|T^*x\| for all x \in \mathcal{H}.

Continuous linear operators with the above equivalent conditions are said to be normal. For example, an orthogonal projection is normal. See w:normal operator for additional examples and the proof of the above lemma.

3 Lemma Let N be a normal operator. If \alpha and \beta are distinct eigenvalues of N, then the respective eigenspaces of \alpha and \beta are orthogonal to each other.
Proof: Let I be the identity operator, and x, y be arbitrary eigenvectors for \alpha, \beta, respectively. Since the adjoint of \alpha I is \bar \alpha I, we have:

0 = \|(N - \alpha I)x\| = \|(N - \alpha I)^*x\| = \|N^*x - \bar \alpha x\|.

That is, N^*x = \bar \alpha x, and we thus have:

\bar \alpha \langle x, y \rangle = \langle N^*x, y \rangle = \langle x, Ny \rangle = \bar \beta \langle x, y \rangle

If \langle x, y \rangle is nonzero, we must have \alpha = \beta. \square

5 Exercise Let \mathcal{H} be a Hilbert space with orthogonal basis e_1, e_2, ..., and x_n be a sequence with \|x_n\| \le K. Prove that there is a subsequence of x_n that converges weakly to some x and that \|x\| \le K. (Hint: Since \langle x_n, e_k \rangle is bounded, by Cantor's diagonal argument, we can find a sequence x_{n_k} such that \langle x_{n_k}, e_k \rangle is convergent for every k.)

5 Theorem (Von Neumann double commutant theorem) M is equal to its double commutant if and only if it is closed in either weak-operator topology or strong-operator topology.
Proof: (see w:Von Neumann bicommutant theorem)

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