Functional Analysis/C*-algebras

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 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 Von Neumann bicommutant theorem)