# Functional Analysis/C*-algebras

 Functional Analysis Chapter 5: C^*-algebras
 (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)