# C*-Algebras

Definition 1.  A Banach algebra is a complex algebra which is a Banach space and satisfies subnormality:

$\|xy\|\leq\|x\|\|y\|$.

A *-algebra is a complex algebra with a unary operation * that satisfies $(x+y)^*=x^*+y^*$, $(xy)^*=y^*x^*$, $x^{**}=x$, and $(\lambda x)^*=\overline{\lambda}x^*$.  A Banach *-algebra is a Banach algebra and a *-algebra that satisfies $\|x^*\|=\|x\|$.  A C*-algebra is a Banach *-algebra that satisfies $\|x^*x\|=\|x\|^2$.  A map between two *-algebras is a *-homomorphism if $f(x^*)=f(x)^*$.

Proposition 2.  Any nonunital Banach algebra $A$ has a unitization $A^\dagger$.

Proof.  Let $A^\dagger=A\oplus\mathbb{C}$.  Let $a\in A$, and send $a\mapsto (a,0)$.  Define $(a,\lambda)+(b,\mu)=(a+b,\lambda+\mu)$, $(a,\lambda)(b,\mu)=(ab+\lambda b+\mu a,\lambda\mu)$, and $\|(a,\lambda)\|=\|a\|+|\lambda|$.  Clearly $A^\dagger$ is a Banach space.  We also have

$\begin{array}{lcl}\|(a,\lambda)(b,\mu)\|&=&\|(ab+\lambda b+\mu a,\lambda\mu)\|\\&=&\|ab+\lambda b+\mu a\|+|\lambda\mu|\\&\leq&\|ab\|+\|\lambda b\|+\|\mu a\|+|\lambda||\mu|\\&\leq&\|a\|\|b\|+|\mu|\|a\|+|\lambda|\|b\|+|\lambda||\mu|\\&=&\|(a,\lambda)\|\|(b,\mu)\|.\end{array}$

The unit is $(0,1)$: $(a,\lambda)(0,1)=(a,\lambda)=(0,1)(a,\lambda)$.

In fact, if $A$ is a nonunital Banach *-algebra, then there is a unitization of $A$ which is also a Banach *-algebra.  We simply use $A^\dagger$ and define $(a,\lambda)^*=(a^*,\overline{\lambda})$.  And if $A$ is a C*-algebra, then we modify the norm on $A^\dagger$ to $\|(a,\lambda)\|=\sup\{\|ab+\lambda b\|:\|b\|=1\}$.

Definition 3.  Let $A$ be a Banach algebra and $x\in A$.  The spectrum of $x$ is the set

$\sigma_A(x)=\{\lambda:x-\lambda\cdot 1\mbox{~is not invertible in }A^\dagger\}$.

Clearly for nonunital $A$, $0\in\sigma_A(x)$ for all $x\in A$.

Lemma 4.  Let $A$ be a unital Banach algebra.  If $\|1-x\|<1$, then $x^{-1}$ exists and is defined by

$\displaystyle x^{-1}=1+\sum_{n=1}^\infty(1-x)^n$.

Proof.  Denote $x$ by $1-(1-x)$.  Then

$\displaystyle\begin{array}{lcl}xx^{-1}&=&\left(1-(1-x)\right)\left(1+\sum_{n=1}^\infty(1-x)^n\right)\\&=&1+(1-x)+(1-x)^2+\cdots-(1-x)-(1-x)^2-\cdots.\end{array}$

Since $\|1-x\|<1$, the series is absolutely convergent and hence convergent.  So the two sums cancel–leaving $1$.

Corollary 5.  If $|\lambda|>\|x\|$, then $\lambda\cdot 1-x$ is a unit.

Proposition 6.  Let $A$ be a unital Banach algebra and $x,y\in A$.

1. $\sigma_A(x)$ is a nonempty compact subset of $\mathbb{C}$.
2. $\rho(x)=\max_{\lambda\in\sigma_A(x)}|\lambda|=\lim_{n\to\infty}\|x^n\|^{1/n}=\inf_{n\in\mathbb{N}} \|x^n\|^{1/n}$.

Proof.  (1) By Heine-Borel, it suffices to show $\sigma_A(x)$ is closed and bounded for all $x\in A$.  Let $U$ denote the set of units of $A$ and a scalar $\lambda$ be called a regular point of $x$ if $\lambda\notin\sigma_A(x)$.  Let $\lambda_0$ be a regular point.  Then

$x_{\lambda_0}=x-\lambda_0\cdot 1\in U$.

Since $U$ is open, there is a neighborhood $B_\varepsilon(x_{\lambda_0})\subset U$.  Moreover, since the function $f:\mathbb{C}\to A$ defined by $f(\lambda)=x-\lambda\cdot 1$ is continuous, $f^{-1}(\lambda_0)$ is open in $\mathbb{C}$.  Thus the set of regular points is open–implying that the spectrum is closed.

Now if $|\lambda|>\|x\|$, then $\lambda$ is a regular point by the previous claim.  Hence $diam(\sigma_A(x))\leq 2\|x\|$.

Nonemptiness and (2) are omitted.

Corollary 7 (Gelfand-Mazur). If $A$ is a Banach division algebra, then $A=\mathbb{C}$.

Proposition 8.  Let $A$ be a C*-algebra and $x$ be self-adjoint ($x=x^*$).  Then $\rho(x)=\|x\|$.

Proof.  From the C*-axiom we have $\|x^*x\|=\|x^2\|=\|x\|^2$.  Iterating this yields $\|x^{2^n}\|=\|x\|^{2^n}$.  Hence by (2) of Proposition 6 and the C*-axiom,

$\displaystyle \rho(x)=\lim_{n\to\infty}\|x^{2^n}\|^{2^{-n}}=\|x\|$.

Since $x^*x$ is self-adjoint for all $x\in A$, this says that the norm of a C*-algebra is uniquely determined:  $\|x\|=\sqrt{\rho(x^*x)}$.

Theorem 9 (Gelfand-Naimark).  Let $A$ be a commutative unital C*-algebra and $\hat{A}$ be a closed subset of the unit ball of the dual space $A^*$.  Then the map $\varphi:A\to C_0(\hat{A})$ defined by $\varphi(x)(f)=f(x)$, called the Gelfand transform, is an isometric *-isomorphism.

[1] Bachman, George and Lawrence Narici.  Functional Analysis.  Dover Publications. 2000.

[2] Blackadar, Bruce.  Operator Algebras.  Encyclopedia of Mathematical Sciences.  Vol. 122.  Springer-Verlag.  2006.