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


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


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.


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