6.1 Banach Algebras

Definition 6.1.1 (Banach Algebra).label Let $A$ be a Banach space over $\com$. If $A$ is an algebra over $\com$ (a vector space equipped with a bilinear multiplication) in which the multiplication satisfies the inequality

\begin{align*}\norm{xy}{}\leq \norm{x}{}\norm{y}{}\end{align*}

then $A$ is called a Banach algebra. The inequality

\begin{align*}\norm{x_1y_1-x_2y_2}{}\leq \norm{x_1}{}\norm{y_1-y_2}{}+\norm{x_1-x_2}{}\norm{y_2}{}\end{align*}

shows that the product $xy$ is a continuous function of two variables $x$ and $y$. If $E$ is a Banach space over $\com$, then the set $\cali{L}(E)$ of all bounded operators on $E$ is a Banach algebra with the natural algebraic operations and norm.

Definition 6.1.2 ($C^{*}$-algebra).label An involution on an algebra $A$ is an anti-automorphism of $A$ of order 2, i.e., a map $x\mto x^{*}$ from $A$ to $A$ satisfying

(1)
$x^{**}=x$
(2)
$(x+y)^{*}=x^{*}+y^{*}$
(3)
$(\al x)^{*}=\cl{\al}x^{*}$
(4)
$(xy)^{*}=y^{*}x^{*}$
(5)
$\norm{x^*}{}=\norm{x}{}$ Takesaki require this but not Folland

for every $x,y\in A,\al\in\com$ then $A$ is called an involutive Banach algebra (or $*$-algebra) and the map $x\mto x^{*}$ the involution (or $*$-operation). If the involution of $A$ satisfies the following additional condition

(1)
$\norm{x^*x}{}=\norm{x^*}{}\norm{x}{},x\in A$

then $A$ is called a $C^{*}$-algebra. If $A$ and $B$ are Banach algebras, a (Banach algebra) homomorphism from $A$ to $B$ is a bounded linear map $\phi:A\to B$ such that $\phi(xy)=\phi(x)\phi(y)$ for all $x,y\in A$. If $A$ and $B$ are $*$-algebras, a $*$-homomorphism from $A$ to $B$ is a homomorphism $\phi$ such that $\phi(x^{*})=\phi(x)^{*}$ for all $x\in A$. If $S$ is a subset of the Banach algebra $A$, we say that $A$ is generated by $S$ if the linear combination of products of elements of $S$ are dense in $A$.

Definition 6.1.3.label Let $\OM$ be a locally compact space. The set $C_{\infty}(\OM)$ of all continuous functions on $\OM$ vanishing at infinity is a $C^{*}$-algebra the following structure

\begin{gather*}(\lam x+\mu y)(\om)=\lam x(\om)+\mu \\ (xy)(\om)=x(\om)y(\om)\\ x^{*}(\om)=\cl{x(\om)}\\ \norm{x}{}=\sup\curl{\abs{x(\om)}:\om\in \OM}\end{gather*}

for every $x,y\in C_{\infty}(\OM)$, $\lam,\mu\in \com$ and $\om\in \OM$. The $C^{*}$-algebra $C_{\infty}(\OM)$ is abelian. The algebra $C_{\infty}(\OM)$ has an identity if and only if $\OM$ is compact and we shall denote $C_{\infty}(\OM)=C(\OM)$.

If $\hi$ is a Hilbert space, then the Banach algebra $\LIN{\hi}$ of all bounded operators on $\hi$ is a $C^{*}$-algebra with the involution $x\mto x^{*}$ defined as the adjoint operator $x^{*}$ of $x$. If the dimension of $\hi$ is greater than one, then $\LINH$ is not abelian.

Proposition 6.1.4.label If $A$ is a Banach algebra with an identity $1$ then there exists a norm $\norm{\cd}{0}$ on $A$ such that

(1)
$\norm{\cd}{0}$ is equivalent to $\norm{\cd}{}$
(2)
$(A,\norm{\cd}{0})$ is a Banach algebra
(3)
$\norm{1}{0}=1$

Proof. For each $x\in A$, let $L_{x}$ denote the operator $y\in A\mto xy\in A$. The map $x\mto L_{x}$ is then injective since $L_{x}1=x$. Put $\norm{x}{0}=\norm{L_x}{}$, $x\in A$. By the inequality $\norm{xy}{}\leq \norm{x}{}\norm{y}{}$, we have $\norm{x}{0}\leq \norm{x}{}$. OTOH, we have

\begin{align*}\norm{x}{0}=\norm{L_x}{}=\sup\curl{\norm{xy}{}:\norm{y}{}\leq 1}\geq\frac{\norm{x}{}}{\norm{1}{}}\end{align*}

Hence the norm $\norm{\cd}{0}$ is equivalent to the original norm. The last 2 assertions are automatic.$\square$

Definition 6.1.5 (Unital Banach Algebra).label A Banach algebra with an identity is said to be unital and by Proposition 6.1.4 we assume that the norm of the identity is one if it exits.

Remark 6.1.6.label If $A$ is a unital involutive Banach algebra then $1^{*}=1$. Furthermore, if $A$ is a unital $C^{*}$-algebra, then the condition $\norm{1}{}=1$ follows form $C^{*}$-algebra postulate.

Lemma 6.1.7.label Let $A$ be a non-unital $*$-algebra then there exists an embedding of $A$ into a unital $*$-algebra $A_{I}$ as an ideal.

Proof. Let $A_{I}\define A\op \com$ as a linear space with the Banach algebra structure

\begin{gather*}(x,\lam)(y,\mu)=(xy+\mu x+\lam y,\lam\mu)\\ (x,\lam)^{*}=(x^{*},\cl{\lam})\\ \norm{(x,\lam)}{}=\norm{x}{}+\abs{\lam}\end{gather*}

for every $(x,\lam),(y,\mu)\in A\op \com$. The map $x\in A\mto (x,0)\in A_{I}$ is an isometric isomorphism and the element $(0,1)$ is the identity of $A_{I}$. Identifying each $x\in A$ and $(x,0)\in A_{I}$, we write $(x,\lam)=x+\lam 1\in A_{I}$. Under the identification, $A$ is an ideal of $A_{I}$.$\square$

Remark 6.1.8.label One may arrive at the Banach algebra structure of $A_{I}$ by requiring

\begin{align*}(x,0)(y,0)=(xy,0)&&(0,\lam)(0,\mu)=(0,\lam\mu)&&(x,0)(0,\mu)=(\mu x,0)&&(0,\lam)(y,0)=(\lam y,0)\end{align*}

Equivalently we may require $(0,1)\in A\op \com$ to be the identity and multiplication is bilinear since

\begin{align*}(0,\lam)(y,0)=\lam(0,1)(y,0)=\lam(y,0)=(\lam y,0)\end{align*}

Lastly we remark that

\begin{align*}(0,\mu)=(0,\mu)+(0,0)\implies (0,\mu)(x,\lam)=(\mu x,\mu\lam)\end{align*}

Definition 6.1.9 (Codimension).label Let $W\suf V$ be vector spaces then the codimension of $W$ in $V$ is $\text{codim}_{V}(W)\define\dim(V/W)$ which is equal to $\text{codim}_{V}(W)=\dim(V)-\dim(W)$ if $W$ is finite dimensional.

Lemma 6.1.10.label Every finite dimensional normed space is complete.

Lemma 6.1.11.label Let $X$ be a normed space and $Y$ a closed subspace. Then $X$ is complete if and only if $Y$ and the quotient $X/Y$ are complete

Lemma 6.1.12.label Let $X$ be a normed space and $Y\suf X$ is a closed subspace of finite codimension then

$X$ is complete $\iff$ $Y$ is complete

In particular if $X=Y\op Z$ then $X/Y\iso Z$ canonically and

$X$ is complete $\iff Y$ is complete and $Z$ is finite dimensional

Proof. Since $Y$ is closed the quotient $X/Y$ is a normed space with $\norm{x+Y}{}=\INF{x\in Y}\norm{x-y}{}$ so

\begin{align*}0=\norm{x+Y}{}=\INF{x\in Y}\norm{x-y}{}\iff x\in Y\iff x+Y=Y\end{align*}

$\dim(X/Y)=\infty$ hence $X/Y$ is complete by Lemma 6.1.10. We have the first assertion by Lemma 6.1.11. Since $X$ decomposes as a direct sum $X/Y\iso Z$ canonically hence we are done by considering $Z$ finite dimensional.$\square$

Proposition 6.1.13.label Using the same notation as in Lemma 6.1.7, if a $C^{*}$-algebra $A$ is not unital, then there exists a norm in $A_{I}$ which makes $A_{I}$ a $C^{*}$-algebra.

Proof. Since $A$ is an ideal of $A_{I}$, we put $L_{x}y=x$ for $x\in A_{I}$ and $y\in A$. By Proposition 6.1.4, we have $\norm{x}{}=\norm{L_x}{}$ for each $x\in A$ (by taking an equivalent norm). Suppose $\norm{L_x}{}=0$ for some $x=x'+\lam 1$, $x'\in A,\lam \in\com,\lam\neq 0$. For every $y\in A$, we have

\begin{align*}0=\frac{1}{\lam}xy=\frac{1}{\lam}x'y+y\end{align*}

so $-\frac{1}{\lam}x'$ is the left identity of $A$; hence $\parens{-\frac{1}{\lam}x'}^{*}$ is a right identity of $A$. Therefore

\begin{align*}-\frac{1}{\lam}x'=\parens{-\frac{1}{\lam}x'}\parens{-\frac{1}{\lam}x'}^{*}=\parens{-\frac{1}{\lam}x'}^{*}\end{align*}

which means that $-\frac{1}{\lam}x'$ must be the identity of $A$, contradicing that $A$ is non-unital. Thus the norm from Proposition 6.1.4 is also a norm for $A_{I}$. Since $A$ is complete and of codimension $1$ in $A_{I}$, by Lemma 6.1.12 $A_{I}$ is complete. For any $x\in A_{I}$ and $\e>0$, there exists $y\in A$ with

\begin{align*}\norm{xy}{}\geq (1-\e)\norm{x}{}&&\norm{y}{}\leq 1\end{align*}

But $xy\in A$ thus

\begin{align*}\norm{x^*x}{}\geq\norm{y^*(x^*x)y}{}=\norm{(xy)^*(xy)}{}=\norm{xy}{}^{2}\geq (1-\e)^{2}\norm{x}{}^{2}\end{align*}

Therefore

\begin{align*}\norm{x}{}^{2}\leq\norm{x^*x}{}\leq\norm{x^*}{}\norm{x}{}=\norm{x}{}^{2}\end{align*}

so $\norm{x^*x}{}=\norm{x^2}{}$$\square$

Remark 6.1.14.label If $A$ is a non-unital $C^{*}$-algebra then denote $A_{I}$ by the unital $C^{*}$-algebra above

Definition 6.1.15 (Invertible).label Let $A$ be a unital Banach algebra, $x\in A$ is invertible or regular if $\exists x\inv$ with $xx\inv =x\inv x=1$. The inverse $x\inv$ of an invertible element if unique. Denote $G(A)\define\curl{x\in A:x\text{ is regular}}$ called the general linear group of A

Proposition 6.1.16.label If $x\in A$ such that $\norm{x-1}{}<1$, then $x$ is invertible and

\begin{align*}x\inv=\sums{\infty}{n=0}(1-x)^{n}\end{align*}

where $a^{0}=1$ for $a\in A$ (here $a=1-x$)

Proof. Since $\abs{x-1}<1$, $\sums{\infty}{n=0}\norm{1-x}{}^{n}<\infty$, the series $\sums{\infty}{n=0}(1-x)^{n}$ converges in norm. Set $x'=\sums{\infty}{n=0}(1-x)^{n}$ then

\begin{align*}x'x&=xx'=(1-(1-x))x'=x'-(1-x)x'\\&=\sums{\infty}{n=0}(1-x)^{n}-\sums{\infty}{n=1}(1-x)^{n}\\&=1\end{align*}

$\square$

Proposition 6.1.17 (General Linear Group is Open).label The group $G(A)$ is an open subset of $A$. More precisely for $x_{0}\in G(A)$ then $B\parens{x_0,\frac{1}{\norm{x_0\inv}{}}}\suf G(A)$ and for each $x\in B\parens{x_0,\frac{1}{\norm{x_0\inv}{}}}$

\begin{align*}x\inv=\parens{\sums{\infty}{n=0}\braks{x_0\inv(x_0-x)}^n}x_{0}\inv\end{align*}

Proof. Let $x_{0}\in G(A)$ and $\norm{x-x_0}{}<\frac{1}{\norm{x_0\inv}{}}$ then

\begin{align*}\norm{1-x_0\inv x}{}=\norm{x_0\inv(x_0-x)}{}\leq \norm{x_0\inv}{}\norm{x_0-x}{}<1\end{align*}

so that $x_{0}\inv x$ has the inverse written by

\begin{align*}(x_{0}\inv x)\inv=\sums{\infty}{n=0}(1-x_{0}\inv x)^{n}=\sums{\infty}{n=0}\braks{x_0\inv(x_0-x)}^{n}\end{align*}

Lastly $x=x_{0}(x_{0}\inv x)$ thus

\begin{align*}x\inv=\sums{\infty}{n=0}\braks{x_0\inv(x_0-x)}^{n}x_{0}\inv=x_{0}\inv+\sums{\infty}{n=1}\braks{x_0\inv(x_0-x)}^{n}x_{0}\inv\end{align*}

$\square$

Lemma 6.1.18 (Continuity of inverse).label The map $x\mto x\inv$ is continuous on $G(A)$

Proof. Using notations as in Proposition 6.1.17 we have

\begin{align*}\norm{x\inv-x_0\inv}{}&=\norm{\sums{\infty}{n=1}\braks{x_0\inv(x_0-x)}^nx_0\inv}{}\\&\leq \norm{x_0\inv}{}\sums{\infty}{n=1}\norm{x_0\inv}{}^{n}\norm{x_0-x}{}^{n}\\&=\norm{x_0\inv}{}^{2}\norm{x_0-x}{}\parens{1-\norm{x_0\inv}{}\norm{x-x_0}{}}\inv\end{align*}

$\square$

The Eigenspace

6.1 Banach Algebras

Direct References

The Eigenspace

Direct References