6.3 Gelfand Representation of Abelian Banach Algebras

Proof. Suppose $\norm{e-x}{}<1$ for some $x\in\frak{m}$ then the power series $y=\sums{\infty}{n=1}(e-x)^{n}$. Since $(e-x)y=\sums{\infty}{n=2}(e-x)^{n}$, we have

hence $e=y-ey+xy+x=-(ey-y)+xy+x$ thus $e\in \frak{m}$. Lastly $\fall a\in\A$, $ea-a\in\frak{m}$ so $a\in\frak{m}$ contradicting $\frak{m}$ is proper.$\square$

Proof. Let $e$ be an identity modulo $\frak{m}$ then by Proposition 6.3.2, $e\nin \cl{\frak{m}}$. $\cl{\frak{m}}$ is regular since $\frak{m}\suf\cl{\frak{m}}$. Lastly $\cl{\frak{m}}$ is an ideal since since multiplication and addtion are continuous in a Banach algebra.$\square$

Proof. Let $e$ be an identity modulo a proper ideal $\frak{m}$ then any ideal containing $\frak{m}$ is regular. The family $\cali{F}$ of all ideals of $A$ containing $\frak{m}$ but not $e$ is an inductive set under inclusion ordering, i.e. every chain $\cali{C}\suf \cali{F}$, the union $\cups{}{A\in\cali{C}}A$ is an upper bound of $\cali{C}$ and belongs to $\cali{F}$. Hence $\cali{F}$ has a maximal element of Zorn’s lemma.$\square$

Proof. $A/\frak{m}$ is a vector space under the standard quotient construction and multiplication is well-defined since $\frak{m}$ is a 2-sided ideal. Since $\frak{m}$ is closed, $A/\frak{m}$ is a Banach space. It remains to show that $A/\frak{m}$ is a Banach algebra.
Let $\pi:A\to A/\frak{m}$ be the quotient map. The norm in $A/\frak{m}$ is given by $\norm{\pi(x)}{}=\INF{m\in\frak{m}}\norm{x+m}{}$. Suppose $\e>0$ then $\exists m,n\in\frak{m}$ satisfying

Thus

Since $\e$ is arbitrary we are done.$\square$

Proof. By assumption, $1\nin Ax$ hence it is a proper ideal. Since $A$ is unital, any ideal is regular so by Proposition 6.3.3 is contained in some maximal ideal $\frak{m}$.$\square$

Proof. Since $\frak{m}$ is maximal regular ideal, Proposition 6.3.3 and Proposition 6.3.5 imply the quotient algebra $A/\frak{m}$ is a unital Banach algebra with no nontrivial proper ideal. Hence by Proposition 6.3.6, it follows that $A/\frak{m}$ is a field. Therefore by Theorem GELFAND MAZUR, $\exists\p:A/\frak{m}\iso \com$ and if $\psi:A/\frak{m}\iso \com$ then $\psi\circ\p\inv$ is a linear automorphism of $\com$ hence the identity thus $\p$ is unique. We extend this to a surjective homomorphism

where $\pi$ is the quotient map $A\to A/\frak{m}$. Since $\p$ is an isomorphism $\om_{\frak{m}}\inv(0)=\pi\inv(0)=\frak{m}$$\square$

Proof. By Lemma 6.3.7 each $\frak{m}\in\cali{M}(A)$ give rise to an $\om_{\frak{m}}\in\OM(A)$ such that $\frak{m}=\om_{\frak{\om}}\inv(0)$. If $\om\in\OM(A)$ then $\frak{m}_{\om}\define \om\inv(0)$ is a maximal ideal of $A$ since $A/\om\inv(0)\iso \com$ which has no nontrivial ideal and regular since $1\in\com\suf \om(A)$. Furthermore, isomorphism $A/\om\inv(0)\iso\com$ is unique; hence $\om=\om_{\frak{m}_\om}$$\square$

Proof. If $\om\in\OM(A)$ then the kernel $\om\inv(0)$ is closed by Proposition 6.3.4. Recall any linear functional on a normed space is continuous if and only if its kernel is closed so $\om$ is continuous hence bounded. Consequently for each $x\in A$

hence $\abs{\om(x)}\leq \limit{n\to\infty}\norm{\om}{}^{1/n}\norm{x^n}{}^{1/n}=\limit{n\to\infty}\norm{x^n}{}^{1/n}=\norm{x}{sp}\leq \norm{x}{}$$\square$

Proof. Let $\OM'(A)=\OM(A)\cup\curl{0}$ then $\OM'(A)\suf S^{*}$ by Proposition 6.3.9. For each $x,y\in A$, the evaluation

is $\s(A^{*},A)$-continuous thus

which is closed. Thus $\OM'(A)$ is compact by Alaoglu’s Theorem. Since $\curl{0}$ is closed in $\OM'(A)$, $\OM(A)$ is an open subset of the compact space $\OM'(A)$ so it is locally compact. Suppose $A$ is unital then $0$ is isolated in $\OM'(A)$ because $\om(1)=1$ for every $\om\in \OM(A)$ so $\OM(A)$ is closed in $\OM'(A)$; hence compact.$\square$

Proof. Since $\ha{x}:\OM(A)\to \com$ is defined by $\ha{x}(\om)=\om(x)$, $\ha{x}$ is $\s(A^{*},A)$-continuous on $\OM(A)$ so

is closed hence compact and we conclude $\ha{x}\in C_{\infty}(\OM(A))$. By definition $\scr{F}$ is linear and multiplicative, hence a homomorphism.
If $A$ is unital then $\frak{m}\define \curl{x-\lam\in A:\lam\in \spec{A}{x}}$ is a maximal regular ideal of $A$ hence

Conversely

Since the quasi spectrum always contain $0$, the assertion for nonunital $A$ follows from above. Thus we conclude

$\square$