Before we start here some notations to have no confusion:

Suppose $A$ is a commutative $C^*$-algebra with unit. $\Sigma(A)$ is the Gelfand spectrum, given by all linear maps $\omega:A\rightarrow\Bbb{C}$ such that $\omega(ab)=\omega(a)\omega(b)$. Also define the Gelfand transform with maps each $a\in A$ to a function $\hat{a}:\Sigma(A)\rightarrow\Bbb{C}$ by $\hat{a}(\omega)=\omega(a)$ ($a\in A$, $\omega\in\Sigma(A)$). We can use the fact that $\left\|\hat{a}\right\|_{\infty}=\left\|a\right\|_{A}$. Now i want to prove the following:

  1. $\left\|\omega\right\|=1$ for all $\omega\in\Sigma(A)$ (Hints: Prove that $\left\|\omega\right\|=1$ is equivalent with $\left\|\omega\right\|\leq1$ and $\omega(1)=1$, show that the negation of $\left\|\omega\right\|\leq1$ is the claim that there exists $a\in A$ with $\left\|a\right\|\leq1$ and $|\omega(a)|=1$. For $\left\|a\right\|<1$ define the von-Neumann series $b=\sum_{n=0}^{\infty}{a^n}$ and show $1-ab=b$ then conclude a contradiction to conclude the result we want to have)
  2. Prove that the Gefland transform is surjective (use that the transform is isometric (i note this). Show that the image of the transform is closed in $C(\Sigma(A))$ and use Stone-Weierstrass-Theorem)

For the first part i can prove $\left\|\omega\right\|\leq1$ and $\omega(1)=1$ implies $\left\|\omega\right\|=1$ but then i can not continued. Can i get hints/solutions?

For the second part i can't imagine how to start with the prove -.-

Thank you for help :)

  • $\begingroup$ I think the difficult proof is the proof of the second part (want how to show that the image is closed?) $\endgroup$
    – user62160
    Feb 13 '13 at 14:50

This is a very long post made for self education reasons and future references when questions about commutative Gelfand's representation theorem will be discussed.

0. Spectral theory and holomorphic calculus for Banach algebras

Results of this section are presented without proofs.

Theorem 0.1 Let $A$ be a unital Banach algebra and $a\in A$, then $\sigma(a)$ is non-empty compact subset of $\mathbb{C}$

Theorem 0.2 Let $A$ be unital involutive Banach algebra and $a\in A$, then there exist unique continuous involutive unital homorphism $\gamma_a:\mathcal{O}(\sigma(a))\to A$ such that $\gamma_a(z)=a$.

Notation 0.3 Homomorphism described above is called holomorphic functional calculus on Banach algebra of element $a$. For a given $f\in\mathcal{O}(\sigma(a))$ we use notation $f(a)$ instead of more rigorous $\gamma_a(f)$

Theorem 0.4. Let $A$ be unital Banach algebra and $a\in A$. Then for all $f\in\mathcal{O}(\sigma(a))$ holds $$ \sigma(f(a))=f(\sigma(a)) $$

Notation 0.5 Let $A$ be unital Banach algebra and $a\in A$. Then the spectrl radius of $a$ is called $$ r(a)=\sup\{|\lambda|:\lambda\in\sigma(a)\} $$

Theoreem 0.6 Let $A$ be unital Banach algebra and $a\in A$, then

  1. $r(a)\leq \Vert a\Vert$
  2. $r(a)=\lim\limits_{n\to\infty}\Vert a^n\Vert^{1/n}$

1. General properties of Banach algebras

Notation 1.1 Let $A$ be involutive unital Banach algebra, then the element $a\in A$ is called

  1. selfadjoint if $a=a^*$
  2. normal if $a^*a=aa^*$
  3. unitary if $a^*=a^{-1}$

Theorem 1.2 Let $A$ be an involutive Banach algebra, then for all $a\in A$ there exist two selfadjoint elements $b,c\in A$ such that $a=b+i c$.

Indeed define $b=\frac{1}{2}(a+a^*)$ and $c=\frac{1}{2i}(a-a^*)$. In this case $c^*=\frac{1}{-2i}(a^*-a)=c$ and $b^*=\frac{1}{2}(a^*+a)=b$, so $b,c$ are seladjoint. And it is easy to check that $a=b+ic$.

Theorem 1.3 Let $A$ be unital involutive algebra, then for all $a\in A$ we have

  1. $(a^*)^{-1}=(a^{-1})^*$
  2. $\sigma(a^*)=\overline{\sigma(a)}$
  3. $\sigma(a^{-1})=\sigma(a)^{-1}$

Let $b=a^{-1}$, then $ab=ba=1$. Now we apply involution to get $b^*a^*=a^*b^*=1^*=1$. Hence $(a^*){-1}=b^*=(a^{-1})^*$.

Let $\lambda\in\mathbb{C}\setminus\sigma(a^*)$, then $a^*-\lambda 1$ is invertible. By previous result $a-\overline{\lambda}1=(a^*-\lambda 1)^*$ is invertible, so $\lambda\in\mathbb{C}\setminus{\overline{\sigma(a)}}$. Since $\lambda$ is arbitrary $\mathbb{C}\setminus\sigma(a^*)\subset\mathbb{C}\setminus{\overline{\sigma(a)}}$ i.e. $\overline{\sigma(a)}\subset\sigma(a^*)$. Similar argument shows that $\sigma(a^*)\subset\overline{\sigma(a)}$. Hence $\sigma(a^*)=\overline{\sigma(a)}$.

Let $\lambda\in\mathbb{C}\setminus\sigma(a^{-1})$, then $a^{-1}-\lambda 1$ is invertible. Hence there exist $b\in A$ such that $(a^{-1}-\lambda 1)b=b(a^{-1}-\lambda 1)=1$, this means that $a^{-1}b-\lambda b=ba^{-1}-\lambda b=1$. Using this properties it is easy to check that $(a-\lambda^{-1} 1)(-\lambda a^{-1}b)=(-\lambda a^{-1}b)(a-\lambda^{-1} 1)=1$. Hence $a-\lambda^{-1} 1$ is invertible and we conclude $\lambda^{-1}\in\mathbb{C}\setminus\sigma(a)$, i.e. $\lambda\in\mathbb{C}\setminus\sigma(a)^{-1}$. Since $\lambda$ was arbitrary $\sigma(a)^{-1}\subset\sigma(a^{-1})$. Now let $\lambda\in\mathbb{C}\setminus\sigma(a)^{-1}$, then $a-\lambda^{-1} 1$ is invertible. Hence there exist $c\in A$ such that $(a-\lambda^{-1} 1)c=c(a-\lambda^{-1} 1)=1$, this means that $ac-\lambda^{-1} c=ca-\lambda^{-1} c=1$. Using this properties it is easy to check that $(a^{-1}-\lambda 1)(-\lambda^{-1} ac)=(-\lambda^{-1} ac)(a^{-1}-\lambda 1)=1$. Hence $a^{-1}-\lambda 1$ is invertible and we conclude $\lambda\in\mathbb{C}\setminus\sigma(a^{-1})$. Since $\lambda$ was arbitrary $\sigma(a^{-1})\subset\sigma(a)^{-1}$. As the result $\sigma(a^{-1})=\sigma(a)^{-1}$.

Theorem 1.4 Let $A$ be unital Banach algebra. If $a,b\in A$ are commute, then

  1. $e^a=\sum\limits_{n=0}^\infty\frac{1}{n!}a^k$
  2. $e^{a+b}=e^a e^b$
  3. $(e^{a})^{-1}=e^{-a}$

Since $\sigma(a)$ is compact, then the function $e^z$ is a limit of polynomials $p_N(z)=\sum\limits_{n=0}^N\frac{1}{n!}z^n$ in $\mathcal{O}(\sigma(a))$. From theorem 0.2 we get $$ e^{a}=\gamma_a(\exp)=\gamma_a\left(\lim\limits_{N\to\infty}p_N\right)=\lim\limits_{N\to\infty}\gamma_a(p_N)=\lim\limits_{N\to\infty}p_N(a)=\lim\limits_{N\to\infty}\sum\limits_{n=0}^N\frac{1}{n!}a^n=\sum\limits_{n=0}^\infty\frac{1}{n!}a^n $$

Since $a$ and $b$ are commute we have $(a+b)^n=\sum\limits_{k=0}^n{n \choose k}a^kb^{n-k}$. In this case $$ e^a e^b =\left(\sum\limits_{k=0}^\infty\frac{1}{k!}a^k\right)\left(\sum\limits_{l=0}^\infty\frac{1}{l!}b^l\right) =\sum\limits_{n=0}^\infty \sum\limits_{k=0}^n\frac{1}{k!}a^k\frac{1}{(n-k)!}b^{n-k} =\sum\limits_{n=0}^\infty\frac{1}{n!}\sum\limits_{k=0}^n{n\choose k}a^{k}b^{n-k} =\sum\limits_{k=0}^n\frac{1}{n!}(a+b)^n =e^{a+b} $$

Obviously $e^0=1$, so $e^{a}e^{-a}=e^{-a}e^{a}=e^0=1$. And we get $(e^{a})^{-1}=e^{-a}$.

Theorem 1.5 Let $A$ be unital involutive Banach algebra. Let $a\in A$, then

  1. $(e^{a})^*=e^{a^*}$
  2. if $a$ is selfadjoint, then $e^{ia}$ is unitary

An easy computation shows $$ (e^a)^* =\left(\sum\limits_{k=0}^\infty\frac{1}{k!}a^k\right)^* =\sum\limits_{k=0}^\infty\frac{1}{k!}(a^k)^* =\sum\limits_{k=0}^\infty\frac{1}{k!}(a^*)^k =e^{a^*} $$

Hence if $a$ is selfadjoint we get $(e^{ia})^*=e^{(ia)^*}=e^{-ia^*}=e^{-ia}=(e^{ia})^{-1}$. So $e^{ia}$ is unitary.

2. Characters of Banach algebras

Notation 2.1 Let $A$ be Banach algebra, then

  1. $X(A)$ denotes the set of its continuous non-zero functionals aka characters
  2. $\Omega(A)$ denotes the set of its maximal closed two-sided ideals
  3. $\mathrm{Inv}(A)$ denotes the set of all invertible elements of $A$.

Theorem 2.2 Let $A$ be unital Banach algebra, and $\chi\in X(A)$, then $\chi(1)=1$.

Note that $1=1^2$ and $1=1^3$, so $\chi(1)=\chi(1)^2$ and $\chi(1)=\chi(1)^3$. Since $\chi\neq 0$, this implies $\chi(1)=1$.

Theorem 2.3 Let $I$ be closed two-sided maximal ideal of commutative unital Banach algebra $A$, then $A/I$ is a field.

Consider Banach factor-algebra $A/I$. Let $a\in A\setminus I$. Since $A$ is commutative and $a\notin I$ the two-sided ideal $aA+I=Aa+I$ is strictly larger than $I$. Since $I$ is maximal we have $aA+I=Aa+I=A$. In particular $ab+c=1$ for some $b\in A$, $c\in I$, hence $(a+I)(b+I)=(b+I)(a+I)=1+I$. This means that $a+I\in\mathrm{Inv}(A/I)$ for all $a\in A\setminus I$. This means that $A/I$ is a field.

Theorem 2.4 Let $A$ be commutative unital Banach algebra, and $\chi\in X(A)$. Then $\mathrm{Ker}(\chi)\in \Omega(A)$ and what is more the map $$ \varkappa: X(A)\to \Omega(A):\chi\mapsto\mathrm{Ker}(\chi) $$ is a bijection.

If $a\in \mathrm{Ker}(\chi)$ and $b\in A$, then $\chi(ab)=\chi(a)\chi(b)=0$ and $\chi(ba)=\chi(b)\chi(a)=0$. This means that $ab,ba\in\mathrm{Ker}(\chi)$. Since $a$ and $b$ are arbitrary $\mathrm{Ker}(\chi)$ is a two-sided ideal. It is closed as kernel of continuous linear functional. Since $\chi$ is a functional, so $\mathrm{dim}(A/\mathrm{Ker}(\chi))=1$. This means that $\mathrm{Ker}(\chi)$ is maximal.

Let $\chi_1,\chi_2\in X(A)$ be two different characters. Assume that $\mathrm{Ker}(\chi_1)=\mathrm{Ker}(\chi_2)$. Denote $I=\mathrm{Ker}(\chi_1)=\mathrm{Ker}(\chi_2)$. Since $\mathrm{dim}(A/I)=1$, then $A=I\oplus\mathrm{span}\{a\}$ for any $a\notin I$. In particular for $a=1\notin I$ and we get $A=I\oplus\mathrm{span}\{1\}$. Since $\chi_1|_I=\chi_2|_I=0$, $\chi_1(1)=1=\chi_2(1)$ and $A=I\oplus\mathrm{span}\{1\}$ we conclude $\chi_1=\chi_2$. Contradiction, so the map $\varkappa$ is injective.

Let $I\in\Omega(A)$, then by theorem 2.3 we have $A/I$ is a field. By Gelfand's-Mazur theorem there exist isomorphism of Banach algebras $i:A/I\to\mathbb{C}$. Thus we get continuous character $\chi:A\to\mathbb{C}:a\mapsto i(a+I)$ with kernel $I$. This means that for each $I\in\Omega(A)$ there exists a character $\chi\in X(A)$ such that $\varkappa(\chi)=I$, i.e. $\varkappa$ is surjective.

Finally $\varkappa$ is bijective.

Theorem 2.5 Let $A$ be a commutative Banach algebra, then $a\in A$ is invertible iff $\chi(a)\neq 0$ for all $\chi\in X(A)$.

Assume $a\in \mathrm{Inv}(A)$, then there exist $b\in A$ such that $ab=1$. For arbitrary $\chi\in X(A)$ we have $\chi(a)\chi(b)=\chi(ab)=\chi(1)=1$, so $\chi(a)\neq 0$. Thus for all $\chi\in X(A)$ holds $\chi(a)\neq 0$.

Assume that for all $\chi\in X(A)$ holds $\chi(a)\neq 0$. Consider two-sided closed ideal $aA$. If $aA$ is proper, then there exist maximal closed two-sided ideal $I$ containing $aA$. Consider respective character $\chi\in X(A)$ such that $\mathrm{Ker}(\chi)=I$. In this case $\chi(a)=0$ because for unital $A$ we have $a\in aA\subset I=\mathrm{Ker}(\chi)$. Contradiction, so $aA=A$. As the consequence $ab=1$ for some $b\in A$, i.e. $a\in \mathrm{Inv}(A)$.

Theorem 2.6 Let $A$ ba a commutative Banach algebra, then for all $a\in A$ we have $$ \sigma(a)=\{\chi(a):\chi\in X(A)\} $$

Using theorem 2.5 we have the following chain of equivalences $$ \begin{align} \lambda\in\mathbb{C}\setminus\sigma(a) &\Longleftrightarrow a-\lambda 1\in\mathrm{Inv}(A)\\ &\Longleftrightarrow \forall\chi\in X(A)\quad\chi(a-\lambda 1)\neq 0\\ &\Longleftrightarrow \forall\chi\in X(A)\quad\lambda\neq\chi(a)\\ &\Longleftrightarrow \lambda\in\mathbb{C}\setminus\{\chi(a):\chi\in X(A)\} \end{align} $$ This means that $\sigma(a)=\{\chi(a):\chi\in X(A)\}$.

Theorem 2.7 Let $A$ be unital Banach algebra, and $\chi\in X(A)$, then $\Vert\chi\Vert=1$.

Assume $\Vert \chi\Vert>1$, then there exist $a''\in A$ such that $\Vert a''\Vert\leq 1$ with $|\chi(a'')|>1$. Consider $a'=a''|\chi(a'')|^{-1}$, then $\Vert a'\Vert<1$ and $|\chi(a')|=1$. Now consider $a=a'\overline{\chi(a')}|\chi(a')|^{-1}$, then we have $a\in A$ such that $\Vert a\Vert<1$ with $\chi(a)=1$. Since $\Vert a\Vert <1$ we have well defined $b=\sum_{n=0}^\infty a^n$. It is straightforward to check that $1+ab=b$. Applying $\chi$ to this equality we get $\chi(1)+\chi(a)\chi(b)=\chi(b)$, which is equivalent to $1+\chi(b)=\chi(b)$. Contradiction, hence $\Vert\chi\Vert\leq 1$. Since $\chi(1)=1$ and $\Vert 1\Vert=1$, then $\Vert\chi\Vert\geq 1$. As we proved later $\Vert\chi\Vert\leq 1$, so $\Vert\chi\Vert=1$.

Theorem 2.8 Let $A$ be a unital Banach Algebra, then $X(A)$ is a weak-$^*$ closed subset of closed unit ball of $A^*$. Moreover $X(A)$ is weak-$^*$ compact.

Let $f\in A^*$ be a touching point of $X(A)$ in the weak-$^*$ topology. Fix $a,b\in A$ and $\varepsilon>0$. By definition of $f$ there is non-empty intersection of $X(A)$ and the weak-$^*$ open set $\{\chi\in A^*:|f(ab)-\chi(ab)|<\varepsilon,\;|f(a)-\chi(a)|<\varepsilon,\;|f(b)-\chi(b)|<\varepsilon\}$. Take some character $\chi$ from this intersection, then $$ |f(ab)-f(a)f(b)| \leq|f(ab)-\chi(ab)|+|\chi(a)\chi(b)-\chi(a)f(b)|+|\chi(a)f(b)-f(a)f(b)| \leq\varepsilon+|\chi(a)|\varepsilon+|f(b)|\varepsilon \leq\varepsilon+\Vert a\Vert\varepsilon+|f(b)|\varepsilon $$ Since $\varepsilon$ is arbitrary we get $|f(ab)-f(a)f(b)|=0$, i.e. $f(ab)=f(a)f(b)$. Since $a$ and $b$ are arbitrary $f$ is a homomorphism of Banach algebras. Similaly there is non zero intersection of $X(A)$ and weak-$^*$ open set $\{\chi\in A^*:|f(1)-\chi(1)|<2^{-1}\}$. Take some character $\chi$ from this intersection, then using $\chi(1)=1$ we get $|f(1)-1|<2^{-1}$. Thus $f(1)\neq 0$ and $f\neq 0$. Thus $f$ is non-zero homomorphism of Banach algebras i. e. a character. Since $f$ is arbitrary $X(A)$ is weak-$^*$ closed. By theorem 2.7 we get $X(A)\subset\mathrm{Ball}_{A^*}(0,1)$. By Banach-Alaoglu theorem $\mathrm{Ball}_{A^*}(0,1)$ is weak-$^*$ compact, but $X(A)$ is its weak-$^*$ closed subset, hence also compact.

Notation 2.9 Topology induced on $X(A)$ by weak-$^*$ topology on $\mathrm{Ball}_{A^*}(0,1)$ is called Gelfands topology.

Remark 2.10 It is worth to mention that bijection $\varkappa$ allows us to turn the space $\Omega(A)$ into compact space, which is not obvious at all.

Remark 2.11 Since $X(A)$ is a compacct the space $C(X(A))$ is a commutative $C^*$- algebra.

3. Properties of $C^*$-algebras

Theorem 3.1 Let $A$ be unital $C^*$-algebra. Let $u\in A$ is a unitary element, then $\Vert u\Vert=\Vert u^{-1}\Vert=1$ and $\sigma(u)\subset \mathbb{T}$

Since $u$ is unitary $u^*=u^{-1}$. Since $A$ is a $C^*$-algebra $$ \Vert u^{-1}\Vert^2=\Vert (u^{-1})^* u^{-1}\Vert=\Vert (u^*)^{-1} u^{-1}\Vert=\Vert(uu^*)^{-1}\Vert=\Vert 1^{-1}\Vert=1\\ \Vert u\Vert^2=\Vert u^*u\Vert=\Vert 1\Vert=1 $$ Hence $\Vert u\Vert=\Vert u^{-1}\Vert=1$. Let $\lambda\in\sigma(u)$, then $|\lambda|\leq\Vert u\Vert=1$. By previous claim $\lambda^{-1}\in\sigma(u^{-1})$, so $|\lambda^{-1}|\leq\Vert u^{-1}\Vert=1$. Thus we conclude that for all $\lambda\in\sigma(u)$ holds $\Vert\lambda\Vert=1$, i.e. $\sigma(u)\subset\mathbb{T}$.

Theorem 3.2 Let $A$ be unital $C^*$-algebra. Let $a\in A$ be selfadjoint element, then $\sigma(a)\subset\mathbb{R}$.

If $\lambda\in \sigma(a)$, then by theorem 0.4 we have $e^{i\lambda}\in \sigma(e^{ia})$. By theorem 1.5 we know that $e^{ia}$ is unitary. Hence by theorem 3.1 we get $\sigma(e^{ia})\subset\mathbb{T}$, so by theorem 0.4 we conclude $e^{i\lambda}\in\mathbb{T}$. This is possible only if $\lambda\in\mathbb{R}$. Since $\lambda$ is arbitrary we get $\sigma(a)\subset\mathbb{R}$.

Theorem 3.3 Let $A$ be a unital $C^*$-algebra and $\chi\in X(A)$, then $\chi$ is involutive character.

Let $a\in A$ and $a=b+ic$ - representation of $a$ given by theorem 1.3. Since $b,c\in A$ are selfadjoint by theorem 3.2 we have $\chi(b)\in\sigma(b)\subset\mathbb{R}$, $\chi(c)\in\sigma(c)\subset\mathbb{R}$. In this case $$ \chi(a^*)=\chi(b^*-i c^*)=\chi(b-ic)=\chi(b)-i\chi(c)=\chi(b)^*-i\chi(c)^*=(\chi(b)+i\chi(c))^*=\chi(b+ic)^*=\chi(a)^* $$

Theorem 3.4 Let $A$ be a unital $C^*$-algebra. Let $a$ be a normal element, then $\Vert a\Vert=r(a)$.

For the begining note that for a selfadjoint element $b$ of $C^*$-algebra we have $\Vert b\Vert^2=\Vert b^* b\Vert=\Vert b^2\Vert$. Then taking $b=a^*a$ we get $\Vert a^*a\Vert^2=\Vert(a^*a)^2\Vert$. Since $a$ is normal we have $$ \Vert a\Vert^4=\Vert a^*a\Vert^2=\Vert (a^*a)^2\Vert=\Vert (a^2)^* a^2\Vert=\Vert a^2\Vert^2 $$ i.e. $\Vert a\Vert=\Vert a^2\Vert^{1/2}$. By induction one can show that $\Vert a\Vert=\Vert a^{2^n}\Vert^{1/2^n}$. Then taking the limit when $n$ tends to infinity we get $$ \Vert a\Vert=\lim\limits_{n\to\infty}\Vert a^{2^n}\Vert^{1/2^n} $$ Since the right hand side of the last equality is the limit of subsequence of the sequence $\{\Vert a^n\Vert^{1/n}: n\in\mathbb{N}\}$. By theorem 0.6 that sequence converges to $r(a)$, hence $$ \Vert a\Vert=\lim\limits_{n\to\infty}\Vert a^{2^n}\Vert^{1/2^n} =\lim\limits_{n\to\infty}\Vert a^n\Vert^{1/n} =r(a) $$

4. Gelfand's transform

Theorem 4.1 Let $A$ be unital commutative Banach algebra and $a\in A$, then the map $\hat{a}:X(A)\to\mathbb{C}:\chi\mapsto(a)$ is continuous. Here we consider $X(A)$ with Gelfand's topology.

Fix $\varepsilon>0$ and $\chi_0\in X(A)$. Consider set $V=\{\chi\in X(A): |\chi(a)-\chi_0(a)|<\varepsilon\}$. It is open in Gelfand's topology and for all $\chi\in V$ we have $|\hat{a}(\chi)-\hat{a}(\chi_0)|=|\chi(a)-\chi_0(a)|<\varepsilon$. Since $\varepsilon>0$ and $\chi_0\in X(A)$ are arbitrary we get $\hat{a}\in C(X(A))$.

Theorem 4.2 Let $A$ be unital involutive commutative Banach algebra, then the map $$ \Gamma:A\mapsto C(X(A)):a\mapsto \hat{a} $$ posess the following properties

  1. $\Vert\Gamma\Vert\leq 1$
  2. $\Gamma(1)=1$
  3. $\Gamma$ is a homomorphism of Banach algebras
  4. $\mathrm{Im}(\Gamma(a))=\sigma(a)$ for all $a\in A$
  5. $\Vert\Gamma(a)\Vert=r(a)$ for all $a\in A$
  6. $\mathrm{Im}(\Gamma)$ separates points in $A$
  7. $\mathrm{Ker}(\Gamma)=\bigcap\{\mathrm{Ker}(\chi):\chi\in X(A)\}=\bigcap\{I:I\in\Omega(A)\}$

By theorem 2.7 for all $\chi\in X(A)$ we have $\Vert \chi\Vert= 1$, hence for all $a\in A$ $$ \Vert\Gamma(a)\Vert=\sup\{|\Gamma(a)(\chi)|:\chi\in X(A)\} =\sup\{|\hat{a}(\chi)|:\chi\in X(A)\} =\sup\{|\chi(a)|:\chi\in X(A)\} \leq\sup\{\Vert\chi\Vert\Vert a\Vert:\chi\in X(A)\} \leq\Vert a\Vert $$ So $\Vert\Gamma\Vert\leq 1$.

Obviously, for all $\chi\in X(A)$ we have $$ \Gamma(1)(\chi)=\hat{1}(\chi)=\chi(1)=1 $$ so $\Gamma(1)=1$

For all $a,b\in A$ and $\chi\in X(A)$ we have $$ \Gamma(ab)(\chi)=\hat{ab}(\chi)=\chi(ab)=\chi(a)\chi(b)=\hat{a}(\chi)\hat{b}(\chi)=\Gamma(a)(\chi)\Gamma(b)(\chi) $$ Hence $\Gamma(ab)=\Gamma(a)\Gamma(b)$. Since $a,b$ are arbitrary we conclude that $\Gamma$ is a homomorphism of Banach algebras.

From theorem 2.6 we get that for all $a\in A$ $$ \sigma(a)=\{\chi(a):\chi\in X(A)\}=\{\Gamma(a)(\chi):\chi\in X(A)\}=\Gamma(a)(X(A))=\mathrm{Im}(\Gamma(a)) $$

As the consequence of previous paragraph we see that for all $a\in A$ $$ \Vert\Gamma(a)\Vert=\sup\{|\Gamma(a)(\chi)|:\chi\in X(A)\} =\sup\{|\lambda|:\lambda\in\sigma(a)\} =r(a) $$

Let $\chi_1,\chi_2\in X(A)$ be two different characters, then there exists $a\in A$ such that $\chi_1(a)\neq\chi_2(a)$. This can be rewritten as $\Gamma(a)(\chi_1)\neq\Gamma(a)(\chi_2)$. This means that $\Gamma(a)$ separates $\chi_1$ and $\chi_2$. Since $\chi_1,\chi_2$ are arbitrary we get that $\mathrm{Im}(\Gamma)$ separates points of $X(A)$.

We have the following chain of equivlences $$ \begin{align} a\in\mathrm{Ker}(\Gamma)&\Longleftrightarrow \Gamma(a)=0\\ &\Longleftrightarrow \forall\chi\in X(A)\quad\Gamma(a)(\chi)=0\\ &\Longleftrightarrow \forall\chi\in X(A)\quad\chi(a)=0\\ &\Longleftrightarrow \forall\chi\in X(A)\quad a\in\mathrm{Ker}(\chi)\\ &\Longleftrightarrow a\in\bigcap\{\mathrm{Ker}(\chi):\chi\in X(A)\} \end{align} $$ So $\mathrm{Ker}(\Gamma)=\bigcap\{\mathrm{Ker}(\chi):\chi\in X(A)\}$. Using bijection given by theorem 2.4 one can also say that $\mathrm{Ker}(\Gamma)=\bigcap\{I:I\in\Omega(A)\}$

Theorem 4.3 Let $A$ be unital commutative $C^*$-algebra, then Gelfand's transform is an involutive isometric isomorphism of Banach algebras.

Let $b\in A$ be some normal element then from theorems 3.4 and 4.2 it follows that $\Vert\Gamma(b)\Vert=r(b)=\Vert b\Vert$. For arbitrary $a\in A$ the element $b=a^*a$ is normal, hence from $C^*$-identity we get $$ \Vert \Gamma(a)\Vert^2=\Vert\Gamma(a)^*\Gamma(a)\Vert=\Vert\Gamma(a^*a)\Vert=\Vert a^*a\Vert=\Vert a\Vert^2 $$ so $\Vert\Gamma(a)\Vert=\Vert a\Vert$. Since $a$ is arbitrary $\Gamma$ is isometric. Since $\Gamma$ is an isometric linear operator between Banach spaces, it image is closed.

From theorem 3.3 it follows that all characters of $A$ preserves involution. Hence for all $a\in A$ and $\chi\in X(A)$ we have $$ \Gamma(a^*)(\chi)=\chi(a^*)=\chi(a)^*=\Gamma(a)(\chi)^* $$ So $\Gamma(a^*)=\Gamma(a)^*$. Since $a$ is arbitrary we see that $\Gamma$ is invloutive.

As the consequence for each $f\in \mathrm{Im}(\Gamma)$ we have $f^*\in\mathrm{Im}(\Gamma)$. Indeed for $f\in \mathrm{Im}(\Gamma)$ we have $f=\Gamma(a)$ for some $a\in A$. Hence $f^*=\Gamma(a)^*=\Gamma(a^*)\in\mathrm{Im}(\Gamma)$. By theorem 4.2 we know that $\mathrm{Im}(\Gamma)$ separates points in $X(A)$ and $1\in\mathrm{Im}(\Gamma)$. Then by Stone-Weierstrass theorem we get that $\mathrm{Im}(\Gamma)$ is dense in $C(X(A))$. But by previous paragraph $\mathrm{Im}(\Gamma)$ is closed. So $\mathrm{Im}(\Gamma)=C(X(A))$, i.e. $\Gamma$ is surjective. Thus we showed that $\Gamma$ is surjective and isometric, hence isometric isomorphism.

  • $\begingroup$ I have a question regarding the proof of the second direction of theorem 1.3 (iii). You write that $$(a-\lambda^{-1} 1)c=c(a-\lambda^{-1} 1)=1$$ implies $$(a^{-1}-\lambda 1)(-\lambda^{-1} ac)=(-\lambda^{-1} ac)(a^{-1}-\lambda 1)=1.$$ But factoring out the second statement I obtain $$- \lambda^{-1}c + ac = - \lambda \color{red}{a c a^{-1}} + ac = 1.$$ Do $c$ and $a^{-1}$ commute? $\endgroup$
    – Ramanujan
    Oct 27 '19 at 14:40
  • 1
    $\begingroup$ They commute. You can derive that from the first equation. $\endgroup$
    – Norbert
    Oct 27 '19 at 17:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.