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:
- $\left\|\omega\right\|=1$ 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)
- 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 :)
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.
1. General properties of Banach algebras
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 $a=b+ic$.
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 $\sigma(a^{-1})=\sigma(a)^{-1}$.
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 $(e^{a})^{-1}=e^{-a}$.
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
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 $\chi(1)=1$.
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.
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.
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. $a\in \mathrm{Inv}(A)$.
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 $\sigma(a)=\{\chi(a):\chi\in X(A)\}$.
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 $\Vert\chi\Vert=1$.
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.
3. Properties of $C^*$-algebras
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. $\sigma(u)\subset\mathbb{T}$.
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 $\sigma(a)\subset\mathbb{R}$.
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 $$ \chi(a^*)=\chi(b^*-i c^*)=\chi(b-ic)=\chi(b)-i\chi(c)=\chi(b)^*-i\chi(c)^*=(\chi(b)+i\chi(c))^*=\chi(b+ic)^*=\chi(a)^* $$
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
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 $\hat{a}\in C(X(A))$.
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 $\mathrm{Ker}(\Gamma)=\bigcap\{I:I\in\Omega(A)\}$
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.