I am reading a proof of the Gelfand-Naimark theorem in Principles of Harmonic Analysis by Anton Deitmar and Siegfried Echterhoff. I have questions about some of the steps.
Theorem. If $A$ is a commutative $C^\ast$-algebra, then the Gelfand transform $a\mapsto \widehat a$ is an isometric $^\ast$-isomorphism
$$A \xrightarrow{\cong} C_0(\Delta_A)$$
So, in particular, one has $\|\widehat a\|_{\Delta_A} = \|a\|$ and $\widehat{a^\ast} = \overline{\widehat{a}}$ for every $a\in A$. Finally, the structure space $\Delta_A$ is compact if and only if $A$ is unital. In this case, the Gelfand transform gives an isomorphism $A \cong C(\Delta_A)$.
Here, $\Delta_A$ is the structure space of $A$, consisting of all non-zero continuous algebra homomorphisms $m:A\to\Bbb C$.
Proof. We show that in the case of a $C^\ast$-algebra the Gelfand map is isometric with a dense image. This will imply that $\widehat{A} = \{\widehat a: a\in A\}$ is a complete, hence closed, dense subalgebra of $C_0(\Delta_A)$, and hence $\widehat{A} = C_0(\Delta_A)$. For $a\in A$, using Theorem $2.5.2$ and Lemma $2.6.4$ we get $\|\widehat a\|_\infty^2 = \|\widehat a \widehat a\|_\infty = \|\widehat{a^\ast a}\|_\infty = r(a^\ast a) = \|a^\ast a\| = \|a^2\|$, so indeed the Gelfand transform is an isometric map. It remains to show that the image is dense.
- Where does $$\|\widehat a\|_\infty^2 = \|\widehat a \widehat a\|_\infty = \|\widehat{a^\ast a}\|_\infty = r(a^\ast a) = \|a^\ast a\| = \|a^2\|$$ come from? Every step (except the last two) seems hard to understand.
The image of the Gelfand map $\widehat A = \{\widehat a: a\in A\} \subset C_0(\Delta_A)$ strictly separates the points of $\Delta_A$: if $m_1,m_2\in \Delta_A$ such that $\widehat{a}(m_1) = \widehat{a}(m_2)$ for every $a\in A$, then we clearly get $m_1 = m_2$, and since we require $m \ne 0$ for the elements $m\in \Delta_A$, we find at least one $a\in A$ with $\widehat a(m) = m(a) \ne 0$ for any given $m \in \Delta_A$. Using this, it is then a direct consequence of the Stone-Weierstraß Theorem $A.10.1$, that $\widehat A$ is dense in $C_0(\Delta_A)$ with respect to the supremum-norm whenever $\widehat A$ is invariant under taking complex conjugates, which is the case thanks to Lemma $2.6.6$. Finally, if $\Delta_A$ is compact, then $A \cong C_0(\Delta_A)$ is unital. The converse
is Theorem $2.4.5(b)$.
- Why is $\widehat A$ invariant under taking complex conjugates? For any $a\in A$ and $m\in \Delta_A$, we have $$\overline{\widehat a(m)} = \overline{m(a)} = m(a^*) $$
Please see this collection of images for the referenced theorems and lemmas. Thank you!
Best Answer
The first equality, in more familiar notation, is $$ (\sup\{|f(t)|:\ t\})^2=\sup\{|f(t)^2|:\ t\} $$ The second equality is $$ \sup\{|f(t)^2|:\ t\}=\sup\{\overline{f(t)}f(t):\ t\}, $$ which comes simply from $|z^2|=|z|^2=\overline z z$.
As for your second question, not entirely sure what the question is, as
shows precisely that $\hat A$ is closed under taking conjugates. In the above, the first equality is the definition of $\hat a$. The second equality is less trivial, it is the fact that an algebra homomorphism between a C$^*$-algebra and $\mathbb C$ preserves adjoints. This is usually done by working in the unitization of $A$, and noting that if $a\in A$ is invertible then $m(a)$ is invertible; this implies that $\sigma(m(a))\subset \sigma(a)$. When $a=a^*$, this means that $m(a)\in\mathbb R$. For arbitrary $a$ we have $a=b+ic$ with $b,c$ selfadjoint, and so $$m(a^*)=m(a-ic)=m(a)-im(c)=\overline{m(a)+im(c)}=\overline{m(a)}.$$ To do all this, one needs to show that $\sigma(a)\subset\mathbb R$. This is typically done with characters, which would be circular, but it can be done using holomorphic functional calculus (this is briefly explained here).