I'm going to find an example of uniform algebra and show that satisfying the definition.
Example: Show that The Gelfand transform $\widehat{f}$ is uniform algebra.
We know that:
A uniform algebra is a closed subalgebra $\mathcal A$ of the complex algebra $C(X)$ that contains the constants and separates points. Here $X$ is a compact Hausdorff space.
The Gelfand transform of $f$ is the function $\widehat{f}$ defined on $M_{\mathcal A}$ in the following way:
$$\begin{align}\widehat{f}: M_{\mathcal A} &\to \Bbb C\\ \varphi &\mapsto \widehat{f}(\varphi)=\varphi(f), \forall \varphi \in M_{\mathcal A} \end{align}$$
How can we show that $\widehat{A}$ such that $3$ conditions above?
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After a period of time to think about my problem which I posted above, I'll write it here:
Let $\widehat{\mathcal A}=\{\widehat{f}: \ f \in \mathcal A \}$
$1$. $\widehat {\mathcal A} \ $ contains the constants
Because $e \in \mathcal A \implies \widehat{e}(\varphi)=\varphi (e)=1, \ \forall \varphi \in M_{\mathcal A}$.
Therefore, $\widehat{(\lambda e)}(\varphi)=\varphi(\lambda e)=\lambda \varphi(e)=\lambda,\ \forall \lambda \in \Bbb C$.
Hence, $\widehat {\mathcal A} \ $ contains the constants
$2$. $\widehat {\mathcal A} \ $ separates points
We assume that $\varphi_1,\ \varphi_2 \in M_{\mathcal A}$ such that $\widehat{f}(\varphi_1)= \widehat{f}(\varphi_2),\ \forall f \in \mathcal A$.
Whence, $\varphi_1(f)= \varphi_2(f),\ \forall f \in \mathcal A$. So $\varphi_1= \varphi_2$.
Hence, $\widehat {\mathcal A} \ $ separates points
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$3$. Now I have stuck when I try to show $\widehat {\mathcal A}$ is a closed subalgebra of algebra Banach $C(M_{\mathcal A})$
I think that we have $\widehat{f}$ is continuous, because $\left |\widehat{f}(\varphi ) \right |=\left | \varphi (f) \right |\le \left \| \varphi \right \|\cdot \left \| f \right \|=\left \| f \right \|$
But How can we prove The first condition (i.e $\widehat {\mathcal A}$ is a closed subalgebra of algebra Banach $C(M_{\mathcal A})$)
I don't remember the definition of closed subalgebra. Can anyone post it help me!
Any help will be appreciated! Thanks!
You know that $$ \|\hat{f}\|_{C(M_A)} = \|f\|_{C(X)} $$ $\|\hat{f}\|_{C(M_A)} \leq \|f\|_{C(X)}$ as you have mentioned in your post, and, for each $x\in X$, define $\varphi \in M_A$ by $f \mapsto f(x)$, then $$ |f(x)| = |\varphi(f)| \leq \|\hat{f}\|_{C(M_A)} $$ This is true for all $x \in X$, and hence $\|f\|_{C(X)} \leq \|\hat{f}\|_{C(M_A)}$
Hence, $f \mapsto \hat{f}$ is an isometry from the Banach space $A$ to the Banach space $C(M_A)$. Hence it must have closed range.