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?


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


$3$. 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!

  • $\begingroup$ What do you mean by "How can we show that $\hat{A}$ such that 3 conditions above?"? $\endgroup$ – Prahlad Vaidyanathan Dec 14 '13 at 4:27
  • $\begingroup$ I mean: $\widehat {\mathcal A}$ such that: contains the constants, closed subalgebra, separates points. Do you understand? $\endgroup$ – kimtahe6 Dec 14 '13 at 4:29
  • $\begingroup$ But I don't remember the definition of closed subalgebra. It's $\alpha f \in \widehat {\mathcal A}$ and $f+g \in \widehat {\mathcal A},\ \forall \alpha \in \Bbb C,\ f,\ g \in \widehat {\mathcal A}$. And $\{f_n\} \subset \widehat {\mathcal A}, \ f_n \to f \in \mathcal A \implies f \in \widehat {\mathcal A}$. Is it correct? $\endgroup$ – kimtahe6 Dec 14 '13 at 4:33
  • 1
    $\begingroup$ You want to show that $\hat{f_n} \to g \in C(M_A)$, then $g \in \hat{A}$ $\endgroup$ – Prahlad Vaidyanathan Dec 14 '13 at 4:37
  • $\begingroup$ Yes, hihi :) , thanks Prahlad Vaidyanathan. Can you help me? $\endgroup$ – kimtahe6 Dec 14 '13 at 4:41

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.

  • $\begingroup$ Thanks Prahlad Vaidyanathan. But I have 2 questions: 1/ I think that $f \mapsto \hat{f}$ is an isometry iff $\left \| f^2 \right \|=\left \| f \right \|^2, \ \forall f \in \mathcal A$? 2/ Since $\left |\widehat{f}(\varphi ) \right |=\left | \varphi (f) \right |\le \left \| \varphi \right \|\cdot \left \| f \right \|=\left \| f \right \|$ then $\left \|\widehat{f} \right \|_{C(M_{\mathcal A})} \le \left \| f \right \|_{C(X)}$. But I don't understand How to take $\varphi$ to be the evaluation maps (proof?) $\endgroup$ – kimtahe6 Dec 14 '13 at 7:44
  • $\begingroup$ Yes. I see :). Since $\|\hat{f}\|_{C(M_A)} = \|f\|_{C(X)}$ then we're done. I'm sorry can you prove $\left \|\widehat{f} \right \|_{C(M_{\mathcal A})} \ge \left \| f \right \|_{C(X)}$?? $\endgroup$ – kimtahe6 Dec 14 '13 at 7:54

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