> [!definition] > > Let $\alg$ be a [[Commutative Banach Algebra|commutative unital Banach algebra]], and $\sigma(\alg)$ be its [[Multiplicative Functional|spectrum]]. For each $x \in \alg$, define $\wh x: \sigma(\alg) \to \complex$ by $h \mapsto h(x)$, then $\wh x$ is continuous since $\sigma(\alg)$ is given the weak*-topology, this map > $ > \Gamma: \alg \to C(\sigma(\alg)) \quad x \mapsto \wh x = \Gamma x = \Gamma_\alg x > $ > is known as the **Gelfand transform** on $\alg$. > > If $\alg$ is a $*$-algebra and $\wh{x^*} = \ol{\wh x}$ for all $x \in \alg$, then $\alg$ is **symmetric**. > [!theorem] > > Let $x \in \alg$, then > 1. $\Gamma: \alg \to C(\sigma(\alg))$ is a Banach algebra homomorphism. > 2. $x$ is invertible if and only if $\Gamma x \ne 0$ everywhere. > 3. $\wh x(\sigma(\alg)) = \sigma(x)$ is the [[Spectrum|spectrum]]. > 4. $\norm{\Gamma x}_u = \rho(x) \le \norm{x}$. > > *Proof*. Let $x, y \in \alg$ and $h \in \sigma(\alg)$, then > $ > \wh{xy}(h) = h(xy) = h(x)h(y) = \wh{x}(h)\wh{y}(h) \quad \wh{e}(h) = h(e) = 1 > $ > Secondly, the following are equivalent: > - $x$ is not invertible. > - The ideal generated by $x$ is proper. > - $x$ is contained in a maximal ideal. > - There exists $h \in \sigma(\alg)$ such that $h(x) = 0$. > - $\wh x$ has a zero. > > Lastly, the following are equivalent: > - $\lambda \in \sigma(x)$ > - $\lambda e - x$ is not invertible. > - There exists $h \in \sigma(\alg)$ such that $\lambda - \wh x(h) = 0$. > - $\lambda \in \wh x(\sigma(\alg))$. > [!theorem] > > Let $\alg$ be a commutative unital Banach $*$-algebra. > 1. $\alg$ is symmetric if and only if $\wh x$ is real valued whenever $x = x^*$. > 2. If $\alg$ is a $C^*$-algebra, then $\alg$ is symmetric. > 3. If $\alg$ is symmetric, then $\Gamma(\alg)$ is dense in $C(\sigma(\alg))$. > > *Proof*. Let $x \in \alg$ such that $x = x^*$. Define $u = (x+x^*)/2$ and $v = (x - x^*)/2i$, then $u = u^*$, $v = v^*$, and $\wh u, \wh v$ are real-valued. From here, > $ > x = u + iv \quad x^* = u - iv \quad \wh{x^*} = \wh u - i\wh{v} = \ol{\wh x} > $ > Now suppose that $\alg$ is a $C^*$-algebra. Let $x \in \alg$ with $x = x^*$, $h \in \sigma(\alg)$ and write $\wh x(h) = h(x) = \alpha + i\beta$. Let $z = x + ite$, then $h(z) = \alpha + i(\beta + t)$ and $z^*z = x^2 + t^2 e$. Since $\abs{h(x)} \le \norm{x}$, > $ > \alpha^2 + (\beta + t)^2 = \abs{h(z)}^2 \le \norm{z}^2 = \norm{z^*z} \le \norm{x^2} + t^2 > $ > as the above holds for all $t \ge 0$, $\beta = 0$, $\wh x$ is real-valued, and $\alg$ is symmetric. > > Lastly, if $\alg$ is symmetric, then $\Gamma(\alg)$ is closed under complex conjugation. It contains constants and separate points. Thus $\Gamma(\alg)$ is dense in $C(\sigma(\alg))$ by the Stone-Weierstrass theorem. > [!theorem] > > Let $x \in \alg$. If any of the following holds: > 1. $\alg$ is generated by $x$ and $e$. > 2. $x$ is invertible and $\alg$ is generated by $x$ and $x^{-1}$. > 3. $\alg$ is symmetric and $\alg$ is generated by $x, x^*$, and $e$. > > then $\wh x: \sigma(\alg) \to \sigma(x)$ is a [[Homeomorphism|homeomorphism]]. > > *Proof*. Since $\sigma(\alg)$ and $\sigma(x)$ are both [[Compactness|compact]] and [[Hausdorff Space|Hausdorff]], it's sufficient to show $\wh x$ is injective. To this end, any $h \in \sigma(\alg)$ is uniquely determined by its actions on $x$ and $e$. In the third case, $\wh {x^*}(h) = \ol{\wh x(h)}$. Therefore $\wh x$ is injective. > [!theorem] > > Let $\alg$ be a commutative unital Banach algebra. > 1. If $x \in \alg$, then $\norm{\wh x}_u = \norm{x}$ if and only if $\norm{x^{2^k}} = \norm{x}^{2^k}$ for all $k \in \nat$. > 2. $\Gamma$ is an isometry if and only if $\norm{x^2} = \norm{x}^2$ for all $x \in \alg$. > > *Proof*. If $\norm{\wh x}_u = \norm{x}$, then > $ > \norm{x^{2^k}} \le \norm{x}^{2^k} = \norm{\wh x}_{u}^{2^k} = \norm{{\wh x}^{2^k}} \le \norm{x^{2^k}} > $ > On the other hand, $\norm{x^{2^k}} = \norm{x}^{2^k}$ for all $k \in \nat$, then $\norm{\wh x}_u = \rho(x) = \norm{x}$. > > If $\norm{x^2} = \norm{x}^2$ for all $x \in \alg$, then $(1)$ holds and $\Gamma$ is an isometry.