Example of a Unital Banach Algebra

While studying spectral theory to prepare for a rigurous treatment of $C^*$-algebras, I ran into the following example of a unital Banach algebra:

Let $S$ be a set. Then $\ell^\infty\left(S\right)$, the space of bounded functions on $S$ taking values in $\mathbb C$, is a unital Banach algebra with respect to the supremum norm. The operations are defined pointwise.

It is clear that $\ell^\infty\left(S\right)$ is an algebra, that $1:S\to\mathbb C$ defined by $s\mapsto1$ is the unit, and that $\left\|1\right\|=1$. Moreover, let $f,g\in\ell^\infty\left(S\right)$. Then

Therefore, $\ell^\infty\left(S\right)$ is a unital normed algebra.

To show completeness, let $\left\{f_n\right\}$ be Cauchy. Then $\left\{f_n\left(x\right)\right\}$ is Cauchy for every $x\in S$ and thus convergent to some $f\left(x\right)$. Define $f:S\to\mathbb C$ by $x\mapsto f\left(x\right)$ and let $\varepsilon>0$. Note that $\left\{f_n\right\}$ is uniformly Cauchy, which implies that it converges uniformly to $f$. Therefore, there is an $N$ such that $\left\|f_n-f\right\|\leq\varepsilon/2<\varepsilon$ whenever $n\geq N$. Moreover, $\left\|f\right\|\leq\left\|f_n-f\right\|+\left\|f_n\right\|<\varepsilon+\left\|f_n\right\|$.

It is a highly instructive exercise to show that a sequence of functions into a Banach space is uniformly convergent if and only if it is uniformly Cauchy.

Let $\Omega$ be a topological space. Then $C_b\left(\Omega\right)$, the set of continuous, bounded functions on $\Omega$ taking values in $\mathbb C$, is a closed subalgebra of $\ell^\infty\left(\Omega\right)$. To see this, let $\left\{f_n\right\}$ be a sequence in $C_b\left(\Omega\right)$ converging to $f\in\ell^\infty\left(\Omega\right)$. Then $\left\{f_n\right\}$ converges uniformly to $f$, which implies that $f$ is continuous. If $\Omega$ is compact, then $C(\Omega)=C_b(\Omega)$.

It is also an instructive exercise to show that if a sequence of continuous functions from a topological space to a normed space is uniformly convergent, then its limit is continuous. The stronger hypothesis that the functions take values in a metric space is not much more difficult to show. In this case, recall that $f:X\to\left(Y,\rho\right)$ is bounded if there are $y\in Y$ and $M\geq0$ such that $\rho\left(f\left(x\right),y\right)\leq M$ for all $x\in X$.

Let $\Omega$ be a locally compact, Hausdorff topological space. Then $C_0\left(\Omega\right)$, the set of continuous functions taking values in $\mathbb C$ that vanish at infinity, is a closed subalgebra of $C_b\left(\Omega\right)$. Recall that $f$ vanishes at infinity if for every $\varepsilon>0$, $\left\{\omega\in\Omega:\left|f\left(\omega\right)\right|\geq\varepsilon\right\}$ is compact. To see that it is closed under addition, let $f,g\in C_0\left(\Omega\right)$, let $\varepsilon>0$, let $A=\left\{\omega\in\Omega:\left|f\left(\omega\right)\right|<\varepsilon/2\right\}$, let $B=\left\{\omega\in\Omega:\left|g\left(\omega\right)\right|<\varepsilon/2\right\}$, and let $\omega\in A\cap B$. Then $\left|f\left(\omega\right)+g\left(\omega\right)\right|\leq\left|f\left(\omega\right)\right|+\left|g\left(\omega\right)\right|<\varepsilon$. Therefore, $A\cap B\subseteq C=\left\{\omega\in\Omega:\left|f\left(\omega\right)+g\left(\omega\right)\right|<\varepsilon\right\}$, which implies that $C^c\subseteq A^c\cup B^c$. Since $C^c$ is closed and $A^c\cup B^c$ is compact, $C^c$ is compact.

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