Let
$$\ell^\infty(\Omega):=\{f:\Omega\to\mathbb C:\|f\|_\infty<\infty\},$$
where $\Omega$ is a set and
$$\|f\|_\infty:=\sup_{\omega\in \Omega}|f(\omega)|.$$
Claim: $\ell^\infty(\Omega)$ is a unital Banach algebra.
Proof: It is clear that that $\ell^\infty(\Omega)$ is unital and an algebra. Let $(f_n)_n$ be Cauchy, let $\epsilon>0$, let $N$ be such that
$$\|f_n-f_m\|_\infty=\sup_{\omega\in\Omega}|f_n(\omega)-f_m(\omega)|<\frac\epsilon2$$
for all $n,m\geq N$, define $f:\Omega\to\mathbb C$ by $\omega\mapsto\lim_nf_n(\omega)$, and note that $f$ is well-defined since $(f_n(\omega))_n$ is Cauchy for all $\omega\in\Omega$. Then
$$\lim_m|f_n(\omega)-f_m(\omega)|=|f_n(\omega)-\lim_mf_m(\omega)|=|f_n(\omega)-f(\omega)|\leq\frac\epsilon2<\epsilon$$
for all $\omega\in\Omega$, meaning that $\|f_n-f\|_\infty<\epsilon$ for all $n\geq N$. Finally,
$$\|f\|_\infty=\|f-f_n+f_n\|_\infty\leq\|f_n-f\|_\infty+\|f_n\|_\infty<\infty.\tag*{$\blacksquare$}$$
Let
$$C_b(\Omega):=\{f\in\ell^\infty(\Omega):f\text{ is continuous}\},$$
where $\Omega$ is a topological space.
Claim: $C_b(\Omega)$ is a unital Banach algebra.
Proof: We show that it is closed: recall that convergence with respect to $\|\cdot\|_\infty$ is equivalent to uniform convergence and that uniformly convergent sequences of continuous functions converge to continuous functions.
$$\tag*{$\blacksquare$}$$
Let
$$C_0(\Omega):=\{f\in C_b(\Omega):f\text{ vanishes at infinity}\},$$
where $\Omega$ is locally compact and Hausdorff.
Recall that $f$ vanishes at infinity if for all $\epsilon>0$, there is $K$ compact such that $|f(\omega)|<\epsilon$ for all $\omega\in K^c$.
Claim: $C_0(\Omega)$ is a Banach algebra.
Proof: We show that it is closed: let $n$ be such that $\|f_n-f\|_\infty<\epsilon/2$ and let $K$ compact be such that $|f_n(\omega)|<\epsilon/2$ for all $\omega\in K^c$. Then
$$|f(\omega)|\leq|f_n(\omega)-f(\omega)|+|f_n(\omega)|<\|f_n-f\|_\infty+\frac\epsilon2<\epsilon$$
for all $\omega\in K^c$.
$$\tag*{$\blacksquare$}$$
$C_0(\Omega)$ is one of the most important examples of a Banach algebra in C*-algebra theory, and it is unital if and only if $\Omega$ is compact, in which case
$$C(\Omega)=C_b(\Omega)=C_0(\Omega).$$
Let $L^\infty(\Omega,\mu)$ be the set of classes of essentially bounded, complex-valued, measurable functions on $\Omega$, where $(\Omega,\mu)$ is a measure space, equipped with the essential supremum norm.
Claim: $L^\infty(\Omega,\mu)$ is a unital Banach algebra.
Let $B_\infty(\Omega):=\{f\in\ell^\infty(\Omega):f\text{ is measurable}\}$, where $\Omega$ is measurable.
Claim: $B_\infty(\Omega)$ is a unital Banach algebra.
Proof: Recall that a point-wise convergent sequence of measurable functions converges to a measurable function.
$$\tag*{$\blacksquare$}$$
Let $A$ be the set of continuous, complex-valued functions on the closed unit disc $D\subset\mathbb C$ that are holomorphic on the interior $D^{\mathrm{o}}$ of $D$.
Claim: $A$ is a unital Banach algebra called the disc algebra.
Proof. If $(f_n)_n$ converges to $f$ with respect to $\|\cdot\|_\infty$, then it converges uniformly and $f$ is thus continuous. Moreover,
$$0=\lim_n\oint_\gamma f_n(z)\,\text{d}z=\oint_\gamma\lim_nf_n(z)\,\text{d}z=\oint_\gamma f(z)\,\text{d}z$$
for all closed, piece-wise $C^1$ curves $\gamma$ on $D^{\mathrm{o}}$. Therefore, by Morera's theorem, $f$ is holomorphic on $D^{\mathrm{o}}$.
$$\tag*{$\blacksquare$}$$
