Let $X$ be a nonempty set, and define

\begin{align*} B(X):=\{f:X\to\mathbb R : \text{there exists } M\in[0,\infty) \text{ such that } |f(x)|\le M \text{ for every } x\in X\}. \end{align*}

On the set of all functions $X\to\mathbb R$, define addition, scalar multiplication, and multiplication pointwise: for functions $f,g:X\to\mathbb R$, a scalar $\lambda\in\mathbb R$, and $x\in X$,

\begin{align*} (f+g)(x)=f(x)+g(x),\qquad (\lambda f)(x)=\lambda f(x),\qquad (fg)(x)=f(x)g(x). \end{align*}

For $f\in B(X)$, define

\begin{align*} \|f\|_\infty:=\sup_{x\in X}|f(x)|. \end{align*}

Then $B(X)$ is closed under pointwise addition, pointwise scalar multiplication, and pointwise multiplication. With these operations and the norm $\|\cdot\|_\infty$, the space $B(X)$ is a unital normed algebra over $\mathbb R$. Its multiplicative identity is the constant-one function $1_X:X\to\mathbb R$, $x\mapsto 1$. Moreover, for all $f,g\in B(X)$ and all $\lambda\in\mathbb R$,

\begin{align*} \|f+g\|_\infty\le \|f\|_\infty+\|g\|_\infty, \end{align*}

\begin{align*} \|\lambda f\|_\infty=|\lambda|\,\|f\|_\infty, \end{align*}

and

\begin{align*} \|fg\|_\infty\le \|f\|_\infty\|g\|_\infty. \end{align*}