The Norm. Alg. ℓ∞(E, A) for a Nonempty Set E and a Normed Algebra A

# The Normed Algebra ℓ∞(E, A) for a Nonempty Set E and a Normed Algebra A

 Proposition 1: Let $E$ be a nonempty set and let $(\mathfrak{A}, \| \cdot \|)$ be a normed algebra over $\mathbf{F}$. Let $\ell^{\infty}(E, \mathfrak{A})$ denote the set of all bounded functions from $E$ to $X$, that is, $f \in \ell^{\infty}(E, \mathfrak{A})$ if and only if there exists an $M > 0$ such that $\| f(e) \| \leq M$ for all $e \in E$. Then $\ell^{\infty}(E, \mathfrak{A})$ with the operations of pointwise function addition, pointwise scalar multiplication, and pointwise function multiplication is a normed algebra with norm defined for all $f \in \ell^{\infty} (E, \mathfrak{A})$ by $\| f \|_{\infty} = \sup \{ \| f(e) \| : e \in E \}$.
• Proof: It is easy to verify that $\ell^{\infty}(E, \mathfrak{A})$ with the operations of pointwise function addition and pointwise scalar multiplication is a linear space. We will first show with the operation of pointwise function multiplication makes $\ell^{\infty}(E, \mathfrak{A})$ an algebra. We will then verify that $\| \cdot \|_{\infty}$ is an algebra norm on $\ell^{\infty}(E, \mathfrak{A})$.
• Showing that $\ell^{\infty}(E, \mathfrak{A})$ is an algebra:
• 1. $f \cdot (g \cdot h) = (f \cdot g) \cdot h$: Let $f, g, h \in \ell^{\infty} (E, \mathfrak{A})$. Then for every $e \in E$ we by the associativity of product in $X$ (since $\mathfrak{A}$ is an algebra) that:
(1)
\begin{align} \quad [f \cdot (g \cdot h)](e) = f(e) \cdot (g(e) \cdot h(e)) = (f(e) \cdot g(e)) \cdot h(e) = [(f \cdot g) \cdot h](e) \end{align}
• Since this holds for all $e \in E$ we see that $f \cdot (g \cdot h) = (f \cdot g) \cdot h$.
• 2. $f \cdot (g + h) = f \cdot g + f \cdot h$: Let $f, g, h \in \ell^{\infty}(E, \mathfrak{A})$. Then for every $e \in E$ we have by the distributivity of product in $\mathfrak{A}$ (again since $\mathfrak{A}$ is an algebra) that:
(2)
\begin{align} \quad [f \cdot (g + h)](e) = f(e)[g(e) + h(e)] = f(e)g(e) + f(e)h(e) = (f \cdot g)(e) + (f \cdot h)(e) \end{align}
• Since this holds for all $e \in E$ we see that $f \cdot (g + h) = f \cdot g + f \cdot h$.
• 3. $(\alpha f) \cdot g = \alpha (f \cdot g) = f \cdot (\alpha g)$: Let $f, g \in \ell^{\infty}(E, \mathfrak{A})$ and let $a \in \mathbf{F}$. Then since $\mathfrak{A}$ is an algebra we have that for every $e \in E$:
(3)
\begin{align} \quad [(\alpha f) \cdot g](e) = (\alpha f(e))g(e) = \alpha (f(e)g(e)) = [\alpha (f \cdot g)](e) \end{align}
• And also:
(4)
\begin{align} \quad [\alpha(f \cdot g)](e) = \alpha(f(e)g(e)) = f(e) (\alpha g(e)) = [f \cdot (\alpha g)](e) \end{align}
• Since this holds for all $e \in E$ we conclude that $(\alpha f) \cdot g = \alpha (f \cdot g) = f \cdot (\alpha g)$. Therefore $\ell^{\infty}(E, \mathfrak{A})$ with pointwise addition, pointwise scalar multiplication, and pointwise product is an algebra.
• Showing that $\| \cdot \|_{\infty}$ is an algebra norm on $\ell^{\infty}(E, \mathfrak{A})$:
• 1. $\| f \|_{\infty} = 0$ if and only if $f = 0$: Suppose that $\| f \|_{\infty} = 0$. Then $\sup \{ \| f(e) \| : e \in E \} = 0$ which implies that $\| f(e) \| \leq 0$ for all $e \in E$, i.e., $\| f(e) \| = 0$ for all $e \in E$. Thus $f(e) = 0$ for all $e \in E$ which tells us that $f = 0$. On the other hand, suppose that $f = 0$. Then $\| f(e) \| = 0$ for all $e \in E$ and thus $\| f \|_{\infty} = \sup \{ \| f(e) \| : e \in E \} = 0$.
• 2. $\| af \|_{\infty} = |a| \| f \|_{\infty}$: Let $f \in \ell^{\infty}(E, \mathfrak{A})$ and let $a \in \mathbf{F}$. Then:
(5)
\begin{align} \quad \| af \|_{\infty} = \sup \{ \| af(e) \| : e \in E \} = \sup \{ |a| \| f(e) \| : x \in E \} = |a| \sup \{ \| f(e) \| : e \in E \} = |a| \| f \|_{\infty} \end{align}
• 3. $\| f + g \|_{\infty} \leq \| f \|_{\infty} + \| g \|_{\infty}$: Let $f, g \in \ell^{\infty}(E, \mathfrak{A})$. Then by the triangle inequality for $\| \cdot \|$ we have that:
(6)
\begin{align} \quad \quad \| f + g \|_{\infty} = \sup \{ \| f(e) + g(e) \| : e \in E \} \leq \sup \{ \| f(e) \| + \| g(e) \| : e \in E \} = \sup \{ \| f(e) \| : e \in E \} + \sup \{ \| g(e) \| : e \in E \} = \| f \|_{\infty} + \| g \|_{\infty} \end{align}
• 4. $\| f \cdot g \|_{\infty} \leq \| f \|_{\infty} \| g \|_{\infty}$: Let $f, g \in \ell^{\infty}(E, \mathfrak{A})$. Then since $\| f(e)g(e) \| \leq \| f(e) \| \| g(e) \|$ for each $e \in E$ (since $\| \cdot \|$ is an algebra norm on $\mathfrak{A}$) we have that:
(7)
\begin{align} \quad \| f \cdot g \|_{\infty} = \sup \{ \| f(e)g(e) \| : e \in E \} \leq \sup \{ \| f(e) \| \| g(e) \| : e \in E \} = \sup \{ \| f(e) \| : e \in E \} \sup \{ \| g(e) \| : e \in E \} = \| f \|_{\infty} \| g \|_{\infty} \end{align}
• We conclude that $\| \cdot \|_{\infty}$ is an algebra norm on $\ell^{\infty} (E, \mathfrak{A})$. $\blacksquare$

The following proposition tells us that when $E$ is a nonempty set and $\mathfrak{A}$ is a Banach algebra then $\ell^{\infty}(E, \mathfrak{A})$ is also a Banach algebra.

 Proposition 2: Let $E$ be a nonempty set. If $\mathfrak{A}$ is a Banach algebra then $\ell^{\infty}(E, \mathfrak{A})$ is a Banach algebra.
• Proof: Let $(f_n)$ be a Cauchy sequence of functions in $\ell^{\infty}(E, \mathfrak{A})$. Then for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then:
(8)
\begin{align} \quad \| f_m - f_n \|_{\infty} = \sup_{e \in E} \| f_m(e) - f_n(e) \| < \epsilon \end{align}
• For each fixed $e \in E$ we have that the sequence $(f_n(e))$ in $\mathfrak{A}$ is Cauchy. Since $\mathfrak{A}$ is a Banach algebra, $(f_n(e))$ converges. Define $f : E \to \mathfrak{A}$ for each $e \in E$ by:
(9)
\begin{align} \quad f(e) = \lim_{n \to \infty} f_n(e) \end{align}
• Then $f$ is a bounded function from $E$ to $\mathfrak{A}$ since for all $\epsilon > 0$ by the Cauchy-ness of $(f_n)$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then $\| f_m - f_n \| < \epsilon$ and thus for each $e \in E$:
(10)
\begin{align} \quad \| f(e) \| = \| f(e) - f_N(e) + f_N(e) \| \leq \| f(e) - f_N(e) \| + \| f_N(e) \| = \| \lim_{n \to \infty} f_n(e) - f_N(e) \| + \| f_N(e) \| = \lim_{n \to \infty} \| f_n(e) - f_N(e) \| + \| f_N(e) \| \leq \epsilon + M \end{align}
• Where $M > 0$ is such that $\| f_N(e) \| \leq M$ for all $e \in E$.
• Furthermore, $(f_n)$ infinity-norm coverges to $f$. To see this, again, let $\epsilon > 0$. Since $(f_n)$ is Cauchy there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then $\| f_m - f_n \|_{\infty} < \epsilon$, i.e., $\| f_m(e) - f_n(e) \| < \epsilon$ for all $e \in E$. But observe that then for all $e \in E$:
(11)
\begin{align} \quad \| f(e) - f_n(e) \| = \| \lim_{m \to \infty} f_m(e) - f_n(e) \| = \lim_{m \to \infty} \| f_m(e) - f_n(e) \| < \epsilon \end{align}
• So $\| f(e) - f_n(e) \|$ for all $e \in E$, and thus, if $n \geq N$ then $\| f - f_n \|_{\infty} < \epsilon$. So $(f_n)$ infinity norm converges to $f$. So every Cauchy sequence in $\ell^{\infty}(E, \mathfrak{A})$ converges and thus $\ell^{\infty}(E, \mathfrak{A})$ is a Banach algebra. $\blacksquare$