Algebras of Functions From a Subset E of a Normed Algebra X

Algebras of Functions From a Subset E of a Normed Algebra X

Definition: Let $(X, \| \cdot \|)$ be a normed linear space and let $E \subseteq X$ with $E \neq \emptyset$. A function $f : E \to X$ is said to be Bounded on $E$ if the set $\{ \| f(x) \| : x \in E \}$ is a bounded set of nonnegative real numbers. The Set of All Bounded Functions on $E$ is denoted $\ell^{\infty}(E, X)$. If $f \in \ell^{\infty}(E, X)$ then the Supremum Norm of $f$ is defined to be $\| f \|_{\infty} = \sup \{ \| f(x) \| : x \in E \}$l.

Note that if $f \in \ell^{\infty} (E, X)$ then $\sup \{ \| f(x) \| : x \in E \}$ exists since $\{ \| f(x) \| : x \in E \}$ is a bounded and nonempty subset of the real numbers and so the supremum norm $\| f \|_{\infty}$ is well-defined.

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