Approximate Identities in a Normed Algebra

# Approximate Identities in a Normed Algebra

 Definition: Let $\mathfrak{A}$ be a normed algebra and let $\{ e(\lambda) \}_{\lambda \in \Lambda}$ be a net in $\mathfrak{A}$. 1) $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is a Left Approximate Identity for $\mathfrak{A}$ if for every $a \in \mathfrak{A}$ we have that $e(\lambda)a \overset{\alpha} \to a$. 2) $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is a Right Approximate Identity for $\mathfrak{A}$ if for every $a \in \mathfrak{A}$ we have that $ae(\lambda) \overset{\alpha} \to a$. 3) $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is a Two-Sided Approximate Identity for $\mathfrak{A}$ if it is both a left approximate identity and a right approximate identity. A left/right/two-sided approximate identity is said to be a Bounded Left/Right/Two-Sided Approximate Identity if it is also bounded as a net, i.e., there exists an $M > 0$ such that $\| e(\lambda) \| \leq M$ for all $\lambda \in \Lambda$.

Equivalently, a net $\{ e(\lambda) \}_{\lambda \in \Lambda}$ in $\mathfrak{A}$ is a left approximate identity for $\mathfrak{A}$ if for each $a \in \mathfrak{A}$ we have that for all $\epsilon > 0$ there exists a $\lambda_0 \in \Lambda$ such that if $\lambda \geq \lambda_0$ then:

(1)
\begin{align} \quad \| a - e(\lambda)a \| < \epsilon \end{align}

And it is a right approximate identity for $\mathfrak{A}$ if for each $a \in \mathfrak{A}$ we have that for all $\epsilon > 0$ there exists a $\lambda_0 \in \Lambda$ such that if $\lambda \geq \lambda_0$ then:

(2)
\begin{align} \quad \| a - ae(\lambda) \| < \epsilon \end{align}

The following proposition gives us a criterion for the existence of a bounded left approximate identity for a normed algebra $\mathfrak{A}$.

 Proposition 1: Let $\mathfrak{A}$ be a normed algebra. If $U \subseteq \mathfrak{A}$ is a bounded subset of $\mathfrak{A}$ such that for any $a \in \mathfrak{A}$ and any $\epsilon > 0$ there exists a $u \in U$ such that $\| a - ua \| < \epsilon$ then $\mathfrak{A}$ has a bounded left approximate identity.
• Proof: Let $U \subseteq \mathfrak{A}$ be a bounded subset of $\mathfrak{A}$. Then there exists an $M > 0$ such that $\| u \| \leq M$ for every $u \in U$.
• Suppose that for any $a \in \mathfrak{A}$ and any $\epsilon > 0$ there exists a $u \in U$ such that $\| a - ua \| < \epsilon$.
• Let $W$ be the set of all quasi-products of elements in $U$, that is:
(3)
\begin{align} \quad W = \{ v \circ u : v, u \in U \} \end{align}
• For each $k \in \mathbb{N}$ let $P(k)$ be the statement that for all finite subsets $F$ of $\mathfrak{A}$ with $|F| = k$, there exists a $w \in W$ such that $\| a - wa \| < \epsilon$ for all $a \in F$.
• $P(1)$ (When $k = 1$): Suppose that $k = 1$. Then $a \in \mathfrak{A}$ and let $F = \{ a \}$. Then by assumption, for $a \in \mathfrak{A}$ and $\epsilon > 0$ there exists a $u \in U$ such that $\| a - ua \| < \frac{\epsilon}{1 + M}$. Let $w = u \circ u \in W$. Then:
(4)
\begin{align} \quad \| a - wa \| &= \| a - (u \circ u)a \| \\ &= \| a - (u + u - uu)a \| \\ &= \| a - ua -ua + uua \| \\ &= \| (a - ua) - u(a - ua) \| \\ &\leq \| a - ua \| + \| u \| \| a - ua \| \\ &\leq [1 + \| u \|] \| a - ua \| \\ &< [1 + M] \frac{\epsilon}{1 + M} \\ &< \epsilon \end{align}
• So $P(1)$ is true.

• $P(2)$ (When $k = 2$): Suppose that $k = 2$. Let $a, b \in \mathfrak{A}$ and let $\mathcal F = \{ a, b \}$.
• By assumption, for $a \in \mathfrak{A}$ and $\frac{\epsilon}{1 + M} > 0$ there exists $u \in U$ such that:
(5)
\begin{align} \quad \| a - ua \| < \frac{\epsilon}{1 + M} \end{align}
• Similarly, for $(b - ub) \in \mathfrak{A}$ and $\epsilon > 0$ there exists $v \in U$ such that:
(6)
\begin{align} \quad \| (b - ub) -v(b - ub) \| < \epsilon \end{align}
• Let $w = v \circ u \in W$. Then:
(7)
\begin{align} \quad \| a - wa \| &= \| a - (v \circ u)a \| \\ &= \| a - (v + u - vu)a \| \\ &= \| a - va -ua + vua \| \\ &= \| (a - ua) - v(a - ua) \| \\ &= \leq \| a - ua \| + \| v \| \| a - ua \| \\ &< \frac{\epsilon}{1 + M} + M \frac{\epsilon}{1 + M} \\ &< \epsilon \end{align}
• And also:
(8)
\begin{align} \quad \| b - wb \| &= \| b - (v \circ u)b \| \\ &= \| b - (v + u - vu)b \| \\ &= \| b - vb - ub + vub \| \\ &= \| (b - ub) - v(b - ub) \| \\ & < \epsilon \end{align}
• So $P(2)$ is true.

• General Case $P(n+1)$ (When $k = n+1$): Suppose that $P(n)$ is true for some $n \in \mathbb{N}$. We want to prove that $P(n+1)$ is true. Let $a_1, a_2, ..., a_{n+1} \in \mathfrak{A}$ and let $F = \{ a_1, a_2, ..., a_{n+1} \}$. By assumption, for the set $\{ a_1, a_2, ..., a_n \}$ of $n$-many elements, there exists a $y \in W$ such that for $\frac{\epsilon}{3(1 + M)^2} > 0$ we have that for all $i \in \{ 1, 2, ..., n \}$ that:
(9)
\begin{align} \quad \| a_i - ya_i \| < \frac{\epsilon}{3(1 + M)^2} \end{align}
• For the point $y$, by the induction hypothesis there exists a $w \in W$ such that:
(10)
\begin{align} \quad \| y - wy \| < \frac{\epsilon}{\max \{ \| a_1 \|, \| a_2 \|, ..., \| a_n \|\}} \end{align}
• The point $w$ can be chosen so that also $\| a_{n+1} - wa_{n+1} \| < \epsilon$
• Then $w \in W$ is such that for each $i \in \{ 1, 2, ..., n \}$ we have that:
(11)
\begin{align} \quad \| a_i - wa_i \| &= \| a_i - ya_i + ya_i - wya_i + wya_i - wa_i \| \\ &= \| (a_i - ya_i) + (y - wy)a_i + w(ya_i - a_i) \| \\ & \leq \| a_i - ya_i \| + \| y - wy \| \| a_i \| + \| w \| \| ya_i - a_i \| \\ & < \frac{\epsilon}{3(1 + M)^2} + \frac{\epsilon}{3\max \{ \| a_1 \|, \| a_2 \|, ..., \| a_n \| \}} \| a_i \| + [2M + M^2]\frac{\epsilon}{3(1 + M)^2} \\ & < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} \\ & < \epsilon \end{align}
• Above, we used the fact that since $w \in W$ there exists $v, u \in U$ such that $w = u \circ v$. So $\| w \|$ can be bounded by:
(12)
\begin{align} \quad \| w \| = \| u + v - uv \| \leq \| u \| + \| v \| + \| u \| \| v \| \leq 2M + M^2 \end{align}
• Therefore $\| a_i - wa_i \| < \epsilon$ for all $i \in \{ 1, 2, ..., n \}$ and by choice of $w$ we also have that $\| a_{n+1} - wa_{n+1} \| < \epsilon$. Thus $P(n+1)$ is true.
• By the principle of mathematical induction, $P(k)$ is true for all $k \in \mathbb{N}$.