Approximate Identities in a Normed Algebra

Approximate Identities in a Normed Algebra

Definition: Let $X$ be a normed algebra and let $\{ e(\lambda) \}_{\lambda \in \Lambda}$ be a net in $X$.
1) $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is a Left Approximate Identity for $X$ if for every $x \in X$ we have that $e(\lambda)x \to x$.
2) $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is a Right Approximate Identity for $X$ if for every $x \in X$ we have that $xe(\lambda)\to x$.
3) $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is a Two-Sided Approximate Identity for $X$ 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 $X$ is a left approximate identity for $X$ if for each $x \in X$ 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 \| e(\lambda)x - x \| < \epsilon \end{align}

And it is a right approximate identity for $X$ if for each $x \in X$ 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 \| xe(\lambda) - x \| < \epsilon \end{align}

The following proposition gives us a criterion for the existence of a bounded left approximate identity for a normed algebra $X$.

Proposition 1: Let $X$ be a normed algebra. If $U \subseteq X$ is a bounded subset of $X$ such that for any $x \in X$ and any $\epsilon > 0$ there exists a $u \in U$ such that $\| x - ux \| < \epsilon$ then $X$ has a bounded left approximate identity.
  • Proof: Let $U \subseteq X$ be a bounded subset of $X$. Then there exists an $M > 0$ such that $\| u \| \leq M$ for every $u \in U$.
  • Suppose that for any $x \in X$ and any $\epsilon > 0$ there exists a $u \in U$ such that $\| x - ux \| < \epsilon$.
  • 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 $X$ with $|F| = k$, there exists a $w \in W$ such that $\| x - wx \| < \epsilon$ for all $x \in F$.
  • $P(1)$ (When $k = 1$): Suppose that $k = 1$. Then $x \in X$ and let $F = \{ x \}$. Then by assumption, for $x \in X$ and $\epsilon > 0$ there exists a $u \in U$ such that $\| x - ux \| < \frac{\epsilon}{1 + M}$. Let $w = u \circ u \in W$. Then:
(4)
\begin{align} \quad \| x - wx \| &= \| x - (u \circ u)x \| \\ &= \| x - (u + u - uu)x \| \\ &= \| x - ux -ux + uux \| \\ &= \| (x - ux) - u(x - ux) \| \\ &\leq \| x - ux \| + \| u \| \| x - ux \| \\ &\leq [1 + \| u \|] \| x - ux \| \\ &< [1 + M] \frac{\epsilon}{1 + M} \\ &< \epsilon \end{align}
  • So $P(1)$ is true.

  • $P(2)$ (When $k = 2$): Suppose that $k = 2$. Let $x, y \in X$ and let $\mathcal F = \{ x, y \}$.
  • By assumption, for $x \in X$ and $\frac{\epsilon}{1 + M} > 0$ there exists $u \in U$ such that:
(5)
\begin{align} \quad \| x - ux \| < \frac{\epsilon}{1 + M} \end{align}
  • Similarly, for $(y - uy) \in X$ and $\epsilon > 0$ there exists $v \in U$ such that:
(6)
\begin{align} \quad \| (y - uy) -v(y - uy) \| < \epsilon \end{align}
  • Let $w = v \circ u \in W$. Then:
(7)
\begin{align} \quad \| x - wx \| &= \| x - (v \circ u)x \| \\ &= \| x - (v + u - vu)x \| \\ &= \| x - vx -ux + vux \| \\ &= \| (x - ux) - v(x - ux) \| \\ &= \leq \| x - ux \| + \| v \| \| x - ux \| \\ &< \frac{\epsilon}{1 + M} + M \frac{\epsilon}{1 + M} \\ &< \epsilon \end{align}
  • And also:
(8)
\begin{align} \quad \| y - wy \| &= \| y - (v \circ u)y \| \\ &= \| y - (v + u - vu)y \| \\ &= \| y - vy - uy + vuy \| \\ &= \| (y - uy) - v(y - uy) \| \\ & < \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 $x_1, x_2, ..., x_{n+1} \in X$ and let $F = \{ x_1, x_2, ..., x_{n+1} \}$. By assumption, for the set $\{ x_1, x_2, ..., x_n \}$ 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 \| x_i - yx_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 \{ \| x_1 \|, \| x_2 \|, ..., \| x_n \|\}} \end{align}
  • The point $w$ can be chosen so that also $\| x_{n+1} - wx_{n+1} \| < \epsilon$
  • Then $w \in W$ is such that $i \in \{ 1, 2, ..., n \}$ we have that:
(11)
\begin{align} \quad \| x_i - wx_i \| &= \| x_i - yx_i + yx_i - wyx_i + wyx_i - wx_i \| \\ &= \| (x_i - yx_i) + (y - wy)x_i + w(yx_i - x_i) \| \\ & \leq \| x_i - yx_i \| + \| y - wy \| \| x_i \| + \| w \| \| yx_i - x_i \| \\ & < \frac{\epsilon}{3(1 + M)^2} + \frac{\epsilon}{3\max \{ \| x_1 \|, \| x_2 \|, ..., \| x_n \| \}} \| x_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 $\| x_i - wx_i \| < \epsilon$ for all $i \in \{ 1, 2, ..., n \}$ and by choice of $w$ we also have that $\| x_{n+1} - wx_{n+1} \| < \epsilon$. Thus $P(n+1)$ is true.
  • By the principle of mathematical induction, $P(k)$ is true for all $k \in \mathbb{N}$.

  • Let $\mathcal F (X)$ be the collection of all finite subsets of $X$ ordered by inclusion $\subseteq$.
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