Approximate Identities in A for X when X is a Banach Left A-Module 1

Definition: Let $\mathfrak{A}$ be a Banach algebra and let $X$ be a Banach left $\mathfrak{A}$-module. A Left Approximate Identity in $\mathfrak{A}$ for $X$ is a net $\{ e(\lambda) \}_{\lambda \in \Lambda}$ in $\mathfrak{A}$ such that $e(\lambda)x \to x$ for every $x \in X$.

Definition: Let $\mathfrak{A}$ be a Banach algebra and let $X$ be a Banach right $\mathfrak{A}$-module. A Right Approximate Identity in $\mathfrak{A}$ for $X$ is a net $\{ e(\lambda) \}_{\lambda \in \Lambda}$ in $\mathfrak{A}$ such that $xe(\lambda) \to x$ for every $x \in X$.

Before we prove a more meaningful result we will need the following lemma:

Lemma 1: Let $\mathfrak{A}$ be a Banach algebra and let $X$ be a Banach left $\mathfrak{A}$-module, and regard $X$ as a Banach left $\mathfrak{A} + \mathbf{F}$ module. Let $e \in \mathfrak{A}$, $C > 1$, $\| e \| \leq C$, and $\gamma = (4C)^{-1}$. Then: a) $1 - \gamma + \gamma e \in \mathrm{Inv}(\mathfrak{A} + \mathbf{F})$. Furthermore, if $f = (1 - \gamma + \gamma e)^{-1}$ then: b) $\| f \| \leq 2$. c) For all $\epsilon > 0$ there exists an $\eta > 0$ such that if $x \in X$ and $\| ex - x \| \leq \eta \| x \|$ then $\| fx - x \| \leq \epsilon \| x \|$.

The proofs of part (a) and (b) are done below. The proof of part (c) can be found on the Approximate Identities in A for X when X is a Banach Left A-Module 2 page.

• Proof of a) Since $\gamma = (4C)^{-1} = \frac{1}{4C}$ and $C > 1$ we have that:

(1)

\begin{align} \quad 0 < \gamma < \frac{1}{4} \quad (*) \end{align}

• And so:

(2)

\begin{align} \quad 1 - \gamma > \frac{3}{4} \quad (**) \end{align}

• And also:

(3)

\begin{align} \quad \frac{\gamma C}{1 - \gamma} = \frac{(4C)^{-1}C}{1 - \gamma} = \frac{1}{4(1 - \gamma)} < \frac{1}{3} \quad \Leftrightarrow \quad \frac{\gamma}{1 - \gamma} < \frac{1}{3C} \end{align}

• Therefore, since $\| e \| \leq C$ by assumption and from $(***)$ we have that:

(4)

\begin{align} \quad \| \gamma(1 - \gamma)^{-1}e \| = \frac{\gamma}{1 - \gamma} \| e \| < \frac{1}{3C} \cdot C = \frac{1}{3} \end{align}

• Since $\mathfrak{A}$ is a Banach algebra, $\mathfrak{A} + \mathbf{F}$ is a Banach algebra with unit, and since $\| \gamma(1 - \gamma)^{-1}e \| < 1$ from above, by the corollary on the Invertibility of 1 - a When r(a) < 1 in a Banach Algebra with Unit page with $1 - x$ being replaced with $\gamma(1 - \gamma)^{-1}e$ we see that $1 - \gamma(1 - \gamma)^{-1}e \in \mathrm{Inv}(\mathfrak{A} + \mathbf{F})$. Thus:

(5)

\begin{align} \quad 1 - \gamma + \gamma e = [1 - \gamma] \cdot (1 + \gamma(1 - \gamma)^{-1}e) \in \mathrm{Inv}(\mathfrak{A} + \mathbf{F}) \quad \blacksquare \end{align}

• Proof of b) Observe that:

(6)

\begin{align} \quad r(-\gamma(1 - \gamma)^{-1}e) \leq \left \| -\gamma(1 - \gamma)^{-1}e \right \| = \gamma(1 - \gamma)^{-1} \| e \| \leq \gamma(1 - \gamma)^{-1}C < \frac{1}{3} \end{align}

• So $r(-\gamma(1 - \gamma)^{-1}e) < 1$. By the theorem on the Invertibility of 1 - a When r(a) < 1 in a Banach Algebra with Unit page we have that $1 - (-\gamma(1 - \gamma)^{-1}e) = 1 + \gamma(1 - \gamma)^{-1}e$ is invertible in $\mathfrak{A} + \mathbf{F}$ and has inverse:

(7)

\begin{align} \quad [1 + \gamma(1 - \gamma)^{-1}e]^{-1} = 1 + \sum_{n=1}^{\infty} [-\gamma(1 - \gamma)^{-1}e]^n \quad \blacksquare & (1*b) \end{align}

(8)

\begin{align} \quad (1 - \gamma + \gamma e)^{-1} = 1 + \sum_{n=1}^{\infty} [\gamma + \gamma e]^n \end{align}

(9)

\begin{align} \quad \| f \| &= \| (1 - \gamma + \gamma e)^{-1} \| \\ &= \left \| \left [ (1 - \gamma)(1 + \gamma(1 - \gamma)^{-1} e) \right ]^{-1} \right \| \quad & \left ( \mathrm{Since \:} 1 - \gamma + \gamma e = (1 - \gamma)(1 + \gamma(1 - \gamma)^{-1}e) \right )\\ &= (1 - \gamma)^{-1} \left \| [1 + \gamma(1 - \gamma)^{-1}e]^{-1} \right \| \\ & = (1 - \gamma)^{-1} \left \| 1 + \sum_{n=1}^{\infty} [-\gamma(1 - \gamma)^{-1}e]^n \right \| & \left ( \mathrm{By \:} (1*b) \right )\\ &\leq (1 - \gamma)^{-1} \left [ \| 1 \| + \sum_{n=1}^{\infty} \| -\gamma(1 - \gamma)^{-1}e \|^n \right ] & \left ( \mathrm{Triangle \: Inequality} \right )\\ & \leq (1 - \gamma)^{-1} \left [ 1 + \sum_{n=1}^{\infty} [\gamma(1 - \gamma)^{-1} \| e \|]^n \right ] \\ & \leq (1 - \gamma)^{-1} \left [1 + \sum_{n=1}^{\infty} [\gamma(1 - \gamma)^{-1} C]^n \right ] & \left ( \mathrm{Since \:} \| e \| \leq C \right )\\ & < (1 - \gamma)^{-1} \left [ 1 + \sum_{n=1}^{\infty} \left [\frac{1}{3} \right ]^n \right ] & \left ( \mathrm{Since \:} \frac{\gamma C}{1 - \gamma} < \frac{1}{3} \right ) \\ & < (1 - \gamma)^{-1} \cdot \frac{3}{2} & \left ( \mathrm{Since \:} \sum_{n=1}^{\infty} \left [ \frac{1}{3} \right ]^n = \frac{1}{2} \right )\\ & < \frac{4}{3} \cdot \frac{3}{2} & \left (\mathrm{Since \:} 1 - \gamma > \frac{3}{4} \: \Leftrightarrow \: (1 - \gamma)^{-1} < \frac{4}{3} \right ) \\ & < 2 \end{align}