Approximate Identities in A for X when X is a Banach Left A-Module 2

We continue the proof of the following lemma. The proof of parts (a) and (b) can be found on the Approximate Identities in A for X when X is a Banach Left A-Module 1 page.

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 $\| fx - x \| \leq \epsilon \| x \|$.

• Proof of c) Let $\epsilon > 0$ be given. Observe that the numerical series $\displaystyle{\sum_{n=1}^{\infty} \gamma^n(1 - \gamma)^{-n}C^n = \sum_{n=1}^{\infty} [\gamma(1 - \gamma)^{-1}C]^n}$ converges by the comparison test since:

(1)

\begin{align} \quad \sum_{n=1}^{\infty} [\gamma(1 - \gamma)^{-1}C]^n \leq \sum_{n=1}^{\infty} \left [ \frac{1}{3} \right ]^n = \frac{1}{2} < \infty \end{align}

• So there exists an $N \in \mathbb{N}$ such that:

(2)

\begin{align} \quad (1 - \gamma)^{-1} \sum_{n=N+1}^{\infty} \gamma^n(1 - \gamma)^{-n}C^n < \frac{\epsilon}{4} \quad (1*c) \end{align}

• Let $\eta > 0$ be such that:

(3)

\begin{align} \quad \eta (1 - \gamma)^{-1} \sum_{n=1}^{N} \gamma^n(1 - \gamma)^{-n}(1 + C + ... + C^{n-1}) < \frac{\epsilon}{2} \quad (2*c) \end{align}

• Now we have already established that $(1 - \gamma + \gamma e)^{-1} = (1 - \gamma)^{-1} \left [ 1 + \sum_{n=1}^{\infty} [-\gamma (1 - \gamma)^{-1}e]^n \right ]$ on the previous page and so:

(4)

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

• Now observe that when $e = 1$ we have that $1 - \gamma + \gamma e = 1$, and so from above we get that:

(5)

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

• Therefore:

(6)

\begin{align} \quad fx - x &= (f - 1)x \\ &= \left \{ (1 - \gamma)^{-1} \left [ 1 + \sum_{n=1}^{\infty} \gamma^n (\gamma - 1)^{-n} e^n \right ] - (1 - \gamma)^{-1} \left [ 1 + \sum_{n=1}^{\infty} \gamma^n (\gamma - 1)^{-n} \right ] \right \} x \\ &= (1 - \gamma)^{-1} \sum_{n=1}^{\infty} \left [ \gamma^n(\gamma - 1)^{-n}e^n - \gamma^n (\gamma - 1)^{-n} \right ] x \\ &= (1 - \gamma)^{-1} \sum_{n=1}^{\infty} \gamma^n(\gamma - 1)^{-n}(e^n - 1)x \\ &= (1 - \gamma)^{-1} \sum_{n=1}^{\infty} \gamma^n(\gamma - 1)^{-n}(e^nx - x) \end{align}

• Taking the norm of the above gives us:

(7)

\begin{align} \quad \| fx - x \| &= \left \| (1 - \gamma)^{-1} \sum_{n=1}^{\infty} \gamma^n(\gamma - 1)^{-n}(e^nx - x) \right \| \\ & \leq (1 - \gamma)^{-1} \sum_{n=1}^{\infty} \| \gamma^n(\gamma - 1)^{-n}(e^nx - x) \| \\ & \leq (1 - \gamma)^{-1} \sum_{n=1}^{\infty} \gamma^n(\gamma - 1)^{-n} \|e^nx - x \| \\ & \leq (1 - \gamma)^{-1} \sum_{n=1}^{N} \gamma^n(1 - \gamma)^{-n} \|e^nx - x \| + (1 - \gamma)^{-1} \sum_{n=N+1}^{\infty} \gamma^n(1 - \gamma)^{-n} \|e^nx - x \| \\ \end{align}

• But observe that:

(8)

\begin{align} \quad (1 - \gamma)^{-1} \sum_{n=N+1}^{\infty} \gamma^n(1 - \gamma)^{-n} \| e^nx - x \| &= (1 - \gamma)^{-1} \sum_{n=N+1}^{\infty} \gamma^n(1 - \gamma)^{-n} \| (e^n - 1)x \| \\ & \leq (1 - \gamma)^{-1} \sum_{n=N+1}^{\infty} \gamma^n (1 - \gamma)^{-n} \| e^n - 1 \| \| x \| \\ & \leq (1 - \gamma)^{-1} \sum_{n=N+1}^{\infty} \gamma^n (1 - \gamma)^{-n} [\| e^n \| + 1] \| x \| \\ & \leq (1 - \gamma)^{-1} \sum_{n=N+1}^{\infty} \gamma^n (1 - \gamma)^{-n} [C^n + 1] \| x \| \\ & \leq (1 - \gamma)^{-1} \sum_{n=N+1}^{\infty} \gamma^n (1 - \gamma)^{-n} C^n \cdot 2\| x \| & \left ( \mathrm{Since \:} 1 < C \leq C^n, \: \mathrm{so \:} C^n + 1 < 2C^n \right ) \\ & < \frac{\epsilon}{4} \cdot 2 \| x \| & \left ( \mathrm{By \:} (1*c) \right )\\ & < \frac{\| x \|}{2} \end{align}

• Suppose that $\| ex - x \| \leq \eta \| x \|$. Then:

(9)

\begin{align} \quad (1 - \gamma)^{-1} \sum_{n=1}^{N} \gamma^n (1 - \gamma)^{-n} \| e^nx - x \| &= (1 - \gamma)^{-1} \sum_{n=1}^{N} \gamma^n (1 - \gamma)^{-n} \| (1 + e + ... + e^{n-1})(ex - x) \| \\ & \leq (1 - \gamma)^{-1} \sum_{n=1}^{N} \gamma^n (1 - \gamma)^{-n} [\| 1 \| + \| e \| + ... + \| e^{n-1} \|]\| (ex - x) \| \\ & \leq (1 - \gamma)^{-1} \sum_{n=1}^{N} \gamma^n (1 - \gamma)^{-n} [1 + C + ... + C^{n-1}] \| (ex - x) \| \\ & \leq \frac{\epsilon}{2 \eta} \| ex - x \| & \left ( \mathrm{By \:} (2*c) \right )\\ & < \frac{\epsilon}{2 \eta} \cdot \eta \| x \| \\ & < \frac{\epsilon}{2} \| x \| & \end{align}

• From the last two sets of inequalities above we see that if $ex - x \| \leq \eta \| x \|$ then $\| fx - x \| < \epsilon \| x \|$. $\blacksquare$