Approximate Identities in A for X when X is a Banach Left A-Module 3

Approximate Identities in A for X when X is a Banach Left A-Module 3

Recall from the Approximate Identities in A for X when X is a Banach Left A-Module 1 and Approximate Identities in A for X when X is a Banach Left A-Module 2 pages the following result:

Let $A$ be a Banach algebra and let $X$ be a Banach left $A$-module and regard $X$ as a Banach left $A + \mathbf{F}$ module. Let $e \in A$, $C > 1$, $\| e \| \leq C$, and $\gamma = (4C)^{-1}$. Then:

  • 1. $1 - \gamma + \gamma e \in \mathrm{Inv} (A + \mathbf{F})$.

Furthermore, if $f = (1 - \gamma + \gamma e)^{-1}$ then:

  • 2. $\| f \| \leq 2$.
  • 3. For all $\epsilon > 0$ there exists an $\eta > 0$ such that if $x \in X$ is such that $\| ex - x \| \leq \eta \| x \|$ then $\| fx - x \| \leq \epsilon \| x \|$.

With this result we state and prove the theorem below. On the Approximate Identities in A for X when X is a Banach Left A-Module 4 we prove some corollaries to this theorem.

Theorem 1: Let $A$ be a Banach algebra and let $X$ be a Banach left $A$-module. If $z \in X$ and $\delta > 0$ then there exists an $a \in A$ and a $y \in X$ such that $z = ay$ and $\| z - y \| \leq \delta$.
  • Proof: Let $\{ e(\lambda) \}_{\lambda \in \Lambda}$ be a bounded approximate identity in $A$ for $X$. Since $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is bounded, there exists a number $C > 1$ such that $\| e(\lambda) \| \leq C$ for all $\lambda \in \Lambda$.
  • Set $\gamma = (4C)^{-1}$. By the lemma mentioned at the top of the page, for each $\lambda \in \Lambda$, since $e(\lambda) \in A$, $C > 1$, $\| e(\lambda) \| \leq C$, and $\gamma = (4C)^{-1}$ we have that:
  • 1. $[1 - \gamma + \gamma e(\lambda)] \in \mathrm{Inv} (A + \mathbf{F})$.
  • And if $f(\lambda) = [1 - \gamma + \gamma e(\lambda)]^{-1}$ then also:
  • 2. $\| f(\lambda) \| < 2$.
  • 3. Given $\epsilon > 0$ there exists an $\eta (\lambda) > 0$ such that if $x \in X$ and $\| e(\lambda)x - x \| \leq \eta(\lambda) \| x \|$ then $\| f(\lambda) x - x \| \leq \epsilon \| x \|$.
  • We aim to construct a sequence $\{ \lambda_n \} \subset \Lambda$ with the following properties: For each $n \in \mathbb{N}$ let:
(1)
\begin{align} \quad b_n = (1 - \gamma)^n + \sum_{k=1}^{n} \gamma(1 - \gamma)^{k - 1} e(\lambda_k) \in \mathrm{Inv}(A + \mathbf{F}) \end{align}
  • And for each $n \in \mathbb{N}$:
(2)
\begin{align} \quad \| t_nz - t_{n-1}z\| \leq \delta 2^{-n} \end{align}
  • Where $t_0 = 1$ and $t_n = b_n^{-1}$ for all $n \in \mathbb{N}$.
  • Constructing $\lambda_1$: Choose $\lambda_1 \in \Lambda$ as follows. Given $\epsilon= \delta 2^{-1} > 0$ there exists an $\eta(\lambda_1) > 0$ such that for $x \in X$ and $\| e(\lambda_1)x - x \| \leq \eta(\lambda_1) \| x \|$ then $\| f(\lambda_1)x - x \| \leq \delta 2^{-1} \| x \|$ from (3). In particular, for $z \in X$, since $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is an approximate identity in $A$ for $X$ we certainly have that there exists an $\eta(\lambda_1) > 0$ such that if $\| e(\lambda)z - z \| \leq \eta$ then $\| f(\lambda_1)z - z \| \leq \delta 2^{-1}$.
  • Here we see that
(3)
\begin{align} \quad b_1 &= (1 - \gamma)^1 + \sum_{k=1}^{1} \gamma(1 - \gamma)^{k-1} e(\lambda_k) \\ &= 1 - \gamma + \gamma e(\lambda_1) \in \mathrm{Inv}(A + \mathbf{F}) & \left ( \mathrm{By \: (1)} \right ) \end{align}
  • And $t_1 = b_1^{-1} = f(\lambda_1)$ is such that:
(4)
\begin{align} \quad \| t_1z - t_0z \| = \| b_1^{-1}z - z \| = \| f(\lambda_1)z - z \| \leq \delta 2^{-1} \end{align}
  • Constructing $\lambda_m$ for $m > 1$: Suppose that $\lambda_1$, $\lambda_2$, …, $\lambda_m$ have been chosen as above. We want to show the existence of $\lambda_{m+1}$. For each $\lambda \in \Lambda$ let:
(5)
\begin{align} \quad u(\lambda) = (1 - \gamma)^m + \sum_{k=1}^{m} \gamma(1 - \gamma)^{k-1} f(\lambda)e(\lambda_k) \end{align}
  • Then for each $\lambda \in \Lambda$ we have that:
(6)
\begin{align} \quad u(\lambda) - b_m &= \left [ (1 - \gamma)^m + \sum_{k=1}^{m} \gamma(1 - \gamma)^{k-1} f(\lambda)e(\lambda_k) \right ] - \left [ (1 - \gamma)^m + \sum_{k=1}^{m} \gamma(1 - \gamma)^{k-1} e(\lambda_k) \right ] \\ &= \sum_{k=1}^{m} \gamma(1 - \gamma)^{k-1}[f(\lambda)e(\lambda_k) - e(\lambda_k)] \end{align}
  • So for each $\lambda \in \Lambda$, $\| u(\lambda) - b_m \|$ depends entirely on the values of $\| f(\lambda)e(\lambda_k) - e(\lambda_k) \|$. So $\| u(\lambda) - b_m \|$ can be made arbitrary small by making $\| f(\lambda)e(\lambda_k) - e(\lambda_k) \|$ sufficiently small for each $k \in \{ 1, 2, ..., m \}$. But each $\| f(\lambda)e(\lambda_k) - e(\lambda_k) \|$ can be made arbitrary small by taking $e(\lambda)e(\lambda_k) - e(\lambda_k) \| \leq \kappa$ for each $k \in \{ 1, 2, ..., m \}$ for some $\kappa$ sufficiently small
  • By the theorem on The Set of Invertible Elements Inv(X) is an Open Subset of X page we know that $\mathrm{Inv}(A + \mathbf{F})$ is an open set in $A + \mathbf{F}$. We know that $ \| u(\lambda) - b_m \| $]] can be made arbitrary small and since $b_m \in \mathrm{Inv}(A + \mathbb{F})$ and from the openness of $\mathrm{Inv}(A + \mathbb{F})$ this tells us that each $u(\lambda)$ is invertible.
  • Let $F : \mathrm{Inv}(A + \mathbf{F}) \to \mathrm{Inv}(A + \mathbf{F})$ be defined for all $x \in \mathrm{Inv}(A + \mathbf{F})$ by $F(x) = x^{-1}$. This function is certainly continuous on $\mathrm{Inv}(A + \mathbf{F})$.
  • So if $\| F(u(\lambda) - F(b_m) \| = \| [u(\lambda)]^{-1} - t_m \|$ can be made arbitrary small whenever $\| u(\lambda) - b_m \|$ can be made sufficiently small. But $\| u(\lambda) - b_m \|$ can be arbitrary small whenever $\| e(\lambda) e(\lambda_k) - e(\lambda_k) \| \leq \kappa$ for all $k \in \{ 1, 2, ..., m \}$ for some $\kappa$ sufficiently small.
  • Let $\lambda_{m+1} \in \Lambda$ be such that $\| e(\lambda_{m+1})z - z \| \leq \kappa$ (which can be done since $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is an approximate identity in $A$ for $X$ and $z \in X$), and with $\kappa$ sufficiently small so that $u(\lambda_{m+1}) \in \mathrm{Inv}(A + \mathbf{F})$ and that:
(7)
\begin{align} 2 \| [u(\lambda_{m+1})]^{-1} - t_m \| \| z \| + \| t_m \| \| f(\lambda_{m+1})z - z \| \leq \delta 2^{-(m+1)} \quad (*) \end{align}
  • The above inequality can hold for some $\lambda_{m+1} \in \Lambda$ and $\kappa$ sufficiently small since $\| [u(\lambda_{m+1})]^{-1} - t_m \|$ can be made arbitrary small by choice of $\kappa$ sufficiently small, and $\| f(\lambda_{m+1})z - z \|$ can be made small by choice of $\lambda_{m+1} \in \Lambda$.
  • Now by definition, since $f(\lambda) = [1 - \gamma + \gamma e(\lambda)]^{-1}$ we have that $[1 - \gamma + \gamma e(\lambda)] f(\lambda) = 1$, and so for all $\lambda \in \Lambda$:
(8)
\begin{align} \quad [1 - \gamma + \gamma e(\lambda)] u(\lambda) &= [1 - \gamma + \gamma e(\lambda)] \left [ (1 - \gamma)^m + \sum_{k=1}^{m} \gamma(1 - \gamma)^{k-1} f(\lambda) e(\lambda_k) \right ] \\ &= (1 - \gamma)^m - \gamma(1 - \gamma)^m + \gamma(1 - \gamma)^m e(\lambda) + \sum_{k=1}^{m} \gamma(1 - \gamma)^{k-1} e(\lambda_k) \\ &= (1 - \gamma)^{m+1} + \gamma(1 - \gamma)^m e(\lambda) + \sum_{k=1}^{m} \gamma(1 - \gamma)^{k-1} e(\lambda_k) \end{align}
  • So plugging in $\lambda = \lambda_{m+1}$ and we get:
(9)
\begin{align} \quad [1 - \gamma + \gamma e(\lambda_{m+1})] u(\lambda_{m+1}) = &= (1 - \gamma)^{m+1} + \gamma(1 - \gamma)^m e(\lambda_{m+1}) + \sum_{k=1}^{m} \gamma(1 - \gamma)^{k-1} e(\lambda_k) \\ &= (1 - \gamma)^{m+1} + \sum_{k=1}^{m+1} \gamma(1 - \gamma)^{k-1} e(\lambda_k) \\ &= b_{m+1} \end{align}
  • So $b_{m+1}$ is a product of two invertible elements in $A + \mathbf{F}$ and so $b_{m+1} \in \mathrm{Inv}(A + \mathbf{F})$, and the inverse of $t_{m+1} := [b_{m+1}]^{-1}$ is given by:
(10)
\begin{align} \quad t_{m+1} = b_{m+1}^{-1} = \left [ [1 - \gamma + \gamma e(\lambda_{m+1})] u(\lambda_{m+1}) \right ]^{-1} = [u(\lambda_{m+1})]^{-1} f(\lambda_{m+1}) \end{align}
  • Hence:
(11)
\begin{align} \quad \| t_{m+1}z - t_mz \| &= \| [u(\lambda_{m+1})]^{-1} f(\lambda_{m+1})z - t_mz \| \\ &= \| [u(\lambda_{m+1})]^{-1}f(\lambda_{m+1})z - t_mf(\lambda_{m+1})z + t_mf(\lambda_{m+1})z - t_mz \| \\ & \leq \| [u(\lambda_{m+1})]^{-1}f(\lambda_{m+1})z - t_mf(\lambda_{m+1})z \| + \| t_mf(\lambda_{m+1})z - t_mz \| \\ & \leq \| \left [ [u(\lambda_{m+1})]^{-1} - t_m \right ] f(\lambda_{m+1})z \| + \| t_m[f(\lambda_{m+1})z - z] \| \\ & \leq \| [u(\lambda_{m+1})]^{-1} - t_m \| \underbrace{\| f(\lambda_{m+1}) \|}_{< \: 2 \: \mathrm{by \:} (2)} \| z \| + \| t_m \| \| f(\lambda_{m+1})z - z \| \\ & < 2 \| [u(\lambda_{m+1})]^{-1} - t_m \| \| z \| + \| t_m \| \| f(\lambda_{m+1}z - z \| \\ & < \delta 2^{-(m+1)} & \left (\mathrm{By \:} (*) \right ) \end{align}
  • Conclusion: Therefore, such a sequence $\{ \lambda_n \}$ exists. For each $n \in \mathbb{N}$ let $y_n = t_n z$. Then for each $n \in \mathbb{N}$, we have that $z = t_n^{-1}y_n = b_ny_n$.
  • Now $\{ y_n \} = \{ t_nz \}$ is a Cauchy sequence since $\| y_{m+1} - y_m \| = \| t_{m+1}z - t_mz \| \leq \delta 2^{-(m + 1)}$ for every $m \in \mathbb{N}$. Since $X$ is a Banach left $A$-module and $\{ y_n \}$ is Cauchy in $X$ we have that $\{ y_n \}$ converges to some $y \in X$. Furthermore, $y \in X$ is such that:
(12)
\begin{align} \quad \| z - y \| = \left \| z - \lim_{n \to \infty} y_n \right \| = \left \| z - \lim_{n \to \infty} t_nz \right \| = \lim_{n \to \infty} \| z - t_nz \| = \lim_{n \to \infty} \| t_0z - t_nz \| \leq \delta \end{align}
  • Where the last inequality above comes from the fact that $\| t_{m+1}z - t_mz \| \leq \delta 2^{-(m+1)}$ for all $m \in \mathbb{N}$ and thus for every $n \in \mathbb{N}$:
(13)
\begin{align} \quad \| t_0z - t_nz \| \leq \sum_{k=1}^{n} \| t_{k-1}z - t_kz \| \leq \delta \sum_{k=1}^{n} 2^{-k} < \delta \end{align}
  • Additionally, the sequence $(b_n)$ to some $a \in A$ since:
(14)
\begin{align} \quad b_n &= \lim_{n \to \infty} \left [ (1 - \gamma)^n + \sum_{k=1}^{n} \gamma(1 - \gamma)^{k-1}e(\lambda_k) \right ] \\ &= \underbrace{\lim_{n \to \infty} (1 - \gamma)^m}_{=0 \: \mathrm{since \:} \frac{3}{4} < 1 - \gamma < 1} + \sum_{k=1}^{\infty} \gamma(1 - \gamma)^{k-1}e(\lambda_k) \\ \end{align}
  • Where the righthand series converges because it is absolutely convergent since:
(15)
\begin{align} \quad \sum_{k=1}^{\infty} \| \gamma(1 - \gamma)^{k-1}e(\lambda_k) \| = \sum_{k=1}^{\infty} \gamma(1 - \gamma)^{k-1} \| e(\lambda_k) \| < 2 \gamma \underbrace{\sum_{k=1}^{\infty} (1 - \gamma)^{k-1}}_{< \infty \: \mathrm{since \: 1 - \gamma < 1}} \end{align}
  • And because $A$ is a Banach algebra, absolute convergence of a series implies convergence of that series. Thus:
(16)
\begin{align} \quad z = \lim_{n \to \infty} b_ny_n = ay \end{align}
  • Where $a \in A$ and $y \in X$ and $\| z - y \| \leq \delta$$\blacksquare$
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