The Quasi-Product of Bounded Nets with Left/Right Approx. Identities

The Quasi-Product of Bounded Nets with Left/Right Approximate Identities

Lemma 1: Let $X$ be a normed algebra.
a) If $\{ f(\mu) \}_{\mu \in M}$ is a bounded net in $X$ and $\{ e(\lambda) \}_{\lambda \in \Lambda}$ be a left approximate identity for $X$ then $\{ f (\mu) \circ e(\lambda) \}_{(\lambda, \mu) \in \Lambda \times M}$ is a left approximate identity for $X$.
b) If $\{ f(\mu) \}_{\mu \in M}$ is a right approximate identity for $X$ and $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is a bounded net then $\{ f (\mu) \circ e(\lambda) \}_{(\lambda, \mu) \in \Lambda \times M}$ is a right approximate identity for $X$.

For each $\lambda \in \Lambda$ and each $\mu \in M$, $f(\mu)$ and $e(\lambda)$ are elements in $X$ and $f(\mu) \circ e(\lambda)$ denotes the quasi-product of these two elements.

  • Proof of a): Let $\Lambda \times M$ be the cartesian product of the sets directed sets $(\Lambda, \leq)$, $(M, \leq)$, where $\Lambda \times M$ is also directed such that if $(\lambda_1, \mu_1), (\lambda_2, \mu_2) \in \Lambda \times M$ is such that $(\lambda_1, \mu_1) \leq (\lambda_2, \mu_2)$ if and only if $\lambda_1 \leq \lambda_2$ and $\mu_1 \leq \mu_2$. (Here, there is some abuse of notation. Of course, the direction relation on $\Lambda$, $M$, and $\Lambda \times M$ is not necessarily the same, but we will use the same symbol, $\leq$, in all three regards.)
  • Since $\{ f(\mu) \}_{\mu \in M}$ is a bounded net, there exists an $K > 0$ such that for every $\mu \in M$, $\| f(\mu) \| \leq K$.
  • Let $x \in X$. Since $\{e(\lambda) \}_{\lambda \in \Lambda}$ is a left approximate identity for $X$ we have that $e(\lambda)x \to x$. So for $\frac{\epsilon}{1 + K} > 0$ there exists a $\lambda_0 \in \Lambda$ such that if $\lambda \geq \lambda_0$ then:
(1)
\begin{align} \quad \| e(\lambda)x - x \| < \frac{\epsilon}{1+K} \end{align}
  • Take any $\mu_0 \in M$. Then if $(\lambda, \mu) \geq (\lambda_0, \mu_0)$ then $\mu \geq \mu_0$ and so:
(2)
\begin{align} \quad \| [f(\mu) \circ e(\lambda)]x - x \| &= \| [f(\mu) + e(\lambda) - f(\mu)e(\lambda)]x - x \| \\ &= \| f(\mu)x + e(\lambda)x - f(\mu)e(\lambda)x - x \| \\ &= \| [e(\lambda)x - x] + f(\mu)[x - e(\lambda)x] \| \\ & \leq \| e(\lambda)x - x \| + \| f(\mu) \| \| x - e(\lambda)x \| \\ & \leq [1 + \| f(\mu) \|] \| e(\lambda)x - x \| \\ & \leq [1 + K] \frac{\epsilon}{1 + K} \\ & < \epsilon \end{align}
  • So $[f(\mu) \circ e(\lambda)]x \to x$ for every $x \in X$. Thus $f(\mu) \circ e(\lambda)$ is a left approximate identity for $X$. $\blacksquare$
  • Proof of b) Since $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is bounded there exists a $K > 0$ such that $\| e(\lambda) \| \leq K$ for all $\lambda \in \Lambda$.
  • Let $x \in X$ and let $\epsilon > 0$ be given. Since $xf(\mu) \to x$, for $\frac{\epsilon}{1 + K} > 0$ there exists a $\mu_0 \in M$ such that if $\mu \geq \mu_0$ then:
(3)
\begin{align} \quad \| xf(\mu) - x \| < \epsilon \end{align}
  • Take any $\lambda_0 \in \Lambda$. Then if $(\lambda, \mu) \geq (\lambda_0, \mu_0)$ then $\mu \geq \mu_0$ and so:
(4)
\begin{align} \quad \| x[f(\mu) \circ e(\lambda)] - x \| &= \| x[f(\mu) + e(\lambda) - f(\mu)e(\lambda)] - x \| \\ &= \| xf(\mu) + xe(\lambda) - xf(\mu)e(\lambda) - x \| \\ &= \| [xf(\mu) - x] + [x - xf(\mu)]e(\lambda) \| \\ &\leq \| xf(\mu) - x \| + \| x - xf(\mu) \| \| e(\lambda) \| \\ &\leq \| xf(\mu) - x \| [1 + \| e(\lambda) \|] \\ &< \frac{\epsilon}{1 + K} [1 + K] \\ &< \epsilon \end{align}
  • Therefore $x[f(\mu) \circ e(\lambda)] \to x$ for every $x \in X$ and so $\{ f(\mu) \circ e(\lambda) \}_{(\lambda, \mu) \in \Lambda \times M}$ is a right approximate identity for $X$. $\blacksquare$
Corollary 2: Let $X$ be a normed algebra.
a) If $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is a bounded left approximate identity for $X$ and $\{ f(\mu) \}_{\mu \in M}$ is a bounded net in $X$. then $\{ f (\mu) \circ e(\lambda) \}_{(\lambda, \mu) \in \Lambda \times M}$ is a bounded left approximate identity for $X$.
b) If $\{ f(\mu) \}_{\mu \in M}$ is a bounded right approximate identity for $X$ and $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is a bounded net in $X$ then $\{ f (\mu) \circ e(\lambda) \}_{(\lambda, \mu) \in \Lambda \times M}$ is a bounded right approximate identity for $X$
  • Proof: By lemma 1 we have that $\{ f(\mu) \circ e(\lambda) \}_{(\lambda, \mu) \in \Lambda \times M}$ is a left approximate identity for $X$. Since $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is bounded there exists an $K_1 > 0$ such that $\| e(\lambda) \| \leq K_1$ for all $\lambda \in \Lambda$, and since $\{ f(\mu) \}_{\mu \in M}$ is bounded there exists an $K_2 > 0$ such that $\| f(\mu) \| \leq K_2$ for all $\mu \in M$.
  • Let $K = K_2 + K_1 + K_2K_1$. Then for all $(\lambda, \mu) \in \Lambda \times M$ we have that:
(5)
\begin{align} \quad \| f(\mu) \circ e(\lambda) \| &= \| f(\mu) + e(\lambda) - f(\mu)e(\lambda) \| \\ & \leq \| f(\mu) \| + \| e(\lambda) \| + \| f(\mu) \| \| e(\lambda) \| \\ & \leq K_2 + K_1 + K_2K_1 \\ & \leq K \end{align}
  • So $\{ f(\mu) \circ e(\lambda) \}_{(\lambda, \mu) \in \Lambda \times M}$ is a bounded left approximate identity for $X$. $\blacksquare$
Corollary 2: Let $X$ be a normed algebra. Let $\{ e(\lambda) \}_{\lambda \in \Lambda}$ be a bounded left approximate identity for $X$ and let $\{ f(\mu) \}_{\mu \in M}$ be a bounded right approximate identity for $X$. Then $\{ f(\mu) \circ e(\lambda) \}_{(\lambda, \mu) \in \Lambda \times M}$ is a bounded two-sided approximate identity for $X$.
  • Proof: Since $\{ f(mu) \}_{\mu \in M}$ is a bounded net and $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is a left approximate identity for $X$ we have by lemma 1 and corollary2 that $\{ f(\mu) \circ e(\lambda) \}_{(\lambda, \mu) \in \Lambda \times M}$ is a bounded left approximate identity for $X$.
  • Similarly, since $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is a bounded net and $\{ f(\mu) \}_{\mu \in M}$ is a right approximate identity for $X$ we again have by lemma 1 and corollary 2 that $\{ f(\mu) \circ e(\lambda) \}_{(\lambda, \mu) \in \Lambda \times M}$ is a bounded right approximate identity for $X$.
  • Thus $\{ f(\mu) \circ e(\lambda) \}_{(\lambda, \mu) \in \Lambda \times M}$ is a bounded two-sided approximate identity for $X$. $\blacksquare$
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