Weak Approximate Identities in a Normed Algebra

# Weak Approximate Identities in a Normed Algebra

Definition: Let $X$ be a normed algebra.1) A net $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is a Weak Left Approximate Identity for $X$ if for every $x \in X$ and for every $f \in X^*$ we have that $f(e(\lambda)x) \to f(x)$.2) A net $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is a Weak Right Approximate Identity for $X$ if for every $x \in X$ and for every $f \in X^*$ we have that $f(xe(\lambda)) \to f(x)$.3) A net $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is a Weak Two-Sided Approximate Identity for $X$ if it is both a weak left approximate identity for $X$ and a weak right approximate identity for $X$. |

Note that if $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is a weak left approximate identity for $X$ then for each $x \in X$ and every $f \in X^*$ 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 |f(e(\lambda)x) - f(x)| < \epsilon \end{align}

Proposition 1: Let $X$ be a normed algebra.a) If $X$ has a bounded weak left approximate identity then $X$ has a bounded left approximate identity.b) If $X$ has a bounded weak right approximate identity then $X$ has a bounded right approximate identity. |

**Proof of a)**Suppose that $X$ has a bounded weak left approximate identity. Then there exists a net $\{ e(\lambda) \}_{\lambda \in \Lambda}$ such that for every $x \in X$ and for every $f \in X^*$ we have that:

\begin{align} \quad f(e(\lambda)x) \to f(x) \end{align}

- Let $U$ be the smallest convex set containing $\{ e(\lambda) : \lambda \in \Lambda \}$, that is, $U$ is the closed convex hull of this set.

- We first show that $U$ is a bounded set. Since $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is a bounded net there exists a $K > 0$ such that $\| e(\lambda) \| \leq K$ for all $\lambda \in \Lambda$. Let $u \in U$. Since $U$ is the convex hull of $\{ e(\lambda) : \lambda \in \Lambda \}$ there exists $\lambda_1, \lambda_2, ..., \lambda_n \in \Lambda$ and $0 \leq \alpha_1, \alpha_2, ..., \alpha_n \leq 1$ with $\alpha_1 + \alpha_2 + ... + \alpha_n = 1$ such that:

\begin{align} \quad u = \alpha_1e(\lambda_1) + \alpha_2e(\lambda_2) + ... + \alpha_ne(\lambda_n) \end{align}

- Taking the norm of $u$ gives us that:

\begin{align} \quad \| u \| \leq \alpha_1 \| e(\lambda_1) \| + \alpha_2 \| e(\lambda_2) \| + ... + \alpha_n \| e(\lambda_n) \| \leq K \sum_{k=1}^{n} \alpha_k = K \end{align}

- So every $u \in U$ is such that $\| u \| \leq K$, i.e., $U$ is a bounded set.

- Lastly, since $e(\lambda)x \to x$ weakly, we have that:

\begin{align} \quad x \in \mathrm{weak-closure} \{ e(\lambda)x : \lambda \in \Lambda \} \end{align}

- Let $U'$ be the convex hull of $\{ e(\lambda)x : \lambda \in \Lambda \}$. Since $\{ e(\lambda)x : \lambda \in \Lambda \} \subseteq U'$, we have that:

\begin{align} \quad \mathrm{weak-closure} \{ e(\lambda)x : \lambda \in \Lambda \} \subseteq \mathrm{weak-closure} (U') \end{align}

- So $x \in \mathrm{weak-closure} (U')$. By Mazur's Theorem, we have that $\mathrm{weak-closure} U' = \mathrm{norm-closure} (U')$. So $x \in \mathrm{norm-closure}(U') = \mathrm{norm-closure}(Ux)$. So for a given $\epsilon > 0$ there exists a $u \in U$ such that:

\begin{align} \quad \| x - ux \| < \epsilon \end{align}

- By the proposition on the Approximate Identities in a Normed Algebra page, since $X$ is a normed algebra and $U$ is a bounded subset of $X$ such that for each $x \in X$ and for every $\epsilon > 0$ there exists a $u \in U$ such that $\| x - ux \| < \epsilon$ we have that $X$ has a bounded left approximate identity. $\blacksquare$