Approximate identity

Remarks

Example: let $A = C_0(\R)$, the continuous functions on $\R$ that tend to $0$ at infinity, with pointwise product and the sup norm. For $n \ge 1$ let $e_n \in A$ be continuous with $0 \le e_n \le 1$, $e_n = 1$ on $[-n, n]$ and $e_n = 0$ outside $[-n-1, n+1]$. For $f \in A$, $\norm{e_n f - f}_\infty \le \sup_{\abs{x} > n} \abs{f(x)} \to 0$. So $(e_n)$ is a bounded approximate identity, with $\norm{e_n} = 1$, although $A$ has no unit.

Non-example: in $L^1(\R)$ with convolution, let $e_n = \tfrac{1}{2n} \mathbf{1}_{[-n, n]}$. For $f \in L^1(\R)$, $\abs{(e_n * f)(x)} \le \norm{f}_1 / (2n)$, so $e_n * f \to 0$ uniformly. If also $e_n * f \to f$ in $L^1$, a subsequence would converge to $f$ almost everywhere, so $f = 0$. Thus the convergence condition fails for every $f \ne 0$. Spikes $u_n = \tfrac{n}{2} \mathbf{1}_{[-1/n, 1/n]}$ do give a bounded approximate identity of $L^1(\R)$.

A unit $1$ of $A$ gives the constant approximate identity $e_\lambda = 1$. Every C*-algebra has an approximate identity of positive elements of norm at most $1$.