Put $\ell^{1}(\mathbb Z)=\{f:\mathbb Z \to \mathbb C: \|f\|_{\ell^{1}}:=\sum_{n\in \mathbb Z}|f(n)|< \infty \}$ and we note that $\ell^{1}(\mathbb Z)$ is an algebra under pointwise multiplication.
My Question is: Does $\ell^{1}(\mathbb Z)$ have a bounded approximate identity with respect to pointwise multiplication? (In other words, does there exist $\{e_{r}\}_{r>0} \subset \ell^{1}$ such that $\|e_{r}\|_{\ell^{1}} \leq C$ for all $r>0$ and some constant $C$ and $\|e_{r}f-f\|_{\ell^{1}} \to 0$ as $r\to 0$ for $f\in \ell^{1}$)
No, this doesn't exists.
Consider the sequences $b^{(i)}i$ such that $b^{(i)}_n = 1$ if $i=n$ and $0$ elsewhere
Take an arbitrary $N \in \Bbb N$. Then as $e^{(r)}$ are an approximation of the identity, there exists $n_0$ such that $\forall n > n_0$, $\forall 1 \leq i \leq N$, $\| e^{(n)} b^{(i)} - b^{(i)} \| \leq \epsilon$
This imply that $\forall i \in [|1,N|]$, you have $1-\epsilon \leq e^{(n)}_i \leq 1+\epsilon$.
Hence $\forall n\geq n_0, \|e^{(n)}\| \geq N$
So the sequence $e^{(n)}$ is unbounded