$L^1[0,1]$ has no unit element with certain convolution, but a net which acts as a unit for $i\in I$

Question

I want to show that the space $L^1[0,1]$ with the product $$ (f*g)(t) = \int_0^t f(t-s) g(s) \, ds,\quad f,g\in L^1[0,1],~t\in[0,1] $$ is a commutative banach algebra (Volterra algebra) and that it has no unit element, but a net $(e_i)_{i\in I}$ such that $$ \lim_{i\in I} e_i * f = f,\quad \sup_{i\in I} \| e_i \| < \infty. $$ Showing that it is indeed a commutative banach algebra is easy, the difficult part are the other two questions. For the unit element, one can use fourier transformation in $L^1(\mathbb{R})$ with the normal convolution to show that it has no unit element. However, this is not possible here, since we are on $[0,1]$ and since our convolution integrates over $[0,t]$. I think one could adapt the proof below, however we cannot use Riemann Lebesgue, since the integral is only on $[0,t]$, does anyone know, how we could still show it? Or maybe something completely different? I also did not manage to find a net for the last statement, so maybe someone has an idea for that too.

Answer 1

The proof from the question works if one takes the absolute value of $e_k*g$ and $\phi_k$, because then we can just let the integral go to 1 instead of $t$.

Answer 2

The Lesbegue Differentiation Theorem implies that $f \ast g$ is differentiable almost everywhere. Not every $f \in L^1$ is almost everywhere equal to an almost everywhere differentiable function. Therefore $f \ast g = f$ cannot hold for every $f \in L^1$.