Let $\{k_{\lambda}\}_{\lambda > 0}$ be an approximate identity on $[0,1]$. By definition, $\{k_{\lambda}\}_{\lambda >0}$ has the following properties:
Normalization: $\int_0^1 k_{\lambda}(x)\,dx = 1$ for every $\lambda > 0$,
$L^1$-boundedness: $\sup_{\lambda}\|k_{\lambda}\|_1 < \infty$,
$L^1$-concentration: for every $0 < \delta <\frac{1}{2}$,
$$ \lim_{\lambda \to \infty}\int_{\delta}^{1-\delta}|k_{\lambda}(x)|\,dx = 0. $$
$\textbf{Conjecture.}$ Let $f,g\in L^1[0,1]$ such that $fg\in L^1[0,1]$. Moreover, let $\{k_{\lambda}\}_{\lambda > 0}$ be an approximate identity on $[0,1]$. Then, $$ \lim_{\lambda \to \infty} \|fg - (f*k_{\lambda})(g*k_{\lambda})\|_1 \to 0. $$
My failed attempt is the following: \begin{align*} \|fg-(f*k_{\lambda})(g*k_{\lambda})\|_1 &= \int_0^1|f(x)g(x) - (f*k_{\lambda})(x)(g*k_{\lambda})(x)|\,dx \\ &=\int_0^1\left|f(x)g(x)\left(\int_0^1k_{\lambda}(t)\,dt\right)\left(\int_0^1k_{\lambda}(\tau)\,d\tau\right)-\int_0^1f(x-t)k_{\lambda}(t)\,dt\int_0^1g(x-\tau)k_{\lambda}(\tau)\,d\tau\right|\,dx \\ &=\int_0^1\left|\int_0^1f(x)k_{\lambda}(t)\,dt\int_0^1g(x)k_{\lambda}(\tau)\,d\tau - \int_0^1f(x-t)k_{\lambda}(t)\,dt\int_0^1g(x-\tau)k_{\lambda}(\tau)\,d\tau\right|\,dx \\ &=\int_0^1\left|\int_0^1\int_0^1f(x)g(x)k_{\lambda}(t)k_{\lambda}(\tau)\,dt\,d\tau - \int_0^1\int_0^1f(x-t)k_{\lambda}(t)g(x-\tau)k_{\lambda}(\tau)\,dt\,d\tau\right|\,dx \\ &=\int_0^1\left|\int_0^1\int_0^1\left(f(x)g(x)-f(x-t)g(x-\tau)\right)k_{\lambda}(t)k_{\lambda}(\tau)\,dt\,d\tau\right|\,dx \\ &\leq \int_0^1\int_0^1\int_0^1\left|f(x)g(x)-f(x-t)g(x-\tau)\right|\,|k_{\lambda}(t)|\,|k_{\lambda}(\tau)|\,dt\,d\tau\,dx \\ &=\int_0^1\int_0^1\int_0^1\left|f(x)g(x)-f(x-t)g(x-\tau)\right|\,|k_{\lambda}(t)|\,|k_{\lambda}(\tau)|\,dx\,dt\,d\tau \\ &=\int_0^1|k_{\lambda}(\tau)|\int_0^1|k_{\lambda}(t)|\int_0^1\left|f(x)g(x)-T_tf(x)T_{\tau}g(x)\right|\,dx\,dt\,d\tau. \end{align*} However, Continuity of product of $L^1$ functions with respect to translation stops this line of reasoning as $T_tfT_{\tau}g$ does not necessarily converge to $fg$. This does not quite disprove the claim, as I'm taking the absolute value inside the integral. However, I'm stuck.
Additionally, this question is similar to When is the convolution of a product the product of convolutions?, which never got answered. Specifically, if $fg\in L^1[0,1]$, we know that $$ \lim_{\lambda \to \infty}\|fg - (fg*k_{\lambda})\|_1 = 0. $$
Any advice or counterexample would be appreciated.
I'm also interested in the slightly weaker property $$ \int_0^1 f(x)g(x)\,dx = \lim_{\lambda\to\infty}\int_0^1(f*k_{\lambda})(x)(g*k_{\lambda})(x)\,dx, $$ if it is more feasible.
