Construct a mollifier operator? Is there a mollifier operator $\mathcal{J}_\epsilon$ such that for any $f$ and $g$
$$\mathcal{J}_\epsilon(fg) = \mathcal{J}_\epsilon(f) \mathcal{J}_\epsilon(g)\  ? $$
If exists, how can we construct it ?
 A: You should probably specify the class of functions to which you want to let $f$ and $g$ belong, and it would be good to precisely define what you want the operator to do.  I'm going to assume that $J_\epsilon$ takes a function and spits out a smooth function, and that $J_\epsilon f \to f$ in $L^1_{loc}$.  Moreover, I'm going to assume that the operator can be applied to all simple functions, i.e. functions of the form $f = \sum_{i=1}^M a_i \chi_{E_i}$ where $a_i \in \mathbb{R}$ and $E_i$ is a measurable set of finite measure.  
Now let $E$ be a measurable set of positive finite measure and consider $f = \chi_E$.  Then $f^2 = f$ and so we can compute
$$
J_\epsilon(f) = J_\epsilon(f^2) = J_\epsilon(f) J_\epsilon(f) = [J_\epsilon(f)]^2.
$$
This means that $J_\epsilon f(x) \in \{0,1\}$ for every $x$, but the assumption that $J_\epsilon f$ is smooth implies that either $J_\epsilon f = 1$ or else $J_\epsilon f =0$.  Our choice of $E$ then means it's impossible to have the convergence $J_\epsilon f \to f$ in $L^1_{loc}$ as $\epsilon \to 0$.   
Note: this argument also works if $J_\epsilon$ just makes functions continuous.  Smoothness is overkill.
