When is $(a_jb_j)_j \subset L^2[0,1]$ complete in $L^2[0,1]$ provided that both $(a_j)_j$ and $(b_j)_j$ are complete? Suppose that $A=(a_j)_j \subset L^4[0,1], B=(b_j)_j \subset L^4[0,1]$ are two complete sequences in $L^2[0,1]$, i.e. whenever $f \in L^2[0,1]$ such that $\langle f,a_j\rangle = 0$ for all $j$ then $f=0$ almost everywhere (the same holds for the sequence $(b_j)_j$).
Problem: When do the above assumptions imply that the sequence $C=(a_jb_j)_j$ is complete in $L^2[0,1]$? Here, $a_jb_j$ is simply the pointwise product of $a_j$ and $b_j$. Note that by Hölder's inequality $a_jb_j$ is in $L^2[0,1]$ since $a_j$ and $b_j$ are in $L^4[0,1]$.
I'm wondering if there are any results (papers, ...) in this direction. Thanks in advance for any help!
 A: This  was a bit too long for a comment.
Let me give an example when this is not  true. For convenience I work  on $[-1,1]$.  Take $a_j=b_j=x^j$, $j=0,1,\dotsc$.  Weierstrass' theorem implies that they are complete systems of any $L^p$, $p\in [1,\infty)$. On the other hand, the collection  $a_jb_j=x^{2j}$ only spans "half" of $L^2[-1,1]$, namely the half consisting of even functions.
Let me give you a modified statement that it is true.  $\newcommand{\eS}{\mathscr{S}}$Suppose that $(\Omega, \eS, \mu)$  is a  finite  measured space such that the sigma algebra $\eS$ is countably  generated. (E.g.,  $\Omega$ could be  a separable metric space and $\eS$ is its Borel sigma-algebra.) Then for any $p\in [1,\infty)$ the space  $L^p(\Omega,\eS,\mu)$ is separable.   Consider  now two  countable  collections$\newcommand{\eA}{\mathscr{A}}\newcommand{\eB}{\mathscr{B}}$
$$
\eA,\eB\subset  L^4(\Omega,\eS,\mu)
$$ such that $\DeclareMathOperator{\spa}{span}$$\spa(\eA)$ and $\spa(\eB)$ are dense in $L^4$. Then  the collection
$$
\eA\ast \eB:=\big\{ ab;\;\; a\in \eA,\;\;b\in \eB\big\}
$$
spans a dense subspace of $L^2$.  To see this it suffices   to show that  for any measurable subset $S\subset \Omega$  its indicator  $\newcommand{\bsI}{\boldsymbol{I}}$$\bsI_S$  belongs to the closure of $\spa(\eA\ast\eB)$. To see this, pick sequences $\alpha_n\in \spa(\eA)$, $\beta_n\in \spa(\eB)$ such that
$$
\lim_{n\to \infty} \bigl\lVert \alpha_n-\bsI_S\bigr\rVert_{L^4}= \lim_{n\to \infty} \bigl\lVert \beta_n-\bsI_S\bigr\rVert_{L^4}=0.
$$
Then $\alpha_n\beta_n\to \bsI_S^2=\bsI_S$ in $L^2$. Clearly $\alpha_n\beta_n\in \spa(\eA\ast\eB)$.
A: Not and answer; another counterexample...  Showing that even assumptions this strong do not suffice.
Let's try the usual trigonometric orthonormal basis.
$$
a_n(t) = e^{2\pi i n t},\qquad n \in \mathbb Z
$$
Then $a_n \in L^\infty$ and continuous on $[0,1]$.  Also, $(a_n)_{n \in \mathbb Z}$ is complete.
Let $b_n = a_n$ for all $n$.  Of course
$(b_n)_{n \in \mathbb Z}$ is complete.
But
$$
a_n(t)b_n(t) = e^{4\pi i n t},\qquad n \in \mathbb Z
$$
and the sequence $(a_nb_n)_{n \in \mathbb Z}$ is not complete.  Indeed, $a_1 = e^{2\pi i t}$ is orthogonal to all $a_nb_n$.
