Multiplicative version of Novikov inequality for Ito integral

Question

It is clear that Ito isometry
$E(∫^t_0fdW)^2=E(∫^t_0f^2dt)$
can be written in the multiplicative form as
$E(∫^t_0fdW\cdot∫^t_0gdW)=E(∫^t_0f⋅gdt).$

Is it possible to obtain the multiplicative version of the Novikov inequality $E(|∫^t_0fdW|^p)≤B_pE(∫^t_0|f|^2dt)^{p/2}?$

It should take a form like:
$E(|∫^t_0fdW|^{p/2}\cdot |∫^t_0gdW|^{p/2})≤B_pE(∫^t_0|f|⋅|g|dt)^{p/2}.$

Even in the case p=2, it is unclear for me if something like
$E(|∫^t_0fdW|⋅|∫^t_0gdW|)≤B_pE(∫^t_0|f|⋅|g|dt)$ can hold thrue.

Of course its weaker version $E(|∫^t_0fdW|⋅|∫^t_0gdW|)≤∥f∥_{L_p}⋅∥g∥_{L_q}$
is obvious.

Answer 1

This cannot be true: just think of the case where $f$ and $g$ have disjoint supports. Then the right hand side vanishes but the left hand side does not.