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.



Multiplicative version of Novikov inequality for Ito integral
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. 


