Let's consider the domain $U=[-\pi,\pi]\times[-1,1]$. Assume that we have two functions $f\in H^2$ and $g\in H^{1/2}$. 


I wonder if the following bound is true:


$$
\|f g\_{x\_1}\|\_{H^{-0.5}(U)}\leq C(\|f\|\_{H^2})\|g\|\_{H^{1/2}}.\quad (1)
$$


I tried using the duality pairing. Then, for a given $h\in H^{1/2}$ with $\|h\|\_{H^{1/2}}\leq 1$, we have
$$
\langle fg\_{x\_1},h\rangle\_{<H^{-1/2},H^{1/2}>}=\int\_{-1}^{1}\int\_{-\pi}^{\pi} fg\_{x\_1}hdx\_1dx\_2=-
\int\_{-1}^{1}\int\_{-\pi}^{\pi} |D|^{1/2}(f h)|D|^{1/2}Hgdx\_1dx\_2,
$$ 
where I used the notation $H$ for the (periodic) Hilbert transform in the variable $x\_1$ and $|D|u=H u\_{x\_1}$, i.e. 
$$
|D|=\sqrt{-\frac{d^2}{dx\_1^2}}.
$$
Then, 
$$
\int\_{-1}^{1}\int\_{-\pi}^{\pi} |D|^{1/2}(f h)|D|^{1/2}Hgdx\_1dx\_2\leq \|fh\|\_{H^{1/2}(U)}\|g\|\_{H^{1/2}(U)}.
$$
So, if multiplication by a $H^2(U)$ function is a continuous operator in $H^{1/2}$, i.e.
$$
\|fh\|\_{H^{1/2}(U)}\leq C\|f\|\_{H^2(U)}\|h\|\_{H^{1/2}(U)},\quad (2)
$$
then, the previous bound (1) would hold.


Consequently, my questions are


A) Is (1) true?


B) Is (2) true?


PD: Note that $f$ is not Lipschitz (otherwise, I will have the desired result). 



A bound in Sobolev spaces of negative order
Yes, it is true. As you already found, it suffices to show that $H^2$ functions are multipliers in $H^{1/2}$. This follows by interpolation, since it is easy to show that $H^2$ functions are multipliers in both $L^2$ and $H^1$.


