A bound in Sobolev spaces of negative order 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?
 A: Yes, (2) is true (which implies (1) as you noted). Here is a more general claim: 
Claim. If $f\in C^\alpha(U)$ with $\alpha>\frac12$ and $h\in H^{1/2}(U)$, then $fh\in H^{1/2}(U)$ and $$\|fh\|_{H^{1/2}} \le  C\|f\|_{C^{\alpha}} \|h\|_{H^{1/2}} \tag1$$
The above claim is dimension-independent. Since in your case $U$ is two-dimensional, $H^{2}(U)$ embeds into $C^\alpha$ for all $\alpha\in (0,1)$. 
Proof of claim follows the proof of Theorem 7.16 in Analysis by Lieb and Loss. In this theorem $f$ is assumed smooth with bounded derivative (which is enough for them), but there is some obvious slack in the proof, removing which gives the claim.
Indeed, in $n$ dimensions  we have $$\|h\|^2_{H^{1/2}} \approx \iint_{U\times U} \frac{|h(x)-h(y)|^2}{|x-y|^{n+1}}\,dx\,dy \tag2$$ (Theorem 7.12 in the same book). Since 
$$|f(x)h(x)-f(y)g(y)| \le |f(x)-f(y)| |h(x)| + |f(y)| |h(x)-h(y)|$$
it follows that 
$$|f(x)h(x)-f(y)g(y)|^2 \le 2|f(x)-f(y)|^2 |h(x)|^2 + |f(y)|^2 |h(x)-h(y)|^2 \tag3$$
Plug  (3)  into (2). The second term of (3) is bounded by $\|f\|_{C^\alpha}^2 |h(x)-h(y)|^2$, so that's good (remember the $C^\alpha$ norm contains the supremum of the function). 
In the  first term of (3) we use the Hölder  estimate $|f(x)-f(y)|^2\le \|f\|_{C^\alpha}^2|x-y|^{2\alpha}$, which leads to
$$\iint_{U\times U} \frac{|h(x)|^2}{|x-y|^{n+1-2\alpha }}\,dx\,dy \tag{4}$$
Integrate (4) over $y$: the result is finite, because $n+1-2\alpha<n$, and is bounded by something dependent only on the diameter of $U$. Then integrate over $x$. $\Box$  
