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).


1 Answer 1


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 $H^1$.


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