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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?

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1 Answer 1

up vote 2 down vote accepted

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 $x$. $\Box$

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