An Sobolev-type inequality on $1D$ torus related to algebra properties for Sobolev spaces

Question

In Kishimoto-Tsutsumi's paper published by Math Research Letter 2018, I see the following inequality in the last line of page 10:

$\| f g \partial_x h \|_{H^{-s}(T)} \lesssim \| f \|_{H^s(T)} \| g \|_{H^s(T)} \| h \|_{H^s(T)}.$

They claimed this is true for $s > \frac{1}{2}.$ But I can only prove this for $s > \frac{3}{4}.$

So I want to know whether this is true for $s \in (\frac{1}{2},\frac{3}{4}]$ and an intuition for this kind of inequality. (On $\mathbb{R}^d,$ scaling is a useful technique to determine the integrability exponent. But on $\mathbb{T}^d,$ I don't have any idea.)

Recall the definitions briefly:

$T = [-\pi,\pi], \widehat{F}(k)$ is the Fourier coefficient of function $F \in L^1(T).$

For $\sigma \in \mathbb{R},$ we define the standard Sobolev norm $\| F \|^2_{H^{\sigma}(T)}:= \sum_k \langle k \rangle^{2s} | \widehat{F} (k) |^2 = \| \langle \cdot \rangle^s \widehat{F} \|_{l^2}^2,$ where $\langle k \rangle:= 1 + |k|.$

The following is my proof for $s > \frac{3}{4}$:

\begin{align*} \| f g \partial_x h \|_{H^{-s}} &:= \Big( \sum_k \langle k \rangle^{-2s} \big| \sum_l \widehat{fg}(k-l) \widehat{\partial_x h}(l) \big|^2 \Big)^{\frac{1}{2}} \\& = \Big( \sum_k \big| \sum_l \frac{\langle l \rangle^s}{\langle k-l \rangle^s \langle k \rangle^s } \langle k-l \rangle^s \widehat{fg}(k-l) \langle l \rangle^{-s} \widehat{\partial_x h}(l) \big|^2 \Big)^{\frac{1}{2}}, \\& \lesssim \Big( \sum_k \big( \sum_l \langle k-l \rangle^s \big| \widehat{fg}(k-l) \big| \langle l \rangle^{-s} \big| \widehat{\partial_x h}(l) \big| \big)^2 \Big)^{\frac{1}{2}}, \\& \lesssim \| \langle \cdot \rangle^s \widehat{fg} \|_{l^2} \| \langle \cdot \rangle^{-s} \widehat{\partial_x h} \|_{l^1}. \end{align*} (Note that the third line is a simple consequence of $|l| \leq |k| + |k-l|$ and the fourth line is Young's convolution inequality.)

Note that since $\langle k \rangle^s \lesssim \langle l \rangle^s + \langle k-l \rangle^s $, by Young's convolution inequality again we have \begin{align*} \| \langle \cdot \rangle^s \widehat{fg} \|_{l^2} \lesssim \| \langle \cdot \rangle^s \widehat{f} \|_{l^2} \| \widehat{g} \|_{l^1} + \| \langle \cdot \rangle^s \widehat{g} \|_{l^2} \| \widehat{f} \|_{l^1}. \end{align*} So $\| fg \|_{H^s} \lesssim \| f \|_{H^s} \| g \|_{H^s}$ since $s > \frac{1}{2}.$ Using the same idea as proving $\| \widehat{g} \|_{l^1} \lesssim \| g \|_{H^s},$ we have \begin{align*} \| \langle \cdot \rangle^{-s} \widehat{\partial_x h} \|_{l^1} &= \sum_k \frac{|k|}{\langle k \rangle^{2s}} \langle k \rangle^s |\widehat{h}(k)| \\& \leq \Big( \sum_k \frac{|k|^2}{\langle k \rangle^{4s}} \Big)^{\frac{1}{2}} \Big( \sum_k \langle k \rangle^{2s} |\widehat{h}(k)|^2 \Big)^{\frac{1}{2}}. \end{align*} Using integral comparison test, we see the first bracket is finite if and only if $2 - 4s + 1 < 0 \Leftrightarrow s > \frac{3}{4}.$

Answer 1

Solved. The range is $s > \frac{1}{2}.$ According to a short discussion with Prof. Tsutsumi, what I miss is that I should use

$\frac{\langle l \rangle^s}{\langle k \rangle^s \langle k-l \rangle^s}$ decays like $\langle k \rangle^{-s}$ if $|l| \gg |k|$ and $\langle k-l \rangle^{-s}$ if $|l| \sim |k|$

instead of just saying that this ratio is bounded. More precisely,

\begin{align*} \| f g \partial_x h \|_{H^{-s}(\mathbb{T})} &:= \Big( \sum_k \langle k \rangle^{-2s} \big| \sum_l \widehat{fg}(k-l) \widehat{\partial_x h}(l) \big|^2 \Big)^{\frac{1}{2}} \\& \lesssim \Big( \sum_k \big| \sum_{|l| \leq 10 |k|} \frac{1}{\langle k-l \rangle^s \langle k \rangle^s} \langle k-l \rangle^s \widehat{fg}(k-l) \widehat{\partial_x h}(l) \big|^2 \Big)^{\frac{1}{2}} \\& \,\, + \Big( \sum_k \langle k \rangle^{-2s} \big| \sum_{|l| > 10 |k|} \frac{\langle l \rangle^s}{\langle k-l \rangle^s} \langle k-l \rangle^s \widehat{fg}(k-l) \langle l \rangle^{-s} \widehat{\partial_x h}(l) \big|^2 \Big)^{\frac{1}{2}} \\& \lesssim \Big( \sum_k \big( \sum_{|l| \leq 10|k|} \langle k-l \rangle^s \big| \widehat{fg}(k-l) \big| \langle l \rangle^{-2s} \big| \widehat{\partial_x h}(l) \big| \big)^2 \Big)^{\frac{1}{2}} \\& \,\, + \Big( \sum_k \langle k \rangle^{-2s} \Big)^{\frac{1}{2}} \| \langle \cdot \rangle^s \widehat{fg} * \langle \cdot \rangle^{-s} \widehat{\partial_x h} \|_{l^\infty} \\& \lesssim \| \langle \cdot \rangle^s \widehat{fg} \|_{l^2} \| \langle \cdot \rangle^{-2s} \widehat{\partial_x h} \|_{l^1} + \Big( \sum_k \langle k \rangle^{-2s} \Big)^{\frac{1}{2}} \| \langle \cdot \rangle^s \widehat{fg} \|_{l^2} \| \frac{|\cdot|}{\langle \cdot \rangle^{2s}}\langle \cdot \rangle^s \widehat{h} \|_{l^2}. \end{align*}