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Let $f, g \in L^2[\mathbb{T}^2]$ be real-valued functions without zero modes. That is, $\int_{\mathbb{T}^2}f=\int_{\mathbb{T}^2}g=0$. Here, ${\mathbb{T}^2}$ is the $2$-dimensional torus $[\mathbb{R}/\mathbb{Z}]^2$.

Now, consider the quantity \begin{equation} (-\Delta)^{-3}\Bigl[ fg- \int_{\mathbb{T}^2} fg \Bigr] \in W^{6,1}[\mathbb{T}^2] \end{equation}

Then I wonder if it is possible to find positive constants $C_i$'s independent of $f,g$ such that \begin{equation} \Bigl \lVert (-\Delta)^{-3}\Bigl[ fg- \int_{\mathbb{T}^2} fg \Bigr] \Bigr \rVert_1 \leq \sum_{i=0}^3 C_i\lVert \Delta^{-3+i} f \rVert_2 \lVert \Delta^{-i} g \rVert_2 \end{equation}

I tried to use the Fourier expansion, but the product turns into convolution and things seem to get tricky...All references I looked for deal only with the Leibniz rule for the fractional Laplacian of positive exponent.

Could anyone help me with the negative exponent case?

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    $\begingroup$ It is hopeless even on $\mathbb T$ and for $(-\Delta)^{-1}$. Take $f,g$ with Fourier coefficients $1/\sqrt{N}$ on $[-2N,-N]\cup[N,2N]$ and $0$ elsewhere. Their $L^2$ norms are about $1$ but the $L^2$-norm of their inverse Laplacians are like $N^{-2}$. However the product $fg$ has Fourier coefficients of size $1$ on the entire interval $[-N/2,N/2]$, say, so just killing the zeroth Fourier coefficient will not bring the $L^1$ (or any other) norm of $(-\Delta)^{-1}(fg)$ down. $\endgroup$
    – fedja
    Commented Jul 25, 2023 at 22:18
  • $\begingroup$ Thank you for your comment. At least isn't it true that the zeroth mode of the Fourier expansion must be eliminated if we want to act on the inverse Laplacian? $\endgroup$
    – Isaac
    Commented Jul 29, 2023 at 12:45
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    $\begingroup$ Of course, that is true. The point is that it still doesn't help much... $\endgroup$
    – fedja
    Commented Jul 29, 2023 at 12:57
  • $\begingroup$ Yes I just wanted to check for sure. Thank you. $\endgroup$
    – Isaac
    Commented Jul 29, 2023 at 12:58

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