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Assume that $f\in W^{\alpha-1,p}(R^n)$ with $0<\alpha<1$ and $p>2n/\alpha$.

Given another function $ g\in W^{\beta,p}(R^n)$ with $\beta>0$.

Under what conditions on $\beta$ can we get that $fg\in L^p(R^n)$?

Many thanks for the answer!

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  • $\begingroup$ Thanks! The problem here is that the function $f$ is in the Sobolev space with negative index. So sobolev embedding theorems seems not to be helpful. $\endgroup$
    – Wenguang Zhao
    Commented Mar 31, 2018 at 20:10
  • $\begingroup$ Yes, sorry, I noticed that afterwards and so deleted my comment. $\endgroup$
    – Nate Eldredge
    Commented Mar 31, 2018 at 20:12
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    $\begingroup$ Not sure how is $fg$ defined: by duality, $(fg,\phi):=(f,g\phi)$? Even if $g$ is infinitely smooth (say, $g(x)=\exp(-|x|^2)$), $fg$ need not be any more regular than $f$: $fg \in L^p$ would mean $|(fg,\phi)|\le C\|\phi\|_q$, that is, $|(f,g\phi)|\le C\|g^{-1}g\phi\|_q$, and so $f\in L^p(|g(x)|^pdx)$. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Mar 31, 2018 at 21:29
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    $\begingroup$ Note that even if $g$ is smooth, rapidly decaying, and equal to $1$ on an open set, it's hard to imagine a reasonable condition for $fg$ to be in $L^p$, except that $f$ is already in $L^p$. $\endgroup$
    – Deane Yang
    Commented Mar 31, 2018 at 21:48
  • $\begingroup$ Thanks for the answer. I am wondering whether it is meaningful in the Bony paraproducts sense ? $\endgroup$
    – Wenguang Zhao
    Commented Apr 1, 2018 at 15:17

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