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I was reading the lecture note

https://perso.math.u-pem.fr/cannone.marco/harmonic_analysis_tools.pdf (Lecture note by Marco Canone)

In his note page 8, (or exactly before section 1.2), he says that

'if two (scalar) functions $f$ and $g$ are in $H^1$, their product only belongs to $H^{1/2}$ and their derivative $\partial(fg)$ is even less regular as it belongs to $H^{−1/2}$.'

I cannot know why it is true.... even we don't know if $fg\in L^2$. Is it actually true?

I guess I need to estimate $$\int (1+|\xi|^2)^{1/2}\hat{f}(\xi)*\hat{g}(\xi)d\xi$$ but I am not sure how to estimate even.

He is usually considering a function in this note as a tempered distribution but that would not affect my question (I guess).

Thank in advance for any help.

Lev Bahn
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  • The situation in unbounded domains is much different to usual results for bounded domains.. – S. Maths Jul 25 '20 at 17:13
  • @S.Maths yeah I think so. Then this case is for unbounded domain since, in the lecture note, the focus is on $\mathbb{R}^3$ – Lev Bahn Jul 25 '20 at 17:31
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    I doubt $fg\in W^{1/2,2}$ is true. The usual condition $k_1+k_2-k\geq n(1/p_1+1/p_2-1/p)\geq 0$ is violated for $k_1=k_2=1$, $k=\frac12$, $p_1=p_2=p=2$, $n=3$. But we do have $fg\in L^2$. – user10354138 Jul 25 '20 at 23:31
  • @user10354138 Thank you for your comment! But I found this link.

    https://math.stackexchange.com/q/2005748/523306

    is the answer there wrong?

     – Lev Bahn Jul 26 '20 at 00:05
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    You don't just have $f,g\in L^2$ here, but $f,g\in W^{1,2}$. Note the example $f(x)=g(x)=x^{-1/4}1_{x\in(0,1)}$ is in $L^2(\mathbb{R})$ but not in $W^{1,2}$. – user10354138 Jul 26 '20 at 00:12
  • @user10354138 Thanks but that sounds still mysterious to me. Is there any reference that I can look up for that? – Lev Bahn Jul 26 '20 at 00:21
  • Look in Haim Brezis, Functional Analysis, Sobolev spaces and PDEs. In corollary 8.10 it says that the product of two functions in $W^{1,p}$ is still in $W^{1,p}$ and the usual formula for differentiation holds. This seems to contradict the claim. – Beni Bogosel Apr 01 '22 at 12:54
  • @BeniBogosel that is for dimension 1 and relies on $W^{1,p}$ embedding into $L^\infty$. This result in higher dimensions is also true if you replace with $W^{k,p}$ where $kp>n$. – Calvin Khor Jan 18 '24 at 19:26

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