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How i can prove that if $u\in H^2(\mathbb{R}^N)$ then $u\in \mathcal{F}(L^{p^*}(\mathbb{R}^N))$, where $1/p+1/{p^*}=1,$ $2\leq p<2N/(N-4)$?

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If $u \in H^2(\mathbb{R}^N)$, then its Fourier transform satisfies $\hat{u} \in L^2$ and $(1 + |\xi|^2) \hat{u} \in L^2$. By Holder inequality you have

$$ \|\hat{u}\|_{q} \leq \| (1 + |\xi|^2)^{-1} \|_{r} \|(1 + |\xi|^2) \hat{u} \|_{2} $$

for appropriate $q^{-1} = r^{-1} + 2^{-1}$. For the $L^r$ integration to be bounded you need $2r > N$. Work through the algebra you get what you want.

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  • $\begingroup$ Sorry, @WillieWong! I dind't understand how $\hat{u}\in L^q$ implies $u\in \mathcal{F}(L^q)$? $\endgroup$
    – Pádua
    Jan 15, 2020 at 15:20
  • $\begingroup$ @Pádua maybe I misunderstood your notation, but I thought you meant by $u\in \mathcal{F}(L^q)$ that $u$ has a Fourier transform that is in $L^q$. Do you mean something else? $\endgroup$ Jan 15, 2020 at 21:30
  • $\begingroup$ Yes, @WillieWong! I wanted to say that $u=\mathcal{F}(v)$ for some $v \in L^{q}\left(\mathbb{R}^{N}\right)$, $1\leq q\leq 2$. $\endgroup$
    – Pádua
    Jan 15, 2020 at 22:35
  • $\begingroup$ @Pádua: then is it not what I just stated? You can easily go from the Fourier transform $\hat{u}$ of the function $u$ to the function $v$ (which is just the inverse Fourier transform of $u$). $\endgroup$ Jan 16, 2020 at 14:00
  • $\begingroup$ Sorry @WillierWong, but i don't know how go from the Fourier transform to inverse. There exists some relation between them? $\endgroup$
    – Pádua
    Jan 16, 2020 at 20:02

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