# Fourier transform for $H^2(\mathbb{R}^N)$, $N\geq 5$

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

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.
• Sorry, @WillieWong! I dind't understand how $\hat{u}\in L^q$ implies $u\in \mathcal{F}(L^q)$? Jan 15, 2020 at 15:20
• @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? Jan 15, 2020 at 21:30
• Yes, @WillieWong! I wanted to say that $u=\mathcal{F}(v)$ for some $v \in L^{q}\left(\mathbb{R}^{N}\right)$, $1\leq q\leq 2$. Jan 15, 2020 at 22:35
• @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 $u$). Jan 16, 2020 at 14:00