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For $\alpha>0^{[1]}$, let $D^\alpha=(-\Delta)^{\frac{\alpha}{2}}$ be the operator defined on all Schwartz class functions $f$ as follows: $$ \widehat{D^\alpha f}(\xi)=\lvert \xi\rvert^\alpha \hat{f}(\xi), $$ the hat denoting Fourier transform. We have the following Sobolev inequality: $$\tag{Sob}\| f\|_{L^p(\mathbb{R}^n)}\le C \| D^\alpha f\|_{L^{q}(\mathbb{R}^n)},\qquad \frac{n}{p}=\frac{n}{q}-\alpha.$$ Now let $$\dot{W}^{\alpha, q}(\mathbb{R}^n)$$ denote the completion of the Schwartz class with respect to the norm $$\| f\|_{\dot{W}^{\alpha, q}(\mathbb{R}^n)}=\| D^\alpha f \|_{L^q(\mathbb{R}^n)}.$$

My question is: is it true that $$\dot{W}^{\alpha, q}(\mathbb{R}^n)=\left\{f\in L^p(\mathbb{R}^n)\ :\ \frac{n}{p}=\frac{n}{q}-\alpha\ \text{and}\ \| D^\alpha f \|_{L^q(\mathbb{R}^n)}<\infty \right\}?$$ Here $D^\alpha$ is taken in the tempered distributional sense.

$^{[1]}$ Let us not consider the case $\alpha <0$ for simplicity.

Giuseppe Negro
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    The quantity $|D^\alpha f|{L^q}$ is a seminorm, not a norm. So, your definition of $\dot W^{\alpha,q}$ identifies functions that differ by a low-degree polynomial. Apart from this technicality, I think the statement is true. The set ${f\in L^p \colon \dots}$ is closed with respect to $|D^\alpha|{L^q}$, and test functions are dense in it. So, it is a concrete model of the abstract completion. –  Aug 01 '14 at 05:33
  • @900sit-upsaday: I must admit that I do not fully understand your remark. Adding a polynomial to a Schwartz class function we do not obtain a Schwartz class function; therefore, $|D^\alpha f|_{L^q}$ defines a norm on the Schwartz class. Am I right? – Giuseppe Negro Aug 01 '14 at 11:03
  • Okay, my first two sentences can be omitted. Is there an issue with the equality of two spaces? Perhaps I am missing some subtlety. –  Aug 02 '14 at 03:43
  • @900sit-upsaday: I don't think there are issues, since it seems to me that your argument perfectly applies. I am asking because I find those "completion" things to be somewhat slippery (see here ). – Giuseppe Negro Aug 02 '14 at 10:01
  • In the case $\alpha=n/2, q=2$ this would imply $\dot{H}^{n/2}\subset L^\infty$ which is not true. It seems that it would be necessary for $\alpha<n/q$. Which is sort of implied by what you wrote. – mathematician Aug 20 '14 at 19:28
  • I have been kindly directed to the following arXiv paper: http://arxiv.org/abs/1202.3970 in which Theorem 2.2 should contain the answer to the present question. – Giuseppe Negro Apr 03 '16 at 15:39
  • They correspond to the spaces if Sobolev functions that vanish at infinity https://math.stackexchange.com/questions/4584970/understanding-the-space-d1-2-mathbbrn/4585092#4585092 – LL 3.14 Nov 06 '23 at 16:23

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