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Given $s\in (0,1)$ and a measurable function $u:\mathbb{R^n}\to\mathbb{C}$, let us define $$\|u\|_{\dot H^s(\mathbb{R}^n)}^2:=\iint\frac{|u(x)-u(y)|^2}{|x-y|^{n+2s}}\,dx\,dy$$ and let $\dot H^s(\mathbb{R}^n)$ denote the completion of $C^\infty_c(\mathbb{R}^n)$ wrt this norm. One can define $H^s(\mathbb{R}^n)$ as the completion of $C^\infty_c(\mathbb{R}^n)$ wrt the stronger norm $\|u\|_{L^2}+\|u\|_{\dot H^s}$, as well.

(1) Clearly, $H^s(\mathbb{R}^n)\subseteq \dot H^s(\mathbb{R}^n)$. Is the inclusion an equality? In light of the open mapping theorem, this is equivalent to ask whether the norms are equivalent on $C^\infty_c(\mathbb{R}^n)$. If not, does $\dot H^s(\mathbb{R}^n)$ contain the constant functions?

(2) If we define instead $\dot H^s(\mathbb{R}^n)$ as the set of measurable functions $u$ such that $\|u\|_{\dot H^s}<\infty$, do we get a larger space?

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  • $\begingroup$ A caveat when dealing with homogeneous Sobolev spaces: the constant function has homogeneous norm 0, so the norm you wrote is not a norm.in the technical sense. $\endgroup$
    – Fan Zheng
    Commented Oct 2, 2016 at 18:07
  • $\begingroup$ Thank you for your comment. I found this definition in some papers; in fact, it is a norm if you restrict to $C^\infty_c$, so that you can still form the completion (but, to be honest, it is not even clear to me if the completion is made of concrete measurable functions...) $\endgroup$
    – Mizar
    Commented Oct 2, 2016 at 19:54

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(1) No, not an equality. Look at the characterization in terms of Fourier transforms (*).

(2) It depends on $s$. For $n=1$ and $\frac12\le s<1$ the completion is a quotient of the semi-Hilbert space defined by the seminorm being finite, with the subspace of constant functions. For all $n$ and $0<s<\frac{n}2$ it is a Hilbert space, and the semi-Hilbert space is the sum of the Hilbert space and the space of constant functions.

(*) $$\|u\|_{\dot H^s(\mathbb{R}^n)}^2:=\iint\frac{|u(x)-u(y)|^2}{|x-y|^{n+2s}}\,dx\,dy\\ =\int|h|^{-n-2s}[\int|u(x+h)-u(x)|^2\,dx]\,dh \\ =\int|h|^{-n-2s}[\int|(e^{ih\cdot\xi}-1)\hat u(\xi)|^2\,d\xi]\,dh \\=\int|h|^{-n-2s}[\int|\frac {e^{ih\xi}-1}{|\xi|}|^2|\xi \hat u(\xi)|^2]\,dh \\=\int|\xi \hat u(\xi)|^2[\int|h|^{-n-2s}|\frac {e^{ih\xi}-1}{|\xi|}|^2\,dh]\,d\xi \\=c\int|\xi \hat u(\xi)|^2|\xi|^{2s-2}\,d\xi$$ with $c=\int|e^{ih_1}-1|^2|h|^{-n-2s}\,dh$

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  • $\begingroup$ For a reference, what I know is what I wrote (40 years ago!) to compute the reproducing kernels of these spaces, among others (with $s\in\mathbb R$). mathoverflow.net/questions/215144/… $\endgroup$
    – Jean Duchon
    Commented Oct 2, 2016 at 21:15
  • $\begingroup$ I am interested in particular in the case $s=\frac 1 2$ and $n=1$. It seems that in this case the continuous embedding $C^\infty_c\hookrightarrow\mathcal S'$ does not extend continuously to the completion. Am I right? If yes, how can we identify the completion with (a quotient of) a suitable function space? $\endgroup$
    – Mizar
    Commented Oct 3, 2016 at 17:01
  • $\begingroup$ @Mizar You're right. Define $V^{1/2}$ as the space of $u\in L^2_{loc}(\mathbb R)$ with $\frac{u(x)-u(y)}{x-y}\in L^2(dx\,dy)$ (this includes constants). The completion is $V^{1/2}/\mathbb R$ (or $/\mathbb C$ if complex). $\endgroup$
    – Jean Duchon
    Commented Oct 4, 2016 at 9:41
  • $\begingroup$ This is interesting: could you give me a reference for the proof of this statement? $\endgroup$
    – Mizar
    Commented Oct 4, 2016 at 10:15
  • $\begingroup$ No, sorry. This is something I worked out for and by myself a long time ago, but didn't actually need. Defining the spaces and using them was enough. I suspect you might not really need this precise result either. Or do you? $\endgroup$
    – Jean Duchon
    Commented Oct 4, 2016 at 15:25

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