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?

  • $\begingroup$ 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 Oct 2 '16 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 Oct 2 '16 at 19:54

(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 $c=\int|e^{ih_1}-1|^2|h|^{-n-2s}\,dh$

| cite | improve this answer | |
  • $\begingroup$ 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 Oct 2 '16 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 Oct 3 '16 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 Oct 4 '16 at 9:41
  • $\begingroup$ This is interesting: could you give me a reference for the proof of this statement? $\endgroup$ – Mizar Oct 4 '16 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 Oct 4 '16 at 15:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.