# Equivalent Norms for the Dual of Sobolev / Bessel Spaces

Using standard notation, we refer to $H^s(\mathbb R) = W^{s,2}(\mathbb R)$ to be the Sobolev Hilbert spaces. As is often the case, it's natural to then consider properties of functions in $H^s(\mathbb R)$ by looking at how it behaves in the Fourier domain, more specifically, we have that for any $s\in \mathbb{R}$,

\begin{align*} H^s(\mathbb{R}) = \{ g \in L_2(\mathbb{R}) : \int (1+t^2)^{s}|\mathcal{F} g(t)|^2dt <\infty\}, \end{align*}

where $\mathcal{F}$ is the Fourier transform. In particular to note is that this works for $s<0$ as well, and so negative order Sobolev spaces are easily defined. In particular, one can define the norm of $H^s(\mathbb{R})$ to be defined by the relation $\|g\|_s = \int (1+t^2)^{s}|\mathcal{F}g(t)|^2dt$.

$\textbf{My first question:}$ I have seen, in particular the book Sobolev Spaces by Adams, Fournier (page 64), that another way to define the norm of the dual space of $H^s$, which appears to be $H^-s$ is given by the following: \begin{align*} \|g\|^*_{-s} = \sup_{h\in H^{s}}\frac{\langle g,h\rangle}{\|h\|_s}, \end{align*}

where $\langle g,h\rangle$ is the standard $L_2$ inner product. I am having difficulty figuring out whether the norm defined by Fourier transformations or the one defined by using the dual are equivalent or not. The furthest I have been able to show is that $\|g\|^*_{-s} \leq (2\pi)^{-1}\|g\|_{-s}$ by Parceval's relation.

$\textbf{My second question:}$ If there appears to be a relationship, then I would like to restrict attention to spaces of the form $H^{s}(A)$ where $A\subset \mathbb{R}$ with defined norm $\|g\|_{s,A} = \inf\{ \|g^{'}\|_s : g^{'}_{|A} = g\}$ for all $g\in H^s(A)$. Then is there a relationship between $\|g\|_{s,A}$ and

\begin{align*} \|g\|_{-s,A}^* = \sup_{h\in H^s(A)}\frac{\langle g,h\rangle_{A}}{\|h\|_{s,A}}, \end{align*}

where $\langle g,h\rangle_{A}$ is the standard $L_2$ inner product restricted to $A$? I am having difficulty trying to relate the two, because there seems to be a natural relationship globally, which I would hope to think there is a local relationship as well.

• First of all, your definition of $H^s$ is only correct for $s \ge 0$. If $s$ is negative, then $H^s$ contains tempered distributions which are not functions. – Mark Meckes Jun 24 '14 at 8:36
• For your first question, you just need to pick an appropriate $h$ to bound $\| g \|_{-s}^*$ from below: use basically $\mathcal{F}h = (1+t^2)^s \mathcal{F} g$ (or something very much like that). – Mark Meckes Jun 24 '14 at 8:39
• I don't know the answer to your second question, but I would guess it's no in general. Derivatives of non-integer order are not local operators. – Mark Meckes Jun 24 '14 at 8:41
• Hi Mark, thanks for all your comments, I do have a couple of comments to make: 1. I am a bit confused by your first comment, seeing as it seems that you can create the Sobolev spaces by looking at an appropriate Hilbert scale, that is to say if we considered the following linear operator $B: L_2(\mathbb{R}) \mapsto L_2(\mathbb{R})$, such that $Bf = f - f^{(2)}$. Then if I have understood correctly, $G_s = Dom(B^{s/2})$ represents the Sobolev spaces $H^s$ under the following inner product, $\langle f,g\rangle_s = \langle B^{s/2}f, B^{s/2}g\rangle$. Unless I am missing something – user61038 Jun 25 '14 at 0:33
• The point of my first comment is that for $s < 0$, you need to replace $L^2$ with $\mathcal{S}'$. For example, $H^{-1}(\mathbb{R})$ contains delta functions, which are not in $L^2$. As for your second question, this is well outside my expertise, so I'm just guessing but: probably you can do something like this when $s$ is an integer, but you may need to assume some kind of regularity of $A$. – Mark Meckes Jun 25 '14 at 6:39

Q1 has been addressed by Mark in the comments (absolute values are missing in your formula), but let me quickly look at this again: Just take Fourier transforms to see that $$\|g\|_{-s}^*=\sup_{\|h\|_s=1} |\langle g,h\rangle | = \sup \left\{ |\langle \widehat{g}, \widehat{h}\rangle | : \|(1+t^2)^{s/2}\widehat{h}\|_2 = 1 \right\} = \sup \left\{ |\langle (1+t^2)^{-s/2}\widehat{g}, (1+t^2)^{s/2}\widehat{h} \rangle | : \ldots\right\}=\|g\|_{-s} .$$ Notice that in the third $\sup$, we can simply interpret $\langle. , .\rangle$ as the standard scalar product of $L^2(\mathbb R)$; this identifies this $\sup$ as the $L^2$ norm of the first argument of the scalar product, which is $\|g\|_{-s}$.
As for Q2, this formula will fail badly for restrictions because (unlike the case $U=\mathbb R$) the action of $H^{-s}(U)$ distributions on test functions does not extend to $H^s(U)$ functions. First of all, we probably want to insist on open sets $A=U$, so that we can meaningfully restrict distributions. (We restrict a distribution to an open set by only applying it to test functions with support in that set.) Then consider, for example, $g(x)=1/x$ on $U=(0,1)$. Then $g\in H^{-1/2-\epsilon}(U)$ according to your definition because $g$ is the restriction of $PV-1/x$ to $U$. However, if we test against an $h\in H^{1/2+\epsilon}(0,1)$ with $h=1$ near zero, then $\langle g, h\rangle$ isn't even defined.
• Two small corrections: firstly absolute values are not required in the formula for the norm and, secondly, one can restrict distributions not just to open sets. In particular to closed intervals, which would, in my opinion, be a more natural setting for this question. (Of course, the OP doesn't specify which kind of $A$ he/she is interested in so that is pure speculation on my part). – blackburne Jun 27 '14 at 7:51
• @blackburne: In my book, spaces are by default over $\mathbb C$, and then we definitely want to insist on absolute values. You are (of course) right that I can omit the absolute value for real spaces. – Christian Remling Jun 27 '14 at 18:37