Sobolev space with negative index

Question

The Dirac delta function is in the Sobolev space $H^{-1/2-\epsilon}(\mathbb{R})=W^{-1/2-\epsilon,2}(\mathbb{R})$ for $\epsilon>0$, but it is a distribution as opposed to a function in the traditional sense. On the other hand we have $H^{0}(\mathbb{R})=L^{2}(\mathbb{R})$, so I presume $H^{s}(\mathbb{R})$ only has traditional functions for $s\geq0$. Now what can I say for $-1/2\leq s<0$? We can naturally expect something larger than an $L^2$ function but I don't know if there are any non-traditional functions there. I tried like $\delta^{1/2}(x)$ but it doesn't really seem to be a distribution.

Answer 1

The following is based on what LL sketched in the comments using the Cantor function, which can be modified to find elements in $H^s$ for each $s<0$ that are not represented by locally integrable functions. My answer may seem a bit unsatisfying as it relies on some technical machinery and is not entirely explicit, but hopefully I can convince you that it is a fairly natural apporach. I will use results from the following reference

Mattila, Pertti, Geometry of sets and measures in Euclidean spaces. Fractals and rectifiability, Cambridge Studies in Advanced Mathematics. 44. Cambridge: Univ. Press. xii, 343 p. (1995). ZBL0819.28004.

Motivation and technical results

I will start with some motivation for the construction, which will also serve as an opportunity for me to state some technical results I will need later. When looking for distributions that are not locally integrable, a natural starting point is to look at Radon measures (zeroth order distributions) which are compactly supported. For such measures $\mu$ the structure of their Fourier-Stieltjes transform gives a close connection to the fractal dimension of its support, which is (sort of) seen in the following result.

Lemma (12.12 from Mattila): Let $\mu$ be a compactly supported Radon measure on $\Bbb R.$ Then for $0<s<1$ there is a constant $c(s)>0$ such that $$ I_s(\mu) = c(s)\int_{\Bbb R} |\xi|^{s-1} |\hat\mu(\xi)|^2 \,\mathrm{d}\xi,$$ where $I_s$ is the $s$-energy defined as $$ I_s(\mu) = \int_{\Bbb R}\int_{\Bbb R} |x-y|^{-s} \,\mathrm{d}\mu(x) \,\mathrm{d}\mu(y). $$

Since $s-1<0,$ we have $(1+|\xi|^2)^{\frac{s-1}2} \leq |\xi|^{s-1}$ and so we evidently have $\mu \in H^{\frac{s-1}2}(\Bbb R)$ provided $I_s(\mu)<\infty.$ This recasts our problem into finding subsets with finite $s$-energy, but what exactly is this quantity? (And what does this have to do with fractal dimension?)

These $s$-energies can be used to define (see 8.4 in Mattila) the Riesz $s$-capacity of a set $A \subset \Bbb R$ as $$ C_s(A) = \sup\left\{ \frac1{I_s(\mu)} : \mu \in \mathcal M(\Bbb R),\, \operatorname{supp} \mu \subset A,\, \mu(\Bbb R) = 1 \right\}.$$ That is, we consider $1/I_s(\mu)$ over all Radon probability measures supported on $A.$ Whether this quantity is non-zero or not is closely related to the fractal dimension of the associated set $A,$ and a consequence of the so-called Frostman's lemma we have the following.

Theorem (8.9 in Mattila): Let $A \subset \Bbb R$ be a Borel set and $s>0.$ If $C_s(A) = 0,$ then $\mathcal H^t(A) = 0$ for all $t>s.$

And thus we have reduced the problem of finding a $H^s$ function to finding a set of suitable fractal dimension. The idea is that if $A$ has Hausdorff dimension $s$ which is close but strictly less that $1,$ then it will lie in a small but negative $H^{-t}$ space but will not be represented by a locally integrable function. Using these results the construction follows fairly easily, and we also see a close connection between these two seemingly unrelated topics.

Main construction

I will show that for all $s \in (0,1/2),$ there exists a Radon measure $\mu \in H^{-s}(\Bbb R)$ which is compactly supported and singular with respect to the Lebesgue measure; that is it is not represented by a locally integrable function.

The construction follows by the two results, and is a matter of putting everything together and by exhibiting sets with the correct fractal dimension. This is where we need the Cantor-type construction, whose construction I will sketch from Section 4.10 in Mattila.

Fix $\lambda \in (0,1/2),$ and let $I_{0,1} = [0,1].$ We inductively define closed intervals $I_{k,1},\dots,I_{k,2^k}$ for each $k\geq 0$ by deleting the middle $(1-2\lambda)$ of each $I_{k-1,j}$ to obtain $I_{k,2j-1}, I_{k,2j}.$ To clarify the construction the first few terms are \begin{align*} &I_{0,1} = [0,1] \\ &I_{1,1} = [0,\lambda], \ \ I_{1,2} = [1-\lambda,1] \\ &I_{2,1} = [0,\lambda^2], \ \ I_{2,2} = [\lambda(1-\lambda),\lambda], \ \ I_{2,3} = [(1-\lambda),\lambda(1-\lambda)], \ \ I_{2,4} = [(1-\lambda)^2,1], \end{align*} from which you can hopefully see the general pattern emerging. We then define $C^{\lambda}_k = \bigcup_{j=1}^{2^k} I_{k,j}$ and define $$ C^{\lambda} = \bigcap_{k \geq 0} C^{\lambda}_k. $$

Evidently this is a closed subset of $[0,1].$ We wish to show this has Hausdorff dimension $s_{\lambda} = \frac{\log 2}{\log(1/\lambda)},$ for which note that $\operatorname{diam}(I_{k,j}) = \lambda^k$ for each $k,j$ so $$ \mathcal H^{s_{\lambda}}(C^{\lambda}) \leq \lim_{k \to \infty}\mathcal H^{s_{\lambda}}_{\lambda^k}(C^{\lambda}_k) = \lim_{k \to \infty} 2^k\lambda^{ks_{\lambda}} = 1, $$ by choice of ${s_{\lambda}}.$ It is further shown in the text that $\mathcal H^{s_{\lambda}}(C^{\lambda}) \geq \frac14;$ it actually equals one but we just need that it is non-zero.

Hence as $\mathcal{H}^{s_{\lambda}}(C^{\lambda})>0,$ we have $C_s(C^{\lambda})) > 0$ for all $s > s_{\lambda}.$ That is, there exists a Radon probability measure $\mu$ supported on $C^{\lambda}$ such that $I_s(\mu) < \infty$ and so $\mu \in H^{\frac{s-1}2}(\Bbb R).$ Note that $s_{\lambda} \to 1$ as $\lambda \to \frac12,$ so we can find such measures in $H^{-t}$ for any $t>0.$ Note also that since $\mu$ is supported on $C^{\lambda}$ which is Lebesgue-null, it is necessarily singular with respect to the Lebesgue measure.

Remark: I believe it is possible, as suggested in the comments, to take $\mu(A) = \mathcal H^{s_{\lambda}}(A \cap C^{\lambda})$ (the uniform measure on $C^{\lambda}$). This lends itself to a more explicit representation, as one can take an analogous staircase function $f_{\lambda},$ whose derivative gives a singular measure coinciding with $\mu$ above. Checking this works is quite technical, but this would provide alternative ways to construct suitable examples.