Does the union of fractional Sobolev spaces fills $L^p$?

Question

Let $\Omega\subset \Bbb R^d$ be any open set. Recall that for $s\in (0,1)$, the fractional Sobolev space $W^{s,p}(\Omega)$ is the collection of function in $L^p(\Omega$ such that \begin{align*} \iint_{\Omega\Omega} \frac{|u(x)-u(y)|^p}{|x-y|^{d+sp}}dy d x <\infty. \end{align*}

I would like to prove or disprove that the inclusion $$ \bigcup_{s\in (0, 1)} W^{s,p}(\Omega) \subset L^p(\Omega)$$ is strict.

In other words there is $u \in L^p(\Omega) \setminus \bigcup_{s\in (0,1)} W^{s,p}(\Omega)$ that is there is $u\in L^p(\Omega) $ such that \begin{align*} \iint_{\Omega\Omega} \frac{|u(x)-u(y)|^p}{|x-y|^{d+sp}}dy d x =\infty \qquad \text{for all $s\in (0,1)$}. \end{align*}

The assertion is true for $d=1$. Indeed it suffices to consider $\Omega= (0,1)$ and $(a_n)_n\subset (0,1)$ with \begin{align*} a_n= \frac{c}{(n+2)\log^2(n+2)}, \end{align*} be such that $\sum_{n=1}^\infty a_n=1.$

Define the partial sum $t_n=\sum_{j=1}^n a_j=1$ so that $0<t_n<t_{n+1}<1$. We consider the open set $O$ be the disjoint union of $(t_{2k}, t_{2k+1})$ that is \begin{align*} O= \bigcup_{k=0}^\infty (t_{2k}, t_{2k+1}) \qquad\text{so that }\qquad (0,1)\setminus O= \bigcup_{k=1}^\infty (t_{2k-1}, t_{2k}). \end{align*}

It is not difficult to see that \begin{align*} \int_0^1\int_0^1 \frac{|1_O(x)-1_O(y)|^p}{|x-y|^{1+sp}}d yd x \geq 2\sum_{k=0}^\infty \int_{t_{2k}}^{t_{2k+1}} \int_{t_{2k+1}}^{t_{2k+2}}\frac{d yd x}{|x-y|^{1+sp}}=\infty. \end{align*}

Does this remains true in higher dimension?

Answer 1

Let us assume that $p=2$, and let us consider $$ \cup_{s>0} H^s(\mathbb R^d)\subset H^0(\mathbb R^d)=L^2(\mathbb R^d). $$ The above inclusion is strict. Let us consider $u\in L^2(\mathbb R^d)$ defined by its Fourier transform, $$ u_\alpha(x)= \int e^{2iπ x\cdot \xi}(1+\vert \xi\vert)^{-\frac d2} \bigl(\log(2+\vert \xi\vert)\bigr)^{-\frac\alpha 2} d\xi, \quad \alpha >1. $$ Now the function $u_\alpha$ does not belong to any $H^s(\mathbb R^d)$ when $s>0$. Otherwise we would have for some $s>0$, $$ (1+\vert \xi\vert)^{-\frac d2+s} \bigl(\log(2+\vert \xi\vert)\bigr)^{-\frac\alpha 2}\in L^2(\mathbb R^d), $$ which is untrue since $ \int_1^{+\infty} r^{2s-1}(\log(2+r))^{-\alpha}dr=+\infty. $