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

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 $$0. We consider the open set $$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?

• Wouldn't (for $\Omega$ being a cube, say) taking $u(x_1, \ldots , x_d) = g(x_1)g(x_2)\ldots g(x_d)$ work, where $g$ is a one-dimensional counterexample? I also think there should be some cheap and dirty argument with the Banach–Steinhaus theorem, but I'm too lazy to figure it out. Commented Mar 25 at 17:14
• @AlekseiKulikov Though I didn't mentioned, I did suspect this extension argument may works. I have not tried it. But I expected some more skilled artefact from big guys in this rooms. Commented Mar 25 at 17:27
• Certainly if you have a spectral characterization (as for $p=2$) as weighted $L^2$ spaces (via Fourier/etc transform), this is easily true by constructing suitable sequences/functions on the spectral side. Commented Mar 25 at 18:11
• Ask your beginning students in functional analysis to prove that no Banach space is the countable union of operator ranges. If necessary, suggest that they prove first that if $T:X\to Y$ is a bounded non surjective linear operator with $X$ and $Y$ Banach, then the closure in $Y$ of the image of the unit ball of $X$ has empty interior in $Y$ (this is sometimes called the "little open mapping theorem"). Commented Mar 25 at 18:25
• One can also choose countably many disjoint open sets $\Omega_j\subset \Omega$. Since the family of spaces $W_0^{s,p}(\Omega_j)$ is strictly monotone wrto s, we can put a function $u_j$ with compact support in each $\Omega_j$, so that the resulting $\sum_j u_j$ is in $L^p$, but in $W^{s,p}(\Omega)$ for no $s\in(0,1)$ Commented Mar 25 at 18:27

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 $$\int_1^{+\infty} r^{2s-1}(\log(2+r))^{-\alpha}dr=+\infty.$$