Sobolev space $H^s(\mathbb{R}), s<s_0$

Question

I read articles where they talk about functions being in $H^s(\mathbb{R})$ for all $s<s_0$ for some specific $s_0$. For example in my case it is 1/2 or 3/2. I see this set can be written as an intersection of infinitely many Sobolev spaces. My question is: is that a Sobolev space? If so what would be the inner product in that space?

Answer 1

No, this is not a Sobolev space.

Take for example $f(x) = \mathcal{F}((1+|y|^2)^{-(s_0+1/2)/2})$ where $\mathcal{F}(u) = \hat{u}$ is the Fourier transform. Then $$ ∫_{\mathbb{R}}|\widehat{f}(x)|^2 \,(1+|x|^2)^s \,\mathrm{d}x = ∫_{\mathbb{R}} (1+|x|^2)^{s-s_0-1/2}\,\mathrm{d}x $$ which is finite if and only if $s<s_0$. So $f∈H^s$ for every $s<s_0$. So if there is a Sobolev space $H^{s_1}$ such that $H^{s_1} = \bigcap_{s<s_0} H^s$, then $s_1<s_0$. But then we can take $s'\in(s_1,s_0)$ and $g(x) = \mathcal{F}((1+|y|^2)^{-(s'+1/2)/2})$ is in $H^{s_1}$ but not in $\bigcap_{s<s_0}H^s$.

Therefore, $\bigcap_{s<s_0} H^s$ is not a Sobolev space.

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Actually, you have some more refined spaces $B^s_{2,q}$ called Besov spaces verifying $$ H^{s_0} = B^s_{2,2} ⊂ B^s_{2,\infty} \subset \bigcap_{s<s_0} H^s ⊂ H^{s_0-\varepsilon} $$ but even here I think the inclusion is strict.