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Let us consider a function $u \in C_c(\mathbb{R}^n)$ compactly supported and consider the fractional sobolev spaces $H^s(\mathbb{R}^n)$ with the norm $$ \lVert u \rVert_{H^s}^2 = \lVert u \rVert_{L^2}^2+C(n,s)\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}{\frac{\lvert u(x)-u(y) \rvert}{\lvert x-y\rvert^{n+2s}}\mathrm{d}x\mathrm{d}y} $$ where $$ C(n,s) = \frac{2^{2s}s \Gamma(n/2+s)}{\pi^{n/2}\Gamma(1-s)}. $$ Suppose now that there exists $s_0 \in (0,1)$ such that, for every $t \in [s_0,1)$ we have $u \in H^t(\mathbb{R}^n)$ and $$ \lVert u \rVert ^2_{H^t} \leq M(n,s_0) $$ for some positive constant $M=M(n,s_0)$. Since we already normalized the $t$-seminorm, can we just conclude that $u \in H^1(\mathbb{R}^n)$?

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    Not familiar with this normalization, but if the norm it gives is comparable to the Fourier transform norm $\|(1+|\xi|^2)^{s/2} \hat f\|_2$, then the conclusion follows from the Monotone Convergence theorem, since $(1+|\xi|^2)^{s/2}$ increases with $s$. – 2017-01-16

1 Answers 1

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Yes you can conclude that $u \in H^1(\mathbb{R}^n)$ since $$ u \in H^1(\mathbb{R}^n) =\bigcap_{0

proof: indeed we have the following embeddings for $0

$$ H^1(\mathbb{R}^n)\subset H^t(\mathbb{R}^n) \subset H^{s_0}(\mathbb{R}^n)\subset H^0(\mathbb{R}^n)=L^2(\mathbb{R}^n)$$ this means that, $$ H^1(\mathbb{R}^n) \subset \bigcap_{0

On the other hand, from Brezis_Bourgain we know that: $$\lim_{s\to1^-}\| u \|_{H^s}^2 =\| u \|_{H^1}^2 $$

Hence, from this we have $$ u \in H^1(\mathbb{R}^n)\supset \bigcap_{0

Therefore, we conclude that,

$$ u \in H^1(\mathbb{R}^n) =\bigcap_{0