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For all $s>0$, $1\le p<+\infty$ and $u\in L^p(\mathrm{d}x)$, we define $$D_{s,p}(u)=\sup_{r>0}\frac{1}{r^{sp+n}}\int_{\mathbb{R}^n}\int_{B(x,r)}|u(y)-u(x)|^p\mathrm{d}y\mathrm{d}x$$ and $$W^{s,p}(\mathrm{d}x)=\left\{u\in L^p(\mathrm{d}x):\lVert u\rVert_p+[D_{s,p}(u)]^{1/p}<+\infty\right\}$$ When $s>1$, $W^{s,p}(\mathrm{d}x)=\{0\}$.

Can someone give the reason for the trivility of $W^{s,p}(\mathrm{d}x)$ for $s>1$?

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    $\begingroup$ Let $u$ be continuously differentiable in a neighbourhood of $x$. Then $u(y) = u(x) + Du(x) \cdot (y-x) + o(|y-x|)$. So $\int_{B(x,r)} |u(y) - u(x)|\mathrm{d}y$ as $r\to 0$ behaves like $C r^{n+p}$. So $D_{s,p}(u) = \infty$ for $s > 1$. $\endgroup$
    – Willie Wong
    Commented Sep 1, 2014 at 9:06
  • $\begingroup$ But we do not have the regularity of $u$. $\endgroup$
    – yangmengqh
    Commented Sep 1, 2014 at 9:26
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    $\begingroup$ No you do not. But you can modify the above proof to show that $D_{s,p} < +\infty$ for $s > 1$ implies that $u$ is a.e. constant for $L^p$ functions. $\endgroup$
    – Willie Wong
    Commented Sep 1, 2014 at 9:44
  • $\begingroup$ If $u$ has some regularity near some points, then we can use the above proof. But for general $L^p$ functions such as $\mathbf{1}_{B_(0,1)}$, how can we use the above idea? The denseness of smooth functions of something? And it is the point I am not clear. $\endgroup$
    – yangmengqh
    Commented Sep 1, 2014 at 10:07
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    $\begingroup$ Try to see if you can estimate the $D_{s,p}$ of a mollified version of $u$ by $D_{s,p}(u)$. It may be convenient to choose a non-negative mollifier. It may also help to use Minkowski's inequality. $\endgroup$
    – Willie Wong
    Commented Sep 1, 2014 at 14:34

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