# If $(m-1)p\geq n$, $u(x)=|x|$ does not belong to $W^{m,p}(B(0,1))$.

Suppose $(m-1)p\geq n$. I tried proof that $u(x)=|x|$ does not belong to $W^{m,p}(B(0,1))$. I was able to calculate

$|D^0u|=|x|$, $|D^1u|=1$ and $|D^2u|=\sqrt{\dfrac{1-|x|_s^2}{|x|^2}}$.

Here $D^ju$ represents the $N^j$-vector of all partial weak derivatives of $u$ of order $j$, $|D^ju|$ stands for the Euclidean norm of this vector and $|x|_s$ stand for the sum norm. But I was not able to calculate $|D^mu|$ and check that $$\int_B|D^mu|^pdx=\infty.$$

The main property of $u$ here is that it's $1$-homogeneous, i.e. $$u(\lambda x) = \lambda u(x) \qquad \text{for every } x \in \mathbb R^n \setminus \{ 0 \}, \ \lambda > 0.$$ Differentiating this identity $m$-times (with $\lambda$ fixed), we deduce that $\nabla^m u$ is $(1-m)$-homogeneous, i.e. $$\nabla^m u(\lambda x) = \lambda^{1-m} \nabla^m u(x) \qquad \text{for every } x \in \mathbb R^n \setminus \{ 0 \}, \ \lambda > 0.$$ Integrating over $B(0,1) \setminus \{ 0 \}$, we obtain \begin{align*} \int_{B(0,1) \setminus \{ 0 \}} |\nabla^m u (x)|^p \, d x & = \int_0^1 r^{n-1} \int_{\partial B(0,1)} |\nabla^m u (r y)|^p \, dy \, dr\\ & = \int_0^1 r^{n-1} \int_{\partial B(0,1)} (r^{1-m}|\nabla^m u (y)|)^p \, dy \, dr, \end{align*} which is up to a constant equal to $$\int_0^1 r^{n-1} (r^{1-m})^p dr.$$ This happens to be infinite for $p(m-1) \ge n$.
On the other hand, if $u$ were to lie in $W^{m,p}(B(0,1))$, its weak derivative $\nabla^m u$ would coincide with the classical derivative in $B(0,1) \setminus \{ 0 \}$, contradicting the calculations above.
• Can you please give more details to obtain $\int_0^1r^{n-1}(r^{1-m})^pdr$? I do not know what to do with $\lambda$. Commented Apr 24, 2018 at 11:07