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I know that this is classical but I have never do the calculations to show that the norms in the sobolev space $W^{k,p}(\Omega)$ \begin{equation} \|u\|_{k,p,\Omega}= \Bigl(\int_{\Omega} \sum_{|\alpha| \le k}|D^{\alpha} u |^{p} dx \Bigr)^{1/p} \end{equation} and \begin{equation} \|u\|_{W^{k.p}(\Omega)} =\sum_{|\alpha|\le k}\|D^{\alpha}u\|_{p} \end{equation} are equivalents.

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up vote 6 down vote accepted

Let $N$ the number of $\alpha$ such that $|\alpha|\leq k$, and enumerate them as $\alpha^{(1)},\dots,\alpha^{(N)}$. Since all the norms are equivalent on $\Bbb R^N$, we can find two constants $C_1,C_2>0$ such that for all $(a_1,\dots,a_N)\in\Bbb R^N$, $$C_1\sum_{j=1}^N|a_j|\leq \left(\sum_{j=1}^N|a_j|^p\right)^{\frac 1p}\leq C_2\sum_{j=1}^N|a_j|.$$ Then apply this inequality to $a_j:=\left(\int_{\Omega}|D^{\alpha^{(j)}}u(x)|^pdx\right)^{\frac 1p}$.

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