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If $u\in W^{k,p}(\Omega)$ we define its norm as $$ \Vert u \Vert_{W^{k,p}(\Omega)}= \begin{cases} \displaystyle \left(\sum_{|\alpha|\le k}\displaystyle\int\limits_\Omega |D^\alpha u|^p\mathrm{d}x\right)^\frac{1}{p} & 1\le p <\infty,\\ \\ \:\:\displaystyle\sum_{|\alpha|\le k} \underset{\Omega}{\mathrm{ess\,sup}}|D^\alpha u| & p=\infty \end{cases} $$

Are these the only norms on a Sobolev space... and how can we say that these norms exists on this space?

Daniele Tampieri
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Inverse Problem
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1 Answers1

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There are other equivalents norm as for example $$\|u\|_{W^{k,p}}=\sum_{|\alpha |\leq k}\|D^\alpha u\|_{L^p}$$ or $$\|u\|_{W^{k,p}}^p=\sum_{|\alpha |\leq k}\|D^\alpha u\|_{L^p}^p.$$

For the existence, what do you mean ? $\|\cdot \|_{W^{k,p}}$ is a norm on $W^{k,p}$, so it exist !

Surb
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  • @surb....i mean we define some norms is those are well defined – Inverse Problem Feb 09 '19 at 14:20
  • @InverseProblem: What do you mean by "well defined" ? That $|u|_{W^{k,p}}<\infty $ ? If yes, then it's well defined on $W^{k,p}$ by definition of $W^{k,p}$. – Surb Feb 09 '19 at 14:25