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I have recently started working with Sobolev Spaces and I wanted to ask the following:

Let $1 \leq p \leq \infty, \Omega \subset \mathbb{R}^d$. Does $ W^{n,p}(\Omega) \subset W^{1,p}(\Omega)$ hold?

Here $W^{n,p}(\Omega)$ is the space of $L^p$ "functions" such that their weak derivatives up to order n are also in $L^p$.
I do not know whether this seems trivial or I am missing some technical details here. Further I have seen that $W^{1,\infty}(\Omega)$ can be described as the set of (locally) Lipschitz functions for bounded $\Omega$. That would also imply that all functions in $W^{n,\infty}$ would have derivatives up to order $n-1$ (including themselves), which are Lipschitz, is that correct?

Thanks for the answers in advance!

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  • $\begingroup$ I can't think of a definition of Sobolev space in which this isn't immediate. What is your definition? $\endgroup$ – user98602 Feb 15 at 17:00
  • $\begingroup$ I guess you mean the subspace of $\mathcal D(\Omega)$ consisting of $L^p$ functions whose first $n$ distributional derivatives are also (identified with) $L^p$ functions. Then certainly $W^{n,p} \subset W^{1,p}$... because $1 \leq n$. Presumably, then, your norm is $$\|f\|_{W^{n,p}} = \sum_{i=0}^n \|f^{(i)}\|_{L^p},$$ and it is clear for the same reason that the inclusion is continuous. $\endgroup$ – user98602 Feb 15 at 18:04
  • $\begingroup$ Thank you, is the rest I have written then also correct? $\endgroup$ – SpiritOfMath Feb 15 at 19:02

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