Sobolev Spaces subsets of each other?

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?

• 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. – user98602 Feb 15 at 18:04