(This is a proof verification, I need to know if the procedure and steps are correct and if the proof is complete)
I want to show that the Sobolev Space $W^{m,p}(I)$ is a closed subspace of $W^{1,p}(I)$. Showing that will automatically yield us that the Space $W^{m,p}(I)$ is complete, reflexive ($1 <p <\infty$) and separable ($1 \leq p < \infty$), provided we have shown $W^{1,p}(I)$ verifies those properties.
Considering the definition of $W^{m,p}$ and $W^{1,p}$ as follows:
$W^{1,p}(I)= \{ u\in L^{p}(I)\ | \ \exists u' \in L^p(I) :\int_{I}^{}u\varphi'= - \int_{I}^{}u' \varphi \}$
$W^{m,p}(I)= \{u \in W^{m-1,p}(I) \ | \ u' \in W^{m-1,p}(I) \}$ $ \ \ m \geq 2$
i.e $u \in W^{m,p}(I)$ if and only if there exist $m$ funcions $u',u'',...,u^{(m)} \in L^{p}(I)$ such that $$\int_{I}u\varphi^{(j)} = (-1)^j \int_I u^{(j)} \varphi \quad \quad \forall \varphi \in C^{\infty}_c(I)$$
For every $j=1,...,m$.
So I want to prove that the closure of $W^{m,p}(I) \subset W^{1,p}(I)$ is $W^{m,p}(I)$.
Which equals to show that any convergent sequence $(u_n) \subset W^{m,p}(I)$ converges to some $u \in W^{m,p}(I)$.
For all $j$ we have: $$u_n^{(j)}\varphi \to \tilde{u}^{(j)} \varphi$$
$$u_n\varphi^{(j)} \to u\varphi^{(j)}$$
(Convergence in $L^p$)
And we need to show that the $\tilde{u}^{(j)}=u^{(j)}$, i.e, they are indeed the (weak) derivatives of the limit function u, that is to say :
$$\int_{I}u\varphi^{(j)} = (-1)^j \int_I \tilde{u}^{(j)} \varphi \quad \quad \forall \varphi \in C^{\infty}_c(I)$$
As for every $u_n$ we have: $$\int_{I}u_n\varphi^{(j)} = (-1)^j \int_I u_n^{(j)} \varphi \quad \quad \forall \varphi \in C^{\infty}_c(I)$$
Then, $$\lim_{n \to \infty} \int_{I}u_n\varphi^{(j)} = \lim_{n \to \infty} (-1)^j \int_I u_n^{(j)} \varphi \quad \quad \forall \varphi \in C^{\infty}_c(I)$$
For both sides of the equality we must show that we can put the limit inside the integral, so I present an example for the Left Hand Side:
$$ | \int_{I} u_n\varphi^{(j)} - \int_{I} u \varphi^{(j)} | \leq | \int_{I} (u_n - u )\varphi^{(j)} | \leq \int_{I} | (u_n - u )\varphi^{(j)} | \leq \\ ||u_n-u||_{p}||\varphi^{(j)}||_{p'}$$
As $u_n \to u$ in $L^{p}$ (Due to the norm $||u||_{W^{m,p}}=||u||_{L^p}+\sum_{n=1}^{m}||u^{(n)}||_{L^p}$), then taking the limit we have the desired result. We proceed the same way for the RHS.