Why is a null sequence in $L^p(M)$ also a null sequence in $\mathscr{C}^p_k(M)$? I am currently reading "Nonlinear analysis on manifolds: Sobolev spaces and inequalities" from Emmanuel Hebey and on page 21, he introduces the Sobolev spaces on Riemannian manifolds. For reference, here's the page my question is about:

Let $(M,g)$ be a Riemannian manifold. For $k$ an integer and $u\in C^\infty(M)$, $\nabla^ku$ denotes the k-th derivative of $u$ (with the convention $\nabla^0 u =u$). As an example, the components of $\nabla u$ in local coordinates are given by $(\nabla u)_i = \partial _i u$, while the components of $\nabla^2 u$ in local coordinates are given by
$$(\nabla^2 u)_{ij} = \partial_{ij} u - \Gamma^k_{ij} \partial _k u.$$
By definition one has that
$$|\nabla^k u|^2 = g^{i_1j_1} \ldots g^{i_kj_k} (\nabla^k u)_{i_1\ldots i_k} (\nabla ^k u)_{j_1\ldots j_k}.$$
For $k$ an integer and $p\geq 1$ real, we denote by $\mathscr{C}^p_k(M)$ the space of smooth functions on $u\in C^\infty(M)$ such that $|\nabla^j u| \in L^p(M)$ for any $j=0, \ldots, k$. Hence,
$$\mathscr{C}^p_k(M) = \left \{u\in C^\infty(M) \text{  s. t. } \forall j=0,\ldots, k, \int _M |\nabla^j u|^p dv(g) < \infty  \right\}$$
where, in local coordinates, $dv(g) = \sqrt{\det(g_{ij})}dx$, and where $dx$ stands for the Lebesgue's volume element of $\mathbb{R}^n, n = \dim M$. If $M$ is compact, one has that $\mathscr{C}^p_k(M) = C^\infty (M)$ for all $k$ and $p\geq 1$. $$ \ $$
Definition 2.1: The Sobolev space $H^p_k(M)$ is the completion of $\mathscr{C}^p_k(M)$ with respect to the norm
$$\| u\|_{H^p_k} = \sum _{j=0}^k \left (\int _M |\nabla^j u|^p dv(g) \right)^{1/p}$$
Noting that a Cauchy sequence in $\mathscr{C}^p_k(M)$ is also a Cauchy sequence in $L^p(M)$, and that a Cauchy sequence in $\mathscr{C}^p_k(M)$ which converges to $0$ in $L^p(M)$ converges to $0$ in $\mathscr{C}^p_k(M)$, the Sobolev spaces $H^p_k(M)$ can be seen as subspaces of $L^p(M)$.

I'm having trouble proving the last statement

$\ldots $ and that a Cauchy sequence in $\mathscr{C}^p_k(M)$ which converges to $0$ in $L^p(M)$ converges to $0$ in $\mathscr{C}^p_k(M)\ldots$

Basically what needs to be proven is that if we have given a Cauchy sequence $u_n$ in $\mathscr{C}^p_k(M)$ with $u_n \to 0$ in $L^p(M)$, then $\| |\nabla^j u| \|_p \to 0$ for $j=0,\ldots k$, where the case $j=0$ is just the convergence in $L^p(M)$.
My approach: The functions $ |\nabla^j u_n|$ converge in $L^p(M)$ as
$$ \|u\|_{H^p_k} \geq \| |\nabla^j u|\|_p.$$
I then tried to somehow multiply the quantity $|\nabla^j u_n|$ by a test function, integrate by parts and go to the limit, as it was done in one of the answers here. However, I just can't get it to work. Any help would be appreciated!
 A: Let me sketch an argument that should work but I haven't checked all the details. Let's concentrate on the case $k = 1$. The basic idea is to use integration by parts, only we have to do it globally on the manifold without resorting to local charts (which defeats the point of Hebey's definition).
Since $(u_k)$ is Cauchy in $\mathscr{C}^p_1(M)$, $|\nabla u_k|$ is Cauchy in $L^p(M)$ so $\nabla u_k$ converges to some $L^p$ vector field $X$ in $L^p$. We need to show that $X = 0$. Choose an arbitrary compactly supported smooth vector field $Z$ on $M$. Then using integration of parts we have:
$$ \left| \int_{M} \left< X, Z \right> \, dv(g) \right| = \left| \int_{M} \left< X - \nabla u_k + \nabla u_k, Z \right> \, dv(g) \right| \leq \\
\left| \int_{M} \left< X - \nabla u_k, Z \right> \, dv(g) \right| + \left| \int_{M} \left< \nabla u_k, Z \right> \, dv(g) \right| \leq \\
\int_{M} \left| X - \nabla u_k \right| \cdot |Z| \, dv(g) + \left| -\int_{M} u_k \, \operatorname{div} Z \, dv(g) \right| \leq \\
\int_{M} \left| X - \nabla u_k \right| \cdot |Z| \, dv(g) + \int_{M} |u_k| \left| \operatorname{div} Z \right| \, dv(g) \leq \\ \| \left| X - \nabla u_k \right| \|_{L^p} \| \left| Z \right| \|_{L^q} + \|u_k\|_{L^p} \| \operatorname{div} Z \|_{L^q}
$$
where $q$ is the conjugate exponent to $p$. Note that the integration of parts is applied only to $\nabla u_k$ and $Z$ which are both smooth and not to $X$. Both terms on the right-hand side are finite (as $Z$ has compact support) and tend to zero so we get that
$$ \int_{M} \left< X, Z \right> \, dv(g) = 0 $$
for all compactly supported $Z$ which implies that $X = 0$.
One should do the case $k > 1$ similarly using an integration by parts formula for higher order tensors (see for example page 29 in these lecture notes).
