equivalence of smooth functions in manifold

Question

Let $(M,g)$ be a smooth Riemannian manifold and $\nabla^ku$ be the covariant derivative of some smooth $u:M\to\mathbb{R}$ . Denote $$\mathcal{C}_p^k(M)=\left\{u\in\mathcal{C}^\infty(M)\,\bigg|\,\int_M|\nabla^ju|^pd\nu(g)<\infty\,\,\forall\,\,j=0,1,,\dots,k\right\}$$ where $d\nu(g)=\sqrt{|g|}\,dx$ in local coordinates . Now it is claimed that if $M$ is compact , then $\mathcal{C}_p^k(M)=\mathcal{C}^\infty(M)$ for all $k$ and $p\geq1$ . I have to somehow use the finite subcover or smooth partition of unity to show this , but the idea is not very clear . Any help is appreciated .

Answer 1

$\mathcal{C}^k_p(M)$ is by definition contained in $\mathcal{C}^{\infty}(M)$. The reverse inclusion follows because $u$ being smooth implies each $|\nabla^ju|$ is a smooth, and in particular continuous, function on $M$, so by compactness of $M$, it is bounded above, and furthermore, $M$ is compact and the Riemannian measure assigns finite values to compact sets, i.e $\nu(M)<\infty$. Hence $|\nabla^ju|\in L^p(\nu)$ (for all $p\in [1,\infty]$ and all $j\geq 0$).