Let $(M,g)$ be a compact Riemannian manifold, $\mu(g)$ the Riemannian Lebesgue measure. Then we can define the usual $L^p$-spaces (lets assume $p<\infty$),
$L^p(M,g):=L^p(M,\mu(g))$.
For $f\in L^p(M,g)$ the $L^p$-norm can be computed as follows: Choose a finite covering $(U_j,x_j)$ of $M$ by charts and a subordinate partition of unity $\varphi_j$. Then
$||f||_p=(\sum_j\int_{x_j(U_j)}(\varphi_j|f|^p)\circ x_j^{-1}G_j)^{1/p}$
where $G_j$ is the determinant of the metric tensor wrt $(U_j,x_j)$ and the integrals are usual Lebesgue integrals.
Another norm on $L^p(M,g)$ is given by
$|f|_p:=\big(\sum_j||(\varphi_jf)\circ x_j^{-1}||_{L^p(x_j(U_j))}^p\big)^{1/p}$
where $||.||_{L^p(x_j(U_j))}$ is the usual $L^p$-norm on $\mathbb{R}^n$, that is
$|f|_p=(\sum_j\int_{x_j(U_j)}(|\varphi_jf|^p)\circ x_j^{-1})^{1/p}$
Notice that $||.||_p$ and $|.|_p$ differ only by the $G_j$ and the power of $\varphi_j$.
I want to prove that $||.||_p$ and $|.|_p$ are equivalent.
My attempt: First of all $\varphi_j$ is zero outside the compact set $supp(\varphi_j)$, so we dont have to worry about the $G_j$ since they are continuous. Since $0\le\varphi_j\le1$ and $p\ge1$ we have that $\varphi_j^p\le\varphi_j$. From that we get $|.|_p\le C ||.||_p$. However, I'm not able to prove $||.||_p\le \tilde{C}|.|_p$ because I don't know how to handle the power of $\varphi_j$ in that case. (Once I have shown that both spaces are complet I could use the bounded inverse theorem to show that both norms are equivalent, but I would prefer a direct calculation.)
Remark: Why do I want to prove the equvialence of these norms? If one wants to define Sobolev spaces on compact manifolds, one way to do it is the following: A measurable $f\colon M\rightarrow\mathbb{R}$ belongs to $W^k_p(M)$ iff $(\varphi_jf)\circ x_j^{-1}$ belongs to $W^k_p(x_j(U_j))$ for all (finitely many) $j$. The norm on $W^k_p(M)$ is then given by
$|f|_{W^k_p(M)}=\big(\sum_j||(\varphi_jf)\circ x_j^{-1}||_{ W^k_p(x_j(U_j))}^p\big)^{1/p}.$
(One can show that choosing different charts and partitions of unity results in equivalent norms.)
Now it would be very desirable that $W^0_p(M)=L^p(M)$ in the sense of equivalent norms. That leads exactly to my question.