Let $(M,g)$ be a closed Riemannian manifold and $D$ denote the Riemannian connection corresponding to $g$. Let $S^2(T^\*M)$ denote the space of $C^2$ symmetric 2-tensor filds on $M$. Let $h\in S^2(T^\*M)$. Then 
 $$D^2\_{x,y}h(z,w)= D\_xD\_yh(z,w)-D\_{D\_xy}h(z,w)$$
where $x,y,z,w$ are vector-fields on $M$. Let $D^\*$ denote the formal adjoint of $D$.\
Define,
 $$\|h\|^2=\int\_M|h|dv\_g$$
where $|h|$ denote the point-wise norm on $h$ and $dv\_g$ denote the volume form defined by $g$.\
 Define,
 $$\|h\|^2\_1=\|D^2h\|^2+\|Dh\|^2+\|h\|^2$$
and 
$$\|h\|^2\_2=\|D^\*Dh\|^2+\|Dh\|^2+\|h\|^2$$
 Are $\|.\|\_1$ and $\|.\|\_2$ equivalent? 



Eqivalency of two norms on the symmetric two tensor-fields on a compact Riemannian manifold.
We have $\|D^2h-DD^\ast h\| \le \|D-D^\*\|\_{L^\infty(M,T^\ast M\otimes End(T^\ast M\otimes TM))} \|Dh\|,$ proving the uniform equivalence of the metrics. Notice that the difference of two connections is an ordinary tensor field, whose $L^\infty$-norm is bounded due to the compactness of $M$.


Another way to see this is that both metrics are locally equivalent to the metric $\xi\mapsto \|\nabla^2\xi\|+\|\xi\|,$ where now $\nabla$ refers to the "flat" connection and the $L^2$-norm refers to the local Lebesgue measure. Since $M$ is compact it can be covered by a finite number of such neighborhoods and in particular both norms are equivalent.


EDIT: I will keep the stuff I wrote at first (I thought of $D^\\*$ as a connection as well, which is not true, as Brian pointed out), but the correct answer involves elliptic regularity. I will consider the following simpler version (which also implies the general case, but once this simpler case is understood the more complicated case becomes easy.). Let $\nabla$ be a connection on $M.$ Then the following two norms are equivalent ($f\in C^\infty(M)$):  

1. $\|f\|\\_{W^{2,2}}, $  

2. $\|\Delta f\|\\_{L^2}+\|f\|\\_{L^2}.$  

Clearly the second norm is dominated by the first. The other inequality follows from the elliptic estimate  

$\|D^2 f\|\\_{L^2}\leq \|\Delta f\|\\_{L^2},$
valid for $f\in C^\infty\\_0(R^n),$ which is actually trivial to prove.


