Q:

Show that $\int_M\Delta f HdV=-\int_M\nabla f\cdot\nabla H dV $?

$M$ is a compact Riemannian  manifold,$f,H$ are functions on $M$,How to show that $\int_M\Delta f HdV=-\int_M\nabla f\cdot\nabla H dV $?Besides,if $M$ is not compact, is the equality right?
Beside, I want to know that, if $F,H$ are field on Riemannian manifold(example vector field),  is $\int_M \nabla F\cdot HdV=-\int_M F\cdot \nabla HdV$ right?
In fact, I am fuzzy with there equality.I think $\cdot$ is Riemannian inner and no matter $f$ is function or field , $\nabla f$ should be treated as operator not gradient of $f$.But,in analysis, there are likely equality,treat them as gradient. So, I think  they  are same no matter treat it as gradient or operator.But I fail to understand the same.
Sorry for my poor English,I don't know whether precisely  I describe my question .

A:

Your statement follows from Stokes's theorem $\int \limits _{\partial M} \omega = \int \limits _M \Bbb d \omega$ and the fact that $M$ has no boundary (i.e. $\partial M = \emptyset$, so the left-hand side is $0$).
Note that $(\Delta f) H \ \Bbb d V = (\text{div} \ \nabla f) H \ \Bbb d V = \text{div} (H \nabla f) \ \Bbb d V - (\nabla f \cdot \nabla H) \ \Bbb d V$. If you define $\omega = i_{H \nabla f} (\Bbb d V)$ (the inner product of a field and a form or, if you prefer, the Lie derivative of $\Bbb d V$ along the field $H \nabla f$), then $\omega$ will be a $n-1$-form on $\partial M$ and it can be shown that $\Bbb d \omega = \text{div} (H \nabla f) \ \Bbb d V$. You have thus obtained that $\Bbb d \omega = (\Delta f) H \ \Bbb d V + (\nabla f \cdot \nabla H) \ \Bbb d V$. Plugging this into Stokes's theorem as described above, the integral of $\Bbb d \omega$ will be $0$, whence follows your formula.
If $M$ is not compact but either $H$ or $f$ has compact support, then the equality remains true. It may fail in the absence of compact support, though.

