Suppose $f$ is a scalar function and $h$ a symmetric $(0,2)$ tensor on a closed Riemannian manifold $(M,g)$. Then is the following equality true?
$$\int_Mf\nabla_i\nabla_jh_{ij}\mathsf{dvol}_g=\int_M(\nabla_i\nabla_j f)h_{ij}\ \mathsf{dvol}_g.$$
I know that $\nabla_i\nabla_jh_{ij}=\mathsf{div}(\nabla_jh_{ij})$ and $f\mathsf{div}(\nabla_jh_{ij})=\mathsf{div}(f\nabla_jh_{ij}) -\langle \nabla f , \nabla_jh_{ij} \rangle$ then integrating over $M$ and using divergence theorem gives:
$$\int_Mf\nabla_i\nabla_jh_{ij}\ \mathsf{dvol}_g=\int_M-\langle \nabla f , \nabla_jh_{ij} \rangle\ \mathsf{dvol}_g$$
The other equation that can help is $\mathsf{div}(\nabla_jfh_{ij})=\nabla_i\nabla_j(fh_{ij})=(\nabla_i\nabla_jf)h_{ij}+f\nabla_i\nabla_jh_{ij}+2\nabla_if\nabla_j h_{ij}$ but I don't know how to relate this to the last equation and deduce the wanted equality.
Best Answer
Hint: The divergence theorem, in abstract index notation, reads $$ \int_M\nabla_iX^idV_g=\int_{\partial M}N_iX^idV_{g|_{\partial M}} $$ This means that on a compact manifold without boundary, you can integrate by parts exactly as you would in single variable calculus, by moving a $\nabla_i$ from one term to another and reversing the sign, regardless of how indices are contracted.