Say you have a complete, noncompact Riemannian manifold without boundary $(M,g)$. I am wondering what condition needs to be placed on the metric $g$ so that when you perform an integration by parts, it simplifies to $$ \int_M v \Delta u\,dV=-\int_M \langle\nabla v,\nabla u\rangle\,dV $$ That is to say, I am looking for a condition on the metric so that the `boundary' integral term at infinity vanishes.

I was trying to think about this problem using an auxiliary function $\eta$ that has compact support (say in $B_{R+1}(p)$) and $\eta\equiv 1$ in $B_R(p)$. Then multiplying the integrand by $\eta$ and integrating by parts (since the integrand has compact support so the boundary term vanishes), you're left with exactly what an error term of the form $$ \int_{B_{R+1}\setminus B_R} v\langle\nabla \eta, \nabla u\rangle \,dV $$ but I am not sure exactly what condition I need to specify to have this term go to 0 as $R\to \infty$.

For instance, is the condition that $(M,g)$ has bounded geometry (ie bounded curvature and positive injectivity radius) sufficient? Or does anyone have a reference for this?

  • $\begingroup$ You need conditions on $u$ and $v$, not merely on the metric. Given any noncompact Riemannian manifold, one can always find functions which are badly behaved at infinity. $\endgroup$
    – Kajelad
    May 14 at 2:14
  • $\begingroup$ @Kajelad sure I totally agree. I’m assuming all these integrals exist at the very least. I’d be willing to assume $u,v\in W^{2,2}$ $\endgroup$ May 14 at 3:06
  • $\begingroup$ I'm not sure that Sobolev-type conditions are sufficient. In $\mathbb{R}^n$, for instance, it's common to use falloff conditions such as $$\exists C>0\text{ s.t. }\|\nabla^ku\|(x)<C\|x\|^{-p}$$Things are more complicated in the general case, but you could probably write something similar e.g. by pulling back by the exponential map. $\endgroup$
    – Kajelad
    May 14 at 6:10


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