$ \lambda \int_M |\text{grad} \; u|^2 dV_g\leq n \int_M |\nabla^2 u|^2 dV_g. $

Question

I am doing problem 5-15 in John Lee's "Introduction to Riemannian Manifolds" and I am somewhat confused as to what the hint is suggesting. The set-up is that $(M,g)$ is a compact Riemannian manifold (without boundary) and $u\in C^\infty (M)$ is an eigenfunction of $M$ (meaning $-\Delta u=\lambda u$) for some constant $\lambda$. We are asked to show that $$ \lambda \int_M |\text{grad} \; u|^2 dV_g\leq n \int_M |\nabla^2 u|^2 dV_g. $$ The hint is to consider the 2-tensor $\nabla^2 u-\frac{1}{n} (\Delta u)g$ and use one of Green's identities. But I am confused how one is supposed to apply Green's identities with this 2-tensor. Green's identities applies to functions and not tensors. Are we supposed to define $v=(\nabla^2 u-\frac{1}{n} (\Delta u)g )(\text{grad} \; u,\text{grad} \; u)$ and apply Green's to $v$?

Answer 1

The missing point is that $n\lvert\nabla^2u\rvert^2\geq(\Delta u)^2$. There are two ways to see that. One is to apply Cauchy--Schwarz to $\langle g,\nabla^2u\rangle$. The other is to set $E=\nabla^2u-\frac{1}{n}(\Delta u)g$ and observe that

$$ 0 \leq \lvert E\rvert^2 = \lvert\nabla^2u\rvert^2 - \frac{1}{n}(\Delta u)^2 . $$

This is, of course, just the usual proof of Cauchy--Schwarz.

Now integrate the inequality $n\lvert\nabla^2u\rvert^2\geq(\Delta u)^2$, use the hypothesis $-\Delta u=\lambda u$, and use the divergence theorem to relate $\int u^2$ and $\int\lvert\nabla u\rvert^2$.