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

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 $$v$$?

• Taking $u=v$ in the Green's identity gives: $$\int_M u\Delta udV_g=-\int_M |\mathrm{grad}\ u|^2dV_g$$ and in other hand: $$\int_M u\Delta udV_g=-\lambda\int_M u^2 dV_g.$$ Commented Mar 3, 2020 at 9:36
• So it is enough to show $$\lambda^2\int_M u^2 dV_g\leq n \int_M |\nabla^2 u|^2dV_g.$$ I don't know how to proceed. Commented Mar 3, 2020 at 9:41

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
$$0 \leq \lvert E\rvert^2 = \lvert\nabla^2u\rvert^2 - \frac{1}{n}(\Delta u)^2 .$$
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 $$\int\lvert\nabla u\rvert^2$$.
• Thank you very much! I was hoping you could clear up one more confusion for me. I've seen many people interchange $\nabla u$ and grad $u$ despite these being very different (one is a form and one is a vector field). Is this just because we have an equality $|$grad $u|^2=|\nabla u|^2$?
• You have to be careful, because some authors use $\nabla u$ to mean the gradient of $u$, while others use that symbol to refer to $du$. Anyway, if a Riemannian metric $g$ is fixed, they are dual to one another by the formula $g(\mathrm{grad}\,u,X)=du(X)$. Commented Mar 3, 2020 at 20:09