Ricci curvature and Bochner identity Let $M ^ n$ a no boundery compact Riemannian manifold and $f \in C^\inf (M)$ a non-constant solution of the equation $\Delta f + \lambda f = 0$  
a)Show the  Green identity (as a result of the divergence theorem) applied to the function $f$ to show that $\lambda> 0$
b) If Ricci curvature of M verify $Ric (v, v)\geq (n-1) | v | ^ 2 $ for all tangent vector from $M$, show using the Bochner identity that $\lambda\geq n$ (enclose the $f$ norm Hessian  by the Laplacian of $f$)
 A: *

*Using the Green identity, the fact that $\Delta f+\lambda f=0$ and that $f$ is non-constant we get
$$0 < \int_{M} |\nabla f|^{2}dv=-\int_{M}f\Delta f dv=\int_{M}\lambda f^{2}dv. \qquad (1)$$
Thus, $\lambda>0$.


*Using the Divergence Theorem, Bochner's formula, the condition about $f$ and the hypothesis about the Ricci curvature, respectively, we obtain
\begin{align*}
    0&=\int_{M} \frac{1}{2}\Delta |\nabla f|^{2} dv \\
    &=\int_{M} (\langle \nabla f,\nabla \Delta f \rangle+|\nabla^{2}f|^{2}+\operatorname{Ric}(\nabla f,\nabla f)) dv \\
    &=\int_{M} (-\lambda|\nabla f|^{2}+|\nabla^{2}f|^{2}+\operatorname{Ric}(\nabla f,\nabla f)) dv \\
    & \geq \int_{M} (-\lambda|\nabla f|^{2}+|\nabla^{2}f|^{2}+(n-1)|\nabla f|^{2}) dv. \qquad (2)
\end{align*}
Now recall that for a linear operator $T:V \to V$ where $V$ is a $n$-dimensional (real) vector space, you have (as a corollary of Cauchy-Schwarz inequality)
$$|T|_{2}^{2}\geq \frac{1}{n}(\operatorname{tr}T)^{2}.$$
In particular, this is true for $\nabla^{2}f:TM \to TM$, and since $\operatorname{tr}\nabla^{2}f=\Delta f$, we get
$$|\nabla^{2}f|^{2} \geq \frac{1}{n} (\Delta f)^{2}.$$
Using this inequality and that $\Delta f+\lambda f=0$, we obtain
$$\int_{M} |\nabla^{2}f|^{2}dv \geq \int_{M}\frac{1}{n}(\Delta f)^{2}dv=\int_{M}\frac{\lambda^{2}}{n}f^{2}dv. \qquad (3)$$
Joining (1), (2) and (3) we find
$$0 \geq \int_{M} \left( -\lambda^{2} +\frac{\lambda^{2}}{n}+\lambda(n-1) \right) f^{2} dv.$$
This implies that the parenthesis in the integrand must be non-positive, that is
$$ -\lambda^{2} +\frac{\lambda^{2}}{n}+\lambda(n-1)\leq 0.$$
Using 1., this is equivalent to
$$\frac{\lambda}{n}+(n-1)-\lambda \leq 0.$$
If you rearrange the terms of this inequality, you obtain what you wanted.
