Let $M$ be a compact Riemannian manifold and $$Lf:=-\operatorname{div} \nabla(f)$$ be the Laplace-Beltrami operator. Let $f$ be a smooth function on $M$. Consider the optimization problem of minimizing
$$\int_ML(f)f$$
under the constraint $\int_M |f|^2=1$. I wonder how to prove that if $f$ minimize the integral $\int_ML(f)f$, then it must be an eigenfunction of $L$, i.e. $L(f)=\lambda f$ for some $\lambda$.
I know it can be shown that $\int_M\|\nabla f\|^2= \int_ML(f)f$ but this may not be helpful.
Note $$ \int_M L(f)f=\lVert\nabla f\rVert^2_{L^2(M)}\geq 0 $$ with equality if and only if $\nabla f=0$ almost everywhere. So any smooth minimizer $f$ has zero derivative everywhere, hence locally constant. But such functions are eigenfunctions of $L$ with eigenvalue $0$.