Are solutions of $\Delta f \leq -a^2 f$ constant on a compact manifold? Let $(M^n,g)$ be a closed (compact, without boundary) smooth Riemannian manifold and let $\Delta = -\operatorname{div} \operatorname{grad}$ be the induced Laplacian on $M$ (so that the eigenvalue problem is written in the form $\Delta u = \lambda u$). My question is simple: let $a \in \mathbb{R}$, $a \neq 0$, and suppose that $f \in C^2(M)$ satisfies the inequality
$$\Delta f(x) \leq -a^2 f(x) \quad x \in M.$$
Question: Does it follow that $f$ is constant? Does it depend on $a$?
What I have done so far: Let $x_0$ be a point of maximum of $f$. It must hold that $\Delta f(x_0) \geq 0$. So,
$$ 0 \leq \Delta f(x_0) \leq -a^2 f(x_0) \implies f(x_0) \leq 0.$$
Thus, $f(x) \leq 0$ for every $x \in M$.
Another thing to notice: no eigenfunction of $\Delta$ satisfies the inequality, unless it is a nonpositive constant.
 A: If one requires that $f$ is nonnegative, then $f\equiv0$ on $M.$ To see this, it suffices to multiply the inequality by $f$ and then integrate by parts. However in general, only the differential inequality cannot ensure  $f$ to be a constant. In fact, given any negative $h\in C^\infty(M),$ we can solve the linear equation $-\Delta_M u+a^2u=h$ on $M.$ To prove this, we consider the energy functional $$E(u)=\int_M\frac{1}{2}|\nabla_Mu|^2+\frac{a^2}{2}u^2-hud\mu_g,\quad u\in X:=W^{1,2}(M),$$ where $X$ is the Banach space with norm $$\|u\|=\left(\int_M|\nabla_Mu|^2+a^2u^2d\mu_g\right)^{1/2}.$$ The (unique) critical point of $E$ gives a solution of above equation. One can easily check that $E\in C^\infty(X,\mathbf R)$ is convex and proper, that is, $$\lim_{\|u\|\to\infty}E(u)=+\infty.$$ Thus, there is some $u_0\in X$ such that $$E(u_0)=\inf_{u\in X}E(u).$$ The elliptic regularity theorem deduces that $u_0\in C^\infty(M)$ if $M$ is smooth. Thus, $u_0$ cannot be constant if $h$ is chosen to be non-constant.
