Let $(M,g)$ be a closed, smooth Riemannian manifold. Let $P$ be a self-adjoint, elliptic differential operator defined on $C^\infty(M)$ with smooth coefficients. Suppose as well that the lowest eigenvalue of $P$ is positive, i.e., $P$ is coercive. Is there a smooth function $u$ such that $u>0$ and $Pu>0$?

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    $\begingroup$ Do you want to require $P$ to be second order? Otherwise I don't see why your claim (the existence of a maximal principle) has to be true. Simply take $P = (\triangle)^2 + k^2$, it is clearly self-adjoint and elliptic, and $\langle Pu,u\rangle = \langle \triangle u + ku,\triangle u + ku\rangle > 0$ provided $k$ is chosen so that $\triangle u + ku = 0$ has no solutions. For this $P$ you have that the function $u = 1$ satisfies $u > 0$ and $Pu = u > 0$. $\endgroup$ – Willie Wong Jul 4 '13 at 12:03
  • $\begingroup$ (I meant to write $P = (\triangle + k)^2$, of course. But one can obviously also just look at $(\triangle)^2 + k^2$ for which the argument is even simpler...) $\endgroup$ – Willie Wong Jul 4 '13 at 12:15
  • $\begingroup$ Willie Wong: No, I don't require that $P$ is second-order. I'm particularly interested in the fourth-order case. $\endgroup$ – Viktor Bundle Jul 5 '13 at 14:39

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