Uniform or Holder estimates for positive Schrodinger operators I am sure this is treated in many standard references, I just can't find them!
Let $V(x)$ be a smooth function on $\mathbb{R}^n$, with $V(x)\geq \epsilon>0$ for all $x$.  Impose other conditions (boundedness, slow growth, etc) on $V$ if required.
Consider the Schrödinger operator $-\Delta+V$ on $\mathbb{R}^n$.  By the standard argument (Hölder's inequality), if $u\in H^{2}$ then
$$
\epsilon\int_{\mathbb{R}^n}u^2dx
\leq\int_{\mathbb{R}^n}[|du|^2+Vu^2]dx
=\int_{\mathbb{R}^n}[(-\Delta+V)u]udx
\leq\sqrt{\int_{\mathbb{R}^n}[(-\Delta+V)u]^2dx}
\sqrt{\int_{\mathbb{R}^n}u^2dx},
$$
so $\epsilon||u||_{L^2}\leq ||(-\Delta+V)u||_{L^2}$, and $-\Delta+V$ is invertible as a map from $H^{2}$ to $H^0$.
Question:  Do similar inequalities hold for $C^{2}$, or for $C^{2,\alpha}$?  That is, is it true, say, that if $u\in C^2$ then
$$ \epsilon\sup |u|\leq \sup|(-\Delta+V)u|?$$
The maximum principle seems the obvious approach, but my naive deployment of it gives only that
$$ \epsilon\sup |u|\leq \epsilon\left[\limsup_{r\to\infty}\sup_{\{|x|=r\}}|u|\right]
+\sup|(-\Delta+V)u|.$$
Question: Even if it's not true, I presume the operator is still invertible, i.e., the inequality holds for some other constant $c$ (just not $c=\epsilon$)?
An ideal answer would work not just on $\mathbb{R}^n$ but on a complete Riemannian manifold.
 A: The main problem is lack of compactness. On a compact (closed) Riemannian manifold, if $u$ is in $C^2$, it attains its maximum. At its maximum you have that $-\Delta u \geq 0$ and the desired result follows. 
Here's what you can do in $\mathbb{R}^n$. Fix $\phi\in C^\infty_0$ some bump function so that $\phi\equiv 1$ in the unit ball, $\phi \equiv 0$ outside the ball of radius 2, and $\phi$ always take values between 0 and 1. 
Let $\phi_\lambda(x) = \phi(\lambda x)$. 
Observe that $\sup |\phi_\lambda u| \leq \sup |u|$, but 
$$ \lim_{\lambda \to 0} \sup |\phi_\lambda u| = \sup |u|. $$
For $\phi_\lambda u$, since it has compact support, it attains its maximum so that the inequality $\epsilon \sup |\phi_\lambda u| \leq \sup |(-\Delta + V)(\phi_\lambda u)|$ holds. 
We compute
$$ -\Delta (\phi_\lambda u) = \phi_\lambda (-\Delta u) - 2\lambda (\nabla \phi)_\lambda \cdot \nabla u  + \lambda^2 (-\Delta \phi)_{\lambda} u $$
So that there exist some constant $C$ such that
$$ |-\Delta (\phi_\lambda u) + V(\phi_\lambda u)| \leq | \phi_\lambda (- \Delta + V)u| + C \lambda |\nabla u| + C \lambda^2 |u| $$
Using that $u\in C^2$, the last two terms are $O(\lambda)$ as $\lambda \to 0$. Therefore we have
$$ \lim_{\lambda \to 0} \sup |-\Delta (\phi_\lambda u) + V(\phi_\lambda u)| \leq \lim_{\lambda \to 0} \sup |\phi_\lambda (-\Delta + V)u| = \sup |-\Delta u + Vu|$$
and the claim is proved. 

The same trick can be performed on a Riemannian manifold $M$ provided you can construct a sequence of functions $\phi_k \in C^\infty_0(M)$ such that 


*

*on any compact $K\subset M$, $\phi_k \to 1$ uniformly. 

*$\sup |\nabla \phi_k| + |\Delta \phi_k| \to 0$. 


such a sequence can certainly be constructed when $M$ has good asymptotic structure. Off the top of my head I don't remember whether geodesic completeness is enough. (For things dealing with first derivatives geodesic completeness is enough since the function $r(x) = d(x,x_0)$ given by the distance function to a fixed point $x_0$ is $1$-Lipschitz, so we can take dilations of a one-dimensional bump function and consider $\phi_\lambda \circ r$. But for this problem you need second derivative estimates of the cut-off and so I am a bit unsure.)
