Let $(M, g_{ij})$ be a compact Riemannian manifold. Let $f \in C^{0,\alpha}(M)$ be a given function. I have the linear operator $L = \Delta h - 12u^2h$, where $u \in C^{2,\alpha}$. I need to show that $L$ is onto; i.e., I need to show that $\forall f \in C^{0,\alpha}$, $\exists h \in C^{2,\alpha}$ such that $Lh=f$.
The only method I have at my disposal for solving $Lh=f$ is the variational method. So, what I need to do first is show that my functional $I(h)=\int \left( \Delta h - 12u^2h^2-fh\right)$ is bounded from below. Since I'm on a compact manifold, I do not have compact support, and thus cannot do this the way I would normally have done it, which would have been the following:
$I(h) = \int\left ( \Delta h - 12u^2h^2-fh\right) = ||Dh||_{L^{2}}^{2} - \int(-12u^{2})(h^{2})-\int fh = ||Dh||_{L^2}^{2} - ||12u||_{L^{2}}^{2}||h||_{L^{2}}^{2}-||f||_{L^{2}}||h||_{L^{2}}^{2} \geq ||Dh||_{L^{2}}^{2} - \left( \frac{\epsilon}{2}||h||_{L^{2}}^{2}+\frac{1}{2\epsilon}||12u||_{L^2}^{2}\right) - \left( \frac{\epsilon}{2}||h||_{L^{2}}^{2}+\frac{1}{2\epsilon}||f||^{2}_{L^{2}}\right) = ||Dh||_{L^{2}}^{2}-\epsilon ||h||_{L^{2}}^{2} - \frac{1}{2\epsilon}\left(||12u||^{2}_{L^{2}} + ||f||_{L^{2}}^{2}\right)$.
Then, since $h$ is the quantity that varies, the only way that $I(h)$ can be bounded from below is for it to be positive, so we need to have $||Dh||_{L^{2}}^{2} \geq \epsilon ||h||_{L^{2}}^{2}$, which is guaranteed by the Poincare Inequality, so $I(h) \geq ||Dh||_{L^{2}}^{2} \geq \epsilon ||h||_{L^{2}}^{2}$, and thus $I(h)$ is bounded below.
But, I cannot use this. In order to use the Poincare Inequality in the space I am in, where I do not have compact support, I need to have $||Dh||_{L^{2}}^{2} \geq \epsilon ||h - \overline{h}||_{L^{2}}^{2}$, where $\overline{h}$ is the average value of the function $h$. However, I don't know at all how to fit this in with what I've done above. Please help!