f=0$). Hence, we can solve $\Delta f=h$ if and only if $h$ is orthogonal to the constants, i.e. $\int_N h\operatorname{vol}_N=0$.

The operator defining the minimal graph equation for a hypersurface is $$P(f)=-\operatorname{div}\left(\frac{\nabla f}{\sqrt{1+|\nabla f|^2}}\right),$$ which is a nonlinear second order operator whose linearisation $L_0P$ at $0$ is $\Delta$. Thus $P$ is a nonlinear elliptic operator at $0$. If we linearise $P$ at $f_0$ we find a more complicated expression depending on $f_0$, but it is still a perturbation of the Laplacian.

Suppose we are on a compact manifold $N$ and we want to solve $P(f)=0$ where $P$ is the minimal graph operator on functions $f$. Let us consider regularity for $f$. We can re-arrange $P(f)=0$ by taking all of the second derivatives to one side as: $$R(x,\nabla f(x))\nabla^2 f(x)=E(x,\nabla f(x))$$ where $x\in N$. Since $L_0P=\Delta$ is elliptic and ellipticity is an open condition we know that the operator $L_f$ (depending on $f$) given by $$L_f(h)(x)=R(x,\nabla f(x))\nabla^2h(x)$$ is a *linear* elliptic operator whenever $\|\nabla f\|_{C^0}$ is small, in particular if $\|f\|_{C^{1,a}}$ is sufficiently small. The operator $L_f$ does not have smooth coefficients, but if $f\in C^{k,a}$ then the coefficients $R\in C^{k-1,a}$.

Suppose that $f\in C^{1,a}$ and $\|f\|_{C^{1,a}}$ is small with $P(f)=0$. 