Then $L_f(f)=E(f)$ and $L_f$ is a linear *second order* elliptic operator with coefficients in $C^{0,a}$ and $E(f)$ is in $C^{0,a}$. So by elliptic regularity we can deduce that $f\in C^{2,a}$. We have gained one degree of regularity, so we can “bootstrap”, i.e. proceed by induction and deduce that any $C^{1,a}$ solution to $P(f)=0$ is smooth.

$C^{1,a}$-minimal submanifolds (and thus calibrated submanifolds) are *smooth*.

More sophisticated techniques can be used to deduce that $C^1$-minimal submanifolds are real analytic [@Morrey]. Notice that elliptic regularity results are *not* valid for $C^k$ spaces, so this result is not obvious.

We can also arrange our simple equation $P(f)=0$ as $\Delta f+Q(\nabla f,\nabla^2f)=0$, where $Q$ is nonlinear but linear in $\nabla^2f$. If we know that $\int_N P(f)\operatorname{vol}_N=0$, i.e. that $P(f)$ is orthogonal to the constants, then we can always solve $\Delta f_0=-Q(\nabla f,\nabla^2f)$. We do know that $\int_NP(f)\operatorname{vol}_N=0$ since $P$ has a divergence form. This means we are in the setting for implementing the Implicit Function Theorem for Banach spaces to conclude that we can always solve $P(f)=0$ for some $f$ near $0$, and $f$ will be smooth by our regularity argument above. In general, we will use the following.

Let $X,Y$ be Banach spaces, let 