# A question about the proof of McLean's theorem: why is the space of $C^{k,\alpha}$ exact $p$-forms a Banach space?

In the famous paper "Deformations of Calibrated Submanifolds" by Robert McLean, he showed that given a smooth compact special Lagrangian $L$ in a Calabi-Yau manifold $(X^{2n},\omega,\Omega)$, there is a diffeomorphism mapping a neighborhood of 0 in the space of harmonic 1-forms on $L$ onto the set of all smooth Lagrangian submanifolds "near" $L$.

The proof goes roughly as follows, he defined the map $F:{\cal U}\subset C^{1,\alpha}(N_L)\rightarrow C^{0,\alpha}(\bigwedge^nT^*L)\oplus C^{0,\alpha}(\bigwedge^2T^*L)$ by sending a small normal vector field $V$ on $L$ to $\big((exp_V)^*(\text{Re}(\Omega)),(exp_V)^*(\omega) \big)$, where ${\cal U}$ is a small neighborhood of $0$ and $N_L$ is the normal bundle of $L$ in $X$ which is the cotangent bundle of $L$. Then $F^{-1}(0)\ni 0$ is the set of normal vector fields which flow $L$ to a special Lagrangian submanifold using the exponential map, and so it is the set we are looking at (modulo a regularity issue). To give it a smooth structure, it suffices to show that $dF(0)$ is surjective (since ${\cal U}$ is allowed to be shrunk as small as we like).

But it turns out that $dF(0)$ is not surjective, unless we change the range of $F$ to the direct sum of the space of exact $C^{0,\alpha}$ $n$-forms and the space of exact $C^{0,\alpha}$ $2$-forms. Here comes my question: In order to apply the Banach space implicit function theorem, the range of $F$ namely, the space of exact forms endowed with Holder norm has to be a Banach space, i.e. it is closed (in the the topological sense) in all $C^{0,\alpha}$ forms. But it is not obvious to me why it is true.

It is clear that the space of closed $C^{k,\alpha}$ $p$-forms is closed as it is the zero set of $d:C^{k,\alpha}(\bigwedge^pT^*L)\rightarrow C^{k-1,\alpha}(\bigwedge^{p+1}T^*L)$. I wonder if the exact $C^{k,\alpha}$ $p$-forms are not so different from the closed $C^{k,\alpha}$ $p$-forms in the sense that their quotient is finite dimensional (which is true when $k=\infty$) so that I may find some hope to solve this question (although this doesn't guarantee that the space of $C^{k,\alpha}$ exact forms is automatically closed).