# Finite dimensionality of the “deRham cohomology” defined using $C^{k,\alpha}$ forms instead of smooth forms.

Let $M$ be a compact manifold. The deRham cohomology group $H^{^\bullet}_{DR}(M)$ is defined to be the cohomology of the deRham complex $\left({\cal C}^{\infty}(\bigwedge^{^\bullet}T^*M),d\right)$. The elliptic PDE theory shows that the deRham cohomology group $H^{^\bullet}_{DR}(M)$ is finite-dimensional.

Now what if we consider forms in $C^{k,\alpha}(\bigwedge^{^\bullet}T^*M)$ instead of smooth forms? Is it true that the quotient \begin{equation} \frac{\text{ker}(d:C^{k,\alpha}(\bigwedge^pT^*M)\longrightarrow C^{k-1,\alpha}(\bigwedge^{p+1}T^*M))}{\text{Im}(d:C^{k+1,\alpha}(\bigwedge^{p-1}T^*M)\longrightarrow C^{k,\alpha}(\bigwedge^pT^*M))} \end{equation} is also finite-dimensional?

The reason why I ask this question is because I want to show that the image of $d:C^{k+1,\alpha}(\bigwedge^{p-1}T^*M)\longrightarrow C^{k,\alpha}(\bigwedge^pT^*M)$ is closed (in the sense of topology). This is an issue I encountered when I was reading McLean's paper "Deformations of Calibrated Submanifolds" (page 722).

If the above quotient is finite-dimensional, then the result follows from Banach open mapping theorem.

Thanks in advance.