$\DeclareMathOperator{im}{im}$

I want to prove that Bott-Chern cohomology group $H^{p,q}_{BC}=\frac{\ker\partial\cap \ker\bar{\partial}}{\im\partial\bar{\partial}}$ has finite dimension via Hodge decomposition like argument. References lead me to this article written by M. Schweitzer.

In 2.b he defines "Bott-Chern Laplacian" as $\Delta^{p,q}_{BC}=(\partial\bar{\partial})(\partial\bar{\partial})^*+\partial^*\partial+\bar{\partial}^*\bar{\partial}.$ He states that if $\Delta^{p,q}_{BC}$ where elliptic, then we would have following decomposition: $$\mathcal{E^{p,q}}=\mathcal{H}^{p,q}_{\Delta_{BC}}\oplus\im\partial\bar{\partial}\oplus (\im\partial^*+\im\bar{\partial}^*).\hspace{20pt}(*)$$ So far so good. Cause $\Delta^{p,q}_{BC}$ is self-adjoint and assuming that $\Delta^{p,q}_{BC}$ is elliptic gives us that $$\mathcal{E^{p,q}}=\mathcal{H}^{p,q}_{\Delta_{BC}}\oplus\im\Delta^{p,q}_{BC}.$$ Repeating standard Hodge decomposition argument we notice that $$\mathcal{H}^{p,q}_{\Delta_{BC}}\perp\im (\partial\bar{\partial})(\partial\bar{\partial})^*\perp \im(\partial^*\partial+\bar{\partial}^*\bar{\partial})\perp\mathcal{H}^{p,q}_{\Delta_{BC}}$$ Additionally we easily verify that $$\mathcal{H}^{p,q}_{\Delta_{BC}}\perp\im (\partial\bar{\partial})^*\perp \im(\partial^*\partial+\bar{\partial}^*\bar{\partial})\hspace{5pt}\text{and} \hspace{5pt}\mathcal{H}^{p,q}_{\Delta_{BC}}\perp( \im\partial^*+\im\bar{\partial}^*)\perp \im (\partial\bar{\partial})(\partial\bar{\partial})^*.$$ So $(*)$ holds (assuming that $\Delta^{p,q}_{BC}$ is elliptic).

Next he shows that $\Delta^{p,q}_{BC}$ is not elliptic. As a result he introduces new operator $$\tilde{\Delta}_{BC}^{p,q}=(\partial\bar\partial)(\partial\bar\partial)^*+(\partial\bar\partial)^*(\partial\bar\partial)+(\bar\partial^*\partial)(\bar\partial^*\partial)^*+(\bar\partial^*\partial)^*(\bar\partial^*\partial)+\bar\partial^*\bar\partial+\partial^*\partial$$ and proves its ellipticity. After that he notice that $$\tilde{\Delta}_{BC}^{p,q}=0\iff \bar\partial=\partial=\bar\partial^*\partial^*=0.$$ Which in fact means that $\ker\Delta_{BC}^{p,q}=\ker\tilde{\Delta}_{BC}^{p,q},$ i.e. $\mathcal{H}^{p,q}_{\Delta_{BC}}=\mathcal{H}^{p,q}_{\tilde{\Delta}_{BC}}.$ As a result of the above he states Theorem 2.2 which claims exactly $(*).$

I do not see how to infer $(*)$. Since $\Delta_{BC}^{p,q}$ is not elliptic, I believe we cannot infer $(*)$ just by the fact that $\mathcal{H}^{p,q}_{\Delta_{BC}}=\mathcal{H}^{p,q}_{\tilde{\Delta}_{BC}}.$

PS. Article is written in French, so I might misunderstood something. Please correct me if I wrote something incorrectly.