Question in proof of Hodge decomposition theorem I am reading this set of notes on Hodge theory: 
https://math.unice.fr/~hoering/hodge/hodge.pdf
In particular the proof of 4.2.6 where the author proves the Hodge decomposition for a compact hermitian manifold. 
The author wrote that 
$\bar{\partial}(C^{\infty}(X,\Omega^{p,q-1}))=\bar{\partial}\bar{\partial^*}(C^{\infty}(X,\Omega^{p,q}))$ 
and this is due to that the fact that $\bar{\partial^2}=0$ and that a form $\alpha$ is $\bar{\partial}$-harmonic iff $\bar{\partial}\alpha=0$ and $\bar{\partial^*}\alpha=0$.
The inclusion $ \bar{\partial}\bar{\partial^*}(C^{\infty}(X,\Omega^{p,q}))\subset\bar{\partial}(C^{\infty}(X,\Omega^{p,q-1}))$ is clear to me however I do not see how the two facts mentioned above helps to prove the equality. 
 A: As the proof states, we have
$$C^{\infty}(X, \Omega^{p,q-1}) = \mathcal{H}^{p,q-1}(X)\oplus\Delta_{\bar{\partial}}(C^{\infty}(X, \Omega^{p,q-1})).$$
If $\alpha \in C^{\infty}(X, \Omega^{p,q-1})$ there are forms $\beta \in \mathcal{H}^{p,q-1}(X)$ and $\gamma \in C^{\infty}(X, \Omega^{p,q-1})$ such that 
$$\alpha = \beta + \Delta_{\bar{\partial}}\gamma = \beta + \bar{\partial}\bar{\partial}^*\gamma + \bar{\partial}^*\bar{\partial}\gamma.$$
Now 
$$\bar{\partial}\alpha = \bar{\partial}\beta + \bar{\partial}\bar{\partial}\bar{\partial}^*\gamma + \bar{\partial}\bar{\partial}^*\bar{\partial}\gamma.$$
Note that $\bar{\partial}\beta = 0$ as $\beta$ is $\bar{\partial}$-harmonic and $\bar{\partial}\bar{\partial}\bar{\partial}^*\gamma = 0$ as $\bar{\partial}^2 = 0$. Therefore we have $\bar{\partial}\alpha = \bar{\partial}\bar{\partial}^*(\bar{\partial}\gamma)$. That is, anything in the image of $\bar{\partial}$ is actually in the image of $\bar{\partial}\bar{\partial}^*$, so $\bar{\partial}C^{\infty}(X, \Omega^{p,q-1}) \subseteq \bar{\partial}\bar{\partial}^*C^{\infty}(X, \Omega^{p,q})$.
