Let $X$ be a compact Kähler manifold of dimension $n$ with Kähler metric $g$ and Kähler form $\omega$.

Step 1: Operators and the Hodge Laplacian

Define the adjoint operators using the metric: $\partial^* = -* \bar{\partial} *, \quad \bar{\partial}^* = -* \partial *$

where $*$ is the Hodge star operator. The Hodge Laplacian is: $\Delta = \partial\partial^* + \partial^*\partial + \bar{\partial}\bar{\partial}^* + \bar{\partial}^*\bar{\partial}$

On a Kähler manifold, we have the key identity: $\Delta = 2(\bar{\partial}\bar{\partial}^* + \bar{\partial}^*\bar{\partial}) = 2(\partial\partial^* + \partial^*\partial)$

This follows from the Kähler identities $[\Lambda, \partial] = -i\bar{\partial}^*$ and $[\Lambda, \bar{\partial}] = -i\partial^*$, where $\Lambda$ is the adjoint of wedging with $\omega$.

Step 2: Ellipticity and Harmonic Forms

The operator $\Delta$ is a second-order elliptic operator. By elliptic theory:

1. The kernel $\ker\Delta$ is finite-dimensional
2. Elements of $\ker\Delta$ are smooth
3. There is an orthogonal decomposition: $\Omega^k(X) = \ker\Delta \oplus \text{Im}\,\Delta$

Define harmonic forms as $\mathcal{H}^k(X) = \ker\Delta \cap \Omega^k(X)$.

Step 3: Harmonic Representatives

For any closed form $\alpha$ with $d\alpha = 0$, consider the minimization problem: $\min\{\|\beta\|_{L^2} : \beta \in [\alpha]\}$

over the cohomology class $[\alpha]$. The minimizer $\beta$ satisfies: $\langle \Delta\beta, \gamma\rangle = 0 \quad \text{for all } \gamma \in \Omega^k(X)$

This implies $\Delta\beta = 0$, so $\beta$ is harmonic. Moreover: $\|\beta\|_{L^2}^2 = \langle\beta, \beta\rangle = \langle d\delta\beta + \delta d\beta, \beta\rangle = \|d\beta\|_{L^2}^2 + \|\delta\beta\|_{L^2}^2$

Since $\beta$ minimizes the $L^2$ norm and represents the same class, we must have $d\beta = \delta\beta = 0$.

Step 4: Type Decomposition

On forms, we have $d = \partial + \bar{\partial}$ and $\delta = \partial^* + \bar{\partial}^*$. For a harmonic form $\alpha$: $0 = \langle\Delta\alpha, \alpha\rangle = \|\partial\alpha\|^2 + \|\partial^*\alpha\|^2 + \|\bar{\partial}\alpha\|^2 + \|\bar{\partial}^*\alpha\|^2$

Each term is non-negative, so each must vanish: $\partial\alpha = \partial^*\alpha = \bar{\partial}\alpha = \bar{\partial}^*\alpha = 0$

Now decompose $\alpha$ into types: $\alpha = \sum_{p+q=k} \alpha^{p,q}$. Since $\bar{\partial}\alpha = 0$: $\sum_{p+q=k} \bar{\partial}\alpha^{p,q} = \sum_{p+q=k} \bar{\partial}\alpha^{p,q} + (\text{mixed types}) = 0$

Comparing types, each $\bar{\partial}\alpha^{p,q} = 0$. Similarly, $\partial\alpha^{p,q} = 0$ and the adjoint conditions hold for each component.

Step 5: Harmonic $(p,q)$-forms

This shows each $\alpha^{p,q}$ is independently harmonic for the Laplacian $\Delta_{\bar{\partial}} = \bar{\partial}\bar{\partial}^* + \bar{\partial}^*\bar{\partial}$ acting on $(p,q)$-forms. Thus: $\mathcal{H}^k(X) = \bigoplus_{p+q=k} \mathcal{H}^{p,q}(X)$

Step 6: Isomorphism with Cohomology

The inclusion $\mathcal{H}^{p,q}(X) \hookrightarrow H^{p,q}(X)$ sending harmonic forms to their cohomology classes is injective (since harmonic forms represent unique classes). For surjectivity, every $\bar{\partial}$-closed form differs from its harmonic projection by a $\bar{\partial}$-exact form.

Therefore: $\mathcal{H}^{p,q}(X) \cong H^{p,q}(X)$

and combining with the de Rham isomorphism $\bigoplus_{p+q=k} H^{p,q}(X) \cong H^k(X,\mathbb{C})$ completes the proof.