[proofplan]
We use the $L^2$ theory associated to the fixed Hermitian metric. The analytic input is the elliptic [Hodge decomposition](/theorems/2745) for the self-adjoint elliptic Dolbeault Laplacian $\Delta_{\bar\partial}$ on the compact manifold $X$, which gives the orthogonal splitting into harmonic, $\bar\partial$-exact, and $\bar\partial^*$-coexact parts. After obtaining this decomposition, the cohomological statement follows by two adjointness computations: exact harmonic forms have zero $L^2$ norm, and every $\bar\partial$-closed form has the same Dolbeault class as its harmonic component.
[/proofplan]
[step:Fix the metric adjoint and the harmonic notation]
Let $dV_g$ denote the Riemannian volume measure induced by the Hermitian metric $g$. For smooth forms $\alpha,\beta\in A^{p,q}(X)$, define the $L^2$ [inner product](/page/Inner%20Product) by
\begin{align*}
(\alpha,\beta)_{L^2}:=\int_X \langle \alpha(x),\beta(x)\rangle_g\,dV_g(x).
\end{align*}
Let
\begin{align*}
\bar\partial^*:A^{p,q+1}(X)\to A^{p,q}(X)
\end{align*}
denote the formal adjoint of
\begin{align*}
\bar\partial:A^{p,q}(X)\to A^{p,q+1}(X),
\end{align*}
so that
\begin{align*}
(\bar\partial\alpha,\eta)_{L^2}=(\alpha,\bar\partial^*\eta)_{L^2}
\end{align*}
for all $\alpha\in A^{p,q}(X)$ and $\eta\in A^{p,q+1}(X)$. The $\bar\partial$-Laplacian on $A^{p,q}(X)$ is
\begin{align*}
\Delta_{\bar\partial}=\bar\partial\bar\partial^*+\bar\partial^*\bar\partial,
\end{align*}
where the first summand uses $\bar\partial^*:A^{p,q}(X)\to A^{p,q-1}(X)$ and the second uses $\bar\partial:A^{p,q}(X)\to A^{p,q+1}(X)$, with the endpoint convention that the missing spaces are zero.
[/step]
[step:Apply elliptic Hodge theory to the Dolbeault Laplacian]
The operator
\begin{align*}
\Delta_{\bar\partial}:A^{p,q}(X)\to A^{p,q}(X)
\end{align*}
is a self-adjoint elliptic differential operator on the smooth vector bundle $\Lambda^{p,q}T^*X$. Since $X$ is compact and has no boundary, the elliptic [Hodge decomposition theorem](/theorems/3941) for the Dolbeault complex applies. Thus
\begin{align*}
A^{p,q}(X)=\mathcal H^{p,q}_{\bar\partial}(X)\oplus \bar\partial A^{p,q-1}(X)\oplus \bar\partial^*A^{p,q+1}(X)
\end{align*}
as an orthogonal direct sum for the $L^2$ inner product. This proves the asserted decomposition. Here the analytic theorem being used is the standard elliptic Hodge decomposition for self-adjoint elliptic complexes on compact manifolds, not yet available as a resolved theorem citation in the wiki.
[guided]
The point of introducing $\Delta_{\bar\partial}$ is that it packages the first-order complex
\begin{align*}
A^{p,q-1}(X)\xrightarrow{\bar\partial} A^{p,q}(X)\xrightarrow{\bar\partial} A^{p,q+1}(X)
\end{align*}
into a second-order [elliptic operator](/page/Elliptic%20Operator) on the single vector bundle $\Lambda^{p,q}T^*X$. The Hermitian metric supplies the $L^2$ inner product and hence the formal adjoint $\bar\partial^*$, so the operator
\begin{align*}
\Delta_{\bar\partial}=\bar\partial\bar\partial^*+\bar\partial^*\bar\partial
\end{align*}
is self-adjoint.
We now invoke the elliptic Hodge decomposition theorem for the Dolbeault complex. Its hypotheses are satisfied: $X$ is compact, the Hermitian metric makes $\Lambda^{p,q}T^*X$ a Hermitian vector bundle, the Dolbeault complex is elliptic, and $\Delta_{\bar\partial}$ is the associated self-adjoint elliptic Laplacian. The theorem gives the $L^2$-[orthogonal decomposition](/theorems/436)
\begin{align*}
A^{p,q}(X)=\ker\Delta_{\bar\partial}\oplus \bar\partial A^{p,q-1}(X)\oplus \bar\partial^*A^{p,q+1}(X).
\end{align*}
By definition,
\begin{align*}
\ker\Delta_{\bar\partial}=\mathcal H^{p,q}_{\bar\partial}(X),
\end{align*}
so this is exactly the asserted decomposition. The endpoint cases $q=0$ and $q=n$ are included by the conventions $A^{p,-1}(X)=0$ and $A^{p,n+1}(X)=0$.
[/guided]
[/step]
[step:Show that harmonic forms define Dolbeault cohomology classes]
Let $h\in\mathcal H^{p,q}_{\bar\partial}(X)$. Since $\Delta_{\bar\partial}h=0$, we pair the equation with $h$ and use the formal adjoint relation in the two adjacent bidegrees:
\begin{align*}
0=(\Delta_{\bar\partial}h,h)_{L^2}=(\bar\partial\bar\partial^*h,h)_{L^2}+(\bar\partial^*\bar\partial h,h)_{L^2}=\|\bar\partial^*h\|_{L^2}^2+\|\bar\partial h\|_{L^2}^2.
\end{align*}
Both summands are non-negative [real numbers](/page/Real%20Numbers), so
\begin{align*}
\bar\partial h=0
\end{align*}
and
\begin{align*}
\bar\partial^*h=0.
\end{align*}
Therefore $h$ is $\bar\partial$-closed, so its Dolbeault class $[h]\in H^{p,q}_{\bar\partial}(X)$ is defined. Hence
\begin{align*}
\Phi:\mathcal H^{p,q}_{\bar\partial}(X)\to H^{p,q}_{\bar\partial}(X),\qquad h\mapsto [h]
\end{align*}
is a well-defined [linear map](/page/Linear%20Map).
[/step]
[step:Prove injectivity by pairing an exact harmonic form with itself]
Suppose $h\in\mathcal H^{p,q}_{\bar\partial}(X)$ and $\Phi(h)=0$. Then there exists $\beta\in A^{p,q-1}(X)$ such that
\begin{align*}
h=\bar\partial\beta.
\end{align*}
If $q=0$, then $A^{p,-1}(X)=0$, so $h=0$. Otherwise, using the formal adjoint relation and the equality $\bar\partial^*h=0$ proved in the preceding step, we get
\begin{align*}
\|h\|_{L^2}^2=(h,h)_{L^2}=(h,\bar\partial\beta)_{L^2}=(\bar\partial^*h,\beta)_{L^2}=0.
\end{align*}
Since the $L^2$ norm is positive definite on smooth forms, $h=0$. Thus $\ker\Phi=\{0\}$, so $\Phi$ is injective.
[/step]
[step:Represent every Dolbeault class by the harmonic part of a closed form]
Let $[\omega]\in H^{p,q}_{\bar\partial}(X)$ be a Dolbeault cohomology class, and choose a representative $\omega\in A^{p,q}(X)$ satisfying
\begin{align*}
\bar\partial\omega=0.
\end{align*}
By the orthogonal decomposition already proved, there are unique forms
\begin{align*}
h\in\mathcal H^{p,q}_{\bar\partial}(X),\qquad \gamma\in A^{p,q-1}(X),\qquad \eta\in A^{p,q+1}(X)
\end{align*}
such that
\begin{align*}
\omega=h+\bar\partial\gamma+\bar\partial^*\eta.
\end{align*}
Apply $\bar\partial$ to this equality. Since $\bar\partial h=0$, $\bar\partial^2\gamma=0$, and $\bar\partial\omega=0$, we obtain
\begin{align*}
\bar\partial\bar\partial^*\eta=0.
\end{align*}
Pair this equation with $\eta$ and use the formal adjoint relation:
\begin{align*}
0=(\bar\partial\bar\partial^*\eta,\eta)_{L^2}=(\bar\partial^*\eta,\bar\partial^*\eta)_{L^2}=\|\bar\partial^*\eta\|_{L^2}^2.
\end{align*}
Hence $\bar\partial^*\eta=0$, and therefore
\begin{align*}
\omega=h+\bar\partial\gamma.
\end{align*}
Thus $[\omega]=[h]=\Phi(h)$ in $H^{p,q}_{\bar\partial}(X)$. Since every Dolbeault class has a harmonic representative, $\Phi$ is surjective. Combining injectivity and surjectivity, $\Phi$ is an isomorphism.
[/step]
