[proofplan]
The proof is the energy identity for the Dolbeault Laplacian. We expand $\Delta_{\bar\partial}$ as $\bar\partial\bar\partial^*+\bar\partial^*\bar\partial$, take the $L^2$ [inner product](/page/Inner%20Product) with $\alpha$, and use the defining adjoint identity for $\bar\partial^*$. This expresses $(\Delta_{\bar\partial}\alpha,\alpha)_{L^2}$ as a sum of two squared $L^2$ norms, so vanishing of the Laplacian forces both first-order terms to vanish; the converse follows immediately from the same formula for $\Delta_{\bar\partial}$.
[/proofplan]
[step:Expand the Dolbeault Laplacian against the form]
Let $(\cdot,\cdot)_{L^2}$ denote the Hermitian $L^2$ inner product on smooth complex-valued forms induced by $g$, linear in the first argument and conjugate-linear in the second. Let
\begin{align*}
\|\beta\|_{L^2}^2 := (\beta,\beta)_{L^2}
\end{align*}
for every smooth form $\beta$ on $X$. By definition, the Dolbeault Laplacian acting on $(p,q)$-forms is
\begin{align*}
\Delta_{\bar\partial}: A^{p,q}(X) \to A^{p,q}(X), \qquad \Delta_{\bar\partial}=\bar\partial\bar\partial^*+\bar\partial^*\bar\partial.
\end{align*}
Thus
\begin{align*}
(\Delta_{\bar\partial}\alpha,\alpha)_{L^2} = (\bar\partial\bar\partial^*\alpha,\alpha)_{L^2}+(\bar\partial^*\bar\partial\alpha,\alpha)_{L^2}.
\end{align*}
[/step]
[step:Convert the two summands into squared norms]
The formal adjoint identity for $\bar\partial^*$ says that, for smooth forms $\beta$ and $\gamma$ of compatible bidegrees,
\begin{align*}
(\bar\partial\beta,\gamma)_{L^2}=(\beta,\bar\partial^*\gamma)_{L^2}.
\end{align*}
Applying this first with $\beta=\bar\partial^*\alpha$ and $\gamma=\alpha$ gives
\begin{align*}
(\bar\partial\bar\partial^*\alpha,\alpha)_{L^2}=(\bar\partial^*\alpha,\bar\partial^*\alpha)_{L^2}=\|\bar\partial^*\alpha\|_{L^2}^2.
\end{align*}
Applying the same adjoint identity in the equivalent form
\begin{align*}
(\bar\partial^*\eta,\theta)_{L^2}=(\eta,\bar\partial\theta)_{L^2}
\end{align*}
with $\eta=\bar\partial\alpha$ and $\theta=\alpha$ gives
\begin{align*}
(\bar\partial^*\bar\partial\alpha,\alpha)_{L^2}=(\bar\partial\alpha,\bar\partial\alpha)_{L^2}=\|\bar\partial\alpha\|_{L^2}^2.
\end{align*}
Therefore
\begin{align*}
(\Delta_{\bar\partial}\alpha,\alpha)_{L^2}=\|\bar\partial\alpha\|_{L^2}^2+\|\bar\partial^*\alpha\|_{L^2}^2.
\end{align*}
[guided]
The point of taking the $L^2$ inner product with $\alpha$ is that the second-order operator $\Delta_{\bar\partial}$ becomes a sum of first-order energies. The operator $\bar\partial^*$ is defined as the formal adjoint of $\bar\partial$ for the fixed Hermitian metric $g$, so for smooth forms $\beta$ and $\gamma$ of compatible bidegrees we have
\begin{align*}
(\bar\partial\beta,\gamma)_{L^2}=(\beta,\bar\partial^*\gamma)_{L^2}.
\end{align*}
We apply this identity to the first term in the expansion of the Laplacian. Since $\alpha\in A^{p,q}(X)$, the form $\bar\partial^*\alpha$ lies in $A^{p,q-1}(X)$, and $\bar\partial(\bar\partial^*\alpha)$ lies in $A^{p,q}(X)$, so the inner product with $\alpha$ is defined. With $\beta=\bar\partial^*\alpha$ and $\gamma=\alpha$, the adjoint identity gives
\begin{align*}
(\bar\partial\bar\partial^*\alpha,\alpha)_{L^2}=(\bar\partial^*\alpha,\bar\partial^*\alpha)_{L^2}.
\end{align*}
By the definition of the $L^2$ norm, this is
\begin{align*}
(\bar\partial\bar\partial^*\alpha,\alpha)_{L^2}=\|\bar\partial^*\alpha\|_{L^2}^2.
\end{align*}
For the second term, use the same adjoint relation with the roles shifted. Since $\bar\partial\alpha\in A^{p,q+1}(X)$, applying $\bar\partial^*$ to $\bar\partial\alpha$ gives a form in $A^{p,q}(X)$. Taking $\eta=\bar\partial\alpha$ and $\theta=\alpha$, the adjoint identity gives
\begin{align*}
(\bar\partial^*\bar\partial\alpha,\alpha)_{L^2}=(\bar\partial\alpha,\bar\partial\alpha)_{L^2}.
\end{align*}
Again by the definition of the $L^2$ norm,
\begin{align*}
(\bar\partial^*\bar\partial\alpha,\alpha)_{L^2}=\|\bar\partial\alpha\|_{L^2}^2.
\end{align*}
Combining the two displayed identities with
\begin{align*}
\Delta_{\bar\partial}\alpha=\bar\partial\bar\partial^*\alpha+\bar\partial^*\bar\partial\alpha
\end{align*}
yields the energy identity
\begin{align*}
(\Delta_{\bar\partial}\alpha,\alpha)_{L^2}=\|\bar\partial\alpha\|_{L^2}^2+\|\bar\partial^*\alpha\|_{L^2}^2.
\end{align*}
This identity is the whole mechanism of the criterion: the left side records harmonicity, while the right side is a sum of nonnegative squared norms.
[/guided]
[/step]
[step:Derive the two first-order equations from harmonicity]
Assume that $\alpha$ is $\bar\partial$-harmonic, meaning
\begin{align*}
\Delta_{\bar\partial}\alpha=0.
\end{align*}
Taking the $L^2$ inner product with $\alpha$ gives
\begin{align*}
0=(\Delta_{\bar\partial}\alpha,\alpha)_{L^2}.
\end{align*}
By the energy identity,
\begin{align*}
0=\|\bar\partial\alpha\|_{L^2}^2+\|\bar\partial^*\alpha\|_{L^2}^2.
\end{align*}
Both summands are nonnegative [real numbers](/page/Real%20Numbers). Hence
\begin{align*}
\|\bar\partial\alpha\|_{L^2}^2=0, \qquad \|\bar\partial^*\alpha\|_{L^2}^2=0.
\end{align*}
The $L^2$ norm is positive definite on smooth forms on the compact manifold $X$, so
\begin{align*}
\bar\partial\alpha=0, \qquad \bar\partial^*\alpha=0.
\end{align*}
[/step]
[step:Use the two first-order equations to recover harmonicity]
Conversely, assume
\begin{align*}
\bar\partial\alpha=0, \qquad \bar\partial^*\alpha=0.
\end{align*}
Using the defining formula for the Dolbeault Laplacian,
\begin{align*}
\Delta_{\bar\partial}\alpha=\bar\partial\bar\partial^*\alpha+\bar\partial^*\bar\partial\alpha.
\end{align*}
Substituting the two assumed vanishing identities gives
\begin{align*}
\Delta_{\bar\partial}\alpha=\bar\partial 0+\bar\partial^*0=0.
\end{align*}
Therefore $\alpha$ is $\bar\partial$-harmonic. This proves the equivalence.
[/step]
