[proofplan]
The proof treats $d$, $\partial$, and $\bar\partial$ by the same Hilbert-space identity. For each operator $D$, the natural Laplacian is $\Delta_D=DD^*+D^*D$, and pairing $\Delta_D\alpha$ with $\alpha$ converts the expression into the sum of the two squared $L^2$ norms $\|D^*\alpha\|_{L^2}^2+\|D\alpha\|_{L^2}^2$. Positive definiteness of the $L^2$ norm then forces both terms to vanish when $\Delta_D\alpha=0$, while the reverse implication follows immediately from substituting $D\alpha=0$ and $D^*\alpha=0$ into the definition of $\Delta_D$.
[/proofplan]
[step:Derive the basic identity for an operator and its formal adjoint]
Let $D$ denote one of the three operators $d$, $\partial$, or $\bar\partial$, acting on the relevant smooth forms, and let $D^*$ denote its formal adjoint with respect to the $L^2$ [inner product](/page/Inner%20Product) induced by $g$. Thus, for smooth forms $\eta$ and $\beta$ of matching degrees,
\begin{align*}
(D\eta,\beta)_{L^2}=(\eta,D^*\beta)_{L^2}.
\end{align*}
Since $X$ is compact and has no boundary, this formal adjoint identity is global and has no boundary correction term.
Let $\alpha$ be a smooth form in the domain of $\Delta_D$. By definition,
\begin{align*}
\Delta_D\alpha=DD^*\alpha+D^*D\alpha.
\end{align*}
Taking the $L^2$ inner product with $\alpha$ and using sesquilinearity gives
\begin{align*}
(\Delta_D\alpha,\alpha)_{L^2}=(DD^*\alpha,\alpha)_{L^2}+(D^*D\alpha,\alpha)_{L^2}.
\end{align*}
Apply the formal adjoint relation to the first term with $\eta=D^*\alpha$ and $\beta=\alpha$:
\begin{align*}
(DD^*\alpha,\alpha)_{L^2}=(D^*\alpha,D^*\alpha)_{L^2}.
\end{align*}
Apply the same adjoint relation to the second term in the equivalent form
\begin{align*}
(D^*\gamma,\alpha)_{L^2}=(\gamma,D\alpha)_{L^2},
\end{align*}
with $\gamma=D\alpha$. This gives
\begin{align*}
(D^*D\alpha,\alpha)_{L^2}=(D\alpha,D\alpha)_{L^2}.
\end{align*}
Therefore
\begin{align*}
(\Delta_D\alpha,\alpha)_{L^2}=\|D^*\alpha\|_{L^2}^2+\|D\alpha\|_{L^2}^2.
\end{align*}
[guided]
The key point is that the Laplacian is built from an operator and its formal adjoint, so its inner product with $\alpha$ becomes a sum of squares. Let $D$ be one of $d$, $\partial$, or $\bar\partial$, and let $D^*$ be the corresponding formal adjoint for the fixed Hermitian metric $g$. The defining property of the formal adjoint is that, whenever the degrees match,
\begin{align*}
(D\eta,\beta)_{L^2}=(\eta,D^*\beta)_{L^2}.
\end{align*}
The compactness and absence of boundary of $X$ are exactly what allow this identity to hold globally without boundary terms.
For a smooth form $\alpha$ in the relevant degree, the natural Laplacian associated to $D$ is
\begin{align*}
\Delta_D\alpha=DD^*\alpha+D^*D\alpha.
\end{align*}
Pairing this expression with $\alpha$ gives
\begin{align*}
(\Delta_D\alpha,\alpha)_{L^2}=(DD^*\alpha,\alpha)_{L^2}+(D^*D\alpha,\alpha)_{L^2}.
\end{align*}
Now use the formal adjoint relation on each term. In the first term, take $\eta=D^*\alpha$ and $\beta=\alpha$. Then
\begin{align*}
(DD^*\alpha,\alpha)_{L^2}=(D^*\alpha,D^*\alpha)_{L^2}=\|D^*\alpha\|_{L^2}^2.
\end{align*}
In the second term, use that $D$ and $D^*$ are formal adjoints of one another, so with $\gamma=D\alpha$,
\begin{align*}
(D^*D\alpha,\alpha)_{L^2}=(D\alpha,D\alpha)_{L^2}=\|D\alpha\|_{L^2}^2.
\end{align*}
Combining the two identities yields
\begin{align*}
(\Delta_D\alpha,\alpha)_{L^2}=\|D^*\alpha\|_{L^2}^2+\|D\alpha\|_{L^2}^2.
\end{align*}
This is the whole mechanism of the theorem: the equation $\Delta_D\alpha=0$ is converted into the vanishing of two nonnegative squared norms.
[/guided]
[/step]
[step:Apply the identity to the de Rham Laplacian]
Let $\alpha\in A^k(X)$ be a smooth complex-valued $k$-form. Apply the identity from the previous step with
\begin{align*}
D=d:A^k(X)\to A^{k+1}(X).
\end{align*}
Then
\begin{align*}
(\Delta_d\alpha,\alpha)_{L^2}=\|d^*\alpha\|_{L^2}^2+\|d\alpha\|_{L^2}^2.
\end{align*}
If $\Delta_d\alpha=0$, then the left-hand side is $0$, so
\begin{align*}
\|d^*\alpha\|_{L^2}^2+\|d\alpha\|_{L^2}^2=0.
\end{align*}
Both summands are nonnegative [real numbers](/page/Real%20Numbers), and the $L^2$ norm is positive definite on smooth forms. Hence
\begin{align*}
d^*\alpha=0, \qquad d\alpha=0.
\end{align*}
Conversely, if $d\alpha=0$ and $d^*\alpha=0$, then substituting into
\begin{align*}
\Delta_d\alpha=dd^*\alpha+d^*d\alpha
\end{align*}
gives $\Delta_d\alpha=0$. Thus
\begin{align*}
\Delta_d\alpha=0 \iff d\alpha=0 \text{ and } d^*\alpha=0.
\end{align*}
[/step]
[step:Apply the same identity to the $\partial$-Laplacian]
Let $\alpha\in A^{p,q}(X)$ be a smooth $(p,q)$-form. Apply the basic identity with
\begin{align*}
D=\partial:A^{p,q}(X)\to A^{p+1,q}(X).
\end{align*}
Then
\begin{align*}
(\Delta_\partial\alpha,\alpha)_{L^2}=\|\partial^*\alpha\|_{L^2}^2+\|\partial\alpha\|_{L^2}^2.
\end{align*}
If $\Delta_\partial\alpha=0$, positive definiteness of the $L^2$ norm gives
\begin{align*}
\partial^*\alpha=0, \qquad \partial\alpha=0.
\end{align*}
Conversely, if $\partial\alpha=0$ and $\partial^*\alpha=0$, then
\begin{align*}
\Delta_\partial\alpha=\partial\partial^*\alpha+\partial^*\partial\alpha=0.
\end{align*}
Therefore
\begin{align*}
\Delta_\partial\alpha=0 \iff \partial\alpha=0 \text{ and } \partial^*\alpha=0.
\end{align*}
[/step]
[step:Apply the same identity to the $\bar\partial$-Laplacian]
Let $\alpha\in A^{p,q}(X)$ be a smooth $(p,q)$-form. Apply the basic identity with
\begin{align*}
D=\bar\partial:A^{p,q}(X)\to A^{p,q+1}(X).
\end{align*}
Then
\begin{align*}
(\Delta_{\bar\partial}\alpha,\alpha)_{L^2}=\|\bar\partial^*\alpha\|_{L^2}^2+\|\bar\partial\alpha\|_{L^2}^2.
\end{align*}
If $\Delta_{\bar\partial}\alpha=0$, positive definiteness of the $L^2$ norm gives
\begin{align*}
\bar\partial^*\alpha=0, \qquad \bar\partial\alpha=0.
\end{align*}
Conversely, if $\bar\partial\alpha=0$ and $\bar\partial^*\alpha=0$, then
\begin{align*}
\Delta_{\bar\partial}\alpha=\bar\partial\bar\partial^*\alpha+\bar\partial^*\bar\partial\alpha=0.
\end{align*}
Therefore
\begin{align*}
\Delta_{\bar\partial}\alpha=0 \iff \bar\partial\alpha=0 \text{ and } \bar\partial^*\alpha=0.
\end{align*}
Together with the de Rham and $\partial$ cases, this proves all three asserted harmonicity criteria.
[/step]
