[proofplan]
The proof uses the Kähler Laplacian identities for the Hermitian metric induced by the Kähler form $\omega$. These identities say, with the stated Laplacian normalization, that
\begin{align*}
\Delta_d=2\Delta_\partial=2\Delta_{\bar\partial}
\end{align*}
on smooth complex-valued forms. Since multiplication of an operator by the positive scalar $2$ does not change its kernel, the three harmonic kernels coincide. Finally, the definition of harmonicity by vanishing of the corresponding Laplacian translates this kernel equality into the stated equivalence.
[/proofplan]
[step:Invoke the Kähler Laplacian identities with the stated normalization]
Let $A^\bullet(X;\mathbb C)$ denote the space of smooth complex-valued differential forms on $X$. We use the standard Kähler Laplacian identities for the Kähler metric induced by $\omega$:
\begin{align*}
\Delta_d = 2\Delta_\partial
\end{align*}
and
\begin{align*}
\Delta_d = 2\Delta_{\bar\partial}
\end{align*}
as operators
\begin{align*}
A^\bullet(X;\mathbb C)\to A^\bullet(X;\mathbb C).
\end{align*}
This is the standard Kähler Laplacian identity for Laplacians with the convention
\begin{align*}
\Delta_D=DD^*+D^*D.
\end{align*}
[guided]
Let $A^\bullet(X;\mathbb C)$ be the space of smooth complex-valued differential forms on $X$. The metric used in all adjoints is the Hermitian metric determined by the Kähler form $\omega$, so the three operators
\begin{align*}
\Delta_d,\qquad \Delta_\partial,\qquad \Delta_{\bar\partial}
\end{align*}
are being compared with respect to the same $L^2$ [inner product](/page/Inner%20Product).
The Kähler condition is used exactly through the standard Kähler Laplacian identities. With the convention
\begin{align*}
\Delta_D=DD^*+D^*D,
\end{align*}
those identities state that, on smooth complex-valued differential forms,
\begin{align*}
\Delta_d = 2\Delta_\partial
\end{align*}
and
\begin{align*}
\Delta_d = 2\Delta_{\bar\partial}.
\end{align*}
Thus the three Laplacians are not merely related abstractly; they are scalar multiples of the same operator. The factor $2$ is a convention-dependent normalization factor, and it is harmless for kernels because $2$ is nonzero. We are invoking this as the standard Kähler Laplacian identity.
[/guided]
[/step]
[step:Identify the three kernels by canceling the scalar factor]
Fix $\alpha\in A^\bullet(X;\mathbb C)$. From $\Delta_d=2\Delta_\partial$, we have
\begin{align*}
\Delta_d\alpha=0 \iff 2\Delta_\partial\alpha=0.
\end{align*}
Since scalar multiplication by $2$ is injective on $A^\bullet(X;\mathbb C)$, this is equivalent to
\begin{align*}
\Delta_\partial\alpha=0.
\end{align*}
Therefore
\begin{align*}
\ker\Delta_d=\ker\Delta_\partial.
\end{align*}
The same argument applied to $\Delta_d=2\Delta_{\bar\partial}$ gives
\begin{align*}
\ker\Delta_d=\ker\Delta_{\bar\partial}.
\end{align*}
Hence
\begin{align*}
\ker\Delta_d=\ker\Delta_\partial=\ker\Delta_{\bar\partial}.
\end{align*}
[/step]
[step:Translate equality of kernels into the harmonicity equivalence]
Since the previous step proves
\begin{align*}
\ker\Delta_d=\ker\Delta_\partial=\ker\Delta_{\bar\partial},
\end{align*}
the fixed form $\alpha$ satisfies
\begin{align*}
\Delta_d\alpha=0 \iff \Delta_\partial\alpha=0 \iff \Delta_{\bar\partial}\alpha=0.
\end{align*}
By the terminology fixed in the statement, $\alpha$ is $d$-harmonic, $\partial$-harmonic, or $\bar\partial$-harmonic precisely when it lies in the kernel of $\Delta_d$, $\Delta_\partial$, or $\Delta_{\bar\partial}$, respectively. Therefore
\begin{align*}
\alpha \text{ is } d\text{-harmonic}
\iff \alpha \text{ is } \partial\text{-harmonic}
\iff \alpha \text{ is } \bar\partial\text{-harmonic}.
\end{align*}
This proves the theorem.
[/step]
