[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*}[/step]

[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: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]