[proofplan] We prove the identity as an equality of operators on complex-valued differential forms. The proof is a direct expansion using the type decomposition $d=\partial+\bar\partial$ and the convention $d^c=i(\bar\partial-\partial)$. The square terms vanish by the [Dolbeault operator relations](/theorems/8046), and the remaining two mixed terms combine because $\partial\bar\partial+\bar\partial\partial=0$. [/proofplan]

[step:Expand $dd^c$ using the definitions of $d$ and $d^c$]Let $\alpha\in A^\bullet(X)$ be a smooth complex-valued differential form, where $A^\bullet(X)=\bigoplus_{k\ge 0}A^k(X;\mathbb C)$ is the graded complex [vector space](/page/Vector%20Space) declared in the statement. By the [type decomposition of the exterior derivative](/theorems/7004) from [citetheorem:7004], the [exterior derivative](/theorems/1525) acts as \begin{align*} d\alpha=(\partial+\bar\partial)\alpha. \end{align*} Using the stated convention $d^c=i(\bar\partial-\partial)$, we compute \begin{align*} dd^c\alpha=(\partial+\bar\partial)i(\bar\partial-\partial)\alpha. \end{align*} Since $i\in\mathbb C$ is constant and the Dolbeault operators are complex-linear on complex-valued forms, this becomes \begin{align*} dd^c\alpha=i(\partial\bar\partial\alpha-\partial^2\alpha+\bar\partial^2\alpha-\bar\partial\partial\alpha). \end{align*}[/step]

[guided]We want an operator identity on $A^\bullet(X)$, so we test both sides on an arbitrary element of that graded complex vector space. Let $\alpha\in A^\bullet(X)$ be a smooth complex-valued differential form, where $A^\bullet(X)=\bigoplus_{k\ge 0}A^k(X;\mathbb C)$. The type decomposition theorem [citetheorem:7004] says that the exterior derivative decomposes on complex-valued forms as \begin{align*} d\alpha=(\partial+\bar\partial)\alpha. \end{align*} The convention in the statement defines the operator $d^c$ by \begin{align*} d^c\alpha=i(\bar\partial-\partial)\alpha. \end{align*} Therefore applying $d$ after $d^c$ gives \begin{align*} dd^c\alpha=(\partial+\bar\partial)i(\bar\partial-\partial)\alpha. \end{align*} Because $i$ is a constant scalar and $\partial,\bar\partial$ are complex-linear operators on complex-valued differential forms, the scalar $i$ factors out. Distributing the composition of operators gives \begin{align*} dd^c\alpha=i(\partial\bar\partial\alpha-\partial^2\alpha+\bar\partial^2\alpha-\bar\partial\partial\alpha). \end{align*} This is the only place where the sign convention matters: if $d^c$ were defined with the opposite sign or a different normalization, the final constant would change.[/guided]

[step:Cancel the square terms and combine the mixed terms] By the Dolbeault operator relations from [citetheorem:8046], \begin{align*} \partial^2\alpha=0,\qquad \bar\partial^2\alpha=0,\qquad \bar\partial\partial\alpha=-\partial\bar\partial\alpha. \end{align*} Substituting these identities into the expansion gives \begin{align*} dd^c\alpha=i(\partial\bar\partial\alpha-\bar\partial\partial\alpha). \end{align*} Using $\bar\partial\partial\alpha=-\partial\bar\partial\alpha$, we obtain \begin{align*} dd^c\alpha=i(\partial\bar\partial\alpha+\partial\bar\partial\alpha)=2i\,\partial\bar\partial\alpha. \end{align*} Since $\alpha\in A^\bullet(X)$ was arbitrary, the operator identity \begin{align*} dd^c=2i\,\partial\bar\partial \end{align*} holds on complex-valued differential forms on $X$. [/step]