[proofplan] We expand the Dolbeault operator on scalar functions in the standard complex coordinates $z_j=x_j+i y_j$. Its vanishing is exactly the system of [Cauchy-Riemann equations](/page/Cauchy-Riemann%20Equations) $\partial f/\partial \bar z_j=0$ for every coordinate direction. Holomorphic functions satisfy these equations by differentiating with respect to the conjugate coordinates. Conversely, the smooth Cauchy-Riemann equations imply holomorphicity by the standard several-variable Cauchy-Riemann characterization. [/proofplan]

[step:Identify $\bar\partial f=0$ with the coordinate Cauchy-Riemann equations] For each $j \in \{1,\dots,n\}$, let $x_j: U \to \mathbb{R}$ and $y_j: U \to \mathbb{R}$ denote the real coordinate functions determined by $z_j=x_j+i y_j$. Since $f:U\to\mathbb{C}$ is smooth as a function on the real manifold $U\subset\mathbb{R}^{2n}$, the real partial derivatives $\partial f/\partial x_j$ and $\partial f/\partial y_j$ exist and are smooth. By definition of the Dolbeault operator on scalar functions, \begin{align*} \bar\partial f = \sum_{j=1}^n \frac{\partial f}{\partial \bar z_j}\, d\bar z_j. \end{align*} The one-forms $d\bar z_1,\dots,d\bar z_n$ form the standard local frame of complex anti-holomorphic one-forms. Hence $\bar\partial f=0$ if and only if every coefficient vanishes: \begin{align*} \frac{\partial f}{\partial \bar z_j}=0 \quad \text{for every } j\in\{1,\dots,n\}. \end{align*} Equivalently, \begin{align*} \frac{1}{2}\left(\frac{\partial f}{\partial x_j}+i\frac{\partial f}{\partial y_j}\right)=0 \quad \text{for every } j\in\{1,\dots,n\}. \end{align*} These are precisely the Cauchy-Riemann equations in each complex coordinate. [/step]

[step:Derive $\bar\partial f=0$ from complex-linearity of the differential] Assume that $f$ is holomorphic on $U$. Fix $a\in U$. By the definition of holomorphicity in the theorem statement, there is a complex-[linear map](/page/Linear%20Map) $L_a: \mathbb{C}^n\to\mathbb{C}$ such that \begin{align*} f(a+h)=f(a)+L_a(h)+o(|h|) \quad \text{as } h\to 0 \text{ in } \mathbb{C}^n. \end{align*} Thus the real differential of $f$ at $a$, denoted $df_a: \mathbb{R}^{2n}\to\mathbb{C}$, is the real-linear map underlying $L_a$. For each $j\in\{1,\dots,n\}$, let $e_j\in\mathbb{C}^n$ denote the $j$-th standard complex coordinate vector. Since multiplication by $i$ sends the real $x_j$-direction $e_j$ to the real $y_j$-direction $i e_j$, complex-linearity gives \begin{align*} \frac{\partial f}{\partial y_j}(a)=df_a(i e_j)=L_a(i e_j)=iL_a(e_j)=i\,df_a(e_j)=i\frac{\partial f}{\partial x_j}(a). \end{align*} Therefore \begin{align*} \frac{\partial f}{\partial \bar z_j}(a)=\frac{1}{2}\left(\frac{\partial f}{\partial x_j}(a)+i\frac{\partial f}{\partial y_j}(a)\right)=\frac{1}{2}\left(\frac{\partial f}{\partial x_j}(a)+i^2\frac{\partial f}{\partial x_j}(a)\right)=0. \end{align*} Because $a\in U$ and $j\in\{1,\dots,n\}$ were arbitrary, every coefficient of $\bar\partial f$ vanishes on $U$. Hence $\bar\partial f=0$. [/step]

[step:Construct the complex differential from the vanishing of $\bar\partial f$]Assume that $\bar\partial f=0$ on $U$. Fix $a\in U$. Since $f$ is smooth as a function on the real [open set](/page/Open%20Set) $U\subset\mathbb{R}^{2n}$, it is real differentiable at $a$. Let $df_a:\mathbb{R}^{2n}\to\mathbb{C}$ denote its real differential at $a$. By the first step, for every $j\in\{1,\dots,n\}$ we have \begin{align*} \frac{\partial f}{\partial x_j}(a)+i\frac{\partial f}{\partial y_j}(a)=0. \end{align*} Equivalently, \begin{align*} \frac{\partial f}{\partial y_j}(a)=i\frac{\partial f}{\partial x_j}(a). \end{align*} Define the map $L_a:\mathbb{C}^n\to\mathbb{C}$ by \begin{align*} L_a(h)=\sum_{j=1}^n h_j\frac{\partial f}{\partial x_j}(a), \quad h=(h_1,\dots,h_n)\in\mathbb{C}^n. \end{align*} This map is complex-linear because each coefficient $\partial f/\partial x_j(a)$ is a complex number and the formula is linear in the complex coordinates $h_j$. Write $h_j=s_j+i t_j$ with $s_j,t_j\in\mathbb{R}$. Using the real differential formula and the relation between the $x_j$- and $y_j$-partials, we get \begin{align*} df_a(h)=\sum_{j=1}^n s_j\frac{\partial f}{\partial x_j}(a)+\sum_{j=1}^n t_j\frac{\partial f}{\partial y_j}(a)=\sum_{j=1}^n \left(s_j+i t_j\right)\frac{\partial f}{\partial x_j}(a)=L_a(h). \end{align*} Therefore the real differentiability expansion becomes \begin{align*} f(a+h)=f(a)+L_a(h)+o(|h|) \quad \text{as } h\to 0 \text{ in } \mathbb{C}^n. \end{align*} Since $L_a$ is complex-linear, $f$ is complex differentiable at $a$. The point $a\in U$ was arbitrary, so $f$ is holomorphic on $U$ in the sense stated in the theorem.[/step]

[guided]We now prove the reverse implication directly, without invoking a separate Cauchy-Riemann characterization. The hypothesis is $\bar\partial f=0$. By the first step, this means that at every point $a\in U$ and for every coordinate index $j\in\{1,\dots,n\}$, \begin{align*} \frac{\partial f}{\partial x_j}(a)+i\frac{\partial f}{\partial y_j}(a)=0. \end{align*} Solving this identity for the $y_j$-partial derivative gives \begin{align*} \frac{\partial f}{\partial y_j}(a)=i\frac{\partial f}{\partial x_j}(a). \end{align*} This relation is the key point: it says that the real derivative in the imaginary coordinate direction is obtained by multiplying the real derivative in the corresponding real coordinate direction by $i$. Fix $a\in U$. Since $f$ is smooth as a function on the real open set $U\subset\mathbb{R}^{2n}$, it is real differentiable at $a$. Let $df_a:\mathbb{R}^{2n}\to\mathbb{C}$ denote the real differential of $f$ at $a$. To prove holomorphicity in the sense of the theorem statement, we must show that this real differential is represented by a complex-linear map $L_a:\mathbb{C}^n\to\mathbb{C}$. Define $L_a:\mathbb{C}^n\to\mathbb{C}$ by \begin{align*} L_a(h)=\sum_{j=1}^n h_j\frac{\partial f}{\partial x_j}(a), \quad h=(h_1,\dots,h_n)\in\mathbb{C}^n. \end{align*} The map $L_a$ is complex-linear because for $\lambda\in\mathbb{C}$ and $h,k\in\mathbb{C}^n$, the coordinate formula gives $L_a(h+k)=L_a(h)+L_a(k)$ and $L_a(\lambda h)=\lambda L_a(h)$. It remains to check that $L_a$ is the real differential of $f$ at $a$. Write each complex coordinate increment as $h_j=s_j+i t_j$, where $s_j,t_j\in\mathbb{R}$. The real differential is computed from the real coordinate directions as \begin{align*} df_a(h)=\sum_{j=1}^n s_j\frac{\partial f}{\partial x_j}(a)+\sum_{j=1}^n t_j\frac{\partial f}{\partial y_j}(a). \end{align*} Substituting the Cauchy-Riemann relation $\partial f/\partial y_j(a)=i\partial f/\partial x_j(a)$ into this expression yields \begin{align*} df_a(h)=\sum_{j=1}^n s_j\frac{\partial f}{\partial x_j}(a)+\sum_{j=1}^n t_j i\frac{\partial f}{\partial x_j}(a)=\sum_{j=1}^n (s_j+i t_j)\frac{\partial f}{\partial x_j}(a)=L_a(h). \end{align*} Thus the real differentiability expansion for $f$ at $a$ can be rewritten as \begin{align*} f(a+h)=f(a)+L_a(h)+o(|h|) \quad \text{as } h\to 0 \text{ in } \mathbb{C}^n. \end{align*} Because $L_a$ is complex-linear, this is precisely [complex differentiability](/page/Complex%20Differentiability) at $a$ in the sense stated in the theorem. Since $a\in U$ was arbitrary, $f$ is holomorphic on $U$.[/guided]

[step:Combine the two implications] The second step proves that holomorphicity of $f$ implies $\bar\partial f=0$. The third step proves that $\bar\partial f=0$ implies holomorphicity of $f$. Therefore $f$ is holomorphic on $U$ if and only if $\bar\partial f=0$ on $U$. [/step]