[proofplan] The assertion is local on $X$, so we work over an [open set](/page/Open%20Set) on which $E$ admits a holomorphic frame. In such a frame, a smooth section is represented by smooth complex-valued coefficient functions, and the bundle Dolbeault operator acts by applying the scalar $\bar{\partial}$ operator to each coefficient. Thus $\bar{\partial}_E s = 0$ is exactly the condition that all local coefficient functions have vanishing scalar $\bar{\partial}$, which is the local definition of holomorphicity for smooth functions. Since holomorphicity of sections is checked in holomorphic frames, the local equivalence patches over $X$. [/proofplan] [step:Reduce the assertion to holomorphic frames on coordinate open sets] Let $p \in X$ be arbitrary. Since $E \to X$ is a holomorphic vector bundle of rank $r$, there exists an open neighbourhood $U \subset X$ of $p$ and holomorphic local sections $e_1: U \to E|_U, \dots, e_r: U \to E|_U$ such that $(e_1(q), \dots, e_r(q))$ is a complex basis of the fiber $E_q$ for every $q \in U$. The restriction $s|_U$ has a unique expression \begin{align*} s|_U = \sum_{i=1}^r s_i e_i \end{align*} where each coefficient function $s_i: U \to \mathbb{C}$ is smooth. By definition of a holomorphic section, $s$ is holomorphic on $U$ if and only if each coefficient function $s_i$ is holomorphic on $U$. Since holomorphicity of a section is local on $X$, it is enough to prove the equivalence on every such $U$. [/step] [step:Compute $\bar{\partial}_E s$ coefficientwise in a holomorphic frame] By the defining local formula for the Dolbeault operator associated to the holomorphic structure on $E$, the expression \begin{align*} s|_U = \sum_{i=1}^r s_i e_i \end{align*} gives \begin{align*} (\bar{\partial}_E s)|_U = \sum_{i=1}^r (\bar{\partial} s_i) \otimes e_i. \end{align*} Here $\bar{\partial} s_i \in \Omega^{0,1}(U)$ is the scalar Dolbeault derivative of the smooth function $s_i: U \to \mathbb{C}$. Because $(e_1,\dots,e_r)$ is a local frame, a section of $\Omega^{0,1}(U,E|_U)$ written in this frame vanishes if and only if all of its coefficient $(0,1)$-forms vanish. Therefore \begin{align*} (\bar{\partial}_E s)|_U = 0 \end{align*} if and only if \begin{align*} \bar{\partial} s_i = 0 \end{align*} for every $i \in \{1,\dots,r\}$. [guided] The purpose of choosing a holomorphic frame is that it converts the bundle-valued question into a coefficientwise scalar question. On the open set $U$, the frame sections $e_1: U \to E|_U, \dots, e_r: U \to E|_U$ give a basis of each fiber. Hence the smooth section $s|_U$ is uniquely represented as \begin{align*} s|_U = \sum_{i=1}^r s_i e_i \end{align*} with smooth coefficient functions $s_i: U \to \mathbb{C}$. The Dolbeault operator $\bar{\partial}_E$ is defined so that, in a holomorphic frame, it differentiates only the coefficient functions in the antiholomorphic directions. Thus \begin{align*} (\bar{\partial}_E s)|_U = \sum_{i=1}^r (\bar{\partial} s_i) \otimes e_i. \end{align*} This formula uses precisely that the frame is holomorphic: no extra connection matrix terms appear in a holomorphic frame. Now we interpret the equation $(\bar{\partial}_E s)|_U = 0$. Since $e_1,\dots,e_r$ are pointwise linearly independent, the representation of an $E$-valued $(0,1)$-form in this frame is unique. Therefore the sum \begin{align*} \sum_{i=1}^r (\bar{\partial} s_i) \otimes e_i \end{align*} vanishes exactly when every coefficient form $\bar{\partial} s_i \in \Omega^{0,1}(U)$ vanishes. Hence \begin{align*} (\bar{\partial}_E s)|_U = 0 \end{align*} if and only if \begin{align*} \bar{\partial} s_i = 0 \end{align*} for all $i \in \{1,\dots,r\}$. [/guided] [/step] [step:Identify vanishing scalar $\bar{\partial}$ with holomorphic coefficients] For each $i \in \{1,\dots,r\}$, the coefficient $s_i: U \to \mathbb{C}$ is smooth. By the scalar Dolbeault criterion for smooth complex-valued functions on a complex manifold, $s_i$ is holomorphic on $U$ if and only if \begin{align*} \bar{\partial} s_i = 0. \end{align*} Combining this scalar criterion with the coefficientwise computation above, we obtain \begin{align*} (\bar{\partial}_E s)|_U = 0 \end{align*} if and only if every coefficient function $s_i$ is holomorphic on $U$. [/step] [step:Patch the local equivalence over $X$] From the definition of holomorphic section in holomorphic local frames, $s$ is holomorphic on $U$ if and only if every coefficient function $s_i$ in the holomorphic frame $(e_1,\dots,e_r)$ is holomorphic on $U$. The previous step shows that this is equivalent to \begin{align*} (\bar{\partial}_E s)|_U = 0. \end{align*} Since $p \in X$ was arbitrary, the same equivalence holds on an open neighbourhood of every point of $X$. Therefore $s$ is holomorphic on all of $X$ if and only if $\bar{\partial}_E s$ vanishes on every such neighbourhood, which is equivalent to \begin{align*} \bar{\partial}_E s = 0 \end{align*} as a global element of $\Omega^{0,1}(X,E)$. This proves the theorem. [/step]