[proofplan] The identity is a direct computation with the product rule for the Wirtinger operators. We apply the two compositions $\partial_{\bar z_k}\delta_j^\varphi$ and $\delta_j^\varphi\partial_{\bar z_k}$ to $u \in C_c^\infty(\Omega)$ and subtract. Three kinds of terms appear: a pair of second-order derivatives of $u$, which cancel because the Wirtinger operators are constant-coefficient and mixed partials of the $C^2$ (indeed $C^\infty$) function $u$ are symmetric; a pair of first-order terms $\varphi_{z_j}\partial_{\bar z_k}u$, which cancel algebraically; and a single zeroth-order term $-(\partial_{\bar z_k}\varphi_{z_j})u$, which equals $-\varphi_{j\bar k}u$ once we use symmetry of the mixed second derivatives of $\varphi$ — the step that consumes the hypothesis $\varphi \in C^2$. A final bookkeeping step records how this commutator enters the Kohn–Morrey–Hörmander energy identity as the positive Levi form. [/proofplan] [step:Record the algebraic structure of the operators and verify the differentiability needed for the product rule] The Wirtinger operators $\partial_{z_j}$ and $\partial_{\bar z_k}$ are, by definition, fixed complex-linear combinations of the real first-order partial derivatives $\partial_{x_i}, \partial_{y_i}$ with constant coefficients. In particular each is a derivation: for functions $f \in C^1(\Omega)$ and $g \in C^1(\Omega)$ the Leibniz rule \begin{align*} \frac{\partial}{\partial \bar z_k}(f g) = \left(\frac{\partial f}{\partial \bar z_k}\right) g + f\,\frac{\partial g}{\partial \bar z_k} \end{align*} holds, since it holds for each real partial $\partial_{x_i}, \partial_{y_i}$ and is preserved under constant-coefficient linear combinations. We check that all expressions below are well defined. Since $\varphi \in C^2(\Omega;\mathbb{R})$, the coefficient $\varphi_{z_j} = \partial_{z_j}\varphi$ belongs to $C^1(\Omega)$. For $u \in C_c^\infty(\Omega)$ we have $\partial_{z_j}u \in C^\infty(\Omega)$ and $\varphi_{z_j} u \in C^1(\Omega)$, so $\delta_j^\varphi u = \partial_{z_j}u - \varphi_{z_j}u \in C^1(\Omega)$ and $\partial_{\bar z_k}(\delta_j^\varphi u) \in C^0(\Omega)$ is defined. Likewise $\partial_{\bar z_k}u \in C^\infty(\Omega)$, so $\delta_j^\varphi(\partial_{\bar z_k}u) \in C^1(\Omega)$ is defined. Hence both compositions, and their difference, are continuous functions on $\Omega$. [guided] Before computing, we make explicit the two structural facts that the whole proof rests on, and we confirm that every derivative we are about to write down actually exists. First, what kind of objects are $\partial_{z_j}$ and $\partial_{\bar z_k}$? By their defining formulas \begin{align*} \frac{\partial}{\partial z_j} = \tfrac{1}{2}\left(\frac{\partial}{\partial x_j} - i\frac{\partial}{\partial y_j}\right), \qquad \frac{\partial}{\partial \bar z_k} = \tfrac{1}{2}\left(\frac{\partial}{\partial x_k} + i\frac{\partial}{\partial y_k}\right), \end{align*} they are linear combinations of the real partial derivatives $\partial_{x_i}, \partial_{y_i}$ with **constant** coefficients $\pm\tfrac12, \pm\tfrac{i}{2}$. Two consequences matter. (i) Each is a derivation, so the Leibniz product rule \begin{align*} \frac{\partial}{\partial \bar z_k}(f g) = \left(\frac{\partial f}{\partial \bar z_k}\right) g + f\,\frac{\partial g}{\partial \bar z_k} \end{align*} holds for $f, g \in C^1(\Omega)$: it is true for each real partial $\partial_{x_i}, \partial_{y_i}$, and taking a constant-coefficient combination of those identities preserves it. (ii) Because the coefficients are constant, questions about commuting $\partial_{z_j}$ with $\partial_{\bar z_k}$ reduce to commuting the underlying real partials — which is exactly what [symmetry of second derivatives](/theorems/332) controls. We will use (i) in the next step and (ii) in the step after. Second, are all the derivatives we plan to take legitimate? This is where we must respect that $\varphi$ is only assumed $C^2$, not $C^\infty$. Differentiating the weight once gives the coefficient $\varphi_{z_j} = \partial_{z_j}\varphi \in C^1(\Omega)$ — it still has one continuous derivative to spare. Now: - $u \in C_c^\infty(\Omega)$ gives $\partial_{z_j}u \in C^\infty(\Omega)$; - the product $\varphi_{z_j}u$ is a product of a $C^1$ function and a $C^\infty$ function, hence $C^1$; - therefore $\delta_j^\varphi u = \partial_{z_j}u - \varphi_{z_j}u \in C^1(\Omega)$, and we may apply $\partial_{\bar z_k}$ to it once, landing in $C^0(\Omega)$; - on the other side, $\partial_{\bar z_k}u \in C^\infty(\Omega)$, so $\delta_j^\varphi(\partial_{\bar z_k}u) \in C^1(\Omega)$. Both compositions are continuous functions on $\Omega$, so the commutator $[\partial_{\bar z_k}, \delta_j^\varphi]u$ is a well-defined element of $C^0(\Omega)$, and the equation we must prove is an equality of continuous functions. If $\varphi$ were only $C^1$, the term $\partial_{\bar z_k}\varphi_{z_j}$ in the next step would not exist — this is the first hint that $C^2$ is exactly the right hypothesis. [/guided] [/step] [step:Expand both compositions with the product rule] Fix $j, k \in \{1, \dots, n\}$ and $u \in C_c^\infty(\Omega)$. Applying $\partial_{\bar z_k}$ to $\delta_j^\varphi u = \partial_{z_j}u - \varphi_{z_j}u$ and using the Leibniz rule from the previous step on the product $\varphi_{z_j}u$, \begin{align*} \frac{\partial}{\partial \bar z_k}\bigl(\delta_j^\varphi u\bigr) &= \frac{\partial}{\partial \bar z_k}\frac{\partial u}{\partial z_j} - \frac{\partial}{\partial \bar z_k}\bigl(\varphi_{z_j}\, u\bigr) \\ &= \frac{\partial}{\partial \bar z_k}\frac{\partial u}{\partial z_j} - \left(\frac{\partial \varphi_{z_j}}{\partial \bar z_k}\right) u - \varphi_{z_j}\,\frac{\partial u}{\partial \bar z_k}. \end{align*} Applying $\delta_j^\varphi$ to $\partial_{\bar z_k}u$ — here $\delta_j^\varphi$ acts on the single function $\partial_{\bar z_k}u$, with no product to differentiate beyond the definition of the operator — \begin{align*} \delta_j^\varphi\!\left(\frac{\partial u}{\partial \bar z_k}\right) = \frac{\partial}{\partial z_j}\frac{\partial u}{\partial \bar z_k} - \varphi_{z_j}\,\frac{\partial u}{\partial \bar z_k}. \end{align*} [guided] We now simply apply each operator and write out every term, deferring all cancellation to the next step. Start with the composition $\partial_{\bar z_k}\delta_j^\varphi$. By definition $\delta_j^\varphi u = \partial_{z_j}u - \varphi_{z_j}u$, so \begin{align*} \frac{\partial}{\partial \bar z_k}\bigl(\delta_j^\varphi u\bigr) = \frac{\partial}{\partial \bar z_k}\frac{\partial u}{\partial z_j} - \frac{\partial}{\partial \bar z_k}\bigl(\varphi_{z_j}\, u\bigr). \end{align*} The first term is a second-order derivative of $u$; leave it as is. The second term is the derivative of a product, and this is where the Leibniz rule established in the previous step is used: with $f = \varphi_{z_j} \in C^1(\Omega)$ and $g = u \in C^\infty(\Omega)$, \begin{align*} \frac{\partial}{\partial \bar z_k}\bigl(\varphi_{z_j}\, u\bigr) = \left(\frac{\partial \varphi_{z_j}}{\partial \bar z_k}\right) u + \varphi_{z_j}\,\frac{\partial u}{\partial \bar z_k}. \end{align*} Note the derivative landed on the coefficient $\varphi_{z_j}$ produces a genuinely new, zeroth-order-in-$u$ term $(\partial_{\bar z_k}\varphi_{z_j})u$ — this is the term that will survive. Substituting, \begin{align*} \frac{\partial}{\partial \bar z_k}\bigl(\delta_j^\varphi u\bigr) = \underbrace{\frac{\partial}{\partial \bar z_k}\frac{\partial u}{\partial z_j}}_{\text{second order}} - \underbrace{\left(\frac{\partial \varphi_{z_j}}{\partial \bar z_k}\right) u}_{\text{survivor}} - \underbrace{\varphi_{z_j}\,\frac{\partial u}{\partial \bar z_k}}_{\text{first order}}. \end{align*} Now the other composition $\delta_j^\varphi\partial_{\bar z_k}$. Here there is no product rule subtlety: we simply feed the function $\partial_{\bar z_k}u$ into the operator $\delta_j^\varphi = \partial_{z_j} - \varphi_{z_j}\cdot$, obtaining \begin{align*} \delta_j^\varphi\!\left(\frac{\partial u}{\partial \bar z_k}\right) = \underbrace{\frac{\partial}{\partial z_j}\frac{\partial u}{\partial \bar z_k}}_{\text{second order}} - \underbrace{\varphi_{z_j}\,\frac{\partial u}{\partial \bar z_k}}_{\text{first order}}. \end{align*} Observe that the first-order term $\varphi_{z_j}\partial_{\bar z_k}u$ appears identically in both expansions; this is the structural reason it will drop out. We are now ready to subtract. [/guided] [/step] [step:Cancel the second-order and first-order terms, leaving the Levi-form term] Subtract the two expansions of the previous step: \begin{align*} \left[\frac{\partial}{\partial \bar z_k},\, \delta_j^\varphi\right] u &= \frac{\partial}{\partial \bar z_k}\bigl(\delta_j^\varphi u\bigr) - \delta_j^\varphi\!\left(\frac{\partial u}{\partial \bar z_k}\right) \\ &= \left(\frac{\partial}{\partial \bar z_k}\frac{\partial u}{\partial z_j} - \frac{\partial}{\partial z_j}\frac{\partial u}{\partial \bar z_k}\right) - \left(\frac{\partial \varphi_{z_j}}{\partial \bar z_k}\right) u + \left(-\varphi_{z_j}\,\frac{\partial u}{\partial \bar z_k} + \varphi_{z_j}\,\frac{\partial u}{\partial \bar z_k}\right). \end{align*} The third parenthesis vanishes identically. For the first parenthesis, the operators $\partial_{z_j}, \partial_{\bar z_k}$ are constant-coefficient combinations of $\partial_{x_i}, \partial_{y_i}$, and $u \in C^\infty(\Omega) \subseteq C^2(\Omega)$; by the [Symmetry of Second Derivatives](/theorems/332), every real mixed second partial of $u$ is independent of the order of differentiation, hence so is every constant-coefficient combination of them, giving \begin{align*} \frac{\partial}{\partial \bar z_k}\frac{\partial u}{\partial z_j} = \frac{\partial}{\partial z_j}\frac{\partial u}{\partial \bar z_k}. \end{align*} Thus the first parenthesis vanishes as well. For the surviving term, $\varphi_{z_j} = \partial_{z_j}\varphi$ with $\varphi \in C^2(\Omega)$, and the same symmetry applied to $\varphi$ gives \begin{align*} \frac{\partial \varphi_{z_j}}{\partial \bar z_k} = \frac{\partial}{\partial \bar z_k}\frac{\partial \varphi}{\partial z_j} = \frac{\partial}{\partial z_j}\frac{\partial \varphi}{\partial \bar z_k} = \frac{\partial^2 \varphi}{\partial z_j\, \partial \bar z_k} = \varphi_{j\bar k}. \end{align*} Therefore \begin{align*} \left[\frac{\partial}{\partial \bar z_k},\, \delta_j^\varphi\right] u = -\,\varphi_{j\bar k}\, u, \end{align*} which is the claimed identity. [guided] Subtracting the two expansions and grouping by type of term: \begin{align*} \left[\frac{\partial}{\partial \bar z_k},\, \delta_j^\varphi\right] u = \underbrace{\left(\frac{\partial}{\partial \bar z_k}\frac{\partial u}{\partial z_j} - \frac{\partial}{\partial z_j}\frac{\partial u}{\partial \bar z_k}\right)}_{(\mathrm{I})\ \text{second order}} \;\underbrace{-\ \left(\frac{\partial \varphi_{z_j}}{\partial \bar z_k}\right) u}_{(\mathrm{II})\ \text{survivor}} \;+\; \underbrace{\left(-\varphi_{z_j}\frac{\partial u}{\partial \bar z_k} + \varphi_{z_j}\frac{\partial u}{\partial \bar z_k}\right)}_{(\mathrm{III})\ \text{first order}}. \end{align*} We dispose of the three groups in turn. Group $(\mathrm{III})$ is the easiest: the two first-order terms are literally identical and appear with opposite signs, so they cancel algebraically with no hypothesis needed. This is the payoff of having noted, at the end of the last step, that $\varphi_{z_j}\partial_{\bar z_k}u$ occurs in both expansions. Group $(\mathrm{I})$ asks: do the Wirtinger operators commute on $u$? Why should they? Because they are constant-coefficient combinations of the real partials $\partial_{x_i}, \partial_{y_i}$, and $u$ is smooth — in particular $u \in C^2(\Omega)$, the exact regularity required by the [Symmetry of Second Derivatives](/theorems/332). That theorem says the real mixed second partials $\partial_{x_i}\partial_{x_\ell}u$, $\partial_{x_i}\partial_{y_\ell}u$, etc., are symmetric in the order of differentiation. Expanding $\partial_{\bar z_k}\partial_{z_j}u$ and $\partial_{z_j}\partial_{\bar z_k}u$ into real partials, the two expansions involve the same real second derivatives of $u$ in opposite orders and with the same constant coefficients; symmetry makes them equal: \begin{align*} \frac{\partial}{\partial \bar z_k}\frac{\partial u}{\partial z_j} = \frac{\partial}{\partial z_j}\frac{\partial u}{\partial \bar z_k}. \end{align*} So $(\mathrm{I})$ vanishes. (If $u$ were merely $C^1$ this could fail — but $u$ is smooth, so there is no obstruction.) Group $(\mathrm{II})$ is the heart of the matter, and the only place the hypothesis on $\varphi$ is consumed. We must identify $\partial_{\bar z_k}\varphi_{z_j}$ with the Levi-form entry $\varphi_{j\bar k} = \partial_{z_j}\partial_{\bar z_k}\varphi$. As written, the derivatives sit in the order "$\partial_{\bar z_k}$ after $\partial_{z_j}$"; the definition of $\varphi_{j\bar k}$ has them in the order "$\partial_{z_j}$ after $\partial_{\bar z_k}$". To swap them we again invoke the [Symmetry of Second Derivatives](/theorems/332), now applied to $\varphi$. Its hypothesis is precisely $\varphi \in C^2(\Omega)$ — which is given — and we use the same reduction to real partials as for $u$: \begin{align*} \frac{\partial \varphi_{z_j}}{\partial \bar z_k} = \frac{\partial}{\partial \bar z_k}\frac{\partial \varphi}{\partial z_j} = \frac{\partial}{\partial z_j}\frac{\partial \varphi}{\partial \bar z_k} = \frac{\partial^2 \varphi}{\partial z_j\, \partial \bar z_k} = \varphi_{j\bar k}. \end{align*} Had $\varphi$ been only $C^1$, this second derivative would not even exist; had it been $C^2$ but with a non-symmetric pathology, the swap would fail — so $C^2$ is exactly what makes group $(\mathrm{II})$ a single well-defined number times $u$. Collecting $(\mathrm{I}) = 0$, $(\mathrm{III}) = 0$, and $(\mathrm{II}) = -\varphi_{j\bar k}u$: \begin{align*} \left[\frac{\partial}{\partial \bar z_k},\, \delta_j^\varphi\right] u = -\,\varphi_{j\bar k}\, u. \end{align*} [/guided] [/step] [step:Identify the Levi-form contribution under the Kohn–Morrey sign convention] The identity just proved says that, as a zeroth-order operator on $C_c^\infty(\Omega)$, \begin{align*} -\left[\frac{\partial}{\partial \bar z_k},\, \delta_j^\varphi\right] = \varphi_{j\bar k}\cdot\,, \end{align*} i.e. minus the commutator is multiplication by the Levi-form entry $\varphi_{j\bar k}$. In the Kohn–Morrey–Hörmander energy identity for the weighted $\bar\partial$-complex, $\delta_j^\varphi$ is the formal adjoint of $\partial/\partial \bar z_j$ with respect to the weighted measure $e^{-\varphi}\, d\mathcal{L}^{2n}$, and integrating by parts converts the difference of the two squared norms into a sum of these commutators paired with the form coefficients. Concretely, for a $(0,1)$-form $v = \sum_{j=1}^n v_j\, d\bar z_j$ with coefficients $v_j \in C_c^\infty(\Omega)$, the curvature term collects the values $\varphi_{j\bar k} = -[\partial_{\bar z_k}, \delta_j^\varphi]$ against $v_j \overline{v_k}$ and sums over $j, k$, producing the pointwise Hermitian form \begin{align*} \sum_{j,k=1}^n \varphi_{j\bar k}\, v_j\, \overline{v_k}. \end{align*} Because $\varphi$ is real-valued, $\overline{\varphi_{j\bar k}} = \overline{\partial_{z_j}\partial_{\bar z_k}\varphi} = \partial_{\bar z_j}\partial_{z_k}\varphi = \varphi_{k\bar j}$, so the coefficient matrix $(\varphi_{j\bar k})_{j,k=1}^n$ is Hermitian and the displayed form is a genuine Hermitian form on $\mathbb{C}^n$ — the Levi form of $\varphi$. This is the positive curvature term in the convention stated, completing the proof. (The full energy identity into which this commutator feeds — the Kohn–Morrey–Hörmander identity — is cited here as context but is a separate result not yet in the wiki; the present theorem establishes only the commutator value $-\varphi_{j\bar k}$ and the resulting sign of the Levi-form contribution.) [/step]