Testing the weighted cofactor against $w^2$ produces a nonnegative gradient integral. Only distributional $\mathrm{dd}^{\mathrm{c}}$-closedness of $hT$ is needed: for a continuous $h>0$, the condition is
of 14.22.1(1).
The product rule gives
A2BD70AF14EC
14.22.5
\begin{equation} \mathrm{dd}^{\mathrm{c}}(w^2)=2w\mathrm{dd}^{\mathrm{c}}w+2\mathrm{d}w\wedge \mathrm d^{\mathrm c}w =2w\mathrm{dd}^{\mathrm{c}}w+\frac{\mathrm{i}}{\pi }\partial w\wedge \bar\partial w. \end{equation}
Approximate $w^2$ in $C^2$ by smooth functions $\varphi $ in (14.22.4). Since $hT$ is continuous and $\mathrm{dd}^{\mathrm{c}}w\wedge T=0$ by (14.20.1), this gives
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14.22.6
\begin{equation} 0=\frac1\pi \int _ X h\, \mathrm{i}\partial w\wedge \bar\partial w\wedge T. \end{equation}
The continuous integrand is nonnegative, so it vanishes identically, which is (14.22.2).
By (14.21.7), (14.22.2) says that $Q(\mathrm{d}w)=0$. For the nonnegative quadratic form $Q$, this implies $Q(\mathrm{d}w,\eta )=0$ for every $\eta $. Therefore
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14.22.7
\begin{equation} \mathrm{d}w_ x\bigl (Q_ x^\sharp \eta \bigr )=0\qquad (x\in X,\ \eta \in T_ x^*X). \end{equation}
By the chain rule and (14.22.7), $w$ is constant along every absolutely continuous curve whose velocity lies in $\operatorname {im}Q^\sharp $ almost everywhere. Thus $w$ is constant on each reachable set, and by continuity one dense reachable set suffices. If instead $Q$ is positive definite on a dense set, (14.22.6) gives $\mathrm{d}w=0$ there, and hence everywhere.
$\square$
The argument uses only the test function $w^2$. For general test functions, integration by parts introduces derivatives of $hT$; the condition $\mathrm{dd}^{\mathrm{c}}(hT)=0$ alone does not make $L\varphi =(\mathrm{dd}^{\mathrm{c}}\varphi \wedge T)/\mathrm{d}V$ a symmetric operator in divergence form.
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