Lemma 8.6.2B8367971A40D

Proof. Fix a smooth $h\geq \varphi $ and a constant

For each positive integer $N$ solve, by the preceding lemma,

At a positive maximum of $w_ N-h$, the form $\omega +\mathrm{dd}^{\mathrm{c}}h$ lies above the strictly admissible form $\omega +\mathrm{dd}^{\mathrm{c}}w_ N$, hence is admissible there. Monotonicity would give $b e^{N(w_ N-h)}\leq H_ m(h)/\omega ^ n<b$, a contradiction. Thus $w_ N\leq h$. The preceding zero-order estimate also gives

Let $x_ N$ maximize $\varphi -(1+N^{-1})w_ N$. The smooth positive form $T_ N=(\omega +\mathrm{dd}^{\mathrm{c}}w_ N)^{m-1}\wedge \omega ^{n-m}$ defines a strictly elliptic linear operator. Admissibility gives $(\omega +\mathrm{dd}^{\mathrm{c}}\varphi )\wedge T_ N\geq 0$ in distributions. Its linear maximum principle, applied to the upper test $(1+N^{-1})w_ N$ plus the maximum value, gives at $x_ N$

This application is to a continuous distributional subsolution of a smooth strictly elliptic operator. It does not assume a second derivative of $\varphi $. Writing $d_ N=H_ m(w_ N)/\omega ^ n$, polarized Gårding positivity yields

Therefore, at this maximum point,

Since $\varphi \leq h$, the maximum of $\varphi -(1+N^{-1})w_ N$ is at most $E_ N+M_ N/N$. It follows on all of $X$ that

Given $\delta >0$, choose an ordinary smooth majorant $\varphi \leq h\leq \varphi +\delta $, and then $N$ so large that $E_ N+2M_ N/N\leq \delta $. The resulting $w_ N$ is strictly admissible and within $\delta $ of $\varphi $. Let $\delta \downarrow 0$. No monotonicity of the chosen approximants is needed. $\square$