PomeLemma 8.12.3.132BB9E1B1F5 For bounded admissible $v$, $0\leq \alpha \leq 1$ and a real constant $b$, the function $v_\alpha =\alpha v+b$ is bounded admissible and
\[ H_ m(v_\alpha )\geq \alpha ^ m H_ m(v)+(1-\alpha )^ m\omega ^ n. \]
Lemma 8.12.3132BB9E1B1F5
Proof. Take smooth decreasing uniformly bounded admissible $v_ j\downarrow v$. Cone convexity makes $\alpha v_ j+b$ admissible because
\[ \omega +\mathrm{dd}^{\mathrm{c}}(\alpha v_ j+b) =\alpha (\omega +\mathrm{dd}^{\mathrm{c}}v_ j)+(1-\alpha )\omega . \]
For $\alpha >0$, an upper test $q$ for $\alpha v+b$ gives the upper test $(q-b)/\alpha $ for $v$. The same convex combination of forms therefore proves admissibility of the limit. For $\alpha =0$ the potential is constant. The pointwise decreasing limit has the displayed canonical affine values. Expand the smooth top product. Every mixed term is positive by polarized Gårding positivity. Keeping the two extreme terms gives
\[ H_ m(\alpha v_ j+b)\geq \alpha ^ m H_ m(v_ j) +(1-\alpha )^ m\omega ^ n. \]
Joint decreasing convergence from 8.11.12 passes this inequality against each nonnegative continuous test to the asserted inequality of positive Radon measures. $\square$
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