Lemma 6.14.16B84DAE83456

Proof. If $K=\varnothing $ we take $C=0$. Otherwise cover $K$ by finitely many compact sets $K_\iota $, each contained in a chart and in the domain of $\alpha $, and write $\alpha =\mathrm i\sum _{j,k}a_{jk}\, \mathrm dz_ j\wedge \mathrm d\bar z_ k$ and $\omega =\mathrm i\sum _{j,k}w_{jk}\, \mathrm dz_ j\wedge \mathrm d\bar z_ k$ there, with Hermitian matrices $(a_{jk})$ and $(w_{jk})$. On $K_\iota $ the smallest eigenvalue of $(w_{jk})$, a continuous positive function, is at least some $c_\iota >0$, and the largest eigenvalue of $-(a_{jk})$ is at most some $M_\iota \geq 0$. Hence $(a_{jk})\geq -M_\iota I\geq -(M_\iota /c_\iota )(w_{jk})$ on $K_\iota $. By the matrix criterion for positivity (section 5.4), it follows that $\alpha \geq -C\omega $ on $K$ with $C=\max _\iota M_\iota /c_\iota $. $\square$