This estimate is contained in the second-named author’s work [38], although it is not explicitly stated there. To see this, start from [38, (3.24)], which gives a constant $C$ so that on $U$ we have

$$C^{-1}(t\omega_{M})\leqslant\tilde{\omega}_{t}.$$

Then use [38, Lemma 3.1] to get

$$C^{-1}\omega_{0}\leqslant\tilde{\omega}_{t},$$

and so adding these two inequalities we get

$$C^{-1}(\omega_{0}+t\omega_{M})\leqslant\tilde{\omega}_{t},$$

or in other words $\textrm{tr}_{\tilde{\omega}_{t}}\omega_{t}\leqslant C$ on $U$, where $\omega_{t}=\omega_{0}+t\omega_{M}$ as before. To get the reverse inequality, we note that on $U$ we have

$$\textrm{tr}_{\omega_{t}}\tilde{\omega}_{t}\leqslant(\textrm{tr}_{\tilde{\omega}_{t}}\omega_{t})^{n-1}\frac{\tilde{\omega}_{t}^{n}}{\omega_{t}^{n}}\leqslant C\frac{\tilde{\omega}_{t}^{n}}{\omega_{t}^{n}}\leqslant C,$$

where the last inequality follows from [38, (3.23)]. We thus get the reverse inequality

$$\tilde{\omega}_{t}\leqslant C(\omega_{0}+t\omega_{M}),$$

thus proving (4.1). ∎