We now show that the measure $\mu_{t}:=f_{t}\omega_{t}^{n}/\penalty Vol_{\omega_{t}}$ are uniformly strongly dominated by the normalized capacity $\mathrm{Cap}_{\omega_{t}}/\penalty Vol_{\omega_{t}}(X).$ It actually follows from a carefull reading of the no parameter proof given in [EGZ], [BGZ].

Proof: Fix a compact set $K\subset X$. Set $V_{t}:=Vol_{\omega_{t}}(X)$. Hölder’s inequality yields

$$\mu_{t}(K)\leq\left(\int_{X}f_{t}^{p^{\prime}}\frac{\omega_{t}^{n}}{V_{t}}\right)^{1/\penalty p^{\prime}}\left(\int_{K}\frac{\omega_{t}^{n}}{V_{t}}\right)^{1/\penalty q^{\prime}}.$$

It remains to dominate uniformly the normalized volume forms $\omega_{t}^{n}/\penalty V_{t}$ by the normalized capacities $\mathrm{Cap}_{\omega_{t}}/\penalty V_{t}$. Fix $\sigma>0$ and observe that for any $t\in]0,1]$,

$$\int_{K}\frac{\omega_{t}^{n}}{V_{t}}\leq\int_{X}e^{-\sigma(V_{K,\omega_{t}}-\max_{X}V_{K,\omega_{t}})}\frac{\omega_{t}^{n}}{V_{t}}T_{\omega_{t}}(K)^{\sigma},$$

where

$$V_{K,\omega_{t}}:=\sup\{\psi\in PSH(X,\omega_{t});\psi\leq 0,\ \mathrm{on}\ K\}$$

is the $\omega_{t}-$extremal function of $K$ and $T_{\omega_{t}}(K):=\exp(-\sup_{X}V_{K,\omega_{t}})$ is the associated $\omega_{t}-$capacity of $K$ (see [GZ 1] for their properties).

Observe that $\omega_{t}^{n}/\penalty V_{t}\leq c_{1}\omega_{1}^{n}$ and $\omega_{t}\leq\omega_{1}$, hence the family of functions $V_{K,\omega_{t}}-\max_{X}V_{K,\omega_{t}}$ is a normalized family of $\omega_{1}-$psh functions. Thus there exists $\sigma>0$ which depends only on $(X,\omega_{1})$ and a constant $B=B(\sigma,X,\omega_{1})$ such that ([Z])

$$\int_{X}e^{-\sigma(V_{K,\omega_{t}}-\max_{X}V_{K,\omega_{t}})}\frac{\omega_{t}^{n}}{V_{t}}\leq B,\forall t\in]0,1].$$

The Alexander-Taylor comparison theorem (see Theorem 7.1 in [GZ 1]) now yields for a constant $C_{3}=C_{3}(\pi,\|F\|_{L^{p}(X)})$

$$\mu_{t}(K)\leq C_{3}\exp\left[-\sigma\left(\frac{V_{t}}{\mathrm{Cap}_{\omega_{t}}(K)}\right)^{1/\penalty n}\right],\forall t\in]0,1].$$

We infer that there is a constant $C_{4}=C_{4}(\pi,\|F\|_{L^{p}(X)})$ such that

$(2)$

$$\mu_{t}(K)\leq C_{4}\left(\frac{\mathrm{Cap}_{\omega_{t}}(K)}{V_{t}}\right)^{2},\forall t\in]0,1].$$