Observe that

$$\int_{X}f_{t}\omega_{t}^{n}=c_{t}t^{n-m}\int_{X}F\omega_{t}^{n}=\int_{X}\omega_{t}^{n}=:Vol_{\omega_{t}}(X),$$

hence $(f_{t})$ is uniformly bounded in $L^{1}(\omega_{t}/V_{t})$, $V_{t}:=Vol_{\omega_{t}}(X)$. We actually need a slightly stronger information.

This shows that the densities $f_{t}$ are uniformly in $L^{1}$ w.r.t. the normalized volume fomrs $\omega_{t}^{n}/\penalty V_{t}$.

Since $J_{\pi}$ is locally given as the square of the modulus of a holomorphic function which does not vanish identically, there exists $\alpha\in]0,1[$ such that $J_{\pi}^{-\alpha}\in L^{1}(X)$. Fix $\beta\in]0,\alpha[$ satisfying the condition $\beta/\penalty p+\beta/\penalty\alpha=1$. It follows from Hölder’s inequality that

$$\int_{X}f_{t}^{\beta}\omega_{X}^{n}\leq\Bigl(\int_{X}F^{p}\omega_{X}^{n}\Bigr)^{\beta/\penalty p}\Bigl(\int_{X}J_{\pi}^{-\alpha}\omega_{X}^{n}\Bigr)^{\beta/\penalty\alpha}.$$

Setting $\varepsilon:=\beta/\penalty q$ and using Hölder’s inequality again , we obtain

$$\int_{X}f_{t}^{1+\varepsilon}\frac{\omega_{t}^{n}}{V_{t}}\leq C_{2}\int_{X}f_{t}^{\varepsilon}F\omega_{X}^{n}.$$

Now applying again Hölder inequality we get

$$\int_{X}f_{t}^{1+\varepsilon}\frac{\omega_{t}^{n}}{V_{t}}\leq C_{2}\Bigl(\int_{X}f_{t}^{\beta}\omega_{X}\Bigr)^{1/\penalty q}\|F\|_{L^{p}(X)}.$$

Therefore denoting by $p^{\prime}:=1+\varepsilon$, we have the following uniform estimate

$$\int_{X}f_{t}^{p^{\prime}}\frac{\omega_{t}^{n}}{V_{t}}\leq C(\pi,\|F\|_{L^{p}(X)}),\forall t\in]0,1],$$

where

$$C(\pi,\|F\|_{L^{p}(X)}):=C_{2}\Bigl(\int_{X}J_{\pi}^{-\alpha}\omega_{X}^{n}\Bigr)^{\beta/\penalty\alpha q}\|F\|_{L^{p}(X)}^{1+\beta/\penalty q}.$$

$\blacktriangleright$

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

$(2)$

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

The comparison principle (see [K] ,[EGZ]) yields for any $s>0$ and $\tau\in[0,1]$

$$\tau^{n}\frac{\mathrm{Cap}_{\omega_{t}}(\{\varphi_{t}\leq-s-\tau\})}{V_{t}}\leq\int_{\{\varphi_{t}\leq-s\}}\frac{(\omega_{t}+dd^{c}\varphi_{t})^{n}}{V_{t}}.$$

It is now an exercise to derive from this inequality an a priori $L^{\infty}-$estimate,

$$\|\varphi_{t}\|_{L^{\infty}(X)}\leq C_{5}+\ {s}_{0}(\omega_{t}),$$

where ${s}_{0}(\omega_{t})$ (see [EGZ],[BGZ]) is the smallest number $s>0$ satisfying the condition $e^{n}C_{0}^{n}\mathrm{Cap}_{\omega_{t}}(\{\psi\leq-s\})/V_{t}\leq 1$ for all $\psi\in PSH(X,\omega_{t})$ such that $\sup_{X}\psi=0$. Recall from ([GZ 1], Prop. 3.6) that

$$\frac{\mathrm{Cap}_{\omega_{t}}(\{\psi\leq-s-\tau\})}{V_{t}}\leq\frac{1}{s}\left(\int_{X}(-\psi)\frac{\omega_{t}^{n}}{V_{t}}+n\right).$$

Since $\frac{\omega_{t}^{n}}{V_{t}}\leq C_{1}\omega_{1}^{n}$, it follows that

$$\frac{\mathrm{Cap}_{\omega_{t}}(\{\psi\leq-s-\tau\})}{V_{t}}\leq\frac{1}{s}\left(C_{1}\int_{X}(-\psi)\omega_{1}^{n}+n\right).$$

Since $\psi$ is $\omega_{1}-$psh and normalized, we know that there is a constant $A=A(X,\omega_{1})>0$ such that $C_{1}\int_{X}(-\psi){\omega_{1}^{n}}\leq A$ for any such $\psi$. Therefore $s_{0}(\omega_{t})\leq s_{0}:=e^{n}C_{0}^{n}(A+n)$ for any $t\in]0,1]$. Finally we obtain the required uniform estimate for all $t\in]0,1]$.