2.1 Preliminary remarks [02G9]

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2.1 Preliminary remarks

Uniform control of $c_{t}$ . Observe that $\omega_{0}^{k}=0$ for $m<k\leq n$ , hence for all $t\in]0,1]$ ,

\omega_{t}^{n}=\sum_{k=1}^{m}{n\choose k}t^{n-k}{\omega_{0}}^{k}\wedge{\omega_{X}}^{n-k}.

Note that $]0,1]\ni t\longmapsto t^{m-n}\omega_{t}^{n}$ is increasing (hence decreases as $t\searrow 0^{+}$ ) and satisfies for $t\in]0,1]$

(1)

{n\choose m}\qquad\frac{\omega_{0}^{m}\wedge\omega_{X}^{n-m}}{\int_{X}\omega_{0}^{m}\wedge\omega_{X}^{n-m}}\leq\frac{\omega_{t}^{n}}{t^{n-m}\int_{X}\omega_{0}^{m}\wedge\omega_{X}^{n-m}}\leq\frac{\omega_{1}^{n}}{\int_{X}\omega_{0}^{m}\wedge\omega_{X}^{n-m}}.

In particular $t\longmapsto c_{t}$ is increasing in $t\in]0,1]$ and

0<{n\choose m}\frac{\int_{X}\omega_{0}^{m}\wedge\omega_{X}^{n-m}}{\int_{X}F\omega_{X}^{n}}=:c_{0}\ \ \leq\ \ c_{t}\leq\ \ c_{1}.

Uniform control of densities. Let $J_{\pi}$ denote the (modulus square) of the Jacobian of the mapping $\pi$ , defined through

\omega_{0}^{m}\wedge\omega_{X}^{n-m}=J_{\pi}\omega_{X}^{n}.

Let us rewrite the equation $(\star)_{t}$ as follows

(\omega_{t}+dd^{c}\varphi_{t})^{n}=f_{t}\omega_{t}^{n},

where for $t\in]0,1]$

0\leq f_{t}:=c_{t}t^{n-m}F\frac{\omega_{X}^{n}}{\omega_{t}^{n}}\leq c_{1}\frac{F}{J_{\pi}}.

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.

Lemma 2.1

There exists $p^{\prime}>1$ and a constant $C=C(\pi,\|F\|_{L^{p}(X)})>0$ such that for all $t\in]0,1]$

\int_{X}f_{t}^{p^{\prime}}\omega_{t}^{n}\leq C\,Vol_{\omega_{t}}(X).

Proof of the lemma. Set $V_{t}:=Vol_{\omega_{t}}=\int_{X}\omega_{t}^{n}$ and observe that

0\leq f_{t}\frac{\omega_{t}^{n}}{V_{t}}\leq c_{1}F\frac{\omega_{X}^{n}}{\int_{X}\omega_{0}^{m}\wedge\omega_{X}^{n-m}}=C_{2}F\omega_{X}^{n},

where $C_{2}:=c_{1}\int_{X}J_{\pi}\omega_{X}^{n}$ .

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}.

$\blacktriangleright$