ScalingStacks

Proof. First notice that

$$\int_{X}\omega_{t}\wedge\omega_{0}^{n-1}=\alpha_{t}\cdot\alpha_{0}^{n-1}\leq C,$$

which gives a uniform $L^{1}$ bound on $\omega_{t}$. Up to covering $X$ by finitely many charts, we may assume that $X=K$ is a compact convex set in $\mathbb{C}^{n}$, and we will denote by $g_{E}$ the Euclidean metric on $K$. If $x_{1},x_{2}\in K$, we denote by $[x_{1},x_{2}]$ the segment joining them in $K$, and we compute the average of the length square of $[x_{1},x_{2}]$ with respect to $\omega_{t}$, when the endpoints vary. Using Fubini’s Theorem and the Cauchy-Schwarz inequality we get

(3.1)

$$\begin{split}&\int_{K\times K}\left(\int_{0}^{1}\sqrt{\omega_{t}((1-t)x_{1}+tx_{2})(x_{2}-x_{1})}dt\right)^{2}dx_{1}dx_{2}\\ &\leq\|x_{2}-x_{1}\|^{2}_{g_{E}}\int_{0}^{1}\int_{K\times K}|\omega_{t}((1-t)x_{1}+tx_{2})|dx_{1}dx_{2}dt\\ &\leq\mathrm{diam}_{g_{E}}^{2}(K)2^{2n}\biggl(\int_{0}^{\frac{1}{2}}\int_{K\times K}|\omega_{t}(y+tx_{2})|dydx_{2}dt\\ &+\int_{\frac{1}{2}}^{1}\int_{K\times K}|\omega_{t}((1-t)x_{1}+y)|dydx_{1}dt\biggr)\\ &\leq\mathrm{diam}_{g_{E}}^{2}(K)2^{2n}\mathrm{Vol}_{g_{E}}(K)\|\omega_{t}\|_{L^{1}(K)}\leq C_{1},\end{split}$$

where $C_{1}$ is a uniform constant, we changed variable $y=(1-t)x_{1}$ if $t\leq\frac{1}{2}$ and $y=tx_{2}$ when $t\geq\frac{1}{2}$ and integrated first with respect to $y$. Then the set $S$ of pairs $(x_{1},x_{2})\in K\times K$ such that the length of $[x_{1},x_{2}]$ with respect to $\omega_{t}$ is more than $(C_{1}/\delta)^{1/2}$ has Euclidean measure less than or equal $\delta$: otherwise

$$\begin{split}&\int_{K\times K}\left(\int_{0}^{1}\sqrt{\omega_{t}((1-t)x_{1}+tx_{2})(x_{2}-x_{1})}dt\right)^{2}dx_{1}dx_{2}\\ &\geq\int_{S}\left(\int_{0}^{1}\sqrt{\omega_{t}((1-t)x_{1}+tx_{2})(x_{2}-x_{1})}dt\right)^{2}dx_{1}dx_{2}\geq\frac{C_{1}}{\delta}\mathrm{Vol}_{g_{E}}(S)\end{split}$$

which is more than $C_{1}$, and this contradicts (3.1). If $x_{1}\in K$ we let $S(x_{1})$ to be the set of the $x_{2}\in K$ such that $(x_{1},x_{2})\in S$, and we let $Q$ to be the set of the $x_{1}\in K$ such that $\mathrm{Vol}_{g_{E}}(S(x_{1}))\geq\frac{1}{2}\mathrm{Vol}_{g_{E}}(K)$ and $R$ to be the set of $(x_{1},x_{2})\in S$ such that $x_{1}\in Q$. Then by Fubini’s Theorem

$$\delta\geq\mathrm{Vol}_{g_{E}}(R)=\int_{R}dx_{2}dx_{1}=\int_{Q}\left(\int_{S(x_{1})}dx_{2}\right)dx_{1}\geq\frac{1}{2}\mathrm{Vol}_{g_{E}}(K)\mathrm{Vol}_{g_{E}}(Q),$$

and so $\mathrm{Vol}_{g_{E}}(Q)\leq\frac{2\delta}{\mathrm{Vol}_{g_{E}}(K)}.$ We let $U_{t,\delta}=K\backslash Q$. Then $U_{t,\delta}$ is open and if $x_{1},x_{2}\in U_{t,\delta}$ then $\mathrm{Vol}_{g_{E}}(S(x_{i}))<\frac{1}{2}\mathrm{Vol}_{g_{E}}(K)$, for $i=1,2$. Hence $\mathrm{Vol}_{g_{E}}((K\backslash S(x_{1}))\cap(K\backslash S(x_{2})))>0$ and so this set is nonempty. If $y$ belongs to it, then $(x_{1},y)$ and $(x_{2},y)$ are not in $S$, which means that the lengths with respect to $\omega_{t}$ of the segments $[x_{1},y]$ and $[y,x_{2}]$ are both less than $(C_{1}/\delta)^{1/2}$. Concatenating these two segments we get a path from $x_{1}$ to $x_{2}$ with length less than $2(C_{1}/\delta)^{1/2}$. We also have that

$$\mathrm{Vol}_{\omega_{0}}(Q)\leq C_{2}\mathrm{Vol}_{g_{E}}(Q)\leq\frac{2C_{2}\delta}{\mathrm{Vol}_{g_{E}}(K)}.$$

Up to adjusting the constants, this is what we want. ∎