ScalingStacks

Thanks to (4.1) we have

(4.4)

$$\begin{split}\int_{\tilde{B}_{t}}\Tr_{\tilde{\omega}_{t}}\omega_{t}d\tilde{\mu}_{t}&=\frac{n\int_{\tilde{B}_{t}}\omega_{t}\wedge\tilde{\omega}_{t}^{n-1}}{\int_{X_{t}}\tilde{\omega}^{n}_{t}}\leqslant C|\log|t||^{n}\int_{X_{t}}\omega_{t}\wedge(\omega^{\prime}_{t}+\omega_{{\rm FS},t})^{n-1}\\ &\leqslant C\int_{X_{t}}c_{1}(L)\cdot c_{1}(\mathfrak{L})^{n-1}\leqslant C.\end{split}$$

We wish to use this to prove that

(4.5)

$$\int_{\tilde{B}_{t}\times\tilde{B}_{t}}\text{dist}_{\omega_{t}}(x,y)d\tilde{\mu}_{t}(x)d\tilde{\mu}_{t}(y)\leqslant C.$$

Indeed, since $|\dot{\gamma}_{x,y}(s)|_{\tilde{\omega}_{t}}\leqslant C$, we can estimate

$$\mathrm{dist}_{\omega_{t}}(x,y)\leqslant C\int_{0}^{1}(\Tr_{\tilde{\omega}_{t}}{\omega_{t}}(sy+(1-s)x))^{\frac{1}{2}}ds,$$

and

$$\int_{\tilde{B}_{t}\times\tilde{B}_{t}}\text{dist}_{\omega_{t}}(x,y)d\tilde{\mu}_{t}(x)d\tilde{\mu}_{t}(y)\leqslant C\int_{\tilde{B}_{t}\times\tilde{B}_{t}}\int_{0}^{1}(\Tr_{\tilde{\omega}_{t}}{\omega_{t}}(sy+(1-s)x))^{\frac{1}{2}}ds\,d\tilde{\mu}_{t}(x)d\tilde{\mu}_{t}(y).$$

We can then argue as in [5, Lemma 1.3], using Fubini

$$\begin{split}&\int_{\tilde{B}_{t}\times\tilde{B}_{t}}\int_{0}^{1}(\Tr_{\tilde{\omega}_{t}}{\omega_{t}}(sy+(1-s)x))^{\frac{1}{2}}ds\,d\tilde{\mu}_{t}(x)d\tilde{\mu}_{t}(y)\\ &=\int_{0}^{1}\int_{\tilde{B}_{t}\times\tilde{B}_{t}}(\Tr_{\tilde{\omega}_{t}}{\omega_{t}}(sy+(1-s)x))^{\frac{1}{2}}d\tilde{\mu}_{t}(x)d\tilde{\mu}_{t}(y)ds\\ &=\int_{0}^{\frac{1}{2}}\int_{\tilde{B}_{t}\times\tilde{B}_{t}}(\Tr_{\tilde{\omega}_{t}}{\omega_{t}}(sy+(1-s)x))^{\frac{1}{2}}d\tilde{\mu}_{t}(x)d\tilde{\mu}_{t}(y)ds\\ &+\int_{\frac{1}{2}}^{1}\int_{\tilde{B}_{t}\times\tilde{B}_{t}}(\Tr_{\tilde{\omega}_{t}}{\omega_{t}}(sy+(1-s)x))^{\frac{1}{2}}d\tilde{\mu}_{t}(y)d\tilde{\mu}_{t}(x)ds\end{split}$$

and in the two innermost integrals we change variable from $x$ (resp. $y$) to $z=sy+(1-s)x$ with $0\leqslant s\leqslant\frac{1}{2}$ (resp. $\frac{1}{2}\leqslant s\leqslant 1$), noting that $\tilde{\mu}_{t}(x)\leqslant C\tilde{\mu}_{t}(z)$ (resp. $\tilde{\mu}_{t}(y)\leqslant C\tilde{\mu}_{t}(z)$). Thus both of these innermost integrals can be bounded above by

$$C\int_{\tilde{B}_{t}}(\Tr_{\tilde{\omega}_{t}}{\omega_{t}}(z))^{\frac{1}{2}}d\tilde{\mu}_{t}(z)\leqslant C\left(\int_{\tilde{B}_{t}}\Tr_{\tilde{\omega}_{t}}{\omega_{t}}(z)d\tilde{\mu}_{t}(z)\right)^{\frac{1}{2}}\leqslant C,$$

by (4.4), and (4.5) follows.