ScalingStacks

We now show the left-hand side inequality in (2.9). To this extent we apply the maximum principle to the quantity

$$K_{1}=e^{-B\sigma^{-\lambda}_{h}}\left(\log(t\cdot\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X})-\frac{A}{t}(\varphi_{t}-\underline{\varphi_{t}})\right),$$

where $A$ is a suitably chosen uniform large constant. The maximum of $K_{1}$ on $X\backslash S$ is obviously achieved, and we will show that $K_{1}\leq C$ for a uniform constant $C$. This together with (3.9) will show that on $X\backslash S$ we have

(3.24)

$$\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}\leq\frac{C}{t}e^{Ce^{B\sigma^{-\lambda}}},$$

which is the other half of (2.9). To prove that $K_{1}\leq C$ we use the maximum principle and, as in (3.16), we compute

(3.25)

$$\begin{split}\Delta_{\tilde{\omega}_{t}}K_{1}&\geq e^{-B\sigma^{-\lambda}}\left(\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}-\frac{C}{t}-C\sigma^{-\lambda}\right)\\ &+\left(\log(t\cdot\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X})-\frac{A}{t}(\varphi_{t}-\underline{\varphi_{t}})\right)\Delta_{\tilde{\omega}_{t}}\left(e^{-B\sigma^{-\lambda}}\right)\\ &+2e^{B\sigma^{-\lambda}}\mathrm{Re}\langle\nabla K_{1},\nabla e^{-B\sigma^{-\lambda}}\rangle_{\tilde{\omega}_{t}}\\ &-2\left(\log(t\cdot\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X})-\frac{A}{t}(\varphi_{t}-\underline{\varphi_{t}})\right)e^{B\sigma^{-\lambda}}|\nabla e^{-B\sigma^{-\lambda}}|^{2}_{\tilde{\omega}_{t}}.\end{split}$$

We estimate this in the same way as before and get

(3.26)

$$\begin{split}\Delta_{\tilde{\omega}_{t}}K_{1}&\geq e^{-B\sigma^{-\lambda}}\left(\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}-\frac{C}{t}-C\sigma^{-\lambda}\right)\\ &-C\frac{e^{-B\sigma^{-\lambda}}}{\sigma^{2\lambda+2}}\log(t\cdot\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X})-\frac{C}{\sigma^{2\lambda+2}}\\ &+2e^{B\sigma^{-\lambda}}\mathrm{Re}\langle\nabla K_{1},\nabla e^{-B\sigma^{-\lambda}}\rangle_{\tilde{\omega}_{t}}.\end{split}$$

At the maximum of $K_{1}$ we get

$$0\geq\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}-\frac{C}{t}-\frac{C}{\sigma^{\lambda}}-\frac{C}{\sigma^{2\lambda+2}}\log(t\cdot\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X})-Ce^{C\sigma^{-\lambda}},$$

and using the inequalities $2ab\leq\varepsilon a^{2}+b^{2}/\varepsilon$ and $(\log x)^{2}\leq x+C$ we get

$$\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}\leq\frac{C}{t}+\frac{C}{\sigma^{\lambda}}+\frac{C}{\sigma^{4\lambda+4}}+Ce^{C\sigma^{-\lambda}}+\frac{1}{2}\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X},$$

whence

$$t\cdot\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}\leq C+tCe^{C\sigma^{-\lambda}}\leq Ce^{C\sigma^{-\lambda}},$$

and so at that point

$$K_{1}\leq C+e^{-B\sigma^{-\lambda}}\log(Ce^{C\sigma^{-\lambda}})\leq C,$$

and we are done. ∎