ScalingStacks

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Lemma 2.10. (cf. [15, Lemma 2.4 and Remark 2.5]) Let f:[t0,∞)→[0,∞)f:[t_{0},\infty)\to[0,\infty) be a nonincreasing right-continuous function, such that

{f⁡(t0)<12​B,tf(τ+t)≤Bf(τ)2,∀τ≥0,0≤t≤1,limt→∞f⁡(t)=0.\begin{cases}f(t_{0})<\frac{1}{2B},\\ tf(\tau+t)\leq Bf(\tau)^{2},\quad\forall\tau\geq 0,\quad 0\leq t\leq 1,\\ \lim_{t\to\infty}f(t)=0.\end{cases}

Then f⁡(t)=0f(t)=0 for t≥t0+4​B​f​(t0)t\geq t_{0}+4Bf(t_{0}).

00PV

Proof. (Thm 2.7) Combining the first two ingredients, the function f⁡(t)=(∫ϕ≤−tωϕnVol​(X))1/2​nf(t)=(\frac{\int_{\phi\leq-t}\omega_{\phi}^{n}}{\text{Vol}(X)})^{1/2n} satisfies

t​f​(t+τ)≤B​f​(τ)2,0≤t≤1,τ≥0,tf(t+\tau)\leq Bf(\tau)^{2},\quad 0\leq t\leq 1,\quad\tau\geq 0,

We conclude that for t>t0+4​B​f​(t0)t>t_{0}+4Bf(t_{0}) the sublevel set {ϕ≤−t}\{\phi\leq-t\} has zero ωϕ\omega_{\phi}-measure, and therefore zero capacity by Lemma 2.8, so ϕ\phi has the lower estimate as claimed in the first statement.

For the second statement, by (2) we have an a priori exponential decay

f(t)≤A1/2​ne−αt/2n,t≥0,f(t)\leq A^{1/2n}e^{-\alpha t/2n},\quad t\geq 0,

which allows us to find an appropriate t0t_{0}. ∎

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