ScalingStacks

Proof. [02DN]

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Proof.

Set M:=‖ψ‖L∞​(X)M:=||\psi||_{L^{\infty}(X)}. Observe that when t≥0t\geq 0, (φ−ψ<−s−t​ψ)⊂(φ−ψ<−s+t​M)(\varphi-\psi<-s-t\psi)\subset(\varphi-\psi<-s+tM), hence it follows from lemma 2.2 that for all s>0s>0, 0≤t≤10\leq t\leq 1,

(3) tn​C​a​pω​(φ−ψ<−s−t)≤(1+M)n​∫(φ−ψ<−s)ωφn,t^{n}Cap_{\omega}(\varphi-\psi<-s-t)\leq(1+M)^{n}\int_{(\varphi-\psi<-s)}\omega_{\varphi}^{n},

using the obvious substitutions s↦s−M​ts\mapsto s-Mt, t↦(1+M)​tt\mapsto(1+M)t. Since μ=ωφn\mu=\omega_{\varphi}^{n} satisfies ℋ⁡(α,A,ω){\mathcal{H}}(\alpha,A,\omega), we infer

tn​C​a​pω​(φ−ψ<−s−t)≤A​(1+M)n​C​a​pω​(φ−ψ<−s)1+α.t^{n}Cap_{\omega}(\varphi-\psi<-s-t)\leq A(1+M)^{n}Cap_{\omega}(\varphi-\psi<-s)^{1+\alpha}.

Consider

f⁡(s):=[C​a​pω​(φ−ψ<−s−ε)]1/n,s>0.f(s):=\left[Cap_{\omega}(\varphi-\psi<-s-\varepsilon)\right]^{1/n},\,s>0.

Then ff satisfies the condition H⁡(α,B)H(\alpha,B) of lemma 2.3 with B=(1+M)​A1/nB=(1+M)A^{1/n}. Assume f⁡(0)=[C​a​pω​(φ−ψ<−ε)]1/n<1(2​B)1/αf(0)=\left[Cap_{\omega}(\varphi-\psi<-\varepsilon)\right]^{1/n}<\frac{1}{(2B)^{1/\alpha}}. It follows in this case from Remarks 2.4 that f⁡(s)=0f(s)=0 for s≥S∞s\geq S_{\infty}, where

S∞≤2​B1−2−α​[C​a​pω​(φ−ψ<−ε)]α/nS_{\infty}\leq\frac{2B}{1-2^{-\alpha}}\left[Cap_{\omega}(\varphi-\psi<-\varepsilon)\right]^{\alpha/n}

Therefore the sets {φ−ψ<−s−ε}\{\varphi-\psi<-s-\varepsilon\} are empty for s>S∞s>S_{\infty}, hence

supX(ψ−φ)≤ε+S∞≤ε+2​(1+M)​A1/n1−2−α​[C​a​pω​(φ−ψ<−ε)]α/n.\sup_{X}(\psi-\varphi)\leq\varepsilon+S_{\infty}\leq\varepsilon+\frac{2(1+M)A^{1/n}}{1-2^{-\alpha}}\left[Cap_{\omega}(\varphi-\psi<-\varepsilon)\right]^{\alpha/n}.

Thus we can take here C≥2​B/(1−2−α)C\geq 2B/(1-2^{-\alpha}).

If f(0)=[Capω(φ−ψ<−ε)]1/n≥(2B)−1/αf(0)=[Cap_{\omega}(\varphi-\psi<-\varepsilon)]^{1/n}\geq(2B)^{-1/\alpha}, then

supX(ψ−φ)≤−infXφ≤C2(α,A,ω),\sup_{X}(\psi-\varphi)\leq-\inf_{X}\varphi\leq C_{2}(\alpha,A,\omega),

by (2), hence it suffices to take C≥2​B​C2C\geq 2BC_{2} to conclude. ∎

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