ScalingStacks

Proof. [040M]

Original official author HTML, exact retained edition. Historical TeX conversion verdicts remain unchanged. Cited-edition alignment and mathematical self-containment are not assessed.

Complete original source context · Original author HTML

Proof.

By scaling analysis we may assume A∼1A\sim 1. Let uu be a T2T^{2}-invariant function with ∫M|∇u|2=1\int_{M}|\nabla u|^{2}=1, so descends to a function on the base ℝ4\mathbb{R}^{4}. Since the weighted Sobolev inequality holds on Euclidean ℝ4\mathbb{R}^{4} (by an interpolation of standard Sobolev inequality and Hardy inequality),

(∫ℝ4|u|2​p​(1+|μ→|a)2​p−4​d​Vola)1/p≤C​∫ℝ4|∇gau|2​d​Vola≤C​∫M|∇g(2)u|2​𝑑Vol≤C,(\int_{\mathbb{R}^{4}}|u|^{2p}(1+|\vec{\mu}|_{a})^{2p-4}d\text{Vol}_{a})^{1/p}\leq C\int_{\mathbb{R}^{4}}|\nabla_{g_{a}}u|^{2}d\text{Vol}_{a}\leq C\int_{M}|\nabla_{g^{(2)}}u|^{2}d\text{Vol}\leq C,

where the second inequality is easily seen using the model metric in Section 2.3. The LHS in this inequality is uniformly equivalent to the LHS in (2.20) except in the region {distga(⋅,𝔇)≤1}\{\text{dist}_{g_{a}}(\cdot,\mathfrak{D})\leq 1\}. So we are left to prove

(∫dist​(⋅,𝔇)≲1|u|2​p​(1+|μ→|a)2​p−4​𝑑Vol)1/p≤C.(\int_{\text{dist}(\cdot,\mathfrak{D})\lesssim 1}|u|^{2p}(1+|\vec{\mu}|_{a})^{2p-4}d\text{Vol})^{1/p}\leq C.

For x∈𝔇1,𝔇2,𝔇3x\in\mathfrak{D}_{1},\mathfrak{D}_{2},\mathfrak{D}_{3}, Sobolev inequality on bounded balls imply

(∫B⁡(x,1)|u−u¯​(x)|2​p)1/p≤C​∫B⁡(x,2)|∇u|2,u¯​(x)=Vol​(B⁡(x,1))−1​∫B⁡(x,1)u(\int_{B(x,1)}|u-\bar{u}(x)|^{2p})^{1/p}\leq C\int_{B(x,2)}|\nabla u|^{2},\quad\bar{u}(x)=\text{Vol}(B(x,1))^{-1}\int_{B(x,1)}u

Furthermore we can find a point x′x^{\prime} with dist​(x,x′)≤3\text{dist}(x,x^{\prime})\leq 3, dist​(x′,𝔇)≳2\text{dist}(x^{\prime},\mathfrak{D})\gtrsim 2, and by Sobolev inequality

(∫B⁡(x′,1)|u−u¯​(x′)|2​p)1/p≤C​∫B⁡(x,5)|∇u|2,u¯​(x′)=Vol​(B⁡(x′,1))−1​∫B⁡(x′,1)u.(\int_{B(x^{\prime},1)}|u-\bar{u}(x^{\prime})|^{2p})^{1/p}\leq C\int_{B(x,5)}|\nabla u|^{2},\quad\bar{u}(x^{\prime})=\text{Vol}(B(x^{\prime},1))^{-1}\int_{B(x^{\prime},1)}u.

By Poincaré inequality

|u¯​(x)−u¯​(x′)|2≤∫B⁡(x,5)|∇u|2.|\bar{u}(x)-\bar{u}(x^{\prime})|^{2}\leq\int_{B(x,5)}|\nabla u|^{2}.

Combining these,

(∫B⁡(x,1)|u|2​p)1/p≤C​(∫B⁡(x′,1)|u|2​p)1/p+C​∫B⁡(x,5)|∇u|2.(\int_{B(x,1)}|u|^{2p})^{1/p}\leq C(\int_{B(x^{\prime},1)}|u|^{2p})^{1/p}+C\int_{B(x,5)}|\nabla u|^{2}.

Multiplying this inequality by (1+|μ→|a​(x))(2​p−4)/p(1+|\vec{\mu}|_{a}(x))^{(2p-4)/p}, and summing over x∈𝔇x\in\mathfrak{D}, we obtain

(∫dist​(⋅,𝔇)≲1|u|2​p​(1+|μ→|a)2​p−4​𝑑Vol)1/p≤C​(∫dist​(⋅,𝔇)≳1|u|2​p​(1+|μ→|a)2​p−4​𝑑Vol)1/p+C​∫|∇u|2≤C\begin{split}&(\int_{\text{dist}(\cdot,\mathfrak{D})\lesssim 1}|u|^{2p}(1+|\vec{\mu}|_{a})^{2p-4}d\text{Vol})^{1/p}\\ \leq&C(\int_{\text{dist}(\cdot,\mathfrak{D})\gtrsim 1}|u|^{2p}(1+|\vec{\mu}|_{a})^{2p-4}d\text{Vol})^{1/p}+C\int|\nabla u|^{2}\leq C\end{split}

as required. ∎

Original mathematics by the credited authors. Source-backed reader collection; mathematical self-containment is not assessed.