ScalingStacks

Proof. [040Y]

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Proof.

The error Δg(2)​(χ1​GTaub​f)−f\Delta_{g^{(2)}}(\chi_{1}G_{\text{Taub}}f)-f comes from two sources: the deviation of the metric gTaubg_{\text{Taub}} from g(2)g^{(2)}, and the cutoff error. The metric deviation error is estimated in Lemma 2.6 which we recall as ‖gTaub−g(2)‖C0,−1k,α≤C.\left\lVert g_{\text{Taub}}-g^{(2)}\right\rVert_{C^{k,\alpha}_{0,-1}}\leq C. In particular for |μ→|a>C3|\vec{\mu}|_{a}>C_{3} and on the support of χ1\chi_{1}, we have ‖gTaub−g(2)‖C0,0k,α≤C​C3−1,\left\lVert g_{\text{Taub}}-g^{(2)}\right\rVert_{C^{k,\alpha}_{0,0}}\leq CC_{3}^{-1}, so the metric deviation error is O⁡(C3−1)O(C_{3}^{-1}).

We turn to the cutoff error. By Lemma 2.20

{‖χ1GTaubf‖C−1+ϵ,τk+2,α≤C,−1<τ<1−ϵ,‖χ1GTaubf‖C0,−2+ϵk+2,α≤C,τ≤−1.\begin{cases}\left\lVert\chi_{1}G_{\text{Taub}}f\right\rVert_{C^{k+2,\alpha}_{-1+\epsilon,\tau}}\leq C,\quad-1<\tau<1-\epsilon,\\ \left\lVert\chi_{1}G_{\text{Taub}}f\right\rVert_{C^{k+2,\alpha}_{0,-2+\epsilon}}\leq C,\quad\tau\leq-1.\end{cases}

which implies

{‖∇2Taub(χ1GTaubf)‖C−3+ϵ,τk+2,α≤C,−1<τ<1−ϵ,‖∇2Taub(χ1GTaubf)‖C−2,−2+ϵk+2,α≤C,τ≤−1.\begin{cases}\left\lVert\nabla^{2}_{\text{Taub}}(\chi_{1}G_{\text{Taub}}f)\right\rVert_{C^{k+2,\alpha}_{-3+\epsilon,\tau}}\leq C,\quad-1<\tau<1-\epsilon,\\ \left\lVert\nabla^{2}_{\text{Taub}}(\chi_{1}G_{\text{Taub}}f)\right\rVert_{C^{k+2,\alpha}_{-2,-2+\epsilon}}\leq C,\quad\tau\leq-1.\end{cases}

so in particular on supp(dχ1)∩{|μ→|a>2C3}\text{supp}(d\chi_{1})\cap\{|\vec{\mu}|_{a}>2C_{3}\} where ℓ∼C3​C1−1\ell\sim C_{3}C_{1}^{-1}, we have

‖∇Taub2(χ1​GTaub​f)‖Cδ,τk,α≤C​C3−ϵ.\left\lVert\nabla^{2}_{\text{Taub}}(\chi_{1}G_{\text{Taub}}f)\right\rVert_{C^{k,\alpha}_{\delta,\tau}}\leq CC_{3}^{-\epsilon}.

By the support assumptions f=0f=0 on supp(dχ1)∩{|μ→|a>2C3}\text{supp}(d\chi_{1})\cap\{|\vec{\mu}|_{a}>2C_{3}\}, hence the cutoff error is O⁡(C3−ϵ)O(C_{3}^{-\epsilon}). Combining the two errors give the claim. ∎

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