ScalingStacks

Proof. [055B]

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Proof.

The proof of the uniform estimate consists of two primary steps: In the first step, we will prove the weighted C1C^{1} and C2C^{2} estimates,

(6.66) ‖∇u‖Cδ,ν+1,μ0​(ℳT)+‖∇2u‖Cδ,ν+2,μ0​(ℳT)≤C⋅‖Δ​u‖Cδ,ν+2,μ0,α​(ℳT).\|\nabla u\|_{C_{\delta,\nu+1,\mu}^{0}(\mathcal{M}_{T})}+\|\nabla^{2}u\|_{C_{\delta,\nu+2,\mu}^{0}(\mathcal{M}_{T})}\leq C\cdot\|\Delta u\|_{C_{\delta,\nu+2,\mu}^{0,\alpha}(\mathcal{M}_{T})}.

Next, based on the above weighted estimate and the weighted Schauder estimate (by Proposition 6.9), we will prove

(6.67) ‖∇u‖Cδ,ν+1,μ0​(ℳT)+‖∇2u‖Cδ,ν+2,μ0​(ℳT)≤C⋅‖Δ​u‖Cδ,ν+2,μ0,α​(ℳT).\|\nabla u\|_{C_{\delta,\nu+1,\mu}^{0}(\mathcal{M}_{T})}+\|\nabla^{2}u\|_{C_{\delta,\nu+2,\mu}^{0}(\mathcal{M}_{T})}\leq C\cdot\|\Delta u\|_{C_{\delta,\nu+2,\mu}^{0,\alpha}(\mathcal{M}_{T})}.

Step 1. (Weighted C1C^{1} and C2C^{2} estimates)

Now we start to prove the estimate (6.66), which will be proved by contradiction. Suppose no such a uniform constant C>0C>0 exists. That is, for fixed parameters

(6.68) −1<ν<0,ν+α<0,0<δ<δN,μ=(1−1n)​(ν+2+α),-1<\nu<0,\quad\nu+\alpha<0,\quad 0<\delta<\delta_{N},\quad\mu=(1-\frac{1}{n})(\nu+2+\alpha),

there are the following contradicting sequences:

  1. (1)

    A sequence of S1S^{1}-invariant Kähler metrics gj=gTjg_{j}=g_{T_{j}} (or ωj=ωTj\omega_{j}=\omega_{T_{j}}) on the neck ℳTj\mathcal{M}_{T_{j}} constructed in Section 4.1 with Tj→+∞T_{j}\to+\infty.

  2. (2)

    A sequence of C2,αC^{2,\alpha}-functions uj∈𝔄u_{j}\in\mathfrak{A} satisfying

    (6.69) ∂uj∂n|ℳj\displaystyle\frac{\partial u_{j}}{\partial n}\Big|_{\mathcal{M}_{j}} =0,\displaystyle=0,
    (6.70) ‖∇uj‖Cδ,ν+1,μ0​(ℳj)+‖∇2uj‖Cδ,ν+2,μ0​(ℳj)\displaystyle\|\nabla u_{j}\|_{C_{\delta,\nu+1,\mu}^{0}(\mathcal{M}_{j})}+\|\nabla^{2}u_{j}\|_{C_{\delta,\nu+2,\mu}^{0}(\mathcal{M}_{j})} =1,\displaystyle=1,
    (6.71) ‖Δ​uj‖Cδ,ν+2,μ0,α​(ℳj)\displaystyle\|\Delta u_{j}\|_{C_{\delta,\nu+2,\mu}^{0,\alpha}(\mathcal{M}_{j})} →0,j→+∞.\displaystyle\to 0,\quad j\to+\infty.

So it follows that either ‖∇uj‖Cδ,ν+1,μ0​(ℳj)≥12\|\nabla u_{j}\|_{C_{\delta,\nu+1,\mu}^{0}(\mathcal{M}_{j})}\geq\frac{1}{2} or ‖∇2uj‖Cδ,ν+2,μ0​(ℳj)≥12\|\nabla^{2}u_{j}\|_{C_{\delta,\nu+2,\mu}^{0}(\mathcal{M}_{j})}\geq\frac{1}{2}. Without loss of generality, we only consider the first case and let 𝒙j∈ℳj\bm{x}_{j}\in\mathcal{M}_{j} satisfy

(6.72) |ρδ,ν+1,μ(0)​(𝒙j)⋅∇uj​(𝒙j)|=‖∇uj‖Cδ,ν+1,μ0​(ℳj)≥12.|\rho_{\delta,\nu+1,\mu}^{(0)}(\bm{x}_{j})\cdot\nabla u_{j}(\bm{x}_{j})|=\|\nabla u_{j}\|_{C_{\delta,\nu+1,\mu}^{0}(\mathcal{M}_{j})}\geq\frac{1}{2}.

Now we renormalize the functions uju_{j} as follows,

(6.73) vj​(𝒙)=uj​(𝒙)−uj​(𝒙j).v_{j}(\bm{x})=u_{j}(\bm{x})-u_{j}(\bm{x}_{j}).

Immediately, vj​(𝒙j)=0v_{j}(\bm{x}_{j})=0, ∂vj∂n|ℳj=0\frac{\partial v_{j}}{\partial n}|_{\mathcal{M}_{j}}=0 and

(6.74) ‖∇vj‖Cδ,ν+1,μ0​(ℳj)\displaystyle\|\nabla v_{j}\|_{C_{\delta,\nu+1,\mu}^{0}(\mathcal{M}_{j})} =‖∇uj‖Cδ,ν+1,μ0​(ℳj)≤1,\displaystyle=\|\nabla u_{j}\|_{C_{\delta,\nu+1,\mu}^{0}(\mathcal{M}_{j})}\leq 1,
(6.75) ‖∇2vj‖Cδ,ν+2,μ0​(ℳj)\displaystyle\|\nabla^{2}v_{j}\|_{C_{\delta,\nu+2,\mu}^{0}(\mathcal{M}_{j})} =‖∇2uj‖Cδ,ν+2,μ0​(ℳj)≤1,\displaystyle=\|\nabla^{2}u_{j}\|_{C_{\delta,\nu+2,\mu}^{0}(\mathcal{M}_{j})}\leq 1,
(6.76) ‖Δ​vj‖Cδ,ν+2,μ0,α​(ℳj)\displaystyle\|\Delta v_{j}\|_{C_{\delta,\nu+2,\mu}^{0,\alpha}(\mathcal{M}_{j})} =‖Δ​uj‖Cδ,ν+2,μ0,α​(ℳj)→0,\displaystyle=\|\Delta u_{j}\|_{C_{\delta,\nu+2,\mu}^{0,\alpha}(\mathcal{M}_{j})}\to 0,
(6.77) ‖vj‖Cδ,ν,μ0​(ℳj)\displaystyle\|v_{j}\|_{C_{\delta,\nu,\mu}^{0}(\mathcal{M}_{j})} ≤C0.\displaystyle\leq C_{0}.

So we are led to apply the weighted Schauder estimate in Proposition 6.9, which gives

(6.78) ‖vj‖Cδ,ν,μ2,α​(ℳj)≤C0.\|v_{j}\|_{C_{\delta,\nu,\mu}^{2,\alpha}(\mathcal{M}_{j})}\leq C_{0}.

Moreover, it is straightforward that

(6.79) |ρδ,ν+1,μ(0)​(𝒙j)⋅∇vj​(𝒙j)|=‖∇vj‖Cδ,ν+1,μ0​(ℳj)\displaystyle|\rho_{\delta,\nu+1,\mu}^{(0)}(\bm{x}_{j})\cdot\nabla v_{j}(\bm{x}_{j})|=\|\nabla v_{j}\|_{C_{\delta,\nu+1,\mu}^{0}(\mathcal{M}_{j})} ≥12.\displaystyle\geq\frac{1}{2}.

We will rescale contradicting spaces (ℳj,gj)(\mathcal{M}_{j},g_{j}) around the above reference points 𝒙j\bm{x}_{j} such that the desired contradiction will arise in the limiting space. Let gjg_{j} be a sequence of contradicting metrics, then we denote the rescaling factors as follows:

  1. (1)

    Rescaling of the metrics:

    Let g~j=λj2⋅gj\tilde{g}_{j}=\lambda_{j}^{2}\cdot g_{j}, then with respect to the fixed reference point 𝒙j∈ℳj\bm{x}_{j}\in\mathcal{M}_{j} picked as the above, we have the convergence,

    (6.80) (ℳj,g~j,𝒙j)→G​H(X∞,d~∞,𝒙∞).(\mathcal{M}_{j},\tilde{g}_{j},\bm{x}_{j})\xrightarrow{GH}(X_{\infty},\tilde{d}_{\infty},\bm{x}_{\infty}).
  2. (2)

    Rescaling of the solutions:

    Let κj>0\kappa_{j}>0 be a sequence of rescaling factors which will be determined later, such that

    (6.81) v~j≡κj⋅vj.\tilde{v}_{j}\equiv\kappa_{j}\cdot v_{j}.
  3. (3)

    Rescaling of the weight functions:

    Denote by ρ~j,δ,ν,μ(k+α)\tilde{\rho}_{j,\delta,\nu,\mu}^{(k+\alpha)} and ρ~∞,δ,ν,μ(k+α)\tilde{\rho}_{\infty,\delta,\nu,\mu}^{(k+\alpha)} the weight functions on the rescaled sequence (ℳj,g~j,𝒙j)(\mathcal{M}_{j},\tilde{g}_{j},\bm{x}_{j}) and the rescaled limit (X∞,g~∞,𝒙∞)(X_{\infty},\tilde{g}_{\infty},\bm{x}_{\infty}) respectively. So we rescale the weight function ρj,δ,ν,μ(k+α)\rho_{j,\delta,\nu,\mu}^{(k+\alpha)} by

    (6.82) ρ~j,δ,ν,μ(k+α)=τj⋅ρj,δ,ν,μ(k+α).\tilde{\rho}_{j,\delta,\nu,\mu}^{(k+\alpha)}=\tau_{j}\cdot\rho_{j,\delta,\nu,\mu}^{(k+\alpha)}.

    Notice that the rescaling factor τj\tau_{j} depends on kk and α\alpha.

In the following, we study the convergence of the renormalized functions v~j∈𝔄\tilde{v}_{j}\in\mathfrak{A}, with respect to the rescaled metrics g~j\tilde{g}_{j}, in each region according to the subdivision given in Section 4.3. The main goal is to show v~∞≡0\tilde{v}_{\infty}\equiv 0 on the rescaled limit X∞X_{\infty} which gives the desired contradiction.

We will produce the desired contradiction in each region of 𝐈𝟏\bf{I}_{1}, 𝐈𝟐\bf{I}_{2}, 𝐈𝟑\bf{I}_{3} on ℳj\mathcal{M}_{j}. Before the detailed contradiction arguments, let us determine the rescaling factors in the following way. First, the scaling invariance requires

(6.83) τj⋅κjλjk+α=1.\frac{\tau_{j}\cdot\kappa_{j}}{\lambda_{j}^{k+\alpha}}=1.

Now we need to combing the regularity scale analysis in Proposition 4.18 and the choice of the weight function in Definition 4.19. So λj\lambda_{j}, τj\tau_{j} and κj\kappa_{j} are determined as follows, which depends on if |z⁡(𝒙j)||z(\bm{x}_{j})| is uniformly bounded: First, if |z⁡(𝒙j)||z(\bm{x}_{j})| is uniformly bounded (corresponding to Region 𝐈𝟏\bf{I}_{1}, 𝐈𝟐\bf{I}_{2} and Case (a) of Region 𝐈𝟑\bf{I}_{3}), we choose

(6.84) {λj=𝔰j−1τj=(𝔰j−1)ν+k+α⋅Tj−μκj=(𝔰j−1)−ν⋅Tjμ.\displaystyle\begin{cases}\lambda_{j}=\mathfrak{s}_{j}^{-1}\\ \tau_{j}=(\mathfrak{s}_{j}^{-1})^{\nu+k+\alpha}\cdot T_{j}^{-\mu}\\ \kappa_{j}=(\mathfrak{s}_{j}^{-1})^{-\nu}\cdot T_{j}^{\mu}.\end{cases}

Next, if |z⁡(𝒙j)|→+∞|z(\bm{x}_{j})|\to+\infty (corresponding to Case (b) and Case (c) of Region 𝐈𝟑\bf{I}_{3}), we choose

(6.85) {λj=𝔰j−1τj=(𝔰j−1)ν+k+α⋅e−Tj⋅Tj−μκj=(𝔰j−1)−ν⋅eTj⋅Tjμ.\displaystyle\begin{cases}\lambda_{j}=\mathfrak{s}_{j}^{-1}\\ \tau_{j}=(\mathfrak{s}_{j}^{-1})^{\nu+k+\alpha}\cdot e^{-T_{j}}\cdot T_{j}^{-\mu}\\ \kappa_{j}=(\mathfrak{s}_{j}^{-1})^{-\nu}\cdot e^{T_{j}}\cdot T_{j}^{\mu}.\end{cases}

In this case, we need to rescale the zz-coordinate in the meanwhile so that the exponential term shows up in the rescaling factors.

Region 𝐈𝟏\bf{I}_{1} (The deepest bubble):

In this case, we consider that the reference points 𝒙j\bm{x}_{j} are in Region 𝐈𝟏\bf{I}_{1}. According to the discussions in Section 4.3, for any 0<γ<10<\gamma<1, (ℳj,g~j,𝒙j)(\mathcal{M}_{j},\tilde{g}_{j},\bm{x}_{j}) converges to the following Riemann product in the C2,γC^{2,\gamma}-topology,

(6.86) (ℳj,g~j,𝒙j)→C2,γ(ℂT​N2×ℂn−2,g~∞,𝒙∞),(\mathcal{M}_{j},\tilde{g}_{j},\bm{x}_{j})\xrightarrow{C^{2,\gamma}}(\mathbb{C}_{TN}^{2}\times\mathbb{C}^{n-2},\tilde{g}_{\infty},\bm{x}_{\infty}),

where g~∞≡gT​N⊕gℂn−2\tilde{g}_{\infty}\equiv g_{TN}\oplus g_{\mathbb{C}^{n-2}} is the product metric of the Taub-NUT metric gT​Ng_{TN} and the Euclidean metric gℂn−2g_{\mathbb{C}^{n-2}}. Moreover, the rescaled weight function will converge to

(6.87) ρ~∞,δ,ν,μ(k+α)​(𝒙)={1,𝒙∈T1​(Σ0),(dg~∞​(𝒙,Σ0))ν+k+α,𝒙∈(ℂT​N2×ℂn−2)∖T1​(Σ0),\displaystyle\tilde{\rho}_{\infty,\delta,\nu,\mu}^{(k+\alpha)}(\bm{x})=\begin{cases}1,&\bm{x}\in T_{1}(\Sigma_{0}),\\ (d_{\tilde{g}_{\infty}}(\bm{x},\Sigma_{0}))^{\nu+k+\alpha},&\bm{x}\in(\mathbb{C}_{TN}^{2}\times\mathbb{C}^{n-2})\setminus T_{1}(\Sigma_{0}),\end{cases}

where Σ0≡{p∞}×ℂn−2⊂ℂT​N2×ℂn−2\Sigma_{0}\equiv\{p_{\infty}\}\times\mathbb{C}^{n-2}\subset\mathbb{C}_{TN}^{2}\times\mathbb{C}^{n-2} for some p∞∈ℂT​N2p_{\infty}\in\mathbb{C}_{TN}^{2}, is the Gromov-Hausdorff limit of the lifted divisor 𝒫≡π−1​(P)⊂ℳj\mathcal{P}\equiv\pi^{-1}(P)\subset\mathcal{M}_{j} with respect to the rescaled metrics g~j\tilde{g}_{j} such that and

(6.88) T1​(Σ0)≡{𝒙∈ℂT​N2×ℂn−2|dg~∞​(𝒙,Σ0)≤1}.T_{1}(\Sigma_{0})\equiv\{\bm{x}\in\mathbb{C}_{TN}^{2}\times\mathbb{C}^{n-2}|d_{\tilde{g}_{\infty}}(\bm{x},\Sigma_{0})\leq 1\}.

It is straightforward that, the rescaled functions v~j\tilde{v}_{j} converge to v~∞\tilde{v}_{\infty} in the C2,α′C^{2,\alpha^{\prime}}-topology for each 0<α′<α0<\alpha^{\prime}<\alpha such that the following properties hold,

  1. (1)

    ‖∇v~∞‖Cδ,ν+1,μ0​(ℂT​N2×ℂn−2,g~∞)+‖∇2v~∞‖Cδ,ν+2,μ0​(ℂT​N2×ℂn−2,g~∞)=1\|\nabla\tilde{v}_{\infty}\|_{C_{\delta,\nu+1,\mu}^{0}(\mathbb{C}_{TN}^{2}\times\mathbb{C}^{n-2},\tilde{g}_{\infty})}+\|\nabla^{2}\tilde{v}_{\infty}\|_{C_{\delta,\nu+2,\mu}^{0}(\mathbb{C}_{TN}^{2}\times\mathbb{C}^{n-2},\tilde{g}_{\infty})}=1,

  2. (2)

    v~∞​(𝒙∞)=0\tilde{v}_{\infty}(\bm{x}_{\infty})=0,

  3. (3)

    Δg~∞​v~∞≡0\Delta_{\tilde{g}_{\infty}}\tilde{v}_{\infty}\equiv 0 on ℂT​N2×ℂn−2\mathbb{C}_{TN}^{2}\times\mathbb{C}^{n-2}.

We will prove that v~∞≡0\tilde{v}_{\infty}\equiv 0 on ℂT​N2×ℂn−2\mathbb{C}_{TN}^{2}\times\mathbb{C}^{n-2}.

To start with, we will show that v~∞\tilde{v}_{\infty} is constant on the Euclidean factor ℂn−2\mathbb{C}^{n-2}. Indeed, we write 𝒙≡(𝒙′,𝒙′′)∈ℂT​N2×ℂn−2\bm{x}\equiv(\bm{x}^{\prime},\bm{x}^{\prime\prime})\in\mathbb{C}_{TN}^{2}\times\mathbb{C}^{n-2}, so it suffices to prove that for every 1≤k≤2​n−41\leq k\leq 2n-4, we have

(6.89) |∇kv~∞|≡0​on​ℂn−2,|\nabla_{k}\tilde{v}_{\infty}|\equiv 0\ \text{on}\ \mathbb{C}^{n-2},

where the partial derivative ∇kv~∞​(𝒙)≡∂v~∞∂xk′′​(𝒙′,𝒙′′)\nabla_{k}\tilde{v}_{\infty}(\bm{x})\equiv\frac{\partial\tilde{v}_{\infty}}{\partial x_{k}^{\prime\prime}}(\bm{x}^{\prime},\bm{x}^{\prime\prime}) is taken in the directions of ℂn−2\mathbb{C}^{n-2}. Now for every 1≤k≤2​n−41\leq k\leq 2n-4,

(6.90) Δg~∞​(∇kv~∞)=ΔℂT​N2​(∇kv~∞)+Δℂn−2​(∇kv~∞).\Delta_{\tilde{g}_{\infty}}(\nabla_{k}\tilde{v}_{\infty})=\Delta_{\mathbb{C}_{TN}^{2}}(\nabla_{k}\tilde{v}_{\infty})+\Delta_{\mathbb{C}^{n-2}}(\nabla_{k}\tilde{v}_{\infty}).

Notice that g~∞=gT​N⊕gℂn−2\tilde{g}_{\infty}=g_{TN}\oplus g_{\mathbb{C}^{n-2}} is a product metric and ∇k\nabla_{k} in effect acts on the Euclidean factor ℂn−2\mathbb{C}^{n-2}, so ∇k\nabla_{k} commutes with both ΔℂT​N2\Delta_{\mathbb{C}_{TN}^{2}} and Δℂn−2\Delta_{\mathbb{C}^{n-2}}. Therefore,

(6.91) Δg~∞​(∇kv~∞)=∇k(ΔℂT​N2​v~∞+Δℂn−2​v~∞)=0.\displaystyle\Delta_{\tilde{g}_{\infty}}(\nabla_{k}\tilde{v}_{\infty})=\nabla_{k}(\Delta_{\mathbb{C}_{TN}^{2}}\tilde{v}_{\infty}+\Delta_{\mathbb{C}^{n-2}}\tilde{v}_{\infty})=0.

The weighted bound implies the estimates

(6.92) {|∇kv~∞​(𝒙)|≤1,dg~∞​(𝒙,Σ0)≤1,|∇kv~∞​(𝒙)|≤dg~∞​(𝒙,Σ0)−(ν+1),dg~∞​(𝒙,Σ0)≥1.\displaystyle\begin{cases}|\nabla_{k}\tilde{v}_{\infty}(\bm{x})|\leq 1,&d_{\tilde{g}_{\infty}}(\bm{x},\Sigma_{0})\leq 1,\\ |\nabla_{k}\tilde{v}_{\infty}(\bm{x})|\leq d_{\tilde{g}_{\infty}}(\bm{x},\Sigma_{0})^{-(\nu+1)},&d_{\tilde{g}_{\infty}}(\bm{x},\Sigma_{0})\geq 1.\end{cases}

Since we have assumed ν∈(−1,0)\nu\in(-1,0), so it is straightforward

(6.93) −(ν+1)∈(−1,0).-(\nu+1)\in(-1,0).

The above implies that |∇kv~∞|≤1|\nabla_{k}\tilde{v}_{\infty}|\leq 1 on ℂT​N2×ℂn−2\mathbb{C}_{TN}^{2}\times\mathbb{C}^{n-2}. Applying Cheng-Yau’s gradient estimate to the harmonic function ∇kv~∞\nabla_{k}\tilde{v}_{\infty} on the Ricci-flat manifold ℂT​N2×ℂn−2\mathbb{C}_{TN}^{2}\times\mathbb{C}^{n-2}, we conclude that ∇kv~∞\nabla_{k}\tilde{v}_{\infty} is constant on ℂT​N2×ℂn−2\mathbb{C}_{TN}^{2}\times\mathbb{C}^{n-2}. By (6.92), ∇kv~∞≡0\nabla_{k}\tilde{v}_{\infty}\equiv 0 for every 1≤k≤2​n−41\leq k\leq 2n-4. Therefore, v~∞\tilde{v}_{\infty} is constant on the Euclidean factor ℂn−2\mathbb{C}^{n-2}.

By the above argument, the limiting function v~∞\tilde{v}_{\infty} can be viewed as a harmonic function on the Ricci-flat Taub-NUT space (ℂT​N2,gT​N)(\mathbb{C}_{TN}^{2},g_{TN}). Now applying Bochner’s formula,

(6.94) 12​ΔgT​N​|∇gT​Nv~∞|2=|∇gT​N2v~∞|2≥0.\frac{1}{2}\Delta_{g_{TN}}|\nabla_{g_{TN}}\tilde{v}_{\infty}|^{2}=|\nabla_{g_{TN}}^{2}\tilde{v}_{\infty}|^{2}\geq 0.

Since v~∞\tilde{v}_{\infty} satisfies the weighted bound

(6.95) ‖∇gT​Nv~∞‖Cδ,ν+1,μ0​(ℂT​N2)+‖∇gT​N2v~∞‖Cδ,ν+2,μ0​(ℂT​N2)=1,\|\nabla_{g_{TN}}\tilde{v}_{\infty}\|_{C_{\delta,\nu+1,\mu}^{0}(\mathbb{C}_{TN}^{2})}+\|\nabla_{g_{TN}}^{2}\tilde{v}_{\infty}\|_{C_{\delta,\nu+2,\mu}^{0}(\mathbb{C}_{TN}^{2})}=1,

so we have for any 𝒙∈ℂT​N2∖B1​(𝒙∞)\bm{x}\in\mathbb{C}_{TN}^{2}\setminus B_{1}(\bm{x}_{\infty}),

(6.96) |∇gT​Nv~∞​(𝒙)|≤dgT​N​(𝒙,𝒙∞)−(ν+1).|\nabla_{g_{TN}}\tilde{v}_{\infty}(\bm{x})|\leq d_{g_{TN}}(\bm{x},\bm{x}_{\infty})^{-(\nu+1)}.

By assumption ν∈(−1,0)\nu\in(-1,0), then |∇gT​Nv~∞|≡0|\nabla_{g_{TN}}\tilde{v}_{\infty}|\equiv 0 on ℂT​N2\mathbb{C}_{TN}^{2} and hence v~∞\tilde{v}_{\infty} is constant on ℂT​N2\mathbb{C}_{TN}^{2}. Notice that v~∞​(𝒙∞)=0\tilde{v}_{\infty}(\bm{x}_{\infty})=0, so we conclude that v~∞​(𝒙∞)≡0\tilde{v}_{\infty}(\bm{x}_{\infty})\equiv 0.

Region 𝐈𝟐\bf{I}_{2} (bubble transformations):

Now we separate the proof in 33 cases:

  1. (a)

    There is some σ0>0\sigma_{0}>0 such that

    (6.97) λj⋅Tj−12≥σ0.\lambda_{j}\cdot T_{j}^{-\frac{1}{2}}\geq\sigma_{0}.
  2. (b)

    Assume that rjr_{j} satisfies the following condition holds,

    (6.98) λj⋅Tj−12→0,λj⋅Tj12→∞.\lambda_{j}\cdot T_{j}^{-\frac{1}{2}}\to 0,\ \lambda_{j}\cdot T_{j}^{\frac{1}{2}}\to\infty.
  3. (c)

    Assume that there is some C0>0C_{0}>0 such that λj⋅Tj12≤C0\lambda_{j}\cdot T_{j}^{\frac{1}{2}}\leq C_{0}.

Case (a):

In this case, the rescaled limit is the Riemann product ℂT​N,σ2×ℂn−2\mathbb{C}_{TN,\sigma}^{2}\times\mathbb{C}^{n-2}, where ℂT​N,σ2\mathbb{C}_{TN,\sigma}^{2} is the Taub-NUT space and the length of the circle fiber at infinity equals σ∈[σ0,1]\sigma\in[\sigma_{0},1]. The remainder of the proof is the same as that in Region 𝐈𝟏\bf{I}_{1}, so we omit it.

Case (b):

In this case, the rescaled spaces (ℳj,g~j,𝒙j)(\mathcal{M}_{j},\tilde{g}_{j},\bm{x}_{j}) converge to the product Euclidean space (ℝ3×ℂn−2,g0,𝒙∞)(\mathbb{R}^{3}\times\mathbb{C}^{n-2},g_{0},\bm{x}_{\infty}) in the pointed Gromov-Hausdorff topology, i.e.,

(6.99) (ℳj,g~j,𝒙j)→G​H(ℝ3×ℂn−2,g0,𝒙∞),(\mathcal{M}_{j},\tilde{g}_{j},\bm{x}_{j})\xrightarrow{GH}(\mathbb{R}^{3}\times\mathbb{C}^{n-2},g_{0},\bm{x}_{\infty}),

where the metric g0g_{0} is the standard Euclidean metric on ℝ3×ℂn−2\mathbb{R}^{3}\times\mathbb{C}^{n-2}. In this rescaled limit, the limiting reference point 𝒙∞\bm{x}_{\infty} satisfies dg0​(𝒙∞,Σ03)=1d_{g_{0}}(\bm{x}_{\infty},\Sigma_{0^{3}})=1 and Σ03≡{03}×ℂn−2⊂ℝ3×ℂn−2\Sigma_{0^{3}}\equiv\{0^{3}\}\times\mathbb{C}^{n-2}\subset\mathbb{R}^{3}\times\mathbb{C}^{n-2} is the singular slice. Moreover, the convergence keeps curvatures uniformly bounded away from the singular slice Σ03\Sigma_{0^{3}}. By passing to the local universal covers, in fact one can show that, away from Σ03⊂ℝ3×ℂn−2\Sigma_{0^{3}}\subset\mathbb{R}^{3}\times\mathbb{C}^{n-2}, the rescaled contradicting functions v~j\tilde{v}_{j} converge to v~∞\tilde{v}_{\infty} in the C2,α′C^{2,\alpha^{\prime}}-topology for each 0<α′<α<10<\alpha^{\prime}<\alpha<1, such that the following properties hold,

  1. (1)

    ‖∇v~∞‖Cδ,ν+1,μ0​(ℝ3×ℂn−2)+‖∇2v~∞‖Cδ,ν+2,μ0​(ℝ3×ℂn−2)=1\|\nabla\tilde{v}_{\infty}\|_{C_{\delta,\nu+1,\mu}^{0}(\mathbb{R}^{3}\times\mathbb{C}^{n-2})}+\|\nabla^{2}\tilde{v}_{\infty}\|_{C_{\delta,\nu+2,\mu}^{0}(\mathbb{R}^{3}\times\mathbb{C}^{n-2})}=1,

  2. (2)

    v~∞​(𝒙∞)=0\tilde{v}_{\infty}(\bm{x}_{\infty})=0,

  3. (3)

    Δg~∞​v~∞≡0\Delta_{\tilde{g}_{\infty}}\tilde{v}_{\infty}\equiv 0 in (ℝ3×ℂn−2)∖Σ03(\mathbb{R}^{3}\times\mathbb{C}^{n-2})\setminus\Sigma_{0^{3}},

where the limiting weight function is

(6.100) ρ∞,δ,ν,μ(k+α)​(𝒙)=(dg0​(𝒙,Σ03))ν+k+α,𝒙∈ℝ3×ℂn−2.\rho_{\infty,\delta,\nu,\mu}^{(k+\alpha)}(\bm{x})=(d_{g_{0}}(\bm{x},\Sigma_{0^{3}}))^{\nu+k+\alpha},\ \bm{x}\in\mathbb{R}^{3}\times\mathbb{C}^{n-2}.

Our goal is to show that v~∞≡0\tilde{v}_{\infty}\equiv 0 on ℝ3×ℂn−2\mathbb{R}^{3}\times\mathbb{C}^{n-2}, which consists of the following ingredients:

First, we will prove that v~∞\tilde{v}_{\infty} in fact globally harmonic in ℝ3×ℂn−2\mathbb{R}^{3}\times\mathbb{C}^{n-2}. To show the singular slice Σ03\Sigma_{0^{3}} is removable, for each q∈Σ03q\in\Sigma_{0^{3}}, we take a unit ball B1​(q)⊂Σ03B_{1}(q)\subset\Sigma_{0^{3}}, and for any r∈(0,1)r\in(0,1), we choose the tubular neighborhood Tr​(B1​(q))⊂ℝ3×ℂn−2T_{r}(B_{1}(q))\subset\mathbb{R}^{3}\times\mathbb{C}^{n-2}. Notice that ∇v~∞\nabla\tilde{v}_{\infty} satisfies the uniform estimate

(6.101) ‖∇v~∞‖Cδ,ν+1,μ0​(ℝ3×ℂn−2)≤1,\|\nabla\tilde{v}_{\infty}\|_{C_{\delta,\nu+1,\mu}^{0}(\mathbb{R}^{3}\times\mathbb{C}^{n-2})}\leq 1,

integrating the above weighted bound, then for any 𝒙∈Tr​(B1​(q))∖B1​(q)\bm{x}\in T_{r}(B_{1}(q))\setminus B_{1}(q),

(6.102) |v~∞​(𝒙)|≤C⋅d​(𝒙,B1​(q))−(ν).|\tilde{v}_{\infty}(\bm{x})|\leq C\cdot d(\bm{x},B_{1}(q))^{-(\nu)}.

By Lemma 6.5, B1​(q)B_{1}(q) is a removable singular set in Tr​(B1​(q))T_{r}(B_{1}(q)) and hence v~∞\tilde{v}_{\infty} is harmonic in Tr​(B1​(q))T_{r}(B_{1}(q)).

Next, we will show that v~∞\tilde{v}_{\infty} is constant in ℂn−2\mathbb{C}^{n-2}. It is straightforward that for each 1≤k≤2​n−41\leq k\leq 2n-4, the partial derivative ∇kv~∞≡∂∂xk′′​v~∞​(𝒙′,𝒙′′)\nabla_{k}\tilde{v}_{\infty}\equiv\frac{\partial}{\partial x_{k}^{\prime\prime}}\tilde{v}_{\infty}(\bm{x}^{\prime},\bm{x}^{\prime\prime}) satisfies

(6.103) Δg0​(∇kv~∞)=0​in​ℝ3×ℂn−2.\Delta_{g_{0}}(\nabla_{k}\tilde{v}_{\infty})=0\ \text{in}\ \mathbb{R}^{3}\times\mathbb{C}^{n-2}.

The weighted condition implies that ∇kv~∞\nabla_{k}\tilde{v}_{\infty} satisfies the uniform estimate,

(6.104) |∇kv~∞|≤d​(𝒙,Σ03)−(ν+1),∀𝒙∈ℝ3×ℂn−2.|\nabla_{k}\tilde{v}_{\infty}|\leq d(\bm{x},\Sigma_{0^{3}})^{-(\nu+1)},\forall\bm{x}\in\mathbb{R}^{3}\times\mathbb{C}^{n-2}.

Since we have assumed ν∈(−1,0)\nu\in(-1,0), Lemma 6.6 implies that |∇kv~∞|≡0|\nabla_{k}\tilde{v}_{\infty}|\equiv 0 on ℝ3×ℂn−2\mathbb{R}^{3}\times\mathbb{C}^{n-2} and hence v~∞\tilde{v}_{\infty} is constant in ℂn−2\mathbb{C}^{n-2}. Therefore, v~∞\tilde{v}_{\infty} can be viewed as a harmonic function in the Euclidean space (ℝ3,gℝ3)(\mathbb{R}^{3},g_{\mathbb{R}^{3}}). By assumption, v~∞\tilde{v}_{\infty} satisfies

(6.105) |v~∞​(𝒙)|≤dgℝ3​(𝒙,03)−ν.|\tilde{v}_{\infty}(\bm{x})|\leq d_{g_{\mathbb{R}^{3}}}(\bm{x},0^{3})^{-\nu}.

Since ν∈(−1,0)\nu\in(-1,0), applying the standard Liouville theorem for sublinear growth harmonic functions on a Euclidean space, we conclude that v~∞\tilde{v}_{\infty} is a constant. The last step is to use the renormalization v~∞​(𝒙∞)=0\tilde{v}_{\infty}(\bm{x}_{\infty})=0, then v~∞≡0\tilde{v}_{\infty}\equiv 0.

Case (c):

The rescaled limit is the cylinder (Q,gc)≡(D×ℝ,gD⊕d​z2)(Q,g_{c})\equiv(D\times\mathbb{R},g_{D}\oplus dz^{2}), where (D,gD)(D,g_{D}) is a closed Calabi-Yau manifold. The limiting solutions v~∞\tilde{v}_{\infty} satisfies

  1. (1)

    ‖∇v~∞‖Cδ,ν+1,μ0​(Q)+‖∇2v~∞‖Cδ,ν+2,μ0​(Q)=1\|\nabla\tilde{v}_{\infty}\|_{C_{\delta,\nu+1,\mu}^{0}(Q)}+\|\nabla^{2}\tilde{v}_{\infty}\|_{C_{\delta,\nu+2,\mu}^{0}(Q)}=1,

  2. (2)

    v~∞​(𝒙∞)=0\tilde{v}_{\infty}(\bm{x}_{\infty})=0,

  3. (3)

    Δg~∞​v~∞≡0\Delta_{\tilde{g}_{\infty}}\tilde{v}_{\infty}\equiv 0 in Q∖PQ\setminus P,

where the limiting weight function is

(6.106) ρ~∞,δ,ν,μ(k+α)​(𝒙)={eδ⋅z⁡(𝒙)⋅𝔯​(𝒙)ν+k+α,z⁡(𝒙)>0e−δ⋅z(𝒙)⋅𝔯(𝒙)ν+k+α,z⁡(𝒙)≤0.\displaystyle\tilde{\rho}_{\infty,\delta,\nu,\mu}^{(k+\alpha)}(\bm{x})=\begin{cases}e^{\delta\cdot z(\bm{x})}\cdot\mathfrak{r}(\bm{x})^{\nu+k+\alpha},&z(\bm{x})>0\\ e^{-\delta\cdot z(\bm{x})}\cdot\mathfrak{r}(\bm{x})^{\nu+k+\alpha},&z(\bm{x})\leq 0.\end{cases}

Similar to Case (b), first we need to extend the limiting function v~∞\tilde{v}_{\infty} across the singular set PP. Integrating ∇v~∞\nabla\tilde{v}_{\infty} around PP, we have that v~∞\tilde{v}_{\infty} satisfies the growth estimate

(6.107) |v~∞​(𝒙)|≤C⋅dgc​(𝒙,P)−ν.|\tilde{v}_{\infty}(\bm{x})|\leq C\cdot d_{g_{c}}(\bm{x},P)^{-\nu}.

Since we have assumed ν∈(−1,0)\nu\in(-1,0), so Lemma 6.5 implies that the singular set PP is removable. Now we have obtained that v~∞\tilde{v}_{\infty} is harmonic on QQ and satisfies

(6.108) |v~∞(𝒙)|≤Ce−δ⋅|z(𝒙)|,|\tilde{v}_{\infty}(\bm{x})|\leq Ce^{-\delta\cdot|z(\bm{x})|},

for |z⁡(𝒙)||z(\bm{x})| large. Therefore, v~∞≡0\tilde{v}_{\infty}\equiv 0 on QQ which completes the proof of Case (c).

Region 𝐈𝟑\bf{I}_{3} (the cylindrical bubble and the boundary behavior):

In this region, the rescaling factors of the metrics gjg_{j} are chosen such that the rescaled Gromov-Hausdorff limit is the cylinder Q≡D×ℝQ\equiv D\times\mathbb{R}. Let ζj≡z⁡(𝒙j)\zeta_{j}\equiv z(\bm{x}_{j}), then there are two different cases to analyze which depends on if the convergence keeps curvatures uniformly bounded.

  1. (a)

    Assume that there is some ζ0>0\zeta_{0}>0 such that |zj|≤ζ0|z_{j}|\leq\zeta_{0}.

  2. (b)

    Assume that zjz_{j} satisfies

    (6.109) |ζj|→∞,Tjn−2nLTj​(zj)→0.|\zeta_{j}|\to\infty,\ \frac{T_{j}^{\frac{n-2}{n}}}{L_{T_{j}}(z_{j})}\to 0.
  3. (c)

    Assume that zjz_{j} satisfies

    (6.110) c0≤Tjn−2nLTj​(zj)≤1.\displaystyle c_{0}\leq\frac{T_{j}^{\frac{n-2}{n}}}{L_{T_{j}}(z_{j})}\leq 1.

Case (a):

So the rescaled spaces (ℳj,g~j,𝒙j)(\mathcal{M}_{j},\tilde{g}_{j},\bm{x}_{j}) converge to the cylinder (Q,gc)=(D2​n×ℝ,gD2​n⊕d​z2)(Q,g_{c})=(D^{2n}\times\mathbb{R},g_{D^{2n}}\oplus dz^{2}) and the sequence has uniformly bounded geometry away from Q∖𝒫Q\setminus\mathcal{P}. Moreover, the weight function in the rescaled limit space is

(6.111) ρ~∞,δ,ν,μ(k+α)​(𝒙)={eδ⋅z⁡(𝒙)⋅𝔯​(𝒙)ν+k+α,z⁡(𝒙)>0e−δ⋅z(𝒙)⋅𝔯(𝒙)ν+k+α,z⁡(𝒙)≤0.\displaystyle\tilde{\rho}_{\infty,\delta,\nu,\mu}^{(k+\alpha)}(\bm{x})=\begin{cases}e^{\delta\cdot z(\bm{x})}\cdot\mathfrak{r}(\bm{x})^{\nu+k+\alpha},&z(\bm{x})>0\\ e^{-\delta\cdot z(\bm{x})}\cdot\mathfrak{r}(\bm{x})^{\nu+k+\alpha},&z(\bm{x})\leq 0.\end{cases}

The rest of the proof is the same as Case (c) in Region II.

Case (b) in Region 𝐈𝟑\bf{I}_{3}

In this case, the reference point 𝒙j\bm{x}_{j} satisfies

(6.112) |z⁡(𝒙j)|→∞,Tjn−2nLTj​(zj)→0.|z(\bm{x}_{j})|\to\infty,\quad\frac{T_{j}^{\frac{n-2}{n}}}{L_{T_{j}}(z_{j})}\to 0.

In addition, we also need to perform the coordinate change centered at the reference point 𝒙j\bm{x}_{j},

(6.113) z⁡(𝒙)=zj+(TjLTj​(zj))n−22​w​(𝒙).z(\bm{x})=z_{j}+\Big(\frac{T_{j}}{L_{T_{j}}(z_{j})}\Big)^{\frac{n-2}{2}}w(\bm{x}).

In the following, we only consider the case zj≪0z_{j}\ll 0. It is shown in Section 4.3 that the rescaled limit is isometric to a cylinder Q=D×ℝQ=D\times\mathbb{R} with a product metric

(6.114) gQ=gD+d​w2.g_{Q}=g_{D}+dw^{2}.

Moreover, as Tj→+∞T_{j}\to+\infty, the rescaled weight function limits to

(6.115) ρ~∞,δ,ν,μ(k+α)(𝒙)=e−δ⋅n⋅k−2⋅w(𝒙).\tilde{\rho}_{\infty,\delta,\nu,\mu}^{(k+\alpha)}(\bm{x})=e^{-\frac{\delta\cdot n\cdot k_{-}}{2}\cdot w(\bm{x})}.

Now the growth condition implies that the limiting function v~∞\tilde{v}_{\infty} satisfies

(6.116) {ΔQ​v~∞​(𝒙)=0,∀𝒙∈Q,|v~∞​(𝒙)|≤eδ⋅n⋅k−2⋅w⁡(𝒙),w∈ℝ.\displaystyle\begin{cases}\Delta_{Q}\tilde{v}_{\infty}(\bm{x})=0,&\forall\bm{x}\in Q,\\ |\tilde{v}_{\infty}(\bm{x})|\leq e^{\frac{\delta\cdot n\cdot k_{-}}{2}\cdot w(\bm{x})},&w\in\mathbb{R}.\end{cases}

By the choice of the parameter δ\delta in (6.12),

(6.117) δ⋅n⋅k−2<λD2.\frac{\delta\cdot n\cdot k_{-}}{2}<\frac{\sqrt{\lambda_{D}}}{2}.

Applying Lemma 6.7, for every 𝒙∈Q\bm{x}\in Q,

(6.118) v~∞​(𝒙)=0.\tilde{v}_{\infty}(\bm{x})=0.

So the proof of Case (b) is done.

Case (c) in Region 𝐈𝟑\bf{I}_{3}

In this case, the reference point 𝒙j\bm{x}_{j} is close to the boundary such that Neumann boundary condition plays a crucial role. Precisely, the scale condition is given by the following: there is some c0>0c_{0}>0 such that

(6.119) c0≤Tjn−2nLTj​(zj)≤1.\displaystyle c_{0}\leq\frac{T_{j}^{\frac{n-2}{n}}}{L_{T_{j}}(z_{j})}\leq 1.

We can assume that zj≪0z_{j}\ll 0 and passing to a subsequence, there is some constant 𝔠0∈[c0,1]\mathfrak{c}_{0}\in[c_{0},1] such that

(6.120) Tjn−2nLTj​(zj)→𝔠0.\frac{T_{j}^{\frac{n-2}{n}}}{L_{T_{j}}(z_{j})}\to\mathfrak{c}_{0}.

For the convenience of the computations, we will perform the coordinate change centered at the boundary slice, that is,

(6.121) z⁡(𝒙)=T−+(TjLTj​(zj))n−22​w​(𝒙).z(\bm{x})=T_{-}+\Big(\frac{T_{j}}{L_{T_{j}}(z_{j})}\Big)^{\frac{n-2}{2}}w(\bm{x}).

We have computed in Section 4.3 that the limit of the rescaled spaces (ℳT,g~j,𝒙j)(\mathcal{M}_{T},\tilde{g}_{j},\bm{x}_{j}) is the Calabi model space (𝒞−n,g𝒞−n,𝒙∞)(\mathcal{C}^{n}_{-},g_{\mathcal{C}^{n}_{-}},\bm{x}_{\infty}). Moreover, the limiting weight function is

(6.122) ρ~∞,δ,ν,μ(k+α)(𝒙)=e−δ⋅𝔠0n2⋅Pn2(w)⋅(Pn2(w))ν+k+α2,\tilde{\rho}_{\infty,\delta,\nu,\mu}^{(k+\alpha)}(\bm{x})=e^{-\delta\cdot\mathfrak{c}_{0}^{\frac{n}{2}}\cdot P_{{\frac{n}{2}}}(w)}\cdot(P_{\frac{n}{2}}(w))^{\frac{\nu+k+\alpha}{2}},

where

(6.123) Pn2​(w)≡(1+k−​𝔠0n2​w)n2.P_{{\frac{n}{2}}}(w)\equiv(1+k_{-}\mathfrak{c}_{0}^{\frac{n}{2}}w)^{\frac{n}{2}}.

Since 𝔠0∈[c0,1]\mathfrak{c}_{0}\in[c_{0},1], so the limiting function v~∞\tilde{v}_{\infty} satisfies

(6.124) {Δg𝒞−n​v~∞=0,𝒙∈𝒞−n,|v~∞​(𝒙)|≤eδ⋅(k−)n2⋅wn2⋅Pn2​(w)−ν2,w⁡(𝒙)≫1,∂v~∞∂w=0,w⁡(𝒙)=w0.\displaystyle\begin{cases}\Delta_{g_{\mathcal{C}_{-}^{n}}}\tilde{v}_{\infty}=0,&\bm{x}\in\mathcal{C}_{-}^{n},\\ |\tilde{v}_{\infty}(\bm{x})|\leq e^{\delta\cdot(k_{-})^{\frac{n}{2}}\cdot w^{\frac{n}{2}}}\cdot P_{\frac{n}{2}}(w)^{-\frac{\nu}{2}},&w(\bm{x})\gg 1,\\ \frac{\partial\tilde{v}_{\infty}}{\partial w}=0,&w(\bm{x})=w_{0}.\end{cases}

In the following, we will prove that v~∞\tilde{v}_{\infty} is vanishing everywhere in the Calabi space 𝒞−n\mathcal{C}_{-}^{n} such that the contradiction arises.

To see this, recall that the incomplete Calabi model space (𝒞−n,g𝒞−)(\mathcal{C}_{-}^{n},g_{\mathcal{C}_{-}}) is diffeomorphic to the topological product [w0,+∞)×Y2​n−1[w_{0},+\infty)\times Y^{2n-1}, where Y2​n−1≡{ρ=ρ0}Y^{2n-1}\equiv\{\rho=\rho_{0}\} is with respect to the boundary slice in the Calabi model (see Section 5 for detailed discussions on it). The above structure leads to a natural coordinate representation 𝒙=(w,𝒚)∈𝒞−n\bm{x}=(w,\bm{y})\in\mathcal{C}_{-}^{n} for each point in the Calabi model space such that the boundary of 𝒞−n\mathcal{C}_{-}^{n} is given by {w=w0}\{w=w_{0}\}, where the coordinate ww is the natural moment map coordinate.

Denote by ΣY2​n−1={Λk}k=0∞\Sigma_{Y^{2n-1}}=\{\Lambda_{k}\}_{k=0}^{\infty} the spectrum of the fiber Y2​n−1Y^{2n-1} with respect to the induced Riemannian metric. Let {φk}k=0∞\{\varphi_{k}\}_{k=0}^{\infty} be the orthonormal basis with respect to the L2L^{2}-inner product on Y2​n−1Y^{2n-1}, such that for each k∈ℕk\in\mathbb{N},

(6.125) −ΔY2​n−1​φk=λk⋅φk.\displaystyle-\Delta_{Y^{2n-1}}\varphi_{k}=\lambda_{k}\cdot\varphi_{k}.

If δ>0\delta>0 is chosen sufficiently small, applying Proposition 5.14, then v~∞\tilde{v}_{\infty} has the expansion

(6.126) v~∞​(w,𝒚)=κ⋅w+ℓ+∑k=1∞ck⋅𝒟k​(w)⋅φk​(𝒚),\tilde{v}_{\infty}(w,\bm{y})=\kappa\cdot w+\ell+\sum\limits_{k=1}^{\infty}c_{k}\cdot\mathcal{D}_{k}(w)\cdot\varphi_{k}(\bm{y}),

where the function 𝒟k​(w)\mathcal{D}_{k}(w) has some definite exponential decaying rate (see Lemma 5.4 and Lemma 5.7 for the accurate rates).

Now we apply the Neumann condition to show that κ=0\kappa=0 and ck=0c_{k}=0 for all k∈ℕk\in\mathbb{N}. In fact,

(6.127) ∂v~∞​(w,𝒚)∂w=κ+∑k=1∞ck⋅𝒟k′​(w)⋅φk​(𝒚).\frac{\partial\tilde{v}_{\infty}(w,\bm{y})}{\partial w}=\kappa+\sum\limits_{k=1}^{\infty}c_{k}\cdot\mathcal{D}_{k}^{\prime}(w)\cdot\varphi_{k}(\bm{y}).

Integrating (6.127) over the boundary slice {w=w0}\{w=w_{0}\},

(6.128) κ⋅Volg𝒞−n⁡(Y2​n−1)=∫Y2​n−1∂v~∞​(w,𝒚)∂w|w=w0=0,\kappa\cdot\Vol_{g_{\mathcal{C}_{-}^{n}}}(Y^{2n-1})=\int_{Y^{2n-1}}\frac{\partial\tilde{v}_{\infty}(w,\bm{y})}{\partial w}\Big|_{w=w_{0}}=0,

which implies

(6.129) κ=0.\kappa=0.

Next, for each fixed k∈ℤ+k\in\mathbb{Z}_{+}, multiplying φk\varphi_{k} on the both sides of (6.127) and integrating over Y2​n−1Y^{2n-1},

(6.130) ck⋅𝒟k′​(w)=∫Y2​n−1φk​(𝒚)⋅∂v~∞​(w,𝒚)∂w|w=w0=0.c_{k}\cdot\mathcal{D}_{k}^{\prime}(w)=\int_{Y^{2n-1}}\varphi_{k}(\bm{y})\cdot\frac{\partial\tilde{v}_{\infty}(w,\bm{y})}{\partial w}\Big|_{w=w_{0}}=0.

The conclusion ck=0c_{k}=0 follows from the claim

(6.131) 𝒟k′​(w)<0,∀w≥w0.\mathcal{D}_{k}^{\prime}(w)<0,\forall\ w\geq w_{0}.

Now we just need to prove the claim. In fact, since 𝒟k′​(w)\mathcal{D}_{k}^{\prime}(w) satisfies the equation

(6.132) 𝒟k′′​(w)=wn−2​(jk2​n24⋅wn+n​λk)​𝒟k​(w)\mathcal{D}_{k}^{\prime\prime}(w)=w^{n-2}(\frac{j_{k}^{2}n^{2}}{4}\cdot w^{n}+n\lambda_{k})\mathcal{D}_{k}(w)

and hence

(6.133) 𝒟k′′​(w)>0.\mathcal{D}_{k}^{\prime\prime}(w)>0.

Notice that 𝒟k​(w)\mathcal{D}_{k}(w) has an exponential decaying rate. This tells us that 𝒟k′′​(w)>0\mathcal{D}_{k}^{\prime\prime}(w)>0 and bounded as w→+∞w\to+\infty. Therefore, 𝒟k′​(w)\mathcal{D}_{k}^{\prime}(w) is increasing and uniformly continuous for w>0w>0. Since 𝒟k​(w)→0\mathcal{D}_{k}(w)\to 0, we conclude that limw→+∞𝒟k′​(w)=0\lim\limits_{w\to+\infty}\mathcal{D}_{k}^{\prime}(w)=0. Therefore, 𝒟k​(w)<0\mathcal{D}_{k}(w)<0 for any w≥w0w\geq w_{0}.

Lastly, v~∞\tilde{v}_{\infty} satisfies the renormalization condition v~∞​(𝒙∞)=0\tilde{v}_{\infty}(\bm{x}_{\infty})=0, immediately, ℓ=0\ell=0. Therefore,

(6.134) v~∞≡0on​𝒞−n.\tilde{v}_{\infty}\equiv 0\quad\text{on}\ \mathcal{C}_{-}^{n}.

The proof is done.

∎

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