Quantitative Unique Continuation for Second Order Elliptic Operators with Singular Coefficients

Abstract

We establish some quantitative strong unique continuation properties for solution of |Pu|≤ C₁|x|^− 1|∇u| + C₀|x|^− 2|u| where P is a second order elliptic operator. As in (Rev. Mat. Iberoam. 27: 475–491, 9), our result is of quantitative nature but requires weaker conditions on the coefficients of P.

1 Introduction

Let $Pu={\sum }_{kl} a_{kl}(x)\partial _{kl} u$ be a second order elliptic operator in $\mathbb {R}^{n}$, where A = (a_kl) is a real, symmetric elliptic matrix. Aronszajn et al. showed in [5] that if A is continuous in B₁ := {x : |x|≤ 1} and for some ε > 0,

$$ |\nabla A(x)| \le C|x|^{-1+\varepsilon} \qquad \text{for a.e. } x\in B_{1}, $$

(1)

then any u that satisfies

$$ |Pu|\leq C|x|^{-1+\varepsilon}|\nabla u|+C|x|^{-2+\varepsilon}|u| $$

(2)

and vanishes to infinite order at 0 must vanish identically in B₁. In other words, (2) has the strong unique continuation property. Earlier results with stronger smoothness assumptions on A were obtained in [4] and [8].

Alinhac and Baouendi [2] proved the same result for complex-valued $A\in C^{\infty }$, provided that A(0) is a multiple of a real positively definite matrix. The necessity of the assumption on A(0) was shown by an example of Alinhac [1]. Subsequently, Hörmander [7] weakened the smoothness assumption on A to (1). The example of Pliš [14] shows that unique continuation may not hold if A is only assumed to belong to the Hölder space C^α with 0 < α < 1. In [14], A is even Lipschitz outside a hypersurface.

In [10], Meshkov showed that instead of (2), it suffices to assume

$$ |Pu|\leq C_{1}|x|^{-1}|\nabla u|+C_{0} |x|^{-2}|u|, $$

(3)

provided C₁ is sufficiently small, depending on P. This result was later reproved in [15] using slightly different Carleman estimates. Meshkov [11] outlined a possible way to construct an example showing the necessity of the smallness condition on C₁. Explicit examples were later given in [3] and [13]. Related work can be found in [6, 12, 16].

Subsequently, more quantitative properties of unique continuation for (3), namely polynomial lower bound and doubling property were proved by Lin, Nakamura and Wang in [9]. A key element in the proof of [9] is a three-ball inequality deduced from the Carleman estimates of [15]. The main result of this note is an improvement on those of [9] and [10], requiring a weaker condition on ∇A. As in [9], we use the same Carleman estimates from [10] and [15]. However, our proof is more direct as it does not use a three-ball inequality. We next state our result, whose proof is contained in Section 3, after some preparation in Section 2.

Theorem 1

Let A ∈ C(B₁) be a symmetric matrix function and suppose that there exist positive constants λ, δ, C_δ so that

$$ \lambda |\xi|^{2} \le \Re \langle A(x)\xi,\xi\rangle \le \lambda^{-1} |\xi|^{2},\quad \forall \xi\in \mathbb{R}^{n}, x\in B_{1}, $$

and

$$ |\nabla A(x)|\le\frac{C_{\delta}}{|x||\log|x||^{2+\delta}}\quad \text{ for a.e. } x\in B_{1}. $$

(4)

Furthermore, assume that A(0) is real and positively definite. Then there exist positive constants R = R(n, λ, δ, C_δ) and C^∗ = C^∗(n, λ) such that if u satisfies

$$ |Pu|\leq\frac{C_{1}}{|x|}|\nabla u|+\frac{C_{0}}{|x|^{2}}|u|, $$

(5)

with C₁ < C^∗ then there exist k, M₁, M₂ > 0 depending on u such that for 0 < r < R,

$$ {\int}_{B_{r}}|u|^{2}\geq M_{1}r^{k} $$

(6)

and

$$ {\int}_{B_{2r}}|u|^{2}\le M_{2}{\int}_{B_{r}}|u|^{2}. $$

(7)

Here, B_r := {x : |x|≤ r}.

We note that the fact that k, M₁, and M₂ depend on u is unavoidable, as the example of spherical harmonics shows. Note also that the properties (6) and (7) are stronger than the strong unique continuation property, as they imply that a solution u of (3) that vanishes to infinite order at 0 must vanish in a neighborhood of 0. Then by using [7, Theorem 2.4], it follows that u vanishes identically in B₁.

2 Preliminaries

We first state the two Carleman estimates that will be used in the proof. To simplify the notation, we assume the constants C₂ in Lemmas 1 and 2 below are the same. A proof of the first estimate can be found in [10]. (It was also reproved in [13] and [15].)

Lemma 1

([10, Theorem 2]) There exists C₂ > 0 depending only on n such that for any $\tau \in \frac {1}{2}+\mathbb {N}$ and $u\in C_{0}^{\infty }(\mathbb {R}^{n}\backslash \{0\})$,

$$ {\sum}_{|\alpha|\le 2}\int \tau^{2-2|\alpha|}|x|^{-2\tau+2|\alpha|-n}|D^{\alpha} u|^{2}\leq C_{2}\int|x|^{-2\tau+4-n}|{\Delta} u|^{2}. $$

The next estimate was proved in [15, Theorem 1.2] under the slightly stronger assumption (1) on ∇A. For the sake of completeness, we provide a quick proof.

Lemma 2

Assume that A satisfies (4) with A(0) = Id. Let $\varphi (x)=\frac {1}{2}|\log |x||^{2}$. Then there exists R₀ ∈ (0,1) and positive constants γ₀ ≥ 2 and C₂ depending only on n, δ and C_δ such that for γ ≥ γ₀ and $u\in C_{0}^{\infty }(B_{R_{0}}\backslash \{0\})$,

$$ \gamma^{3}\int |x|^{-n}|\log |x||^{2} e^{2\gamma\varphi}u^{2}+\gamma\int |x|^{-n+2} e^{2\gamma\varphi}|\nabla u|^{2}\leq C_{2}\int |x|^{-n+4}e^{2\gamma\varphi}|Pu|^{2}. $$

For the proof of this lemma, we shall need the following elliptic estimate.

Lemma 3

Suppose the assumptions of Lemma 2 hold. Then there exists R₀ ∈ (0,1) and positive constants γ₀ and C depending only on n, δ and C_δ such that for any γ > γ₀ and $u\in C_{0}^{\infty }(B_{R_{0}}\backslash \{0\})$,

$$ \begin{array}{@{}rcl@{}} \int |x|^{-n+4}|\log|x||^{-2-\delta}e^{2\gamma\varphi}|\nabla^{2} u|^{2} &\leq& 2 \int |x|^{-n+4}|\log|x||^{-2-\delta}e^{2\gamma\varphi}|Pu|^{2} \\ &&+C \gamma^{2}\int |x|^{-n+2} |\log|x||^{-\delta} e^{2\gamma\varphi}|\nabla u|^{2}. \end{array} $$

(8)

Here $\nabla ^{2} u =(\partial _{ij} u)_{i,j=1}^{n}$ is the Hessian of u.

Proof

Since $|A(x)-Id|\le C_{\delta } |\log |x||^{-1-\delta }$, by triangle inequality, it suffices to prove (8) with Δu in place of Pu on the right-hand side. By splitting u into real and imaginary parts, we can further assume u is real-valued. Integrating by parts twice gives

$$ \int |x|^{-n+4}|\log|x||^{-2-\delta}e^{2\gamma\varphi} \partial_{ii} u \partial_{jj} u =\int |x|^{-n+4}|\log|x||^{-2-\delta}e^{2\gamma\varphi} |\partial_{ij} u|^{2} + E, $$

where

$$ \begin{array}{@{}rcl@{}} |E| &\le& C\gamma \int |x|^{-n+3}|\log|x||^{-1-\delta}e^{2\gamma\varphi} |\nabla^{2} u||\nabla u| \\ & \le& \frac{1}{2n^{2}}\int |x|^{-n+4}|\log|x||^{-2-\delta}e^{2\gamma\varphi} |\nabla^{2} u|^{2} + C\gamma^{2} \int |x|^{-n+2}|\log|x||^{-\delta}e^{2\gamma\varphi} |\nabla u|^{2}. \end{array} $$

Summing over i and j, we obtain the desired inequality. □

Proof of Lemma 2

In view of Lemma 3, to prove Lemma 2, it suffices to show

$$ \begin{array}{@{}rcl@{}} &&\gamma^{3}\int |x|^{-n}|\log|x||^{2}e^{2\gamma\varphi}u^{2}+\gamma\int |x|^{-n+2} e^{2\gamma\varphi}|\nabla u|^{2} \\ &&\quad\leq C_{2}\int |x|^{-n+4}e^{2\gamma\varphi}|Pu|^{2} + C_{2} \gamma^{-1}\int |x|^{-n+4}|\log|x||^{-2-\delta}e^{2\gamma\varphi}|\nabla^{2} u|^{2}. \end{array} $$

(9)

Let v = ue^γφ then e^γφPu = P_γv where

$$ P_{\gamma} v = e^{\gamma \varphi}P(e^{-\gamma \varphi}v). $$

It is easy to see that (9) follows from

$$ \begin{array}{@{}rcl@{}} &&\gamma^{3}\int |x|^{-n}|\log|x||^{2}v^{2}+\gamma\int |x|^{-n+2} |\nabla v|^{2} \\ &&\quad\leq C_{2}\int |x|^{-n+4}|P_{\gamma} v|^{2} + C_{2} \gamma^{-1}\int |x|^{-n+4}|\log|x||^{-2-\delta}|\nabla^{2} v|^{2}. \end{array} $$

(10)

Let ω = x/|x| and $t = \log |x|$ for x≠ 0, i.e., x = e^tω. Then

$$ \partial_{j}=e^{-t}(\omega_{j} \partial_{t} +{\Omega}_{j}), $$

where Ω_j are vector fields on $\mathbb {S}^{n-1}$ satisfying

$$ \sum\limits_{j} \omega_{j}{\Omega}_{j} =0, \quad\sum\limits_{j} {\Omega}_{j}\omega_{j} =n-1,\quad {\Omega}_{j}^{\ast}=(n-1)\omega_{j}-{\Omega}_{j}. $$

Let (D₀,…, D_n) = (i∂_t, iΩ₁,…, iΩ_n). We denote by Dv the vector (D₀v,…, D_nv) and by D²v the matrix D_jD_kv, 0 ≤ j, k ≤ n. Then (10) takes the form

$$ \gamma^{3} \int |v|^{2} t^{2} dt d\omega + \gamma \int |D v|^{2} dt d\omega \le C_{2} \int \left( |e^{2t}P_{\gamma} v|^{2} +\gamma^{-1}t^{-2-\delta}|D^{2} v|^{2}\right)dt d\omega. $$

(11)

We have

$$ \begin{array}{@{}rcl@{}} Pu&=&e^{-2t}\sum\limits_{k,l=1}^{n} a_{kl}(e^{t}\omega)(\omega_{k}\partial_{t}-\omega_{k}+{\Omega}_{k})(\omega_{l}\partial_{t}+{\Omega}_{l})u \\ &=&e^{-2t}\left[\partial_{t}^{2} u +(n-2)\partial_{t} u +{\Delta}_{\omega} u + \sum\limits_{j+|\alpha|\le 2} C_{j,\alpha}(t,\omega){\partial_{t}^{j}} {\Omega}^{\alpha} u\right], \end{array} $$

and consequently,

$$ \begin{array}{@{}rcl@{}} e^{2t}P_{\gamma} v &=& (\partial_{t}-\gamma t)^{2} v +(n-2)(\partial_{t}-\gamma t)v+{\Delta}_{\omega} v+ \sum\limits_{j+|\alpha|\le 2} C_{j,\alpha}(t,\omega)(\partial_{t}-\gamma t)^{j} {\Omega}^{\alpha} v \\ &=& {\partial^{2}_{t}} v+{\Delta}_{\omega} v + [(n-2)-2\gamma t]\partial_{t} v + [\gamma^{2} t^{2}-(n-2)\gamma t -\gamma] v \\ &&+ {\sum}_{j+|\alpha|\le 2} C_{j,\alpha}(t,\omega)(\partial_{t}-\gamma t)^{j} {\Omega}^{\alpha} v. \end{array} $$

Let

$$ Q v = {\partial^{2}_{t}} v+{\Delta}_{\omega} v -2\gamma t \partial_{t} v + \left[\gamma^{2} t^{2}- 2 \gamma\right] v + \sum\limits_{j+|\alpha|= 2} C_{j,\alpha}(t,\omega)(\partial_{t}-\gamma t)^{j} {\Omega}^{\alpha} v. $$

Since by (4), C_{j, α} are bounded, it follows that

$$ |e^{2t}P_{\gamma} v - Qv|\le C(\gamma |tv|+|D v|). $$

Thus, by triangle inequality, it suffices to prove (11) with Qv in place of e^2tP_γv on the right-hand side. The last term of Qv can be written as

$$ \sum\limits_{j+|\alpha|= 2} C_{j,\alpha}(t,\omega)(\partial_{t}-\gamma t)^{j} {\Omega}^{\alpha} v = \sum\limits_{|\alpha|\le 2} (V_{\alpha,0}+iV_{\alpha,1})(t,\omega)(\gamma t)^{2-|\alpha|} D^{\alpha} v, $$

where the real-valued functions V_{α, k}’s are linear combinations of C_{j, α}’s.

Let

$$ Mv = {\partial_{t}^{2}} v +{\Delta}_{\omega} v +\gamma^{2} t^{2} v + \sum\limits_{|\alpha|\le 2} V_{\alpha,0}(t,\omega)(\gamma t)^{2-|\alpha|} D^{\alpha} v $$

and

$$ Nv = -2\gamma t \partial_{t} v - 2\gamma v + \sum\limits_{|\alpha|\le 2} i V_{\alpha,1}(t,\omega)(\gamma t)^{2-|\alpha|} D^{\alpha} v. $$

Then $\|Qv\|^{2}_{L^{2}}=\|Mv+Nv\|^{2}_{L^{2}} \ge 2 \Re \langle Mv,Nv\rangle$. The right-hand side consists of the following terms:

$$ \begin{array}{@{}rcl@{}} T_{1}&=& 2{\Re\langle\partial_{t}^{2}} v, -2\gamma t \partial_{t} v -2\gamma v\rangle = 6\gamma \|\partial_{t} v\|^{2}, \\ T_{2}&=&2\Re\langle{\Delta}_{\omega} v, -2\gamma t \partial_{t} v -2\gamma v\rangle= 2\gamma \|{\Omega} v\|^{2}, \\ T_{3}&=&2\Re\langle\gamma^{2} t^{2} v, -2\gamma t \partial_{t} v-2\gamma v\rangle= 2\gamma^{3} \|tv\|^{2}, \\ T_{4}&=&2\Re\left< \sum\limits_{|\alpha|\le 2} V_{\alpha,0}(t,\omega)(\gamma t)^{2-|\alpha|} D^{\alpha} v, -2\gamma v\right> \\ &\ge& -\frac{1}{2}\gamma^{3} \|tv\|^{2} - C \sum\limits_{|\alpha|\le 2} \gamma^{3-2|\alpha|}\left\|t^{-\delta-|\alpha|}D^{\alpha} v\right\|^{2}. \end{array} $$

Here we have used Cauchy–Schwarz at the last line. The remaining terms have the form

$$ \Im \left< W_{\alpha,\upbeta} (t,\omega)(\gamma t)^{4-|\alpha|-|\upbeta|} D^{\alpha} v, D^{\upbeta} v\right>, $$

where W_α,β are real-valued function satisfying |W_α,β| + |DW_α,β| = O(|t|^− 1−δ). Integrating by parts |α| + |β| times gives

$$ \begin{array}{@{}rcl@{}} \Im \left< W_{\alpha,\upbeta}(\gamma t)^{4-|\alpha|-|\upbeta|} D^{\alpha} v, D^{\upbeta} v\right>&=& \Im \left< W_{\alpha,\upbeta}(\gamma t)^{4-|\alpha|-|\upbeta|} D^{\upbeta} v, D^{\alpha} v\right> \\ &&+ \underset{<|\alpha|+|\upbeta|}{\underset{|\alpha^{\prime}|+|{\upbeta}^{\prime}|}{\sum}} \Im \left< Z_{\alpha^{\prime},{\upbeta}^{\prime}}(\gamma t)^{4-|\alpha|-|\upbeta|} D^{\alpha^{\prime}} v, D^{{\upbeta}^{\prime}} v\right>. \end{array} $$

Hence,

$$ \Im \left< W_{\alpha,\upbeta} (\gamma t)^{4-|\alpha|-|\upbeta|} D^{\alpha} v, D^{\upbeta} v\right> = \frac{1}{2} \underset{<|\alpha|+|\upbeta|}{\underset{|\alpha^{\prime}|+|{\upbeta}^{\prime}|}{\sum}} \Im \left< Z_{\alpha^{\prime},{\upbeta}^{\prime}} (\gamma t)^{4-|\alpha|-|\upbeta|} D^{\alpha^{\prime}} v, D^{{\upbeta}^{\prime}} v\right>. $$

Here, $Z_{\alpha ^{\prime },{\upbeta }^{\prime }} =O(|t|^{-1-\delta })$. Applying Cauchy–Schwarz, we see that the right-hand side is bounded in absolute value by

$$ C {\sum}_{|\alpha|\le 2} \gamma^{3-2|\alpha|}\|t^{\frac{2-\delta}{2} -|\alpha|}D^{\alpha} v\|^{2}. $$

Summing up all the terms, we obtain

$$ 2 \Re \langle Mv,Nv\rangle \ge \gamma^{3} \|tv\|^{2} + \gamma \|D v\|^{2} - C \gamma^{-1}\|t^{-1-\frac{\delta}{2} }D^{2} v\|^{2}. $$

This gives the desired inequality (11). □

We will also need the following Caccioppoli type estimate. Since the proof follows standard arguments, we will skip it. For 0 < a < b let $\mathcal {A}(a,b)=\{x:a\le |x|\le b\}$.

Lemma 4

There exist C = C(n, λ) > 0 such that if $C_{3}=C(1+C_{0}+{C_{1}^{2}}+C_{\delta }^{2})$ then for any u satisfying (5) and $0<r<\frac {1}{2}$,

$$ {\int}_{\mathcal{A}(5r/4,7r/4)} |x|^{-n+2}|\nabla u|^{2}dx \le C_{3} {\int}_{\mathcal{A}(r,2r)} |x|^{-n}|u|^{2}dx. $$

3 Proof of Theorem 1

In this proof, C denotes a constant depending only on n and λ, whose value may change from line to line. By a change of variable which may change the values of R and C^∗ by a factor of λ, we can assume A(0) = Id. Under this additional assumption, we will prove Theorem 1 with $C^{\ast } = \frac {1}{4\sqrt {C_{2}}}$. By using [7, Theorem 2.4], it suffices to consider the case u does not vanish identically on any balls in B₁.

For 0 < r < R₀/8 where R₀ is the constant which appears in Lemma 2, let ζ be a smooth cut-off function satisfying $\chi _{\mathcal {A}(7r/4,5R_{0}/8)}\le \zeta \le \chi _{\mathcal {A}(5r/4,7R_{0}/8)}$ and $| \partial ^{\alpha }\zeta (x)|\le 10|x|^{-|\alpha |}, \forall x\in \mathbb {R}^{n}$ and |α|≤ 2. Here, χ_E denotes the characteristic function of the set E.

Let v = ζu and $E=\mathcal {A}(5r/4,7r/4)\cup \mathcal {A}(5R_{0}/8,7R_{0}/8)$. Then

$$ \begin{array}{@{}rcl@{}} |Pv|&=&|\zeta Pu+ 2a_{jk}(\partial_{j} \zeta)(\partial_{k} u) + a_{jk}\partial_{jk} \zeta u| \\ &\le& \zeta\left( C_{1}|x|^{-1}|\nabla u|+C_{0}|x|^{-2}|u|\right) + C\left( |x|^{-1}|\nabla u|+|x|^{-2}|u|\right)\chi_{E}\\ &\le& C_{1}|x|^{-1}|\nabla v| + C_{0}|x|^{-2}|v| + C\left( |x|^{-1}|\nabla u|+|x|^{-2}|u|\right)\chi_{E}. \end{array} $$

Applying Lemma 2 to v and using the above inequality, we have

$$ \begin{array}{@{}rcl@{}} &&\gamma^{3}\int |x|^{-n}e^{2\gamma \varphi}|v|^{2}dx + \gamma \int |x|^{-n+2}e^{2\gamma\varphi}|\nabla v|^{2}dx \\ &&\leq C_{2} \int |x|^{-n+4}e^{2\gamma\varphi}|Pv|^{2} dx \\ &&\leq 4 C_{2} \int |x|^{-n+4}e^{2\gamma\varphi}\left( {C_{0}^{2}}|x|^{-4}|v|^{2}+{C_{1}^{2}}|x|^{-2}|\nabla v|^{2}\right)dx \\ &&\quad + 4C_{2}C^{2} {\int}_{E} |x|^{-n+4}e^{2\gamma\varphi}\left( |x|^{-4}|u|^{2}+|x|^{-2}|\nabla u|^{2}\right)dx. \end{array} $$

(12)

Assuming $\gamma \ge \gamma _{1} := \max \limits \{\gamma _{0},2C_{0}^{2/3}C_{2}^{1/3},8{C_{1}^{2}} C_{2}\}$, the first term on the right-hand side of (12) can be absorbed by its left-hand side. Thus, we deduce that

$$ 2 \int |x|^{-n}e^{2\gamma \varphi}|v|^{2}dx \leq 4C_{2}C^{2} {\int}_{E} |x|^{-n+4}e^{2\gamma\varphi}\left( |x|^{-4}|u|^{2}+|x|^{-2}|\nabla u|^{2}\right)dx. $$

Using Lemma 4 to bound the gradient terms on the right-hand side, we get

$$ \begin{array}{@{}rcl@{}} 2{\int}_{\mathcal{A}(2r,\frac{R_{0}}{2})} |x|^{-n}e^{2\gamma \varphi}|u|^{2}dx &\leq& C_{4}e^{2\gamma \varphi(r)}{\int}_{\mathcal{A}(r,2r)} |x|^{-n}|u|^{2} dx\\ && + C_{4}e^{2\gamma\varphi\left( \frac{R_{0}}{2}\right)} {\int}_{\mathcal{A}\left( \frac{R_{0}}{2},R_{0}\right)}|x|^{-n}|u|^{2} dx, \end{array} $$

(13)

where C₄ = 32C₂C²(C₃ + 1).

We now fix

$$ \gamma=\max\left\{\gamma_{1},\frac{\log\left( C_{4}{\int}_{\mathcal{A}\left( \frac{R_{0}}{2},R_{0}\right)}|x|^{-n}|u|^{2}/{\int}_{\mathcal{A}\left( \frac{R_{0}}{4},\frac{R_{0}}{3}\right)}|x|^{-n}|u|^{2}\right)}{2\varphi\left( \frac{R_{0}}{3}\right)-2\varphi\left( \frac{R_{0}}{2}\right)}\right\}. $$

For this choice of γ,

$$ C_{4}e^{2\gamma\varphi\left( \frac{R_{0}}{2}\right)}{\int}_{\mathcal{A}\left( \frac{R_{0}}{2},R_{0}\right)}|x|^{-n}|u|^{2} \le {\int}_{\mathcal{A}\left( \frac{R_{0}}{4},\frac{R_{0}}{3}\right)}|x|^{-n}e^{2\gamma\varphi}|u|^{2}, $$

hence the last term on the right-hand side of (13) can be absorbed by the left-hand side, giving

$$ {\int}_{\mathcal{A}\left( 2r,\frac{R_{0}}{2}\right)}|x|^{-n}e^{2\gamma\varphi}|u|^{2}\le C_{4} e^{2\gamma\varphi(r)}{\int}_{\mathcal{A}(r,2r)} |x|^{-n}|u|^{2}. $$

(14)

Note that this would give a lower bound that is worse than polynomial. We will use Lemma 1 to improve upon (14) to reach the conclusion. Let R₁ ∈ (0, R₀/8] satisfy

$$ |\log R_{1}| \ge \max\left\{\left( 2\sqrt{C_{2}}C_{\delta}(5\gamma+ \log C_{4})\right)^{\frac{1}{\delta}},(8C_{0}C_{2}C_{\delta})^{\frac{1}{1+\delta}}\right\} $$

and η be a smooth cut-off function such that $\chi _{\mathcal {A}(7r/4,5R_{1}/8)}\le \eta \le \chi _{\mathcal {A}(5r/4,7R_{1}/8)}$ and $| \partial ^{\alpha }\eta (x)|\le 10|x|^{-|\alpha |},~\forall x\in \mathbb {R}^{n}$ and |α|≤ 2.

From $|\nabla A(x)|\leq C_{\delta } |x|^{-1}|\log |x||^{-2-\delta }$ and A(0) = Id, we see that

$$ |A(x)-\text{Id}|\le C_{\delta}|\log|x||^{-1-\delta} \le C_{\delta}|\log R_{1}|^{-1-\delta}\qquad \forall x\in B_{R_{1}}. $$

Appplying Lemma 1 to w = ηu, we obtain

$$ \begin{array}{@{}rcl@{}} {\sum}_{j=0}^{2}\int\tau^{2-2j}|x|^{-2\tau+2j-n}|\nabla^{j}w|^{2} &\leq& C_{2} \int|x|^{-2\tau+4-n}|{\varDelta} w|^{2} \\ &\leq& 2C_{2} \int|x|^{-2\tau+4-n}|Pw|^{2} \\ &&+2C_{2}C_{\delta}^{2} |\log R_{1} |^{-2-2\delta} \int |x|^{-2\tau+4-n} |\nabla^{2}w|^{2}. \end{array} $$

Choosing $\tau =\lfloor \frac {1}{2\sqrt {C_{2}}C_{\delta }}|\log R_{1}|^{1+\delta }\rfloor$, the last term can be absorbed by the left-hand side, hence we obtain

$$ \tau^{2}\int|x|^{-2\tau-n}|w|^{2}+\int|x|^{-2\tau+2-n}|\nabla w|^{2}\leq 2C_{2} \int|x|^{-2\tau+4-n}|Pw|^{2}. $$

(15)

Note that by our choice of C^∗, R₁, and τ, we have $\tau ^{2}\ge 16 {C_{0}^{2}} C_{2}$ and $1\ge 16{C_{1}^{2}} C_{2}$. Hence, using the same arguments that lead to (13) from Lemma 2, we obtain from (15) that

$$ 2{\int}_{\mathcal{A}(2r,R_{1})}|x|^{-2\tau-n}|u|^{2} \leq C_{4}r^{-2\tau}{\int}_{\mathcal{A}(r,2r)}|x|^{-n}|u|^{2} +C_{4}R_{1}^{-2\tau}{\int}_{\mathcal{A}(R_{1},2R_{1})}|x|^{-n}|u|^{2}. $$

(16)

From our choice of R₁ and τ, we have

$$ \tau \ge (5\gamma+\log C_{4})|\log R_{1}|, $$

which implies that for $R_{2}=\frac {1}{2}{R_{1}^{2}}$,

$$ (2R_{2})^{-2\tau}e^{-2\gamma\varphi(R_{2})+2\gamma\varphi(2R_{1})} \geq {C_{4}^{2}}R_{1}^{-2\tau}. $$

Hence, for 0 < r < R₂/2, using (14) we have

$$ \begin{array}{@{}rcl@{}} {\int}_{\mathcal{A}(2r,R_{1})}|x|^{-2\tau-n}|u|^{2} &\geq& (2R_{2})^{-2\tau}{\int}_{\mathcal{A}(R_{2},2R_{2})} |x|^{-n}|u|^{2}\\ & \geq& C_{4}^{-1}(2R_{2})^{-2\tau}e^{-2\gamma\varphi(R_{2})}{\int}_{\mathcal{A}\left( 2R_{2},\frac{R_{0}}{2}\right)} |x|^{-n}e^{2\gamma\varphi}u^{2}\\ & \geq& C_{4}^{-1}(2R_{2})^{-2\tau}e^{-2\gamma\varphi(R_{2})+2\gamma\varphi(2R_{1})}{\int}_{\mathcal{A}(R_{1},2R_{1})} |x|^{-n}|u|^{2}\\ & \geq& C_{4}R_{1}^{-2\tau}{\int}_{\mathcal{A}(R_{1},2R_{1})} |x|^{-n}|u|^{2}. \end{array} $$

Thus, the last term of (16) can be absorbed by its left-hand side. Hence, for r < R₂/2,

$$ {\int}_{\mathcal{A}(2r,R_{1})}|x|^{-2\tau-n}|u|^{2}\leq C_{4}r^{-2\tau}{\int}_{\mathcal{A}(r,2r)}|x|^{-n}|u|^{2}. $$

(17)

From this, (6) follows with k = 2τ + n and

$$ M_{1}=C_{4}^{-1}{\int}_{\mathcal{A}(R_{1}/2,R_{1})}|x|^{-2\tau-n}|u|^{2}. $$

Moreover, we can deduce from (17) that

$$ {\int}_{\mathcal{A}(2r,4r)} |u|^{2} \le 4^{2\tau+n}C_{4}{\int}_{\mathcal{A}(r,2r)}|u|^{2}. $$

Adding ${\int \limits }_{B_{2r}}|u|^{2}$ to both sides, it follows that

$$ {\int}_{B_{2r}}|u|^{2}\ge \frac{1}{4^{2\tau+n}C_{4}+1}{\int}_{B_{4r}}|u|^{2}. $$

Thus, (7) follows with M₂ = 4^2τ+nC₄ + 1

To finish the proof, note that for r ∈ (R₂/2, R₀/8], (6) and (7), possibly with different M₁ and M₂, follow from (14).