BMO Solvability and $A_{\infty }$ Condition of the Elliptic Measures in Uniform Domains

Abstract

We consider the Dirichlet boundary value problem for divergence form elliptic operators with bounded measurable coefficients. We prove that for uniform domains with Ahlfors regular boundary, the BMO solvability of such problems is equivalent to a quantitative absolute continuity of the elliptic measure with respect to the surface measure, i.e., $\omega _L\in A_{\infty }(\sigma )$. This generalizes a previous result on Lipschitz domains by Dindos, Kenig, and Pipher (see Dindos et al. in J Geom Anal 21:78–95, 2011).

Appendices

Appendix 1: Proof of Lemma 5.3: Properties of $f_{\epsilon }$

The function $f_{\epsilon }$ as in (5.27) is well defined because

$$\begin{aligned} \int _{y\in \partial \Omega } \varphi _{\epsilon }(x-y)\mathrm{d}\sigma (y) \ge \frac{1}{\epsilon ^{n-1}} \int _{y\in \Delta \left( x, \frac{\epsilon }{2}\right) } \mathrm{d}\sigma (y) \ge C_1 > 0. \end{aligned}$$

(7.1)

We also have

$$\begin{aligned} \int _{y\in \partial \Omega } \varphi _{\epsilon }(x-y)\mathrm{d}\sigma (y) \le \frac{1}{\epsilon ^{n-1}} \int _{y\in \Delta (x, \epsilon )} \mathrm{d}\sigma (y) \le C_2 . \end{aligned}$$

(7.2)

The constants $C_1$ and $C_2$ are independent of $\epsilon$.

Proof of (1)

For any surface ball $\Delta _0=\Delta (x_0, r_0)$, we denote $\Delta _0^{\epsilon } = \Delta (x_0,r_0+\epsilon )$. Since f is supported in $2\Delta$, each $f_{\epsilon }$ is supported in $\left( 2\Delta \right) ^{\epsilon }$. Thus, all $f_{\epsilon }$’s are supported in $3\Delta$ if $\epsilon < r$, the radius of $\Delta$. $\square$

Note that

$$\begin{aligned}&\left| \int _{\partial \Omega } \varphi _{\epsilon }(x-y) \mathrm{d}\sigma (y) -\int _{\partial \Omega } \varphi _{\epsilon }(\tilde{x}-y) \mathrm{d}\sigma (y) \right| \nonumber \\&\quad = \left| \int _{\partial \Omega } \int _0^1 \frac{d}{ds} \varphi _{\epsilon }((1-s)\tilde{x}+s x-y) ds \mathrm{d}\sigma (y) \right| \nonumber \\&\quad \le \frac{|x-\tilde{x}|}{\epsilon ^n} \int _0^1 \int _{y\in \partial \Omega } \left| \nabla \varphi \left( \dfrac{(1-s)\tilde{x}+s x-y}{\epsilon } \right) \right| \mathrm{d}\sigma (y) ds. \end{aligned}$$

(7.3)

Since $\Vert \nabla \varphi \Vert _{L^{\infty }} \le C$, for any $w\in \mathbb {R}^n$ we have

$$\begin{aligned} \int _{y\in \partial \Omega } \left| \nabla \varphi \left( \frac{w-y}{\epsilon }\right) \right| \mathrm{d}\sigma (y) \le C \sigma \left( B(w,\epsilon )\cap \partial \Omega \right) \le C\epsilon ^{n-1}. \end{aligned}$$

(7.4)

Combining (7.3) and (7.4),

$$\begin{aligned} \left| \int _{\partial \Omega } \varphi _{\epsilon }(x-y) \mathrm{d}\sigma (y) -\int _{\partial \Omega } \varphi _{\epsilon }(\tilde{x}-y) \mathrm{d}\sigma (y) \right| \le C\frac{|x-\tilde{x}|}{\epsilon }, \end{aligned}$$

so for any $\epsilon$ fixed, the map $x\in \partial \Omega \mapsto \int _{\partial \Omega } \varphi _{\epsilon }(x-y) \mathrm{d}\sigma (y)$ is continuous. Since $0\le f \le 1$, we can prove similarly $\int _{\partial \Omega } f(y) \varphi _{\epsilon }(x-y) \mathrm{d}\sigma (y)$ is also continuous. Thus, $f_{\epsilon }(x)$ is continuous.

Proof of (2)

Fix $\epsilon >0$. Let $\widetilde{\Delta }=\Delta (x_0, r_0)$ be an arbitrary surface ball. Let $\lambda$ be a real number to be determined later. We consider two cases. $\square$

Case 1 If $r_0 \ge \epsilon /2$, by the definition (5.27) and the estimates (7.1), (7.2),

$$\begin{aligned}&\int _{\widetilde{\Delta }} \left| f_{\epsilon }(x) -\lambda \right| \mathrm{d}\sigma (x) \\&\quad \le \frac{1}{C_1} \int _{\widetilde{\Delta }}\left| \int _{\partial \Omega } f(y) \varphi _{\epsilon }(x-y) \mathrm{d}\sigma (y) - \lambda \int _{\partial \Omega } \varphi _{\epsilon }(x-y) \mathrm{d}\sigma (y) \right| \mathrm{d}\sigma (x) \\&\quad \le \widetilde{C}_1 \int _{x\in \widetilde{\Delta }} \int _{y\in \Delta (x,\epsilon ) } |f(y) -\lambda | \varphi _{\epsilon }(x-y) \mathrm{d}\sigma (y) \mathrm{d}\sigma (x) \\&\quad \le \widetilde{C}_1 \int _{y\in \widetilde{\Delta }^{\epsilon }} |f(y) -\lambda | \int _{x\in \partial \Omega } \varphi _{\epsilon }(x-y) \mathrm{d}\sigma (x) \mathrm{d}\sigma (y) \\&\quad \le \widetilde{C}_1 C_2 \int _{y\in \widetilde{\Delta }^{\epsilon }} |f(y) -\lambda | \mathrm{d}\sigma (y) \\&\quad \le C' \sigma (\widetilde{\Delta }^{\epsilon }) \Vert f\Vert _{\mathrm{BMO}(\sigma )} . \end{aligned}$$

The last inequality is true if we choose $\lambda = \lambda (\widetilde{\Delta }, \epsilon )$ be the constant satisfying . Thus,

Case 2 If $r_0<\epsilon /2$, by the definition (5.27) and the estimate (7.1),

$$\begin{aligned} \int _{\widetilde{\Delta }} \left| f_{\epsilon }(x) -\lambda \right| \mathrm{d}\sigma (x)&\le \widetilde{C}_1 \int _{x\in \widetilde{\Delta }} \int _{y\in \Delta (x,\epsilon ) } |f(y) -\lambda | \varphi _{\epsilon }(x-y) \mathrm{d}\sigma (y) \mathrm{d}\sigma (x) \nonumber \\&\le \widetilde{C}_1 \iint _{y\in \widetilde{\Delta }^{\epsilon }} |f(y) -\lambda | \int _{x\in \widetilde{\Delta }} \varphi _{\epsilon }(x-y) \mathrm{d}\sigma (x) \mathrm{d}\sigma (y). \end{aligned}$$

(7.5)

Note

$$\begin{aligned} \int _{x\in \widetilde{\Delta }} \varphi _{\epsilon }(x-y) \mathrm{d}\sigma (x) \le \frac{1}{\epsilon ^{n-1}} \int _{x\in \widetilde{\Delta }} \mathrm{d}\sigma (x)=\frac{\sigma (\widetilde{\Delta })}{\epsilon ^{n-1}} , \end{aligned}$$

it follows from (7.5) that

We have proved the following: for any $\epsilon$ and any surface ball $\widetilde{\Delta }$, one can find a constant $\lambda =\lambda (\widetilde{\Delta }, \epsilon )$ such that . The constant C does not depend on either $\epsilon$ or $\widetilde{\Delta }$, therefore, $\Vert f_{\epsilon }\Vert _{\mathrm{BMO}(\sigma )} \le C\Vert f\Vert _{\mathrm{BMO}(\sigma )}$ for all $\epsilon$.

Proof of (3)

Fix $x\in \partial \Omega$. If $f(x)=0$, then obviously $f(x) \le \liminf _{\epsilon \rightarrow 0} f_{\epsilon }(x)$. For any arbitrary $\lambda >0$ such that $\lambda < f(x)$, there exists $\epsilon _0>0$ such that $f(x) > \lambda + \epsilon _0$. It means

$$\begin{aligned} \sup _{\Delta '\ni x} \frac{\sigma (E\cap \Delta ')}{\sigma (\Delta ')} = M_{\sigma }\chi _E(x) > e^{\frac{\lambda + \epsilon _0 - 1}{\delta }}. \end{aligned}$$

In particular, there is some surface ball $\Delta ' \ni x$ such that

$$\begin{aligned} \frac{\sigma (E\cap \Delta ')}{\sigma (\Delta ')} > e^{\frac{\lambda + \epsilon _0 - 1}{\delta }}. \end{aligned}$$

Then for any point $y\in \Delta '$, we also have $M_{\sigma }\chi _E(y) > \exp {(\lambda +\epsilon _0-1)/\delta }$, and thus $f(y) > \lambda + \epsilon _0$. Consider all $f_{\epsilon }$ with $\epsilon < {{\mathrm{dist}}}(x, \partial \Omega {\setminus }\Delta ')$, we have $\Delta (x,\epsilon )\subset \Delta '$, hence

$$\begin{aligned} f_{\epsilon }(x)&= \frac{ \int _{y\in \Delta (x,\epsilon )} f(y) \varphi _{\epsilon } (x-y) \mathrm{d}\sigma (y)}{\int _{y\in \Delta (x,\epsilon )} \varphi _{\epsilon }(x-y)\mathrm{d}\sigma (y)} \\&\ge (\lambda +\epsilon _0) \frac{ \int _{y\in \Delta (x,\epsilon )} \varphi _{\epsilon } (x-y) \mathrm{d}\sigma (y)}{\int _{y\in \Delta (x,\epsilon )} \varphi _{\epsilon }(x-y)\mathrm{d}\sigma (y)} = \lambda + \epsilon _0. \end{aligned}$$

Therefore, $\liminf _{\epsilon \rightarrow 0} f_{\epsilon }(x) > \lambda$ for all $\lambda < f(x)$. Thus, $\liminf _{\epsilon \rightarrow 0} f_{\epsilon }(x) \ge f(x)$. $\square$

Appendix 2: Properties of the Truncated Square Function

1.1 Proof of Lemma 6.4

Assume $Q\in \partial \Omega$ satisfies $S^2_r u(Q) = \iint _{\Gamma _r(Q)} |\nabla u|^2 \delta (X)^{2-n} \mathrm{d}X > \lambda ^2$ and is finite, then there exists $\eta <r$ such that

$$\begin{aligned} \iint _{\Gamma _r(Q){\setminus }B(Q,\eta ) } |\nabla u|^2 \delta (X)^{2-n} \mathrm{d}X > \left( \frac{ S_r u(Q) + \lambda }{2}\right) ^2. \end{aligned}$$

Fix $\eta$, we claim there exists $\epsilon >0$ such that $S_r u(P)>\lambda$ for any $P\in B(Q,\epsilon \eta )\cap \partial \Omega$. In fact,

$$\begin{aligned}&\left| \iint _{\Gamma _r(Q){\setminus }B(Q,\eta ) } |\nabla u|^2 \delta (X)^{2-n} \mathrm{d}X - \iint _{\Gamma _r(P){\setminus }B(P,\eta ) } |\nabla u|^2 \delta (X)^{2-n} \mathrm{d}X\right| \nonumber \\&\quad \le \iint _{D} |\nabla u|^2 \delta (X)^{2-n} \mathrm{d}X \end{aligned}$$

(8.1)

where D is the set difference between $\Gamma _r(Q){\setminus }B(Q,\eta )$ and $\Gamma _r(P){\setminus }B(P,\eta )$.

Assume $X\in \Gamma _r(Q){\setminus }B(Q,\eta )$, then $|X-Q|\le \alpha \delta (X)$ and $\eta \le |X-Q|<r$. In particular, $\eta \le \alpha \delta (X)$. If in addition $X\notin \Gamma _r(P){\setminus }B(P,\eta )$ for some $P \in B(Q,\epsilon \eta )$, then X falls in one of the following three categories:

• $|X-P|< \eta$, then $|X-Q|\le |X-P|+|P-Q| <(1+\epsilon )\eta$, in particular $\eta \le |X-Q| <(1+\epsilon )\eta$;

• $|X-P|\ge r$, then $|X-Q|\ge |X-P|-|P-Q| > r-\epsilon \eta$, in particular $r-\epsilon \eta<|X-Q|<r$;

• $|X-P|> \alpha \delta (X)$, then $|X-Q|\ge |X-P|-|P-Q|> (1-\epsilon ) \alpha \delta (X)$, in particular $(1-\epsilon ) \alpha \delta (X) < |X-Q| \le \alpha \delta (X)$.

Similarly, the points in $\left( \Gamma _r(P){\setminus }B(P,\eta )\right) {\setminus }\left( \Gamma _r(Q){\setminus }B(Q,\eta ) \right)$ also fall in three categories, just with Q replaced by P. Therefore, D, the set difference between $\left( \Gamma _r(Q){\setminus }B(Q,\eta )\right) {\setminus }\left( \Gamma _r(P){\setminus }B(P,\eta ) \right)$ and $\left( \Gamma _r(P){\setminus }B(P,\eta )\right) {\setminus }\left( \Gamma _r(Q){\setminus }B(Q,\eta ) \right)$, is contained in the union of three sets (corresponding to the above three cases):

$$\begin{aligned} V_1= & {} \big \{X\in \Omega : (1-\epsilon )\eta< |X-Q|<(1+2\epsilon )\eta , ~\delta (X) \ge \eta /\alpha \big \}\\ V_2= & {} \big \{X\in \Omega : r-2\epsilon \eta< |X-Q|< r+\epsilon \eta , ~\delta (X) \ge \eta /\alpha \big \}\\ V_3= & {} \big \{X\in \Omega :(1-2\epsilon ) \alpha \delta (X) < |X-Q| \le (1+\epsilon )\alpha \delta (X), ~\delta (X) \ge \eta /\alpha \big \}. \end{aligned}$$

Since $\delta (X) \ge \eta /\alpha$ in D,

$$\begin{aligned} \iint _{D} |\nabla u|^2 \delta (X)^{2-n} \mathrm{d}X \le \left( \frac{\alpha }{\eta }\right) ^{n} \iint _{V_1\cup V_2 \cup V_3} |\nabla u|^2 \delta (X)^2 \mathrm{d}X. \end{aligned}$$

(8.2)

Note that $u\in W^{1,2}(\Omega )$, we have

$$\begin{aligned} \iint _{\Omega } |\nabla u|^2 \delta (X)^2 \mathrm{d}X \le {{\mathrm{\,diam\,}}}(\Omega )^2 \iint _{\Omega } |\nabla u|^2 \mathrm{d}X <\infty . \end{aligned}$$

Hence, the integral $\iint _{V_1\cup V_2 \cup V_3} |\nabla u|^2 \delta (X)^2 \mathrm{d}X$ is small as long as the Lebesgue measures of $V_1$, $V_2$ and $V_3$ are small enough. Both $V_1$ and $V_2$ are contained in annuli of radius $3\epsilon \eta$, so their Lebesgue measures are small if we choose $\epsilon$ small enough (depending on $\eta$). Rewrite $V_3$ as

$$\begin{aligned} V_3 = \bigg \{X\in \Omega : \frac{1}{ (1+\epsilon ) \alpha }< \frac{\delta (X)}{|X-Q|} \le \frac{1}{(1-2\epsilon )\alpha }, ~ \delta (X) \ge \frac{\eta }{\alpha }\bigg \}. \end{aligned}$$

Away from Q, say in $\Omega {\setminus }B(Q,\eta /2)$, the function $F(X) = \delta (X)/|X-Q|$ is Lipschitz, and $0\le F\le 1$. Choose $\epsilon <1/4$, then for any $X\in V_3$, $|X-Q|\ge (1-2\epsilon )\alpha \delta (X) \ge \eta /2$. So $V_3\subset \Omega {\setminus }B(Q,\eta /2)$, and thus F is Lipschitz on $V_3$. By the coarea formula,

$$\begin{aligned} \mathcal {H}^n(V_3) = \int _{\frac{1}{(1+\epsilon )\alpha } }^{\frac{1}{(1-2\epsilon )\alpha }} \int _{F^{-1}(t)} \frac{1}{JF} \chi _{V_3} \mathrm{d}\mathcal {H}^{n-1} \mathrm{d}t. \end{aligned}$$

(8.3)

On the other hand,

$$\begin{aligned} \int _{0}^1 \int _{F^{-1}(t)} \frac{1}{JF} \chi _{V_3} \mathrm{d}\mathcal {H}^{n-1} \mathrm{d}t&\le \int _{0}^1 \int _{F^{-1}(t)} \frac{1}{JF} \chi _{\Omega {\setminus }B(Q,\eta /2) } \mathrm{d}\mathcal {H}^{n-1} \mathrm{d}t \\&=\mathcal {H}^n\big (\Omega {\setminus }B(Q,\eta /2) \big ) \end{aligned}$$

is finite. Therefore, by (8.3), we may choose $\epsilon$ small enough (depending on $\alpha$) such that $\mathcal {H}^n(V_3)$ is small, which in turn implies $\iint _{V_3} |\nabla u|^2 \delta (X)^2 \mathrm{d}X$ is small. To sum up, we have shown that one can choose $\epsilon = \epsilon (\delta ,\alpha ,\eta ,r)$ small enough such that

$$\begin{aligned} \left( \frac{\alpha }{\eta }\right) ^{n} \iint _{V_1\cup V_2 \cup V_3} |\nabla u|^2 \delta (X)^2 \mathrm{d}X< \delta < \left( \dfrac{S_r u(Q) + \lambda }{2} \right) ^2 -\lambda ^2. \end{aligned}$$

Therefore, we conclude from (8.1) and (8.2) that

$$\begin{aligned}&\iint _{\Gamma _r(P){\setminus }B(P,\eta ) } |\nabla u|^2 \delta (X)^{2-n} \mathrm{d}X \\&\quad \ge \iint _{\Gamma _r(Q){\setminus }B(Q,\eta ) } |\nabla u|^2 \delta (X)^{2-n} \mathrm{d}X - \iint _{D } |\nabla u|^2 \delta (X)^{2-n} \mathrm{d}X\\&\quad> \left( \frac{ S_r u(Q) + \lambda }{2}\right) ^2 - \delta \\&\quad >\lambda ^2. \end{aligned}$$

Hence, $S_r u(P) \ge \left( \iint _{\Gamma _r(P){\setminus }B(P,\eta ) } |\nabla u|^2 \delta (X)^{2-n} \mathrm{d}X \right) ^{1/2} >\lambda$, for all $P\in B(Q,\epsilon \eta )\cap \partial \Omega$. This finishes the proof that $\big \{Q\in \partial \Omega : S_r u(Q)>\lambda \big \}$ is open in $\partial \Omega$. $\square$

1.2 Proof of Lemma 6.5

We prove the estimate by duality: let r be the conjugate of q / 2, namely $1/r+2/q = 1$, then

$$\begin{aligned} \left( \int _{\Delta } |S_r^{\overline{\alpha }} u(Q) |^q \mathrm{d}\sigma (Q)\right) ^{2/q} = \sup \left\{ \int _{\Delta } |S_r^{\overline{\alpha }}u(Q)|^2 \psi (Q) \mathrm{d}\sigma (Q): \Vert \psi \Vert _{L^r(\Delta )} = 1\right\} . \end{aligned}$$

(8.4)

Recall $\Delta =\Delta (Q_0,r)$. Extending $\psi$ to all of $\partial \Omega$ by setting it to zero outside of $\Delta$. For any X, let $Q_X\in \partial \Omega$ be a boundary point such that $|X-Q_X| = \delta (X)$. By Fubini’s theorem,

$$\begin{aligned}&\int _{\Delta } |S_r^{\overline{\alpha }}u(Q)|^2 \psi (Q) \mathrm{d}\sigma (Q) \nonumber \\&\quad = \int _{\Delta } \left( \iint _{\Gamma ^{\overline{\alpha }}_r(Q)} |\nabla u|^2 \delta (X)^{2-n} \mathrm{d}X \right) \psi (Q) \mathrm{d}\sigma (Q) \nonumber \\&\quad \le \iint _{B(Q_0,2r)\cap \Omega } |\nabla u|^2 \delta (X)^{2-n} \int _{|Q-Q_X|\le (\overline{\alpha }+1)\delta (X)} \psi (Q) \mathrm{d}\sigma (Q) \mathrm{d}X \nonumber \\&\quad \lesssim \iint _{B(Q_0,2r)\cap \Omega } |\nabla u|^2 \delta (X) A_{(\overline{\alpha }+1)\delta (X)}\psi (Q_X) \mathrm{d}X, \end{aligned}$$

(8.5)

where $A_{s}\psi (Q)$ is defined as $A_{s}\psi (Q) = \frac{1}{s^{n-1}} \int _{\Delta (Q,s)} \psi (P) \mathrm{d}\sigma (P)$. Let $\beta >1$, simply calculations show that

$$\begin{aligned}&A_{s} \left( A_{\beta s} \psi \right) (Q) \\&\quad = \frac{1}{s^{n-1}} \int _{\Delta (Q, s)} \left( \frac{1}{(\beta s)^{n-1}} \int _{\Delta (P,\beta s)} \psi (P') \mathrm{d}\sigma (P') \right) \mathrm{d}\sigma (P) \\&\quad \ge \frac{1}{s^{n-1}} \int _{\Delta (Q, s)} \left( \frac{1}{(\beta s)^{n-1}} \int _{\Delta (Q,(\beta -1)s)} \psi (P') \mathrm{d}\sigma (P') \right) \mathrm{d}\sigma (P) \\&\quad \gtrsim \left( \frac{(\beta -1)s}{\beta s}\right) ^{n-1} A_{(\beta -1)s} \psi (Q). \end{aligned}$$

Let $s=(\alpha -1) \delta (X)$, $\beta -1=\left( \overline{\alpha }+1\right) /\left( \alpha -1\right)$, then

$$\begin{aligned} A_{(\overline{\alpha }+1)\delta (X)}\psi (Q) \lesssim _{\alpha ,\overline{\alpha }} A_{(\alpha -1)\delta (X)}\left( A_{\beta s}\psi \right) (Q) \lesssim A_{(\alpha -1) \delta (X)}M\psi (Q). \end{aligned}$$

For the last inequality, we use $|A_{\beta s}\psi (Q)| \le C \left( M\psi (Q)\right)$, where $M\psi$ is the Hardy–Littlewood maximal function of $\psi$ with respect to $\sigma$, and the constant C only depend on the Ahlfors regularity of $\sigma$. Thus it follows from (8.5) that

$$\begin{aligned}&\int _{\Delta } |S_r^{\overline{\alpha }}u(Q)|^2 \psi (Q) \mathrm{d}\sigma (Q)\nonumber \\&\quad \lesssim \iint _{B(Q_0,2r)\cap \Omega } |\nabla u|^2 \delta (X) A_{(\alpha -1)\delta (X)}M\psi (Q_X) \mathrm{d}X \nonumber \\&\quad \lesssim \iint _{B(Q_0,2r)\cap \Omega } |\nabla u|^2 \delta (X)^{2-n} \left( \int _{\Delta (Q_X, (\alpha -1)\delta (X))}M\psi (Q)\mathrm{d}\sigma (Q)\right) \mathrm{d}X . \end{aligned}$$

(8.6)

By switching the order of integration, we can bound the right hand side by:

$$\begin{aligned}&\int _{\Delta } |S_r^{\overline{\alpha }}u(Q)|^2 \psi (Q) \mathrm{d}\sigma (Q) \nonumber \\&\quad \lesssim \int _{\Delta (Q_0, 2(\alpha +1)r)} M\psi (Q) \iint _{\Gamma _{2\alpha r} (Q)} |\nabla u|^2\delta (X)^{2-n} \mathrm{d}X \mathrm{d}\sigma (Q) \nonumber \\&\quad = \int _{\Delta (Q_0, 2(\alpha +1)r)} M\psi (Q) |S_{2\alpha r} u(Q)|^2 \mathrm{d}\sigma (Q) \nonumber \\&\quad \le \Vert M\psi \Vert _{L^r(\Delta (Q_0,2(\alpha +1)r))} \left( \int _{\Delta (Q_0, 2(\alpha +1)r)}|S_{2\alpha r} u(Q)|^q \mathrm{d}\sigma (Q)\right) ^{2/q}. \end{aligned}$$

(8.7)

Since $1<r<\infty$, we have

$$\begin{aligned} \Vert M\psi \Vert _{L^r(\Delta (Q_0,2(\alpha +1)r))} \le C \Vert \psi \Vert _{L^r(\Delta (Q_0,2(\alpha +1)r))} = C \Vert \psi \Vert _{L^r(\Delta )}= C. \end{aligned}$$

(8.8)

By (8.7), (8.8) and the definition (8.4), we conclude

$$\begin{aligned} \int _{\Delta } |S_r^{\overline{\alpha }} u(Q) |^q \mathrm{d}\sigma (Q)&\le C\int _{\Delta (Q_0, 2(\alpha +1)r)}|S_{2\alpha r} u(Q)|^q \mathrm{d}\sigma (Q)\\&\le C\int _{\Delta (Q_0, 2(\alpha +1)r)}|S_{2(\alpha +1) r} u(Q)|^q \mathrm{d}\sigma (Q) . \end{aligned}$$

This finishes the proof of Lemma 6.5. $\square$