Proposition 1
Let
\(\theta > 0\)
and
\(1 \leq s <\infty \). If
\(h \in \operatorname{Lip}_{\nu }^{ \theta }(G)\), then there exists a positive constant
C
such that, for all
\(B = B(x, r)\)
with
\(x\in G\)
and
\(r > 0\),
$$ \biggl( \frac{1}{\vert B\vert } \int_{B}\bigl\vert h(y)-h_{B}\bigr\vert ^{s} \,\mathrm{d}y \biggr) ^{1/s}\leq C\Vert h\Vert _{\operatorname{Lip} _{\nu } ^{\theta }(G)}r^{\nu } \biggl( 1+\frac{r}{\rho (x)} \biggr) ^{(l_{0}+1)\theta }. $$
Proof
Since \(h \in \operatorname{Lip}_{\nu }^{\theta }(G)\), then
$$\begin{aligned} & \biggl( \frac{1}{\vert B(x,r)\vert } \int_{ B(x,r)}\bigl\vert h(y)-h _{B}\bigr\vert ^{s} \,\mathrm{d}y \biggr) ^{\frac{1}{s}} \\ &\quad \leq \biggl( \frac{1}{\vert B(x,r)\vert } \int_{ B(x,r)} \bigl\vert h(y)-h(x)\bigr\vert ^{s} \,\mathrm{d}y \biggr) ^{\frac{1}{s}}+ \biggl( \frac{1}{ \vert B(x,r)\vert } \int_{ B(x,r)}\bigl\vert h(x)-h_{B}\bigr\vert ^{s} \,\mathrm{d}y \biggr) ^{\frac{1}{s}} \\ &\quad \leq 2C \biggl( \frac{1}{\vert B(x,r)\vert } \int_{ B(x,r)} \biggl\vert \Vert h\Vert _{\operatorname{Lip} _{\nu }^{\theta }(G)}d^{\nu }(x,y) \biggl( 1+\frac{d(x,y)}{ \rho (x)}+\frac{d(x,y)}{\rho (y)} \biggr) ^{\theta }\biggr\vert ^{s} \, \mathrm{d}y \biggr) ^{\frac{1}{s}} \\ &\quad \leq C\Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)}r^{\nu } \biggl( 1+ \frac{r}{ \rho (x)}+\frac{r}{\rho (y)} \biggr) ^{\theta } \\ &\quad \leq C\Vert h\Vert _{\operatorname{Lip} _{\nu }^{\theta }(G)}r^{\nu } \biggl( 1+ \frac{r}{ \rho (x)} \biggr) ^{ ( l_{0}+1 ) \theta }, \end{aligned}$$
where we have used Lemma 6 in the penultimate inequality. □
Similar to the proof of Proposition 1, we immediately get the following.
Lemma 10
Let
\(h \in \operatorname{Lip}_{\nu }^{\theta }(G)\), \(B = B(x, r)\)
and
\(s \geq 1\). Then there exists a positive constant
C
such that, for all
\(k\in N\),
$$ \biggl( \frac{1}{\vert 2^{k}B\vert } \int_{2^{k}B}\bigl\vert h(y)-h _{B}\bigr\vert ^{s} \,\mathrm{d}y \biggr) ^{\frac{1}{s}}\leq C\Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)}2^{k\nu }r^{\nu } \biggl( 1+\frac{2^{k}r}{ \rho (x)} \biggr) ^{ ( l_{0}+1 ) \theta }. $$
Proposition 2
Let
\(W\in B_{q_{1}}\)
for
\(q_{1} \geq \frac{D}{2}\). Let
$$ \frac{1}{q_{2}}= \textstyle\begin{cases} 1, & \textit{if } q_{1}\geq D, \\ 1 - \frac{1}{q_{1}}+\frac{1}{D}, & \textit{if } \frac{D}{2}\leq q_{1}< D. \end{cases} $$
Then there exists
\(C>0\)
such that, for any
\(f \in C_{0}^{\infty }(G)\),
$$ \bigl\Vert \mathcal{R}f(x)\bigr\Vert _{L^{p}(G)} \leq C \Vert f \Vert _{L^{p}(G)}, $$
where
\(1< p< q'_{2}\).
Proposition 3
Let
\(W\in B_{q_{1}}\)
for
\(q_{1} \geq \frac{D}{2}\). Let
$$ \frac{1}{q_{2}}= \textstyle\begin{cases} 1,& \textit{if } q_{1}\geq D, \\ 1 - \frac{1}{q_{1}}+\frac{1}{D},&\textit{if } \frac{D}{2}\leq q_{1}< D. \end{cases} $$
Then there exists
\(C>0\)
such that, for any
\(f \in C_{0}^{\infty }(G)\),
$$ \bigl\Vert \tilde{\mathcal{R}}f(x)\bigr\Vert _{L^{p}(G)} \leq C \Vert f\Vert _{L^{p}(G)}, $$
where
\(q_{2}< p<\infty \).
For the proofs of Proposition 2 and Proposition 3, one can refer to [3].
Proposition 4
There exists a sequence of points
\(\{x_{k}\}_{k=1}^{\infty }\)
in
G, so that the set of critical balls
\(Q_{k} = B(x_{k}, \rho (x_{k}))\), \(k \geq 1\), satisfies
-
(i)
\(\bigcup_{k} Q_{k} = G\);
-
(ii)
There exists
N
such that, for every
\(k\in N\), \(\sharp \{j\colon 4Q_{j}\cap 4Q_{k}\neq \varnothing \}\leq N\).
Proposition 5
For
\(1 < p < \infty \), there exist positive constants
C, α
and
β
such that if
\(\{Q_{k}\}_{k=1}^{\infty }\)
is a sequence of balls as in Proposition
4, then
$$ \int_{G}\bigl\vert M_{\rho ,\alpha }f(x)\bigr\vert ^{p}\,\mathrm{d}x \leq C \int_{G}\bigl\vert M_{\rho ,\beta }^{\sharp }f(x) \bigr\vert ^{p}\,\mathrm{d}x+C\sum_{k} \vert Q_{k}\vert \biggl( \frac{1}{\vert Q _{k}\vert } \int_{2Q_{k}}\bigl\vert f(x)\bigr\vert \,\mathrm{d}x \biggr) ^{p} $$
for all
\(f\in L_{\mathrm{loc}}^{1}(G)\).
The above propositions have been proved in [9] and [10] in the case of a homogeneous space, respectively.
Lemma 11
Let
\(W\in B_{q_{1}}\)
for
\(q_{1} \geq \frac{D}{2}\), and let
$$ \frac{1}{q_{2}}= \textstyle\begin{cases} 1,& \textit{if }q_{1}\geq D, \\ 1 - \frac{1}{q_{1}}+\frac{1}{D}, & \textit{if }\frac{D}{2}\leq q_{1}< D \end{cases} $$
and
\(h\in \operatorname{Lip}_{\nu }^{\theta }(G)\). Then, for
\(q_{2}< m<\infty \), there exists a positive constant
C
such that
$$ \frac{1}{\vert Q\vert } \int_{Q}\bigl\vert [h, \tilde{\mathcal{R}}]f(x)\bigr\vert \,\mathrm{d}x\leq C\Vert h \Vert _{\operatorname{Lip}_{\nu } ^{\theta }(G)}\inf_{y\in Q} \bigl\{ M_{m\nu }\bigl(\vert f\vert ^{m}\bigr) (y)\bigr\} ^{ \frac{1}{m}} $$
holds true for all
\(f\in L_{\mathrm{loc}}^{m}(G)\)
and every ball
\(Q=B(x_{0}, \rho (x_{0}))\), where
\(M_{m\nu }\)
is a fractional maximal operator.
Proof
Throughout the proof of the lemma, we always assume \(\frac{D}{2}\leq q_{1}< D\). Let \(f\in L^{p}(G)\) and \(Q=B(x_{0},\rho (x _{0}))\). For
$$ [h,\tilde{\mathcal{R}}]f=(h-h_{Q})\tilde{ \mathcal{R}}f- \tilde{\mathcal{R}}\bigl((h-h_{Q})f\bigr), $$
(21)
we need to consider the average on Q for each term. By the Hölder inequality with \(m>q_{2}\) and Proposition 1,
$$\begin{aligned} \frac{1}{\vert Q\vert } \int_{Q}\bigl\vert (h-h_{Q}) \tilde{ \mathcal{R}}f(x)\bigr\vert \,\mathrm{d}x &\leq \biggl( \frac{1}{ \vert Q\vert } \int_{Q}\bigl\vert \bigl( h(x)-h_{Q} \bigr) \bigr\vert ^{m^{\prime}} \,\mathrm{d}x \biggr) ^{1/m^{\prime}} \biggl( \frac{1}{\vert Q\vert } \int_{Q}\bigl\vert \tilde{\mathcal{R}}f(x)\bigr\vert ^{m}\, \mathrm{d}x \biggr) ^{1/m} \\ &\leq C\Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)} \bigl( \rho (x_{0}) \bigr) ^{ \nu } \biggl( \frac{1}{\vert Q\vert } \int_{Q}\bigl\vert \tilde{\mathcal{R}}f(x)\bigr\vert ^{m}\,\mathrm{d}x \biggr) ^{1/m}. \end{aligned}$$
If we write \(f = f_{1} + f_{2}\) with \(f_{1} = f_{\chi_{2Q}}\), due to Proposition 3, we get
$$\begin{aligned} \bigl(\rho (x_{0})\bigr)^{\nu } \biggl( \frac{1}{\vert Q\vert } \int_{Q} \bigl\vert \tilde{\mathcal{R}}f_{1}(x) \bigr\vert ^{m}\,\mathrm{d}x \biggr) ^{1/m} &\leq C\bigl( \rho (x_{0})\bigr)^{\nu } \biggl( \frac{1}{\vert 2Q\vert } \int_{2Q}\bigl\vert f(x)\bigr\vert ^{m}\, \mathrm{d}x \biggr) ^{1/m} \\ & \leq C \inf_{y\in Q} \bigl\{ M_{m\nu }\bigl(\vert f \vert ^{m}\bigr) (y) \bigr\} ^{\frac{1}{m}} \end{aligned}$$
(22)
for \(x\in Q\), and using (17) in Lemma 9, we split \(\tilde{\mathcal{R}}f_{2}(x)\) into two parts
$$\begin{aligned} \bigl\vert \tilde{\mathcal{R}}f_{2}(x)\bigr\vert &=\biggl\vert \int _{d(x_{0},z) >2\rho (x_{0})}\tilde{\mathcal{\mathcal{K}}}(x,z)f(z) \,\mathrm{d}z \biggr\vert \leq C\bigl\{ I_{1}(x)+I_{2}(x)\bigr\} , \end{aligned}$$
where
$$ I_{1}(x)= \int_{ d(x_{0},z) >2\rho (x_{0})}\frac{\vert f(z)\vert }{V(d(x,z))(1+d(x,z) / \rho (x))^{N}} \,\mathrm{d}z $$
and
$$ I_{2}(x)= \int_{ d(x_{0},z) >2\rho (x_{0})}\frac{d(x,z)\vert f(z)\vert }{V(d(x,z))(1+d(x,z) / \rho (x))^{N}} \int _{B(z,\frac{d(x,z)}{4})}\frac{d(u,z)W(u)}{V(d(u,z))} \, \mathrm{d}u \,\mathrm{d}z. $$
To deal with \(I_{1}(x)\), noting that \(\rho (x) \sim \rho (x_{0})\) and \(d(z,x) \sim d(z,x_{0})\), we split \(d(z,x_{0})>2 \rho (x_{0})\) into annuli to obtain
$$\begin{aligned} \bigl( \rho (x_{0}) \bigr) ^{\nu }I_{1}(x) &= \bigl( \rho (x_{0}) \bigr) ^{\nu } \int_{ d(x_{0},z) >2\rho (x_{0})}\frac{\vert f(z)\vert }{V(d(x,z))(1+d(x,z) / \rho (x))^{N}} \,\mathrm{d}z \\ & \leq C \sum_{k \geq 1}\frac{2^{-Nk} ( \rho (x_{0}) ) ^{ \nu }}{V(2^{k}\rho (x_{0}))} \int_{d(x_{0},z) < 2^{k}\rho (x _{0})} \bigl\vert f(z)\bigr\vert \,\mathrm{d}z \\ & \leq C \inf_{y\in Q} \bigl\{ M_{m\nu } \bigl( \vert f \vert ^{m} \bigr) ( y ) \bigr\} ^{\frac{1}{m}}. \end{aligned}$$
(23)
Secondly, we consider the term \(I_{2}(x)\). We have, for \(x \in Q\),
$$\begin{aligned} \bigl( \rho (x_{0}) \bigr) ^{\nu }I_{2}(x) &= \bigl( \rho (x_{0}) \bigr) ^{\nu } \int_{ d(x_{0},z) >2\rho (x_{0})}\frac{d(x,z)\vert f(z)\vert }{V(d(x,z))(1+d(x,z) / \rho (x))^{N}} \\ &\quad {}\times\int _{B(z,\frac{d(x,z)}{4})}\frac{d(u,z)W(u)}{V(d(u,z))} \, \mathrm{d}u \,\mathrm{d}z \\ & \leq C \bigl( \rho (x_{0}) \bigr) ^{\nu } \int _{ d(x_{0},z) >2\rho (x_{0})}\frac{d(x,z)\vert f(z)\vert }{V(d(x,z))(1+d(x,z) / \rho (x))^{N}} \\ &\quad {}\times \int_{B(z,4d(x_{0},z))} \frac{d(u,z)W(u)}{V(d(u,z))} \,\mathrm{d}u \,\mathrm{d}z \\ & \leq C \bigl( \rho (x_{0}) \bigr) ^{\nu } \sum _{k \geq 1}\frac{2^{-Nk}2^{k} \rho (x_{0})}{V(2^{k}\rho (x_{0}))} \int _{d(x_{0},z)< 2^{k+1}\rho (x_{0})} \bigl\vert f(z)\bigr\vert \\ &\quad {}\times \int _{B(z,2^{k+3}d(x_{0},z))}\frac{d(u,z)W(u)}{V(d(u,z))} \, \mathrm{d}u \,\mathrm{d}z \\ & \leq C \bigl( \rho (x_{0}) \bigr) ^{\nu } \sum _{k \geq 1}\frac{2^{-Nk}2^{k} \rho (x_{0})}{V(2^{k}\rho (x_{0}))} \\ &\quad {}\times \int _{d(x_{0},z)< 2^{k+1}\rho (x_{0})} \bigl\vert f(z)\bigr\vert \mathcal{I}_{1}(W_{\chi_{B(x_{0},2^{k}\rho (x_{0}))}}) (z) \, \mathrm{d} z. \end{aligned}$$
Let \(q_{2}< m< D\). Using the Hölder inequality and the boundedness of the fractional integral \(\mathcal{I}_{1} \colon L^{m'} \mapsto L^{q _{1}}\) with
$$ \frac{1}{q_{1}}= \frac{1}{m^{\prime}}-\frac{1}{\beta }, $$
where
$$ \beta = \textstyle\begin{cases} d, & \textit{if } 2^{k}\rho (x_{0})< 1, \\ D, & \textit{if } 2^{k}\rho (x_{0}) \geq 1 \end{cases} $$
(cf. Theorem 1.6 in [11]), we obtain
$$\begin{aligned} &\int_{d(x_{0},z)< 2^{k+1}\rho (x_{0})} \bigl\vert f(z)\bigr\vert \mathcal{I}_{1}(W \chi_{B(x_{0},2^{k}\rho (x_{0}))}) (z) \,\mathrm{d}z \\ &\quad \leq \Vert f\chi_{B(x_{0},2^{k}\rho (x_{0}))}\Vert _{m}\bigl\Vert \mathcal{I} _{1}(W\chi_{B(x_{0},2^{k}\rho (x_{0}))}) \bigr\Vert _{m^{\prime}} \\ &\quad \leq \Vert f\chi_{B(x_{0},2^{k}\rho (x_{0}))}\Vert _{m}\Vert W \chi_{B(x_{0},2^{k}\rho (x_{0}))}\Vert _{q_{1}}. \end{aligned}$$
Since \(W\in B_{q_{1}}\), we obtain
$$\begin{aligned} \Vert W\chi_{B(x_{0},2^{k}\rho (x_{0}))}\Vert _{q_{1}} &\leq C \bigl(V \bigl(2^{k}\rho (x_{0})\bigr) \bigr)^{-\frac{1}{q_{1}^{\prime}}} \int_{B(x_{0},2^{k}\rho (x_{0}))} W \\ &\leq C\bigl(2^{k}\rho (x_{0})\bigr)^{-2} \bigl(V\bigl(2^{k}\rho (x_{0})\bigr) \bigr)^{1-\frac{1}{q _{1}^{\prime}}} \frac{(2^{k}\rho (x_{0}))^{2}}{V(2^{k}\rho (x_{0}))} \int_{B(x_{0},2^{k}\rho (x_{0}))} W \\ &\leq C\bigl(2^{k}\bigr)^{l_{1}-2} \bigl(V\bigl(2^{k} \rho (x_{0})\bigr) \bigr)^{1-\frac{1}{q _{1}^{\prime}}}\bigl(\rho (x_{0}) \bigr)^{-2}, \end{aligned}$$
where in the last two inequalities we have used doubling measure and the definition of ρ, respectively. Therefore,
$$ I_{2}(x)\leq V\bigl(\rho (x_{0})\bigr)^{-1/q_{1}^{\prime}-1}\sum _{k\geq 1}2^{k(-N+l _{1}-1-d/q_{1}^{\prime})}\Vert f\chi_{B(x_{0},2^{k}\rho (x_{0}))} \Vert _{m}. $$
Finally, observing that
$$ \Vert f\chi_{B(x_{0},2^{k}\rho (x_{0}))}\Vert _{m}\leq \bigl(V \bigl(2^{k} \rho (x_{0})\bigr)\bigr)^{\frac{1}{m}}\inf _{y\in Q}M_{m}f(y) $$
and using that \(\frac{1}{q_{1}^{\prime}}-\frac{1}{m}= \frac{1}{D}\) or \(\frac{1}{d}\), we have
$$ \bigl(\rho (x_{0})\bigr)^{\nu }I_{2}(x)\leq C\inf _{y\in Q}\bigl\{ M_{m\nu }\bigl(\vert f\vert ^{m}\bigr) (y)\bigr\} ^{\frac{1}{m}} $$
by choosing N large enough.
So far, we have solved the term \((h-h_{Q})\mathcal{R}f\), now we want to control \(\int \mathcal{R}[(h-h_{Q})f]\,\mathrm{d}x\) by the term \(C \inf_{y\in Q}\{M_{m\nu }(\vert f \vert ^{m})(y)\}^{\frac{1}{m}}\). We still split \(f=f_{1}+f_{2}\). Choose \(q_{2}<\tilde{m}<m\) and set \(t=\frac{\tilde{m}m}{m-\tilde{m}}\). Using the boundedness of \(\tilde{\mathcal{R}}\) on \(L^{\tilde{m}}(G)\) and the Hölder inequality, we get
$$\begin{aligned} &\frac{1}{\vert Q\vert } \int_{Q}\bigl\vert \tilde{\mathcal{R}}(h-h _{Q})f_{1}(x)\bigr\vert \,\mathrm{d}x \\ &\quad \leq \biggl( \frac{1}{\vert Q\vert } \int_{Q}\bigl\vert \tilde{\mathcal{R}}(h-h_{Q})f_{1}(x) \bigr\vert ^{\tilde{m}} \,\mathrm{d}x \biggr) ^{1/\tilde{m}} \\ & \quad \leq C \biggl( \frac{1}{\vert Q\vert } \int_{Q}\bigl\vert (h-h _{Q})f(x)\bigr\vert ^{\tilde{m}} \,\mathrm{d}x \biggr) ^{1/\tilde{m}} \\ &\quad \leq C \biggl( \frac{1}{\vert Q \vert } \int_{2Q}\bigl\vert f(x)\bigr\vert ^{m} \, \mathrm{d}x \biggr) ^{1/m} \biggl( \frac{1}{\vert Q \vert } \int_{2Q}\bigl\vert h(x)-h _{Q}\bigr\vert ^{t}\,\mathrm{d}x \biggr) ^{1/t} \\ &\quad \leq C \biggl( \frac{(\rho (x_{0}))^{m\nu }}{\vert Q \vert } \int_{2Q} \bigl\vert f(x)\bigr\vert ^{m} \, \mathrm{d}x \biggr) ^{1/m} \Vert h\Vert _{\operatorname{Lip} _{\nu }^{\theta }(G)} \biggl( 1+ \frac{\rho (x_{0})}{\rho (x_{0})} \biggr) ^{(l_{0}+1)\theta } \\ & \quad \leq C \Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)}\inf_{y\in Q} \bigl\{ M_{m\nu } \bigl( \vert f \vert ^{m} \bigr) ( y ) \bigr\} ^{ \frac{1}{m}}, \end{aligned}$$
where we have applied Proposition 1 to the last but one inequality. Similarly, for \(x\in Q\) and using (17) in Lemma 9, we have
$$\begin{aligned} \bigl\vert \tilde{R}(h-h_{Q})f_{2}(x) \bigr\vert &= \biggl\vert \int _{d(x_{0},z)>2\rho (x_{0})} \tilde{\mathcal{\mathcal{K}}}(x,z) \bigl(h(z)-h _{Q}\bigr)f(z)\,\mathrm{d}z\biggr\vert \leq C \bigl\{ \tilde{I}_{1}(x)+\tilde{I}_{2}(x) \bigr\} , \end{aligned}$$
where
$$ \tilde{I}_{1}(x)= \int_{ d(x_{0},z) >2\rho (x_{0})}\frac{\vert (h-h_{Q})f(z)\vert }{V(d(x,z))(1+d(x,z) / \rho (x))^{N}} \, \mathrm{d}z $$
and
$$ \tilde{I}_{2}(x)= \int_{ d(x_{0},z) >2\rho (x_{0})}\frac{\vert (h-h_{Q})f(z)\vert }{V(d(x,z))(1+d(x,z) / \rho (x))^{N}} \int _{B(z,\frac{d(x,z)}{4})}\frac{d(u,z)W(u)}{V(d(u,z))} \, \mathrm{d}u \,\mathrm{d}z. $$
We start by observing that for \(1\leq \tilde{m}< m\), \(t= \frac{\tilde{m}m}{m- \tilde{m}}\), and by Lemma 6,
$$\begin{aligned} &\bigl\Vert \bigl[(h-h_{Q})f\bigr] { \chi_{B(x_{0},2^{k}\rho (x_{0}))}}\bigr\Vert _{ \tilde{m}} \\ & \quad \leq \Vert f_{\chi_{B(x_{0},2^{k}\rho (x_{0}))}}\Vert _{m}\bigl\Vert (h-h _{Q}){\chi_{B(x_{0},2^{k}\rho (x_{0}))}}\bigr\Vert _{t} \\ & \quad \leq C \biggl( \int_{B(x_{0},2^{k}\rho (x_{0}))}\bigl\vert f(y)\bigr\vert ^{m}\, \mathrm{d}y \biggr) ^{1/m} \Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)} \bigl(2^{k}\rho (x_{0})\bigr)^{ \nu } \bigl(1+2^{k}\bigr)^{(l_{0}+1)\theta }V^{1/t}\bigl(2^{k} \rho (x_{0})\bigr) \\ & \quad \leq C 2^{k(l_{0}+1)\theta }V^{1/\tilde{m}}\bigl(2^{k}\rho (x_{0})\bigr)\Vert h \Vert _{\operatorname{Lip} _{\nu }^{\theta }(G)}\inf _{y\in Q} \bigl\{ M_{m\nu } \bigl( \vert f \vert ^{m} \bigr) ( y ) \bigr\} ^{\frac{1}{m}}. \end{aligned}$$
(24)
For \(\tilde{I}_{1}(x)\), using (24) with \(\tilde{m}=1\), we have
$$\begin{aligned} \tilde{I}_{1}(x) & \leq C\sum_{k \geq 1} \frac{2^{-Nk}}{V(2^{k}\rho (x _{0}))} \int_{d(x_{0},z)>2^{k}\rho (x_{0})}\bigl\vert h(z)-h_{Q}\bigr\vert \bigl\vert f(z) \bigr\vert \,\mathrm{d}z \\ & \leq C \Vert h \Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)}\inf_{y\in Q} \bigl\{ M_{m \nu } \bigl( \vert f \vert ^{m} \bigr) ( y ) \bigr\} ^{\frac{1}{m}} \sum_{k \geq 1}2^{k(-N+(l_{0}+1)\theta )} \\ & \leq C \Vert h \Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)}\inf_{y\in Q} \bigl\{ M_{m \nu } \bigl( \vert f \vert ^{m} \bigr) ( y ) \bigr\} ^{\frac{1}{m}} \end{aligned}$$
if we choose N sufficiently large.
To deal with \(\tilde{I}_{2}(x)\), we discuss as in the estimate for \(I_{2}(x)\) with \((h-h_{Q})f\) instead of f and m̃ and \(\tilde{q_{1}}\) instead of m and \(q_{1}\), but we cannot avoid to discuss the different cases where \(2^{k}\rho (x_{0}) \geq 1\) and \(2^{k}\rho (x_{0})< 1\). Let
$$ \frac{1}{\tilde{m}}= \textstyle\begin{cases} \frac{1}{\tilde{q_{1}}}-\frac{1}{d}, & \textit{if } 2^{k}\rho (x_{0})< 1, \\ \frac{1}{\tilde{q_{1}}}-\frac{1}{D}, & \textit{if } 2^{k}\rho (x_{0}) \geq 1. \end{cases} $$
Using (24), we similarly have
$$\begin{aligned} \tilde{I}_{2}(x) \leq C \Vert h \Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)}\inf _{y\in Q} \bigl\{ M_{m\nu } \bigl( \vert f \vert ^{m} \bigr) ( y ) \bigr\} ^{ \frac{1}{m}}, \end{aligned}$$
where we choose N large enough to ensure the above series converges. □
Lemma 12
Let
\(\tilde{\mathcal{R}}=(-\Delta +W)^{-\frac{1}{2}}\nabla \)
be the adjoint operator of the Riesz transform
\(\mathcal{R}\). Then there exists
\(C>0\)
such that, for any
\(f\in L_{\mathrm{loc}}^{m}(G)\)
and
\(h \in \operatorname{Lip}_{\nu } ^{\theta }(G)\),
$$ M_{\rho ,\alpha }^{\sharp }\bigl([h,\tilde{R}]f\bigr) (x) \leq C \Vert h \Vert _{\operatorname{Lip}_{ \nu }^{\theta }(G)}\bigl( \bigl\{ M_{m\nu } \bigl( \vert f \vert ^{m} \bigr) ( y ) \bigr\} ^{\frac{1}{m}}+\bigl\{ M_{m\nu}\bigl(| \tilde{\mathcal{R}}f| ^{m}\bigr)(x)\bigr\} ^{\frac{1}{m}}\bigr). $$
(25)
Proof
Let \(f\in L_{\mathrm{loc}}^{m}(G)\), \(x\in G\) and a ball \(B=B(x_{0},r)\) with \(x\in B\) and \(r<\epsilon \rho (x_{0})\), \(\epsilon >0\), we need to control \(J=\frac{1}{\vert B \vert }\int_{B}\vert [h,\tilde{\mathcal{R}}]f(y)-c \vert \,\mathrm{d}y\) by the right-hand side of (25) for some constant c, which will be designated later. Let \(f=f_{1}+f_{2}\), where \(f_{1}=f_{\chi_{2B}}\) and \(f_{2}=f-f_{1}\). Then
$$\begin{aligned}{} [h,\tilde{\mathcal{R}}]f &=[h-h_{2B},\tilde{\mathcal{R}}]f \\ &=(h-h_{2B}) \tilde{\mathcal{R}}f-\tilde{\mathcal{R}}(h-h_{2B})f_{1}- \tilde{\mathcal{R}}(h-h_{2B})f_{2} \\ & \doteq A_{1}f+A_{2}f+A_{3}f. \end{aligned}$$
Take \(c=\int_{d(x_{0},z)\geq 2r}\tilde{\mathcal{K}}(x_{0},z)(h(z)-h _{2B})f(z)\,\mathrm{d}z\). Then we have
$$\begin{aligned} J & \leq \frac{1}{\vert B \vert } \int_{B}\bigl\vert A_{1}f(y)\bigr\vert \, \mathrm{d}y + \frac{1}{\vert B \vert } \int_{B}\bigl\vert A_{2}f(y)\bigr\vert \, \mathrm{d}y+ \frac{1}{\vert B \vert } \int_{B}\bigl\vert A_{3}f(y)-c\bigr\vert \, \mathrm{d}y \\ & \doteq J_{1}+J_{2}+J_{3}. \end{aligned}$$
At first, we consider \(J_{1}\). Note that \(d(x,x_{0})\leq r<\epsilon \rho (x_{0})\) implies \(\rho (x)\sim \rho (x_{0})\). By the Hölder inequality, Proposition 1 and Proposition 3, we have
$$\begin{aligned} J_{1} &= \frac{1}{\vert B \vert } \int_{B}\bigl\vert A_{1}f(y)\bigr\vert \, \mathrm{d}y \\ &= \frac{1}{\vert B \vert } \int_{B}\bigl\vert \bigl(h(y)-h_{2B}\bigr) \tilde{\mathcal{R}}f(y)\bigr\vert \,\mathrm{d}y \\ &\leq \frac{C}{\vert B \vert } \biggl[ \int_{B}\bigl\vert h(y)-h_{2B} \bigr\vert ^{\frac{m}{m-1}} \,\mathrm{d}y \biggr] ^{\frac{m-1}{m}} \biggl[ \int_{B}\bigl\vert \tilde{\mathcal{R}}f(y) \bigr\vert ^{m}\,\mathrm{d}y \biggr] ^{\frac{1}{m}} \\ &\leq \frac{C}{\vert B \vert } \biggl[ \int_{2B}\bigl\vert h(y)-h_{2B} \bigr\vert ^{ \frac{m}{m-1}}\,\mathrm{d}y \biggr] ^{\frac{m-1}{m}} \biggl[ \int _{B}\bigl\vert \tilde{\mathcal{R}}f(y) \bigr\vert ^{m}\,\mathrm{d}y \biggr] ^{\frac{1}{m}} \\ &\leq \frac{C}{\vert B \vert }\Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)} \biggl( 1+ \frac{2r}{ \rho (x_{0})} \biggr) ^{(l_{0}+1)\theta } r^{\nu }\vert B\vert ^{ \frac{m-1}{m}} \biggl[ \int_{B}\bigl\vert \tilde{\mathcal{R}}f(y) \bigr\vert ^{m}\, \mathrm{d}y \biggr] ^{\frac{1}{m}} \\ &\leq \frac{C}{\vert B \vert }\Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)} \biggl( 1+ \frac{2 \epsilon \rho (x_{0})}{\rho (x_{0})} \biggr) ^{(l_{0}+1)\theta } r^{ \nu }\vert B\vert ^{\frac{m-1}{m}} \biggl[ \int_{B}\bigl\vert \tilde{\mathcal{R}}f(y) \bigr\vert ^{m}\,\mathrm{d}y \biggr] ^{\frac{1}{m}} \\ &\leq C^{\prime} \Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)} \biggl( \frac{r ^{m\nu }}{\vert B \vert } \int_{B}\bigl\vert \tilde{\mathcal{R}} f(y) \bigr\vert ^{m} \, \mathrm{d}y \biggr) ^{ \frac{1}{m}} \\ & \leq C^{\prime} \Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)} \bigl\{ M_{m \nu } \bigl( \vert \tilde{\mathcal{R}} f \vert ^{m} \bigr) ( x ) \bigr\} ^{\frac{1}{m}} \end{aligned}$$
for \(m>q_{2}\).
For \(J_{2}\), by the Hölder inequality and Proposition 3,
$$\begin{aligned} J_{2} &\leq \frac{1}{\vert B \vert } \int_{B} \bigl\vert A_{2}f(y) \bigr\vert \, \mathrm{d}y \\ & \leq \biggl( \frac{1}{\vert B \vert } \int_{B} \bigl\vert A_{2}f(y) \bigr\vert ^{\tilde{m}}\, \mathrm{d}y \biggr) ^{\frac{1}{\tilde{m}}} \\ & \leq C \biggl( \frac{1}{\vert B \vert } \biggr) ^{\frac{1}{\tilde{m}}} \biggl( \int_{2B}\bigl\vert \bigl(h(y)-h_{2B}\bigr)f(y) \bigr\vert ^{\tilde{m}}\, \mathrm{d}y \biggr) ^{\frac{1}{\tilde{m}}} \\ & \leq \frac{C}{\vert B \vert ^{\frac{1}{\tilde{m}}}} \biggl( \int_{2B} \bigl\vert h(y)-h_{2B}\bigr\vert ^{\frac{m\tilde{m}}{m-\tilde{m}}} \, \mathrm{d}y \biggr) ^{\frac{1}{\tilde{m}}-\frac{1}{m}} \biggl( \int _{2B}\bigl\vert f(y) \bigr\vert ^{m}\, \mathrm{d}y \biggr) ^{\frac{1}{m}} \\ & \leq C \Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)}r^{\nu } \biggl( 1+ \frac{2r}{ \rho (x_{0})} \biggr) ^{(l_{0}+1)\theta } \biggl( \frac{1}{\vert B \vert } \biggr) ^{\frac{1}{m}} \biggl( \int_{2B}\bigl\vert f(y) \bigr\vert ^{m}\, \mathrm{d}y \biggr) ^{\frac{1}{m}} \\ & \leq C \Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)}r^{\nu } \biggl( 1+ \frac{2 \epsilon \rho (x_{0})}{\rho (x_{0})} \biggr) ^{(l_{0}+1)\theta } \biggl( \frac{1}{\vert B \vert } \biggr) ^{\frac{1}{m}} \biggl( \int_{2B}\bigl\vert f(y) \bigr\vert ^{m}\, \mathrm{d}y \biggr) ^{\frac{1}{m}} \\ & \leq C^{\prime} \Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)} \biggl( \frac{r ^{m\nu }}{\vert B \vert } \int_{2B}\bigl\vert f(y) \bigr\vert ^{m}\, \mathrm{d}y \biggr) ^{ \frac{1}{m}} \\ & \leq C^{\prime} \Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)} \biggl( \frac{(2r)^{m \nu }}{\vert 2B \vert } \int_{2B}\bigl\vert f(y) \bigr\vert ^{m}\, \mathrm{d}y \biggr) ^{ \frac{1}{m}} \\ & \leq C^{\prime} \Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)} \bigl\{ M_{m \nu } \bigl( \vert f \vert ^{m} \bigr) ( y ) \bigr\} ^{\frac{1}{m}}, \end{aligned}$$
where \(q_{2}<\tilde{m}<m\).
Finally, we consider \(J_{3}\).
Case of
\(q_{1} \geq D\): By Lemmas 9 and 6, we have
$$\begin{aligned} & \biggl( \int_{2^{k}r\leq d(x_{0},z)< 2^{k+1}r}\bigl\vert \tilde{\mathcal{K}}(y,z)- \tilde{ \mathcal{K}}(x_{0},z)\bigr\vert ^{m^{\prime}}\,\mathrm{d}z \biggr) ^{\frac{1}{m ^{\prime}}} \biggl( \int_{2^{k}r\leq d(x_{0},z)< 2^{k+1}r}\bigl\vert h(z)-h _{2B}\bigr\vert ^{s^{\prime}}\,\mathrm{d}z \biggr) ^{\frac{1}{s^{\prime}}} \\ & \leq \frac{C_{N}r^{\delta }}{ ( 1+2^{k}r/\rho (x_{0}) ) ^{N}} \biggl( \int_{2^{k}r\leq d(x_{0},z)< 2^{k+1}r}\frac{1}{d ^{m^{\prime}\delta }(x_{0},z)V^{m^{\prime}}(d(x_{0},z))}\,\mathrm{d}z \biggr) ^{\frac{1}{m^{\prime}}} \\ & \quad \times \biggl( \int_{d(x_{0},z)< 2^{k+1}r}\bigl\vert h(z)-h _{2B}\bigr\vert ^{s^{\prime}}\,\mathrm{d}z \biggr) ^{\frac{1}{s^{\prime}}} \\ & \leq \frac{C_{N}}{ ( 1+2^{k}r/\rho (x_{0}) ) ^{N}}\frac{r ^{\delta }}{(2^{k}r)^{\delta }V^{(m^{\prime}-1)/m}(2^{k}r)} \Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)}2^{k\nu }r^{\nu } \biggl( 1+\frac{2^{k}r}{ \rho (x_{0})} \biggr) ^{(l_{0}+1)\theta } V^{\frac{1}{s}}\bigl(2^{k}r\bigr) \\ & \leq \Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)}2^{k\nu }r^{\nu } \frac{C _{N}}{ ( 1+2^{k}r/\rho (x_{0}) ) ^{N-(l_{0}+1)\theta }} \frac{r ^{\delta }}{(2^{k}r)^{\delta }V^{1/m}(2^{k}r)}, \end{aligned}$$
(26)
where \(\frac{1}{m}+\frac{1}{m^{\prime}}+\frac{1}{s}=1\) and \(\frac{1}{s^{\prime}}+\frac{1}{s}=1\). Therefore, via the Hölder inequality,
$$\begin{aligned} J_{3} &= \frac{1}{\vert B \vert } \int_{B} \bigl\vert A_{3}f(y)-c \bigr\vert \, \mathrm{d}y \\ & \leq \frac{1}{\vert B \vert } \int_{B} \biggl\vert \int_{d(x_{0},z)>2r} \bigl(\tilde{\mathcal{K}}(y,z)-\tilde{ \mathcal{K}}(x_{0},z)\bigr) \bigl(h(z)-h_{2Q}\bigr)f(z) \, \mathrm{d}z\biggr\vert \,\mathrm{d}y \\ & \leq \frac{1}{\vert B \vert } \int_{B} \sum_{k=1}^{\infty } \int _{2^{k}r\leq d(x_{0},z)< 2^{k+1}r} \bigl\vert \bigl(\tilde{\mathcal{K}}(y,z)- \tilde{\mathcal{K}}(x_{0},z)\bigr) \bigl(h(z)-h_{2Q} \bigr)f(z)\bigr\vert \,\mathrm{d}z\, \mathrm{d}y \\ & \leq \frac{\Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)}}{\vert B \vert } \int _{B} \sum_{k=1}^{\infty }2^{k\nu }r^{\nu } \frac{C}{ ( 1+2^{k}r/ \rho (x_{0}) ) ^{N-(l_{0}+1)\theta }} \frac{r^{\delta }}{(2^{k}r)^{ \delta }V^{\frac{1}{m}-\frac{1}{s}}(2^{k}r)} \\ & \quad \times \biggl( \int_{d(x_{0},z)< 2^{k+1}r}\bigl\vert f(z) \bigr\vert ^{m}\, \mathrm{d}z \biggr) ^{\frac{1}{m}} \\ & \leq C \Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)} \sum_{k=1}^{ \infty } \frac{C_{N}}{ ( 1+2^{k}r/\rho (x_{0}) ) ^{N-(l_{0}+1) \theta }} \frac{r^{\delta }}{(2^{k}r)^{\delta }V^{1/m}(2^{k}r)}\frac{(2^{k}r)^{ \nu }}{(2^{k+1}r)^{\nu }}V^{\frac{1}{m}} \bigl(2^{k+1}r\bigr) \\ & \quad \times \biggl( \frac{(2^{k+1}r)^{m\nu }}{V(2^{k+1}r)} \int _{d(x_{0},z)< 2^{k+1}r}\bigl\vert f(z) \bigr\vert ^{m}\, \mathrm{d}z \biggr) ^{ \frac{1}{m}} \\ & \leq C \Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)} \sum_{k=1}^{ \infty }2^{-k\delta } \frac{C_{N}}{ ( 1+2^{k}r/\rho (x_{0}) ) ^{N-(l_{0}+1)\theta }} \bigl\{ M_{m\nu } \bigl( \vert f \vert ^{m} \bigr) ( y ) \bigr\} ^{\frac{1}{m}} \\ & \leq C \Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)} \bigl\{ M_{m \nu } \bigl( \vert f \vert ^{m} \bigr) ( y ) \bigr\} ^{\frac{1}{m}} \end{aligned}$$
if we choose N sufficiently large.
Case of
\(\frac{D}{2} \leq q_{1} < D\): By Lemmas 9 and 6, we have
$$\begin{aligned} & \biggl( \int_{2^{k}r\leq d(x_{0},z)< 2^{k+1}r}\bigl\vert \tilde{\mathcal{K}}(y,z)- \tilde{ \mathcal{K}}(x_{0},z)\bigr\vert ^{p}\,\mathrm{d}z \biggr) ^{ \frac{1}{p}} \biggl( \int_{2^{k}r\leq d(x_{0},z)< 2^{k+1}r}\bigl\vert h(z)-h _{2B}\bigr\vert ^{t}\,\mathrm{d}z \biggr) ^{\frac{1}{t}} \\ & \quad \leq \frac{C_{N}r^{\delta }(2^{k}r)^{1-\delta }}{V(2^{k}r) ( 1+2^{k}r/ \rho (x_{0}) ) ^{N}} \biggl( \int _{2^{k}r\leq d(x_{0},z)< 2^{k+1}r}\biggl\vert \int_{B(z,2^{k+3}r)} \frac{d(z,y)W(y)}{V(z,y)} \,\mathrm{d}y\biggr\vert ^{p}\,\mathrm{d}z \biggr) ^{\frac{1}{p}} \\ & \quad \quad {}\times \biggl( \int_{2^{k}r\leq d(x_{0},z)< 2^{k+1}r} \bigl\vert h(z)-h_{2B}\bigr\vert ^{t}\,\mathrm{d}z \biggr) ^{\frac{1}{t}} \\ &\quad \quad {}+ \frac{C _{N}\Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)}2^{l\nu }r^{\nu }}{ ( 1+2^{k}r/\rho (x_{0}) ) ^{N-(l_{0}+1)\theta }} \frac{r^{ \delta }}{(2^{k}r)^{\delta }V^{1/m}(2^{k}r)} \\ & \quad \leq \frac{C_{N}\Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)}2^{l \nu }r^{\nu }}{ ( 1+2^{k}r/\rho (x_{0}) ) ^{N-(l_{0}+1)\theta }} \\ & \quad \quad {}\times \biggl[ \frac{r^{\delta }(2^{k}r)^{1-\delta }}{V^{ \frac{t-1}{t}}(2^{k}r)} \biggl( \int_{B(x_{0},2^{k+3}r)}W^{q_{1}}(z) \,\mathrm{d}z \biggr) ^{\frac{1}{q_{1}}}+\frac{r^{\delta }}{(2^{k}r)^{ \delta }V^{1/m}(2^{k}r)} \biggr] \\ & \quad \leq \frac{C_{N}\Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)}2^{l \nu }r^{\nu }}{ ( 1+2^{k}r/\rho (x_{0}) ) ^{N-(l_{0}+1)\theta }} \\ & \quad \quad{} \times \biggl[ \frac{r^{\delta }V^{\frac{1}{q_{1}}}(2^{k}r)(2^{k}r)^{-1- \delta }}{V^{\frac{t-1}{t}}(2^{k}r)} \biggl( \frac{(2^{k+3}r)^{2}}{V(2^{k+3}r)} \int_{B(x_{0},2^{k+3}r)}W(z) \,\mathrm{d}z \biggr) +\frac{r^{ \delta }}{(2^{k}r)^{\delta }V^{1/m}(2^{k}r)} \biggr] \\ & \quad \leq \frac{C_{N}\Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)}2^{l \nu }r^{\nu }}{ ( 1+2^{k}r/\rho (x_{0}) ) ^{N-l_{1}-(l_{0}+1) \theta }} \biggl( \frac{r^{\delta }V^{\frac{1}{q_{1}}}(2^{k}r)(2^{k}r)^{-1- \delta }}{V^{\frac{t-1}{t}}(2^{k}r)} +\frac{r^{\delta }}{(2^{k}r)^{ \delta }V^{1/m}(2^{k}r)} \biggr) \\ & \quad \leq \frac{C\Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)}2^{l\nu }r ^{\nu }}{ ( 1+2^{k}r/\rho (x_{0}) ) ^{N-l_{1}-(l_{0}+1)\theta }} \frac{r^{\delta }}{(2^{k}r)^{\delta }V^{1/m}(2^{k}r)}, \end{aligned}$$
where \(\frac{1}{m}+\frac{1}{p}+\frac{1}{t}=1\) and \(\frac{1}{q_{1}}= \frac{1}{p}+\frac{1}{D}\). Therefore, for \(m>q_{2}\),
$$\begin{aligned} J_{3} &= \frac{1}{\vert B \vert } \int_{B} \bigl\vert A_{3}f(y)-c \bigr\vert \, \mathrm{d}y \\ & \leq \frac{1}{\vert B \vert } \int_{B} \biggl\vert \int_{d(x_{0},z)>2r} \bigl(\tilde{\mathcal{K}}(y,z)-\tilde{ \mathcal{K}}(x_{0},z)\bigr) \bigl(h(z)-h_{2Q}\bigr)f(z) \, \mathrm{d}z\biggr\vert \,\mathrm{d}y \\ & \leq \frac{1}{\vert B \vert } \int_{B} \sum_{k=1}^{\infty } \int _{2^{k}r\leq d(x_{0},z)< 2^{k+1}r} \bigl\vert \bigl(\tilde{\mathcal{K}}(y,z)- \tilde{\mathcal{K}}(x_{0},z)\bigr) \bigl(h(z)-h_{2Q} \bigr)f(z)\bigr\vert \,\mathrm{d}z\, \mathrm{d}y \\ & \leq \frac{\Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)}}{\vert B \vert } \int _{B} \sum_{k=1}^{\infty }2^{k\nu }r^{\nu } \frac{C}{ ( 1+2^{k}r/ \rho (x_{0}) ) ^{N-l_{1}-(l_{0}+1)\theta }} \frac{r^{\delta }}{(2^{k}r)^{ \delta }V^{\frac{1}{m}-\frac{1}{s}}(2^{k}r)} \\ & \quad \times \biggl( \int_{d(x_{0},z)< 2^{k+1}r}\bigl\vert f(z) \bigr\vert ^{m}\, \mathrm{d}z \biggr) ^{\frac{1}{m}} \\ & \leq C \Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)} \sum_{k=1}^{ \infty } \frac{C_{N}}{ ( 1+2^{k}r/\rho (x_{0}) ) ^{N-l_{1}-(l _{0}+1)\theta }} \frac{r^{\delta }}{(2^{k}r)^{\delta }V^{1/m}(2^{k}r)}\frac{(2^{k}r)^{ \nu }}{(2^{k+1}r)^{\nu }}V^{\frac{1}{m}} \bigl(2^{k+1}r\bigr) \\ & \quad \times \biggl( \frac{(2^{k+1}r)^{m\nu }}{V(2^{k+1}r)} \int _{d(x_{0},z)< 2^{k+1}r}\bigl\vert f(z) \bigr\vert ^{m}\, \mathrm{d}z \biggr) ^{ \frac{1}{m}} \\ & \leq C \Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)} \sum_{k=1}^{ \infty }2^{-k\delta } \frac{C_{N}}{ ( 1+2^{k}r/\rho (x_{0}) ) ^{N-l_{1}-(l_{0}+1)\theta }} \bigl\{ M_{m\nu } \bigl( \vert f \vert ^{m} \bigr) ( y ) \bigr\} ^{\frac{1}{m}} \\ & \leq C \Vert h\Vert _{\operatorname{Lip}_{\nu }^{\theta }(G)} \bigl\{ M_{m \nu } \bigl( \vert f \vert ^{m} \bigr) ( y ) \bigr\} ^{\frac{1}{m}}, \end{aligned}$$
if we choose N sufficiently large. □