让我们先证明定理3.1。
对于任何固定\(\gamma\in\mathbb{Z}\)和\(a\in\mathbb{Q}^{n}_{p} \),很容易看出
$$开始{对齐}和\frac{1}{ω{\gamma}(a))}\biggl(a) })(x)\bigr\vert^{q}\,dx\biggr^{c}_{\gamma}(a)})(x)\bigr\vert^{q}\,dx\biggr)^{1/q}\\&&\ quad:=I+II,\end{aligned}$$
(4.1)
哪里\(B)^{c}_{\伽马}(a)\)是对的补充\(B_{\gamma}(a)\)在里面\(\mathbb{Q}^{n}_{p} \)。
使用引理2.2和(3.1),因此
$$开始{对齐}I\lesssim&\frac{1}{\omega(B_{\gamma}(a)(a)}{ω}\int_{B_{\gamma}(a)}\bigl\vertf(x)\bigr\vert^{q}\,dx\biggr)^{1/q}\\lesssim&\Vertf\vert_{GM_{q,\nu}}。\结束{对齐}$$
(4.2)
对于二,让我们先估计一下\(|T_{k}(f\chi_{B^{c}_{\gamma}(a)})(x)|\)。
自\(x\在B_{\gamma}(a)中\)和\(L^{infty}中的\varOmega\(\mathbb{Q}_{p}^{n})\),我们有
$$\开始{aligned}\bigl\vert T_{k}(f\chi_{B^{c}_{\gamma}(a)})(x)\bigr\vert=&\biggl\vert\int_{\vert y\vert_{p}>p^{k}}(f\chi_{B^{c}_{\gamma}(a)})(x-y)\frac{\varOmega(y)}{\vert y\vert_{p}^{n}}\,dy\biggr\vert\=&\biggl\vert_int_{\vert-x-z\vert_}p}>p^{k}}(f\chi_{B^{c}_{\gamma}(a)})(z)\frac{\varOmega(x-z)}{\vert x-z\vert_{p}^{n}}\,dz\biggr\vert\\\lesssim&\int_{B^{c}_{\gamma}(a)}\frac{\vertf(z)\vert}{\vert x-z\vert_{p}^{n}}\,dz\\lesssim&\sum_{j=\gamma+1}^{\infty}\int_{S_{j}(a)}p^{-jn}\bigl\vert f(y)\bigr\vert\,dy\\leq&\sum_{j=\ gamma+1{{\inffy}p^{jn}\ biggl(int_{B_{j}(a)}\bigl\vertf(y)\bigr\vert^{q}\,dy\biggr)^{1/q}\bigle\vertB_{j}(a)\biger\vert_{H}^{1-1/q}\\=&\Vertf\Vert_{GM_{q,\nu}}\sum_{j=\gamma+1}^{infty}\nu\bigl(B_{j}(a)\bigr)。\结束{对齐}$$
(4.3)
因此,从(3.1)和(4.3)由此可见
$$\beart{aligned}II=&\frac{1}{\omega(B_{\gamma}(a))}\biggl(\frac{1}{\vert B_{\gamma}(a)\vert _{H}}\int _{B_{\gamma}(a)}\bigl\vert T_{k}(f\chi _{B^{c}_{\gamma}(a)})(x)\bigr\vert^{q}\,dx\biggr)^{1/q}\\lesssim&\Vertf\vert_{GM_{q,nu}}\sum_{j=\gamma+1}^{infty}\nu\bigl数字}}。\结束{对齐}$$
(4.4)
结合以下估算(4.1), (4.2),以及(4.4),我们有
$$\frac{1}{\omega(B_{\gamma}(a))}\biggl(\frac{1}}{\vertB_{\ gamma}(a)\vert_{H}}\int_{B_{gamma}-(a)}\bigl\vertT_{k}(f)(x)\bigr\vert^{q}\,dx\biggr)^{1/q}\lesssim\vert f\vert_{GM_{q,\nu}}$$
也就是说\(T_{k}\)以为界\(GM_{q,\nu}\)到\(GM_{q,\omega}\)为所有人\(k\in\mathbb{Z}\)。
此外,从引理2.2以及\(GM_{q,\omega}(\mathbb{问}_{p} ^{n})\)很明显\(T(f)=\lim_{k\rightarrow-\infty}T_{k}(f)\)存在于\(GM_{q,\omega}\)和操作员T型以为界\(GM_{q,\nu}\)到\(GM_{q,\omega}\)。
对于任何\(x\in\mathbb{Q}^{n}_{p} \),自\(L^{infty}中的\varOmega\(\mathbb{问}_{p} ^{n})\)和\(b_{i}\在{varLambda{beta{i}}}\中),\(i=1,2,\点,m\),很容易看出
$$\begin{aligned}和\bigl\vert T_{k}^{\vec{b}}f(x)\bigr\vert\和\quad\leq\int_{vert y\vert_{p}>p^{k}}\prod_{i=1}^{m}\bigl\ vert b{i}\vert y\vert_{p}^{n}}\,dy\\&\quad\lesssim\int_{mathbb{问}_{p} ^{n}}\frac{\vertf(z)\vert}{\vertx-z\vert_{p}^{n-\beta}}\,dz\\&\quad\lesssim I^{beta}_{p}\bigl(\vertf\vert\bigr)(x)。\结束{对齐}$$
因此从引理2.3很明显,换向器\(T_{k}^{\vec{b}})以为界\(GM_{q,\nu}\)到\(GM_{r,\omega}\)为所有人\(k\in\mathbb{Z}\)。
此外,从\(GM_{q,\omega}(\mathbb{问}_{p} ^{n})\)很明显\(T^{\vec{b}}(f)=\lim_{k\rightarrow-\infty}存在于\(GM_{q,\omega}\)和换向器\(T^{\vec{b}}\)以为界\(GM_{q,\nu}\)到\(GM_{q,\omega}\)。 □
在不失一般性的情况下,我们只需要证明这个结论适用于\(m=2\)。
对于任何固定\(\gamma\in\mathbb{Z}\)和\(a\in\mathbb{Q}^{n}_{p} \),我们写\(f^{0}=f\chi_{B_{gamma}(a)}\)和\(f^{infty}=f\chi_{B^{c}_{\伽马}(a)}\).然后
$$开始{对齐}和\frac{1}{\omega(B_{\gamma}(a)\quad\leq\frac{1}{\omega(B_{\gamma}(a)-(b{1}){b{\gamma}(a)}\biger)\bigl(b{2}(x)-biggl(\frac{1}{\vertB_{\gamma}(a)\vert_{H}}\int_{b_{gamma}(a)}\bigl\vert\bigl(b_1}(x)-(b_1{})_{b{\garma}_{2}-(b_{2})_{b_{\gamma}(a)}\bigr)f^{0}\biger)bigl\vert\bigl(b_{2}(x)-(b_2})_{b_{gamma}(a)}\bigr)T_{k}\bigl_{1}-(b_{1})_{b_{\gamma}(a)}\bigr)f^{0}\biger)转换T_{k}\bigl(_{1}-(b{1}){b{\gamma}(a)}\bigr)\bigl(b_{2}-(b_{2})_{b_{\gamma}(a)}\bigr)f^{0}\biger)b_{\gamma}(a)}\bigl\vert\bigl\bigr)(x)\bigr\vert^{r}\,dx\biggr)^{1/r}\\&\qquad{}+\frac{1}{\omega(B_{gamma}(a))}\\&\ qquad}}\times\biggl(\frac}{\vertB_{gamma}{1}){B_{gamma}(a)}_{2}-(b_{2})_{b_{\gamma}(a)}\bigr)f^{\infty}\biger)(x)\bigr\vert^{r}\,dx\biggr{b_{γ}(a)}_{1}-(b_{1})_{b_{\gamma}(a)}\bigr)f^{\infty}\biger))}\bigl\vert T_{k}\bigle(\bigl(b)_{1}-(b{1}){b{\gamma}(a)}\bigr)\bigl(b_{2}-(b_{2})_{b_{\gamma}(a)}\bigr)f^{\infty}\biger)(x)\bigr\vert^{r}\,dx\biggr)^{1/r}\\&\quad=:E_{1}+E_{2}+E_}3}+E_5}+E_6}+E_7}+E_8}。\结束{对齐}$$
(4.5)
我们进一步估计了每一部分。
自\(1/r=1/q+1/q{1}+1/q{2}\),来自Hölder不等式,引理2.2(i)由此得出
$$\begin{aligned}E_{1}=&\frac{1}{\omega(B_{\gamma}(a))}\\&{}\times\biggl(\frac{1}{\vertB_{\ gamma}(a)\vert_{H}}\int_{B_{(a{2}(x)-(B_2}){B_{\gamma}(a)}\bigr)T_{k}\bigl(f^{0}\biger)(x)\bigr\vert^{r}\,dx\biggr)^{1/r}\\leq&\frac{1}{\omega(B_{\gamma}(a))\vert B_{\gamma}{\gamma}(a)}\bigl\vert T _{k}\bigl(f^{0}\bigr)(x)\bigr\vert ^{q}\,dx\biggr)^{1/{q}}\\lesssim&\frac{\nu _{1}(B_{\gamma}(a)){{2}l \vertf(x)\bigr\vert^{q},dx\biggr)^{1/q}\\leq&\frac{nu(a) )}\prod_{i=1}^{2}\Vertb_{i}\Vert_{GC_{q{i},\nu_{i{}}\Vertf\Vert_{GM_{q,\nu}}\\lesssim&\prod_{i=1{2}\ Vertb_a{i}\ Vert_GC_{q{i},\ nu_{i}}\ Vertf\Vert_GM_{q,\ nu}}。\结束{对齐}$$
让\(1/\bar{q}=1/q+1/q{2}).然后\(1/r=1/q{1}+1/\bar{q}\)因此,从Hölder不等式出发,引理2.2和(i)我们获得
$$\begin{aligned}E_{2}=&\frac{1}{\omega(B_{\gamma}(a))}\biggl(B)_{2}-(b_{2})_{b_{\gamma}(a)}\bigr)f^{0}\biger){1}(x)-(b_1}){b_{\gamma}(a)}\bigr\vert^{q_1}\,dx\biggr)^{1/q_{1}}\\&{}\ times\biggl(int_{b_}\gamma(a\bigl(\bigl_{2}-(b _{2})_{b_{\gamma}(a)}\bigr)f^{0}\bigr)(x)\bigr \vert^{\bar{q}}\,dx\biggr)^{1/{\bar{q}}}}\\lesssim&&frac{1}{\omega(b_{\gamma}(a))\vert b_{\gamma}(a)\vert _{H}^{1/r}-(b_{1})_{b_{\gamma}(a)}\bigr\vert^{q_{1}}\,dx\biggr)^{1/q_{1}}}\&{}\times\biggl(\ int _{b_{\gamma}(a)}\bigl\vert\bigl(b_{2}(x)-(b_}2})_{b_{gamma}(a)}\biger)f(x)\bigr\vert^{bar{q}}\,dx\biggr)^{1/{bar{q}}}\\leq&\frac{1}{omega(b_{gamma}i=1}^{2}\biggl(int_{b_{gamma}(a)}\bigl\vertb_{i}}\biggl(\int _{B_{\gamma}(a)}\bigl\vert f(x)\bigr\vert ^{q}\,dx\biggr)^{1/{q}}\\leq&&frac{\nu(B_{\gamma}(a))\nu _{1}(B_{\gamma}(a))\nu _{2}(B_{\gamma}(a))}{GC_{q_{i},\nu _{i}}}}\vert f\vert _{GM_{q,\nu}}\\lesssim&\prod_{i=1}^{2}\vert B_{i}\vert _{GC_{q_{i},\nu _{i}}}\垂直f\垂直_{GM_{q,\nu}}。\结束{对齐}$$
同样,
$$E_{3}\lesssim\prod_{i=1}^{2}\Vertb_{i}\Vert_{GC{q{i},\nu{i}}\Vertf\Vert_{GM{q,\nu}}$$
对于\(E_{4}\),来自引理2.2,Hölder不等式,(i)我们得到
$$\begin{aligned}E_{4}=&\frac{1}{\omega(B_{\gamma}(a))}\biggl(\frac{1}}{\vertB_{\ gamma}(a)\vert_{H}}\int_{B_{gamma}[a)}\bigl\vertT_{k}\bigl(B)_{1}-(b_{1})_{b_{\gamma}(a)}\bigr)\bigl(b_{2}-(b_{2})_{b_{\gamma}(a)}\bigr)f^{0})(x)\bigr\vert^{r}\,dx\biggr)^{1/r}\\lesssim&\frac{1}{omega(b_{γ}(a))\vertB_{\ gamma}(a)\vert_{H}^{1/r}\biggl(b_1}(x)-(b_1{1}){b_{\gamma}(a)}\bigr)\bigl(b_2}(x)-}\\leq&\frac{1}{omega(B_{gamma}(a))\vertB_{gamma}biggr)^{1/q_{i}}\biggl(int_{B_{\gamma}(a)}\bigl\vertf(x)\bigr\vert^{q}\,dx\biggr(B_{\gamma}(a)){{2}\Vert_{GC_{q_{i},\nu_{i{}}\Vertf\Vert_{GM_{q,\nu}}。\结束{对齐}$$
估计\(E_{5}\),我们首先需要考虑\(|T_{k}(f^{\infty})(x)|\)事实上(4.3)很容易看出这一点
$$\bigl\vert T_{k}\bigl(f^{infty}\bigr)(x)\bigr\vert\lesssim\vert f\vert_{GM_{q,\nu}}\sum_{j=\gamma+1}^{inffy}\nu\bigle(B_{j}(a)\biger)$$
(4.6)
因此,从霍尔德不等式来看(4.6)和(ii)我们得到
$$\begin{aligned}E_{5}=&\frac{1}{\omega(B_{\gamma}(a))}\\&{}\times\biggl(\frac{1}{\vertB_{\ gamma}(a)\vert_{H}}\int_{B_{(a{2}(x)-(B_2}){B_{\gamma}(a)}\bigr)T_{k}\bigl(f^{\infty}\biger)(x)\bigr\vert^{r}\,dx\biggr)^{1/r}\\leq&\frac{1}{ω(B_{\gamma}(a))\vertB_{\γ}(a)\vert_{H}^{1/r}}\prod_{i=1}^{2}\biggl}\\&{}\times\biggl(int_{B_{gamma}(a)}\bigl\vert T_{k}\bigle(f^{infty}\biger)(x)f(x)\bigr\vert^{q}\,dx\biggr)^{1/{q}}\\lesssim&\sum_{j=\gamma+1}^{\infty}\frac{\nu(B_{j}(a))\nu_{1}垂直f\Vert_{GM_{q,nu}}\\lesssim&\prod_{i=1}^{2}\Vertb_{i}\Vert_{GC_{q_{i{,nu{i}}\垂直f\Vert_{GM_{q、nu}}。\结束{对齐}$$
与估计值类似(4.3)的\(x \在B_{\gamma}(a)\中).签署人\(L^{infty}中的\varOmega\(\mathbb{问}_{p} ^{n})\)和(2.2)我们可以推断出
$$\开始{aligned}&\big|T_{k}\bigl(b_{2}-(b_{2})_{b_{\gamma}(a)}\bigr)f^{\infty})(x)\big|\\&\quad=\biggl\vert\int_{verty\vert_p}>p^{k}}\bigl(b_2}(x-y)-^{c}_{\gamma}(a)}(x-y)\frac{\varOmega(y)}{\vert y\vert_{p}^{n}}\,dy\biggr\vert\\&\quad\leq\int_{B^{c}_{\gamma}}\bigl\vert b_{2}(z)-(b_2})_{b_{\gama}(a)}\bigr\vert\bigl\overt f(z)\bigr\ vert\frac{\vert\varOmega(x-z)\vert}{\vert x-z\vert_{p}^{n}}\,dz\\&\quad\lesssim\int_{b^{c}_{\gamma}}(z)-(b{2}){b_{gamma}(a)}\vert\vert f(z)\vert}{\vert x-z\vert_p}^n}}{2})_{b_{\gamma}(a)}\bigr\vert\bigl\vertf(y)\bigr\ vert\,dy\\&\quad=\sum_{j=\gamma+1}^{\infty}p^{-jn}\bigl\ vertB_{j}(a)\bigr\vert_{H}^{1-1/q-1/q_2}}\biggl(\int_{S_j}(a)}\bigl\vert f(y)\bigr\ vert^{q}\,dy\biggr)^{1/{q}}\biggl 2},dy\biggr)^{1/q_{2}}\\&\quad\leq\Vertf\vert_{GM_{q,\nu}}\sum_{j=\gamma+1}^{infty}p^{-jn}\bigl\vertB_{j}(a)\bigr\vert_{H}^{1-1/q_{2]}\nu\bigl(B_{j}(a)\bigr)\biggl(int _{B_{j}(a)}\bigl\vert B_{2}(y)-(B_{2})_{B _{gamma}(B)}\bigr\vert ^ q_{2}}\,dy\biggr}}\Vertf\vert_{GM_{q,\nu}}\sum_{j=\gamma+1}^{\infty}(j+1-\gamma)\nu\bigl(B_{j}(a)\bigr)\nu_{2}\bigl。\结束{对齐}$$
(4.7)
让\(1/\bar{q}=1/q+1/q{2}).然后\(1/r=1/q_{1}+1/\bar{q}\)因此,根据Hölder不等式(4.7)以及(ii)因此
$$\begin{aligned}E_{6}=&\frac{1}{\omega(B_{\gamma}(a))}\biggl(\frac{1}}{\vertB_{\ gamma}(a)\vert_H}}\int_{B_{gamma}[a)}\bigl\vert\bigl(B_{1}(x)-(B)_{2}-(b_{2})_{b_{\gamma}(a)}\bigr)f^{\infty}\biger)(x)\bigr\vert^{r}\,dx\biggr vertb_{1}(x)-(b_1})_{b_{\gamma}(a)}\bigr\vert^{q_1}}\,dx\biggr)^{1/q_{1{}}\\&{}\ times\biggl(int_{b{\garma}(a)}\bigl\vertT_{k}\bigl(\bigl_{2}-(b_{2})_{b_{\gamma}(a)}\bigr)f^{\infty}\biger)(x)\bigr\vert^{\bar{q}}\,dx\biggr)^{1/{\bar}q}}\\leq&\prod_{i=1}^{2}\Vertb_{i}\vert_GC_{q{i},nu_{i{}}\vert f\vert_{GM_{q,\nu}}\frac{1}{\omega(b_{\gamma}(a))}\sum_{j=\gamma+1}^{\infty}(j+1-\gamma)\nu\bigl\bigl(B_{\gamma}(a)\biger)\\lesssim&\prod_{i=1}^{2}\Vertb_{i}\Vert_{GC_{q_{i{,nu_{ineneneep}\Vertf\Vert_{GM_{q,nu}}。\结束{对齐}$$
类似估算\(E_{6}\),我们获得
$$E_{7}\lesssim\prod_{i=1}^{2}\Vertb_{i}\Vert_{GC{q{i},\nu{i}}\Vertf\Vert_{GM{q,\nu}}$$
此外,由于\(L^{infty}中的\varOmega\(\mathbb{问}_{p} ^{n})\),由(2.2)很容易看出这一点
$$\开始{对齐}&\bigl\vert T_{k}\bigl(_{1}-(b_{1})_{b_{\gamma}(a)}\bigr)\bigl(b_{2}-(b_2})_{b_{\gamma}(a)}\bigr)f^{\infty}\biger)(x)\bigr\vert\\&\quad=\biggl\vert\int_{\vert x-z\vert_p}>p^{k}}\bigl(b_1}(z)-(b_1{})\gamma}(a)}\biger)f\chi_{b^{c}_{\gamma}(a)}(z)\frac{\varOmega(x-z)}{\vert x-z\vert_{p}^{n}}\,dz\biggr\vert\&\quad\leq\int_{B^{c}_{\gamma}}\bigl\vert b_{1}(z)-(b_1})_{b_{\gama}(a)}\bigr\vert\bigl\overtb_{2}&\quad\lesssim\sum_{j=\gamma+1}^{\infty}\int_{S_{j}(a)}p^{-jn}\bigl\vertb_{1}(z)-(b_1})_{b_{\gamma}(a)}\bigr\vert\bigl\vert b_{2}(z)-(b_2})_{b_{gamma}_{1}-1/q_{2}}\biggl(int_{S_j}(a)}\bigl\vertf(y)\bigr\vert^{q}\,dy\biggr)^{1/{q}}\\&\qquad}\times\biggl(int_[S_{j},dy\biggr)^{1/q_{1}}\\&\qquad{}\times\biggl,dy\biggr)^{1/q_{2}}\\&\quad\lesssim\prod_{i=1}^{2}\Vertb_{i}\Vert_{GC_{q_{i{,\nu_{ineneneep}\Vertf\Vert_{GM_{q,\nu}}\sum_{j=\gamma+1}^{\infty}(j+1-\gamma)^{2{nu\bigl(b_{j}(a)\biger)\nu{1}\bigl(b_{j}(a)\bigr)。\结束{对齐}$$
(4.8)
因此,从(4.8)(ii)我们明白了
$$\begin{aligned}E_{8}=&&frac{1}{\omega(B_{\gamma}(a))}\bigl(\frac{1}{B_{\gamma}(a)| _{H}}\ int _{B}\bigl \ vert T_{k}\bigl(\bigl(B_{1}-(b_{1})_{b_{\gamma}(a)}\bigr)\bigl(b_{2}-(b_{2})_{b_{\gamma}(a)}\bigr)f^{\infty}\biger)(x)\bigr\vert^{r}\,dx\biggr{ω(b_{\gamma}(a))}\sum_{j=\gamma+1}^{\infty}(j+1-\gamma)^{2}\nu\bigl\较大)\\较小&\prod_{i=1}^{2}\Vert b_{i}\Vert_{GC_{q{i},\nu_{i{}}\ Vert f\Vert{GM_{q,\nu}}。\结束{对齐}$$
组合(4.5)以及对\(E_{1},E_{2},\点,E_{8}\),我们有
$$\frac{1}{\omega(B_{\gamma}(a 1}^{2}\Vertb_{i}\vert_{GC_{q_{i{},\nu_{ineneneep}\Vertf\vert_{GM_{q,\nu}}$$
这意味着换向器\(T_{k}^{(b{1},b{2})}\)以为界\(GM_{q,\nu}\)到\(GM_{r,\omega}\)。
此外,通过引理2.2以及\(GM_{q,\omega}(\mathbb{问}_{p} ^{n})\)很明显,换向器\(T^{\vec{b}}(f)=\lim_{k\rightarrow-\infty}存在于\(GM_{q,\omega}\)、和\(T^{\vec{b}}\)以为界\(GM_{q,\nu}\)到\(GM_{q,\omega}\)。
因此,定理的证明3.3已完成。 □