72.26999pt\,+\Big{(}-\frac{1}{4}\lambda_{1}^{-3/2}\lambda_{2}^{-1}+\frac{1}{2}\lambda_{1}^{-1}\lambda_{2}^{-1/2}\Big{)}e^{\lambda_{1}+\lambda_{2}}\frac{\partial^{3}F}{\partial\lambda_{1}^{2}\partial\lambda_{2}}(\sqrt{\lambda_{1}},\sqrt{\lambda_{2}})$
$\displaystyle\hskip
72.26999pt\quad\;\,+\Big{(}\frac{1}{2}\lambda_{1}^{-1/2}\lambda_{2}^{-1}-\frac{1}{4}\lambda_{1}^{-1}\lambda_{2}^{-3/2}\Big{)}e^{\lambda_{1}+\lambda_{2}}\frac{\partial^{3}F}{\partial\lambda_{1}\partial\lambda_{2}^{2}}(\sqrt{\lambda_{1}},\sqrt{\lambda_{2}})$
$\displaystyle\hskip
144.54pt\qquad\;+\frac{1}{4}\lambda_{1}^{-1}\lambda_{2}^{-1}e^{\lambda_{1}+\lambda_{2}}\frac{\partial^{4}F}{\partial\lambda_{1}^{2}\partial\lambda_{2}^{2}}(\sqrt{\lambda_{1}},\sqrt{\lambda_{2}})$

and using that $F\in\mathcal{C}^{4}(\mathbb{R}^{2})$ and $F$ is compactly
supported we get $J_{44}=O(\xi_{1}^{1/2}\xi_{2}^{1/2})$. This along with the
similar estimates for $J_{42},J_{43}$ and the estimate $|J_{41}|\leq c$ imply
(4.93). In turn, (4.92), (4.93), and the similar estimates for the integrals
over $A_{2}$ and $A_{3}$ yield (4.88). Finally, (4.88) and (4.89) imply
(4.87).

From (4.87) it follows that $\widehat{H}\in L^{1}(\mathbb{R}^{2})$ and
obviously $H\in L^{1}(\mathbb{R}^{2})$. Then the Fourier inversion formula
gives

(4.94)
$H({\bm{\lambda}})=\frac{1}{(2\pi)^{2}}\int_{\mathbb{R}^{2}}\widehat{H}({\bm{\xi}})e^{i{\bm{\lambda}}\cdot{\bm{\xi}}}d{\bm{\xi}}.$

Assuming that $\varphi,\psi\in L^{2}(X)\cap L^{1}(X)$ we get using that
$F({\mathbf{t}}\sqrt{L})=H({\mathbf{t}}^{2}L)e^{-{\mathbf{t}}^{2}L}$ and
(4.94)

$\displaystyle\langle F({\mathbf{t}}\sqrt{L})\varphi,$
$\displaystyle\psi\rangle$
$\displaystyle=\int_{[0,\infty)^{2}}H(t_{1}^{2}\lambda_{1},t_{2}^{2}\lambda_{2})e^{-t_{1}^{2}\lambda_{1}-t_{2}^{2}\lambda_{2}}d\langle
E_{(\lambda_{1},\lambda_{2})}\varphi,\psi\rangle$
$\displaystyle=\int_{[0,\infty)^{2}}\Big{(}\frac{1}{(2\pi)^{2}}\int_{\mathbb{R}^{2}}\widehat{H}({\bm{\xi}})e^{i(t_{1}^{2}\lambda_{1}\xi_{1}+t_{2}^{2}\lambda_{2}\xi_{2})}d{\bm{\xi}}\Big{)}e^{-t_{1}^{2}\lambda_{1}-t_{2}^{2}\lambda_{2}}d\langle
E_{(\lambda_{1},\lambda_{2})}\varphi,\psi\rangle$
$\displaystyle=\frac{1}{(2\pi)^{2}}\int_{\mathbb{R}^{2}}\widehat{H}({\bm{\xi}})\Big{(}\int_{[0,\infty)^{2}}e^{-t_{1}^{2}\lambda_{1}(1-i\xi_{1})-t_{2}^{2}\lambda_{2}(1-i\xi_{2})}d\langle
E_{(\lambda_{1},\lambda_{2})}\varphi,\psi\rangle\Big{)}d{\bm{\xi}}$
$\displaystyle=\frac{1}{(2\pi)^{2}}\int_{\mathbb{R}^{2}}\widehat{H}({\bm{\xi}})\Big{\langle}e^{-t_{1}^{2}\lambda_{1}(1-i\xi_{1})L_{1}-t_{2}^{2}\lambda_{2}(1-i\xi_{2})L_{2}}\varphi,\psi\Big{\rangle}d{\bm{\xi}}.$

Above the shift of order of integration is justified by Fubini’s theorem and
the fact that

$\displaystyle\int_{\mathbb{R}^{2}}|\widehat{H}({\bm{\xi}})|$
$\displaystyle\Big{(}\int_{[0,\infty)^{2}}\big{|}e^{-t_{1}^{2}\lambda_{1}(1-i\xi_{1})-t_{2}^{2}\lambda_{2}(1-i\xi_{2})}\big{|}|d\langle
E_{(\lambda_{1},\lambda_{2})}\varphi,\psi\rangle|\Big{)}d{\bm{\xi}}$
$\displaystyle\leq\int_{\mathbb{R}^{2}}|\widehat{H}({\bm{\xi}})|\Big{(}\int_{[0,\infty)^{2}}|d\langle
E_{(\lambda_{1},\lambda_{2})}\varphi,\psi\rangle|\Big{)}d{\bm{\xi}}$
$\displaystyle\leq\int_{\mathbb{R}^{2}}|\widehat{H}({\bm{\xi}})|d{\bm{\xi}}\Big{(}\int_{[0,\infty)^{2}}d\|E_{(\lambda_{1},\lambda_{2})}\varphi\|^{2}\Big{)}^{1/2}\Big{(}\int_{[0,\infty)^{2}}d\|E_{(\lambda_{1},\lambda_{2})}\psi\|^{2}\Big{)}^{1/2}$
$\displaystyle\leq\|\widehat{H}\|_{L^{1}}\|\varphi\|_{L^{2}}\|\psi\|_{L^{2}}<\infty.$

In going further we get from above

$\displaystyle\langle F({\mathbf{t}}\sqrt{L})\varphi,\overline{\psi}\rangle$
$\displaystyle=\frac{1}{(2\pi)^{2}}\int_{\mathbb{R}^{2}}\widehat{H}({\bm{\xi}})\Big{\langle}e^{-t_{1}^{2}\lambda_{1}(1-i\xi_{1})L_{1}-t_{2}^{2}\lambda_{2}(1-i\xi_{2})L_{2}}\varphi,\overline{\psi}\Big{\rangle}d\xi$
$\displaystyle=\frac{1}{(2\pi)^{2}}\int_{\mathbb{R}^{2}}\widehat{H}({\bm{\xi}})\int_{X}\int_{X}p_{(t_{1}^{2}(1-i\xi_{1}),t_{2}^{2}(1-i\xi_{2}))}({\mathbf{x}},{\mathbf{y}})\varphi({\mathbf{y}})\psi({\mathbf{x}})d\mu({\mathbf{y}})d\mu({\mathbf{x}})d{\bm{\xi}}$
$\displaystyle=\int_{X}\int_{X}\Big{(}\frac{1}{(2\pi)^{2}}\int_{\mathbb{R}^{2}}\widehat{H}({\bm{\xi}})p_{(t_{1}^{2}(1-i\xi_{1}),t_{2}^{2}(1-i\xi_{2}))}({\mathbf{x}},{\mathbf{y}})d{\bm{\xi}}\Big{)}\varphi({\mathbf{y}})\psi({\mathbf{x}})d\mu({\mathbf{y}})d\mu({\mathbf{x}})$

To justify the shift of order of integration above we apply Fubini’s theorem
taking into account that

$\displaystyle\int_{\mathbb{R}^{2}}\int_{X}\int_{X}$
$\displaystyle|\widehat{H}({\bm{\xi}})||p_{(t_{1}^{2}(1-i\xi_{1}),t_{2}^{2}(1-i\xi_{2}))}({\mathbf{x}},{\mathbf{y}})||\varphi({\mathbf{y}})||\psi({\mathbf{x}})|d\mu({\mathbf{y}})d\mu({\mathbf{x}})d{\bm{\xi}}$
$\displaystyle\leq
c\|\widehat{H}\|_{L^{1}}\|\varphi\|_{L^{1}}\|\psi\|_{L^{1}}<\infty.$

From above it follows that

${\mathscr{K}}_{F({\mathbf{t}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})=\frac{1}{(2\pi)^{2}}\int_{\mathbb{R}^{2}}\widehat{H}(\xi)p_{(t_{1}^{2}(1-i\xi_{1}),t_{2}^{2}(1-i\xi_{2}))}({\mathbf{x}},{\mathbf{y}})d{\bm{\xi}},$

implying

$\displaystyle\int_{X}{\mathscr{K}}_{F({\mathbf{t}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})d\mu({\mathbf{y}})$
$\displaystyle=\frac{1}{(2\pi)^{2}}\int_{\mathbb{R}^{2}}\widehat{H}({\bm{\xi}})\int_{X}p_{(t_{1}^{2}(1-i\xi_{1}),t_{2}^{2}(1-i\xi_{2}))}({\mathbf{x}},{\mathbf{y}})d{\mathbf{y}}d\xi$
$\displaystyle=\frac{1}{(2\pi)^{2}}\int_{\mathbb{R}^{2}}\widehat{H}({\bm{\xi}})d{\bm{\xi}}=H({\bm{0}})=F({\bm{0}}),$

where we used (2.8), (2.7), (4.33), and (4.94). The shift of the order of
integration above is easily justified by applying Fubini’s theorem. The proof
of (4.67) is complete. ∎

We next extend Theorem 4.18 from the case of compactly supported functions to
the case of smooth ones with sufficiently fast decay on ${\mathbb{R}}^{2}$.

######  Theorem 4.21.

Let $F\in\mathcal{C}^{k_{1}+k_{2}}({\mathbb{R}}^{2})$, $k_{i}\in{\mathbb{N}}$,
$k_{i}\geq 3d_{i}/2$, $i=1,2$, $k_{1}+k_{2}\geq 4$, be real-valued and obey
the conditions: $F(\pm\lambda_{1},\pm\lambda_{2})=F(\lambda_{1},\lambda_{2})$
for $(\lambda_{1},\lambda_{2})\in{\mathbb{R}}^{2}$ and for
${\bm{\beta}}\in{\mathbb{N}}_{0}^{2}$ such that $|{\bm{\beta}}|\leq
k_{1}+k_{2}$

$|\partial^{{\bm{\beta}}}F({\bm{\lambda}})|\leq
C_{\mathbf{k}}(1+|{\bm{\lambda}}|)^{-r},\quad r>d_{1}+d_{2}+k_{1}+k_{2}.$

Then $F({\bm{\delta}}\sqrt{L})$,
${\bm{\delta}}=(\delta_{1},\delta_{2})>{\bm{0}}$, is an integral operator with
kernel ${\mathscr{K}}_{F({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})$
satisfying

(4.95)
$|{\mathscr{K}}_{F({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})|\leq
c_{\mathbf{k}}{\mathfrak{D}}_{{\bm{\delta}},{\mathbf{k}}}({\mathbf{x}},{\mathbf{y}})\quad\text{and}$
(4.96)
$|{\mathscr{K}}_{F({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})-{\mathscr{K}}_{F({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}}^{\prime})|\leq
c_{\mathbf{k}}^{\prime}\sum_{i=1,2}\Big{(}\frac{\rho_{i}(y_{i},y^{\prime}_{i})}{\delta_{i}}\Big{)}^{\alpha_{i}}{\mathfrak{D}}_{{\bm{\delta}},{\mathbf{k}}}({\mathbf{x}},{\mathbf{y}})$

if $\rho_{i}(y_{i},y_{i}^{\prime})\leq\delta_{i}$, $i=1,2$, where
${\mathbf{k}}=(k_{1},k_{2})$ and the constants
$c_{\mathbf{k}},c_{{\mathbf{k}}^{\prime}}$ depend on $k_{1}$, $k_{2}$,
${\mathbf{d}}$ and $C_{\mathbf{k}}$. Moreover,

(4.97)
$\int_{X}{\mathscr{K}}_{F({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})d\mu({\mathbf{y}})=F({\bm{0}}),\quad\forall{\mathbf{x}}\in
X.$

###### Proof.

Let $\varphi_{0},\varphi\in\mathcal{C}^{\infty}(\mathbb{R}^{2})$ be functions
satisfying conditions (LABEL:dec-unity). Set
$\varphi_{j}({\bm{\lambda}}):=\varphi(2^{-j}{\bm{\lambda}})$, $j\geq 1$, and
$F_{j}({\bm{\lambda}}):=F({\bm{\lambda}})\varphi_{j}({\bm{\lambda}})$, $j\geq
0$. Then

(4.98) $F({\bm{\delta}}\sqrt{L})=\sum_{j\geq
0}F({\bm{\delta}}\sqrt{L})\varphi_{j}({\bm{\delta}}\sqrt{L})=\sum_{j\geq
0}F_{j}({\bm{\delta}}\sqrt{L}),$

where the convergence is strong in $L^{2}$. Observe that
$F_{j}(\pm\lambda_{1},\pm\lambda_{2})=F_{j}(\lambda_{1},\lambda_{2})$ and for
$j\geq 1$ we have
$\operatorname{supp}F_{j}\subset\Omega_{j}:=\\{{\bm{\lambda}}:2^{j-1}\leq|{\bm{\lambda}}|\leq
2^{j+1}\\}$ and

$\|F_{j}\|_{L^{\infty}}\leq\|F\|_{L^{\infty}(\Omega_{j})}\leq
c2^{-jr},\quad\|F_{j}\|_{W^{k_{1}+k_{2}}_{\infty}}\leq
c\|F\|_{W^{k_{1}+k_{2}}_{\infty}(\Omega_{j})}\leq c2^{-jr}.$

We now apply Theorem 4.18 to $F_{j}$ with $R=2^{j+1}$ to obtain

$\displaystyle|{\mathscr{K}}_{F({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})|\leq\sum_{j\geq
0}|{\mathscr{K}}_{F_{j}({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})|$
(4.99) $\displaystyle\leq c\sum_{j\geq
0}2^{j(d_{1}+d_{2})}\big{(}\|F_{j}\|_{L^{\infty}}+2^{j(k_{1}+k_{2})}\|F_{j}\|_{W^{k_{1}+k_{2}}_{\infty}}\big{)}{\mathfrak{D}}_{{\bm{\delta}},{\mathbf{k}}}({\mathbf{x}},{\mathbf{y}})$
$\displaystyle\leq c\sum_{j\geq
0}2^{j(d_{1}+d_{2})}2^{-jr}2^{j(k_{1}+k_{2})}{\mathfrak{D}}_{{\bm{\delta}},k}({\mathbf{x}},{\mathbf{y}})\leq
c{\mathfrak{D}}_{{\bm{\delta}},{\mathbf{k}}}({\mathbf{x}},{\mathbf{y}}),$

where for the last inequality we used that $r>(d_{1}+d_{2})+(k_{1}+k_{2})$.
Therefore,
${\mathscr{K}}_{F({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})$
satisfies (4.95).

The proof that
${\mathscr{K}}_{F({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})$ obeys
(4.96) is similar. We omit it.

To prove (4.97) we use (4.98) and (4.7) to write

${\mathscr{K}}_{F({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})=\sum_{j\geq
0}{\mathscr{K}}_{F_{j}({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})$

and integrating we get

$\int_{X}{\mathscr{K}}_{F({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})d\mu({\mathbf{y}})=\sum_{j\geq
0}\int_{X}{\mathscr{K}}_{F_{j}({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})d\mu({\mathbf{y}})=\sum_{j\geq
0}F_{j}({\bm{1}})=F({\bm{1}}).$

Here we applied (4.67) to each of the kernels
${\mathscr{K}}_{F_{j}({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})$
using the fact that $F_{j}$ is compactly supported. This completes the proof
of the theorem. ∎

We will frequently need the decay of kernels of operators of the form

$L^{{\bm{\nu}}}F({\bm{\delta}}\sqrt{L}):=(L_{1}^{\nu_{1}}\otimes
L_{2}^{\nu_{2}})F({\bm{\delta}}\sqrt{L}),\quad{\bm{\nu}}=(\nu_{1},\nu_{2})\in{\mathbb{N}}_{0}^{2}.$

In this regard we have the following corollary of Theorem 4.21.

######  Corollary 4.22.

Let $F\in\mathcal{C}^{\infty}({\mathbb{R}}^{2})$ be real-valued,
$F(\pm\lambda_{1},\pm\lambda_{2})=F(\lambda_{1},\lambda_{2})$ for
$(\lambda_{1},\lambda_{2})\in{\mathbb{R}}^{2}$, and

$|\partial^{{\bm{\beta}}}F({\bm{\lambda}})|\leq
c_{r,{\bm{\beta}}}(1+|{\bm{\lambda}}|)^{-r},\quad\forall
r>0,\;{\bm{\beta}}\in{\mathbb{N}}_{0}^{2}.$

Then for any ${\bm{\nu}}\in{\mathbb{N}}_{0}^{2}$ and ${\bm{\delta}}>{\bm{0}}$,
$L^{{\bm{\nu}}}F({\bm{\delta}}\sqrt{L})$ is an integral operator with kernel
${\mathscr{K}}_{L^{{\bm{\nu}}}F({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})$
satisfying the following inequalities: For any
${\bm{\sigma}}\in\mathbb{R}^{2}_{+}$ there exists a constant
$c_{\bm{\sigma}}>0$ such that

(4.100)
$\big{|}{\mathscr{K}}_{L^{{\bm{\nu}}}F({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})\big{|}\leq
c_{\bm{\sigma}}{\bm{\delta}}^{-2{\bm{\nu}}}{\mathfrak{D}}_{{\bm{\delta}},{\bm{\sigma}}}({\mathbf{x}},{\mathbf{y}})$

and

(4.101)
$\big{|}{\mathscr{K}}_{L^{{\bm{\nu}}}F({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})-{\mathscr{K}}_{L^{{\bm{\nu}}}F({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}}^{\prime})\big{|}\leq
c_{\bm{\sigma}}{\bm{\delta}}^{-2{\bm{\nu}}}\sum_{i=1,2}\Big{(}\frac{\rho_{i}(y_{i},y^{\prime}_{i})}{\delta_{i}}\Big{)}^{\alpha_{i}}{\mathfrak{D}}_{{\bm{\delta}},{\bm{\sigma}}}({\mathbf{x}},{\mathbf{y}})$

if $\rho_{i}(y_{i},y^{\prime}_{i})\leq\delta_{i}$, $i=1,2$.

###### Proof.

Set
$G(\lambda_{1},\lambda_{2}):=\lambda_{1}^{2\nu_{1}}\lambda_{2}^{2\nu_{2}}F(\lambda_{1},\lambda_{2})$
and observe that

$G({\bm{\delta}}\sqrt{L})=\delta_{1}^{2\nu_{1}}\delta_{2}^{2\nu_{2}}L^{{\bm{\nu}}}F({\bm{\delta}}\sqrt{L}).$

It is easily seen that $G$ satisfies the assumptions of Theorem 4.21 and,
therefore, applying that theorem to $G$ the result follows. ∎

###  4.8. Maximal operators

The Hardy-Littlewood maximal operator is a handful tool in proving estimates
in Harmonic analysis. We will use it extensively in what follows.

In our setting the (strong) maximal operator $\mathcal{M}_{r}$ with parameter
$r>0$ is defined by

(4.102) $\mathcal{M}_{r}f({\mathbf{x}}):=\sup_{B_{1}\times
B_{2}\ni{\mathbf{x}}}\Big{(}\frac{1}{\mu(B_{1}\times B_{2})}\int_{B_{1}\times
B_{2}}|f({\mathbf{y}})|^{r}d\mu({\mathbf{y}})\Big{)}^{1/r},\quad{\mathbf{x}}\in
X,$

where the $\sup$ is taken over all balls $B_{1}\subset X_{1}$, $B_{2}\subset
X_{2}$ such that ${\mathbf{x}}\in B_{1}\times B_{2}$.

The maximal operator $\mathscr{M}_{1}$ in the direction of $x_{1}\in X_{1}$ is
defined by

$\mathscr{M}_{1}f({\mathbf{x}})=\sup_{B_{1}\ni
x_{1}}\frac{1}{\mu_{1}(B_{1})}\int_{B_{1}}|f(y_{1},x_{2})|d\mu_{1}(y_{1}),\;\;{\mathbf{x}}\in
X,$

where the $\sup$ is taken over all balls $B_{1}$ in $X_{1}$ containing
$x_{1}$. The maximal operator $\mathscr{M}_{2}$ in the direction of $x_{2}\in
X_{2}$ is defined similarly. It is easy to see that this inequality is valid:

(4.103)
$\mathcal{M}_{r}f({\mathbf{x}})\leq\left(\mathscr{M}_{2}(\mathscr{M}_{1}|f|^{r})\right)^{1/r}({\mathbf{x}}),\quad{\mathbf{x}}\in
X.$

We will utilize the following version of the Fefferman-Stein vector-valued
maximal inequality: If $p\in(0,\infty),\;q\in(0,\infty]$, and
$0<r<\min\\{p,q\\}$, then for any sequence of functions $\\{f_{j}\\}$ on $X$

(4.104)
$\Big{\|}\Big{(}\sum_{j}|\mathcal{M}_{r}f_{j}|^{q}\Big{)}^{1/q}\Big{\|}_{p}\leq
c\Big{\|}\Big{(}\sum_{j}|f_{j}|^{q}\Big{)}^{1/q}\Big{\|}_{p}.$

From the fact that the Radon measure $\mu_{i}$ ($i=1,2$) obeys the doubling
condition (2.1) it follows that the maximal inequality is valid in
$(X_{i},\mu_{i})$ (see [47] and also [34]) and then (4.104) follows by
iteration.

For future use we next prove the following

######  Lemma 4.23.

Let ${\bm{\delta}},{\bm{\tau}}\in(0,\infty)^{2}$, ${\bm{\tau}}>{\mathbf{d}}$,
and $r>0$. Then there exists a constant $c>0$ such that for any measurable
function $f$ on $X$ we have

(4.105)
$\int_{X}\frac{|f({\mathbf{y}})|^{r}}{V({\mathbf{x}},{\bm{\delta}})\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho_{i}(x_{i},y_{i})\big{)}^{\tau_{i}}}d\mu({\mathbf{y}})\leq
c[\mathcal{M}_{r}f({\mathbf{x}})]^{r},\quad{\mathbf{x}}\in X.$

###### Proof.

We introduce the abbreviated notation

$F({\mathbf{x}},{\mathbf{y}}):=\frac{|f({\mathbf{y}})|^{r}}{V({\mathbf{x}},{\bm{\delta}})\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho_{i}(x_{i},y_{i})\big{)}^{\tau_{i}}}$

and $B_{i,j}:=B_{i}(x_{i},2^{j}\delta_{i})$, $i=1,2$. Set

$W_{i,0}:=B_{i,0}\quad\hbox{and}\quad W_{i,j}:=B_{i,j}\setminus
B_{i,j-1},\quad j\geq 1.$

Note that $X=\cup_{j,k\geq 0}W_{1,j}\times W_{2,k}$ and $W_{1,j}\times
W_{2,k}\subset B_{1,j}\times B_{2,k}$. Therefore,

(4.106) $\int_{X}F({\mathbf{x}},{\mathbf{y}})d\mu({\mathbf{y}})=\sum_{j,k\geq
0}\int_{W_{1,j}\times W_{2,k}}F({\mathbf{x}},{\mathbf{y}})d\mu({\mathbf{y}}).$

Observe that

$F({\mathbf{x}},{\mathbf{y}})\leq
c2^{-j\tau_{1}-k\tau_{2}}V({\mathbf{x}},{\bm{\delta}})^{-1}|f({\mathbf{y}})|^{r},\quad{\mathbf{y}}\in
W_{1,j}\times W_{2,k},$

and hence

$\displaystyle\int_{W_{1,j}\times
W_{2,k}}F({\mathbf{x}},{\mathbf{y}})d\mu({\mathbf{y}})\leq
c2^{-j\tau_{1}-k\tau_{2}}V({\mathbf{x}},{\bm{\delta}})^{-1}\int_{B_{1,j}\times
B_{2,k}}|f({\mathbf{y}})|^{r}d\mu({\mathbf{y}})$ (4.107)
$\displaystyle=c2^{-j\tau_{1}-k\tau_{2}}\frac{\mu(B_{1,j}\times
B_{2,k})}{V({\mathbf{x}},{\bm{\delta}})}\frac{1}{\mu(B_{1,j}\times
B_{2,k})}\int_{B_{1,j}\times B_{2,k}}|f({\mathbf{y}})|^{r}d\mu({\mathbf{y}})$
$\displaystyle\leq
c2^{-j(\tau_{1}-d_{1})-k(\tau_{2}-d_{2})}\mathcal{M}_{r}f({\mathbf{x}})^{r},$

where we used that in light of (2.9) and (3.9)

$\displaystyle\mu(B_{1,j}\times B_{2,k})$
$\displaystyle=\mu_{1}(B_{1}(x_{1},2^{j}\delta_{1}))\mu_{2}(B_{2}(x_{2},2^{k}\delta_{2}))$
$\displaystyle\leq
c2^{jd_{1}+kd_{2}}\mu_{1}(B_{1}(x_{1},\delta_{1}))\mu_{2}(B_{2}(x_{2},\delta_{2}))=c2^{jd_{1}+kd_{2}}V({\mathbf{x}},{\bm{\delta}}).$

Finally, we insert (4.8) in (4.106) and sum up using that $\tau_{i}>d_{i}$ to
obtain (4.105). ∎

##  5\. Distributions associated to operators

Distributions associated to operators have been introduced in [39]. Here we
develop distributions in the two-parameter setting of this article.

###  5.1. Test functions and distributions

We will operate in the setting described in Subsection 2.1. We define the
class of test functions $\mathcal{S}:=\mathcal{S}(L_{1},L_{2})$ as the set of
all function

$\phi\in\cap_{{\bm{\nu}}\in{\mathbb{N}}_{0}^{2}}D(L^{{\bm{\nu}}}),\quad
L^{{\bm{\nu}}}:=L_{1}^{\nu_{1}}\otimes
L_{2}^{\nu_{2}},\;{\bm{\nu}}=(\nu_{1},\nu_{2}),$

with $D(L^{{\bm{\nu}}})$ being the domain of the operator $L^{{\bm{\nu}}}$,
such that

(5.1) $\mathcal{P}_{m,k}(\phi):=\sup_{{\mathbf{x}}\in
X}\prod_{i=1,2}\big{(}1+\rho_{i}(x_{i},x_{0i})\big{)}^{k}\max_{0\leq\nu_{i}\leq
m}\big{|}L^{{\bm{\nu}}}\phi({\mathbf{x}})\big{|}<\infty,\quad\forall\,k,m\in{\mathbb{N}}_{0}.$

Here the point ${\mathbf{x}}_{0}=(x_{01},x_{02})\in X$ is selected arbitrarily
and fixed once and for all. Clearly, the definition of the class $\mathcal{S}$
is independent of the choice of ${\mathbf{x}}_{0}$.

It should be pointed out that the term
$\prod_{i=1,2}\big{(}1+\rho_{i}(x_{i},x_{0i})\big{)}^{k}$ in (5.1) can be
removed if $\mu(X_{i})<\infty$ for $i=1$ or $i=2$ because $\mu(X_{i})<\infty$
if and only if $\operatorname{diam}(X_{i})<\infty$, see [12, Proposition 2.1].

######  Proposition 5.1.

The vector space $\mathcal{S}:=\mathcal{S}(L_{1},L_{2})$ with the natural
topology induced by the norms $\mathcal{P}_{m,k}(\phi)$,
$m,k\in{\mathbb{N}}_{0}$, is a Fréchet space.

###### Proof.

Clearly, $\\{\mathcal{P}_{m,k}(\phi)\\}_{m,k\geq 0}$ is a countable family of
norms and hence $\mathcal{S}$ is a metrizable locally convex space. Therefore,
to prove that $\mathcal{S}$ is Fréchet we only need to show that it is
complete. Assume that $\\{\phi_{n}\\}_{n\geq 1}$ is a Cauchy sequence in
$\mathcal{S}$, i.e.
$\lim_{n,j\rightarrow\infty}\mathcal{P}_{m,k}(\phi_{n}-\phi_{j})=0$ for all
$m,k\geq 0$. Fix $k\geq(d_{1}\vee d_{2}+1)/2$. Then for any
${\bm{\nu}}\in{\mathbb{N}}_{0}^{2}$ with $0\leq\nu_{i}\leq m$ we have

$\displaystyle\big{\|}L^{{\bm{\nu}}}(\phi_{n}-\phi_{j})\big{\|}_{2}$
$\displaystyle\leq\mathcal{P}_{m,k}(\phi_{n}-\phi_{j})\int_{X_{1}\times
X_{2}}\prod_{i=1,2}\big{(}1+\rho_{i}(x_{i},x_{0i})\big{)}^{-2k}d\mu_{1}\times
d\mu_{2}$ $\displaystyle\leq
cV_{1}(x_{01},1)V_{2}(x_{02},1)\mathcal{P}_{m,k}(\phi_{n}-\phi_{j}),$

where we used (2.12). Therefore, $\\{L^{\bm{\nu}}\phi_{n}\\}_{n\geq 1}$ is a
Cauchy sequence in $L^{2}(X,d\mu)$ and hence there exists $\psi_{\bm{\nu}}\in
L^{2}(X,d\mu)$ such that
$\lim_{n\to\infty}\|L^{{\bm{\nu}}}\phi_{n}-\psi_{\bm{\nu}}\|_{2}=0$.

Denote $\phi:=\psi_{(0,0)}$ and let ${\bm{\nu}}\in{\mathbb{N}}_{0}^{2}$. From

$\lim_{n\to\infty}\|\phi_{n}-\phi\|_{2}=0\quad\hbox{and}\quad\lim_{n\to\infty}\|L^{{\bm{\nu}}}\phi_{n}-\psi_{\bm{\nu}}\|_{2}=0$

and the fact that the operator $L^{{\bm{\nu}}}=L_{1}^{\nu_{1}}\otimes
L_{2}^{\nu_{2}}$ is closed (as a self-adjoint operator, see Proposition 4.1)
it follows that $\phi\in D(L^{\bm{\nu}})$ and

(5.2) $\lim_{n\to\infty}\|L^{{\bm{\nu}}}\phi_{n}-L^{{\bm{\nu}}}\phi\|_{2}=0.$

Furthermore,
$\big{\|}L^{{\bm{\nu}}}\phi_{n}-L^{{\bm{\nu}}}\phi_{j}\big{\|}_{\infty}\leq\mathcal{P}_{m,0}(\phi_{n}-\phi_{j})\to
0$ as $n,j\to\infty$. From this, (5.2) and the completeness of $L^{\infty}$ it
follows that

$\lim_{n\to\infty}\|L^{{\bm{\nu}}}\phi_{n}-L^{{\bm{\nu}}}\phi\|_{\infty}=0,\quad\forall{\bm{\nu}}\in{\mathbb{N}}_{0}^{2}.$

In turn, this and the fact that
$\lim_{n,j\rightarrow\infty}\mathcal{P}_{m,k}(\phi_{n}-\phi_{j})=0$ for all
$m,k\geq 0$ imply that
$\lim_{n\rightarrow\infty}\mathcal{P}_{m,k}(\phi_{n}-\phi)=0$ for all $m,k\geq
0$, which completes the proof. ∎

######  Definition 5.2.

The space $\mathcal{S}^{\prime}=\mathcal{S}^{\prime}(L_{1},L_{2})$ of
distributions associated to $L_{1}$, $L_{2}$ is defined as the set of all
continuous linear functionals on $\mathcal{S}=\mathcal{S}(L_{1},L_{2})$. The
action of $f\in\mathcal{S}^{\prime}$ on $\overline{\phi}\in\mathcal{S}$ will
be denoted by $\langle f,\phi\rangle:=f(\overline{\phi})$, which is consistent
with the inner product in $L^{2}(X,d\mu)$.

Clearly, for any $f\in\mathcal{S}^{\prime}$ there exist constants
$m,k\in{\mathbb{N}}_{0}$ and $c>0$ such that

(5.3) $|\langle f,\phi\rangle|\leq
c\mathcal{P}_{m,k}(\phi),\quad\forall\phi\in\mathcal{S}.$

We next show that there are sufficiently many test functions.

######  Proposition 5.3.

Let $\varphi\in\mathcal{S}({\mathbb{R}}^{2})$ be real-valued and
$\varphi(\pm\lambda_{1},\pm\lambda_{2})=\varphi(\lambda_{1},\lambda_{2})$ for
$(\lambda_{1},\lambda_{2})\in{\mathbb{R}}^{2}$. Recall the notation
$\varphi(\sqrt{L}):=\varphi(\sqrt{L_{1}},\sqrt{L_{2}})$ and
$\mathcal{S}:=\mathcal{S}(L_{1},L_{2})$. Then:

$(i)$ $\varphi(\sqrt{L})$ is an integral operator whose kernel
${\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},\cdot)\in\mathcal{S}$,
$\forall{\mathbf{x}}\in X$, and
${\mathscr{K}}_{\varphi(\sqrt{L})}(\cdot,{\mathbf{y}})\in\mathcal{S}$,
$\forall{\mathbf{y}}\in X$.

$(ii)$ For each $\phi\in\mathcal{S}$ we have
$\varphi(\sqrt{L})\phi\in\mathcal{S}$. Furthermore, the spectral multiplier
$\varphi(\sqrt{L})$ maps continuously $\mathcal{S}$ into $\mathcal{S}$.

###### Proof.

(i) Since
${\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},{\mathbf{y}})={\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{y}},{\mathbf{x}})$
for ${\mathbf{x}},{\mathbf{y}}\in X$ ($\varphi$ is real-valued) it suffices to
only show that
${\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},\cdot)\in\mathcal{S}$. To this
end we need to show that

(5.4)
$\mathcal{P}_{m,k}({\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},\cdot))<\infty,\quad\forall
m,k\in{\mathbb{N}}_{0}.$

Fix ${\mathbf{x}}\in X$. From Corollary 4.22 it readily follows that for any
$m,k\in{\mathbb{N}}_{0}$ there exists a constant $c>0$ such that for all
${\bm{\nu}}\in{\mathbb{N}}_{0}^{2}$ with $\nu_{i}\leq m$

$\sup_{{\mathbf{y}}\in
X}\prod_{i=1,2}(1+\rho(x_{i},y_{i})^{k}|{\mathscr{K}}_{L^{\bm{\nu}}\varphi(\sqrt{L})}({\mathbf{x}},{\mathbf{y}})|\leq
cV({\mathbf{x}},{\bm{1}})^{-1}.$

Therefore, to show that
${\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},\cdot)\in\mathcal{S}$ it
suffices to show that

(5.5)
${\mathscr{K}}_{L^{\bm{\nu}}\varphi(\sqrt{L})}({\mathbf{x}},{\mathbf{y}})=L^{\bm{\nu}}[{\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},\cdot)]({\mathbf{y}}),\quad{\mathbf{x}},{\mathbf{y}}\in
X.$

For this we first need to verify that
${\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},\cdot)\in D(L^{\bm{\nu}})$.
Let $g\in D(L^{\bm{\nu}})$. As is well known (e.g. [55, Chapter XI, Section
12, Theorem 3, (iv)]) $L^{\bm{\nu}}g\in D(\varphi(\sqrt{L}))$ if and only if
$g\in D(\varphi(\sqrt{L})L^{\bm{\nu}})$, and

(5.6)
$\varphi(\sqrt{L})(L^{\bm{\nu}}g)=[\varphi(\sqrt{L})L^{\bm{\nu}}]g=[L^{\bm{\nu}}\varphi(\sqrt{L})]g.$

But $\varphi(\sqrt{L})$ is a bounded operator on $L^{2}(X,d\mu)$ and hence
$L^{\bm{\nu}}g\in D(\varphi(\sqrt{L}))$, which implies that $g\in
D(\varphi(\sqrt{L})L^{\bm{\nu}})$ and (5.6) is valid. Hence, for almost all
${\mathbf{x}}\in X$

(5.7)
$\int_{X}{\mathscr{K}}_{L^{\bm{\nu}}\varphi(\sqrt{L})}({\mathbf{x}},{\mathbf{y}})g({\mathbf{y}})d\mu({\mathbf{y}})=\int_{X}{\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},{\mathbf{y}})L^{\bm{\nu}}g({\mathbf{y}})d\mu({\mathbf{y}}).$

But the kernels
${\mathscr{K}}_{L^{\bm{\nu}}\varphi(\sqrt{L})}({\mathbf{x}},{\mathbf{y}})$ and
${\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},{\mathbf{y}})$ are continuous
and hence (5.7) is valid for all ${\mathbf{x}}\in X$. Applying the Cauchy-
Schwarz inequality we obtain from (5.7)

$\Big{|}\int_{X}L^{\bm{\nu}}g({\mathbf{y}}){\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},{\mathbf{y}})d\mu({\mathbf{y}})\Big{|}\leq\|{\mathscr{K}}_{L^{\bm{\nu}}\varphi(\sqrt{L})}({\mathbf{x}},\cdot)\|_{2}\|g\|_{2}\leq
c\|g\|_{2},$

where we used Corollary 4.22 and (3.19). Thus

$|\big{\langle}L^{\bm{\nu}}g,{\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},\cdot)\big{\rangle}|\leq
c\|g\|_{2},\quad\forall g\in D(L^{\bm{\nu}}),$

which leads to the conclusion that
${\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},\cdot)\in
D((L^{\bm{\nu}})^{*})$. But $D(L^{\bm{\nu}})=D((L^{\bm{\nu}})^{*})$ because
$L^{\bm{\nu}}$ is self-adjoint. Hence
${\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},\cdot)\in D(L^{\bm{\nu}})$.

We now show (5.5). Appealing to (5.7) and the fact that $L^{\bm{\nu}}$ is
self-adjoint we obtain for all $g\in D(L^{\bm{\nu}})$ and ${\mathbf{x}}\in X$

$\displaystyle\big{\langle}{\mathscr{K}}_{L^{\bm{\nu}}\varphi(\sqrt{L})}({\mathbf{x}},\cdot),\overline{g}\big{\rangle}$
$\displaystyle=\big{\langle}L^{\bm{\nu}}g,{\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},\cdot)\big{\rangle}=\big{\langle}g,(L^{\bm{\nu}})^{*}[{\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},\cdot)]\big{\rangle}$
$\displaystyle=\big{\langle}g,L^{\bm{\nu}}[{\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},\cdot)]\big{\rangle}=\big{\langle}L^{\bm{\nu}}[{\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},\cdot)],\overline{g}\big{\rangle},$

which implies (5.5).

(ii) Let $m,k\in{\mathbb{N}}_{0}$ and assume that
${\bm{\nu}}\in{\mathbb{N}}_{0}^{2}$ with ${\bm{\nu}}_{i}\leq m$ and
${\bm{\sigma}}\in(0,\infty)^{2}$ is such that $\sigma_{i}\geq k+2d_{i}+1$,
$i=1,2$. Denote $\tilde{d}:=\lfloor d_{1}\vee d_{2}\rfloor+1$. Using (4.100)
and (3.20) we get for $\phi\in\mathcal{S}$

$\displaystyle\big{|}L^{{\bm{\nu}}}\varphi(\sqrt{L})$
$\displaystyle\phi({\mathbf{x}})\big{|}=\big{|}\varphi(\sqrt{L})L^{{\bm{\nu}}}\phi({\mathbf{x}})\big{|}\leq\int_{X}|{\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},{\mathbf{y}})||L^{{\bm{\nu}}}\phi({\mathbf{y}})|d\mu({\mathbf{y}})$
$\displaystyle\leq
c\int_{X}{\mathfrak{D}}_{{\bm{1}},{\bm{\sigma}}}({\mathbf{x}},{\mathbf{y}})|L^{{\bm{\nu}}}\phi({\mathbf{y}})|d\mu({\mathbf{y}})$
$\displaystyle\leq
c\mathcal{P}_{m,k+\tilde{d}+1}(\phi)\int_{X}V({\mathbf{x}},{\bm{1}})^{-1}{\mathfrak{D}}_{{\bm{1}},\bar{k}+{\mathbf{d}}+{\bm{1}}}^{*}({\mathbf{x}},{\mathbf{y}}){\mathfrak{D}}_{{\bm{1}},\bar{k}+{\mathbf{d}}+{\bm{1}}}^{*}({\mathbf{y}},{\mathbf{x}}_{0})d\mu({\mathbf{y}})$
$\displaystyle\leq
c\mathcal{P}_{m,k+\tilde{d}+1}(\phi){\mathfrak{D}}_{{\bm{1}},\bar{k}+{\bm{1}}}^{*}({\mathbf{x}},{\mathbf{x}}_{0})\leq
c\mathcal{P}_{m,k+\tilde{d}+1}(\phi){\mathfrak{D}}_{{\bm{1}},\bar{k}}^{*}({\mathbf{x}},{\mathbf{x}}_{0}),$

where $\bar{k}=:(k,k)$ and ${\bm{1}}:=(1,1)$. From above and (3.10) it readily
follows that $\varphi(\sqrt{L})\phi\in\mathcal{S}$ and
$\mathcal{P}_{m,k}((\varphi\sqrt{L})\phi)\leq
c\mathcal{P}_{m,k+\tilde{d}+1}(\phi)$, which leads to the conclusion that
$\varphi(\sqrt{L})$ maps continuously $\mathcal{S}$ into $\mathcal{S}$. ∎

We use Proposition 5.3 (ii) to make the following

######  Definition 5.4.

Let $\varphi\in\mathcal{S}({\mathbb{R}}^{2})$ be real-valued and
$\varphi(\pm\lambda_{1},\pm\lambda_{2})=\varphi(\lambda_{1},\lambda_{2})$ for
$(\lambda_{1},\lambda_{2})\in{\mathbb{R}}^{2}$. For any
$f\in\mathcal{S}^{\prime}=\mathcal{S}^{\prime}(L_{1},L_{2})$ we define
$\varphi(\sqrt{L})f$ by

(5.8) $\langle\varphi(\sqrt{L})f,\phi\rangle=\langle
f,\varphi(\sqrt{L})\phi\rangle,\quad\forall\phi\in\mathcal{S}.$

######  Proposition 5.5.

Let $\varphi$ be as in Definition 5.4. Then for any
$f\in\mathcal{S}^{\prime}=\mathcal{S}^{\prime}(L_{1},L_{2})$

(5.9) $\varphi(\sqrt{L})f({\mathbf{x}}):=\langle
f,{\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},\cdot)\rangle\quad\hbox{for}\quad{\mathbf{x}}\in
X.$

###### Proof.

Let $f\in\mathcal{S}^{\prime}$. Just as in (5.3) there exist constants
$m,k\in{\mathbb{N}}_{0}$ and $c>0$ such that

(5.10) $|\langle f,\phi\rangle|\leq
c\mathcal{P}_{m,k}(\phi),\quad\forall\phi\in\mathcal{S}.$

Fix $\phi\in\mathcal{S}$. Clearly,

$\varphi(\sqrt{L})\phi({\mathbf{x}})=\int_{X}{\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},{\mathbf{y}})\phi({\mathbf{y}})d\mu({\mathbf{y}}),\quad{\mathbf{x}}\in
X.$

We will regard the above integral as a Bochner integral over the Banach space

$V_{m,k}:=\\{g\in\cap_{0\leq\nu_{i}\leq
m}D(L^{\bm{\nu}}):\|g\|_{V_{m,k}}:=\mathcal{P}_{m,k}(g)<\infty\\}$

with $\mathcal{P}_{m,k}$ defined in (5.1), see e.g. [55], pp. 131–133. Just as
in the proof of Proposition 5.1 it follows that the space $V_{m,k}$ is
complete and hence it is a Banach space. By the Hahn-Banach theorem and (5.10)
the continuous linear functional $f$ can be extended to $V_{m,k}$ with the
same norm. We denote this extension by $f$ again.

Denote
$F_{\mathbf{y}}({\mathbf{x}}):={\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},{\mathbf{y}})\phi({\mathbf{y}})$.
Then

$\displaystyle\|F_{\mathbf{y}}\|_{V_{m,k}}$
$\displaystyle=\sup_{{\mathbf{x}}\in
X}\prod_{i=1,2}\big{(}1+\rho_{i}(x_{i},x_{0i})\big{)}^{k}\max_{0\leq{\bm{\nu}}_{i}\leq
m}\big{|}L^{{\bm{\nu}}}{\mathscr{K}}_{\varphi(\sqrt{L})}({\mathbf{x}},{\mathbf{y}})\phi({\mathbf{y}})\big{|}$
$\displaystyle=\sup_{{\mathbf{x}}\in
X}\prod_{i=1,2}\big{(}1+\rho_{i}(x_{i},x_{0i})\big{)}^{k}\max_{0\leq{\bm{\nu}}_{i}\leq
m}\big{|}{\mathscr{K}}_{L^{{\bm{\nu}}}\varphi(\sqrt{L})}({\mathbf{x}},{\mathbf{y}})\phi({\mathbf{y}})\big{|},$

where we used (5.5). Clearly, the function $\varphi$ obeys the conditions of
Corollary 4.22 and hence there exists a constant $c_{k}>0$ such that

$|{\mathscr{K}}_{L^{{\bm{\nu}}}\varphi(\sqrt{L})}({\mathbf{x}},{\mathbf{y}})|\leq
c_{k}V({\mathbf{y}},{\bm{1}})^{-1}\prod_{i=1,2}\big{(}1+\rho_{i}(x_{i},y_{i})\big{)}^{-k},\quad
0\leq\nu_{i}\leq m.$

On the other hand, since $\phi\in\mathcal{S}$ for any $N>0$ we have

$|\phi({\mathbf{y}})|\leq
c_{N}\mathcal{P}_{0,N}(\phi)\prod_{i=1,2}\big{(}1+\rho_{i}(y_{i},x_{0i})\big{)}^{-N}.$

We choose $N=k+2(d_{1}\vee d_{2})+1$. Putting the above estimates together we
get

$\displaystyle\|F_{\mathbf{y}}\|_{V_{m,k}}\leq\frac{c\mathcal{P}_{0,N}(\phi)}{V({\mathbf{y}},{\bm{1}})}\sup_{{\mathbf{x}}\in
X}\frac{\prod_{i=1,2}\big{(}1+\rho_{i}(x_{i},x_{0i})\big{)}^{k}}{\prod_{i=1,2}\big{(}1+\rho_{i}(x_{i},y_{i})\big{)}^{k}\prod_{i=1,2}\big{(}1+\rho_{i}(y_{i},x_{0i})\big{)}^{N}}.$

Using the obvious inequality
$1+\rho_{i}(x_{i},x_{0i})\leq(1+\rho_{i}(x_{i},y_{i}))(1+\rho_{i}(y_{i},x_{0i}))$
and (3.13) we obtain

$\displaystyle\|F_{\mathbf{y}}\|_{V_{m,k}}\leq\frac{c\mathcal{P}_{0,N}(\phi)}{V({\mathbf{x}}_{0},{\bm{1}})\prod_{i=1,2}\big{(}1+\rho_{i}(y_{i},x_{0i})\big{)}^{d_{i}+1}}.$

From this and (3.19) it follows that

$\int_{X}\|{\mathscr{K}}_{\varphi(\sqrt{L})}(\cdot,{\mathbf{y}})\phi({\mathbf{y}})\|_{V_{m,k}}d\mu({\mathbf{y}})\leq
c\mathcal{P}_{0,N}(\phi)<\infty.$

Now, applying the theory of Bochner’s integral we obtain

$\displaystyle\langle
f,\varphi(\sqrt{L})\phi\rangle=\Big{\langle}f,\int_{X}{\mathscr{K}}_{\varphi(\sqrt{L})}(\cdot,{\mathbf{y}})\phi({\mathbf{y}})d\mu({\mathbf{y}})\Big{\rangle}=\int_{X}\big{\langle}f,{\mathscr{K}}_{\varphi(\sqrt{L})}(\cdot,{\mathbf{y}})\big{\rangle}\overline{\phi({\mathbf{y}})}d\mu({\mathbf{y}}).$

This coupled with (5.8) implies (5.9). ∎

######  Remark 5.6.

From Proposition 5.3 (ii) it readily follows that the operator
$\varphi(\sqrt{L})$ from above maps continuously $\mathcal{S}^{\prime}$ into
$\mathcal{S}^{\prime}$.

Our next step is to show that $\varphi(\sqrt{L})f$ is a slowly growing
continuous function for smooth $\varphi$ and $f\in\mathcal{S}^{\prime}$.

######  Proposition 5.7.

Let $\varphi\in\mathcal{S}({\mathbb{R}}^{2})$ be real-valued and
$\varphi(\pm\lambda_{1},\pm\lambda_{2})=\varphi(\lambda_{1},\lambda_{2})$ for
$(\lambda_{1},\lambda_{2})\in{\mathbb{R}}^{2}$. Let
${\bm{\delta}}=(\delta_{1},\delta_{2})>{\bm{0}}$. Then for any
$f\in\mathcal{S}^{\prime}=\mathcal{S}^{\prime}(L_{1},L_{2})$,
$\varphi(\delta\sqrt{L})f$ is a slowly growing continuous function. More
explicitly, there exist constants $N,c>0$ $($depending on $f$$)$ such that

(5.11) $|\varphi({\bm{\delta}}\sqrt{L})f({\mathbf{x}})|\leq
cV(x_{0},{\bm{\delta}})^{-1}\prod_{i=1,2}(\delta_{i}^{N}+\delta_{i}^{-N})\big{(}1+\rho_{i}(x_{i},x_{0i})\big{)}^{N}\quad\hbox{and}$
$\displaystyle|\varphi({\bm{\delta}}\sqrt{L})f({\mathbf{x}})$
$\displaystyle-\varphi({\bm{\delta}}\sqrt{L})f({\mathbf{x}}^{\prime})|$ (5.12)
$\displaystyle\leq
cV({\mathbf{x}}_{0},{\bm{\delta}})^{-1}\prod_{i=1,2}(\delta_{i}^{N}+\delta_{i}^{-N})\big{(}1+\rho_{i}(x_{i},x_{0i})\big{)}^{N}\sum_{i=1,2}\rho_{i}(x_{i},x^{\prime}_{i})^{\alpha_{i}}$

if $\rho_{i}(x_{i},x^{\prime}_{i})\leq\delta_{i}$. Here $\alpha_{1}$,
$\alpha_{2}$ are from (2.4).

###### Proof.

We will only prove (5.7); the proof of (5.11) is easier and will be omitted.
Let $f\in\mathcal{S}^{\prime}$. Just as in (5.3) there exist
$m,k\in{\mathbb{N}}_{0}$ such that

$|\langle f,\phi\rangle|\leq
c\mathcal{P}_{m,k}(\phi),\quad\forall\phi\in\mathcal{S}.$

We use this, (5.8), and (5.5) to obtain

$\displaystyle|\varphi({\bm{\delta}}\sqrt{L})f({\mathbf{x}})-\varphi({\bm{\delta}}\sqrt{L})f({\mathbf{x}}^{\prime})|$
$\displaystyle=\big{|}\big{\langle}f,{\mathscr{K}}_{\varphi({\bm{\delta}}\sqrt{L})}({\mathbf{x}},\cdot)-{\mathscr{K}}_{\varphi({\bm{\delta}}\sqrt{L})}({\mathbf{x}}^{\prime},\cdot)\big{\rangle}\big{|}\leq
c\mathcal{P}_{m,k}\big{(}{\mathscr{K}}_{\varphi({\bm{\delta}}\sqrt{L})}({\mathbf{x}},\cdot)-{\mathscr{K}}_{\varphi({\bm{\delta}}\sqrt{L})}({\mathbf{x}}^{\prime},\cdot)\big{)}$
$\displaystyle=c\sup_{{\mathbf{y}}\in
X}\prod_{i=1,2}\big{(}1+\rho_{i}(y_{i},x_{0i})\big{)}^{k}$
$\displaystyle\hskip 101.17755pt\times\max_{0\leq\nu_{i}\leq
m}\big{|}L^{{\bm{\nu}}}[{\mathscr{K}}_{\varphi({\bm{\delta}}\sqrt{L})}({\mathbf{x}},\cdot)]({\mathbf{y}})-L^{{\bm{\nu}}}[{\mathscr{K}}_{\varphi({\bm{\delta}}\sqrt{L})}({\mathbf{x}}^{\prime},\cdot)]({\mathbf{y}})\big{|}$
$\displaystyle\leq c\max_{0\leq\nu_{i}\leq m}\sup_{{\mathbf{y}}\in
X}\prod_{i=1,2}\big{(}1+\rho_{i}(y_{i},x_{0i})\big{)}^{k}\big{|}{\mathscr{K}}_{L^{\bm{\nu}}\varphi({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})-{\mathscr{K}}_{L^{\bm{\nu}}\varphi({\bm{\delta}}\sqrt{L})}({\mathbf{x}}^{\prime},{\mathbf{y}})\big{|}.$

The function $\varphi$ obeys the conditions of Corollary 4.22 and hence if
$\rho_{i}(x_{i},x^{\prime}_{i})\leq\delta_{i}$, then

$\displaystyle\big{|}{\mathscr{K}}_{L^{\bm{\nu}}\varphi({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})-{\mathscr{K}}_{L^{\bm{\nu}}\varphi({\bm{\delta}}\sqrt{L})}({\mathbf{x}}^{\prime},{\mathbf{y}})\big{|}$
$\displaystyle\leq
cV({\mathbf{x}},{\bm{\delta}})^{-1}{\bm{\delta}}^{-2{\bm{\nu}}}\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho_{i}(x_{i},y_{i})\big{)}^{-k}\sum_{i=1,2}(\rho_{i}(x_{i},x^{\prime}_{i})/\delta_{i})^{\alpha_{i}}$
$\displaystyle\leq
cV({\mathbf{x}},{\bm{\delta}})^{-1}\prod_{i=1,2}(1+\delta_{i})^{k}(1+\delta_{i}^{-1})^{2\nu_{i}+\alpha_{i}}\big{(}1+\rho_{i}(x_{i},y_{i})\big{)}^{-k}\sum_{i=1,2}\rho_{i}(x_{i},x^{\prime}_{i})^{\alpha_{i}}.$

Now, we use that
$1+\rho_{i}(y_{i},x_{0i})\leq(1+\rho_{i}(x_{i},y_{i}))(1+\rho_{i}(x_{i},x_{0i}))$
and

$\displaystyle V({\mathbf{x}}_{0},{\bm{\delta}})$ $\displaystyle\leq
c\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},x_{0i}))^{d_{i}}V({\mathbf{x}},{\bm{\delta}})$
$\displaystyle\leq
c\prod_{i=1,2}(1+\delta_{i}^{-1})^{d_{i}}(1+\rho_{i}(x_{i},x_{0i}))^{d_{i}}V({\mathbf{x}},{\bm{\delta}}),$

which follows from (3.13) to obtain

$\displaystyle\big{|}{\mathscr{K}}_{L^{\bm{\nu}}\varphi({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})-{\mathscr{K}}_{L^{\bm{\nu}}\varphi({\bm{\delta}}\sqrt{L})}({\mathbf{x}}^{\prime},{\mathbf{y}})\big{|}$
$\displaystyle\leq
c\frac{\prod_{i=1,2}(1+\delta_{i})^{k}(1+\delta_{i}^{-1})^{2\nu_{i}+\alpha_{i}+d_{i}}\big{(}1+\rho_{i}(x_{i},x_{0i})\big{)}^{k+d_{i}}}{V({\mathbf{x}}_{0},{\bm{\delta}})\prod_{i=1,2}\big{(}1+\rho_{i}(y_{i},x_{0i})\big{)}^{k}}\sum_{i=1,2}\rho_{i}(x_{i},x^{\prime}_{i})^{\alpha_{i}}.$

Putting all of the above together we obtain (5.7) with $N$ sufficiently large.
∎

###  5.2. Decomposition

We next establish a basic convergence theorem and a Calderón type reproducing
formula, which will play an important role in this study. We begin with the
following

######  Theorem 5.8.

Let $\varphi\in\mathcal{S}(\mathbb{R}^{2})$ be a real-valued function that
satisfies the conditions: $\varphi({\bm{0}})=1$,
$\partial^{\bm{\beta}}\varphi({\bm{0}})=0$ for all
${\bm{\beta}}\in{\mathbb{N}}_{0}^{2}$ with $|{\bm{\beta}}|\geq 1$,
${\bm{0}}:=(0,0)$, and
$\varphi(\pm\lambda_{1},\pm\lambda_{2})=\varphi(\lambda_{1},\lambda_{2})$ for
$(\lambda_{1},\lambda_{2})\in\mathbb{R}^{2}$. Then for any
$\phi\in\mathcal{S}$

(5.13)
$\phi=\lim_{{\bm{\delta}}\to{\bm{0}}}\varphi({\bm{\delta}}\sqrt{L})\phi\quad\mbox{
in }\;\mathcal{S}$

and, therefore, for any $f\in\mathcal{S}^{\prime}$

(5.14)
$f=\lim_{{\bm{\delta}}\to{\bm{0}}}\varphi({\bm{\delta}}\sqrt{L})f\quad\mbox{
in }\;\mathcal{S}^{\prime}.$

###### Proof.

We need to show that for any $m,k\in{\mathbb{N}}_{0}$ and $\phi\in\mathcal{S}$

(5.15) $\displaystyle\lim_{{\bm{\delta}}\rightarrow{\bm{0}}}$
$\displaystyle\mathcal{P}_{m,k}(\phi-\varphi({\bm{\delta}}\sqrt{L})\phi)$
$\displaystyle=\lim_{{\bm{\delta}}\rightarrow{\bm{0}}}\sup_{{\mathbf{x}}\in
X}\prod_{i=1,2}\big{(}1+\rho_{i}(x_{i},x_{0i})\big{)}^{k}\max_{0\leq\nu_{i}\leq
m}\big{|}L^{{\bm{\nu}}}\big{(}\phi-\varphi({\bm{\delta}}\sqrt{L})\phi\big{)}({\mathbf{x}})|=0.$

Choose ${\mathbf{n}}=(n_{1},n_{2})\in{\mathbb{N}}^{2}$, $n_{i}>k+5d_{i}/2$,
and $r\in{\mathbb{N}},r>(d_{1}+d_{2}+n_{1}+n_{2})/2$. We define
$\omega({\bm{\lambda}}):=(1-\varphi({\bm{\lambda}}))|{\bm{\lambda}}|^{-2r}$.
Clearly,
$I-\varphi({\bm{\delta}}\sqrt{L})=\omega({\bm{\delta}}\sqrt{L})(\delta_{1}^{2}L_{1}+\delta_{2}^{2}L_{2})^{r}$.
Since $|{\bm{\lambda}}|^{2r}=(\lambda_{1}^{2}+\lambda_{2}^{2})^{r}$ is a
radial function from the definition of $\varphi$ it follows that $\omega$
satisfies the assumptions of Theorem 4.21. Hence

(5.16)
${\mathscr{K}}_{\omega({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})\leq
c_{\mathbf{n}}{\mathfrak{D}}_{{\bm{\delta}},{\mathbf{n}}}({\mathbf{x}},{\mathbf{y}}),\quad{\mathbf{x}},{\mathbf{y}}\in
X.$

From the representation
$(\delta_{1}^{2}L_{1}+\delta_{2}^{2}L_{2})^{r}=\sum_{\ell=0}^{r}\binom{r}{\ell}\delta_{1}^{2\ell}\delta_{2}^{2(r-\ell)}L_{1}^{\ell}L_{2}^{r-\ell}$
and assuming $0\leq\nu_{i}\leq m$, we get

(5.17) $\displaystyle
L^{{\bm{\nu}}}\big{(}\phi-\varphi({\bm{\delta}}\sqrt{L})\phi\big{)}({\mathbf{x}})$
$\displaystyle=L^{{\bm{\nu}}}\omega({\bm{\delta}}\sqrt{L})(\delta_{1}^{2}L_{1}+\delta_{2}^{2}L_{2})^{r}\phi({\mathbf{x}})$
$\displaystyle=\sum_{\ell=0}^{r}\binom{r}{\ell}\delta_{1}^{2\ell}\delta_{2}^{2(r-\ell)}\omega(\delta\sqrt{L})L^{(\nu_{1}+\ell,\nu_{2}+r-\ell)}\phi({\mathbf{x}}).$

Pick $N\geq(n_{1}-d_{1}/2)\vee(n_{2}-d_{2}/2)$ and assume
${\bm{0}}<{\bm{\delta}}\leq{\bm{1}}$, ${\bm{1}}:=(1,1)$. Then using
(5.16)–(5.17) we obtain

$\displaystyle|L^{{\bm{\nu}}}$
$\displaystyle\big{(}\phi-\varphi({\bm{\delta}}\sqrt{L})\phi\big{)}({\mathbf{x}})|\leq
c\sum_{\ell=0}^{r}\delta_{1}^{2\ell}\delta_{2}^{2(r-\ell)}\int_{X}{\mathfrak{D}}_{{\bm{\delta}},{\mathbf{n}}}({\mathbf{x}},{\mathbf{y}})|L^{(\nu_{1}+\ell,\nu_{2}+r-\ell)}\phi({\mathbf{y}})|d\mu({\mathbf{y}})$
$\displaystyle\leq
c\sum_{\ell=0}^{r}\delta_{1}^{2\ell}\delta_{2}^{2(r-\ell)}\mathcal{P}_{m+r,N}(\phi)\int_{X}{\mathfrak{D}}_{{\bm{\delta}},{\mathbf{n}}}({\mathbf{x}},{\mathbf{y}}){\mathfrak{D}}^{*}_{{\bm{1}},{\mathbf{n}}-{\mathbf{d}}/2}({\mathbf{y}},{\mathbf{x}}_{0})d\mu({\mathbf{y}})$
$\displaystyle\leq
c\sum_{\ell=0}^{r}\delta_{1}^{2\ell}\delta_{2}^{2(r-\ell)}\mathcal{P}_{m+r,N}(\phi)\int_{X}\frac{1}{V({\mathbf{y}},{\bm{\delta}})}{\mathfrak{D}}^{*}_{{\bm{\delta}},{\mathbf{n}}-{\mathbf{d}}/2}({\mathbf{x}},{\mathbf{y}}){\mathfrak{D}}^{*}_{{\bm{1}},{\mathbf{n}}-{\mathbf{d}}/2}({\mathbf{y}},{\mathbf{x}}_{0})d\mu({\mathbf{y}})$
$\displaystyle\leq
c\sum_{\ell=0}^{r}\delta_{1}^{2\ell}\delta_{2}^{2(r-\ell)}\mathcal{P}_{m+r,N}(\phi){\mathfrak{D}}^{*}_{{\bm{1}},{\mathbf{n}}-{\mathbf{d}}/2}({\mathbf{x}},{\mathbf{x}}_{0}),$

where for the third inequality we used (3.14) and for the last inequality we
used (3.22) and that $n_{i}-d_{i}/2>2d_{i}$. Finally, putting all together we
obtain

$\displaystyle\mathcal{P}_{m,k}(\phi-\varphi({\bm{\delta}}\sqrt{L})\phi)$
$\displaystyle\leq
c\sum_{\ell=0}^{r}\delta_{1}^{2\ell}\delta_{2}^{2(r-\ell)}\mathcal{P}_{m+r,N}(\phi)\sup_{{\mathbf{x}}\in
X}\prod_{i=1,2}\frac{\big{(}1+\rho_{i}(x_{i},x_{0i})\big{)}^{k}}{\big{(}1+\rho_{i}(x_{i},x_{0i})\big{)}^{n_{i}-d_{i}/2}}$
$\displaystyle\leq
c\mathcal{P}_{m+r,N}(\phi)\sum_{\ell=0}^{r}\delta_{1}^{2\ell}\delta_{2}^{2(r-\ell)}\rightarrow
0\;\text{ as }\;\delta_{1},\delta_{2}\rightarrow 0,$

where we used that $n_{i}\geq k+d_{i}/2$. The proof of (5.15) is complete. ∎

We next derive a Calderón type reproducing formula in the setting of this
article. (Compare with the result from Lemma 4.15.) It relies on two pairs of
functions $\varphi^{i}_{0},\varphi^{i}\in\mathcal{C}^{\infty}(\mathbb{R})$,
$i=1,2$, with the properties:

(5.18) $\displaystyle(i)\;\;\hbox{$\varphi^{i}_{0},\varphi^{i}$ are real-
valued and even,}$ $\displaystyle(ii)\
\operatorname{supp}\varphi^{i}_{0}\subset[-2,2],\
\operatorname{supp}\varphi^{i}\subset[2^{-1},2]\setminus[-1/2,1/2],$

and

(5.19) $\varphi^{i}_{0}(t)+\sum_{n\geq 1}\varphi^{i}(2^{-n}t)=1,\quad\forall
t\in\mathbb{R}.$

It is easy to construct functions
$\varphi^{i}_{0},\varphi^{i}\in\mathcal{C}^{\infty}(\mathbb{R})$, $i=1,2$,
with the above properties. Indeed, let
$\varphi_{0}^{i}\in\mathcal{C}^{\infty}(\mathbb{R})$ ($i=1,2$) be an even
real-valued function satisfying the conditions: (a) $\varphi_{0}^{i}(t)=1$,
$t\in[-1,1]$, (b) $\operatorname{supp}\varphi_{0}^{i}\subset[-2,2]$, (c)
$0\leq\varphi_{0}^{i}\leq 1$, and define
$\varphi^{i}(t):=\varphi^{i}_{0}(t)-\varphi_{0}^{i}(2t)$. Then it follows
immediately that $\varphi^{i}_{0},\varphi^{i}$ satisfy conditions
(5.18)–(5.19).

Using the standard notation
$\varphi^{i}_{n}(t):=\varphi^{i}(2^{-n}t),n\in{\mathbb{N}}$, for any
${\mathbf{j}}=(j_{1},j_{2})\in{\mathbb{N}}^{2}_{0}$ we set
$\varphi_{\mathbf{j}}:=\varphi^{1}_{j_{1}}\otimes\varphi^{2}_{j_{2}}$. Then
from (5.19) it readily follows that

(5.20)
$\sum_{{\mathbf{j}}\in{\mathbb{N}}_{0}^{2}}\varphi_{\mathbf{j}}({\bm{\lambda}})=1,\quad\forall{\bm{\lambda}}\in\mathbb{R}^{2}.$

######  Corollary 5.9.

Let
$\varphi^{1}_{0},\varphi^{2}_{0},\varphi^{1},\varphi^{2}\in\mathcal{C}^{\infty}(\mathbb{R})$
be functions satisfying $(\ref{phi-phi0})$–$(\ref{1dCalderon})$ and let
$\varphi_{\mathbf{j}}$ be as above. Then, for any $\phi\in\mathcal{S}$

(5.21)
$\phi=\sum_{{\mathbf{j}}\in{\mathbb{N}}_{0}^{2}}\varphi_{\mathbf{j}}(\sqrt{L})\phi\quad\mbox{
in }\;\mathcal{S}$

and, therefore, for any $f\in\mathcal{S}^{\prime}$

(5.22)
$f=\sum_{{\mathbf{j}}\in{\mathbb{N}}_{0}^{2}}\varphi_{\mathbf{j}}(\sqrt{L})f\quad\mbox{
in }\;\mathcal{S}^{\prime}.$

###### Proof.

Let $f\in\mathcal{S}$ (resp. $f\in\mathcal{S}^{\prime}$). We set
$\theta^{i}(t):=\varphi^{i}_{0}(t)+\varphi^{i}(2^{-1}t)$, $i=1,2$, and observe
that (5.18)–(5.19) imply $\operatorname{supp}\theta^{i}\subset[-4,4]$ and
$\theta^{i}(t)=1$ for $t\in[-2,2]$. Moreover from the definition of
$\theta^{i}$ and (5.19) we have

$\sum_{n=0}^{\ell+1}\varphi^{i}_{n}(t)=\theta^{i}\big{(}2^{-\ell}t\big{)},\quad\ell\in{\mathbb{N}}_{0}.$

Set $\theta:=\theta^{1}\otimes\theta^{2}$. Then it is easily seen that for any
${\mathbf{k}}:=(k_{1},k_{2})\in{\mathbb{N}}^{2}$,

$\sum_{j_{1}=0}^{k_{1}+1}\sum_{j_{2}=0}^{k_{2}+1}\varphi_{{\mathbf{j}}}({\bm{\lambda}})=(\theta^{1}\otimes\theta^{2})(b_{1}^{-k_{1}}\lambda_{1},b_{2}^{-k_{2}}\lambda_{2})=\theta\big{(}b^{-{\mathbf{k}}}{\bm{\lambda}}\big{)}.$

Finally, applying Theorem 5.8 to $\theta$ we obtain

$\displaystyle\sum_{{\mathbf{j}}\in{\mathbb{N}}_{0}^{2}}\varphi_{\mathbf{j}}(\sqrt{L})f$
$\displaystyle=\lim_{k_{1},k_{2}\rightarrow\infty}\sum_{j_{1}=0}^{k_{1}+1}\sum_{j_{2}=0}^{k_{2}+1}\big{(}\varphi^{1}_{j_{1}}(\sqrt{L_{1}})\otimes\varphi^{2}_{j_{2}}(\sqrt{L_{2}})\big{)}(f)$
$\displaystyle=\lim_{k_{1},k_{2}\rightarrow\infty}\theta(b^{-{\mathbf{k}}}\sqrt{L})f=f,$

which completes the proof. ∎

Our next step is to establish our principle convergence result.

######  Theorem 5.10.

Let $\varphi\in\mathcal{S}(\mathbb{R}^{2})$ be a real-valued function
satisfying the conditions: $\varphi({\bm{0}})=1$ and
$\varphi(\pm\lambda_{1},\pm\lambda_{2})=\varphi(\lambda_{1},\lambda_{2})$ for
$(\lambda_{1},\lambda_{2})\in\mathbb{R}^{2}$. Then for any
$\phi\in\mathcal{S}$

(5.23)
$\phi=\lim_{{\bm{\delta}}\to{\bm{0}}}\varphi({\bm{\delta}}\sqrt{L})\phi\quad\mbox{
in }\;\mathcal{S}$

and, therefore, for any $f\in\mathcal{S}^{\prime}$

(5.24)
$f=\lim_{{\bm{\delta}}\to{\bm{0}}}\varphi({\bm{\delta}}\sqrt{L})f\quad\mbox{
in }\;\mathcal{S}^{\prime}.$

Furthermore, (5.24) holds in $L^{p}$ for any $f\in L^{p}$, $1\leq
p\leq\infty$, where $L^{\infty}={\rm UCB}$ is the space of all uniformly
continuous and bounded functions on $X$.

For the proof of this theorem we need the following

######  Lemma 5.11.

Let ${\bm{\sigma}}>{\bm{0}}$, $k\geq(\sigma_{1}+d_{1})\vee(\sigma_{2}+d_{2})$,
and $m>\frac{d_{1}+\alpha_{1}}{2}\vee\frac{d_{2}+\alpha_{2}}{2}$ with
${\bm{\alpha}}=(\alpha_{1},\alpha_{2})$ from $(\ref{lip})$. Then there exists
a constant $c>0$ such that for any $\phi\in\mathcal{S}$ and
${\mathbf{x}},{\mathbf{y}}\in X$

(5.25) $|\phi({\mathbf{x}})-\phi({\mathbf{y}})|\leq
c\Big{(}\sum_{i=1,2}\rho_{i}(x_{i},y_{i})^{\alpha_{i}}\Big{)}\mathcal{P}_{m,k}(\phi)\big{[}{\mathfrak{D}}^{*}_{{\bm{1}},{\bm{\sigma}}}({\mathbf{x}},{\mathbf{x}}_{0})+{\mathfrak{D}}^{*}_{{\bm{1}},{\bm{\sigma}}}({\mathbf{y}},{\mathbf{x}}_{0})].$

###### Proof.

Let
$\varphi^{1}_{0},\varphi^{2}_{0},\varphi^{1},\varphi^{2}\in\mathcal{C}^{\infty}(\mathbb{R})$
be functions satisfying $(\ref{phi-phi0})$–$(\ref{1dCalderon})$ and let
$\varphi_{\mathbf{j}}:=\varphi^{1}_{j_{1}}\otimes\varphi^{2}_{j_{2}}$ be just
as in Corollary 5.9. Then by Corollary 5.9

(5.26)
$\phi=\sum_{{\mathbf{j}}\in{\mathbb{N}}_{0}^{2}}\varphi_{\mathbf{j}}(\sqrt{L})\phi\quad\mbox{
in }\;\mathcal{S}.$

We write

(5.27) $\displaystyle\phi({\mathbf{x}})-\phi({\mathbf{y}})$
$\displaystyle=\sum_{{\mathbf{j}}\in{\mathbb{N}}_{0}^{2}}\big{(}\varphi_{\mathbf{j}}(\sqrt{L})\phi({\mathbf{x}})-\varphi_{\mathbf{j}}(\sqrt{L})\phi({\mathbf{y}})\big{)}=:\sum_{k=1}^{4}\sum_{{\mathbf{j}}\in\mathcal{J}_{k}}U_{{\mathbf{j}}}({\mathbf{x}},{\mathbf{y}}),$

where $\mathcal{J}_{1}:={\mathbb{N}}^{2}$, $\mathcal{J}_{2}:=\\{(0,0)\\}$,
$\mathcal{J}_{3}:={\mathbb{N}}\times\\{0\\}$ and
$\mathcal{J}_{4}:=\\{0\\}\times{\mathbb{N}}$.

We next estimate
$\sum_{{\mathbf{j}}\in\mathcal{J}_{1}}|U_{{\mathbf{j}}}({\mathbf{x}},{\mathbf{y}})|$.
Let ${\mathbf{m}}:=(m,m)$ and set

$\omega(\lambda_{1},\lambda_{2}):=|\lambda_{1}|^{-2m}|\lambda_{2}|^{-2m}\varphi^{1}(\lambda_{1})\varphi^{2}(\lambda_{2}).$

Clearly,
$\omega(\pm\lambda_{1},\pm\lambda_{2})=\omega(\lambda_{1},\lambda_{2})$ and
$\operatorname{supp}\omega\subset[-2,2]\setminus[-1,1]$. For
${\mathbf{j}}\in\mathcal{J}_{1}={\mathbb{N}}^{2}$ and $\phi\in\mathcal{S}$ we
have
$\varphi_{\mathbf{j}}(\sqrt{L})\phi=2^{-2m(j_{1}+j_{2})}\omega(2^{-{\mathbf{j}}}\sqrt{L})L^{\mathbf{m}}\phi$
and, therefore,

$\displaystyle\varphi_{\mathbf{j}}(\sqrt{L})\phi({\mathbf{x}})-$
$\displaystyle\varphi_{\mathbf{j}}(\sqrt{L})\phi({\mathbf{y}})$ (5.28)
$\displaystyle=2^{-2m|{\mathbf{j}}|}\int_{X}\big{(}{\mathscr{K}}_{\omega(2^{-{\mathbf{j}}}\sqrt{L})}({\mathbf{x}},{\mathbf{z}})-{\mathscr{K}}_{\omega(2^{-{\mathbf{j}}}\sqrt{L})}({\mathbf{y}},{\mathbf{z}})\big{)}L^{{\mathbf{m}}}\phi({\mathbf{z}})d\mu({\mathbf{z}}).$

By Theorem 4.18 it readily follows that

$\displaystyle|{\mathscr{K}}_{\omega(2^{-{\mathbf{j}}}\sqrt{L})}({\mathbf{x}},{\mathbf{z}})|$
$\displaystyle\leq
c_{{\bm{\sigma}}}{\mathfrak{D}}_{2^{-{\mathbf{j}}},{\bm{\sigma}}+3{\mathbf{d}}/2}({\mathbf{x}},{\mathbf{z}})\leq
cV({\mathbf{x}},2^{-{\mathbf{j}}})^{-1}{\mathfrak{D}}_{2^{-{\mathbf{j}}},{\bm{\sigma}}+{\mathbf{d}}}^{*}({\mathbf{x}},{\mathbf{z}})$
(5.29) $\displaystyle\leq
c2^{{\mathbf{j}}\cdot{\mathbf{d}}}V({\mathbf{x}},{\bm{1}})^{-1}{\mathfrak{D}}_{{\bm{1}},{\bm{\sigma}}+{\mathbf{d}}}^{*}({\mathbf{x}},{\mathbf{z}}),$

in light of (3.14) and (3.12). Moreover, if $\rho_{i}(x_{i},y_{i})\leq
2^{-j_{i}}$, $i=1,2$, then

$\displaystyle|{\mathscr{K}}_{\omega(2^{-{\mathbf{j}}}\sqrt{L})}({\mathbf{x}},{\mathbf{z}})-$
$\displaystyle{\mathscr{K}}_{\omega(2^{-{\mathbf{j}}}\sqrt{L})}({\mathbf{y}},{\mathbf{z}})|$
(5.30) $\displaystyle\leq
c\Big{(}\sum_{i=1,2}(2^{j_{i}}\rho_{i}(x_{i},y_{i}))^{\alpha_{i}}\Big{)}2^{{\mathbf{j}}\cdot{\mathbf{d}}}V({\mathbf{x}},{\bm{1}})^{-1}{\mathfrak{D}}_{{\bm{1}},{\bm{\sigma}}+{\mathbf{d}}}^{*}({\mathbf{x}},{\mathbf{z}}),$

Note also that, since $\phi\in\mathcal{S}$,

(5.31)
$|L^{{\mathbf{m}}}\phi({\mathbf{z}})|\leq\mathcal{P}_{m,k}(\phi){\mathfrak{D}}^{*}_{{\bm{1}},{\bm{\sigma}}+{\mathbf{d}}}({\mathbf{z}},{\mathbf{x}}_{0}).$

Combining (5.2), (5.30) and (5.31) we conclude that if
$\rho_{i}(x_{i},y_{i})\leq 2^{-j_{i}}$, $i=1,2$, then

$\displaystyle|\varphi_{\mathbf{j}}(\sqrt{L})\phi({\mathbf{x}})$
$\displaystyle-\varphi_{\mathbf{j}}(\sqrt{L})\phi({\mathbf{y}})|\leq
c2^{-2m|{\mathbf{j}}|}2^{{\mathbf{j}}\cdot{\mathbf{d}}}\Big{(}\sum_{i=1,2}(2^{j_{i}}\rho_{i}(x_{i},y_{i}))^{\alpha_{i}}\Big{)}\mathcal{P}_{m,k}(\phi)$
(5.32) $\displaystyle\times
V({\mathbf{x}},{\bm{1}})^{-1}\int_{X}{\mathfrak{D}}_{{\bm{1}},{\bm{\sigma}}+{\mathbf{d}}}^{*}({\mathbf{x}},{\mathbf{z}}){\mathfrak{D}}^{*}_{{\bm{1}},{\bm{\sigma}}+{\mathbf{d}}}({\mathbf{z}},{\mathbf{x}}_{0})d{\mathbf{z}}$
$\displaystyle\leq
c\mathcal{P}_{m,k}(\phi)2^{-2m|{\mathbf{j}}|}2^{{\mathbf{j}}\cdot{\mathbf{d}}}2^{{\mathbf{j}}\cdot{\bm{\alpha}}}\Big{(}\sum_{i=1,2}\rho_{i}(x_{i},y_{i})^{\alpha_{i}}\Big{)}{\mathfrak{D}}_{{\bm{1}},{\bm{\sigma}}}^{*}({\mathbf{x}},{\mathbf{x}}_{0}),$

where we used (3.20).

Assume $\rho_{\iota}(x_{\iota},y_{\iota})>2^{-j_{\iota}}$ for some
$\iota\in\\{1,2\\}$. Then by (5.29), (5.31) and (3.20) we obtain, for
${\mathbf{x}}\in X$,

$\displaystyle|\varphi_{\mathbf{j}}(\sqrt{L})\phi({\mathbf{x}})|\leq
2^{-2m|{\mathbf{j}}|}\int_{X}|{\mathscr{K}}_{\omega(2^{-{\mathbf{j}}}\sqrt{L})}({\mathbf{x}},{\mathbf{z}})||L^{{\mathbf{m}}}\phi({\mathbf{z}})|d{\mathbf{z}}$
$\displaystyle\leq
c2^{-2m|{\mathbf{j}}|}2^{{\mathbf{j}}\cdot{\mathbf{d}}}\mathcal{P}_{m,k}(\phi)(2^{\iota}\rho_{\iota}(x_{\iota},y_{\iota}))^{\alpha_{\iota}}V({\mathbf{x}},{\bm{1}})^{-1}\int_{X}{\mathfrak{D}}_{{\bm{1}},{\bm{\sigma}}+{\mathbf{d}}}^{*}({\mathbf{x}},{\mathbf{z}}){\mathfrak{D}}^{*}_{{\bm{1}},{\bm{\sigma}}+{\mathbf{d}}}({\mathbf{z}},{\mathbf{x}}_{0})d{\mathbf{z}}$
$\displaystyle\leq
c\mathcal{P}_{m,k}(\phi)2^{-2m|{\mathbf{j}}|}2^{{\mathbf{j}}\cdot{\mathbf{d}}}2^{{\mathbf{j}}\cdot{\bm{\alpha}}}\Big{(}\sum_{i=1,2}\rho_{i}(x_{i},y_{i})^{\alpha_{i}}\Big{)}{\mathfrak{D}}_{{\bm{1}},{\bm{\sigma}}}^{*}({\mathbf{x}},{\mathbf{x}}_{0}),$

implying

$\displaystyle|\varphi_{\mathbf{j}}(\sqrt{L})\phi({\mathbf{x}})-\varphi_{\mathbf{j}}(\sqrt{L})\phi({\mathbf{y}})|\leq|\varphi_{\mathbf{j}}(\sqrt{L})\phi({\mathbf{x}})|+|\varphi_{\mathbf{j}}(\sqrt{L})\phi({\mathbf{y}})|$
$\displaystyle\leq
c2^{-2m|{\mathbf{j}}|}2^{{\mathbf{j}}\cdot{\mathbf{d}}}2^{{\mathbf{j}}\cdot{\bm{\alpha}}}\Big{(}\sum_{i=1,2}\rho_{i}(x_{i},y_{i})^{\alpha_{i}}\Big{)}\mathcal{P}_{m,k}(\phi)\big{[}{\mathfrak{D}}^{*}_{{\bm{1}},{\bm{\sigma}}}({\mathbf{x}},{\mathbf{x}}_{0})+{\mathfrak{D}}_{{\bm{1}},{\bm{\sigma}}}^{*}({\mathbf{y}},{\mathbf{x}}_{0})\big{]}.$

Summing up (5.2) and the above inequalities over
${\mathbf{j}}\in\mathcal{J}_{1}={\mathbb{N}}^{2}$, using the assumption that
$m>\frac{d_{1}+\alpha_{1}}{2}\vee\frac{d_{2}+\alpha_{2}}{2}$, we get

$\displaystyle\sum_{{\mathbf{j}}\in\mathcal{J}_{1}}$
$\displaystyle|U_{{\mathbf{j}}}({\mathbf{x}},{\mathbf{y}})|$ (5.33)
$\displaystyle\leq
c\mathcal{P}_{m,k}(\phi)\Big{(}\sum_{i=1,2}\rho_{i}(x_{i},y_{i})^{\alpha_{i}}\Big{)}\big{[}{\mathfrak{D}}^{*}_{{\bm{1}},{\bm{\sigma}}}({\mathbf{x}},{\mathbf{x}}_{0})+{\mathfrak{D}}_{{\bm{1}},{\bm{\sigma}}}^{*}({\mathbf{y}},{\mathbf{x}}_{0})\big{]}.$

The estimation of the sums
$\sum_{{\mathbf{j}}\in\mathcal{J}_{i}}|U_{{\mathbf{j}}}({\mathbf{x}},{\mathbf{y}})|$,
$i=2,3,4$, is similar (simpler) and we omit it. The desired estimate (5.25)
follows from these estimates, (5.27), and (5.2). ∎

###### Proof of Theorem 5.10.

We will prove (5.23). Then (5.24) follows by duality. We need to show that for
any $n,\ell\in{\mathbb{N}}_{0}$ and $\phi\in\mathcal{S}$

(5.34)
$\lim_{{\bm{\delta}}\to{\bm{0}}}\mathcal{P}_{n,\ell}(\phi-\varphi({\bm{\delta}}\sqrt{L})\phi)=0.$

By Theorem 4.21 it follows that for any ${\bm{\sigma}}>{\bm{0}}$ there exists
a constant $c_{{\bm{\sigma}}}>0$ such that

(5.35)
$|{\mathscr{K}}_{\varphi({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})|\leq
c_{\bm{\sigma}}V({\mathbf{x}},{\bm{\delta}})^{-1}{\mathfrak{D}}_{{\bm{\delta}},{\bm{\sigma}}}^{*}({\mathbf{x}},{\mathbf{y}})$

and

(5.36)
$\int_{X}{\mathscr{K}}_{\varphi({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})d\mu({\mathbf{y}})=\varphi({\bm{0}})=1.$

Fix $n,\ell\in{\mathbb{N}}_{0}$ and let ${\bm{\ell}}:=(\ell,\ell)$ and
${\bm{\nu}}\in{\mathbb{N}}_{0}^{2}$ be so that $0\leq\nu_{i}\leq n$, $i=1,2$.
We choose ${\bm{\sigma}}>{\mathbf{d}}+{\bm{\ell}}+{\bm{\alpha}}$. Also, let
$m,k\in{\mathbb{N}}_{0}$ be just as in Lemma 5.11, i.e.
$m>\frac{d_{1}+\alpha_{1}}{2}\vee\frac{d_{2}+\alpha_{2}}{2}$ and
$k\geq(\sigma_{1}+d_{1})\vee(\sigma_{2}+d_{2})$. To prove (5.34) we have to
deal with

$\displaystyle{\mathfrak{D}}_{{\bm{1}},{\bm{\ell}}}^{*}$
$\displaystyle({\mathbf{x}},{\mathbf{x}}_{0})^{-1}|L^{{\bm{\nu}}}(\phi-\varphi({\bm{\delta}}\sqrt{L})\phi)({\mathbf{x}})|$
$\displaystyle\leq{\mathfrak{D}}_{{\bm{1}},{\bm{\ell}}}^{*}({\mathbf{x}},{\mathbf{x}}_{0})^{-1}\int_{X}|{\mathscr{K}}_{\varphi({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})||L^{{\bm{\nu}}}\phi({\mathbf{x}})-L^{{\bm{\nu}}}\phi({\mathbf{y}})|d\mu({\mathbf{y}})$
(5.37) $\displaystyle\leq
c{\mathfrak{D}}_{{\bm{1}},{\bm{\ell}}}^{*}({\mathbf{x}},{\mathbf{x}}_{0})^{-1}\int_{X}\frac{{\mathfrak{D}}_{{\bm{\delta}},{\bm{\sigma}}}^{*}({\mathbf{x}},{\mathbf{y}})}{V({\mathbf{x}},{\bm{\delta}})}|L^{{\bm{\nu}}}\phi({\mathbf{x}})-L^{{\bm{\nu}}}\phi({\mathbf{y}})|d\mu({\mathbf{y}})$
$\displaystyle\leq
c{\mathfrak{D}}_{{\bm{1}},{\bm{\ell}}}^{*}({\mathbf{x}},{\mathbf{x}}_{0})^{-1}\int_{Q({\mathbf{x}},{\bm{1}})}\dots+c{\mathfrak{D}}_{{\bm{1}},{\bm{\ell}}}^{*}({\mathbf{x}},{\mathbf{x}}_{0})^{-1}\int_{X\setminus
Q({\mathbf{x}},{\bm{1}})}\dots$
$\displaystyle=:W_{1}({\mathbf{x}})+W_{2}({\mathbf{x}}),$

where we used (5.35) and (5.36). Recall that
$Q({\mathbf{x}},{\bm{1}}):=B_{1}(x_{1},1)\times B_{2}(x_{2},1)$, see (3.8).

We focus first on the estimation of $W_{1}({\mathbf{x}})$. As
$\phi\in\mathcal{S}$ we have $L^{{\bm{\nu}}}\phi\in\mathcal{S}$, and in light
of (5.1) $\mathcal{P}_{m,k}(L^{{\bm{\nu}}}\phi)\leq\mathcal{P}_{m+n,k}(\phi)$.
We use this and Lemma 5.11 to obtain

$\displaystyle W_{1}({\mathbf{x}})$ $\displaystyle\leq
c\mathcal{P}_{m+n,k}(\phi){\mathfrak{D}}_{{\bm{1}},{\bm{\ell}}}^{*}({\mathbf{x}},{\mathbf{x}}_{0})^{-1}$
$\displaystyle\times\int_{Q({\mathbf{x}},{\bm{1}})}\Big{(}\sum_{i=1,2}\rho_{i}(x_{i},y_{i})^{\alpha_{i}}\Big{)}\big{[}{\mathfrak{D}}^{*}_{{\bm{1}},{\bm{\sigma}}}({\mathbf{x}},{\mathbf{x}}_{0})+{\mathfrak{D}}^{*}_{{\bm{1}},{\bm{\sigma}}}({\mathbf{y}},{\mathbf{x}}_{0})]\frac{{\mathfrak{D}}_{{\bm{\delta}},{\bm{\sigma}}}^{*}({\mathbf{x}},{\mathbf{y}})}{V({\mathbf{x}},{\bm{\delta}})}d\mu({\mathbf{y}})$
(5.38)
$\displaystyle=c\mathcal{P}_{m+n,k}(\phi)\Big{(}\int_{Q({\mathbf{x}},{\bm{1}})}{\mathfrak{D}}^{*}_{{\bm{1}},{\bm{\sigma}}}({\mathbf{x}},{\mathbf{x}}_{0})\cdots+\int_{Q({\mathbf{x}},{\bm{1}})}{\mathfrak{D}}^{*}_{{\bm{1}},{\bm{\sigma}}}({\mathbf{y}},{\mathbf{x}}_{0})\cdots\Big{)}$
$\displaystyle=:c\mathcal{P}_{m+n,k}(\phi)(J_{1}({\mathbf{x}})+J_{2}({\mathbf{x}})).$

To estimate $J_{1}({\mathbf{x}})$ we note first that
$\rho_{i}(x_{i},y_{i})^{\alpha_{i}}<\delta_{i}^{\alpha_{i}}(1+\delta_{i}^{-1}\rho_{i}(x_{i},y_{i}))^{\alpha_{i}}$,
$i=1,2$, and hence

(5.39)
$\sum_{i=1,2}\rho_{i}(x_{i},y_{i})^{\alpha_{i}}<\frac{\delta_{1}^{\alpha_{1}}+\delta_{2}^{\alpha_{2}}}{{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\alpha}}}({\mathbf{x}},{\mathbf{y}})}.$

From the choice of ${\bm{\sigma}}$ it follows that
${\bm{\sigma}}\geq{\bm{\ell}}$ and ${\bm{\sigma}}>{\mathbf{d}}+{\bm{\alpha}}$.
Then using these, (5.39), and (3.19)) we get

(5.40) $\displaystyle J_{1}({\mathbf{x}})$ $\displaystyle\leq
c(\delta_{1}^{\alpha_{1}}+\delta_{2}^{\alpha_{2}})V({\mathbf{x}},{\bm{\delta}})^{-1}\int_{X}{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\sigma}}-{\bm{\alpha}}}({\mathbf{x}},{\mathbf{y}})d\mu({\mathbf{y}})\leq
c(\delta_{1}^{\alpha_{1}}+\delta_{2}^{\alpha_{2}}).$

We next estimate $J_{2}({\mathbf{x}})$. We may assume
${\bm{\delta}}<{\bm{1}}$. The obvious inequalities
$1+\rho_{i}(x_{i},x_{0i})\leq(1+\delta_{i}^{-1}\rho_{i}(x_{i},y_{i}))(1+\rho_{i}(y_{i},x_{0i}))$,
$i=1,2$, implies that

${\mathfrak{D}}_{{\bm{1}},{\bm{\ell}}}^{*}({\mathbf{x}},{\mathbf{x}}_{0})^{-1}\leq{\mathfrak{D}}_{{\bm{\delta}},{\bm{\ell}}}^{*}({\mathbf{x}},{\mathbf{y}})^{-1}{\mathfrak{D}}_{{\bm{1}},{\bm{\ell}}}^{*}({\mathbf{y}},{\mathbf{x}}_{0})^{-1}.$

From this and (5.39) it follows that

$\displaystyle\frac{(\sum_{i=1,2}\rho_{i}(x_{i},y_{i})^{\alpha_{i}}){\mathfrak{D}}^{*}_{{\bm{1}},{\bm{\sigma}}}({\mathbf{y}},{\mathbf{x}}_{0}){\mathfrak{D}}_{{\bm{\delta}},{\bm{\sigma}}}^{*}({\mathbf{x}},{\mathbf{y}})}{{\mathfrak{D}}_{{\bm{1}},{\bm{\ell}}}^{*}({\mathbf{x}},{\mathbf{x}}_{0})}$
(5.41) $\displaystyle\hskip
108.405pt\leq(\delta_{1}^{\alpha_{1}}+\delta_{2}^{\alpha_{2}}){\mathfrak{D}}_{{\bm{\delta}},{\bm{\sigma}}-{\bm{\ell}}-{\bm{\alpha}}}^{*}({\mathbf{x}},{\mathbf{y}}){\mathfrak{D}}_{{\bm{1}},{\bm{\sigma}}-{\bm{\ell}}}^{*}({\mathbf{y}},{\mathbf{x}}_{0})$
$\displaystyle\hskip
108.405pt\leq(\delta_{1}^{\alpha_{1}}+\delta_{2}^{\alpha_{2}}){\mathfrak{D}}_{{\bm{\delta}},{\bm{\sigma}}-{\bm{\ell}}-{\bm{\alpha}}}^{*}({\mathbf{x}},{\mathbf{y}}),$

where we used that ${\bm{\sigma}}>{\bm{\ell}}$. We use (5.2) and (3.19) and
the fact that ${\bm{\sigma}}>{\mathbf{d}}+{\bm{\ell}}+{\bm{\alpha}}$ to derive

$J_{2}({\mathbf{x}})\leq
c(\delta_{1}^{\alpha_{1}}+\delta_{2}^{\alpha_{2}})V({\mathbf{x}},{\bm{\delta}})^{-1}\int_{X}{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\sigma}}-{\bm{\alpha}}}({\mathbf{x}},{\mathbf{y}})d\mu({\mathbf{y}})\leq
c(\delta_{1}^{\alpha_{1}}+\delta_{2}^{\alpha_{2}}).$

From this, (5.2), and (5.40) it follows that

(5.42) $W_{1}(x)\leq
c\mathcal{P}_{m+n,k}(\phi)(\delta_{1}^{\alpha_{1}}+\delta_{2}^{\alpha_{2}}).$

We next estimate $W_{2}({\mathbf{x}})$. We split it into two:

$\displaystyle W_{2}({\mathbf{x}})$ $\displaystyle\leq
c{\mathfrak{D}}_{{\bm{1}},{\bm{\ell}}}^{*}({\mathbf{x}},{\mathbf{x}}_{0})^{-1}\int_{X\setminus
Q({\mathbf{x}},{\bm{1}})}\frac{{\mathfrak{D}}_{{\bm{\delta}},{\bm{\sigma}}}^{*}({\mathbf{x}},{\mathbf{y}})}{V({\mathbf{x}},{\bm{\delta}})}|L^{{\bm{\nu}}}(\phi)({\mathbf{x}})|d\mu({\mathbf{y}})$
(5.43)
$\displaystyle+c{\mathfrak{D}}_{{\bm{1}},{\bm{\ell}}}^{*}({\mathbf{x}},{\mathbf{x}}_{0})^{-1}\int_{X\setminus
Q({\mathbf{x}},{\bm{1}})}\frac{{\mathfrak{D}}_{{\bm{\delta}},{\bm{\sigma}}}^{*}({\mathbf{x}},{\mathbf{y}})}{V({\mathbf{x}},{\bm{\delta}})}|L^{{\bm{\nu}}}(\phi)({\mathbf{y}})|d\mu({\mathbf{y}})$
$\displaystyle=:W_{21}({\mathbf{x}})+W_{22}({\mathbf{x}}).$

Note that because $0\leq\nu_{i}\leq n$ and $k>\sigma_{1}\vee\sigma_{2}$ we
have

(5.44)
$|L^{{\bm{\nu}}}\phi({\mathbf{z}})|\leq\mathcal{P}_{n,k}(\phi){\mathfrak{D}}^{*}_{{\bm{1}},{\bm{\sigma}}}({\mathbf{z}},{\mathbf{x}}_{0}),\quad\forall{\mathbf{z}}\in
X,$

and using that ${\bm{\sigma}}>{\bm{\ell}}$ we obtain

(5.45) $W_{21}({\mathbf{x}})\leq\mathcal{P}_{n,k}(\phi)\int_{X\setminus
Q({\mathbf{x}},{\bm{1}})}\frac{{\mathfrak{D}}_{{\bm{\delta}},{\bm{\sigma}}}^{*}({\mathbf{x}},{\mathbf{y}})}{V({\mathbf{x}},{\bm{\delta}})}d\mu({\mathbf{y}}).$

Clearly, for any ${\mathbf{y}}\in X\setminus Q({\mathbf{x}},{\bm{1}})$, there
exists $\iota\in\\{1,2\\}$ such that $\rho_{\iota}(x_{\iota},y_{\iota})>1$ and
hence

(5.46)
$1<\delta_{\iota}^{\alpha_{\iota}}(1+\delta_{\iota}^{-1}\rho_{\iota}(x_{\iota},y_{\iota}))^{\alpha_{\iota}}<\frac{\delta_{1}^{\alpha_{1}}+\delta_{2}^{\alpha_{2}}}{{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\alpha}}}({\mathbf{x}},{\mathbf{y}})}.$

Combining (5.45), (5.46) and using (3.19) and the fact that
${\bm{\sigma}}>{\mathbf{d}}+{\bm{\alpha}}$ we get

(5.47) $\displaystyle W_{21}({\mathbf{x}})$ $\displaystyle\leq
c\mathcal{P}_{n,k}(\phi)(\delta_{1}^{\alpha_{1}}+\delta_{2}^{\alpha_{2}})\int_{X}\frac{{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\sigma}}-{\bm{\alpha}}}({\mathbf{x}},{\mathbf{y}})}{V({\mathbf{x}},{\bm{\delta}})}d\mu({\mathbf{y}})$
$\displaystyle\leq
c\mathcal{P}_{n,k}(\phi)(\delta_{1}^{\alpha_{1}}+\delta_{2}^{\alpha_{2}}).$

Similarly

$\displaystyle W_{22}({\mathbf{x}})$ $\displaystyle\leq
c\mathcal{P}_{n,k}(\phi)(\delta_{1}^{\alpha_{1}}+\delta_{2}^{\alpha_{2}})\int_{X}\frac{{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\sigma}}-{\bm{\alpha}}}({\mathbf{x}},{\mathbf{y}}){\mathfrak{D}}^{*}_{{\bm{1}},{\bm{\sigma}}}({\mathbf{y}},{\mathbf{x}}_{0})}{{\mathfrak{D}}^{*}_{{\bm{1}},{\mathbf{m}}}({\mathbf{x}},{\mathbf{x}}_{0})V({\mathbf{x}},{\bm{\delta}})}d\mu({\mathbf{y}})$
(5.48) $\displaystyle\leq
c\mathcal{P}_{n,k}(\phi)(\delta_{1}^{\alpha_{1}}+\delta_{2}^{\alpha_{2}})\int_{X}\frac{{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\sigma}}-{\bm{\ell}}-{\bm{\alpha}}}({\mathbf{x}},{\mathbf{y}})}{V({\mathbf{x}},{\bm{\delta}})}d\mu({\mathbf{y}})$
$\displaystyle\leq
c\mathcal{P}_{n,k}(\phi)(\delta_{1}^{\alpha_{1}}+\delta_{2}^{\alpha_{2}}),$

where we used (3.19) and the fact that
${\bm{\sigma}}>{\mathbf{d}}+{\bm{\ell}}+{\bm{\alpha}}$. Combining (5.2),
(5.47), and (5.2) we get

(5.49) $W_{2}(x)\leq
c\mathcal{P}_{n,k}(\phi)(\delta_{1}^{\alpha_{1}}+\delta_{2}^{\alpha_{2}}).$

Finally, (5.42) and (5.49) yield

${\mathfrak{D}}_{{\bm{1}},{\bm{\ell}}}^{*}({\mathbf{x}},{\mathbf{x}}_{0})^{-1}|L^{{\bm{\nu}}}(\phi-\varphi({\bm{\delta}}\sqrt{L})\phi)({\mathbf{x}})|\leq
c\mathcal{P}_{m+n,k}(\phi)(\delta_{1}^{\alpha_{1}}+\delta_{2}^{\alpha_{2}}),$

and hence

$\mathcal{P}_{n,\ell}\big{(}\phi-\varphi({\bm{\delta}}\sqrt{L})\phi\big{)}\leq
c\mathcal{P}_{m+n,k}(\phi)(\delta_{1}^{\alpha_{1}}+\delta_{2}^{\alpha_{2}}).$

In turn this yields (5.34).

The convergence in (5.24) in $L^{p}$ for $f\in L^{p}$ follows by a standard
argument using (5.35) and (5.36). ∎

##  6\. Spectral spaces

In this section we deal with spectral spaces, which can be viewed as a
generalization of the band limited functions in the classical setting on
$\mathbb{R}^{n}$. Our main goal here is to establish Peetre and Nikolski type
inequalities that will be used in what follows.

######  Definition 6.1.

The spectral space $\Sigma_{\mathbf{t}}$,
${\mathbf{t}}=(t_{1},t_{2})\in(0,\infty)^{2}$, is defined by

(6.1)
$\Sigma_{\mathbf{t}}:=\\{g\in\mathcal{S}^{\prime}:\theta(\sqrt{L})g=g,\text{
for all
}\theta\in\mathcal{A}_{\mathbf{t}}\\},\quad\mathcal{S}^{\prime}:=\mathcal{S}^{\prime}(L_{1},L_{2}),$

where

$\mathcal{A}_{\mathbf{t}}:=\\{\theta\in\mathcal{C}^{\infty}_{0}({\mathbb{R}}^{2}):\theta(\pm\lambda_{1},\pm\lambda_{2})=\theta(\lambda_{1},\lambda_{2})\text{
and }\theta\equiv 1\text{ on }[0,t_{1}]\times[0,t_{2}]\\}$

with $\mathcal{C}^{\infty}_{0}({\mathbb{R}}^{2})$ being the set of all
compactly supported $\mathcal{C}^{\infty}$ functions on $\mathbb{R}^{2}$.

The above definition and Proposition 5.7 imply that each
$g\in\Sigma_{\mathbf{t}}$ is a slowly growing continuous function.

######  Proposition 6.2.

Each $g\in\Sigma_{\mathbf{t}}$, ${\mathbf{t}}\in(0,\infty)^{2}$, is a slowly
growing continuous function in the sense that there exist constants $N,c>0$
$($depending on $g$$)$ such that

(6.2) $\displaystyle|g({\mathbf{x}})|$ $\displaystyle\leq
cV({\mathbf{x}}_{0},1)^{-1}\prod_{i=1,2}(t_{i}^{N}+t_{i}^{-N})\big{(}1+\rho_{i}(x_{i},x_{0i})\big{)}^{N}\quad\hbox{and}$
(6.3) $\displaystyle|g({\mathbf{x}})-g({\mathbf{x}}^{\prime})|$
$\displaystyle\leq
cV({\mathbf{x}}_{0},1)^{-1}\prod_{i=1,2}(t_{i}^{N}+t_{i}^{-N})\big{(}1+\rho_{i}(x_{i},x_{0i})\big{)}^{N}\sum_{i=1,2}\rho_{i}(x_{i},x^{\prime}_{i})^{\alpha_{i}}$

if $\rho_{i}(x_{i},x^{\prime}_{i})\leq t_{i}^{-1}$, $i=1,2$. Here
$\alpha_{1},\alpha_{2}$ are from (2.4).

The following simple claim will be needed:

######  Proposition 6.3.

Let $\varphi\in\mathcal{C}^{\infty}_{0}({\mathbb{R}}^{2})$ be real-valued,
$\varphi(\pm\lambda_{1},\pm\lambda_{2})=\varphi(\lambda_{1},\lambda_{2})$, and
$\operatorname{supp}\varphi\subset[-t_{1},t_{1}]\times[-t_{2},t_{2}]$,
$t_{1},t_{2}>0$. Then for any
$f\in\mathcal{S}^{\prime}=\mathcal{S}^{\prime}(L_{1},L_{2})$ we have
$\varphi(\sqrt{L})f\in\Sigma_{\mathbf{t}}$.

###### Proof.

Let $\theta$ be a function as in Definition 6.1. Then
$\theta({\bm{\lambda}})\varphi({\bm{\lambda}})=\varphi({\bm{\lambda}})$ and in
light of Proposition 5.5 for any $f\in\mathcal{S}^{\prime}$ we have
$\theta(\sqrt{L})\varphi(\sqrt{L})f=\varphi(\sqrt{L})f,$ which completes the
proof. ∎

###  6.1. Peetre type maximal inequality

For reader’s convenience we recall the following compact notation from (3.15);
it will be used in what follows:

(6.4)
$V({\mathbf{x}},{\mathbf{t}}^{-1})^{{\bm{\gamma}}}:=V_{1}(x_{1},t_{1}^{-1})^{\gamma_{1}}V_{2}(x_{2},t_{2}^{-1})^{\gamma_{2}},\quad{\mathbf{t}}\in{\mathbb{R}}^{2}_{+},{\bm{\gamma}}\in{\mathbb{R}}^{2}.$

######  Theorem 6.4.

Let ${\bm{\gamma}},{\bm{\tau}}\in\mathbb{R}^{2}$, ${\bm{\tau}}>2{\mathbf{d}}$,
$r>0$, and ${\bm{\nu}}\in{\mathbb{N}}_{0}^{2}$. Then there exists a constant
$c>0$ such that for any $g\in\Sigma_{\mathbf{t}}$ with
${\mathbf{t}}\in[1,\infty)^{2}$ we have

(6.5) $\displaystyle{\mathbf{t}}^{-2{\bm{\nu}}}\sup_{{\mathbf{y}}\in
X}\frac{V({\mathbf{y}},{\mathbf{t}}^{-1})^{{\bm{\gamma}}}|L^{\bm{\nu}}g({\mathbf{y}})|}{\prod_{i=1,2}\big{(}1+t_{i}\rho(x_{i},y_{i})\big{)}^{\tau_{i}/r}}$
$\displaystyle\quad\leq c\sup_{{\mathbf{y}}\in
X}\frac{V({\mathbf{y}},{\mathbf{t}}^{-1})^{{\bm{\gamma}}}|g({\mathbf{y}})|}{\prod_{i=1,2}\big{(}1+t_{i}\rho(x_{i},y_{i})\big{)}^{\tau_{i}/r}}\leq
c\mathcal{M}_{r}\big{(}V(\cdot,{\mathbf{t}}^{-1})^{{\bm{\gamma}}}g(\cdot)\big{)}({\mathbf{x}}),\;\;{\mathbf{x}}\in
X.$

###### Proof.

We first establish the left-hand side inequality in (LABEL:Peetre-max).

Let $g\in\Sigma_{\mathbf{t}}$, ${\mathbf{t}}\in[1,\infty)^{2}$, and set
${\bm{\delta}}:={\mathbf{t}}^{-1}=(t_{1}^{-1},t_{2}^{-1})$. Let
$\theta\in\mathcal{C}^{\infty}_{0}({\mathbb{R}}^{2})$,
$\theta(\pm\lambda_{1},\pm\lambda_{2})=\theta(\lambda_{1},\lambda_{2})$ and
$\theta({\bm{\lambda}})=1$ for ${\bm{\lambda}}\in[0,1]^{2}$. Then by (6.1)
$g=\theta({\bm{\delta}}\sqrt{L})g$. Let
$\omega(\lambda_{1},\lambda_{2}):=\lambda_{1}^{2\nu_{1}}\lambda_{2}^{2\nu_{2}}\theta(\lambda_{1},\lambda_{2})$.
Clearly,
$\omega({\bm{\delta}}\sqrt{L})={\bm{\delta}}^{2{\bm{\nu}}}L^{{\bm{\nu}}}\theta({\bm{\delta}}\sqrt{L})$
and hence

(6.6)
$L^{\bm{\nu}}g({\mathbf{y}})=L^{\bm{\nu}}\theta({\bm{\delta}}\sqrt{L})g({\mathbf{y}})={\bm{\delta}}^{-2{\bm{\nu}}}\omega({\bm{\delta}}\sqrt{L})g({\mathbf{y}})={\bm{\delta}}^{-2{\bm{\nu}}}\int_{X}{\mathscr{K}}_{\omega({\bm{\delta}}\sqrt{L})}({\mathbf{y}},{\mathbf{z}})g({\mathbf{z}})d\mu({\mathbf{z}}).$

We use Corollary 4.22 and (3.14) to conclude that for any
${\bm{\sigma}}\in(0,\infty)^{2}$

(6.7)
$|{\mathscr{K}}_{\omega({\bm{\delta}}\sqrt{L})}({\mathbf{y}},{\mathbf{z}})|\leq
c_{\bm{\sigma}}V({\mathbf{x}},{\bm{\delta}})^{-1}{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\sigma}}}({\mathbf{y}},{\mathbf{z}}).$

Therefore,

(6.8) ${\bm{\delta}}^{2{\bm{\nu}}}|L^{\bm{\nu}}g({\mathbf{y}})|\leq
cV({\mathbf{y}},{\bm{\delta}})^{-1}\int_{X}{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\sigma}}}({\mathbf{y}},{\mathbf{z}})|g({\mathbf{z}})|d\mu({\mathbf{z}}).$

We know from (3.15) that

(6.9) $V({\mathbf{y}},{\bm{\delta}})^{{\bm{\gamma}}}\leq
cV({\mathbf{z}},{\bm{\delta}})^{{\bm{\gamma}}}\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho_{i}(y_{i},z_{i})\big{)}^{d_{i}|\gamma_{i}|}.$

and obviously

(6.10)
$1+\delta_{i}^{-1}\rho_{i}(x_{i},z_{i})\leq(1+\delta_{i}^{-1}\rho_{i}(x_{i},y_{i}))(1+\delta_{i}^{-1}\rho_{i}(y_{i},z_{i})),\quad
i=1,2.$

We use these inequalities and (6.8) with
$\sigma_{i}:=\tau_{i}/r+d_{i}|\gamma_{i}|+d_{i}+1$ to obtain

$\displaystyle\frac{{\bm{\delta}}^{2{\bm{\nu}}}V({\mathbf{y}},{\bm{\delta}})^{{\bm{\gamma}}}|L^{\bm{\nu}}g({\mathbf{y}})|}{\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho(x_{i},y_{i})\big{)}^{\tau_{i}/r}}$
$\displaystyle\leq
c\int_{X}\frac{V({\mathbf{y}},{\bm{\delta}})^{-1}V({\mathbf{z}},{\bm{\delta}})^{{\bm{\gamma}}}\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho_{i}(y_{i},z_{i})\big{)}^{d_{i}|\gamma_{i}|}|g({\mathbf{z}})|}{\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho(x_{i},y_{i})\big{)}^{\tau_{i}/r}\big{(}1+\delta_{i}^{-1}\rho(y_{i},z_{i})\big{)}^{\tau_{i}/r+d_{i}|\gamma_{i}|+d_{i}+1}}d\mu({\mathbf{z}})$
$\displaystyle\leq
cV({\mathbf{y}},{\bm{\delta}})^{-1}\int_{X}\frac{V({\mathbf{z}},{\bm{\delta}})^{{\bm{\gamma}}}|g({\mathbf{z}})|}{\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho(x_{i},z_{i})\big{)}^{\tau_{i}/r}\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho(y_{i},z_{i})\big{)}^{d_{i}+1}}d\mu({\mathbf{z}})$
$\displaystyle\leq c\sup_{{\mathbf{z}}\in
X}\frac{V({\mathbf{z}},{\bm{\delta}})^{{\bm{\gamma}}}|g({\mathbf{z}})|}{\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho(x_{i},z_{i})\big{)}^{\tau_{i}/r}}\int_{X}\frac{V({\mathbf{y}},{\bm{\delta}})^{-1}}{\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho(y_{i},z_{i})\big{)}^{d_{i}+1}}d\mu({\mathbf{z}})$
$\displaystyle\leq c\sup_{{\mathbf{z}}\in
X}\frac{V({\mathbf{z}},{\bm{\delta}})^{{\bm{\gamma}}}|g({\mathbf{z}})|}{\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho(x_{i},z_{i})\big{)}^{\tau_{i}/r}},$

which implies the left-hand side inequality in (LABEL:Peetre-max). Above for
the last inequality we used (3.19).

To prove the right-hand side inequality in (LABEL:Peetre-max) we consider
three cases.

Case 1: $1<r<\infty$. With the notation from above it follows from Corollary
4.22 and (3.14) that for any ${\bm{\sigma}}\in(0,\infty)^{2}$

(6.11)
$|{\mathscr{K}}_{\theta({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})|\leq
c_{\bm{\sigma}}V({\mathbf{x}},{\bm{\delta}})^{-1}{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\sigma}}}({\mathbf{x}},{\mathbf{y}})$

and hence

(6.12) $|g({\mathbf{y}})|=|\theta({\bm{\delta}}\sqrt{L})({\mathbf{y}})|\leq
cV({\mathbf{y}},{\bm{\delta}})^{-1}\int_{X}{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\sigma}}}({\mathbf{y}},{\mathbf{z}})|g({\mathbf{z}})|d\mu({\mathbf{z}}).$

Choose $\sigma_{i}:=\tau_{i}/r+d_{i}|\gamma_{i}|+(d_{i}+1)/r^{\prime}$,
$i=1,2$, where $1/r+1/r^{\prime}=1$. Then just as above we use (6.9) and
(6.10) to obtain

$\displaystyle\frac{V({\mathbf{y}},{\bm{\delta}})^{{\bm{\gamma}}}|g({\mathbf{y}})|}{\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho(x_{i},y_{i})\big{)}^{\tau_{i}/r}}$
$\displaystyle\leq
c\int_{X}\frac{V({\mathbf{y}},{\bm{\delta}})^{-1}V({\mathbf{z}},{\bm{\delta}})^{{\bm{\gamma}}}\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho_{i}(y_{i},z_{i})\big{)}^{d_{i}|\gamma_{i}|}|g({\mathbf{z}})|}{\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho(x_{i},y_{i})\big{)}^{\tau_{i}/r}\big{(}1+\delta_{i}^{-1}\rho(y_{i},z_{i})\big{)}^{\tau_{i}/r+d_{i}|\gamma_{i}|+(d_{i}+1)/r^{\prime}}}d\mu({\mathbf{z}})$
$\displaystyle\leq
c\int_{X}\frac{V({\mathbf{y}},{\bm{\delta}})^{-1}V({\mathbf{z}},{\bm{\delta}})^{{\bm{\gamma}}}|g({\mathbf{z}})|}{\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho(x_{i},z_{i})\big{)}^{\tau_{i}/r}\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho(y_{i},z_{i})\big{)}^{(d_{i}+1)/r^{\prime}}}d\mu({\mathbf{z}})$

Applying now Hölder’s inequality we get

$\displaystyle\frac{V({\mathbf{y}},{\bm{\delta}})^{{\bm{\gamma}}}|g({\mathbf{y}})|}{\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho(x_{i},y_{i})\big{)}^{\tau_{i}/r}}$
$\displaystyle\qquad\leq
c\Big{(}\int_{X}\frac{[V({\mathbf{z}},{\bm{\delta}})^{{\bm{\gamma}}}|g({\mathbf{z}})|]^{r}}{V({\mathbf{y}},{\bm{\delta}})\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho(x_{i},z_{i})\big{)}^{\tau_{i}}}d\mu({\mathbf{z}})\Big{)}^{1/r}$
$\displaystyle\hskip
93.95122pt\times\Big{(}\int_{X}\frac{1}{V({\mathbf{y}},{\bm{\delta}})\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho(y_{i},z_{i})\big{)}^{d_{i}+1}}d\mu({\mathbf{z}})\Big{)}^{1/r^{\prime}}$
$\displaystyle\qquad\leq
c\mathcal{M}_{r}\big{(}V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}}g(\cdot)\big{)}({\mathbf{x}}),$

where for the last inequality we used Lemma 4.23 and (3.19). The right-hand
side inequality in (LABEL:Peetre-max) follows from above.

Case 2: $r=1$. The proof in this case is similar to the above proof but is
easier; we omit it.

Case 3: $0<r<1$. To prove the right-hand side inequality in (LABEL:Peetre-max)
in this case we first show that it is valid with a constant $c=c(g)$ depending
on $g$.

The function $\theta({\bm{\lambda}})$ from above obeys the assumptions of
Proposition 5.7 and hence there exist constants $N,c>0$, depending on $g$,
such that

(6.13) $\displaystyle|g({\mathbf{x}})|$
$\displaystyle=|\theta({\bm{\delta}}\sqrt{L})g({\mathbf{x}})|$
$\displaystyle\leq
c(g)V({\mathbf{x}}_{0},{\bm{\delta}})^{-1}\prod_{i=1,2}(\delta_{i}^{N}+\delta_{i}^{-N})\big{(}1+\rho_{i}(x_{i},x_{0i})\big{)}^{N}.$

We may assume $N\geq(\tau_{1}\vee\tau_{2})/r$. On the other hand, from (3.17)
we obtain

(6.14) $\displaystyle V({\mathbf{x}},{\bm{\delta}})^{\bm{\gamma}}$
$\displaystyle\leq
cV({\mathbf{x}}_{0},{\bm{\delta}})^{\bm{\gamma}}\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho_{i}(x_{i},x_{0i})\big{)}^{d_{i}|\gamma_{i}|}$
$\displaystyle\leq
cV({\mathbf{x}}_{0},{\bm{\delta}})^{\bm{\gamma}}\prod_{i=1,2}(1+\delta_{i}^{-1})^{d_{i}|\gamma_{i}|}\big{(}1+\rho_{i}(x_{i},x_{0i})\big{)}^{d_{i}|\gamma_{i}|}.$

This coupled with (6.13) implies

(6.15) $V({\mathbf{x}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{x}})|\leq
c(g,{\bm{\delta}})\prod_{i=1,2}\big{(}1+\rho_{i}(x_{i},x_{0i})\big{)}^{\tilde{N}},$

where $\tilde{N}:=N+|\gamma_{1}|d_{1}+|\gamma_{2}|d_{2}$. In turn, (6.15) with
${\mathbf{x}}$ replaced by ${\mathbf{z}}$ and the fact that
$\delta_{i}=t_{i}^{-1}\leq 1$ lead to the estimates

$\displaystyle\frac{V({\mathbf{z}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{z}})|}{\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho_{i}(x_{i},z_{i})\big{)}^{\tilde{N}}}\leq\frac{c\prod_{i=1,2}\big{(}1+\rho_{i}(z_{i},x_{0i})\big{)}^{\tilde{N}}}{\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho_{i}(x_{i},z_{i})\big{)}^{\tilde{N}}}$
$\displaystyle\leq\frac{c\prod_{i=1,2}\big{(}1+\rho_{i}(x_{i},z_{i})\big{)}^{\tilde{N}}\prod_{i=1,2}\big{(}1+\rho_{i}(x_{i},x_{0i})\big{)}^{\tilde{N}}}{\prod_{i=1,2}\big{(}1+\rho_{i}(x_{i},z_{i})\big{)}^{\tilde{N}}},$

implying

(6.16) $\sup_{{\mathbf{z}}\in
X}\frac{V({\mathbf{z}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{z}})|}{\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho_{i}(x_{i},z_{i})\big{)}^{\tilde{N}}}\leq
c(g,{\bm{\delta}})\prod_{i=1,2}\big{(}1+\rho_{i}(x_{i},x_{0i})\big{)}^{\tilde{N}}.$

Choose $\sigma_{i}:=\tilde{N}+|\gamma_{1}|d_{1}+|\gamma_{2}|d_{2}$ and set
$\tilde{\sigma}_{i}:=\tilde{N}$, $i=1,2$. As before using (6.11), (6.9), and
(6.10) we obtain

$\displaystyle V({\mathbf{z}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{z}})|$
$\displaystyle=V({\mathbf{z}},{\bm{\delta}})^{\bm{\gamma}}\Big{|}\int_{X}{\mathscr{K}}_{\theta({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})g({\mathbf{y}})d\mu({\mathbf{y}})\Big{|}$
$\displaystyle\leq
c\int_{X}V({\mathbf{z}},{\bm{\delta}})^{\bm{\gamma}}V({\mathbf{y}},{\bm{\delta}})^{-1}{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\sigma}}}({\mathbf{z}},{\mathbf{y}})|g({\mathbf{y}})|d\mu({\mathbf{y}})$
$\displaystyle\leq
c\int_{X}V({\mathbf{y}},{\bm{\delta}})^{{\bm{\gamma}}}V({\mathbf{y}},{\bm{\delta}})^{-1}{\mathfrak{D}}^{*}_{{\bm{\delta}},\tilde{{\bm{\sigma}}}}({\mathbf{z}},{\mathbf{y}})|g({\mathbf{y}})|d\mu({\mathbf{y}})$

and hence

$\displaystyle\frac{V({\mathbf{z}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{z}})|}{\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},z_{i}))^{\tilde{N}}}$
$\displaystyle\qquad\leq
c\int_{X}\frac{V({\mathbf{y}},{\bm{\delta}})^{\bm{\gamma}}V({\mathbf{y}},{\bm{\delta}})^{-1}|g({\mathbf{y}})|}{\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},z_{i}))^{\tilde{N}}(1+\delta_{i}^{-1}\rho_{i}(z_{i},y_{i}))^{\tilde{N}}}d\mu({\mathbf{y}})$
$\displaystyle\qquad\leq
c\int_{X}\frac{V({\mathbf{y}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{y}})|}{V({\mathbf{y}},{\bm{\delta}})\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},y_{i}))^{\tilde{N}}}d\mu({\mathbf{y}})$
$\displaystyle\qquad\leq c\Big{(}\sup_{{\mathbf{y}}\in
X}\frac{V({\mathbf{y}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{y}})|}{\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},y_{i}))^{\tilde{N}}}\Big{)}^{1-r}$
$\displaystyle\hskip
115.63243pt\times\int_{X}\frac{[V({\mathbf{y}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{y}})|]^{r}}{V({\mathbf{y}},{\bm{\delta}})\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},y_{i}))^{\tilde{N}r}}d\mu({\mathbf{y}}).$

We now take $\sup_{{\mathbf{z}}\in X}$ above and use the fact that by (6.16)

$\sup_{{\mathbf{y}}\in
X}\frac{V({\mathbf{y}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{y}})|}{\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},y_{i}))^{\tilde{N}}}\leq
c(g,{\bm{\delta}},{\mathbf{x}})<\infty,$

which allows us to divide and obtain

(6.17) $\Big{(}\sup_{{\mathbf{z}}\in
X}\frac{V({\mathbf{z}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{z}})|}{\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},z_{i}))^{\tilde{N}}}\Big{)}^{r}\\\
\leq
c\int_{X}\frac{[V({\mathbf{y}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{y}})|]^{r}}{V({\mathbf{y}},{\bm{\delta}})\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},y_{i}))^{\tilde{N}r}}d\mu({\mathbf{y}}).$

Taking ${\mathbf{z}}={\mathbf{x}}$ on the left we get

$[V({\mathbf{x}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{x}})|]^{r}\leq
c\int_{X}\frac{[V({\mathbf{y}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{y}})|]^{r}}{V({\mathbf{y}},{\bm{\delta}})\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},y_{i}))^{\tilde{N}r}}d\mu({\mathbf{y}}).$

Using this estimate with ${\mathbf{z}}$ in the place of ${\mathbf{x}}$ and the
fact that $\tilde{N}r\geq\tau_{1}\vee\tau_{2}$ we obtain

$\displaystyle\frac{[V({\mathbf{z}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{z}})|]^{r}}{\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},z_{i}))^{\tau_{i}}}$
$\displaystyle\leq
c\int_{X}\frac{[V({\mathbf{y}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{y}})|]^{r}}{V({\mathbf{y}},{\bm{\delta}})\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},z_{i}))^{\tau_{i}}(1+\delta_{i}^{-1}\rho_{i}(z_{i},y_{i}))^{\tau_{i}}}d\mu({\mathbf{y}})$
$\displaystyle\leq
c\int_{X}\frac{[V({\mathbf{y}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{y}})|]^{r}}{V({\mathbf{y}},{\bm{\delta}})\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},y_{i}))^{\tau_{i}}}d\mu({\mathbf{y}}).$
$\displaystyle\leq
c\int_{X}\frac{[V({\mathbf{y}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{y}})|]^{r}}{V({\mathbf{x}},{\bm{\delta}})\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},y_{i}))^{\tau_{i}-d_{i}}}d\mu({\mathbf{y}}),$

where for the first inequality we used (6.10) and for the last (3.17).
Applying Lemma 4.23 using that $\tau_{i}>2d_{i}$ we obtain

(6.18)
$\frac{[V({\mathbf{z}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{z}})|]^{r}}{\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},z_{i}))^{\tau_{i}}}\leq
c(g)\big{[}\mathcal{M}_{r}\big{(}V(\cdot,{\bm{\delta}})^{\bm{\gamma}}g(\cdot)\big{)}({\mathbf{x}})\big{]}^{r},$

which implies the right-hand side inequality in (LABEL:Peetre-max). Here the
constant depends on $g$ because $N$ depends on $g$.

We are now prepared to complete the proof of the right-hand inequality in
(LABEL:Peetre-max). Recall the notation (see (3.8))

$Q({\mathbf{x}},{\bm{\delta}}):=B_{1}(x_{1},\delta_{1})\times
B_{2}(x_{2},\delta_{2})\quad\hbox{and}\quad{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\sigma}}}({\mathbf{x}},{\mathbf{y}}):=\prod_{i=1,2}(1+\delta_{i}^{-1}\rho(x_{i},y_{i}))^{-\sigma_{i}}$

and observe that
$\mu(Q({\mathbf{x}},{\bm{\delta}}))=V({\mathbf{x}},{\bm{\delta}})$, see (3.9).

Using the operator $\theta({\bm{\delta}}\sqrt{L})$ from above we have

(6.19)
$g({\mathbf{x}})=\theta({\bm{\delta}}\sqrt{L})g({\mathbf{x}})=\int_{M}{\mathscr{K}}_{\theta({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})g({\mathbf{y}})d\mu({\mathbf{y}}).$

Let ${\mathbf{y}},{\mathbf{z}}\in X$ and assume ${\mathbf{u}}\in
Q({\mathbf{y}},{\bm{\delta}})$. By Theorem 4.18 and (3.14) it follows that for
any ${\bm{\sigma}}\in(0,\infty)^{2}$ there exist a constant $c>0$ such that

(6.20)
$\big{|}{\mathscr{K}}_{\theta({\bm{\delta}}\sqrt{L})}({\mathbf{y}},{\mathbf{z}})-{\mathscr{K}}_{\theta({\bm{\delta}}\sqrt{L})}({\mathbf{u}},{\mathbf{z}})\big{|}\leq
cV({\mathbf{y}},{\bm{\delta}})^{-1}\sum_{i=1,2}\big{(}\rho_{i}(y_{i},u_{i})/\delta_{i}\big{)}^{\alpha_{i}}{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\sigma}}}({\mathbf{y}},{\mathbf{z}}).$

Fix $0<\varepsilon<1$. Then

$|g({\mathbf{y}})|\leq\inf_{u\in
Q({\mathbf{y}},\varepsilon{\bm{\delta}})}|g({\mathbf{u}})|+\sup_{{\mathbf{u}}\in
Q({\mathbf{y}},\varepsilon{\bm{\delta}})}|g({\mathbf{y}})-g({\mathbf{u}})|.$

Denote

$g^{*}({\mathbf{x}}):=\sup_{{\mathbf{y}}\in
X}\frac{V({\mathbf{y}},{\bm{\delta}})^{{\bm{\gamma}}}|g({\mathbf{y}})|}{\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho(x_{i},y_{i})\big{)}^{\tau_{i}/r}}=\sup_{{\mathbf{y}}\in
X}V({\mathbf{y}},{\bm{\delta}})^{{\bm{\gamma}}}{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\tau}}/r}({\mathbf{x}},{\mathbf{y}})|g({\mathbf{y}})|.$

Then

$\displaystyle g^{*}({\mathbf{x}})$ $\displaystyle\leq\sup_{{\mathbf{y}}\in
X}V({\mathbf{y}},{\bm{\delta}})^{{\bm{\gamma}}}{\mathfrak{D}}^{*}_{\delta,{\bm{\tau}}/r}({\mathbf{x}},{\mathbf{y}})\inf_{{\mathbf{u}}\in
Q({\mathbf{y}},\varepsilon{\bm{\delta}})}|g({\mathbf{u}})|$
$\displaystyle+\sup_{{\mathbf{y}}\in
X}V({\mathbf{y}},{\bm{\delta}})^{{\bm{\gamma}}}{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\tau}}/r}({\mathbf{x}},{\mathbf{y}})\sup_{{\mathbf{u}}\in
Q({\mathbf{y}},\varepsilon{\bm{\delta}})}|g({\mathbf{y}})-g({\mathbf{u}})|$
(6.21) $\displaystyle=:\sup_{{\mathbf{y}}\in
X}g^{*}_{1}({\mathbf{y}})+\sup_{{\mathbf{y}}\in X}g^{*}_{2}({\mathbf{y}}).$

We first estimate $g^{*}_{1}({\mathbf{y}})$, ${\mathbf{y}}\in X$. Clearly,

$\inf_{{\mathbf{u}}\in
Q({\mathbf{y}},\varepsilon{\bm{\delta}})}|g({\mathbf{u}})|\leq\Big{(}\frac{1}{\mu(Q({\mathbf{y}},\varepsilon{\bm{\delta}}))}\int_{Q({\mathbf{y}},\varepsilon{\bm{\delta}})}|g({\mathbf{u}})|^{r}d\mu({\mathbf{u}})\Big{)}^{1/r},$

and hence

(6.22) $\displaystyle g^{*}_{1}({\mathbf{y}})$
$\displaystyle\leq\mu(Q({\mathbf{y}},{\bm{\delta}}))^{{\bm{\gamma}}}{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\tau}}/r}({\mathbf{x}},{\mathbf{y}})\Big{(}\frac{1}{\mu(Q({\mathbf{y}},\varepsilon{\bm{\delta}}))}\int_{Q({\mathbf{y}},\varepsilon{\bm{\delta}})}|g({\mathbf{u}})|^{r}d\mu({\mathbf{u}})\Big{)}^{1/r}$
$\displaystyle\leq
c{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\tau}}/r}({\mathbf{x}},{\mathbf{y}})\Big{(}\frac{1}{\mu(Q({\mathbf{y}},\varepsilon{\bm{\delta}}))}\int_{Q({\mathbf{y}},\varepsilon{\bm{\delta}})}\mu(Q({\mathbf{u}},{\bm{\delta}}))^{{\bm{\gamma}}}|g({\mathbf{u}})|\bigr{)}^{r}d\mu({\mathbf{u}})\Big{)}^{1/r},$

where in the last inequality we used that
$\mu(Q({\mathbf{y}},{\bm{\delta}}))\sim\mu(Q({\mathbf{u}},{\bm{\delta}}))$
since for every ${\mathbf{u}}\in Q({\mathbf{y}},\varepsilon{\bm{\delta}})$,
$Q({\mathbf{y}},{\bm{\delta}})\subset Q({\mathbf{u}},2{\bm{\delta}})$ and
$Q({\mathbf{u}},{\bm{\delta}})\subset Q({\mathbf{y}},2{\bm{\delta}})$.

We introduce the abbreviated notation:

$\displaystyle Q$
$\displaystyle:=Q({\mathbf{y}},\varepsilon{\bm{\delta}}),\quad\tilde{Q}:=B_{1}\big{(}x_{1},\rho_{1}(x_{1},y_{1})+\varepsilon\delta_{1}\big{)}\times
B_{2}\big{(}x_{2},\rho_{2}(x_{2},y_{2})+\varepsilon\delta_{2}\big{)},$
$\displaystyle Q^{*}$
$\displaystyle:=B_{1}\big{(}y_{1},2\rho_{1}(x_{1},y_{1})+\varepsilon\delta_{1}\big{)}\times
B_{2}\big{(}y_{2},2\rho_{2}(x_{2},y_{2})+\varepsilon\delta_{2}\big{)}.$

From (6.22) taking into account that $Q\subset\tilde{Q}$ we get

(6.23) $\displaystyle g^{*}_{1}({\mathbf{y}})$ $\displaystyle\leq
c\Big{(}\frac{\mu(\tilde{Q})}{\mu(Q)}{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\tau}}}({\mathbf{x}},{\mathbf{y}})\Big{)}^{1/r}\Big{(}\frac{1}{\mu(\tilde{Q})}\int_{Q}\big{(}V({\mathbf{u}},{\bm{\delta}})^{{\bm{\gamma}}}|g({\mathbf{u}})|\big{)}^{r}d\mu({\mathbf{u}})\Big{)}^{1/r}$
$\displaystyle\leq
c\Big{(}\frac{\mu(\tilde{Q})}{\mu(Q)}{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\tau}}}({\mathbf{x}},{\mathbf{y}})\Big{)}^{1/r}\Big{(}\frac{1}{\mu(\tilde{Q})}\int_{\tilde{Q}}\big{(}V({\mathbf{u}},{\bm{\delta}})^{{\bm{\gamma}}}|g({\mathbf{u}})|\big{)}^{r}d\mu({\mathbf{u}})\Big{)}^{1/r}$
$\displaystyle\leq
c\Big{(}\frac{\mu(\tilde{Q})}{\mu(Q)}{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\tau}}}({\mathbf{x}},{\mathbf{y}})\Big{)}^{1/r}\mathcal{M}_{r}\big{(}V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}}g(\cdot)\big{)}({\mathbf{x}}).$

By (3.12) and the fact that $0<\varepsilon<1$ we get

$\mu(\tilde{Q})\leq\mu(Q^{*})\leq
c\mu(Q)\prod_{i=1,2}\Big{(}1+\frac{2\rho_{i}(x_{i},y_{i})}{\varepsilon\delta_{i}}\Big{)}^{d_{i}}\leq
c\varepsilon^{-|{\mathbf{d}}|}\mu(Q){\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\tau}}}({\mathbf{x}},{\mathbf{y}})^{-1}$

and hence

(6.24)
$\Big{(}\frac{\mu(\tilde{Q})}{\mu(Q)}{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\tau}}}({\mathbf{x}},{\mathbf{y}})\Big{)}^{1/r}\leq
c\varepsilon^{-|{\mathbf{d}}|/r}.$

This inequality coupled with (6.23) yields

(6.25) $g^{*}_{1}({\mathbf{y}})\leq
c\varepsilon^{-|{\mathbf{d}}|/r}\mathcal{M}_{r}\big{(}V(\cdot,\delta)^{{\bm{\gamma}}}g(\cdot)\big{)}({\mathbf{x}}).$

We now estimate $g^{*}_{2}({\mathbf{y}})$. Using that
$\theta(\delta\sqrt{L})g=g$ and (6.20) we obtain

$\displaystyle\sup_{{\mathbf{u}}\in Q}|g({\mathbf{y}})-g({\mathbf{u}})|$
$\displaystyle\leq\sup_{{\mathbf{u}}\in
Q}\int_{X}|{\mathscr{K}}_{\theta(\delta\sqrt{L})}({\mathbf{y}},{\mathbf{z}})-{\mathscr{K}}_{\theta(\delta\sqrt{L})}({\mathbf{u}},{\mathbf{z}})||g({\mathbf{z}})|d\mu({\mathbf{z}})$
$\displaystyle\leq c\sup_{{\mathbf{u}}\in
Q}V({\mathbf{y}},{\bm{\delta}})^{-1}\sum_{i=1,2}\big{(}\rho_{i}(y_{i},u_{i})/\delta_{i}\big{)}^{\alpha_{i}}\int_{X}{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\sigma}}}({\mathbf{y}},{\mathbf{z}})|g({\mathbf{z}})|d\mu({\mathbf{z}})$
$\displaystyle\leq
c(\varepsilon^{\alpha_{1}}+\varepsilon^{\alpha_{2}})V({\mathbf{y}},{\bm{\delta}})^{-1}\int_{X}{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\sigma}}}({\mathbf{y}},{\mathbf{z}})|g({\mathbf{z}})|d\mu({\mathbf{z}}).$

Choose $\sigma_{i}:=\tau_{i}/r+d_{i}|\gamma_{i}|+d_{i}+1$, $i=1,2$. From
above, the definition of $g^{*}_{2}({\mathbf{y}})$, (6.9), and (6.10) we
obtain

$\displaystyle g^{*}_{2}({\mathbf{y}})$
$\displaystyle\leq\frac{c(\varepsilon^{\alpha_{1}}+\varepsilon^{\alpha_{2}})}{V({\mathbf{y}},{\bm{\delta}})}\int_{X}\frac{V({\mathbf{y}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{z}})|}{\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},y_{i}))^{\tau_{i}/r}(1+\delta_{i}^{-1}\rho_{i}(y_{i},z_{i}))^{\sigma_{i}}}d\mu({\mathbf{z}})$
$\displaystyle\leq\frac{c(\varepsilon^{\alpha_{1}}+\varepsilon^{\alpha_{2}})}{V({\mathbf{y}},{\bm{\delta}})}\int_{X}\frac{V({\mathbf{z}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{z}})|\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(y_{i},z_{i}))^{d_{i}|\gamma_{i}|}}{\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},y_{i}))^{\tau_{i}/r}(1+\delta_{i}^{-1}\rho_{i}(y_{i},z_{i}))^{\sigma_{i}}}d\mu({\mathbf{z}})$
$\displaystyle\leq\frac{c(\varepsilon^{\alpha_{1}}+\varepsilon^{\alpha_{2}})}{V({\mathbf{y}},{\bm{\delta}})}\int_{X}\frac{V({\mathbf{z}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{z}})|}{\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},y_{i}))^{\tau_{i}/r}(1+\delta_{i}^{-1}\rho_{i}(y_{i},z_{i}))^{\tau_{i}/r+d_{i}+1}}d\mu({\mathbf{z}})$
$\displaystyle\leq
c(\varepsilon^{\alpha_{1}}+\varepsilon^{\alpha_{2}})\sup_{{\mathbf{z}}\in
X}\frac{V({\mathbf{z}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{z}})|}{\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},z_{i}))^{\tau_{i}/r}}$
$\displaystyle\hskip
108.405pt\times\frac{1}{V({\mathbf{y}},{\bm{\delta}})}\int_{X}\frac{1}{\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(y_{i},z_{i}))^{d_{i}+1}}d\mu({\mathbf{z}})$
$\displaystyle\leq
c(\varepsilon^{\alpha_{1}}+\varepsilon^{\alpha_{2}})g^{*}({\mathbf{x}}).$

Thus

$g^{*}_{2}({\mathbf{y}})\leq\tilde{c}(\varepsilon^{\alpha_{1}}+\varepsilon^{\alpha_{2}})g^{*}({\mathbf{x}}),$

where the constant $\tilde{c}$ is independent of $\varepsilon$. We choose
$\varepsilon$ so that
$\tilde{c}(\varepsilon^{\alpha_{1}}+\varepsilon^{\alpha_{2}})=1/2$. Then
$\sup_{{\mathbf{y}}\in X}g^{*}_{2}({\mathbf{y}})\leq g^{*}({\mathbf{x}})/2$
which combined with (6.25) leads to

$\displaystyle g^{*}({\mathbf{x}})\leq$ $\displaystyle\sup_{{\mathbf{y}}\in
X}g^{*}_{1}({\mathbf{y}})+\sup_{{\mathbf{y}}\in X}g^{*}_{2}({\mathbf{y}})$
$\displaystyle\leq
c\varepsilon^{-|{\mathbf{d}}|/r}\mathcal{M}_{r}\big{(}V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}}g(\cdot)\big{)}({\mathbf{x}})+g^{*}({\mathbf{x}})/2.$

This implies (LABEL:Peetre-max) using that by (6.18)
$g^{*}({\mathbf{x}})<\infty$ if
$\mathcal{M}\big{(}V(\cdot,{\bm{\delta}})^{\bm{\gamma}}g(\cdot)\big{)}({\mathbf{x}})<\infty$.
∎

###  6.2. Nikolski type inequality

We will often compare different $L^{p}$-norms of spectral functions using
Nikolski type inequalities.

######  Theorem 6.5.

$(i)$ Let $0<p\leq q\leq\infty$, ${\bm{\nu}}\in{\mathbb{N}}_{0}^{2}$ and
${\bm{\gamma}}\in{\mathbb{R}}^{2}$. Then there exists a constant $c>0$ such
that for any $g\in\Sigma_{\mathbf{t}}$ with ${\mathbf{t}}\in[1,\infty)^{2}$
and ${\bm{\nu}}\in{\mathbb{N}}_{0}^{2}$ we have

(6.26)
$\big{\|}V(\cdot,{\mathbf{t}}^{-1})^{{\bm{\gamma}}}L^{{\bm{\nu}}}g(\cdot)\big{\|}_{q}\leq
c{\mathbf{t}}^{2{\bm{\nu}}}\big{\|}V(\cdot,{\mathbf{t}}^{-1})^{{\bm{\gamma}}+(\frac{1}{q}-\frac{{\bm{1}}}{p}){\bm{1}}}g(\cdot)\big{\|}_{p},\quad{\mathbf{t}}^{2{\bm{\nu}}}:=t_{1}^{2\nu_{1}}t_{2}^{2\nu_{2}}.$

$(ii)$ If in addition we assume the non-collapsing condition $\eqref{non-
collapsing}$ for the coordinate spaces $X_{1},X_{2}$, then for any
$g\in\Sigma_{\mathbf{t}}$, ${\mathbf{t}}\in[1,\infty)^{2}$, and
${\bm{\nu}}\in{\mathbb{N}}_{0}^{2}$

(6.27) $\|L^{\bm{\nu}}g\|_{q}\leq
c{\mathbf{t}}^{2{\bm{\nu}}+(\frac{1}{p}-\frac{1}{q}){\mathbf{d}}}\|g\|_{p}.$

Recall that ${\mathbf{d}}=(d_{1},d_{2})$, ${\bm{1}}:=(1,1)$, and for the
definition of $V(\cdot,{\mathbf{t}}^{-1})^{{\bm{\gamma}}}$,
${\bm{\gamma}}\in\mathbb{R}^{2}$, see $\eqref{V-V12}$.

###### Proof.

(i) Let $g\in\Sigma_{\mathbf{t}}$, ${\mathbf{t}}\in[1,\infty)^{2}$, and set
${\bm{\delta}}:={\mathbf{t}}^{-1}=(t_{1}^{-1},t_{2}^{-1})$. Let
$\theta\in\mathcal{C}^{\infty}_{0}({\mathbb{R}}^{2})$,
$\theta(\pm\lambda_{1},\pm\lambda_{2})=\theta(\lambda_{1},\lambda_{2})$ and
$\theta({\bm{\lambda}})=1$ for ${\bm{\lambda}}\in[0,1]^{2}$. Then
$g=\theta({\bm{\delta}}\sqrt{L})g$. Let
$\omega(\lambda_{1},\lambda_{2}):=\lambda_{1}^{2\nu_{1}}\lambda_{2}^{2\nu_{2}}\theta(\lambda_{1},\lambda_{2})$.
Then
$\omega({\bm{\delta}}\sqrt{L})={\bm{\delta}}^{2{\bm{\nu}}}L^{{\bm{\nu}}}\theta({\bm{\delta}}\sqrt{L})$
and hence

(6.28)
$L^{\bm{\nu}}g({\mathbf{x}})=L^{\bm{\nu}}\theta({\bm{\delta}}\sqrt{L})g({\mathbf{x}})={\bm{\delta}}^{-2{\bm{\nu}}}\omega({\bm{\delta}}\sqrt{L})g({\mathbf{x}})={\bm{\delta}}^{-2{\bm{\nu}}}\int_{X}{\mathscr{K}}_{\omega({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})g({\mathbf{y}})d\mu({\mathbf{y}}).$

We use Corollary 4.22 and (3.14) to conclude that for any
${\bm{\sigma}}\in(0,\infty)^{2}$

(6.29)
$|{\mathscr{K}}_{\omega({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})|\leq
c_{\bm{\sigma}}V({\mathbf{x}},{\bm{\delta}})^{-1}{\mathfrak{D}}^{*}_{{\bm{\delta}},{\bm{\sigma}}}({\mathbf{x}},{\mathbf{y}}).$

We next show that inequality (6.26) is valid when $q=\infty$, i.e.

(6.30)
$\big{\|}V(\cdot,{\mathbf{t}}^{-1})^{{\bm{\gamma}}}L^{{\bm{\nu}}}g\big{\|}_{\infty}\leq
c{\mathbf{t}}^{2{\bm{\nu}}}\big{\|}V(\cdot,{\mathbf{t}}^{-1})^{{\bm{\gamma}}-\frac{{\bm{1}}}{p}}g\big{\|}_{p}.$

Case 1: $1<p<\infty$. We use (6.28) and Hölder’s inequality to obtain

$\displaystyle{\bm{\delta}}^{2{\bm{\nu}}}|L^{\bm{\nu}}g({\mathbf{x}})|$
$\displaystyle\leq\big{\|}V(\cdot,\delta)^{{\bm{\gamma}}-\frac{{\bm{1}}}{p}}g\big{\|}_{p}\Big{(}\int_{X}\big{(}|{\mathscr{K}}_{\omega({\bm{\delta}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})|V({\mathbf{y}},{\bm{\delta}})^{-{\bm{\gamma}}+\frac{{\bm{1}}}{p}}\big{)}^{p^{\prime}}d\mu({\mathbf{y}})\Big{)}^{1/p^{\prime}}$
(6.31)
$\displaystyle=:\big{\|}V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}-\frac{{\bm{1}}}{p}}g\big{\|}_{p}\mathcal{Q}.$

By (3.17) we have

(6.32) $V({\mathbf{y}},{\bm{\delta}})^{-{\bm{\gamma}}+\frac{{\bm{1}}}{p}}\leq
cV({\mathbf{x}},{\bm{\delta}})^{-{\bm{\gamma}}+\frac{{\bm{1}}}{p}}\prod_{i=1,2}\big{(}1+\delta_{i}^{-1}\rho_{i}(x_{i},y_{i})\big{)}^{d_{i}|-\gamma_{i}+\frac{1}{p}|}.$

Choose $\sigma_{i}:=d_{i}|-\gamma_{i}+\frac{1}{p}|+(d_{i}+1)/p^{\prime}$ and
set
$\tilde{\sigma}_{i}:=p^{\prime}(\sigma_{i}-d_{i}|-\gamma_{i}+\frac{1}{p}|)$,
$i=1,2$. Clearly $\tilde{\sigma}_{i}=d_{i}+1$. Using (6.29), (6.32), and
(3.19) we obtain

$\mathcal{Q}\leq
V({\mathbf{x}},{\bm{\delta}})^{-{\bm{\gamma}}-(1-\frac{1}{p}){\bm{1}}}\Big{(}\int_{X}{\mathfrak{D}}^{*}_{{\bm{\delta}},\tilde{{\bm{\sigma}}}}({\mathbf{x}},{\mathbf{y}})d\mu({\mathbf{y}})\Big{)}^{1/p^{\prime}}\leq
cV({\mathbf{x}},{\bm{\delta}})^{-{\bm{\gamma}}}.$

This coupled with (6.31) implies (6.30).

Case 2: $p=1$. The proof of inequality (6.30) in this case is similar to the
above proof but easier; we omit it.

Case 3: $0<p<1$. We first show that inequality (6.30) holds in this case when
${\bm{\nu}}=(0,0)$, i.e.

(6.33)
$\big{\|}V(\cdot,{\mathbf{t}}^{-1})^{{\bm{\gamma}}}g\big{\|}_{\infty}\leq
c\big{\|}V(\cdot,{\mathbf{t}}^{-1})^{{\bm{\gamma}}-\frac{{\bm{1}}}{p}}g\big{\|}_{p}.$

To this end we first prove that this inequality is valid with a constant
$c=c(g)$ depending on $g$. The function $\theta({\bm{\lambda}})$ from above
obeys the assumptions of Proposition 5.7 and hence there exist constants
$N,c>0$, depending on $g$, such that

(6.34) $\displaystyle|g({\mathbf{x}})|$
$\displaystyle=|\theta({\bm{\delta}}\sqrt{L})g({\mathbf{x}})|$
$\displaystyle\leq
c(g)V({\mathbf{x}}_{0},\delta)^{-1}\prod_{i=1,2}(\delta_{i}^{N}+\delta_{i}^{-N})\big{(}1+\rho_{i}(x_{i},x_{0i})\big{)}^{N}.$

Now, precisely as in the proof of Theorem 6.4 (cf. (6.17)) with $r$ replaced
by $p$ we obtain

$\Big{(}\sup_{{\mathbf{z}}\in
X}\frac{V({\mathbf{z}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{z}})|}{\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},z_{i}))^{\tilde{N}}}\Big{)}^{p}\\\
\leq
c\int_{X}\frac{[V({\mathbf{y}},{\bm{\delta}})^{\bm{\gamma}}|g({\mathbf{y}})|]^{p}}{V({\mathbf{y}},{\bm{\delta}})\prod_{i=1,2}(1+\delta_{i}^{-1}\rho_{i}(x_{i},y_{i}))^{\tilde{N}p}}d\mu({\mathbf{y}})\\\
\leq
c\int_{X}[V({\mathbf{y}},{\bm{\delta}})^{{\bm{\gamma}}-(1/p){\bm{1}}}|g({\mathbf{y}})|]^{p}d\mu({\mathbf{y}}).$

With ${\mathbf{z}}={\mathbf{x}}$ the above implies

(6.35) $\|V(\cdot,{\bm{\delta}})^{\bm{\gamma}}g(\cdot)\|_{\infty}\leq
c(g)\|V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}-(1/p){\bm{1}}}g(\cdot)\|_{p}.$

Here the constant depends on $g$ because $N$ depends on $g$.

We are now ready to complete the proof of (6.33). Applying (6.30) with $p=2$
and ${\bm{\nu}}=(0,0)$ we get

$\displaystyle\big{\|}V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}}g\big{\|}_{\infty}$
$\displaystyle\leq
c\big{\|}V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}-(1/2){\bm{1}}}g\big{\|}_{2}$
(6.36)
$\displaystyle=c\Big{(}\int_{X}\big{(}V({\mathbf{x}},{\bm{\delta}})^{{\bm{\gamma}}}|g({\mathbf{x}})|\big{)}^{2-p}\big{(}V({\mathbf{x}},{\bm{\delta}})^{{\bm{\gamma}}-(1/p){\bm{1}}}|g({\mathbf{x}})|\big{)}^{p}d\mu({\mathbf{x}})\Big{)}^{1/2}$
$\displaystyle\leq
c\big{\|}V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}}g(\cdot)\big{\|}_{\infty}^{1-p/2}\big{\|}V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}-(1/p){\bm{1}}}g(\cdot)\big{\|}_{p}^{p/2}.$

If
$\|V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}-\frac{{\bm{1}}}{p}}g(\cdot)\|_{p}=\infty$,
then there is nothing to prove. If
$\|V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}-\frac{{\bm{1}}}{p}}g(\cdot)\|_{p}<\infty$,
then by (6.35)
$\|V(\cdot,{\bm{\delta}})^{\bm{\gamma}}g(\cdot)\|_{\infty}<\infty$ and we can
divide in (6.2) by
$\big{\|}V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}}g\big{\|}_{\infty}^{1-p/2}$ to
obtain (6.33).

Finally, using (6.30) with $p=2$, (6.2), and (6.33) we get

$\displaystyle{\bm{\delta}}^{2{\bm{\nu}}}\big{\|}V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}}L^{{\bm{\nu}}}g(\cdot)\big{\|}_{\infty}$
$\displaystyle\leq
c\big{\|}V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}-\frac{{\bm{1}}}{2}}g\big{\|}_{2}$
$\displaystyle\leq
c\big{\|}V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}}g(\cdot)\big{\|}_{\infty}^{1-p/2}\big{\|}V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}-\frac{{\bm{1}}}{p}}g(\cdot)\big{\|}_{p}^{p/2}$
$\displaystyle\leq
c\big{\|}V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}-\frac{{\bm{1}}}{p}}g\big{\|}_{p},$

which completes the proof of (6.30).

We next prove (6.26) in the case $0<p<q<\infty$. From Theorem 6.4 it follows
that for any ${\bm{\gamma}}\in{\mathbb{R}}^{2}$

$\displaystyle{\bm{\delta}}^{2{\bm{\nu}}}V({\mathbf{x}},\delta)^{{\bm{\gamma}}}|L^{\bm{\nu}}g({\mathbf{x}})|\leq
c\mathcal{M}_{r}\big{(}V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}}g(\cdot)\big{)}({\mathbf{x}}).$

We choose $0<r<p$ and apply the maximal inequality (4.104) to obtain

(6.37)
$\delta^{2{\bm{\nu}}}\|V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}}L^{\bm{\nu}}g(\cdot)\|_{p}\leq
c\|\mathcal{M}_{r}\big{(}V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}}g(\cdot)\big{)}\|_{p}\leq
c\|V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}}g(\cdot)\|_{p}.$

Finally, using (6.30) with ${\bm{\gamma}}$ replaced by
${\bm{\gamma}}+\frac{{\bm{1}}}{q}$ and (6.37) with ${\bm{\gamma}}$ replaced by
${\bm{\gamma}}+(\frac{1}{q}-\frac{1}{p}){\bm{1}}$ we get

$\displaystyle{\bm{\delta}}^{2{\bm{\nu}}}$
$\displaystyle\big{\|}V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}}L^{\bm{\nu}}g(\cdot)\big{\|}_{q}$
$\displaystyle\leq\big{(}\delta^{2{\bm{\nu}}}\big{\|}V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}+\frac{{\bm{1}}}{q}}L^{\bm{\nu}}g\big{\|}_{\infty}\big{)}^{1-\frac{p}{q}}\big{(}{\bm{\delta}}^{2{\bm{\nu}}}\big{\|}V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}+(\frac{1}{q}-\frac{1}{p}){\bm{1}}}L^{\bm{\nu}}g(\cdot)\big{\|}_{p}\big{)}^{\frac{p}{q}}$
$\displaystyle\leq
c\big{\|}V(\cdot,{\bm{\delta}})^{{\bm{\gamma}}+(\frac{1}{q}-\frac{1}{p}){\bm{1}}}g(\cdot)\big{\|}_{p},$

which confirms (6.26).

(ii) From (i) with ${\bm{\gamma}}=(0,0)$ we derive

${\bm{\delta}}^{2{\bm{\nu}}}\|L^{\bm{\nu}}g\|_{q}\leq
c\big{\|}V(\cdot,{\bm{\delta}})^{(\frac{1}{q}-\frac{1}{p}){\bm{1}}}g(\cdot)\big{\|}_{p}$

and using (3.18), which is a consequence of the non-collapsing condition
(2.6), we arrive at (6.27). ∎

##  7\. Product Hardy spaces

Hardy spaces $H^{p}$, $0<p\leq 1$, play a fundamental role in Harmonic
analysis, see [19] and [47] and the references therein. R. Coifman and G.
Weiss [11] introduced and pioneered the theory of Hardy spaces on spaces of
homogeneous type. Later Hardy spaces have been extensively studied on
Euclidean [18] and metric spaces associated with operators [35]. Recently, in
[15] the equivalence of Hardy spaces defined by several maximal operators as
well as atomic Hardy spaces were obtained in the setting of a doubling metric
measure space $(X,\rho,\mu)$ in the presence of a non-negative self-adjoint
operator $L$ whose heat kernel has Gaussian localization. In this setting the
Hardy space $H^{p}$, $0<p\leq 1$, is defined as the set of distributions
$f\in\mathcal{S}^{\prime}$ such that

(7.1)
$\|f\|_{H^{p}}:=\big{\|}\sup_{t>0}|e^{-t^{2}L}f(\cdot)|\big{\|}_{p}<\infty.$

Hardy spaces $H^{p}$, $0<p<\infty$, in the two-parameter classical setting
have been developed mainly by Sun-Yung Chang and R. Fefferman, see [6, 7, 8,
20] and the references therein. In the anisotropic setting the spaces have
been studied in [4, 5]. Hardy spaces $H^{p}$, $p\geq 1$, have been introduced
via area functions on product of spaces of homogeneous type associated with
operators [9].

In the present section we initiate the development of Hardy spaces $H^{p}$,
$0<p<\infty$, in the two-parameter setting on product domains of this article.
We will introduce the bi-variable version of (7.1) and establish equivalent
characterizations via several maximal operators.

We will operate in the setting of this article (see Section 2), assuming in
addition that $\mu_{i}(X_{i})=\infty$, $i=1,2$. We start with the following

######  Definition 7.1.

The product Hardy space $H^{p}=H^{p}(L_{1},L_{2})$, $0<p\leq 1$, is defined as
the set of all $f\in\mathcal{S}^{\prime}=\mathcal{S}^{\prime}(L_{1},L_{2})$
such that

(7.2)
$\|f\|_{H^{p}}:=\big{\|}\sup_{{\mathbf{t}}>{\bm{0}}}|e^{-t_{1}^{2}L_{1}-t_{2}^{2}L_{2}}f(\cdot)|\big{\|}_{p}<\infty.$

The bi-variate nature of these spaces is exhibited in the vector-dilation of
the parameter ${\mathbf{t}}=(t_{1},t_{2})$.

###  7.1. Maximal operators and the main result

We next introduce several maximal operators that will be used in
characterizing the two-parameter (product) Hardy spaces $H^{p}$. We first
introduce the class of admissible functions on ${\mathbb{R}}^{2}$, suitable
for the theory in the product setting.

######  Definition 7.2.

We say that a function $\varphi\in\mathcal{S}({\mathbb{R}}^{2})$ is admissible
if $\varphi$ is real-valued, $\varphi(0,0)\neq 0$, and

(7.3)
$\varphi(\pm\lambda_{1},\pm\lambda_{2})=\varphi(\lambda_{1},\lambda_{2}),\quad\forall(\lambda_{1},\lambda_{2})\in{\mathbb{R}}^{2}.$

######  Definition 7.3.

Let $\varphi\in\mathcal{S}({\mathbb{R}}^{2})$ be an admissible function. We
define the following maximal operators: Assuming $f\in\mathcal{S}^{\prime}$
and ${\mathbf{x}}\in X$, we set

(7.4)
$M(f;\varphi)({\mathbf{x}}):=\sup_{{\mathbf{t}}>{\bm{0}}}|\varphi({\mathbf{t}}\sqrt{L})f({\mathbf{x}})|,$
(7.5)
$M^{*}_{\mathbf{a}}(f;\varphi)({\mathbf{x}}):=\sup_{{\mathbf{t}}>{\bm{0}}}\sup_{\\{{\mathbf{y}}\in
X:\rho_{i}(x_{i},y_{i})\leq
a_{i}t_{i}\\}}|\varphi({\mathbf{t}}\sqrt{L})f({\mathbf{y}})|,\quad{\mathbf{a}}\geq{\bm{0}},$

and

(7.6)
$M^{**}_{\bm{\gamma}}(f;\varphi)({\mathbf{x}}):=\sup_{{\mathbf{t}}>{\bm{0}}}\sup_{{\mathbf{y}}\in
X}\frac{|\varphi({\mathbf{t}}\sqrt{L})f({\mathbf{y}})|}{\prod_{i=1,2}(1+t_{i}^{-1}\rho_{i}(x_{i},y_{i}))^{\gamma_{i}}},\quad{\bm{\gamma}}>{\bm{0}}.$

Using the compact notation
${\mathfrak{D}}^{*}_{{\mathbf{t}},{\bm{\gamma}}}({\mathbf{x}},{\mathbf{y}}):=\prod_{i=1,2}(1+t_{i}^{-1}\rho_{i}(x_{i},y_{i}))^{-\gamma_{i}}$,
see (3.10), this can be written in the form

(7.7)
$M^{**}_{\bm{\gamma}}(f;\varphi)({\mathbf{x}}):=\sup_{{\mathbf{t}}>{\bm{0}}}\sup_{{\mathbf{y}}\in
X}|\varphi({\mathbf{t}}\sqrt{L})f({\mathbf{y}})|{\mathfrak{D}}^{*}_{{\mathbf{t}},{\bm{\gamma}}}({\mathbf{x}},{\mathbf{y}}).$

Obviously,

(7.8) $M(f;\varphi)({\mathbf{x}})\leq
M^{*}_{\mathbf{a}}(f;\varphi)({\mathbf{x}})\leq({\bm{1}}+{\mathbf{a}})^{\bm{\gamma}}M^{**}_{\bm{\gamma}}(f;\varphi)({\mathbf{x}}),\quad\forall
f\in\mathcal{S}^{\prime},\;{\mathbf{x}}\in X.$

We next introduce the grand maximal operator.

######  Definition 7.4.

Let $\varphi\in\mathcal{S}(\mathbb{R}^{2})$ be admissible and
$N\in{\mathbb{N}}$. We define

(7.9)
$\mathcal{N}_{N}(\varphi):=\sup_{{\bm{\lambda}}\in{\mathbb{R}}^{2}}(1+|{\bm{\lambda}}|)^{N}\max_{0\leq|{\bm{\beta}}|\leq
N}|\partial^{{\bm{\beta}}}\varphi({\bm{\lambda}})|$

and set

$\mathcal{F}_{N}:=\\{\varphi\in\mathcal{S}({\mathbb{R}}^{2}):\varphi\;{\rm
is\;admissible\;and}\;\;\mathcal{N}_{N}(\varphi)\leq 1\\}.$

The grand maximal operator is defined by

(7.10)
$\mathscr{M}_{N}(f)({\mathbf{x}}):=\sup_{\varphi\in\mathcal{F}_{N}}M^{*}_{\bm{1}}(f;\varphi)({\mathbf{x}}),\quad
f\in\mathcal{S}^{\prime},\;{\mathbf{x}}\in X,$

that is,

(7.11)
$\mathscr{M}_{N}(f)({\mathbf{x}}):=\sup_{\varphi\in\mathcal{F}_{N}}\sup_{{\mathbf{t}}>{\bm{0}}}\sup_{\\{{\mathbf{y}}\in
X:\rho_{i}(x_{i},y_{i})\leq
t_{i}\\}}|\varphi({\mathbf{t}}\sqrt{L})f({\mathbf{y}})|.$

Observe that for any admissible function $\varphi$,
${\mathbf{a}}\geq{\bm{1}}$, and $N\in{\mathbb{N}}$,

(7.12)
$M^{*}_{\mathbf{a}}(f;\varphi)({\mathbf{x}})\leq|{\mathbf{a}}|^{N}\mathcal{N}_{N}(\varphi)\mathscr{M}_{N}(f)(x),\quad\forall
f\in\mathcal{S}^{\prime},\;{\mathbf{x}}\in X.$

Indeed, consider
$\phi(\lambda_{1},\lambda_{2}):=A^{-1}\varphi(a_{1}^{-1}\lambda_{1},a_{2}^{-1}\lambda_{2})$,
where $A:=|{\mathbf{a}}|^{N}\mathcal{N}_{N}(\varphi)$. It is readily seen that
$M^{*}_{\mathbf{a}}(f;\varphi)({\mathbf{x}})=AM^{*}_{\bm{1}}(f;\phi)({\mathbf{x}})$
and a simple manipulation shows that $\mathcal{N}_{N}(\phi)\leq 1$. Then
(7.12) follows.

The main result in the section is the following

######  Theorem 7.5.

Let $0<p\leq 1$. Assume $N>6|{\mathbf{d}}|/p+3|{\mathbf{d}}|/2+3$,
${\mathbf{a}}\geq{\bm{1}}$, ${\bm{\gamma}}>(2/p){\mathbf{d}}$ and let
$\varphi\in\mathcal{S}({\mathbb{R}}^{2})$ be admissible. Then for any
$f\in\mathcal{S}^{\prime}$

(7.13)
$\|f\|_{H^{p}}\sim\|\mathscr{M}_{N}(f)\|_{L^{p}}\sim\|M(f;\varphi)\|_{L^{p}}\sim\|M^{*}_{\mathbf{a}}(f;\varphi)\|_{L^{p}}\sim\|M^{**}_{\bm{\gamma}}(f;\varphi)\|_{L^{p}}.$

Here the equivalence constants depend on ${\mathbf{a}}$, ${\bm{\gamma}}$, $N$,
and $\varphi$, respectively.

For the proof of this theorem we need some preparation. We first establish an
useful decomposition identity.

######  Lemma 7.6.

Let $\varphi\in\mathcal{S}({\mathbb{R}}^{2})$ be admissible with
$\varphi(0,0)=1$ and let $N\in{\mathbb{N}}_{0}$. Then there exist an
admissible function $\psi\in\mathcal{S}({\mathbb{R}}^{2})$ with $\psi(0,0)=1$
and a function $\Psi\in\mathcal{S}({\mathbb{R}}^{2})$ with the properties
$\Psi(\pm\lambda_{1},\pm\lambda_{2})=\Psi(\lambda_{1},\lambda_{2})$ for
${\bm{\lambda}}\in\mathbb{R}^{2}$ and $\partial^{{\bm{\beta}}}\Psi(0,0)=0$ for
$|{\bm{\beta}}|\leq N$ such that for any $f\in\mathcal{S}^{\prime}$ and
${\mathbf{j}}\in{\mathbb{Z}}^{2}$

(7.14)
$f=\psi(2^{-{\mathbf{j}}}\sqrt{L})\varphi(2^{-{\mathbf{j}}}\sqrt{L})f+\sum_{k=1}^{\infty}\Psi(2^{-{\mathbf{j}}-{\mathbf{k}}}\sqrt{L})\big{(}\varphi(2^{-{\mathbf{j}}-{\mathbf{k}}}\sqrt{L})-\varphi(2^{-{\mathbf{j}}-{\mathbf{k}}+{\bm{1}}}\sqrt{L})\big{)}f,$

where the convergence is in $\mathcal{S}^{\prime}$ and ${\mathbf{k}}:=(k,k)$,
${\bm{1}}:=(1,1)$.

###### Proof.

Let $\varphi$ be from the hypothesis of the lemma and assume $N\geq 2$. From
the binomial formula it readily follows that

(7.15)
$\sum_{m=1}^{N}{\textstyle\binom{N}{m}}\varphi({\bm{\lambda}})^{2m}(1-\varphi({\bm{\lambda}})^{2})^{N-m}=1-(1-\varphi({\bm{\lambda}})^{2})^{N}$

and for $k\geq 1$

$\displaystyle\sum_{m=1}^{N}{\textstyle\binom{N}{m}}$
$\displaystyle(\varphi(2^{-k}{\bm{\lambda}})^{2}-\varphi(2^{-k+1}{\bm{\lambda}})^{2})^{m}(1-\varphi(2^{-k}{\bm{\lambda}})^{2})^{N-m}$
$\displaystyle=(1-\varphi(2^{-k+1}{\bm{\lambda}})^{2})^{N}-(1-\varphi(2^{-k}{\bm{\lambda}})^{2})^{N}.$

Summing up the above identities and using that $\varphi(0,0)=1$ we get

$\displaystyle\sum_{k=1}^{\infty}\sum_{m=1}^{N}{\textstyle\binom{N}{m}}$
$\displaystyle(\varphi(2^{-k}{\bm{\lambda}})^{2}-\varphi(2^{-k+1}{\bm{\lambda}})^{2})^{m}(1-\varphi(2^{-k}{\bm{\lambda}})^{2})^{N-m}=(1-\varphi({\bm{\lambda}})^{2})^{N}.$

Combining this with (7.15) we find

$\displaystyle 1$
$\displaystyle=\sum_{m=1}^{N}{\textstyle\binom{N}{m}}\varphi({\bm{\lambda}})^{2m}(1-\varphi({\bm{\lambda}})^{2})^{N-m}$
$\displaystyle+\sum_{k=1}^{\infty}\sum_{m=1}^{N}{\textstyle\binom{N}{m}}(\varphi(2^{-k}{\bm{\lambda}})^{2}-\varphi(2^{-k+1}{\bm{\lambda}})^{2})^{m}(1-\varphi(2^{-k}{\bm{\lambda}})^{2})^{N-m},$

which takes the form

(7.16)
$1=\psi({\bm{\lambda}})\varphi({\bm{\lambda}})+\sum_{k=1}^{\infty}\Psi(2^{-k}{\bm{\lambda}})(\varphi(2^{-k}{\bm{\lambda}})-\varphi(2^{-k+1}{\bm{\lambda}})),$

where

$\psi({\bm{\lambda}}):=\sum_{m=1}^{N}{\textstyle\binom{N}{m}}\varphi({\bm{\lambda}})^{2m-1}(1-\varphi({\bm{\lambda}})^{2})^{N-m}$

and

$\Psi({\bm{\lambda}}):=(\varphi({\bm{\lambda}})+\varphi(2{\bm{\lambda}}))\sum_{m=1}^{N}{\textstyle\binom{N}{m}}(\varphi({\bm{\lambda}})^{2}-\varphi(2{\bm{\lambda}})^{2})^{m-1}(1-\varphi({\bm{\lambda}})^{2})^{N-m}.$

Note that since $\varphi$ is admissible, then $\psi$ and $\Psi$ are admissible
as well. Also by the construction $\psi(0,0)=1$ and
$\partial^{\beta}\Psi(0,0)=0$ for $|\beta|\leq N-2$. We now replace
${\bm{\lambda}}$ by $2{\bm{\lambda}}$ in (7.16) and subtract the resulting
identity from (7.16) to obtain

$\Psi({\bm{\lambda}})(\varphi({\bm{\lambda}})-\varphi(2{\bm{\lambda}}))=\psi({\bm{\lambda}})\varphi({\bm{\lambda}})-\psi(2{\bm{\lambda}})\varphi(2{\bm{\lambda}}).$

From this it readily follows that for any $m\in{\mathbb{N}}$

$\displaystyle\sum_{k=1}^{m}\Psi(2^{-{\mathbf{j}}-{\mathbf{k}}}{\bm{\lambda}})$
$\displaystyle(\varphi(2^{-{\mathbf{j}}-{\mathbf{k}}}{\bm{\lambda}})-\varphi(2^{-{\mathbf{j}}-{\mathbf{k}}+1}{\bm{\lambda}}))$
$\displaystyle=\psi(2^{-{\mathbf{j}}-{\mathbf{m}}}{\bm{\lambda}})\varphi(2^{-{\mathbf{j}}-{\mathbf{m}}}{\bm{\lambda}})-\psi(2^{-{\mathbf{j}}}{\bm{\lambda}})\varphi(2^{-{\mathbf{j}}}{\bm{\lambda}}),\quad{\mathbf{m}}:=(m,m).$

and hence for any $f\in\mathcal{S}^{\prime}$

$\displaystyle\psi(2^{-{\mathbf{j}}}\sqrt{L})\varphi(2^{-{\mathbf{j}}}\sqrt{L})f$
$\displaystyle+\sum_{k=1}^{m}\Psi(2^{-{\mathbf{j}}-k}\sqrt{L})\big{(}\varphi(2^{-{\mathbf{j}}-k}\sqrt{L})-\varphi(2^{-{\mathbf{j}}-k+1}\sqrt{L})\big{)}f$
(7.17)
$\displaystyle=\psi(2^{-{\mathbf{j}}-{\mathbf{m}}}\sqrt{L})\varphi(2^{-{\mathbf{j}}-{\mathbf{m}}}\sqrt{L})f.$

From Theorem 5.10 it follows that
$\lim_{m\to\infty}\psi(2^{-{\mathbf{j}}-{\mathbf{m}}}\sqrt{L})\varphi(2^{-{\mathbf{j}}-{\mathbf{m}}}\sqrt{L})f=f$
in $\mathcal{S}^{\prime}$ and passing to the limit in (7.1) as $m\to\infty$ we
obtain (7.14). Finally, the lemma follows from the above considerations by
replacing $N$ with $N+2$. ∎

Our next step is to connect different maximal operators from Section 7.1 by
means of the Hardy-Littlewood maximal operator.

######  Proposition 7.7.

Let $\varphi\in\mathcal{S}({\mathbb{R}}^{2})$ be admissible and $0<\theta\leq
1$. Let also ${\bm{\gamma}}=(\gamma_{1},\gamma_{2})>{\bm{0}}$ be so that
${\bm{\gamma}}>(2/\theta){\mathbf{d}}$. Then there exists a constant $c>0$
such that

(7.18) $M_{{\bm{\gamma}}}^{**}(f;\varphi)({\mathbf{x}})\leq
c\mathcal{M}_{\theta}(M(f;\varphi))({\mathbf{x}}),\quad\text{for}\;f\in\mathcal{S}^{\prime},\;{\mathbf{x}}\in
X,$

where $\mathcal{M}_{\theta}$ is the Hardy-Littlewood maximal operator, see
(4.102).

###### Proof.

Let $\varphi,\theta$ and ${\bm{\gamma}}$ be from the hypothesis of the
proposition. Without loss of generality we may assume that $\varphi(0,0)=1$;
otherwise we use $\varphi(0,0)^{-1}\varphi$ instead. Fix
$f\in\mathcal{S}^{\prime}$ and ${\mathbf{x}}\in X$.

Part 1. We first show that inequality (7.18) is valid under the additional
assumption

(7.19) $M_{{\bm{\gamma}}}^{**}(f;\varphi)({\mathbf{x}})<\infty.$

By Lemma 7.6 for any $N\in{\mathbb{N}}_{0}$ there exists an admissible
function $\psi\in\mathcal{S}({\mathbb{R}}^{2})$ with $\psi(0,0)=1$ and a
function $\Psi\in\mathcal{S}({\mathbb{R}}^{2})$ with the properties
$\Psi(\pm\lambda_{1},\pm\lambda_{2})=\Psi(\lambda_{1},\lambda_{2})$ for
${\bm{\lambda}}\in\mathbb{R}^{2}$, and $\partial^{{\bm{\beta}}}\Psi(0,0)=0$
for $|{\bm{\beta}}|\leq N$ such that

$f=\psi(2^{-{\mathbf{j}}}\sqrt{L})\varphi(2^{-{\mathbf{j}}}\sqrt{L})f+\sum_{k=1}^{\infty}\Psi(2^{-{\mathbf{j}}-{\mathbf{k}}}\sqrt{L})(\varphi(2^{-{\mathbf{j}}-{\mathbf{k}}}\sqrt{L})-\varphi(2^{-{\mathbf{j}}-{\mathbf{k}}+{\bm{1}}}\sqrt{L}))f,$

where ${\mathbf{k}}:=(k,k)$. We choose $N:=\lfloor
3|{\bm{\gamma}}|+3|{\mathbf{d}}|/2+4\rfloor$.

Fix ${\mathbf{t}}>{\bm{0}}$ and let ${\mathbf{j}}\in{\mathbb{Z}}^{2}$ be such
that $2^{-{\mathbf{j}}}\leq{\mathbf{t}}<2^{-{\mathbf{j}}+{\bm{1}}}$. In light
of (7.7) we need to deal with
$|\varphi({\mathbf{t}}\sqrt{L})f({\mathbf{y}})|{\mathfrak{D}}^{*}_{{\mathbf{t}},{\bm{\gamma}}}({\mathbf{x}},{\mathbf{y}})$.
From above, using that $t_{i}\sim 2^{-j_{i}}$, we get

(7.20)
$\displaystyle|\varphi({\mathbf{t}}\sqrt{L})f({\mathbf{y}})|{\mathfrak{D}}^{*}_{{\mathbf{t}},{\bm{\gamma}}}({\mathbf{x}},{\mathbf{y}})\leq
c|\varphi({\mathbf{t}}\sqrt{L})\psi(2^{-{\mathbf{j}}}\sqrt{L})\varphi(2^{-{\mathbf{j}}}\sqrt{L})f({\mathbf{y}})|{\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}},{\bm{\gamma}}}({\mathbf{x}},{\mathbf{y}})$
$\displaystyle+c\sum_{k=1}^{\infty}|\varphi({\mathbf{t}}\sqrt{L})\Psi(2^{-{\mathbf{j}}-{\mathbf{k}}}\sqrt{L})\big{[}\varphi(2^{-{\mathbf{j}}-{\mathbf{k}}}\sqrt{L})-\varphi(2^{-{\mathbf{j}}-{\mathbf{k}}+{\bm{1}}}\sqrt{L})\big{]}f({\mathbf{y}})|{\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}},{\bm{\gamma}}}({\mathbf{x}},{\mathbf{y}}).$

Let
$\omega({\bm{\lambda}}):=\varphi({\mathbf{t}}2^{{\mathbf{j}}}{\bm{\lambda}})\Psi(2^{-k}{\bm{\lambda}})$.
Then
$\varphi({\mathbf{t}}\sqrt{L})\Psi(2^{-{\mathbf{j}}-{\mathbf{k}}}\sqrt{L})=\omega(2^{-{\mathbf{j}}}\sqrt{L})$.
Choose
$m:=\lfloor\gamma_{1}+d_{1}/2+1\rfloor+\lfloor\gamma_{2}+d_{2}/2+1\rfloor$ and
$r:=m+|{\mathbf{d}}|+1$. Since $\varphi$ and $\Psi$ are admissible, then
$\omega\in\mathcal{S}(\mathbb{R}^{2})$ and
$\omega(\pm\lambda_{1},\pm\lambda_{2})=\omega(\lambda_{1},\lambda_{2})$. Also,
using that ${\mathbf{t}}\sim 2^{-{\mathbf{j}}}$ and $k\geq 1$ we get

$|\partial^{{\bm{\beta}}}\omega({\bm{\lambda}})|\leq
c(1+|{\bm{\lambda}}|)^{-N},\quad{\bm{\lambda}}\in\mathbb{R}^{2}.$

From this it readily follow that

$|\partial^{{\bm{\beta}}}\omega({\bm{\lambda}})|\leq
c2^{-k(N-r)/2}(1+|{\bm{\lambda}}|)^{-r},\quad\hbox{if $\;|{\bm{\lambda}}|\geq
2^{k/2}$, $|{\bm{\beta}}|\leq N$,}$

and since $N\geq 3|{\bm{\gamma}}|+3|{\mathbf{d}}|/2+3+2\tilde{{\varepsilon}}$
for some $\tilde{{\varepsilon}}>0$ it follows that

(7.21) $|\partial^{{\bm{\beta}}}\omega({\bm{\lambda}})|\leq
c2^{-k(|{\bm{\gamma}}|+\tilde{{\varepsilon}})}(1+|{\bm{\lambda}}|)^{-r},\quad|{\bm{\lambda}}|\geq
2^{k/2},|{\bm{\beta}}|\leq N.$

On the other hand, since $\partial^{{\bm{\beta}}}\Psi(0,0)=0$ for
$|{\bm{\beta}}|\leq N$, Taylor’s formula implies that
$|\partial^{{\bm{\beta}}}\Psi({\bm{\lambda}})|\leq
c|{\bm{\lambda}}|^{N-|{\bm{\beta}}|}$ for $|{\bm{\lambda}}|\leq 1$ and
$|{\bm{\beta}}|\leq N$. Then for $|{\bm{\lambda}}|<2^{k/2}$, we get

$\displaystyle|\partial^{{\bm{\beta}}}(\Psi(2^{-k}{\bm{\lambda}}))|=2^{-k|{\bm{\beta}}|}|(\partial^{{\bm{\beta}}}\Psi)(2^{-k}{\bm{\lambda}})|\leq
c2^{-k|{\bm{\beta}}|}(2^{-k}|{\bm{\lambda}}|)^{N-|{\bm{\beta}}|}\leq
c2^{-kN/2}.$

From this and the fact that $\varphi\in\mathcal{S}(\mathbb{R}^{2})$ it follows
that

$|\partial^{{\bm{\beta}}}\omega({\bm{\lambda}})|\leq
c2^{-kN/2}(1+|{\bm{\lambda}}|)^{-N}\leq
c2^{-k(|{\bm{\gamma}}|+\tilde{{\varepsilon}})}(1+|{\bm{\lambda}}|)^{-r},\quad|{\bm{\lambda}}|<2^{k/2}.$

Combining this estimate with (7.21) leads to

(7.22) $|\partial^{{\bm{\beta}}}\omega({\bm{\lambda}})|\leq
c2^{-k(|{\bm{\gamma}}|+\tilde{{\varepsilon}})}(1+|{\bm{\lambda}}|)^{-r},\quad{\bm{\lambda}}\in\mathbb{R}^{2},|{\bm{\beta}}|\leq
m.$

We use this and apply Theorem 4.21 with the roles of $k_{1},k_{2}$ played by
$\lfloor\gamma_{i}+d_{i}/2+1\rfloor$, $i=1,2$, to obtain

(7.23)
$\displaystyle|{\mathscr{K}}_{\varphi({\mathbf{t}}\sqrt{L})\Psi(2^{-{\mathbf{j}}-{\mathbf{k}}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})|=|{\mathscr{K}}_{\omega(2^{-{\mathbf{j}}}\sqrt{L})}({\mathbf{x}},{\mathbf{y}})|$
$\displaystyle\leq
c2^{-k(|{\bm{\gamma}}|+\tilde{{\varepsilon}})}{\mathfrak{D}}_{2^{-{\mathbf{j}}},{\bm{\gamma}}+{\mathbf{d}}/2}({\mathbf{x}},{\mathbf{y}})\leq
c2^{-k(|{\bm{\gamma}}|+\tilde{{\varepsilon}})}V({\mathbf{x}},2^{-{\mathbf{j}}})^{-1}{\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}},{\bm{\gamma}}}({\mathbf{x}},{\mathbf{y}}),$

where for the last inequality we used (3.13). In turn this estimate implies

$\displaystyle\big{|}\varphi({\mathbf{t}}\sqrt{L})$
$\displaystyle\Psi(2^{-{\mathbf{j}}-{\mathbf{k}}}\sqrt{L})(\varphi(2^{-{\mathbf{j}}-{\mathbf{k}}}\sqrt{L})-\varphi(2^{-{\mathbf{j}}-{\mathbf{k}}+{\bm{1}}}\sqrt{L}))f({\mathbf{y}})\big{|}{\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}},{\bm{\gamma}}}({\mathbf{x}},{\mathbf{y}})$
(7.24)
$\displaystyle\leq{\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}},{\bm{\gamma}}}({\mathbf{x}},{\mathbf{y}})\int_{X}|{\mathscr{K}}_{\omega(2^{-{\mathbf{j}}}\sqrt{L})}({\mathbf{y}},{\mathbf{z}})|$
$\displaystyle\hskip
72.26999pt\;\;\times(|\varphi(2^{-{\mathbf{j}}-{\mathbf{k}}}\sqrt{L})f({\mathbf{z}})|+|\varphi(2^{-{\mathbf{j}}-{\mathbf{k}}+{\bm{1}}}\sqrt{L})f({\mathbf{z}})|)d\mu({\mathbf{z}})$
$\displaystyle\leq
c2^{-k(|{\bm{\gamma}}|+\tilde{{\varepsilon}})}{\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}},{\bm{\gamma}}}({\mathbf{x}},{\mathbf{y}})\int_{X}\frac{{\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}},{\bm{\gamma}}}({\mathbf{y}},{\mathbf{z}})}{V({\mathbf{z}},2^{-{\mathbf{j}}})}(|F_{{\mathbf{j}},k}({\mathbf{z}})|+|F_{{\mathbf{j}},k-1}({\mathbf{z}})|)d\mu({\mathbf{z}}),$

where we used the short-hand notation
$F_{{\mathbf{j}},k}({\mathbf{z}}):=\varphi(2^{-{\mathbf{j}}-{\mathbf{k}}}\sqrt{L})f({\mathbf{z}})$.

On the other hand,
$1+2^{j_{i}}\rho_{i}(x_{i},z_{i})\leq(1+2^{j_{i}}\rho_{i}(x_{i},y_{i}))(1+2^{j_{i}}\rho_{i}(y_{i},z_{i}))$
and hence

$2^{-k|{\bm{\gamma}}|}{\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}},{\bm{\gamma}}}({\mathbf{x}},{\mathbf{y}}){\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}},{\bm{\gamma}}}({\mathbf{y}},{\mathbf{z}})\leq
2^{-k|{\bm{\gamma}}|}{\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}},{\bm{\gamma}}}({\mathbf{x}},{\mathbf{z}})\leq{\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}-{\mathbf{k}}},{\bm{\gamma}}}({\mathbf{x}},{\mathbf{z}}).$

With $0<\theta\leq 1$ from the hypothesis of the proposition this leads to

$\displaystyle
2^{-k(|{\bm{\gamma}}|+\tilde{{\varepsilon}})}{\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}},{\bm{\gamma}}}({\mathbf{x}},{\mathbf{y}})$
$\displaystyle{\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}},{\bm{\gamma}}}({\mathbf{y}},{\mathbf{z}})|F_{{\mathbf{j}},k}({\mathbf{z}})|$
$\displaystyle\leq
c2^{-k\tilde{{\varepsilon}}}M_{{\bm{\gamma}}}^{**}(f;\varphi)({\mathbf{x}})^{1-\theta}{\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}-{\mathbf{k}}},\theta{\bm{\gamma}}}({\mathbf{x}},{\mathbf{z}})|F_{{\mathbf{j}},k}({\mathbf{z}})|^{\theta}.$

Combining this with (7.1) we get

$\displaystyle|\varphi({\mathbf{t}}\sqrt{L})\Psi(2^{-{\mathbf{j}}-{\mathbf{k}}}\sqrt{L})(\varphi(2^{-{\mathbf{j}}-{\mathbf{k}}}\sqrt{L})-\varphi(2^{-{\mathbf{j}}-{\mathbf{k}}+{\bm{1}}}\sqrt{L}))f({\mathbf{y}})|{\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}},{\bm{\gamma}}}({\mathbf{x}},{\mathbf{y}})$
$\displaystyle\leq
c2^{-k\tilde{{\varepsilon}}}M_{{\bm{\gamma}}}^{**}(f;\varphi)({\mathbf{x}})^{1-\theta}\int_{X}{\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}-k},\theta{\bm{\gamma}}}({\mathbf{x}},{\mathbf{z}})\frac{|F_{{\mathbf{j}},k}({\mathbf{z}})|^{\theta}+|F_{{\mathbf{j}},k-1}({\mathbf{z}})|^{\theta}}{V({\mathbf{z}},2^{-{\mathbf{j}}})}d\mu({\mathbf{z}}).$

In the same way we obtain

$\displaystyle|\varphi({\mathbf{t}}\sqrt{L})\psi(2^{-{\mathbf{j}}}\sqrt{L})$
$\displaystyle\varphi(2^{-{\mathbf{j}}}\sqrt{L})f({\mathbf{y}})|{\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}},{\bm{\gamma}}}({\mathbf{x}},{\mathbf{y}})$
$\displaystyle\leq
cM_{{\bm{\gamma}}}^{**}(f;\varphi)({\mathbf{x}})^{1-\theta}\int_{X}{\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}},\theta{\bm{\gamma}}}({\mathbf{x}},{\mathbf{z}})\frac{|F_{{\mathbf{j}},0}({\mathbf{z}})|^{\theta}}{V({\mathbf{z}},2^{-{\mathbf{j}}})}d\mu({\mathbf{z}}).$

Combining the above estimates with (7.20) we obtain

$\displaystyle|\varphi({\mathbf{t}}\sqrt{L})f({\mathbf{y}})|$
$\displaystyle{\mathfrak{D}}^{*}_{{\mathbf{t}},{\bm{\gamma}}}({\mathbf{x}},{\mathbf{y}})$
(7.25) $\displaystyle\leq
cM_{{\bm{\gamma}}}^{**}(f;\varphi)({\mathbf{x}})^{1-\theta}\sum_{k=0}^{\infty}2^{-k\tilde{{\varepsilon}}}\int_{X}{\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}-k},\theta{\bm{\gamma}}}({\mathbf{x}},{\mathbf{z}})\frac{|F_{{\mathbf{j}},k}({\mathbf{z}})|^{\theta}}{V({\mathbf{z}},2^{-{\mathbf{j}}})}d\mu({\mathbf{z}}).$

By (3.13) it follows that

(7.26) $V({\mathbf{z}},2^{-{\mathbf{j}}})^{-1}\leq
cV({\mathbf{x}},2^{-{\mathbf{j}}})^{-1}{\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}},{\mathbf{d}}}({\mathbf{x}},{\mathbf{z}})^{-1}\leq
cV({\mathbf{x}},2^{-{\mathbf{j}}})^{-1}{\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}-{\mathbf{k}}},{\mathbf{d}}}({\mathbf{x}},{\mathbf{z}})^{-1},$

which applied in (7.1) leads to

$\displaystyle|\varphi({\mathbf{t}}\sqrt{L})f({\mathbf{y}})|{\mathfrak{D}}^{*}_{{\mathbf{t}},{\bm{\gamma}}}({\mathbf{x}},{\mathbf{y}})$
$\displaystyle\leq
c[M_{{\bm{\gamma}}}^{**}(f;\varphi)({\mathbf{x}})]^{1-\theta}\sum_{k=0}^{\infty}2^{-k\tilde{{\varepsilon}}}\int_{X}V({\mathbf{x}},2^{-{\mathbf{j}}}){\mathfrak{D}}^{*}_{2^{-{\mathbf{j}}-{\mathbf{k}}},\theta{\bm{\gamma}}-{\mathbf{d}}}({\mathbf{x}},{\mathbf{z}})|F_{{\mathbf{j}},k}({\mathbf{z}})|^{\theta}d\mu({\mathbf{z}})$
$\displaystyle\leq
c[M_{{\bm{\gamma}}}^{**}(f;\varphi)({\mathbf{x}})]^{1-\theta}\sum_{\nu=0}^{\infty}2^{-\nu\epsilon}\mathcal{M}_{\theta}(F_{{\mathbf{j}},k})({\mathbf{x}})^{\theta}$
$\displaystyle\leq
c[M_{{\bm{\gamma}}}^{**}(f;\varphi)({\mathbf{x}})]^{1-\theta}[\mathcal{M}_{\theta}(M(f;\varphi))({\mathbf{x}})]^{\theta},$

where for the former inequality we used Lemma 4.23 and the fact that
$\theta{\bm{\gamma}}-{\mathbf{d}}>{\mathbf{d}}$ due to
${\bm{\gamma}}>(2/\theta){\mathbf{d}}$. Therefore, using the definition of
$M_{{\bm{\gamma}}}^{**}(f;\varphi)$ (see (7.7)) we get

$M_{{\bm{\gamma}}}^{**}(f;\varphi)(x)\leq
c[M_{{\bm{\gamma}}}^{**}(f;\varphi)({\mathbf{x}})]^{1-\theta}[\mathcal{M}_{\theta}(M(f;\varphi))({\mathbf{x}})]^{\theta}.$

At this point we use our assumption that
$M_{{\bm{\gamma}}}^{**}(f;\varphi)({\mathbf{x}})<\infty$ from (7.19) to obtain
the desired estimate (7.18).

Part 2. Our next step is to show that under the hypothesis of the proposition
if $f\in\mathcal{S}^{\prime}$ and
$M_{\theta}(M(f;\varphi))({\mathbf{x}})<\infty$, then
$M_{\bm{\gamma}}^{**}(f;\varphi)({\mathbf{x}})<\infty$, i.e. (7.19) is valid.
As we know from Theorem 5.7 since $f\in\mathcal{S}^{\prime}$, then
$\varphi({\mathbf{t}}\sqrt{L})f$ is slowly growing (and continuous). More
explicitly, there exist constants $K,c>0$ depending on $f$ such that

(7.27) $|\varphi({\mathbf{t}}\sqrt{L})f({\mathbf{y}})|\leq
c\prod_{i=1,2}(t_{i}^{-K}+t_{i}^{K})(1+\rho(y_{i},x_{0i}))^{K},\quad\forall{\mathbf{y}}\in
X,{\mathbf{t}}>{\bm{0}}.$

With $K$ from above and $0<{\varepsilon}<1$,
${\bm{\varepsilon}}:=({\varepsilon},{\varepsilon})$, we introduce the
following auxiliary maximal operator

$\displaystyle M_{\bm{\gamma}}^{**}$
$\displaystyle(f;\varphi)^{{\varepsilon},K}({\mathbf{x}})$
$\displaystyle:=\sup_{{\bm{0}}<{\mathbf{t}}\leq\frac{{\bm{1}}}{{\bm{\varepsilon}}}}\sup_{{\mathbf{y}}\in
X}\frac{|\varphi({\mathbf{t}}\sqrt{L})f({\mathbf{y}})|}{\prod_{i=1,2}(1+t_{i}^{-1}\rho(x_{i},y_{i}))^{\gamma_{i}}(1+{\varepsilon}\rho(x_{i},y_{i}))^{K}(1+{\varepsilon}t_{i}^{-K})}.$

It is easy to see that

(7.28) $M_{\bm{\gamma}}^{**}(f;\varphi)^{{\varepsilon},K}({\mathbf{x}})\leq
c({\mathbf{x}},{\varepsilon},K)<\infty.$

Indeed, from (7.27) it follows that for
${\bm{0}}<{\mathbf{t}}\leq\frac{{\bm{1}}}{{\bm{\varepsilon}}}$

$\displaystyle\frac{|\varphi(t\sqrt{L})f({\mathbf{y}})|}{\prod_{i=1,2}(1+t_{i}^{-1}\rho(x_{i},y_{i}))^{\gamma_{i}}(1+{\varepsilon}\rho(x_{i},y_{i}))^{K}(1+{\varepsilon}t_{i}^{-K})}$
$\displaystyle\qquad\leq\frac{c\prod_{i=1,2}(t_{i}^{-K}+t_{i}^{K})(1+\rho(y_{i},x_{0i}))^{K}}{\prod_{i=1,2}(1+t_{i}^{-1}\rho(x_{i},y_{i}))^{\gamma_{i}}(1+{\varepsilon}\rho(x_{i},y_{i}))^{K}(1+{\varepsilon}t_{i}^{-K})}$
$\displaystyle\qquad\leq\frac{c\prod_{i=1,2}(1+t_{i}^{2K})(1+\rho(y_{i},x_{i}))^{K}(1+\rho(x_{i},x_{0i}))^{K}}{{\varepsilon}^{2K+2}\prod_{i=1,2}(1+\rho(x_{i},y_{i}))^{K}}$
$\displaystyle\qquad\leq
c{\varepsilon}^{-2K-2}(1+{\varepsilon}^{-2K})^{2}\prod_{i=1,2}(1+\rho(x_{i},x_{0i}))^{K},$

which confirms (7.28).

Our next goal is to establish the inequality

(7.29)
$M_{\bm{\gamma}}^{**}(f;\varphi)^{{\varepsilon},K}({\mathbf{x}})\leq\tilde{c}\mathcal{M}_{\theta}(M(f;\varphi))({\mathbf{x}}),$

where the constant $\tilde{c}=\tilde{c}(f)$ is allowed to depend on $f$, but
is independent of ${\varepsilon}$. Then passing to the limit as
${\varepsilon}\to 0$ it will follow that
$M_{\bm{\gamma}}^{**}(f;\varphi)({\mathbf{x}})\leq\tilde{c}\mathcal{M}_{\theta}(M(f;\varphi))({\mathbf{x}})$,
which in turn shows that (7.19) holds and the proof will be complete.

The proof of inequality (7.29) will follow very closely in the footsteps of
the proof from Part 1. For this reason we will only indicated the main
differences and omit most of the details.

Just as in Part 1 we apply Lemma 7.6 but this time with

(7.30) $N:=\lfloor 3|{\bm{\gamma}}|+6K+3|{\mathbf{d}}|/2+4\rfloor,$

where $K$ is from (7.27). Fix $0<{\varepsilon}\leq 1$. Let
${\bm{0}}<{\mathbf{t}}\leq{\bm{1}}/{\bm{\varepsilon}}$ and choose
${\mathbf{j}}\in{\mathbb{Z}}^{2}$ so that
$2^{-{\mathbf{j}}}\leq{\mathbf{t}}<2^{-{\mathbf{j}}+{\bm{1}}}$. Hence,
$2^{\mathbf{j}}\geq{\bm{1}}/{\mathbf{t}}\geq{\bm{\varepsilon}}$. By Lemma 7.6
there exist an admissible function $\psi\in\mathcal{S}({\mathbb{R}}^{2})$ with
$\psi(0,0)=1$ and a function $\Psi\in\mathcal{S}({\mathbb{R}}^{2})$ with the
properties $\Psi(\pm\lambda_{1},\pm\lambda_{2})=\Psi(\lambda_{1},\lambda_{2})$
for ${\bm{\lambda}}\in\mathbb{R}^{2}$ and $\partial^{{\bm{\beta}}}\Psi(0,0)=0$
for $|{\bm{\beta}}|\leq N$ such that for any $f\in\mathcal{S}^{\prime}$ and
${\mathbf{j}}\in{\mathbb{Z}}^{2}$ identity (7.14) is valid. Therefore,

$\displaystyle\frac{|\varphi({\mathbf{t}}\sqrt{L})f({\mathbf{y}})|}{\prod_{i=1,2}(1+t_{i}\rho(x_{i},y_{i}))^{\gamma_{i}}(1+{\varepsilon}\rho(x_{i},y_{i}))^{K}(1+{\varepsilon}t_{i}^{-K})}$
$\displaystyle\qquad\leq
c\frac{|\varphi({\mathbf{t}}\sqrt{L})\psi(2^{-{\mathbf{j}}}\sqrt{L})\varphi(2^{-{\mathbf{j}}}\sqrt{L})f({\mathbf{y}})|}{\prod_{i=1,2}(1+2^{j_{i}}\rho(x_{i},y_{i}))^{\gamma_{i}}(1+{\varepsilon}\rho(x_{i},y_{i}))^{K}(1+{\varepsilon}t_{i}^{-K})}$
$\displaystyle\qquad+c\sum_{k=1}^{\infty}\frac{|\varphi({\mathbf{t}}\sqrt{L})\Psi(2^{-{\mathbf{j}}-{\mathbf{k}}}\sqrt{L})[\varphi(2^{-{\mathbf{j}}-{\mathbf{k}}}\sqrt{L})-\varphi(2^{-{\mathbf{j}}-{\mathbf{k}}+{\bm{1}}}\sqrt{L})]f(y)|}{\prod_{i=1,2}(1+2^{j_{i}}\rho(x_{i},y_{i}))^{\gamma_{i}}(1+{\varepsilon}\rho(x_{i},y_{i}))^{K}(1+{\varepsilon}t_{i}^{-K})}.$

Let
$\omega({\bm{\lambda}}):=\varphi({\mathbf{t}}2^{\mathbf{j}}{\bm{\lambda}})\Psi(2^{-{\mathbf{k}}}{\bm{\lambda}})$.
Then
$\varphi({\mathbf{t}}\sqrt{L})\Psi(2^{-{\mathbf{j}}-{\mathbf{k}}}\sqrt{L})=\omega(2^{-{\mathbf{j}}}\sqrt{L})$.
Choose
$m:=\lfloor\gamma_{1}+K+d_{1}/2+1\rfloor+\lfloor\gamma_{2}+K+d_{2}/2+1\rfloor$
and $r:=m+|{\mathbf{d}}|+1$. Now just as in the derivation of (7.22) for some
constant $\tilde{{\varepsilon}}>0$

(7.31) $|\partial^{\bm{\beta}}\omega({\bm{\lambda}})|\leq
c2^{-k(|{\bm{\gamma}}|+2K+\tilde{{\varepsilon}})}(1+|{\bm{\lambda}}|)^{-r},\;\;{\bm{\lambda}}\in\mathbb{R}^{2},\;\;|{\bm{\beta}}|\leq
m.$

Applying Theorem 4.21 to $\omega(2^{-{\mathbf{j}}}\sqrt{L})$ with
$\lfloor\gamma_{i}+K+d_{i}/2+1\rfloor$, $i=1,2$, playing the role of
$k_{1},k_{2}$ and using that $2^{j_{i}}\geq{\varepsilon}$, $i=1,2$, we get

$\displaystyle|{\mathscr{K}}_{\varphi({\mathbf{t}}\sqrt{L})\Psi(2^{-{\mathbf{j}}-{\mathbf{k}}}\sqrt{L})}({\mathbf{y}},{\mathbf{z}})|$
$\displaystyle\leq\frac{c2^{-k(|{\bm{\gamma}}|+2K+\tilde{{\varepsilon}})}}{V({\mathbf{z}},2^{-{\mathbf{j}}})\prod_{i=1,2}\big{(}1+2^{j_{i}}\rho(y_{i},z_{i})\big{)}^{\gamma_{i}+K}}$
(7.32)
$\displaystyle\leq\frac{c2^{-k(|{\bm{\gamma}}|+2K+\tilde{{\varepsilon}})}}{V({\mathbf{z}},2^{-{\mathbf{j}}})\prod_{i=1,2}\big{(}1+2^{j_{i}}\rho(y_{i},z_{i})\big{)}^{\gamma_{i}}\big{(}1+{\varepsilon}\rho(y_{i},z_{i})\big{)}^{K}},$

This and the inequalities
$1+2^{j_{i}}\rho(x_{i},z_{i})\leq(1+2^{j_{i}}\rho(x_{i},y_{i}))(1+2^{j_{i}}\rho(y_{i},z_{i}))$
and
$1+{\varepsilon}\rho(x_{i},z_{i})\leq(1+{\varepsilon}\rho(x_{i},y_{i}))(1+{\varepsilon}\rho(y_{i},z_{i}))$
lead to

$\displaystyle\frac{|\varphi({\mathbf{t}}\sqrt{L})\Psi(2^{-{\mathbf{j}}-{\mathbf{k}}}\sqrt{L})\varphi(2^{-{\mathbf{j}}-{\mathbf{k}}}\sqrt{L})f({\mathbf{y}})|}{\prod_{i=1,2}(1+2^{j_{i}}\rho(x_{i},y_{i}))^{\gamma_{i}}(1+{\varepsilon}\rho(x_{i},y_{i}))^{K}(1+{\varepsilon}t_{i}^{-K})}$
$\displaystyle\leq\frac{c2^{-k(|{\bm{\gamma}}|+2K+\tilde{{\varepsilon}})}}{\prod_{i=1,2}(1+2^{j_{i}}\rho(x_{i},y_{i}))^{\gamma_{i}}(1+{\varepsilon}\rho(x_{i},y_{i}))^{K}(1+{\varepsilon}t_{i}^{-K})}$
$\displaystyle\hskip