$r\leq|x-z|\leq\frac{\sqrt{5}}{2}r\leq\frac{\sqrt{5}}{4}|x-y|,\qquad(1-\frac{\sqrt{5}}{4})|x-y|\leq|y-z|\leq(1+\frac{\sqrt{5}}{4})|x-y|,$
(6.9)

we see that

$\displaystyle A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,z)$
$\displaystyle\asymp\bigg{(}1\wedge\frac{x_{d}\vee
t^{1/\alpha}}{r}\bigg{)}^{\beta_{1}}\bigg{(}1\wedge\frac{(x_{d}+r)\vee
t^{1/\alpha}}{r}\bigg{)}^{\beta_{2}}$
$\displaystyle\quad\times\log^{\beta_{3}}\bigg{(}e+\frac{((x_{d}+r)\vee
t^{1/\alpha})\wedge r}{(x_{d}\vee t^{1/\alpha})\wedge
r}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{r}{((x_{d}+r)\vee
t^{1/\alpha})\wedge r}\bigg{)}$
$\displaystyle\asymp\bigg{(}1\wedge\frac{x_{d}\vee
t^{1/\alpha}}{r}\bigg{)}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{r}{(x_{d}\vee
t^{1/\alpha})\wedge r}\bigg{)}$ (6.10) $\displaystyle\asymp
A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x(r))$

and

$\displaystyle A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,z,y)$
$\displaystyle\asymp\bigg{(}1\wedge\frac{y_{d}\vee
t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}}\bigg{(}1\wedge\frac{x_{d}+r}{|x-y|}\bigg{)}^{\beta_{2}}\log^{\beta_{3}}\bigg{(}e+\frac{x_{d}+r}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{|x-y|}{x_{d}+r}\bigg{)}$
(6.11) $\displaystyle\asymp
A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x(r),y).$

Thus, using the semigroup property, Proposition 6.2 and (6.9), we get

$\displaystyle p(t,x,y)\geq\int_{K}p(t/2,x,z)p(t/2,z,y)dz$ $\displaystyle\geq
c_{2}^{2}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\int_{K}\frac{tA_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,z)}{|x-z|^{d+\alpha}}\,\frac{tA_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,z,y)}{|y-z|^{d+\alpha}}dz$
$\displaystyle\geq
c_{3}t^{2}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}$
$\displaystyle\quad\times\int_{(y_{d}\vee
t^{1/\alpha})\wedge(|x-y|/4)}^{|x-y|/2}\frac{A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x(r))}{r^{d+\alpha}}\frac{A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x(r),y)}{|x-y|^{d+\alpha}}\int_{K(r)}d\widetilde{z}dr$
$\displaystyle\geq
c_{4}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\frac{t}{|x-y|^{d+\alpha}}$
$\displaystyle\quad\times t\int_{(y_{d}\vee
t^{1/\alpha})\wedge(|x-y|/4)}^{|x-y|/2}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x(r))\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x(r),y)\frac{dr}{r^{\alpha+1}}.$

$\Box$

Note that, for any $x,y\in{\mathbb{R}}^{d}_{+}$ and $|x-y|/4\leq
r\leq|x-y|/2$,

$\displaystyle
A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+r{\mathbf{e}}_{d})$
$\displaystyle\asymp\bigg{(}1\wedge\frac{x_{d}\vee
t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{(x_{d}\vee
t^{1/\alpha})\wedge|x-y|}\bigg{)}$ $\displaystyle\asymp
A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+\frac{|x-y|}{2}{\mathbf{e}}_{d})$
(6.12)

and, since $y_{d}\leq x_{d}+4r$,

$\displaystyle
A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+r{\mathbf{e}}_{d},y)$
$\displaystyle\asymp\bigg{(}1\wedge\frac{y_{d}\vee
t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{(y_{d}\vee
t^{1/\alpha})\wedge|x-y|}\bigg{)}$ $\displaystyle\asymp
A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+\frac{|x-y|}{2}{\mathbf{e}}_{d},y)$
$\displaystyle\asymp
A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,y,y+\frac{|x-y|}{2}{\mathbf{e}}_{d}).$
(6.13)

In particular, we have

$\displaystyle
A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+\frac{|x-y|}{2}{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+\frac{|x-y|}{2}{\mathbf{e}}_{d},y)$
$\displaystyle\asymp
A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+\frac{|x-y|}{2}{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,y,y+\frac{|x-y|}{2}{\mathbf{e}}_{d})$
$\displaystyle\asymp
A_{\beta_{1},\beta_{1},0,\beta_{3}}(t,x,y)\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{((x_{d}\wedge
y_{d})\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}.$ (6.14)

By (6)–(6), we have that for all $t>0$ and $x,y\in{\mathbb{R}}^{d}_{+}$,

$\displaystyle\int_{|x-y|/4}^{|x-y|/2}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+r{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+r{\mathbf{e}}_{d},y)\frac{dr}{r^{\alpha+1}}$
$\displaystyle\asymp
A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+\frac{|x-y|}{2}{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+\frac{|x-y|}{2}{\mathbf{e}}_{d},y)\int_{|x-y|/4}^{|x-y|/2}\frac{dr}{r^{\alpha+1}}$
$\displaystyle\asymp|x-y|^{-\alpha}A_{\beta_{1},\beta_{1},0,\beta_{3}}(t,x,y)\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{((x_{d}\wedge
y_{d})\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}.$ (6.15)

######  Remark 6.5.

Here we give a proof of the comparability of (1.6) and (1.7). Let $t>0$ and
$x,y\in{\overline{\mathbb{R}}}^{d}_{+}$ be such that $|x-y|>6t^{1/\alpha}$.
For any $z\in B(x+2^{-1}|x-y|{\mathbf{e}}_{d},4^{-1}|x-y|)$, by using the
triangle inequality several times, we have

$\displaystyle z_{d}-t^{1/\alpha}\asymp z_{d}\asymp x_{d}\wedge
y_{d}+|x-y|\asymp x_{d}\vee y_{d}+|x-y|$ (6.16)

and

$\displaystyle|x+t^{1/\alpha}{\mathbf{e}}_{d}-z|\asymp|y+t^{1/\alpha}{\mathbf{e}}_{d}-z|\asymp|x-z|\asymp|y-z|\asymp|x-y|.$
(6.17)

For any $t>0$ and $x,y\in{\overline{\mathbb{R}}}^{d}_{+}$ with
$|x-y|>6t^{1/\alpha}$, if $x_{d}\wedge y_{d}\geq|x-y|/4$, then by (A3), (6.3),
(6.4), (6.16) and (6.17),

$\displaystyle
t|x-y|^{d+\alpha}\int_{B(x+2^{-1}|x-y|{\mathbf{e}}_{d},\,4^{-1}|x-y|)}J(x+t^{1/\alpha}{\mathbf{e}}_{d},z)\,J(z,y+t^{1/\alpha}{\mathbf{e}}_{d})dz$
$\displaystyle\asymp\frac{t}{|x-y|^{d+\alpha}}\int_{B(x+2^{-1}|x-y|{\mathbf{e}}_{d},\,4^{-1}|x-y|)}dz\asymp\frac{t}{|x-y|^{\alpha}}$
$\displaystyle\asymp\bigg{(}\\!1\wedge\frac{t}{|x-y|^{\alpha}}\\!\bigg{)}B_{\beta_{1},\beta_{1},0,\beta_{3}}(x+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})\log^{\beta_{3}}\\!\bigg{(}\\!e\\!+\\!\frac{|x-y|}{((x_{d}\wedge
y_{d})+t^{1/\alpha})\wedge|x-y|}\bigg{)},$

Otherwise, if $x_{d}\wedge y_{d}<|x-y|/4$, then $x_{d}\vee y_{d}\leq
x_{d}\wedge y_{d}+|x-y|<5|x-y|/4$ so that

$z_{d}-t^{1/\alpha}\asymp z_{d}\asymp|x-y|\quad\text{for $z\in
B(x+2^{-1}|x-y|{\mathbf{e}}_{d},\,4^{-1}|x-y|)$}$

by (6.16). Using this, (6.17) and (6.3), we get that for $z\in
B(x+2^{-1}|x-y|{\mathbf{e}}_{d},\,4^{-1}|x-y|)$,

$\displaystyle
B_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(x+t^{1/\alpha}{\mathbf{e}}_{d},z)\asymp
B_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(x+t^{1/\alpha}{\mathbf{e}}_{d},z+t^{1/\alpha}{\mathbf{e}}_{d})$
$\displaystyle\asymp A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,z)\asymp
A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+2^{-1}|x-y|{\mathbf{e}}_{d})$
(6.18)

and

$\displaystyle
B_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(z,y+t^{1/\alpha}{\mathbf{e}}_{d})\asymp
B_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(z+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})$
$\displaystyle\asymp A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,z,z)\asymp
A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+2^{-1}|x-y|{\mathbf{e}}_{d},y).$
(6.19)

By (A3), (6.17), (6.5), (6.5) and (6), we arrive at

$\displaystyle
t|x-y|^{d+\alpha}\int_{B(x+2^{-1}|x-y|{\mathbf{e}}_{d},\,4^{-1}|x-y|)}J(x+t^{1/\alpha}{\mathbf{e}}_{d},z)\,J(z,y+t^{1/\alpha}{\mathbf{e}}_{d})dz$
$\displaystyle\asymp\frac{tA_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+2^{-1}|x-y|{\mathbf{e}}_{d})A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+2^{-1}|x-y|{\mathbf{e}}_{d},y)}{|x-y|^{d+\alpha}}\int_{B(x+2^{-1}|x-y|{\mathbf{e}}_{d},\,4^{-1}|x-y|)}\\!\\!\\!\\!\\!\\!\\!\\!dz$
$\displaystyle\asymp\frac{t}{|x-y|^{\alpha}}A_{\beta_{1},\beta_{1},0,\beta_{3}}(t,x,y)\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{((x_{d}\wedge
y_{d})\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}$
$\displaystyle\asymp\bigg{(}\\!1\wedge\frac{t}{|x-y|^{\alpha}}\\!\bigg{)}B_{\beta_{1},\beta_{1},0,\beta_{3}}(x+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})\log^{\beta_{3}}\\!\bigg{(}\\!e\\!+\\!\frac{|x-y|}{((x_{d}\wedge
y_{d})+t^{1/\alpha})\wedge|x-y|}\bigg{)}.$

Hence, (1.6) and (1.7) are comparable.

Combining (6), Proposition 6.2 and Lemma 6.4, we get the following result.

######  Proposition 6.6.

There exists a constant $C>0$ such that for all $t>0$ and
$x,y\in{\mathbb{R}}^{d}_{+}$,

$\displaystyle p(t,x,y)$ $\displaystyle\geq
C\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right)\bigg{[}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y)$
$\displaystyle\quad+\bigg{(}1\wedge\frac{t}{|x-y|^{\alpha}}\bigg{)}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+\frac{|x-y|}{2}{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+\frac{|x-y|}{2}{\mathbf{e}}_{d},y)\bigg{]}$
$\displaystyle\asymp\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right)\bigg{[}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y)$
$\displaystyle\quad+\bigg{(}1\wedge\frac{t}{|x-y|^{\alpha}}\bigg{)}A_{\beta_{1},\beta_{1},0,\beta_{3}}(t,x,y)\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{((x_{d}\wedge
y_{d})\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}\bigg{]}.$

######  Remark 6.7.

If $\beta_{2}<\alpha+\beta_{1}$, then there is a constant $C>0$ such that for
all $t>0$ and $x,y\in{\mathbb{R}}^{d}$,

$\displaystyle\bigg{(}1\wedge\frac{t}{|x-y|^{\alpha}}\bigg{)}A_{\beta_{1},\beta_{1},0,\beta_{3}}(t,x,y)\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{((x_{d}\wedge
y_{d})\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}\leq
CA_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y).$

Indeed, for ${\varepsilon}:=(\alpha+\beta_{1}-\beta_{2})/2>0$, using (10.1),
we get that for all $t>0$ and $x,y\in{\mathbb{R}}^{d}$,

$\displaystyle\bigg{(}1\wedge\frac{t}{|x-y|^{\alpha}}\bigg{)}A_{\beta_{1},\beta_{1},0,\beta_{3}}(t,x,y)\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{((x_{d}\wedge
y_{d})\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}$ $\displaystyle\leq
c_{1}\bigg{(}1\wedge\frac{t}{|x-y|^{\alpha}}\bigg{)}\bigg{(}1\wedge\frac{x_{d}\vee
t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}-{\varepsilon}}\bigg{(}1\wedge\frac{y_{d}\vee
t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}-{\varepsilon}}$
$\displaystyle=c_{1}\bigg{(}1\wedge\frac{t^{1/\alpha}}{|x-y|}\bigg{)}^{{\varepsilon}}\bigg{(}1\wedge\frac{x_{d}\vee
t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}-{\varepsilon}}\bigg{(}1\wedge\frac{t^{1/\alpha}}{|x-y|}\bigg{)}^{-\beta_{1}+\beta_{2}+{\varepsilon}}\bigg{(}1\wedge\frac{y_{d}\vee
t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}-{\varepsilon}}$ $\displaystyle\leq
c_{1}\bigg{(}1\wedge\frac{x_{d}\vee
t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}}\bigg{(}1\wedge\frac{y_{d}\vee
t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{2}}\leq
c_{1}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y).$

Therefore, in view of Proposition 6.2, Proposition 6.6 is relevant only if
$\beta_{2}\geq\alpha+\beta_{1}$.

######  Lemma 6.8.

Suppose that $\beta_{2}=\alpha+\beta_{1}$. Then for all $t>0$ and
$x,y\in{\mathbb{R}}^{d}_{+}$,

$\displaystyle\big{(}|x-y|^{\alpha}\wedge t\big{)}\int_{(x_{d}\vee y_{d}\vee
t^{1/\alpha})\wedge(|x-y|/4)}^{|x-y|/2}\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+r{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+r{\mathbf{e}}_{d},y)\frac{dr}{r^{\alpha+1}}$
$\displaystyle\asymp\bigg{(}1\wedge\frac{t}{|x-y|^{\alpha}}\bigg{)}A_{\beta_{1},\beta_{1},0,\beta_{3}+\beta_{4}+1}(t,x,y)\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{((x_{d}\wedge
y_{d})\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}.$

Proof. Without loss of generality, we assume $x_{d}\leq y_{d}$.

Assume first that $y_{d}\vee t^{1/\alpha}<|x-y|/4$. Using
$\beta_{2}=\alpha+\beta_{1}$ and (6)–(6), since $x_{d}\leq y_{d}\vee
t^{1/\alpha}$, we get

$\displaystyle\int_{y_{d}\vee
t^{1/\alpha}}^{|x-y|/2}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+r{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+r{\mathbf{e}}_{d},y)\frac{dr}{r^{\alpha+1}}$
$\displaystyle\asymp\int_{y_{d}\vee
t^{1/\alpha}}^{|x-y|/2}\bigg{(}\frac{x_{d}\vee
t^{1/\alpha}}{r}\bigg{)}^{\beta_{1}}\bigg{(}\frac{y_{d}\vee
t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}}\bigg{(}\frac{r}{|x-y|}\bigg{)}^{\alpha+\beta_{1}}$
$\displaystyle\qquad\times\log^{\beta_{3}}\bigg{(}e+\frac{r}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{r}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{|x-y|}{r}\bigg{)}\frac{dr}{r^{\alpha+1}}$
$\displaystyle\asymp\bigg{(}\frac{x_{d}\vee
t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}}\bigg{(}\frac{y_{d}\vee
t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}}|x-y|^{-\alpha}$
$\displaystyle\qquad\times\int_{y_{d}\vee
t^{1/\alpha}}^{|x-y|/2}\log^{\beta_{3}}\bigg{(}e+\frac{r}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{r}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{|x-y|}{r}\bigg{)}\frac{dr}{r}.$
(6.20)

By a change of the variables and Lemma 10.11, we see that

$\displaystyle\int_{y_{d}\vee
t^{1/\alpha}}^{|x-y|/2}\log^{\beta_{3}}\bigg{(}e+\frac{r}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{r}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{|x-y|}{r}\bigg{)}\frac{dr}{r}$
$\displaystyle=\int_{4(y_{d}\vee
t^{1/\alpha})/|x-y|}^{2}\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|s}{4(x_{d}\vee
t^{1/\alpha})}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|s}{4(y_{d}\vee
t^{1/\alpha})}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{4}{s}\bigg{)}\frac{ds}{s}$
$\displaystyle\asymp\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{4(x_{d}\vee
t^{1/\alpha})}\bigg{)}\log^{\beta_{3}+\beta_{4}+1}\bigg{(}e+\frac{|x-y|}{4(y_{d}\vee
t^{1/\alpha})}\bigg{)}.$ (6.21)

By (6)–(6), since $y_{d}\vee t^{1/\alpha}<|x-y|/4$, we obtain

$\displaystyle\big{(}|x-y|^{\alpha}\wedge t\big{)}\int_{y_{d}\vee
t^{1/\alpha}}^{|x-y|/2}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+r{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+r{\mathbf{e}}_{d},y)\frac{dr}{r^{\alpha+1}}$
$\displaystyle\asymp\bigg{(}1\wedge\frac{t}{|x-y|^{\alpha}}\bigg{)}\bigg{(}\frac{x_{d}\vee
t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}}\bigg{(}\frac{y_{d}\vee
t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}}$
$\displaystyle\quad\times\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{4(x_{d}\vee
t^{1/\alpha})}\bigg{)}\log^{\beta_{3}+\beta_{4}+1}\bigg{(}e+\frac{|x-y|}{4(y_{d}\vee
t^{1/\alpha})}\bigg{)}.$ (6.22)

If $y_{d}\vee t^{1/\alpha}\geq|x-y|/4$, then
$\log\big{(}e+\frac{|x-y|}{(y_{d}\vee t^{1/\alpha})\wedge|x-y|}\big{)}\asymp
1$. Thus, by (6),

$\displaystyle\big{(}|x-y|^{\alpha}\wedge
t\big{)}\int_{|x-y|/4}^{|x-y|/2}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+r{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+r{\mathbf{e}}_{d},y)\frac{dr}{r^{\alpha+1}}$
$\displaystyle\asymp\bigg{(}1\wedge\frac{t}{|x-y|^{\alpha}}\bigg{)}A_{\beta_{1},\beta_{1},0,\beta_{3}+\beta_{4}+1}(t,x,y)\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{(x_{d}\vee
t^{1/\alpha})\wedge|x-y|}\bigg{)}.$ (6.23)

The proof is now complete. $\Box$

Combining Proposition 6.2, Lemma 6.4 and Lemma 6.8. we get the following

######  Proposition 6.9.

Suppose that $\beta_{2}=\alpha+\beta_{1}$. There exists a constant $C>0$ such
that for all $t>0$ and $x,y\in{\mathbb{R}}^{d}_{+}$,

$\displaystyle p(t,x,y)$ $\displaystyle\geq
C\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right)\bigg{[}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y)$
$\displaystyle\quad+\bigg{(}1\wedge\frac{t}{|x-y|^{\alpha}}\bigg{)}A_{\beta_{1},\beta_{1},0,\beta_{3}+\beta_{4}+1}(t,x,y)\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{((x_{d}\wedge
y_{d})\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}\bigg{]}.$

######  Remark 6.10.

In this section, for $\overline{Y}$, (A3)(II) is only used to get (IUBS).
Therefore, the results of Propositions 6.6 and 6.9 hold under weaker
assumptions (A1), (A3)(I), (A4) and (IUBS).

##  7\. Sharp upper bounds of heat kernels

In this section we prove the sharp upper bounds of heat kernels. The key
results are Theorem 7.5 and its Corollary 7.9 which deals with the case
$\beta_{2}<\alpha+\beta_{1}$, and Theorem 7.10 which deals with the case
$\beta_{2}\geq\alpha+\beta_{1}$.

Recall that
$U(r)=\\{x=(\widetilde{x},x_{d})\in{\mathbb{R}}^{d}:\,|\widetilde{x}|<r/2,\,0\leq
x_{d}<r/2\\}$ for $r>0$, $d\geq 2$ and $U(r)=[0,r/2)$ for $d=1$, and that the
function $A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y)$ is defined by
(6). We also recall that we use $p(t,x,y)$ for both $\overline{p}(t,x,y)$ and
$p^{\kappa}(t,x,y)$. We remind the readers that, for an open set
$D\subset{\overline{\mathbb{R}}}^{d}_{+}$ relative to the topology on
${\overline{\mathbb{R}}}^{d}_{+}$,
$\tau_{D}=\overline{\tau}_{D}=\inf\\{t>0:\overline{Y}_{t}\notin D\\}$ when we
consider $\overline{Y}$, and
$\tau_{D}=\tau^{\kappa}_{D}=\inf\\{t>0:Y^{\kappa}_{t}\notin
D\cap{\mathbb{R}}^{d}_{+}\\}$ when we consider $Y^{\kappa}$.

######  Lemma 7.1.

Let $b_{1},b_{3}\geq 0$ be constants with $b_{1}>0$ if $b_{3}>0$. Suppose that
there exists a constant $C_{0}>0$ such that for all $t>0$ and
$z,y\in{\mathbb{R}}^{d}_{+}$,

$p(t,z,y)\leq
C_{0}\left(1\wedge\frac{z_{d}}{t^{1/\alpha}}\right)^{q}A_{b_{1},0,b_{3},0}(t,z,y)\left(t^{-d/\alpha}\wedge\frac{t}{|z-y|^{d+\alpha}}\right).$
(7.1)

Then there exists a constant $C=C(C_{0})>0$ such that for all $t>0$ and
$x=(\widetilde{0},x_{d})\in{\mathbb{R}}^{d}_{+}$ with $x_{d}\leq 2^{-5}$,

${\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta)\leq
Ct\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee
t^{1/\alpha})^{b_{1}}\log^{b_{3}}\bigg{(}e+\frac{1}{x_{d}\vee
t^{1/\alpha}}\bigg{)}.$ (7.2)

Proof. We first note that (7.1) implies that for all $t>0$ and
$y\in{\mathbb{R}}^{d}_{+}$,

$\displaystyle{\mathbb{P}}_{y}(|Y_{t}-y|>2^{-3},\,t<\zeta)=\int_{z\in{\mathbb{R}}^{d}_{+},\,|z-y|>2^{-3}}p(t,y,z)dz$
$\displaystyle\leq
c_{1}t\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\int_{z\in{\mathbb{R}}^{d}_{+},\,|z-y|>2^{-3}}\bigg{(}\frac{y_{d}\vee
t^{1/\alpha}}{|z-y|}\wedge
1\bigg{)}^{b_{1}}\log^{b_{3}}\bigg{(}e+\frac{|z-y|}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\frac{dz}{|z-y|^{d+\alpha}}$ $\displaystyle\leq
c_{2}t\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(y_{d}\vee
t^{1/\alpha})^{b_{1}}\log^{b_{3}}\bigg{(}e+\frac{2^{-3}}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\int_{z\in{\mathbb{R}}^{d}_{+},\,|z-y|>2^{-3}}\frac{dz}{|z-y|^{d+\alpha/2+b_{1}}}$
$\displaystyle\leq
c_{3}t\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(y_{d}\vee
t^{1/\alpha})^{b_{1}}\log^{b_{3}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)},$ (7.3)

where in the first inequality above we used (7.1) and Lemma 10.2, and in the
second we used (10.2).

By Proposition 2.7 (see also Remark 3.12), we have

$\sup_{s\leq t,\,y\in{\mathbb{R}}^{d}_{+}}{\mathbb{P}}_{y}\big{(}|Y_{s}-y|\geq
2^{-2},\,s<\zeta\big{)}\leq c_{1}\sup_{s\leq
t,\,y\in{\mathbb{R}}^{d}_{+}}\int_{z\in{\mathbb{R}}^{d}_{+},\,|z-y|\geq
2^{-2}}\frac{s}{|z-y|^{d+\alpha}}dz\leq c_{2}t.$ (7.4)

If $t\geq 1/(2c_{2})$, then (7.2) follows from Corollary 5.2.

Let $t<1/(2c_{2})$. By the strong Markov property and (7.4), we have

$\displaystyle{\mathbb{P}}_{x}\big{(}\tau_{U(1)}<t<\zeta,\,|Y_{t}-Y_{\tau_{U(1)}}|\geq
2^{-2}\big{)}={\mathbb{E}}_{x}\left[{\mathbb{P}}_{Y_{\tau_{U(1)}}}\left(|Y_{t-\tau_{U(1)}}-Y_{0}|\geq
2^{-2}\right):\tau_{U(1)}<t<\zeta\right]$
$\displaystyle\leq{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta)\sup_{s\leq
t,\,y\in{\mathbb{R}}^{d}_{+}}{\mathbb{P}}_{y}(|Y_{s}-y|\geq
2^{-2},\,s<\zeta)\leq 2^{-1}{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta).$

Thus,

$\displaystyle{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta)$
$\displaystyle=2\big{(}{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta)-2^{-1}{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta)\big{)}$
$\displaystyle\leq
2\big{(}{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta)-{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta,\,|Y_{t}-Y_{\tau_{U(1)}}|\geq
2^{-2})\big{)}$
$\displaystyle=2{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta,\,|Y_{t}-Y_{\tau_{U(1)}}|<2^{-2}).$
(7.5)

Note that by the triangle inequality, for any
$y\in{\mathbb{R}}^{d}_{+}\setminus U(1)$ and $z\in B(y,2^{-2})$, we have
$|z-x|\geq|y-x|-|y-z|>15/32-1/4>2^{-3}$. Therefore using (7) and (7), we have

$\displaystyle{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta)\leq
2{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta,\,|Y_{t}-Y_{\tau_{U(1)}}|<2^{-2})$
$\displaystyle\leq 2{\mathbb{P}}_{x}(|Y_{t}-x|>2^{-3},\,t<\zeta)\leq
c_{4}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee
t^{1/\alpha})^{b_{1}}\log^{b_{3}}\bigg{(}e+\frac{1}{x_{d}\vee
t^{1/\alpha}}\bigg{)}.$

The proof is complete. $\Box$

Note that for any $t,k,r>0$ and $a\geq 0$,

$\bigg{(}1\wedge\frac{r}{t^{1/\alpha}}\bigg{)}^{a}(r\vee
t^{1/\alpha})^{k}=r^{k}\bigg{(}1\wedge\frac{r}{t^{1/\alpha}}\bigg{)}^{a-k}.$
(7.6)

######  Lemma 7.2.

There exists a constant $C>0$ such that for all $t>0$ and
$x,y\in{\mathbb{R}}^{d}_{+}$,

$p(t,x,y)\leq C\left(1\wedge\frac{x_{d}\wedge
y_{d}}{t^{1/\alpha}}\right)^{q}A_{\beta_{1},0,\beta_{3},0}(t,x,y)\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right).$

Proof. By Proposition 5.1, the lemma holds for $\beta_{1}=0$.

We assume $\beta_{1}>0$ and set $a_{n}=\beta_{1}\wedge\frac{n\alpha}{2}$ for
$n\geq 0$. Below, we prove by induction that for any $n\geq 0$, there exists a
constant $C>0$ such that for all $t>0$ and $x,y\in{\mathbb{R}}^{d}_{+}$,

$\displaystyle p(t,x,y)\leq C\left(1\wedge\frac{x_{d}\wedge
y_{d}}{t^{1/\alpha}}\right)^{q}A_{a_{n},0,\beta_{3},0}(t,x,y)\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right).$
(7.7)

The lemma is a direct consequence of (7.7).

By Proposition 5.1 and the fact that the logarithmic term in
$A_{\beta_{1},0,\beta_{3},0}(t,x,y)$ is always larger than 1, (7.7) holds for
$n=0$. Suppose (7.7) holds for $n-1$. By symmetry and (3.30), we can assume
without loss of generality that $x_{d}\leq y_{d}$, $\widetilde{x}=0$ and
$|x-y|=5$. If $t>1$ or $x_{d}>2^{-5}$, then (7.7) follows from Proposition 5.1
and (6.4).

Let $t\leq 1$ and $x_{d}\leq 2^{-5}$. Then $y_{d}\leq x_{d}+|x-y|\leq
4+2^{-5}$ by the triangle inequality. Our goal is to show that there exists a
constant $c_{1}>0$ independent of $t,x,y$ such that

$p(t,x,y)\leq
c_{1}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee
t^{1/\alpha})^{a_{n}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}.$ (7.8)

Set $V_{1}=U(1)$, $V_{3}=B(y,2)\cap{\overline{\mathbb{R}}}^{d}_{+}$ and
$V_{2}={\overline{\mathbb{R}}}^{d}_{+}\setminus(V_{1}\cup V_{3})$. Similarly
to (5.10) and (5.11), we get from Proposition 2.7 and the triangle inequality
that

$\sup_{s\leq t,\,z\in V_{2}}p(s,z,y)\leq c_{2}\sup_{s\leq
t,\,z\in{\mathbb{R}}^{d}_{+},|z-y|\geq 2}\frac{s}{|z-y|^{d+\alpha}}\leq
2^{-d-\alpha}c_{2}t$ (7.9)

and

${\rm dist}(V_{1},V_{3})\geq\sup_{u\in V_{1},\,w\in V_{3}}(4-|x-u|-|y-w|)\geq
1.$ (7.10)

By the induction hypothesis, condition (7.1) in Lemma 7.1 holds with
$b_{1}=a_{n-1}$ and $b_{3}=\beta_{3}$. Thus, since
$a_{n}-a_{n-1}\leq\alpha/2$, we get from Lemma 7.1 and (7.9) that

$\displaystyle{\mathbb{P}}_{x}(\tau_{V_{1}}<t<\zeta)\sup_{s\leq t,\,z\in
V_{2}}p(s,z,y)$ $\displaystyle\leq
c_{3}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee
t^{1/\alpha})^{a_{n-1}}(t^{1/\alpha})^{\alpha/2}t^{1/2}\log^{\beta_{3}}\bigg{(}e+\frac{1}{t^{1/\alpha}}\bigg{)}\,$
$\displaystyle\leq
c_{3}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee
t^{1/\alpha})^{a_{n-1}}(t^{1/\alpha})^{a_{n}-a_{n-1}}\,\left(\sup_{s\leq
1}s^{1/2}\log^{\beta_{3}}\bigg{(}e+\frac{1}{s^{1/\alpha}}\bigg{)}\right)$
$\displaystyle\leq
c_{4}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee
t^{1/\alpha})^{a_{n}}.$ (7.11)

In order to apply Lemma 3.15 and get the desired result, it remains to bound
$\int_{0}^{t}\int_{V_{3}}\int_{V_{1}}p^{V_{1}}(t-s,x,u){\cal
B}(u,w)p(s,y,w)dudwds$. We consider the following two cases separately.

(Case 1) $q\geq\alpha+a_{n}$ and $10x_{d}<t^{1/\alpha}$.

Pick ${\varepsilon}\in(0,\beta_{1})$ such that
$q<\alpha+\beta_{1}-{\varepsilon}$. Using (A3)(II), Lemmas 10.1(i)–(ii), 10.2
(see Remark 10.3), and (7.10), we see that for all $u\in V_{1}$ and $w\in
V_{3}$,

$\displaystyle{\cal B}(u,w)\leq
c_{5}B_{\beta_{1}-{\varepsilon},0,0,0}(u,w)\leq
c_{6}\left(\frac{u_{d}}{|u-w|}\right)^{\beta_{1}-{\varepsilon}}\leq
c_{6}u_{d}^{\beta_{1}-{\varepsilon}}.$ (7.12)

By (7.12) and Lemma 5.12, we have

$\displaystyle\int_{0}^{t}\int_{V_{3}}\int_{V_{1}}p^{V_{1}}(t-s,x,u){\cal
B}(u,w)p(s,y,w)dudwds$ $\displaystyle\leq
c_{6}\int_{0}^{t}\int_{V_{1}}p^{V_{1}}(t-s,x,u)u_{d}^{\beta_{1}-{\varepsilon}}du\int_{V_{3}}p(s,y,w)dwds\leq
c_{6}\int_{0}^{\infty}\int_{V_{1}}p^{V_{1}}(s,x,u)u_{d}^{\beta_{1}-{\varepsilon}}duds$
$\displaystyle\leq
c_{7}x_{d}^{q}=c_{7}t\left(\frac{x_{d}}{t^{1/\alpha}}\right)^{q}(t^{1/\alpha})^{q-\alpha}\leq
c_{8}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee
t^{1/\alpha})^{a_{n}}.$ (7.13)

In the last inequality above, we used the facts that $q-\alpha\geq a_{n}$,
$10x_{d}<t^{1/\alpha}$ and $x_{d}\vee t^{1/\alpha}\leq 1$. Now, using Lemma
3.15, (7) and (7), we get (7.8) in this case.

(Case 2) $q<\alpha+a_{n}$ or $10x_{d}\geq t^{1/\alpha}$.

By (A3)(II), (7.10), Lemma 10.1(ii) and Lemma 10.2 (see Remark 10.3), it holds
that for all $u\in V_{1}$ and $w\in V_{3}$,

$\displaystyle{\cal B}(u,w)$ $\displaystyle\leq
c_{9}B_{\beta_{1},0,\beta_{3},0}(u,w)\leq
c_{10}\left(\frac{u_{d}}{|u-w|}\right)^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}\wedge|u-w|}{u_{d}\wedge|u-w|}\bigg{)}$
$\displaystyle\leq
c_{10}u_{d}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}.$
(7.14)

By (10.2) (or using that $u\mapsto u^{\beta_{1}}\log^{\beta_{3}}(e+t/u)$ is
almost increasing) and Corollary 5.2, since $\beta_{1}>0$ and
$a_{n}\leq\beta_{1}$, we get that for any $0<s<t$ and $w\in V_{3}$,

$\displaystyle\int_{u\in
V_{1}:u_{d}<x_{d}}p(s,x,u)u_{d}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}du\leq\int_{u\in
V_{1}:u_{d}<x_{d}\vee
s^{1/\alpha}}p(s,x,u)u_{d}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}du$
$\displaystyle\leq c_{11}(x_{d}\vee
s^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{x_{d}\vee
s^{1/\alpha}}\bigg{)}\int_{u\in V_{1}:u_{d}<x_{d}\vee s^{1/\alpha}}p(s,x,u)du$
$\displaystyle\leq c_{11}{\mathbb{P}}_{x}(\zeta>s)(x_{d}\vee
s^{1/\alpha})^{a_{n}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{x_{d}\vee
s^{1/\alpha}}\bigg{)}$ $\displaystyle\leq
c_{12}\bigg{(}1\wedge\frac{x_{d}}{s^{1/\alpha}}\bigg{)}^{q}(x_{d}\vee
s^{1/\alpha})^{a_{n}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{x_{d}\vee
s^{1/\alpha}}\bigg{)}.$ (7.15)

Next, using the induction hypothesis and Lemma 10.10, since
$a_{n}\leq\beta_{1}$ and $a_{n}<\alpha+a_{n-1}$, we get that for any $0<s<t$
and $w\in V_{3}$,

$\displaystyle\int_{u\in V_{1}:u_{d}\geq
x_{d}}p(s,x,u)u_{d}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}du$
$\displaystyle\leq
c_{13}\bigg{(}1\wedge\frac{x_{d}}{s^{1/\alpha}}\bigg{)}^{q}\int_{u\in
V_{1}:u_{d}\geq x_{d}}\bigg{(}\frac{x_{d}\vee s^{1/\alpha}}{|x-u|}\wedge
1\bigg{)}^{a_{n-1}}\log^{\beta_{3}}\bigg{(}e+\frac{|x-u|}{(x_{d}\vee
s^{1/\alpha})\wedge|x-u|}\bigg{)}$ $\displaystyle\hskip
156.49014pt\times\left(s^{-d/\alpha}\wedge\frac{s}{|x-u|^{d+\alpha}}\right)u_{d}^{a_{n}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}du$
$\displaystyle\leq
c_{14}\bigg{(}1\wedge\frac{x_{d}}{s^{1/\alpha}}\bigg{)}^{q}(x_{d}\vee
s^{1/\alpha})^{a_{n}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{x_{d}\vee
s^{1/\alpha}}\bigg{)}.$ (7.16)

Similarly, again splitting the integration into two parts $w_{d}<y_{d}$ and
$w_{d}\geq y_{d}$, and using the induction hypothesis and Lemma 10.10 again,
we also get that for any $0<s<t$,

$\displaystyle\int_{V_{3}}p(s,y,w)\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}dw$
$\displaystyle\leq
c_{15}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee
s^{1/\alpha}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}$ $\displaystyle\leq
c_{15}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}.$ (7.17)

By (7)–(7) and (7.6), we have

$\displaystyle\int_{0}^{t}\int_{V_{3}}\int_{V_{1}}p^{V_{1}}(t-s,x,u){\cal
B}(u,w)p(s,y,w)dudwds$ $\displaystyle\leq
c_{10}\int_{0}^{t}\int_{V_{3}}p(s,y,w)\int_{V_{1}}p(t-s,x,u)u_{d}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}dudwds$
$\displaystyle\leq
c_{16}\int_{0}^{t}\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q}(x_{d}\vee(t-s)^{1/\alpha})^{a_{n}}\int_{V_{3}}p(s,y,w)\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}dwds$
$\displaystyle\leq
c_{17}x_{d}^{a_{n}}\int_{0}^{t}\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-a_{n}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}ds=:I.$ (7.18)

When $q<\alpha+a_{n}$, we get from Lemma 10.4 that

$\displaystyle I\leq
c_{18}tx_{d}^{a_{n}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-a_{n}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}.$ (7.19)

When $10x_{d}\geq t^{1/\alpha}$, we also get from Lemma 10.4 that

$\displaystyle I$ $\displaystyle\leq
c_{17}x_{d}^{a_{n}}\int_{0}^{t}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}ds\leq
c_{19}tx_{d}^{a_{n}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}$ $\displaystyle\leq
10^{q-a_{n}}c_{19}tx_{d}^{a_{n}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-a_{n}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}.$ (7.20)

Now, (7.8) follows from (7.6), (7), (7), (7) and Lemma 3.15. The proof is
complete. $\Box$

Now we use Lemma 7.2 to improve the bound in (7).

######  Lemma 7.3.

Let $0<t\leq 1$ and $x,y\in{\mathbb{R}}^{d}_{+}$ be such that
$\widetilde{x}=\widetilde{0}$, $x_{d}\leq 2^{-5}$ and $|x-y|=5$. Set
$V_{1}=U(1)$, $V_{3}=B(y,2)\cap{\overline{\mathbb{R}}}^{d}_{+}$ and
$V_{2}={\overline{\mathbb{R}}}^{d}_{+}\setminus(V_{1}\cup V_{3})$. There
exists a constant $C>0$ independent of $t,x$ and $y$ such that

$\displaystyle{\mathbb{P}}_{x}(\tau_{V_{1}}<t<\zeta)\sup_{s\leq t,\,z\in
V_{2}}p(s,z,y)$ $\displaystyle\leq
Ct^{2}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee
t^{1/\alpha})^{\beta_{1}}(y_{d}\vee t^{1/\alpha})^{\beta_{1}}$
$\displaystyle\quad\times\log^{\beta_{3}}\bigg{(}e+\frac{1}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}.$

Proof. Note that, since $y_{d}\leq x_{d}+|x-y|\leq 5+2^{-5}<6$, for any
$0<s\leq 1$ and $z\in{\mathbb{R}}^{d}_{+}$ with $|z-y|\geq 2$,

$\displaystyle(y_{d}\vee s^{1/\alpha})\wedge|z-y|\geq\frac{1}{3}(y_{d}\vee
s^{1/\alpha}).$ (7.21)

By Lemmas 7.2 and 10.2 and applying (7.21),

$\displaystyle\sup_{s\leq t,\,z\in V_{2}}p(s,z,y)\leq c_{1}\sup_{s\leq
t,\,z\in{\mathbb{R}}^{d}_{+},|z-y|\geq 2}\left(1\wedge\frac{z_{d}\wedge
y_{d}}{s^{1/\alpha}}\right)^{q}A_{\beta_{1},0,\beta_{3},0}(s,z,y)\frac{s}{|z-y|^{d+\alpha}}$
$\displaystyle\leq c_{1}\sup_{s\leq t,\,z\in{\mathbb{R}}^{d}_{+},|z-y|\geq
2}\left(1\wedge\frac{y_{d}}{s^{1/\alpha}}\right)^{q}\bigg{(}\frac{y_{d}\vee
s^{1/\alpha}}{|z-y|}\bigg{)}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{3|z-y|}{y_{d}\vee
s^{1/\alpha}}\bigg{)}\frac{s}{|z-y|^{d+\alpha}}.$ (7.22)

Using that $u\mapsto u^{\beta_{1}}\log^{\beta_{3}}(e+1/u)$ is almost
increasing, we have that for any $0<s\leq t\leq 1$ and
$z\in{\mathbb{R}}^{d}_{+}$ with $|z-y|\geq 2$,

$\displaystyle\left(1\wedge\frac{y_{d}}{s^{1/\alpha}}\right)^{q}\bigg{(}\frac{y_{d}\vee
s^{1/\alpha}}{|z-y|}\bigg{)}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{3|z-y|}{y_{d}\vee
s^{1/\alpha}}\bigg{)}\frac{s}{|z-y|^{d+\alpha}}$ $\displaystyle\leq
c_{2}\left(1\wedge\frac{y_{d}}{s^{1/\alpha}}\right)^{q}(y_{d}\vee
s^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee
s^{1/\alpha}}\bigg{)}\frac{s}{|z-y|^{d+\alpha}}$ $\displaystyle\leq
2^{-d-\alpha}c_{2}s\left(1\wedge\frac{y_{d}}{s^{1/\alpha}}\right)^{q}(y_{d}\vee
s^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee
s^{1/\alpha}}\bigg{)}.$ (7.23)

Let ${\varepsilon}>0$ be such that $q<\alpha+\beta_{1}-{\varepsilon}$. Using
(7.6), we see that

$\displaystyle s\left(1\wedge\frac{y_{d}}{s^{1/\alpha}}\right)^{q}(y_{d}\vee
s^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee
s^{1/\alpha}}\bigg{)}$
$\displaystyle=sy_{d}^{\beta_{1}-{\varepsilon}}\left(1\wedge\frac{y_{d}}{s^{1/\alpha}}\right)^{q-\beta_{1}+{\varepsilon}}(y_{d}\vee
s^{1/\alpha})^{{\varepsilon}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee
s^{1/\alpha}}\bigg{)}$
$\displaystyle=s^{(\alpha+\beta_{1}-q-{\varepsilon})/\alpha}y_{d}^{\beta_{1}-{\varepsilon}}(y_{d}\wedge
s^{1/\alpha})^{q-\beta_{1}+{\varepsilon}}(y_{d}\vee
s^{1/\alpha})^{{\varepsilon}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee
s^{1/\alpha}}\bigg{)}.$

Thus, the map $s\mapsto
s\big{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\big{)}^{q}(y_{d}\vee
s^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\big{(}e+\frac{1}{y_{d}\vee
s^{1/\alpha}}\big{)}$ is almost increasing on $(0,t]$ by (10.2). Using this
and (7)–(7), we get that

$\displaystyle\sup_{s\leq t,\,z\in V_{2}}p(s,z,y)$ $\displaystyle\leq
c_{3}\sup_{s\leq
t}s\left(1\wedge\frac{y_{d}}{s^{1/\alpha}}\right)^{q}(y_{d}\vee
s^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee
s^{1/\alpha}}\bigg{)}$ $\displaystyle\leq
c_{4}t\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(y_{d}\vee
t^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}.$ (7.24)

Note that (7.1) is satisfied with $a_{1}=\beta_{1}$ and $a_{3}=\beta_{3}$ by
Lemma 7.2. Thus, by Lemma 7.1 and (7) we obtain

$\displaystyle{\mathbb{P}}_{x}(\tau_{V_{1}}<t<\zeta)\sup_{s\leq t,\,z\in
V_{2}}p(s,z,y)$ $\displaystyle\leq
c_{4}t^{2}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee
t^{1/\alpha})^{\beta_{1}}(y_{d}\vee t^{1/\alpha})^{\beta_{1}}$
$\displaystyle\quad\times\log^{\beta_{3}}\bigg{(}e+\frac{1}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}.$

$\Box$

######  Lemma 7.4.

Let $\eta_{1},\eta_{2},\gamma\geq 0$. There exists a constant $C>0$ such that
for any $x\in{\mathbb{R}}^{d}_{+}$ and any $s,k,l>0$,

$\displaystyle\int_{B_{+}(x,2)}p(s,x,z)z_{d}^{\gamma}\log^{\eta_{1}}\bigg{(}e+\frac{k}{z_{d}}\bigg{)}\log^{\eta_{2}}\bigg{(}e+\frac{z_{d}}{l}\bigg{)}dz$
$\displaystyle\leq
Cx_{d}^{\gamma}\bigg{(}1\wedge\frac{x_{d}}{s^{1/\alpha}}\bigg{)}^{q-\gamma}\log^{\eta_{1}}\bigg{(}e+\frac{k}{x_{d}\vee
s^{1/\alpha}}\bigg{)}\log^{\eta_{2}}\bigg{(}e+\frac{x_{d}\vee
s^{1/\alpha}}{l}\bigg{)}$ $\displaystyle\quad+C{\bf
1}_{\\{\gamma>\alpha+\beta_{1}\\}}sx_{d}^{\beta_{1}}\bigg{(}1\wedge\frac{x_{d}}{s^{1/\alpha}}\bigg{)}^{q-\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{2}{x_{d}\vee
s^{1/\alpha}}\bigg{)}\log^{\eta_{1}}(e+k)\log^{\eta_{2}}\bigg{(}e+\frac{1}{l}\bigg{)}$
$\displaystyle\quad+C{\bf 1}_{\\{\gamma=\alpha+\beta_{1},\,x_{d}\vee
s^{1/\alpha}<2\\}}sx_{d}^{\beta_{1}}\bigg{(}1\wedge\frac{x_{d}}{s^{1/\alpha}}\bigg{)}^{q-\beta_{1}}$
$\displaystyle\qquad\times\int_{x_{d}\vee
s^{1/\alpha}}^{2}\log^{\beta_{3}}\bigg{(}e+\frac{r}{x_{d}\vee
s^{1/\alpha}}\bigg{)}\log^{\eta_{1}}\bigg{(}e+\frac{k}{r}\bigg{)}\log^{\eta_{2}}\bigg{(}e+\frac{r}{l}\bigg{)}\frac{dr}{r}.$

Proof. For any $x,z\in{\mathbb{R}}^{d}_{+}$ and $s>0$, by Lemmas 7.2 and 10.2,

$\displaystyle p(s,x,z)\leq
c_{1}\left(1\wedge\frac{x_{d}}{s^{1/\alpha}}\right)^{q}A_{\beta_{1},0,\beta_{3},0}(s,x,z)\left(s^{-d/\alpha}\wedge\frac{s}{|x-z|^{d+\alpha}}\right)$
$\displaystyle\leq
c_{2}\left(1\wedge\frac{x_{d}}{s^{1/\alpha}}\right)^{q}\bigg{(}1\wedge\frac{x_{d}\vee
s^{1/\alpha}}{|x-z|}\bigg{)}^{\beta_{1}}\log^{\beta_{3}}\left(e+\frac{|x-z|}{(x_{d}\vee
s^{1/\alpha})\wedge|x-z|}\right)\left(s^{-d/\alpha}\wedge\frac{s}{|x-z|^{d+\alpha}}\right).$

Now combining (7.6) and Lemma 10.10, we get the desired result. $\Box$

We now state the first main result of this section.

######  Theorem 7.5.

For any ${\varepsilon}\in(0,\alpha/2]$, there exists a constant $C>0$ such
that

$\displaystyle p(t,x,y)$ $\displaystyle\leq
C\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}A_{\beta_{1},\beta_{2}\wedge(\alpha+\beta_{1}-{\varepsilon}),\beta_{3},\beta_{4}}(t,x,y)\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right),$

for all $t>0$ and $x,y\in{\mathbb{R}}^{d}_{+}$.

This theorem will proved by using several lemmas. We first introduce some
additional notation: For $b_{1},b_{2},b_{3},b_{4}\geq 0$, $t>0$ and
$x,y\in{\mathbb{R}}^{d}_{+}$, define

$\displaystyle{\cal I}_{t,1}(x,y;b_{1},b_{2},b_{3},b_{4})$
$\displaystyle:=\int_{0}^{t/2}\int_{B_{+}(y,2)}\int_{B_{+}(x,2)}{\bf
1}_{\\{u_{d}\leq w_{d}\\}}p(t-s,x,u)p(s,y,w)$ $\displaystyle\hskip
56.9055pt\times
u_{d}^{b_{1}}w_{d}^{b_{2}}\log^{b_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{w_{d}}\bigg{)}\,du\,dw\,ds,$
(7.25) $\displaystyle{\cal I}_{t,2}(x,y;b_{1},b_{2},b_{3},b_{4})$
$\displaystyle:=\int_{0}^{t/2}\int_{B_{+}(y,2)}\int_{B_{+}(x,2)}{\bf
1}_{\\{w_{d}\leq u_{d}\\}}p(t-s,x,u)p(s,y,w)$ $\displaystyle\hskip
56.9055pt\times
u_{d}^{b_{2}}w_{d}^{b_{1}}\log^{b_{3}}\bigg{(}e+\frac{u_{d}}{w_{d}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{u_{d}}\bigg{)}\,du\,dw\,ds.$
(7.26)

######  Lemma 7.6.

Let $b_{1},b_{2},b_{3},b_{4}\geq 0$. Let $b_{1}^{\prime}\in[0,b_{1}]$ and
$b_{2}^{\prime}:=b_{1}+b_{2}-b_{1}^{\prime}$. Then

${\cal I}_{t,i}(x,y;b_{1},b_{2},b_{3},b_{4})\leq{\cal
I}_{t,i}(x,y;b_{1}^{\prime},b_{2}^{\prime},b_{3},b_{4}),\quad i=1,2.$

Proof. If $u_{d}\leq w_{d}$, then $u_{d}^{b_{1}}w_{d}^{b_{2}}\leq
u_{d}^{b_{1}^{\prime}}w_{d}^{b_{2}^{\prime}}$, implying that ${\cal
I}_{t,1}(x,y;b_{1},b_{2},b_{3},b_{4})\leq{\cal
I}_{t,1}(x,y;b_{1}^{\prime},b_{2}^{\prime},\linebreak b_{3},b_{4}).$ The other
inequality is analogous. $\Box$

######  Lemma 7.7.

Let $b_{1},b_{2},b_{3},b_{4}\geq 0$ with $b_{1}\vee b_{2}<\alpha+\beta_{1}$,
$x,y\in{\mathbb{R}}^{d}_{+}$ with $|x-y|=5$, and $t\in(0,1]$.

(i) If $b_{2}>q-\alpha$ or $y_{d}\geq t^{1/\alpha}$, then there exists a
constant $C>0$ independent of $t,x,y$ such that

$\displaystyle{\cal I}_{t,1}(x,y;b_{1},b_{2},b_{3},b_{4})$ $\displaystyle\leq
Ctx_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}.$

(ii) If $b_{1}>q-\alpha$ or $y_{d}\geq t^{1/\alpha}$, then there exists a
constant $C>0$ independent of $t,x,y$ such that

$\displaystyle{\cal I}_{t,2}(x,y;b_{1},b_{2},b_{3},b_{4})$ $\displaystyle\leq
Ctx_{d}^{b_{2}}y_{d}^{b_{1}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\log^{b_{3}}\bigg{(}e+\frac{x_{d}\vee
t^{1/\alpha}}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{x_{d}\vee
t^{1/\alpha}}\bigg{)}.$

Proof. We give the proof for (i). (ii) can be proved similarly.

By using Lemma 7.4 together with the fact that $t-s\asymp t$ if $0\leq s\leq
t/2$, we see that for $0\leq s\leq t/2$,

$\displaystyle\int_{B_{+}(x,2)}p(t-s,x,u)u_{d}^{b_{1}}\log^{b_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}du\leq
c_{1}x_{d}^{b_{1}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\log^{b_{3}}\bigg{(}e+\frac{w_{d}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}.$

Thus, using Lemma 7.4 again, we get that for $0\leq s\leq t/2$,

$\displaystyle\int_{B_{+}(y,2)}\int_{B_{+}(x,2)}p(t-s,x,u)p(s,y,w)u_{d}^{b_{1}}w_{d}^{b_{2}}\log^{b_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{w_{d}}\bigg{)}\,du\,dw$
$\displaystyle\leq
c_{1}x_{d}^{b_{1}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\int_{B_{+}(y,2)}p(s,y,w)w_{d}^{b_{2}}\log^{b_{3}}\bigg{(}e+\frac{w_{d}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{w_{d}}\bigg{)}dw$
$\displaystyle\leq
c_{2}x_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
s^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
s^{1/\alpha}}\bigg{)}$

From this and Lemma 10.5, we get that

$\displaystyle{\cal I}_{t,1}(x,y;b_{1},b_{2},b_{3},b_{4})$ $\displaystyle\leq
c_{2}x_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\int_{0}^{\frac{t}{2}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
s^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
s^{1/\alpha}}\bigg{)}ds$ $\displaystyle\leq
c_{2}x_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\int_{0}^{t}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
s^{1/\alpha}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
s^{1/\alpha}}\bigg{)}ds$ $\displaystyle\leq
c_{3}tx_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}.$

$\Box$

######  Lemma 7.8.

Let $b_{1},b_{2},b_{3},b_{4}\geq 0$ be such that $b_{1}>q-\alpha$ and
$b_{1}\vee b_{2}<\alpha+\beta_{1}$. Assume that $b_{2}>0$ if $b_{4}>0$. Then,
there exists a constant $C>0$ such that for all $x,y\in{\mathbb{R}}^{d}_{+}$
with $x_{d}\leq y_{d}$ and $|x-y|=5$, and all $t\in(0,1]$,

$\displaystyle{\cal I}_{t,1}(x,y;b_{1},b_{2},b_{3},b_{4})+{\cal
I}_{t,2}(x,y;b_{1},b_{2},b_{3},b_{4})$ $\displaystyle\leq
Ctx_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}$ (7.27)

and

$\displaystyle{\cal I}_{t,1}(y,x;b_{1},b_{2},b_{3},b_{4})+{\cal
I}_{t,2}(y,x;b_{1},b_{2},b_{3},b_{4})$ $\displaystyle\leq
Ctx_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}.$ (7.28)

Proof. Let $\delta\in(0,1)$ be such that

$\displaystyle b_{1}\vee b_{2}+(1-\delta)(b_{1}\wedge
b_{2})<\alpha+\beta_{1},$ (7.29)

and let $b_{1}^{\prime}:=\delta(b_{1}\wedge b_{2})$ and
$b_{2}^{\prime}:=b_{1}+b_{2}-b_{1}^{\prime}$. Then we see that
$b_{1}^{\prime}\in[0,b_{1}\wedge b_{2}]$, $b_{2}^{\prime}\geq b_{1}>q-\alpha$
and

$\displaystyle b_{1}^{\prime}\leq b_{2}^{\prime}=b_{1}\vee
b_{2}+(1-\delta)(b_{1}\wedge b_{2})<\alpha+\beta_{1}$ (7.30)

by (7.29). Moreover, since $x_{d}\leq y_{d}$ and
$b_{2}^{\prime}-b_{1}=b_{2}-b_{1}^{\prime}$, we see that

$\displaystyle(x_{d}\vee t^{1/\alpha})^{b_{2}^{\prime}-b_{1}}(y_{d}\vee
t^{1/\alpha})^{b_{1}^{\prime}-b_{2}}\log^{b_{4}}\bigg{(}e+\frac{1}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{-b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\leq c_{1}.$ (7.31)

Indeed, when $b_{4}=0$, (7.31) clearly holds with $c_{1}=1$. If $b_{4}>0$,
then $b_{2}>0$ so that $b_{2}>\delta b_{2}\geq b_{1}^{\prime}$. Hence, we get
(7.31) from (10.2).

(i) We first prove (7.8). For this, we distinguish between two cases:
$y_{d}\geq t^{1/\alpha}$ and $y_{d}<t^{1/\alpha}$.

Assume first that $y_{d}\geq t^{1/\alpha}$. The desired bound for ${\cal
I}_{t,1}(x,y;b_{1},b_{2},b_{3},b_{4})$ follows from Lemma 7.7(i). On the other
hand, by using Lemma 7.6 in the first inequality, Lemma 7.7(ii) in the second
inequality (which uses (7.30) and $y_{d}\geq t^{1/\alpha}$), (7.6) in the
equality, and (7.31) in the last inequality, we get that

$\displaystyle{\cal I}_{t,2}(x,y;b_{1},b_{2},b_{3},b_{4})\leq{\cal
I}_{t,2}(x,y;b_{1}^{\prime},b_{2}^{\prime},b_{3},b_{4})$ $\displaystyle\leq
c_{2}tx_{d}^{b_{2}^{\prime}}y_{d}^{b_{1}^{\prime}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}^{\prime}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}^{\prime}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{x_{d}\vee
t^{1/\alpha}}\bigg{)}$
$\displaystyle=c_{2}tx_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}$ $\displaystyle\quad\;\times(x_{d}\vee
t^{1/\alpha})^{b_{2}^{\prime}-b_{1}}(y_{d}\vee
t^{1/\alpha})^{b_{1}^{\prime}-b_{2}}\log^{b_{4}}\bigg{(}e+\frac{1}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{-b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}$ $\displaystyle\leq
c_{1}c_{2}tx_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}.$

Assume now that $y_{d}<t^{1/\alpha}$. Using Lemma 7.6 and Lemma 7.7(i) (which
uses $b_{2}^{\prime}>q-\alpha$), we get

$\displaystyle{\cal I}_{t,1}(x,y;b_{1},b_{2},b_{3},b_{4})\leq{\cal
I}_{t,1}(x,y;b_{1}^{\prime},b_{2}^{\prime},b_{3},b_{4})$ $\displaystyle\leq
c_{3}tx_{d}^{b_{1}^{\prime}}y_{d}^{b_{2}^{\prime}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}^{\prime}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}^{\prime}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}.$ (7.32)

Also, since $b_{1}>q-\alpha$, we get from Lemma 7.7(ii) that

$\displaystyle{\cal I}_{t,2}(x,y;b_{1},b_{2},b_{3},b_{4})$ $\displaystyle\leq
c_{4}tx_{d}^{b_{2}}y_{d}^{b_{1}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\\!\log^{b_{3}}\bigg{(}e+\frac{x_{d}\vee
t^{1/\alpha}}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{x_{d}\vee
t^{1/\alpha}}\bigg{)}.$ (7.33)

Since $x_{d}\leq y_{d}<t^{1/\alpha}$ and
$b_{1}^{\prime}+b_{2}^{\prime}=b_{1}+b_{2}$, it holds that $x_{d}\vee
t^{1/\alpha}=y_{d}\vee t^{1/\alpha}=t^{1/\alpha}$ and

$\displaystyle
x_{d}^{b_{1}^{\prime}}y_{d}^{b_{2}^{\prime}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}^{\prime}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}^{\prime}}=x_{d}^{b_{2}}y_{d}^{b_{1}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}$
$\displaystyle=\frac{x_{d}^{q}y_{d}^{q}}{t^{(2q-b_{1}-b_{2})/\alpha}}=x_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}.$

Thus, (7.8) follows from (7) and (7).

(ii) Now, we prove (7.8). By using Lemma 7.6 in the first inequality, Lemma
7.7(i) in the second inequality (which uses (7.30) and
$b_{2}^{\prime}>q-\alpha$), (7.6) together with the fact that $x_{d}\leq
y_{d}$ in the third inequality, (7.31) in the last inequality, we obtain

$\displaystyle{\cal I}_{t,1}(y,x;b_{1},b_{2},b_{3},b_{4})\leq{\cal
I}_{t,1}(y,x;b_{1}^{\prime},b_{2}^{\prime},b_{3},b_{4})$ $\displaystyle\leq
c_{5}tx_{d}^{b_{2}^{\prime}}y_{d}^{b_{1}^{\prime}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}^{\prime}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}^{\prime}}\log^{b_{3}}\bigg{(}e+\frac{x_{d}\vee
t^{1/\alpha}}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{x_{d}\vee
t^{1/\alpha}}\bigg{)}$ $\displaystyle\leq
c_{5}tx_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}$ $\displaystyle\quad\times(x_{d}\vee
t^{1/\alpha})^{b_{2}^{\prime}-b_{1}}(y_{d}\vee
t^{1/\alpha})^{b_{1}^{\prime}-b_{2}}\log^{b_{4}}\bigg{(}e+\frac{1}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{-b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}$ $\displaystyle\leq
c_{1}c_{5}tx_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}.$

On the other hand, by Lemma 7.7(ii), it holds that

$\displaystyle{\cal I}_{t,2}(y,x;b_{1},b_{2},b_{3},b_{4})$ $\displaystyle\leq
c_{3}tx_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}.$

The proof is complete. $\Box$

Proof of Theorem 7.5. Set
$\widehat{\beta}_{2}:=\beta_{2}\wedge(\alpha+\beta_{1}-{\varepsilon})$. As in
the proof of Lemma 7.2, by symmetry, Proposition 5.1, (3.30), and (6.4), we
can assume without loss of generality that $x_{d}\leq y_{d}\wedge 2^{-5}$,
$\widetilde{x}=0$ and $|x-y|=5$, and then it is enough to show that there
exists a constant $c_{1}>0$ independent of $x$ and $y$ such that for any
$t\leq 1$,

$\displaystyle p(t,x,y)$ $\displaystyle\leq
c_{1}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee
t^{1/\alpha})^{\beta_{1}}(y_{d}\vee t^{1/\alpha})^{\widehat{\beta}_{2}}$
$\displaystyle\quad\times\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\,\log^{\beta_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}.$ (7.34)

Let $t\leq 1$. Set $V_{1}=U(1)$,
$V_{3}=B(y,2)\cap{\overline{\mathbb{R}}}^{d}_{+}$ and
$V_{2}={\overline{\mathbb{R}}}^{d}_{+}\setminus(V_{1}\cup V_{3})$. By Lemma
7.3,

$\displaystyle{\mathbb{P}}_{x}(\tau_{V_{1}}<t<\zeta)\sup_{s\leq t,\,z\in
V_{2}}p(s,z,y)$ $\displaystyle\leq
c_{2}t^{2}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee
t^{1/\alpha})^{\beta_{1}}(y_{d}\vee
t^{1/\alpha})^{\beta_{1}}\log^{2\beta_{3}}\bigg{(}e+\frac{1}{t^{1/\alpha}}\bigg{)}$
$\displaystyle=c_{2}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee
t^{1/\alpha})^{\beta_{1}}(y_{d}\vee
t^{1/\alpha})^{\alpha+\beta_{1}-{\varepsilon}}(y_{d}\vee
t^{1/\alpha})^{-\alpha+{\varepsilon}}t\log^{2\beta_{3}}\bigg{(}e+\frac{1}{t^{1/\alpha}}\bigg{)}$
$\displaystyle\leq
c_{2}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee
t^{1/\alpha})^{\beta_{1}}(y_{d}\vee
t^{1/\alpha})^{\widehat{\beta}_{2}}t^{{\varepsilon}/\alpha}\log^{2\beta_{3}}\bigg{(}e+\frac{1}{t^{1/\alpha}}\bigg{)}$
$\displaystyle\leq
c_{3}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee
t^{1/\alpha})^{\beta_{1}}(y_{d}\vee t^{1/\alpha})^{\widehat{\beta}_{2}},$
(7.35)

where in the last inequality above we used (10.1).

Next, we show that there exists a constant $C^{\prime}>0$ such that

$\displaystyle I$
$\displaystyle:=\int_{0}^{t}\int_{V_{3}}\int_{V_{1}}p^{V_{1}}(t-s,x,u){\cal
B}(u,w)p(s,y,w)du\,dw\,ds$ $\displaystyle\leq
C^{\prime}tx_{d}^{\beta_{1}}y_{d}^{\widehat{\beta}_{2}}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q-\beta_{1}}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q-\widehat{\beta}_{2}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}.$ (7.36)

Once we get (7), by (7.6) and (7), we can apply Lemma 3.15 to get (7) and
finish the proof.

By (A3)(II), since $|u-w|\asymp 1$ and $u_{d}\vee w_{d}\leq y_{d}+2\leq
7+2^{-5}$ for $u\in V_{1}$ and $w\in V_{3}$, we have

$\displaystyle I$ $\displaystyle\leq
c_{4}\int_{0}^{t}\int_{V_{3}}\int_{V_{1}}{\bf 1}_{\\{u_{d}\leq
w_{d}\\}}p(t-s,x,u)p(s,y,w)u_{d}^{\beta_{1}}w_{d}^{\widehat{\beta}_{2}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{w_{d}}\bigg{)}du\,dw\,ds$
$\displaystyle\;+c_{4}\int_{0}^{t}\int_{V_{3}}\int_{V_{1}}{\bf
1}_{\\{w_{d}\leq
u_{d}\\}}p(t-s,x,u)p(s,y,w)u_{d}^{\widehat{\beta}_{2}}w_{d}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{u_{d}}{w_{d}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{u_{d}}\bigg{)}du\,dw\,ds$
$\displaystyle=c_{4}\bigg{(}\big{(}\int_{0}^{t/2}+\int_{t/2}^{t}\big{)}\int_{V_{3}}\int_{V_{1}}{\bf
1}_{\\{u_{d}\leq
w_{d}\\}}\dots+\big{(}\int_{0}^{t/2}+\int_{t/2}^{t}\big{)}\int_{V_{3}}\int_{V_{1}}{\bf
1}_{\\{w_{d}\leq u_{d}\\}}\dots\bigg{)}.$

Thus, by the change of variable $\widetilde{s}=t-s$ in integrals
$\int_{t/2}^{t}$,

$\displaystyle I\leq c_{4}\sum_{i=1}^{2}{\cal
I}_{t,i}(x,y;\beta_{1},\widehat{\beta}_{2},\beta_{3},\beta_{4})+c_{4}\sum_{i=1}^{2}{\cal
I}_{t,i}(y,x;\beta_{1},\widehat{\beta}_{2},\beta_{3},\beta_{4}),$

where the functions ${\cal
I}_{t,i}(x,y;\beta_{1},\widehat{\beta}_{2},\beta_{3},\beta_{4})$, $1\leq i\leq
2$, are defined in (7.25)–(7.26). Then by using Lemma 7.6(i) and Lemma 7.8, we
conclude that (7) holds. The proof is complete. $\Box$

As an immediate consequence of Theorem 7.5, we obtain the following sharp
upper bound of the heat kernel for $\beta_{2}<\alpha+\beta_{1}$.

######  Corollary 7.9.

Suppose that $\beta_{2}<\alpha+\beta_{1}$. There exists a constant $C>0$ such
that

$\displaystyle p(t,x,y)$ $\displaystyle\leq
C\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y)\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right),$

for all $t>0$ and $x,y\in{\mathbb{R}}^{d}_{+}$.

Here is the second main result of the section.

######  Theorem 7.10.

Suppose that $\beta_{2}\geq\alpha+\beta_{1}$. There exists a constant $C>0$
such that

$\displaystyle p(t,x,y)$ $\displaystyle\leq
C\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right)\bigg{[}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y)$
$\displaystyle\quad+\bigg{(}1\wedge\frac{t}{|x-y|^{\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{((x_{d}\wedge
y_{d})+t^{1/\alpha})\wedge|x-y|}\bigg{)}$
$\displaystyle\quad\quad\times\bigg{(}{\bf
1}_{\\{\beta_{2}>\alpha+\beta_{1}\\}}A_{\beta_{1},\beta_{1},0,\beta_{3}}(t,x,y)+{\bf
1}_{\\{\beta_{2}=\alpha+\beta_{1}\\}}A_{\beta_{1},\beta_{1},0,\beta_{3}+\beta_{4}+1}(t,x,y)\bigg{)}\bigg{]}$

for all $t>0$ and $x,y\in{\mathbb{R}}^{d}_{+}$.

Again, we first introduce additional notation and prove a lemma. For
$b_{1},b_{2},b_{3},b_{4}\geq 0$, $t>0$ and $x,y\in{\mathbb{R}}^{d}_{+}$,
define

$\displaystyle{\cal I}_{t}(x,y;b_{1},b_{2},b_{3},b_{4})$
$\displaystyle:=\int_{0}^{t}\int_{B_{+}(y,2)}\int_{B_{+}(x,2)}p(t-s,x,u)p(s,y,w)$
$\displaystyle\qquad\times
u_{d}^{b_{1}}w_{d}^{b_{2}}\log^{b_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{w_{d}}\bigg{)}\,du\,dw\,ds.$
(7.37)

######  Lemma 7.11.

Let $b_{1},b_{2},b_{3},b_{4}\geq 0$ be such that $b_{1}\wedge b_{2}>q-\alpha$
and $b_{1}<\alpha+\beta_{1}$. There exists a constant $C>0$ such that for all
$x,y\in{\mathbb{R}}^{d}_{+}$ with $|x-y|=5$, and all $t\in(0,1]$,

$\displaystyle{\cal I}_{t}(x,y;b_{1},b_{2},b_{3},b_{4})$ $\displaystyle\leq
Ctx_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\\!\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e\\!+\\!\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\\!\log^{b_{4}}\bigg{(}e\\!+\\!\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}$ $\displaystyle+{\bf
1}_{\\{b_{2}>\alpha+\beta_{1}\\}}Ct^{2}x_{d}^{b_{1}}y_{d}^{\beta_{1}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\\!\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-\beta_{1}}\\!\log^{b_{3}}\bigg{(}e\\!+\\!\frac{1}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\\!\log^{b_{3}}\bigg{(}e\\!+\\!\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}$ $\displaystyle+{\bf
1}_{\\{b_{2}=\alpha+\beta_{1},y_{d}<2\\}}Ctx_{d}^{b_{1}}y_{d}^{\beta_{1}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\\!\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-\beta_{1}}$
$\displaystyle\quad\times\int_{y_{d}}^{2}(r^{\alpha}\wedge
t)\log^{b_{3}}\bigg{(}e+\frac{r}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{r}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}.$

Proof. By using Lemma 7.4 in the first inequality (integration with respect to
$u$; note that $b_{1}<\alpha+\beta_{1}$) and the second inequality
(integration with respect to $w$), we get

$\displaystyle{\cal I}_{t}(x,y;b_{1},b_{2},b_{3},b_{4})\leq
c_{1}x_{d}^{b_{1}}\int_{0}^{t}\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}$
$\displaystyle\hskip
71.13188pt\times\int_{B_{+}(y,2)}p(s,y,w)w_{d}^{b_{2}}\log^{b_{3}}\bigg{(}e+\frac{w_{d}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{w_{d}}\bigg{)}dwds$
$\displaystyle\leq c_{2}\left(x_{d}^{b_{1}}y_{d}^{b_{2}}I_{1}+{\bf
1}_{\\{b_{2}>\alpha+\beta_{1}\\}}x_{d}^{b_{1}}y_{d}^{\beta_{1}}I_{2}+{\bf
1}_{\\{b_{2}=\alpha+\beta_{1},y_{d}<2\\}}x_{d}^{b_{1}}y_{d}^{\beta_{1}}I_{3}\right),$

where

$I_{1}:=\int_{0}^{t}\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}\\!\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
s^{1/\alpha}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
s^{1/\alpha}}\bigg{)}ds,$
$I_{2}:=\int_{0}^{t}s\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-\beta_{1}}\\!\log^{b_{3}}\bigg{(}e+\frac{2}{y_{d}\vee
s^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{1}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}ds,$

and

$\displaystyle I_{3}:=$
$\displaystyle\int_{0}^{t}s\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-\beta_{1}}$
$\displaystyle\hskip 36.98866pt\times\int_{y_{d}\vee
s^{1/\alpha}}^{2}\log^{b_{3}}\bigg{(}e+\frac{r}{y_{d}\vee
s^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{r}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}\,ds.$

Applying Lemma 10.5 to $I_{1}$ and $I_{2}$, Lemma 10.6 to $I_{3}$, and (10.2),
wee see that

$I_{1}\leq
c_{3}t\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)},$ $I_{2}\leq
c_{3}t^{2}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-\beta_{1}}\log^{b_{3}}\bigg{(}e+\frac{1}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)},$

and

$\displaystyle I_{2}\leq$ $\displaystyle
c_{3}t\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-\beta_{1}}$
$\displaystyle\quad\times\int_{y_{d}}^{2}(r^{\alpha}\wedge
t)\log^{b_{3}}\bigg{(}e+\frac{r}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{r}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}.$

This proves the lemma. $\Box$

Proof of Theorem 7.10. As in the proof of Lemma 7.2, by symmetry, (6),
Proposition 5.1, (3.30) and (6.4), we can assume without loss of generality
that $x_{d}\leq y_{d}\wedge 2^{-5}$, $\widetilde{x}=0$ and $|x-y|=5$, and then
it is enough to show that there exists a constant $c_{1}>0$ independent of $x$
and $y$ such that for any $t\leq 1$,

$\displaystyle p(t,x,y)\leq
c_{1}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee
t^{1/\alpha})^{\beta_{1}}$ $\displaystyle\qquad\times\bigg{[}(y_{d}\vee
t^{1/\alpha})^{\beta_{2}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\,\log^{\beta_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}$ (7.38) $\displaystyle\quad\qquad+{\bf
1}_{\\{\beta_{2}>\alpha+\beta_{1}\\}}t(y_{d}\vee
t^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}$ $\displaystyle\quad\qquad+{\bf
1}_{\\{\beta_{2}=\alpha+\beta_{1}\\}}t(y_{d}\vee
t^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{3}+\beta_{4}+1}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\bigg{]}.$ (7.39)

Let $t\leq 1$. Set $V_{1}=U(1)$,
$V_{3}=B(y,2)\cap{\overline{\mathbb{R}}}^{d}_{+}$ and
$V_{2}={\overline{\mathbb{R}}}^{d}_{+}\setminus(V_{1}\cup V_{3})$. By Lemmas
3.15 and 7.3 it remains to prove that
$I:=\int_{0}^{t}\int_{V_{3}}\int_{V_{1}}p^{V_{1}}(t-s,x,u){\cal
B}(u,w)p(s,y,w)dudwds$ is bounded above by the right-hand side of (7.39).

By (A3)(II), since $|u-w|\asymp 1$ for $u\in V_{1}$ and $w\in V_{3}$, using
the change of variables $\widetilde{s}=t-s$ we have

$\displaystyle I\leq c_{2}\left({\cal
I}_{t}(x,y;\beta_{1},\beta_{2},\beta_{3},\beta_{4})+{\cal
I}_{t}(y,x;\beta_{1},\beta_{2},\beta_{3},\beta_{4})\right),$ (7.40)

where the functions ${\cal
I}_{t}(x,y;\beta_{1},\beta_{2},\beta_{3},\beta_{4})$ is defined in (7.37). By
Lemma 7.11 and (7.6), the right hand side of (7.40) is less than or equal to
$c_{3}t(1\wedge\frac{x_{d}}{t^{1/\alpha}})^{q}(1\wedge\frac{y_{d}}{t^{1/\alpha}})^{q}$
times

$\displaystyle(x_{d}\vee t^{1/\alpha})^{\beta_{1}}(y_{d}\vee
t^{1/\alpha})^{\beta_{2}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\,\log^{\beta_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}$ (7.41) $\displaystyle\quad+(x_{d}\vee
t^{1/\alpha})^{\beta_{2}}(y_{d}\vee
t^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{x_{d}\vee
t^{1/\alpha}}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{x_{d}\vee
t^{1/\alpha}}\bigg{)}$ (7.42) $\displaystyle\quad+{\bf
1}_{\\{\beta_{2}>\alpha+\beta_{1}\\}}t(x_{d}\vee
t^{1/\alpha})^{\beta_{1}}(y_{d}\vee
t^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}$ (7.43) $\displaystyle\quad+{\bf
1}_{\\{\beta_{2}=\alpha+\beta_{1}\\}}(x_{d}\vee
t^{1/\alpha})^{\beta_{1}}(y_{d}\vee t^{1/\alpha})^{\beta_{1}}$
$\displaystyle\qquad\;\times\int_{x_{d}}^{2}(r^{\alpha}\wedge
t)\log^{\beta_{3}}\bigg{(}e+\frac{r}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{r}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}.$
(7.44)

In the last line we used $x_{d}\leq y_{d}\wedge 2$ to get that ${\bf
1}_{\\{y_{d}<2\\}}\int_{y_{d}}^{2}\leq\int_{x_{d}}^{2}$.

Since $\beta_{2}>\beta_{1}$ and $x_{d}\leq y_{d}$, clearly

$\displaystyle(x_{d}\vee
t^{1/\alpha})^{\beta_{2}}\log^{\beta_{3}}\bigg{(}e+\frac{x_{d}\vee
t^{1/\alpha}}{y_{d}\vee t^{1/\alpha}}\bigg{)}\leq(x_{d}\vee
t^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)},$ (7.45)

and by using (10.2) (with $\varepsilon=(\beta_{2}-\beta_{1})/\beta_{4})$), we
get

$\displaystyle(y_{d}\vee
t^{1/\alpha})^{\beta_{1}}\log^{\beta_{4}}\bigg{(}e+\frac{1}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\leq c_{4}(y_{d}\vee
t^{1/\alpha})^{\beta_{2}}\log^{\beta_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}.$ (7.46)

Applying (7.45) and (7.46) to (7.42) and combining it with (7.41) and (7.43),
we arrive at the result in case $\beta_{2}>\alpha+\beta_{1}$.

Assume now that $\beta_{2}=\alpha+\beta_{1}$. From the above argument, to
prove the result, in view of (7.38), (7.39) and (7.44), it suffices to show
that

$\displaystyle\int_{x_{d}}^{2}(r^{\alpha}\wedge
t)\log^{\beta_{3}}\bigg{(}e+\frac{r}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{r}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}$
$\displaystyle\leq c_{5}t\log^{\beta_{3}}\bigg{(}e+\frac{1}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{3}+\beta_{4}+1}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}$ $\displaystyle\quad+c_{5}(y_{d}\vee
t^{1/\alpha})^{\beta_{2}-\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}.$ (7.47)

By Lemma 10.11 (with $b_{1}=\beta_{3}$, $b_{2}=\beta_{4}$, $k=x_{d}\vee
t^{1/\alpha}$ and $l=y_{d}\vee t^{1/\alpha}$), it holds that

$\displaystyle\int_{y_{d}\vee t^{1/\alpha}}^{2}(r^{\alpha}\wedge
t)\log^{\beta_{3}}\bigg{(}e+\frac{r}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{r}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}$
$\displaystyle\leq t\int_{y_{d}\vee
t^{1/\alpha}}^{2}\log^{\beta_{3}}\bigg{(}e+\frac{r}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{r}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}$
$\displaystyle\leq c_{6}t\log^{\beta_{3}}\bigg{(}e+\frac{1}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{3}+\beta_{4}+1}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}.$ (7.48)

On the other hand, using (10.2), we see that

$\displaystyle\int_{x_{d}}^{y_{d}\vee t^{1/\alpha}}(r^{\alpha}\wedge
t)\log^{\beta_{3}}\bigg{(}e+\frac{r}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{r}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}$
$\displaystyle\leq\log^{\beta_{3}}(e+1)\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\int_{x_{d}}^{y_{d}\vee
t^{1/\alpha}}\log^{\beta_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r^{1-\alpha}}$
$\displaystyle\leq c_{7}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\int_{x_{d}}^{y_{d}\vee
t^{1/\alpha}}\bigg{(}\frac{y_{d}\vee
t^{1/\alpha}}{r}\bigg{)}^{\alpha/2}\frac{dr}{r^{1-\alpha}}$ $\displaystyle\leq
c_{8}(y_{d}\vee
t^{1/\alpha})^{\alpha}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}$ $\displaystyle=c_{8}(y_{d}\vee
t^{1/\alpha})^{\beta_{2}-\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}.$ (7.49)

Combining (7)–(7), we show that (7) holds true. The proof is complete. $\Box$

##  8\. Proofs of Theorems 1.1 and 1.2

Proof of Theorem 1.1. Since $\overline{p}(t,x,y)$ is jointly continuous (see
Remark 3.12), it suffices to prove that (1.1)–(1.1) hold for
$(t,x,y)\in(0,\infty)\times{\mathbb{R}}^{d}_{+}\times{\mathbb{R}}^{d}_{+}$.

We first note that by (A3) and (6.4),

$\displaystyle
t^{-d/\alpha}\wedge\big{(}tJ(x+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})\big{)}\asymp
t^{-d/\alpha}\wedge\frac{tB_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(x+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})}{|x-y|^{d+\alpha}}$
$\displaystyle\asymp\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right)B_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(x+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d}),$
(8.1)

which implies the second comparison in (1.1).

(i) Using (6.3), we get the lower heat kernel estimate in the first comparison
in (1.1) from Proposition 6.2 and the upper heat kernel estimate from
Corollary 7.9.

(ii) For (1.4), using (6.3), we get the lower heat kernel estimate from
Proposition 6.6 (see Remark 6.7) and the upper heat kernel estimate from
Theorem 7.10.

(iii) Using (6.3), we get the lower heat kernel estimate in (1.1) from
Proposition 6.9. The upper heat kernel estimate in (1.1) follows from Theorem
7.10.

From the comparisons in (i)-(iii) and (6.4), we have that
$\overline{p}(t,x,y)\asymp t^{-d/\alpha}$ when $t^{1/\alpha}\geq|x-y|/8$ and
this implies that (1.1) holds for $t^{1/\alpha}\geq|x-y|/8$. Moreover, (1.1)
for $\beta_{2}<\alpha+\beta_{1}$ follows from (1.1), (6.3) and (8).

By (A3) and (6.3), we have that when $t^{1/\alpha}<|x-y|/8$,

$\displaystyle t\int_{(x_{d}\vee y_{d}\vee
t^{1/\alpha})\wedge(|x-y|/4)}^{|x-y|/2}J(x+t^{1/\alpha}{\mathbf{e}}_{d},x+r{\mathbf{e}}_{d})J(x+r{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})r^{d-1}dr$
$\displaystyle\asymp\frac{t}{|x-y|^{d+\alpha}}\int_{(x_{d}\vee y_{d}\vee
t^{1/\alpha})\wedge(|x-y|/4)}^{|x-y|/2}\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+r{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+r{\mathbf{e}}_{d},y)\frac{dr}{r^{\alpha+1}}.$
(8.2)

Thus, we see that, for $\beta_{2}=\alpha+\beta_{1}$ and
$t^{1/\alpha}<|x-y|/8$, (1.1) follows from (1.1), (8) and Lemma 6.8, and that,
for $\beta_{2}>\alpha+\beta_{1}$ and $t^{1/\alpha}<|x-y|/8$, the lower bound
in (1.1) follows from (8) and Proposition 6.2 and Lemma 6.4.

We have from (8), (6) and (6.3) that, when $t^{1/\alpha}<|x-y|/8$,

$\displaystyle t\int_{(x_{d}\vee y_{d}\vee
t^{1/\alpha})\wedge(|x-y|/4)}^{|x-y|/2}J(x+t^{1/\alpha}{\mathbf{e}}_{d},x+r{\mathbf{e}}_{d})J(x+r{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})r^{d-1}dr$
$\displaystyle\geq\frac{c_{1}t}{|x-y|^{d+2\alpha}}B_{\beta_{1},\beta_{1},0,\beta_{3}}(x+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{((x_{d}\wedge
y_{d})\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}.$ (8.3)

Now, for $\beta_{2}>\alpha+\beta_{1}$ and $t^{1/\alpha}<|x-y|/8$, the upper
bound in (1.1) follows from (8), (8) and the upper bound in (1.4).

Finally, from the joint continuity of $\overline{p}(t,x,y)$ and upper heat
kernel estimates, we deduce that $\overline{Y}$ is a Feller process and finish
the proof by Remark 3.12. $\Box$

Proof of Theorem 1.2. The second comparison in (1.9) follows from Corollary
6.3. By (6.3), Theorem 1.1 and Propositions 6.2, 6.6 and 6.9,
$p^{\kappa}(t,x,y)\geq
c_{1}\big{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\big{)}^{q_{\kappa}}\big{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\big{)}^{q_{\kappa}}\overline{p}(t,x,y)$
for all
$(t,x,y)\in(0,\infty)\times{\mathbb{R}}^{d}_{+}\times{\mathbb{R}}^{d}_{+}$. On
the other hand, by (6.3), Theorem 1.1, Corollary 7.9, and Theorem 7.10,
$p^{\kappa}(t,x,y)\leq
c_{2}\big{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\big{)}^{q_{\kappa}}\big{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\big{)}^{q_{\kappa}}\overline{p}(t,x,y)$
for all
$(t,x,y)\in(0,\infty)\times{\mathbb{R}}^{d}_{+}\times{\mathbb{R}}^{d}_{+}$ and
hence (1.9) holds true.

Note that for each $(t,y)\in(0,\infty)\times{\mathbb{R}}^{d}_{+}$, the map
$x\mapsto p^{\kappa}(t,x,y)$ vanishes continuously on
$\partial{\mathbb{R}}^{d}_{+}$. Hence, using the joint continuity of
$p^{\kappa}(t,x,y)$ and upper heat kernel estimates, we deduce that
$Y^{\kappa}$ is a Feller process. By Remark 3.12, the proof is complete.
$\Box$

##  9\. Green function estimates

In this section, we give proofs of Theorems 1.3 and 1.4.

Proof of Theorem 1.3. When $d>\alpha$, we get the upper bound of (1.11) from
Corollary 3.13. On the other hand, by Lemma 3.7 and Remark 3.12, we have

$\displaystyle\overline{G}(x,y)\geq\int_{|x-y|^{\alpha}}^{\infty}\overline{p}(t,x,y)dt\geq
c_{1}\int_{|x-y|^{\alpha}}^{\infty}t^{-d/\alpha}dt=\begin{cases}\frac{c_{1}\alpha}{d-\alpha}|x-y|^{-d+\alpha}\;\;&\mbox{if
}d>\alpha;\\\\[4.0pt] \infty\;\;&\mbox{if }d\leq\alpha.\end{cases}$

The proof is complete. $\Box$

In the remainder of this section, we assume the setting of Theorem 1.4 holds
and denote by $q$ the constant $q_{\kappa}$ in (1.8), which is strictly
positive.

Let $x,y\in{\mathbb{R}}^{d}_{+}$ be such that $x_{d}\leq y_{d}$ and $|x-y|=1$.
From Theorem 7.5, Proposition 6.2 and (6.4), we have for $t\leq 1$,

$\displaystyle p^{\kappa}(t,x,y)$ $\displaystyle\leq
Ct\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\big{(}(x_{d}\vee
t^{1/\alpha})\wedge 1\big{)}^{\beta_{1}}\big{(}(y_{d}\vee
t^{1/\alpha}\big{)}\wedge 1\big{)}^{\beta_{2}\wedge(\alpha/2+\beta_{1})}$
$\displaystyle\quad\times\log^{\beta_{3}}\bigg{(}e+\frac{(y_{d}\vee
t^{1/\alpha})\wedge 1}{(x_{d}\vee t^{1/\alpha})\wedge
1}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{(y_{d}\vee t^{1/\alpha})\wedge
1}\bigg{)},$ (9.1) $\displaystyle p^{\kappa}(t,x,y)$ $\displaystyle\geq
ct\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\big{(}(x_{d}\vee
t^{1/\alpha})\wedge 1\big{)}^{\beta_{1}}\big{(}(y_{d}\vee
t^{1/\alpha}\big{)}\wedge 1\big{)}^{\beta_{2}}$
$\displaystyle\quad\times\log^{\beta_{3}}\bigg{(}e+\frac{(y_{d}\vee
t^{1/\alpha})\wedge 1}{(x_{d}\vee t^{1/\alpha})\wedge
1}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{(y_{d}\vee t^{1/\alpha})\wedge
1}\bigg{)},$ (9.2)

and for $t>1$,

$\displaystyle p^{\kappa}(t,x,y)\asymp
t^{-d/\alpha}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}.$
(9.3)

######  Lemma 9.1.

Let $x,y\in{\mathbb{R}}^{d}_{+}$ be such that $x_{d}\leq y_{d}$ and $|x-y|=1$.
Set $\widehat{q}:=2\alpha+\beta_{1}+\beta_{2}-q$. Then we have

$\int_{0}^{1}p^{\kappa}(t,x,y)dt\asymp\begin{cases}(x_{d}\wedge
1)^{q}(y_{d}\wedge 1)^{q}&\mbox{ if }q<\widehat{q},\\\\[5.0pt]
\displaystyle(x_{d}\wedge 1)^{q}(y_{d}\wedge
1)^{q}\log^{\beta_{4}+1}\left(e+\frac{1}{y_{d}\wedge 1}\right)&\mbox{ if
}q=\widehat{q},\\\\[8.0pt] \displaystyle(x_{d}\wedge 1)^{q}(y_{d}\wedge
1)^{\widehat{q}}\log^{\beta_{4}}\left(e+\frac{1}{y_{d}\wedge 1}\right)&\mbox{
if }q>\widehat{q},\end{cases}$

where the comparison constant is independent of $x,y$.

Proof. Set $G_{1}:=\int_{0}^{1}p^{\kappa}(t,x,y)dt$ and
$\widehat{\beta}_{2}:=\beta_{2}\wedge(\alpha/2+\beta_{1})$.

(Case 1) $x_{d}\geq 1$: We have from (9) and (9) that
$G_{1}\asymp\int_{0}^{1}tdt\asymp 1.$

(Case 2) $y_{d}\geq 1>x_{d}$: By (9) and (7.6),

$\displaystyle G_{1}$ $\displaystyle\geq
c_{1}x_{d}^{\beta_{1}}\int_{1/2}^{1}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q-\beta_{1}}dt\geq
2^{-1}c_{1}x_{d}^{\beta_{1}}\int_{1/2}^{1}(1\wedge
x_{d})^{q-\beta_{1}}dt=2^{-2}c_{1}x_{d}^{q}.$

Besides, since $\alpha+\beta_{1}-q>0$, we get from (9) and (7.6) that

$\displaystyle G_{1}$ $\displaystyle\leq
c_{2}x_{d}^{\beta_{1}}\int_{0}^{1}t\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{t^{1/\alpha}}\bigg{)}dt$
$\displaystyle\leq
c_{3}x_{d}^{q}\int_{0}^{1}t^{(\alpha+\beta_{1}-q)/\alpha}\log^{\beta_{3}}\left(e+\frac{1}{t^{1/\alpha}}\right)dt\leq
c_{4}x_{d}^{q}.$

(Case 3) $1>y_{d}\geq x_{d}$: Note that

$\displaystyle\int_{y_{d}^{\alpha}}^{1}t^{-1+(\widehat{q}-q)/\alpha}\log^{\beta_{4}}\bigg{(}e+\frac{1}{t^{1/\alpha}}\bigg{)}dt\asymp\begin{cases}1&\mbox{
if }\,q<\widehat{q};\\\\[2.0pt]
\displaystyle\log^{\beta_{4}+1}(e+1/y_{d})&\mbox{ if
}\,q=\widehat{q};\\\\[2.0pt] \displaystyle
y_{d}^{\widehat{q}-q}\log^{\beta_{4}}(e+1/y_{d})&\mbox{ if
}\,q>\widehat{q}.\end{cases}$ (9.4)

For the lower bound, we get from (9) and (9.4) that

$\displaystyle G_{1}$ $\displaystyle\geq
c_{5}x_{d}^{q}y_{d}^{q}\int_{y_{d}^{\alpha}}^{1}t^{(\alpha+\beta_{1}+\beta_{2}-2q)/\alpha}\log^{\beta_{4}}\bigg{(}e+\frac{1}{t^{1/\alpha}}\bigg{)}dt$
$\displaystyle=c_{5}x_{d}^{q}y_{d}^{q}\int_{y_{d}^{\alpha}}^{1}t^{-1+(\widehat{q}-q)/\alpha}\log^{\beta_{4}}\bigg{(}e+\frac{1}{t^{1/\alpha}}\bigg{)}dt$
$\displaystyle\asymp x_{d}^{q}y_{d}^{q}\times\begin{cases}1&\mbox{ if
}\,q<\widehat{q};\\\\[2.0pt] \displaystyle\log^{\beta_{4}+1}(e+1/y_{d})&\mbox{
if }\,q=\widehat{q};\\\\[2.0pt] \displaystyle
y_{d}^{\widehat{q}-q}\log^{\beta_{4}}(e+1/y_{d})&\mbox{ if
}\,q>\widehat{q}.\end{cases}$

For the upper bound, we see from (9) that

$\displaystyle G_{1}$ $\displaystyle\leq
c_{6}x_{d}^{\beta_{1}}y_{d}^{\widehat{\beta}_{2}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}}{x_{d}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{y_{d}}\bigg{)}\int_{0}^{x_{d}^{\alpha}}tdt$
$\displaystyle\quad+c_{6}x_{d}^{q}y_{d}^{\widehat{\beta}_{2}}\log^{\beta_{4}}\left(e+\frac{1}{y_{d}}\right)\int_{x_{d}^{\alpha}}^{y_{d}^{\alpha}}t^{(\alpha+\beta_{1}-q)/\alpha}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}}{t^{1/\alpha}}\bigg{)}dt$
$\displaystyle\quad+c_{6}x_{d}^{q}y_{d}^{q}\int_{y_{d}^{\alpha}}^{1}t^{(\alpha+\beta_{1}+\widehat{\beta}_{2}-2q)/\alpha}\log^{\beta_{4}}\left(e+\frac{1}{t^{1/\alpha}}\right)dt$
$\displaystyle=:c_{6}(I+II+III).$

Using (10.1), we obtain

$\displaystyle I\leq
c_{7}x_{d}^{2\alpha+\beta_{1}}y_{d}^{\widehat{\beta}_{2}}\bigg{(}\frac{y_{d}}{x_{d}}\bigg{)}^{2\alpha+\beta_{1}-q}\log^{\beta_{4}}\left(e+\frac{1}{y_{d}}\right)=c_{7}x_{d}^{q}y_{d}^{2\alpha+\beta_{1}+\widehat{\beta}_{2}-q}\log^{\beta_{4}}\left(e+\frac{1}{y_{d}}\right).$
(9.5)

Since $\alpha+\beta_{1}>q$, the map $t\mapsto
t^{(\alpha+\beta_{1}-q)/\alpha}\log^{\beta_{3}}(e+y_{d}/t^{1/\alpha})$ is
almost increasing. Therefore,

$\displaystyle II\leq
c_{8}x_{d}^{q}y_{d}^{\alpha+\beta_{1}+\widehat{\beta}_{2}-q}\log^{\beta_{4}}\left(e+\frac{1}{y_{d}}\right)\int_{x_{d}^{\alpha}}^{y_{d}^{\alpha}}dt\leq
c_{8}x_{d}^{q}y_{d}^{2\alpha+\beta_{1}+\widehat{\beta}_{2}-q}\log^{\beta_{4}}\left(e+\frac{1}{y_{d}}\right).$
(9.6)

We consider the cases $q<\widehat{q}$ and $q\geq\widehat{q}$ separately.
First, suppose that $q<\widehat{q}$, which is equivalent to
$q<\alpha+(\beta_{1}+\beta_{2})/2$. Since $q<\alpha+\beta_{1}$, it follows
that $q<\alpha+(\beta_{1}+\widehat{\beta}_{2})/2$. Using (10.1), since
$y_{d}<1$, we see from (9.5)–(9.6) that

$\displaystyle I+II\leq
c_{9}x_{d}^{q}y_{d}^{q}y_{d}^{2\alpha+\beta_{1}+\widehat{\beta}_{2}-2q}\log^{\beta_{4}}\left(e+\frac{1}{y_{d}}\right)\leq
c_{10}x_{d}^{q}y_{d}^{q}.$

Moreover, since $(\alpha+\beta_{1}+\widehat{\beta}_{2}-2q)/\alpha>-1$, we get

$\displaystyle III\leq
c_{7}x_{d}^{q}y_{d}^{q}\int_{0}^{1}t^{(\alpha+\beta_{1}+\widehat{\beta}_{2}-2q)/\alpha}\log^{\beta_{4}}\left(e+\frac{1}{t^{1/\alpha}}\right)dt\leq
c_{11}x_{d}^{q}y_{d}^{q}.$

Therefore, we arrive at the result in this case.

Suppose that $q\geq\widehat{q}$. In this case, we have
$2\alpha+\beta_{1}+\beta_{2}=q+\widehat{q}\leq 2q<2\alpha+2\beta_{1}$. Thus,
$\beta_{2}<\beta_{1}$ and $\widehat{\beta}_{2}=\beta_{2}$. Then we deduce the
desired upper bound from (9.4)–(9.6). The proof is complete. $\Box$

######  Lemma 9.2.

Let $x,y\in{\mathbb{R}}^{d}_{+}$ be such that $x_{d}\leq y_{d}$ and $|x-y|=1$.
Then we have

$\int_{1}^{\infty}p^{\kappa}(t,x,y)dt\asymp\begin{cases}(x_{d}\wedge
1)^{q}(y_{d}\wedge 1)^{q}&\mbox{ if }d>\alpha,\\\\[5.0pt] (x_{d}\wedge
1)^{q}(y_{d}\wedge 1)^{q}\log\big{(}e+(x_{d}\vee 1)\big{)}&\mbox{ if
}d=1=\alpha,\\\\[5.0pt] \displaystyle(x_{d}\wedge 1)^{q}(y_{d}\wedge
1)^{q}(x_{d}\vee 1)^{\alpha-1}&\mbox{ if }d=1<\alpha,\end{cases}$

where the comparison constant is independent of $x,y$.

Proof. By (9.3), we have

$\int_{1}^{\infty}p^{\kappa}(t,x,y)dt\asymp\int_{1}^{\infty}t^{-d/\alpha}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}dt=:G_{2}.$

If $d>\alpha$, then by Lemma 10.12, $G_{2}\asymp(x_{d}\wedge
1)^{q}(y_{d}\wedge 1)^{q}$.

If $d=1=\alpha$, then using Lemma 10.12, we get

$\displaystyle G_{2}$ $\displaystyle\asymp\int_{1}^{x_{d}\vee
1}t^{-1}dt+x_{d}^{q}\int_{x_{d}\vee
1}^{\infty}t^{-1-q}\left(1\wedge\frac{y_{d}}{t}\right)^{q}dt$
$\displaystyle\asymp{\bf 1}_{\\{x_{d}>1\\}}\log(x_{d})+x_{d}^{q}(x_{d}\vee
1)^{-q}\left(1\wedge\frac{y_{d}}{x_{d}\vee 1}\right)^{q}$
$\displaystyle\asymp\begin{cases}\log(e+x_{d})&\text{ if }x_{d}>1\\\
x_{d}^{q}(y_{d}\wedge 1)^{q}&\text{ if }x_{d}\leq
1\end{cases}\asymp(x_{d}\wedge 1)^{q}(y_{d}\wedge 1)^{q}\log(e+(x_{d}\vee
1)).$

If $d=1<\alpha$, then since $y_{d}>2$ implies $x_{d}\geq y_{d}-|x-y|\geq
y_{d}/2>1$, and $\alpha-q-1\leq 0$, we get

$\displaystyle G_{2}$ $\displaystyle\asymp{\bf
1}_{\\{y_{d}>2\\}}\bigg{(}\int_{1}^{x_{d}^{\alpha}}t^{-1/\alpha}dt+x_{d}^{q}\int_{x_{d}^{\alpha}}^{(2y_{d})^{\alpha}}t^{-(q+1)/\alpha}dt+x_{d}^{q}y_{d}^{q}\int_{(2y_{d})^{\alpha}}^{\infty}t^{-(2q+1)/\alpha}dt\bigg{)}$
$\displaystyle\quad+{\bf 1}_{\\{y_{d}\leq
2\\}}x_{d}^{q}y_{d}^{q}\int_{1}^{\infty}t^{-(2q+1)/\alpha}dt$
$\displaystyle\asymp{\bf
1}_{\\{y_{d}>2\\}}\big{(}x_{d}^{\alpha-1}+x_{d}^{q+\alpha-(q+1)}+x_{d}^{q}y_{d}^{\alpha-q-1}\big{)}+{\bf
1}_{\\{y_{d}\leq 2\\}}x_{d}^{q}y_{d}^{q}$ $\displaystyle\asymp{\bf
1}_{\\{y_{d}>2\\}}x_{d}^{\alpha-1}+{\bf 1}_{\\{y_{d}\leq
2\\}}x_{d}^{q}y_{d}^{q}\asymp(x_{d}\wedge 1)^{q}(y_{d}\wedge 1)^{q}(x_{d}\vee
1)^{\alpha-1}.$

The proof is complete. $\Box$

Proof of Theorem 1.4. From scaling property (3.30), we obtain

$\displaystyle
G^{\kappa}(x,y)=|x-y|^{-d+\alpha}\,G^{\kappa}(x/|x-y|,y/|x-y|),\quad
x,y\in{\mathbb{R}}^{d}_{+}.$ (9.7)

Using (9.7), symmetry and Lemmas 9.1 and 9.2, we deduce the desired result.
$\Box$

##  10\. Appendix

Note that for any ${\varepsilon}>0$,

$\log(e+r)<(2+{\varepsilon}^{-1})r^{\varepsilon}\quad\text{for all}\;r\geq 1,$
(10.1)
$\frac{\log(e+ar)}{\log(e+r)}<(1+{\varepsilon}^{-1})a^{{\varepsilon}}\quad\quad\text{for
all }a\geq 1\text{ and }r>0.$ (10.2)

Recall the definition of $A_{b_{1},b_{2},b_{3},b_{4}}(t,x,y)$ from (6).

######  Lemma 10.1.

Let $b_{1},b_{2},b_{3},b_{4}\geq 0$.

(i) If $b_{1}>0$, then for any ${\varepsilon}\in(0,b_{1}]$, there exists
$c_{1}>0$ such that

$A_{b_{1},b_{2},b_{3},b_{4}}(t,x,y)\leq
c_{1}A_{b_{1}-{\varepsilon},b_{2},0,b_{4}}(t,x,y)\quad\text{ for all }t\geq
0,\,x,y\in{\mathbb{R}}^{d}_{+}.$

(ii) If $b_{2}>0$, then for any ${\varepsilon}\in(0,b_{2}]$, there exists
$c_{2}>0$ such that

$A_{b_{1},b_{2},b_{3},b_{4}}(t,x,y)\leq
c_{2}A_{b_{1},b_{2}-{\varepsilon},b_{3},0}(t,x,y)\quad\text{ for all }t\geq
0,\,x,y\in{\mathbb{R}}^{d}_{+}.$

Proof. The results follow from (10.1). $\Box$

######  Lemma 10.2.

Let $b_{1},b_{3}\geq 0$. Suppose that $b_{1}>0$ if $b_{3}>0$. Then there
exists a constant $C>0$ such that for all $t\geq 0$ and
$x,y\in{\mathbb{R}}^{d}_{+}$,

$\displaystyle A_{b_{1},0,b_{3},0}(t,x,y)$ $\displaystyle\leq
C\,\bigg{(}\frac{x_{d}\vee t^{1/\alpha}}{|x-y|}\wedge
1\bigg{)}^{b_{1}}\log^{b_{3}}\bigg{(}e+\frac{(y_{d}\vee
t^{1/\alpha})\wedge|x-y|}{(x_{d}\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}.$

Proof. When $x_{d}\leq y_{d}$, the result is clear. Assume that $x_{d}>y_{d}$.
Set $r=\frac{x_{d}\vee t^{1/\alpha}}{|x-y|}\wedge 1$ and $s=\frac{y_{d}\vee
t^{1/\alpha}}{|x-y|}\wedge 1$. Then $0<s\leq r\leq 1$ and the desired
inequality is equivalent to

$\log^{b_{3}}(e+r/s)\log^{-b_{3}}(e+s/r)\leq C(r/s)^{b_{1}}.$ (10.3)

If $b_{3}=0$, then (10.3) clearly holds with $C=1$. If $b_{3}>0$ and
$b_{1}>0$, then we get from (10.1) that

$\log^{b_{3}}(e+r/s)\log^{-b_{3}}(e+s/r)\leq\log^{b_{3}}(e+r/s)\leq
c_{1}(r/s)^{b_{1}}.$

This proves the lemma. $\Box$

######  Remark 10.3.

Recall that
$B_{b_{1},b_{2},b_{3},b_{4}}(x,y)=A_{b_{1},b_{2},b_{3},b_{4}}(0,x,y)$ for
$x,y\in{\mathbb{R}}^{d}_{+}$. Therefore, the results of Lemmas 10.1 and 10.2
hold true with $B_{b_{1},0,b_{3},0}(x,y)$ instead of
$A_{b_{1},0,b_{3},0}(t,x,y)$.

######  Lemma 10.4.

Let $\gamma\in{\mathbb{R}}$, $b\geq 0$ and $t,k,l>0$. Suppose that either
$\gamma<\alpha$ or $k\geq t^{1/\alpha}$. Then we have

$\int_{0}^{t}\bigg{(}1\wedge\frac{k}{s^{1/\alpha}}\bigg{)}^{\gamma}\log^{b}\bigg{(}e+\frac{l}{k\vee
s^{1/\alpha}}\bigg{)}ds\leq
Ct\bigg{(}1\wedge\frac{k}{t^{1/\alpha}}\bigg{)}^{\gamma}\log^{b}\bigg{(}e+\frac{l}{k\vee
t^{1/\alpha}}\bigg{)},$

where $C>0$ is a constant which depends only on $\gamma$ and $b$.

Proof. If $k\geq t^{1/\alpha}$, then the desired inequality holds since the
left hand side is $t\log^{b}(e+l/k)$.

Suppose that $k<t^{1/\alpha}$ and $\gamma<\alpha$. Let ${\varepsilon}>0$ be
such that $\gamma+b{\varepsilon}<\alpha$. Then the left hand side is

$\displaystyle\log^{b}\bigg{(}e+\frac{l}{k}\bigg{)}\int_{0}^{k^{\alpha}}ds+k^{\gamma}\int_{k^{\alpha}}^{t}\frac{1}{s^{\gamma/\alpha}}\log^{b}\bigg{(}e+\frac{l}{s^{1/\alpha}}\bigg{)}ds$
$\displaystyle\leq
k^{\alpha}\log^{b}\bigg{(}e+\frac{l}{k}\bigg{)}+c_{1}k^{\gamma}\log^{b}\bigg{(}e+\frac{l}{t^{1/\alpha}}\bigg{)}\int_{k^{\alpha}}^{t}\frac{t^{b{\varepsilon}/\alpha}}{s^{\gamma/\alpha+b{\varepsilon}/\alpha}}ds$
$\displaystyle\leq
c_{2}k^{\alpha}\bigg{(}\frac{t^{1/\alpha}}{k}\bigg{)}^{\alpha-\gamma}\log^{b}\bigg{(}e+\frac{l}{t^{1/\alpha}}\bigg{)}+c_{2}k^{\gamma}t^{1-\gamma/\alpha}\log^{b}\bigg{(}e+\frac{l}{t^{1/\alpha}}\bigg{)}$
$\displaystyle=2c_{2}t\bigg{(}\frac{k}{t^{1/\alpha}}\bigg{)}^{\gamma}\log^{b}\bigg{(}e+\frac{l}{t^{1/\alpha}}\bigg{)}=2c_{2}t\bigg{(}1\wedge\frac{k}{t^{1/\alpha}}\bigg{)}^{\gamma}\log^{b}\bigg{(}e+\frac{l}{t^{1/\alpha}}\bigg{)}.$

We used (10.2) in both inequalities above. $\Box$

######  Lemma 10.5.

Let $b_{1},b_{2}\in{\mathbb{R}}$, $b_{3},b_{4}\geq 0$ and $t,x_{d},y_{d}>0$.
Suppose that (1) either $b_{1}>q-\alpha$ or $x_{d}\geq t^{1/\alpha}$, and (2)
either $b_{2}>q-\alpha$ or $y_{d}\geq t^{1/\alpha}$. Then we have

$\displaystyle\int_{0}^{t}\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
s^{1/\alpha}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
s^{1/\alpha}}\bigg{)}ds$ $\displaystyle\leq
Ct\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}$

where $C>0$ is a constant which depends only on $b_{1},b_{2},b_{3}$ and
$b_{4}$.

Proof. Observe that

$\displaystyle\int_{0}^{\frac{t}{2}}\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
s^{1/\alpha}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
s^{1/\alpha}}\bigg{)}ds$ $\displaystyle\leq
c_{1}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\int_{0}^{\frac{t}{2}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
s^{1/\alpha}}\bigg{)}ds$

and

$\displaystyle\int_{\frac{t}{2}}^{t}\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
s^{1/\alpha}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
s^{1/\alpha}}\bigg{)}ds$ $\displaystyle\leq
c_{2}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\int_{0}^{\frac{t}{2}}\bigg{(}1\wedge\frac{x_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{1}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee
t^{1/\alpha}}{x_{d}\vee s^{1/\alpha}}\bigg{)}ds.$

Using Lemma 10.4 twice, we arrive at the result. $\Box$

######  Lemma 10.6.

Let $b_{1},b_{2}\in{\mathbb{R}}$, $b_{3},b_{4}\geq 0$ and $t,x_{d},y_{d}>0$.
Suppose that (1) either $b_{1}>q-\alpha$ or $x_{d}\geq t^{1/\alpha}$, and (2)
either $b_{2}>q-\alpha$ or $y_{d}\geq t^{1/\alpha}$. Then we have that for
$y_{d}\vee t^{1/\alpha}<2$,

$\displaystyle\int_{0}^{t}s\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}$
$\displaystyle\quad\times\int_{y_{d}\vee
s^{1/\alpha}}^{2}\log^{b_{3}}\bigg{(}e+\frac{r}{y_{d}\vee
s^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{r}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}\,ds$
$\displaystyle\leq
Ct\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}$
$\displaystyle\quad\times\int_{y_{d}}^{2}(r^{\alpha}\wedge
t)\log^{b_{3}}\bigg{(}e+\frac{r}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{r}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r},$

where $C>0$ is a constant which depends only on $b_{1},b_{2},b_{3}$ and
$b_{4}$.

Proof. Using Fubini’s theorem and Lemma 10.4 twice as in the proof of Lemma
10.5, we get

$\displaystyle\int_{0}^{t}s\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}$
$\displaystyle\quad\times\int_{y_{d}\vee
s^{1/\alpha}}^{2}\log^{b_{3}}\bigg{(}e+\frac{r}{y_{d}\vee
s^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{r}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}\,ds$
$\displaystyle=$ $\displaystyle\int_{y_{d}}^{2}\int_{0}^{r^{\alpha}\wedge
t}s\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}$
$\displaystyle\quad\times\log^{b_{3}}\bigg{(}e+\frac{r}{y_{d}\vee
s^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{r}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\,ds\frac{dr}{r}$
$\displaystyle\leq$ $\displaystyle\int_{y_{d}}^{2}(r^{\alpha}\wedge
t)\int_{0}^{t}\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}$
$\displaystyle\quad\times\log^{b_{3}}\bigg{(}e+\frac{r}{y_{d}\vee
s^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{r}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\,ds\frac{dr}{r}$
$\displaystyle\leq$
$\displaystyle\,Ct\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}$
$\displaystyle\quad\times\int_{y_{d}}^{2}(r^{\alpha}\wedge
t)\log^{b_{3}}\bigg{(}e+\frac{r}{x_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{r}{y_{d}\vee
t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}.$

$\Box$

######  Lemma 10.7.

There is a constant $C>0$ such that for all $x\in{\mathbb{R}}^{d}_{+}$ and
$A>0$,

$\displaystyle\int_{B_{+}(x,A)}z_{d}^{-1/2}dz\leq CA^{d}(x_{d}\vee A)^{-1/2}.$

Proof. We have

$\displaystyle\int_{B_{+}(x,A)}z_{d}^{-1/2}dz\leq\int_{\widetilde{z}\in{\mathbb{R}}^{d-1},\,|\widetilde{x}-\widetilde{z}|<A}d\widetilde{z}\int_{(x_{d}-A)\vee
0}^{x_{d}+A}z_{d}^{-1/2}dz_{d}\leq c_{1}A^{d-1}\int_{(x_{d}-A)\vee
0}^{x_{d}+A}z_{d}^{-1/2}dz_{d}.$ (10.4)

If $x_{d}\geq 2A$, then

$\displaystyle\int_{(x_{d}-A)\vee
0}^{x_{d}+A}z_{d}^{-1/2}dz_{d}\leq\frac{1}{(x_{d}/2)^{1/2}}\int_{x_{d}-A}^{x_{d}+A}dz_{d}=2^{3/2}Ax_{d}^{-1/2}.$
(10.5)

If $x_{d}<2A$, then

$\displaystyle\int_{(x_{d}-A)\vee
0}^{x_{d}+A}z_{d}^{-1/2}dz_{d}\leq\int_{0}^{4A}z_{d}^{-1/2}dz_{d}=4A^{1/2}.$
(10.6)

Combining (10.4) with (10.5)–(10.6), we arrive at the result. $\Box$

######  Lemma 10.8.

(i) There is a constant $C>0$ such that for all $x\in{\mathbb{R}}^{d}_{+}$ and
$0<A<x_{d}$,

$\displaystyle\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\geq|x-z|>A}\frac{dz}{z_{d}^{1/2}|x-z|^{d+\alpha}}\leq
Cx_{d}^{-1/2}A^{-\alpha}.$

(ii) Let ${\varepsilon}\in(0,1)$ and $\delta>0$. There is a constant
$C^{\prime}>0$ such that for all $x\in{\mathbb{R}}^{d}_{+}$ and $A\geq x_{d}$,

$\displaystyle\int_{z\in{\mathbb{R}}^{d}_{+},\,|x-z|>A}\frac{dz}{z_{d}^{{\varepsilon}}\,|x-z|^{d+\delta}}\leq
C^{\prime}A^{-{\varepsilon}-\delta}.$

Proof. Without loss of generality, we assume $x=(\widetilde{0},x_{d})$.

(i) Note that

$\displaystyle\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\geq|x-z|>A}\frac{dz}{z_{d}^{1/2}|x-z|^{d+\alpha}}$
$\displaystyle=\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\geq|x-z|>A,\,|\widetilde{z}|\leq|x_{d}-z_{d}|}\frac{dz}{z_{d}^{1/2}|x-z|^{d+\alpha}}+\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\geq|x-z|>A,\,|\widetilde{z}|>|x_{d}-z_{d}|}\frac{dz}{z_{d}^{1/2}|x-z|^{d+\alpha}}$
$\displaystyle=:I+II.$

First, we have

$\displaystyle I$
$\displaystyle\leq\int_{x_{d}\geq|x_{d}-z_{d}|>\frac{A}{2}}\frac{1}{z_{d}^{1/2}|x_{d}-z_{d}|^{d+\alpha}}\int_{\widetilde{z}\in{\mathbb{R}}^{d-1},\,|\widetilde{z}|\leq|x_{d}-z_{d}|}d\widetilde{z}dz_{d}$
$\displaystyle\leq
c_{1}\bigg{(}\int_{0}^{\frac{x_{d}}{2}}\frac{dz_{d}}{z_{d}^{1/2}|x_{d}-z_{d}|^{1+\alpha}}+\int_{\frac{x_{d}}{2}}^{x_{d}-\frac{A}{2}}\frac{dz_{d}}{z_{d}^{1/2}|x_{d}-z_{d}|^{1+\alpha}}+\int_{x_{d}+\frac{A}{2}}^{2x_{d}}\frac{dz_{d}}{z_{d}^{1/2}|x_{d}-z_{d}|^{1+\alpha}}\bigg{)}$
$\displaystyle\leq
c_{1}\bigg{(}\frac{2^{1+\alpha}}{x_{d}^{1+\alpha}}\int_{0}^{\frac{x_{d}}{2}}{z_{d}^{-1/2}}dz_{d}+\frac{1}{(x_{d}/2)^{1/2}}\int_{\frac{x_{d}}{2}}^{x_{d}-\frac{A}{2}}\frac{dz_{d}}{|x_{d}-z_{d}|^{1+\alpha}}+{x_{d}^{-1/2}}\int_{x_{d}+\frac{A}{2}}^{2x_{d}}\frac{dz_{d}}{|x_{d}-z_{d}|^{1+\alpha}}\bigg{)}$
$\displaystyle\leq
c_{2}\left(x_{d}^{-1/2-\alpha}+x_{d}^{-1/2}A^{-\alpha}+x_{d}^{-1/2}A^{-\alpha}\right)\leq
c_{3}x_{d}^{-1/2}A^{-\alpha}.$

On the other hand, we see that for any $z\in{\mathbb{R}}^{d}_{+}$ with
$x_{d}\geq|x-z|>A$ and $|\widetilde{z}|>|x_{d}-z_{d}|$,

$\displaystyle z_{d}\geq x_{d}-|x_{d}-z_{d}|\geq
x_{d}-\frac{1}{2}(|\widetilde{z}|+|x_{d}-z_{d}|)\geq
x_{d}-\frac{1}{\sqrt{2}}|x-z|\geq\left(1-\frac{1}{\sqrt{2}}\right)x_{d}.$

Hence, we also have that

$\displaystyle II$ $\displaystyle\leq
c_{4}x_{d}^{-1/2}\int_{z\in{\mathbb{R}}^{d}_{+},\,|x-z|>A}\frac{dz}{|x-z|^{d+\alpha}}\leq
c_{5}x_{d}^{-1/2}A^{-\alpha}.$

(ii) Observe that

$\displaystyle\int_{z\in{\mathbb{R}}^{d}_{+},\,|x-z|>A}\frac{dz}{z_{d}^{{\varepsilon}}\,|x-z|^{d+\delta}}$
$\displaystyle\leq\int_{z\in{\mathbb{R}}^{d}_{+},\,|x_{d}-z_{d}|\geq|\widetilde{z}|\vee\frac{A}{2}}\frac{d\widetilde{z}\,dz_{d}}{z_{d}^{{\varepsilon}}\,|x_{d}-z_{d}|^{d+\delta}}+\int_{z\in{\mathbb{R}}^{d}_{+},\,|\widetilde{z}|\geq|x_{d}-z_{d}|\vee\frac{A}{2}}\frac{d\widetilde{z}\,dz_{d}}{z_{d}^{{\varepsilon}}\,|\widetilde{z}|^{d+\delta}}=:I+II.$

Using Fubini’s theorem, since $A\geq x_{d}$, we see that

$\displaystyle II$
$\displaystyle\leq\int_{\widetilde{z}\in{\mathbb{R}}^{d-1},\,|\widetilde{z}|\geq\frac{A}{2}}\frac{1}{|\widetilde{z}|^{d+\delta}}\int_{0}^{|\widetilde{z}|+x_{d}}z_{d}^{-{\varepsilon}}dz_{d}\,d\widetilde{z}\leq
c_{1}\int_{\widetilde{z}\in{\mathbb{R}}^{d-1},\,|\widetilde{z}|\geq\frac{A}{2}}\frac{(|\widetilde{z}|+A)^{1-{\varepsilon}}}{|\widetilde{z}|^{d+\delta}}d\widetilde{z}$
$\displaystyle\leq
c_{2}\int_{\widetilde{z}\in{\mathbb{R}}^{d-1},\,|\widetilde{z}|\geq\frac{A}{2}}\frac{d\widetilde{z}}{|\widetilde{z}|^{d-1+{\varepsilon}+\delta}}\leq
c_{3}A^{-{\varepsilon}-\delta}.$

On the other hand, using the fact that
$\int_{\widetilde{z}\in{\mathbb{R}}^{d-1},\,|\widetilde{z}|\leq|x_{d}-z_{d}|}d\widetilde{z}\leq
c_{4}|x_{d}-z_{d}|^{d-1}$, we also see that

$\displaystyle I$ $\displaystyle\leq
c_{5}\bigg{(}\int_{0}^{(x_{d}-\frac{A}{2})\vee
0}\frac{dz_{d}}{z_{d}^{{\varepsilon}}\,|x_{d}-z_{d}|^{1+\delta}}+\int_{x_{d}+\frac{A}{2}}^{\infty}\frac{dz_{d}}{z_{d}^{{\varepsilon}}\,|x_{d}-z_{d}|^{1+\delta}}\bigg{)}$
$\displaystyle\leq
c_{6}\bigg{(}\frac{1}{A^{1+\delta}}\int_{0}^{(x_{d}-\frac{A}{2})\vee
0}z_{d}^{-{\varepsilon}}dz_{d}+\frac{1}{A^{{\varepsilon}}}\int_{x_{d}+\frac{A}{2}}^{\infty}\frac{dz_{d}}{|x_{d}-z_{d}|^{1+\delta}}\bigg{)}$
$\displaystyle\leq
c_{7}\bigg{(}\frac{1}{A^{1+\delta}}\Big{(}\big{(}x_{d}-\frac{A}{2}\big{)}\vee
0\Big{)}^{1-{\varepsilon}}+\frac{1}{A^{{\varepsilon}+\delta}}\bigg{)}\leq\frac{c_{6}}{A^{{\varepsilon}+\delta}},$

where we used the fact that $A\geq x_{d}$ in the last inequality. The proof is
complete. $\Box$

For $\gamma,\eta_{1},\eta_{2}\geq 0$ and $k,l>0$, define

$f_{\gamma,\eta_{1},\eta_{2},k,l}(r):=r^{\gamma}\log^{\eta_{1}}\bigg{(}e+\frac{k}{r}\bigg{)}\log^{\eta_{2}}\bigg{(}e+\frac{r}{l}\bigg{)}.$

######  Lemma 10.9.

Let $\gamma,\eta_{1},\eta_{2}\geq 0$.

(i) For any ${\varepsilon}>0$, there exist constants $C,C^{\prime}>0$ such
that for any $k,l,r>0$ and any $a\geq 1$,

$Ca^{\gamma-{\varepsilon}}\leq\frac{f_{\gamma,\eta_{1},\eta_{2},k,l}(ar)}{f_{\gamma,\eta_{1},\eta_{2},k,l}(r)}\leq
C^{\prime}a^{\gamma+{\varepsilon}}.$ (10.7)

(ii) Assume that $\gamma>0$. Then there exists a constant $C>0$ such that for
any $k,l,r>0$ and any $a\geq 1$,

$\frac{f_{\gamma,\eta_{1},\eta_{2},k,l}(ar)}{f_{\gamma,\eta_{1},\eta_{2},k,l}(r)}\geq
C.$

Proof. (i) If $\eta_{2}=0$, the second inequality in (10.7) is true with any
${\varepsilon}\geq 0$. In case $\eta_{2}>0$, for any given ${\varepsilon}>0$,
let ${\varepsilon}^{\prime}:={\varepsilon}/\eta_{2}$. We get from (10.2) that
for all $a\geq 1$ and $r>0$,

$\displaystyle\frac{f_{\gamma,\eta_{1},\eta_{2},k,l}(ar)}{f_{\gamma,\eta_{1},\eta_{2},k,l}(r)}\leq
a^{\gamma}\left(\frac{\log(e+ar/l)}{\log(e+r/l)}\right)^{\eta_{2}}\leq
a^{\gamma}\big{(}(1+1/{\varepsilon}^{\prime})a^{{\varepsilon}^{\prime}}\big{)}^{\eta_{2}}=c({\varepsilon},\eta_{2})a^{\gamma+{\varepsilon}}.$

The first inequality can be proved by a similar argument.

(ii) The desired result follows from the first inequality in (10.7) with
${\varepsilon}=\gamma$. $\Box$

######  Lemma 10.10.

Let $b_{1},b_{2},\eta_{1},\eta_{2},\gamma\geq 0$. There exists a constant
$C>0$ such that for any $x\in{\mathbb{R}}^{d}_{+}$ and $s,k,l>0$,

$\displaystyle\int_{B_{+}(x,2)}\bigg{(}1\wedge\frac{x_{d}\vee
s^{1/\alpha}}{|x-z|}\bigg{)}^{b_{1}}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{(x_{d}\vee
s^{1/\alpha})\wedge|x-z|}\bigg{)}$ $\displaystyle\hskip
51.21504pt\times\left(s^{-d/\alpha}\wedge\frac{s}{|x-z|^{d+\alpha}}\right)z_{d}^{\gamma}\log^{\eta_{1}}\bigg{(}e+\frac{k}{z_{d}}\bigg{)}\log^{\eta_{2}}\bigg{(}e+\frac{z_{d}}{l}\bigg{)}dz$
$\displaystyle\leq C(x_{d}\vee
s^{1/\alpha})^{\gamma}\log^{\eta_{1}}\bigg{(}e+\frac{k}{x_{d}\vee
s^{1/\alpha}}\bigg{)}\log^{\eta_{2}}\bigg{(}e+\frac{x_{d}\vee
s^{1/\alpha}}{l}\bigg{)}$ $\displaystyle\quad+C{\bf 1}_{\\{x_{d}\vee
s^{1/\alpha}<2\\}}s(x_{d}\vee s^{1/\alpha})^{b_{1}}\bigg{[}{\bf
1}_{\\{\gamma>\alpha+b_{1}\\}}\log^{b_{2}}\bigg{(}e+\frac{2}{x_{d}\vee
s^{1/\alpha}}\bigg{)}\log^{\eta_{1}}(e+k)\log^{\eta_{2}}\bigg{(}e+\frac{1}{l}\bigg{)}$
$\displaystyle\hskip 88.2037pt+{\bf
1}_{\\{\gamma=\alpha+b_{1}\\}}\int_{x_{d}\vee
s^{1/\alpha}}^{2}\log^{b_{2}}\bigg{(}e+\frac{r}{x_{d}\vee
s^{1/\alpha}}\bigg{)}\log^{\eta_{1}}\bigg{(}e+\frac{k}{r}\bigg{)}\log^{\eta_{2}}\bigg{(}e+\frac{r}{l}\bigg{)}\frac{dr}{r}\bigg{]}.$

Proof. Using the triangle inequality, we see that for any
$z\in{\mathbb{R}}^{d}_{+}$,

$z_{d}\leq x_{d}+|x-z|\leq 2(x_{d}\vee|x-z|).$ (10.8)

Therefore, using Lemma 10.9(i)-(ii) and Lemma 10.7, we get that

$\displaystyle\int_{B_{+}(x,s^{1/\alpha})}\bigg{(}1\wedge\frac{x_{d}\vee
s^{1/\alpha}}{|x-z|}\bigg{)}^{b_{1}}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{(x_{d}\vee
s^{1/\alpha})\wedge|x-z|}\bigg{)}\,s^{-d/\alpha}f_{\gamma,\eta_{1},\eta_{2},k,l}(z_{d})dz$
$\displaystyle=(\log^{b_{2}}(e+1))s^{-d/\alpha}\int_{B_{+}(x,s^{1/\alpha})}z_{d}^{-1/2}f_{\gamma+\frac{1}{2},\eta_{1},\eta_{2},k,l}(z_{d})dz$
$\displaystyle\leq
c_{1}s^{-d/\alpha}f_{\gamma+\frac{1}{2},\eta_{1},\eta_{2},k,l}(2(x_{d}\vee
s^{1/\alpha}))\int_{B_{+}(x,s^{1/\alpha})}z_{d}^{-1/2}dz$ $\displaystyle\leq
c_{2}(x_{d}\vee
s^{1/\alpha})^{-1/2}f_{\gamma+\frac{1}{2},\eta_{1},\eta_{2},k,l}(x_{d}\vee
s^{1/\alpha})=c_{2}f_{\gamma,\eta_{1},\eta_{2},k,l}(x_{d}\vee s^{1/\alpha}).$

When $x_{d}>s^{1/\alpha}$, we get from (10.8), Lemma 10.9(i)-(ii) and Lemma
10.8(i) that

$\displaystyle
s\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\geq|x-z|>s^{1/\alpha}}\bigg{(}1\wedge\frac{x_{d}\vee
s^{1/\alpha}}{|x-z|}\bigg{)}^{b_{1}}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{(x_{d}\vee
s^{1/\alpha})\wedge|x-z|}\bigg{)}\frac{f_{\gamma,\eta_{1},\eta_{2},k,l}(z_{d})}{|x-z|^{d+\alpha}}dz$
$\displaystyle=(\log^{b_{2}}(e+1))s\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\geq|x-z|>s^{1/\alpha}}\frac{f_{\gamma+\frac{1}{2},\eta_{1},\eta_{2},k,l}(z_{d})}{z_{d}^{1/2}|x-z|^{d+\alpha}}dz$
$\displaystyle\leq
c_{3}sf_{\gamma+\frac{1}{2},\eta_{1},\eta_{2},k,l}(2x_{d})\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\geq|x-z|>s^{1/\alpha}}\frac{1}{z_{d}^{1/2}|x-z|^{d+\alpha}}dz$
$\displaystyle\leq
c_{4}x_{d}^{-1/2}f_{\gamma+\frac{1}{2},\eta_{1},\eta_{2},k,l}(x_{d})=c_{4}f_{\gamma,\eta_{1},\eta_{2},k,l}(x_{d}\vee
s^{1/\alpha}).$

It remains to bound the integral over $\\{z\in{\mathbb{R}}^{d}_{+}:x_{d}\vee
s^{1/\alpha}<|x-z|<2\\}$ under the assumption $x_{d}\vee s^{1/\alpha}<2$. For
this, we consider the following three cases separately.

(i) Case $\gamma<\alpha+b_{1}$: Fix ${\varepsilon}\in(0,1)$ such that
$\gamma+3{\varepsilon}<\alpha+b_{1}$. Using (10.8), (10.1), Lemma 10.9(i)-(ii)
and Lemma 10.8(ii), we get

$\displaystyle s\int_{z\in{\mathbb{R}}^{d}_{+},\,|x-z|>x_{d}\vee
s^{1/\alpha}}\bigg{(}1\wedge\frac{x_{d}\vee
s^{1/\alpha}}{|x-z|}\bigg{)}^{b_{1}}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{(x_{d}\vee
s^{1/\alpha})\wedge|x-z|}\bigg{)}\frac{f_{\gamma,\eta_{1},\eta_{2},k,l}(z_{d})}{|x-z|^{d+\alpha}}dz$
$\displaystyle=s(x_{d}\vee
s^{1/\alpha})^{b_{1}}\int_{z\in{\mathbb{R}}^{d}_{+},\,|x-z|>x_{d}\vee
s^{1/\alpha}}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{x_{d}\vee
s^{1/\alpha}}\bigg{)}\frac{f_{\gamma+{\varepsilon},\eta_{1},\eta_{2},k,l}(z_{d})}{z_{d}^{{\varepsilon}}|x-z|^{d+\alpha+b_{1}}}dz$
$\displaystyle\leq c_{5}s(x_{d}\vee
s^{1/\alpha})^{b_{1}}\int_{z\in{\mathbb{R}}^{d}_{+},\,|x-z|>x_{d}\vee
s^{1/\alpha}}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{x_{d}\vee
s^{1/\alpha}}\bigg{)}\frac{f_{\gamma+{\varepsilon},\eta_{1},\eta_{2},k,l}(2|x-z|)}{z_{d}^{{\varepsilon}}|x-z|^{d+\alpha+b_{1}}}dz$
$\displaystyle\leq c_{6}s(x_{d}\vee
s^{1/\alpha})^{b_{1}}\int_{z\in{\mathbb{R}}^{d}_{+},\,|x-z|>x_{d}\vee
s^{1/\alpha}}\bigg{(}\frac{|x-z|}{x_{d}\vee
s^{1/\alpha}}\bigg{)}^{{\varepsilon}+\gamma+2{\varepsilon}}\frac{f_{\gamma+{\varepsilon},\eta_{1},\eta_{2},k,l}(x_{d}\vee
s^{1/\alpha})}{z_{d}^{{\varepsilon}}|x-z|^{d+\alpha+b_{1}}}dz$
$\displaystyle=c_{6}s(x_{d}\vee
s^{1/\alpha})^{b_{1}-\gamma-3{\varepsilon}}f_{\gamma+{\varepsilon},\eta_{1},\eta_{2},k,l}(x_{d}\vee
s^{1/\alpha})\int_{z\in{\mathbb{R}}^{d}_{+},\,|x-z|>x_{d}\vee
s^{1/\alpha}}\frac{1}{z_{d}^{\varepsilon}|x-z|^{d+\alpha+b_{1}-\gamma-3{\varepsilon}}}dz$
$\displaystyle\leq c_{7}s(x_{d}\vee s^{1/\alpha})^{-\alpha}(x_{d}\vee
s^{1/\alpha})^{-{\varepsilon}}f_{\gamma+{\varepsilon},\eta_{1},\eta_{2},k,l}(x_{d}\vee
s^{1/\alpha})\leq c_{7}f_{\gamma,\eta_{1},\eta_{2},k,l}(x_{d}\vee
s^{1/\alpha}).$

(ii) Case $\gamma>\alpha+b_{1}$: Fix ${\varepsilon}>0$ such that
$\gamma-{\varepsilon}>\alpha+b_{1}$. Using (10.8), Lemma 10.9(i)-(ii) and
(10.2), we get

$\displaystyle s\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\vee
s^{1/\alpha}<|x-z|<2}\bigg{(}1\wedge\frac{x_{d}\vee
s^{1/\alpha}}{|x-z|}\bigg{)}^{b_{1}}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{(x_{d}\vee
s^{1/\alpha})\wedge|x-z|}\bigg{)}\frac{f_{\gamma,\eta_{1},\eta_{2},k,l}(z_{d})}{|x-z|^{d+\alpha}}dz$
$\displaystyle\leq c_{8}s(x_{d}\vee
s^{1/\alpha})^{b_{1}}\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\vee
s^{1/\alpha}<|x-z|<2}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{x_{d}\vee
s^{1/\alpha}}\bigg{)}\frac{f_{\gamma,\eta_{1},\eta_{2},k,l}(2|x-z|)}{|x-z|^{d+\alpha+b_{1}}}dz$
$\displaystyle\leq c_{9}s(x_{d}\vee
s^{1/\alpha})^{b_{1}}f_{\gamma,\eta_{1},\eta_{2},k,l}(4)\log^{b_{2}}\bigg{(}e+\frac{2}{x_{d}\vee
s^{1/\alpha}}\bigg{)}\int_{z\in{\mathbb{R}}^{d}_{+},\,|x-z|<2}\frac{dz}{|x-z|^{d+\alpha+b_{1}-\gamma+{\varepsilon}}}$
$\displaystyle\leq c_{10}s(x_{d}\vee
s^{1/\alpha})^{b_{1}}f_{\gamma,\eta_{1},\eta_{2},k,l}(1)\log^{b_{2}}\bigg{(}e+\frac{2}{x_{d}\vee
s^{1/\alpha}}\bigg{)}.$

(iii) Case $\gamma=\alpha+b_{1}$: In this case, we see that

$\displaystyle s\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\vee
s^{1/\alpha}<|x-z|<2}\bigg{(}1\wedge\frac{x_{d}\vee
s^{1/\alpha}}{|x-z|}\bigg{)}^{b_{1}}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{(x_{d}\vee
s^{1/\alpha})\wedge|x-z|}\bigg{)}\frac{f_{\gamma,\eta_{1},\eta_{2},k,l}(z_{d})}{|x-z|^{d+\alpha}}dz$
$\displaystyle\leq c_{11}s(x_{d}\vee
s^{1/\alpha})^{b_{1}}\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\vee
s^{1/\alpha}<|x-z|<2}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{x_{d}\vee
s^{1/\alpha}}\bigg{)}\frac{f_{\gamma,\eta_{1},\eta_{2},k,l}(|x-z|)}{|x-z|^{d+\alpha+b_{1}}}dz$
$\displaystyle=c_{11}s(x_{d}\vee s^{1/\alpha})^{b_{1}}$
$\displaystyle\times\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\vee
s^{1/\alpha}<|x-z|<2}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{x_{d}\vee
s^{1/\alpha}}\bigg{)}\log^{\eta_{1}}\bigg{(}e+\frac{k}{|x-z|}\bigg{)}\log^{\eta_{2}}\bigg{(}e+\frac{|x-z|}{l}\bigg{)}\frac{dz}{|x-z|^{d}}$
$\displaystyle\leq c_{12}s(x_{d}\vee s^{1/\alpha})^{b_{1}}\int_{x_{d}\vee
s^{1/\alpha}}^{2}\log^{b_{2}}\bigg{(}e+\frac{r}{x_{d}\vee
s^{1/\alpha}}\bigg{)}\log^{\eta_{1}}\bigg{(}e+\frac{k}{r}\bigg{)}\log^{\eta_{2}}\bigg{(}e+\frac{r}{l}\bigg{)}\frac{dr}{r}.$

The proof is complete. $\Box$

######  Lemma 10.11.

Let $b_{1},b_{2}\geq 0$. For any $0<k\leq l<1$,

$\int_{l}^{2}\log^{b_{1}}\bigg{(}e+\frac{r}{k}\bigg{)}\log^{b_{1}}\bigg{(}e+\frac{r}{l}\bigg{)}\log^{b_{2}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}\asymp\log^{b_{1}}\bigg{(}e+\frac{1}{k}\bigg{)}\log^{b_{1}+b_{2}+1}\bigg{(}e+\frac{1}{l}\bigg{)},$

with comparison constants independent of $k$ and $l$.

Proof. Note that

$\log\bigg{(}e+\frac{1}{r}\bigg{)}\asymp\log\bigg{(}\frac{2e}{r}\bigg{)}\asymp\log\bigg{(}e+\frac{1}{\sqrt{r}}\bigg{)},\quad
0<r<2.$ (10.9)

Hence, we get

$\displaystyle\int_{l}^{2}\log^{b_{1}}\bigg{(}e+\frac{r}{k}\bigg{)}\log^{b_{1}}\bigg{(}e+\frac{r}{l}\bigg{)}\log^{b_{2}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}$
$\displaystyle\geq
c_{1}\int_{\sqrt{l}}^{2}\log^{b_{1}}\bigg{(}\frac{2er}{k}\bigg{)}\log^{b_{1}}\bigg{(}\frac{2er}{l}\bigg{)}\log^{b_{2}}\bigg{(}\frac{2e}{r}\bigg{)}\frac{dr}{r}$
$\displaystyle\geq
c_{1}\log^{b_{1}}\bigg{(}\frac{2e}{\sqrt{k}}\bigg{)}\log^{b_{1}}\bigg{(}\frac{2e}{\sqrt{l}}\bigg{)}\int_{\sqrt{l}}^{2}\log^{b_{2}}\bigg{(}\frac{2e}{r}\bigg{)}\frac{dr}{r}$
$\displaystyle=\frac{c_{1}}{b_{2}+1}\log^{b_{1}}\bigg{(}\frac{2e}{\sqrt{k}}\bigg{)}\log^{b_{1}}\bigg{(}\frac{2e}{\sqrt{l}}\bigg{)}\bigg{(}\log^{b_{2}+1}\bigg{(}\frac{2e}{\sqrt{l}}\bigg{)}-1\bigg{)}$
$\displaystyle\geq
c_{2}\log^{b_{1}}\bigg{(}e+\frac{1}{k}\bigg{)}\log^{b_{1}+b_{2}+1}\bigg{(}e+\frac{1}{l}\bigg{)}.$

On the other hand, using (10.9) and (10.2), we get

$\displaystyle\int_{l}^{2}\log^{b_{1}}\bigg{(}e+\frac{r}{k}\bigg{)}\log^{b_{1}}\bigg{(}e+\frac{r}{l}\bigg{)}\log^{b_{2}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}$
$\displaystyle\leq
c_{3}\log^{b_{1}}\bigg{(}e+\frac{2}{k}\bigg{)}\log^{b_{1}}\bigg{(}e+\frac{2}{l}\bigg{)}\int_{l}^{2}\log^{b_{2}}\bigg{(}\frac{2e}{r}\bigg{)}\frac{dr}{r}\leq
c_{4}\log^{b_{1}}\bigg{(}e+\frac{1}{k}\bigg{)}\log^{b_{1}+b_{2}+1}\bigg{(}e+\frac{1}{l}\bigg{)}.$

$\Box$

######  Lemma 10.12.

Let $\gamma>1$ and $b_{1},b_{2}\geq 0$. For any $a,k,l>0$,

$\int_{a}^{\infty}t^{-\gamma}\bigg{(}1\wedge\frac{k}{t}\bigg{)}^{b_{1}}\bigg{(}1\wedge\frac{l}{t}\bigg{)}^{b_{2}}dt\asymp
a^{1-\gamma}\bigg{(}1\wedge\frac{k}{a}\bigg{)}^{b_{1}}\bigg{(}1\wedge\frac{l}{a}\bigg{)}^{b_{2}},$

with comparison constants independent of $a,k$ and $l$.

Proof. We have

$\displaystyle\int_{a}^{\infty}t^{-\gamma}\bigg{(}1\wedge\frac{k}{t}\bigg{)}^{b_{1}}\bigg{(}1\wedge\frac{l}{t}\bigg{)}^{b_{2}}dt\leq\bigg{(}1\wedge\frac{k}{a}\bigg{)}^{b_{1}}\bigg{(}1\wedge\frac{l}{a}\bigg{)}^{b_{2}}\int_{a}^{\infty}t^{-\gamma}dt\leq\frac{a^{1-\gamma}}{\gamma-1}\bigg{(}1\wedge\frac{k}{a}\bigg{)}^{b_{1}}\bigg{(}1\wedge\frac{l}{a}\bigg{)}^{b_{2}}$

and

$\displaystyle\int_{a}^{\infty}t^{-\gamma}\bigg{(}1\wedge\frac{k}{t}\bigg{)}^{b_{1}}\bigg{(}1\wedge\frac{l}{t}\bigg{)}^{b_{2}}dt$
$\displaystyle\geq\bigg{(}1\wedge\frac{k}{2a}\bigg{)}^{b_{1}}\bigg{(}1\wedge\frac{l}{2a}\bigg{)}^{b_{2}}\int_{a}^{2a}t^{-\gamma}dt$
$\displaystyle\geq\frac{(1-2^{1-\gamma})a^{1-\gamma}}{2^{b_{1}+b_{2}}(\gamma-1)}\bigg{(}1\wedge\frac{k}{a}\bigg{)}^{b_{1}}\bigg{(}1\wedge\frac{l}{a}\bigg{)}^{b_{2}}.$

$\Box$

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