c_{1,d,\alpha}r^{-\frac{d+\alpha}{2}}$. Similarly, if $r>1$

$c_{2,d,\alpha}:=\sup_{r\geq 1}r^{1/2}e^{r}K_{\frac{d+\alpha}{2}}(r),$

is finite by the first relation in (2.3). Thus we have

$\kappa_{m}(r)\leq
c_{2,d,\alpha}m^{\frac{d+\alpha+1}{2\alpha}}r^{\frac{1}{2}}e^{m^{\frac{1}{\alpha}}r}=c_{2,d,\alpha}r^{-\frac{d+\alpha}{2}}\left(m^{\frac{1}{\alpha}}r\right)^{\frac{d+\alpha+1}{2}}e^{m^{\frac{1}{\alpha}}r}\leq
c_{2,d,\alpha}\left(\frac{d+\alpha+1}{2}\right)^{\frac{d+\alpha+1}{2}}r^{-\frac{d+\alpha}{2}}.$

∎

Since the function

$\Phi^{m}_{d,\alpha}(r)=\frac{r^{\alpha/2}}{\kappa_{m}(r)^{\alpha/(d+\alpha)}},\quad
r>0$

will play a prominent role in what follows, we show some of its elementary
properties.

######  Lemma 6.6.

The following hold:

  1. (1)

For $r\downarrow 0$ we have

$\Phi^{m}_{d,\alpha}(r)\sim\frac{2^{\frac{\alpha}{d+\alpha}-\frac{\alpha}{2}}}{\Gamma^{\frac{\alpha}{d+\alpha}}\left(\frac{d+\alpha}{2}\right)}r^{\alpha}.$

As a consequence, for all $R>0$ there exists a constant $C_{d,\alpha,m,R}$
such that for all $r\in(0,R]$

$\Phi^{m}_{d,\alpha}(r)\leq C_{d,\alpha,m,R}r^{\alpha}.$ (6.7)

  2. (2)

As $r\uparrow\infty$ we have

$\Phi^{m}_{d,\alpha}(r)\sim\frac{2^{\frac{\alpha}{2(d+\alpha)}}}{m^{\frac{1}{2}-\frac{1}{2(d+\alpha)}}\pi^{\frac{\alpha}{2(d+\alpha)}}}r^{\frac{\alpha}{2}+\frac{\alpha}{2(d+\alpha)}}e^{\frac{\alpha}{d+\alpha}m^{1/\alpha}r}.$

  3. (3)

The function $\Phi^{m}_{d,\alpha}(r)$ is strictly increasing.

  4. (4)

For every fixed $r>0$ we have

$\lim_{m\downarrow
0}\Phi^{m}_{d,\alpha}(r)=\frac{2^{1-\frac{\alpha}{2}}}{\Gamma\Big{(}\frac{d+\alpha}{2}\Big{)}^{d/(d+\alpha)}}r^{\alpha}.$

  5. (5)

There exists a constant $C_{d,\alpha}$ such that for all $r>0$

$\Phi^{m}_{d,\alpha}(r)\geq C_{d,\alpha}r^{\alpha}.$ (6.8)

###### Proof.

Using (2.3), properties (1)-(2) are straightforward. (3) is also immediate
since $\Phi^{m}_{d,\alpha}(r)$ is the product of the two strictly increasing
positive functions $r\in(0,\infty)\mapsto m^{-1/2}r^{\alpha/2}$ and
$r\in(0,\infty)\mapsto\big{(}K_{\frac{d+\alpha}{2}}(m^{1/\alpha}r)\big{)}^{-\alpha/(d+\alpha)}$.
Finally, (4) and (5) are consequences of (6.5) and (6.6), respectively. ∎

####  6.2.2. Proof of estimate (2.14) in the $m>0$ case

######  Proposition 6.7.

Let $\alpha\in(0,2)$, $d\geq 3$. There exists a constant $C_{d,\alpha}>0$ such
that for all $m>0$, $\lambda\geq 0$ and $x\in\mathbb{R}^{d}\setminus\\{0\\}$,

$R^{m}_{\lambda}(x)\leq
C_{d,\alpha}\left(\frac{1}{|x|^{d-\alpha}}+\frac{m^{\frac{2}{\alpha}-1}}{|x|^{d-2}}\right).$

###### Proof.

While in the massless case we split $R_{\lambda}(x)$ into two integrals, here
we have to deal with four different integrals depending on the range of $|x|$.
Indeed, by (2.12) it follows for $m>0$ that there exists a constant $c_{1}>0$
such that

$\displaystyle p^{m}_{t}(x)\leq\begin{cases}c_{1}t|x|^{-d-\alpha}&\mbox{if
$t<\Phi_{d,\alpha}^{m}(|x|)\wedge\frac{1}{m}$}\\\ \\\
c_{1}t^{-d/\alpha}&\mbox{if $\Phi_{d,\alpha}^{m}(|x|)<t\leq\frac{1}{m}$}\\\
\\\ c_{1}m^{d/\alpha-d/2}t^{-d/2}e^{-c(m^{1/\alpha}|x|)}&\mbox{if
$\frac{1}{m}<t<m^{(1-\alpha)/2}|x|$}\\\ \\\
c_{1}m^{d/\alpha-d/2}t^{-d/2}e^{-c(m^{2/\alpha-1}|x|^{2}/t)}&\mbox{if
$m^{(1-\alpha)/2}|x|\vee\frac{1}{m}\leq t$}.\end{cases}$ (6.9)

According to this split, we define

$\displaystyle
I_{1}^{m}:=\int_{0}^{\Phi_{d,\alpha}^{m}(|x|)\wedge\frac{1}{m}}e^{-\lambda
t}p_{t}^{m}(x)dt,\qquad
I_{2}^{m}:=\begin{cases}\displaystyle\int_{\Phi_{d,\alpha}^{m}(|x|)}^{1/m}e^{-\lambda
t}p_{t}^{m}(x)dt&\mbox{if $\Phi_{d,\alpha}^{m}(|x|)<\frac{1}{m}$}\\\
0&\mbox{otherwise}\end{cases}$ $\displaystyle
I_{3}^{m}:=\begin{cases}\displaystyle\int_{1/m}^{m^{(1-\alpha)/\alpha}|x|}e^{-\lambda
t}p_{t}^{m}(x)dt&\mbox{if $|x|^{\alpha}>\frac{1}{m}$}\\\
0&\mbox{otherwise,}\end{cases}\qquad
I_{4}^{m}:=\int_{m^{(1-\alpha)/\alpha}|x|\vee\frac{1}{m}}^{\infty}e^{-\lambda
t}p_{t}^{m}(x)dt,$

so that

$R_{\lambda}(x)=I_{1}^{m}+I_{2}^{m}+I_{3}^{m}+I_{4}^{m}.$

We will use (6.9) to estimate each integral for above. To estimate $I_{1}^{m}$
we need to distinguish two cases. If
$\Phi^{m}_{d,\alpha}(|x|)\geq\frac{1}{m}$, we have

$I_{1}^{m}\leq
c_{1}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)\int_{0}^{\frac{1}{m}}te^{-\lambda
t}dt\leq\frac{c_{1}}{2m^{2}}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|).$

However, since $\frac{1}{m^{2}}\leq\left(\Phi^{m}_{d,\alpha}(|x|)\right)^{2}$,
we get

$I_{1}^{m}\leq\frac{c_{1}}{2}|x|^{-\frac{d-\alpha}{2}}\kappa^{\frac{d-\alpha}{d+\alpha}}_{m}(|x|)\leq
C_{d,\alpha}\frac{c_{1}}{2}|x|^{-(d-\alpha)},$

where we also used (6.6). In case $\Phi^{m}_{d,\alpha}(|x|)<\frac{1}{m}$, we
get

$I_{1}^{m}\leq
c_{1}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)\int_{0}^{\Phi^{m}_{d,\alpha}(|x|)}te^{-\lambda
t}dt\leq\frac{c_{1}}{2}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)(\Phi_{m}^{d,\alpha}(|x|))^{2}\leq
C_{d,\alpha}\frac{c_{1}}{2}|x|^{-(d-\alpha)}.$

We only need to estimate $I_{2}^{m}$ in case
$\Phi^{m}_{d,\alpha}(|x|)<\frac{1}{m}$. Since $d\geq 3>\alpha$, we have

$I_{2}^{m}\leq
c_{1}\int_{\Phi_{d,\alpha}^{m}(|x|)}^{\frac{1}{m}}t^{-\frac{d}{\alpha}}e^{-\lambda
t}dt\leq
c_{1}\int_{\Phi_{d,\alpha}^{m}(|x|)}^{\infty}t^{-\frac{d}{\alpha}}dt=\frac{\alpha
c_{1}}{d-\alpha}\left(\Phi_{d,\alpha}^{m}(|x|)\right)^{-\frac{d-\alpha}{\alpha}}\leq
C_{d,\alpha}\frac{\alpha c_{1}}{d-\alpha}|x|^{-(d-\alpha)}.$

Next we estimate $I_{3}^{m}$. We only need to consider the case
$|x|^{\alpha}\geq\frac{1}{m}$ giving

$\displaystyle I_{3}^{m}$ $\displaystyle\leq
c_{1}m^{\frac{d}{\alpha}-\frac{d}{2}}e^{-cm^{\frac{1}{\alpha}}|x|}\int_{\frac{1}{m}}^{m^{\frac{1-\alpha}{\alpha}}|x|}e^{-\lambda
t}t^{-\frac{d}{2}}dt\leq
c_{1}m^{\frac{d}{\alpha}-\frac{d}{2}}e^{-cm^{\frac{1}{\alpha}}|x|}\int_{\frac{1}{m}}^{\infty}t^{-\frac{d}{2}}dt=\frac{2c_{1}}{d-2}m^{\frac{d}{\alpha}-1}e^{-cm^{\frac{1}{\alpha}}|x|}$
$\displaystyle=\frac{2c_{1}}{(d-2)c^{d-\alpha}|x|^{d-\alpha}}\left(cm^{\frac{1}{\alpha}}|x|\right)^{d-\alpha}e^{-cm^{\frac{1}{\alpha}}|x|}\leq\frac{2c_{1}(d-\alpha)^{d-\alpha}}{(d-2)e^{d-\alpha}c^{d-\alpha}|x|^{d-\alpha}}.$

Finally, consider $I_{4}^{m}$. With $|x|^{\alpha}\leq\frac{1}{m}$ we obtain

$\displaystyle I_{4}^{m}\leq
c_{1}m^{\frac{d}{\alpha}-\frac{d}{2}}\int_{\frac{1}{m}}^{\infty}e^{-\lambda
t}t^{-\frac{d}{2}}e^{-cm^{\frac{2}{\alpha}-1}|x|^{2}/t}dt\leq
c_{1}m^{\frac{d}{\alpha}-\frac{d}{2}}\int_{\frac{1}{m}}^{\infty}t^{-\frac{d}{2}}e^{-cm^{\frac{2}{\alpha}-1}|x|^{2}/t}dt.$

The change of variable $s=cm^{\frac{2}{\alpha}-1}|x|^{2}/t$ gives

$\displaystyle
I_{4}^{m}\leq\frac{c_{1}m^{\frac{2}{\alpha}-1}}{c^{\frac{d-2}{2}}|x|^{d-2}}\int_{cm^{\frac{2}{\alpha}}|x|^{2}}^{\infty}s^{\frac{d}{2}-2}e^{-s}ds\leq\frac{c_{1}m^{\frac{2}{\alpha}-1}}{c^{\frac{d-2}{2}}|x|^{d-2}}\Gamma\Big{(}\frac{d}{2}-1\Big{)}.$

∎

####  6.2.3. Proof of estimate (2.15)

Since

$|x|^{\alpha-d}+m^{2/\alpha-1}|x|^{2-d}\leq 2(1\vee
m^{2/\alpha-1})(|x|^{\alpha-d}\mathbf{1}_{\\{|x|\leq
1\\}}+|x|^{2-d}\mathbf{1}_{\\{|x|>1\\}}),$

the above also shows the upper bound in (2.15). The lower bound is shown in
the next statement.

######  Lemma 6.8.

Let $\alpha\in(0,2)$ and $d\geq 3$. Then, for all $m>0$ there exists a
constant $C_{d,\alpha,m}>0$ such that for all
$x\in\mathbb{R}^{d}\setminus\\{0\\}$

$R_{0}^{m}(x)\geq C_{d,\alpha,m}(|x|^{-(d-\alpha)}\mathbf{1}_{\\{|x|\leq
1\\}}+|x|^{-(d-2)}\mathbf{1}_{\\{|x|>1\\}}).$

###### Proof.

Now we use the lower bounds in (2.12). Thus there exists a constant $c_{2}>0$
such that

$\displaystyle p^{m}_{t}(x)\geq\begin{cases}c_{2}t|x|^{-d-\alpha}&\mbox{if
$t<\Phi_{d,\alpha}^{m}(|x|)\wedge\frac{1}{m}$}\\\ \\\
c_{2}t^{-d/\alpha}&\mbox{if $\Phi_{d,\alpha}^{m}(|x|)<t\leq\frac{1}{m}$}\\\
\\\ c_{2}m^{d/\alpha-d/2}t^{-d/2}e^{-c(m^{1/\alpha}|x|)}&\mbox{if
$\frac{1}{m}<t<m^{(1-\alpha)/2}|x|$}\\\ \\\
c_{2}m^{d/\alpha-d/2}t^{-d/2}e^{-c(m^{2/\alpha-1}|x|^{2}/t)}&\mbox{if
$m^{(1-\alpha)/2}|x|\vee\frac{1}{m}\leq t$}.\end{cases}$ (6.10)

Consider the case $|x|>1$ and let first $|x|^{\alpha}>\frac{1}{m}$. We obtain

$I_{3}^{m}\geq
c_{2}m^{\frac{d}{\alpha}-\frac{d}{2}}e^{-cm^{\frac{1}{\alpha}}|x|}\int_{\frac{1}{m}}^{m^{\frac{1-\alpha}{\alpha}}|x|}t^{-\frac{d}{2}}dt\geq
c_{2}m^{\frac{d}{\alpha}-\frac{d}{2}}\int_{\frac{1}{m}}^{m^{\frac{1-\alpha}{\alpha}}|x|}t^{-\frac{d}{2}}e^{-cm^{\frac{2}{\alpha}-1}|x|^{2}/t}dt,$

since for $t\leq m^{\frac{1-\alpha}{\alpha}}|x|$ we clearly have
$e^{-cm^{\frac{1}{\alpha}}|x|}\geq e^{-cm^{\frac{2}{\alpha}-1}|x|^{2}/t}$.
Hence for all $|x|>1$ we get

$R_{0}^{m}(x)\geq I_{3}^{m}+I_{4}^{m}\geq
c_{2}m^{\frac{d}{\alpha}-\frac{d}{2}}\int_{\frac{1}{m}}^{\infty}t^{-\frac{d}{2}}e^{-cm^{\frac{2}{\alpha}-1}|x|^{2}/t}dt.$

The substitution $s=\frac{cm^{\frac{2}{\alpha}-1}|x|^{2}}{t}$ then gives

$R_{0}^{m}(x)\geq
c_{2}m^{\frac{2}{\alpha}-1}c^{-\frac{d-2}{2}}|x|^{-(d-2)}\int_{0}^{cm^{\frac{2}{\alpha}}|x|^{2}}s^{\frac{d}{2}-2}e^{-s}ds\geq
c_{2}m^{\frac{2}{\alpha}-1}c^{-\frac{d-2}{2}}|x|^{-(d-2)}\int_{0}^{cm^{\frac{2}{\alpha}}}s^{\frac{d}{2}-2}e^{-s}ds$

where the integral at the right-hand side is finite since $d\geq 3$. Next
consider $|x|\leq 1$. If $\Phi_{d,\alpha}^{m}(|x|)\leq\frac{1}{m}$, then

$\displaystyle R_{0}^{m}(x)\geq I_{1}^{m}\geq
c_{2}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)\int_{0}^{\Phi_{d,\alpha}^{m}(|x|)}t\,dt=\frac{c_{2}}{2}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)(\Phi_{d,\alpha}^{m}(|x|)^{2}\geq
C_{d,\alpha}|x|^{-(d-\alpha)},$

where we used (6.8) and the fact that $\kappa_{m}(|x|)\geq
C_{d,\alpha}|x|^{-\frac{d+\alpha}{2}}$ since $|x|\leq 1$. If
$\Phi_{d,\alpha}^{m}(|x|)>\frac{1}{m}$, we have

$\displaystyle R_{0}^{m}(x)$ $\displaystyle\geq I_{1}^{m}\geq
c_{2}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)\int_{0}^{\frac{1}{m}}t\,dt=\frac{c_{2}}{2m^{2}}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)$
$\displaystyle\geq\frac{c_{2}}{2m^{2}(\Phi_{d,\alpha}^{m}(1))^{2}}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)(\Phi_{d,\alpha}^{m}(|x|))^{2}\geq
c_{d,\alpha,m}|x|^{-(d-\alpha)}.$

∎

######  Remark 6.9.

Notice that, up to a multiplicative constant, the term $|x|^{-(d-2)}$
represents the $0$-resolvent kernel of the Laplacian. Indeed, we could say
that the $0$-potential $R_{0}^{m}$ interpolates between the massless case
$R_{0}(x)=C_{d,\alpha}|x|^{-(d-\alpha)}$ and the classical Newtonian case
(i.e. $\alpha=2$) that we denote here by $R_{0}^{\rm N}(x)=C_{d}|x|^{-(d-2)}$.
To observe this, we introduce a constant $c_{\ell}$ and write $c_{\ell}^{2}m$
in place of $m$; this constant is physically motivated and it stands for the
speed of light (which was set to $1$ in units used in mathematical physics).
Taking $m\downarrow 0$ and using [6, Lem. 3.2] it follows that
$[\cdot]_{0,\alpha}={\rm M}-\lim_{m\to 0}[\cdot]_{c_{\ell}^{2}m,\alpha}$,
which implies by an application of [6, Prop. 2.1] the strong resolvent
convergence of the operators $-L_{c_{\ell}^{2}m,\alpha}$ to $-L_{0,\alpha}$ as
$m\downarrow 0$. On the other hand, if we consider the classical limit
$c_{\ell}\to\infty$, we obtain again by [6, Lem. 3.2] that $[\cdot]_{0,2}={\rm
M}-\lim_{c_{\ell}\to\infty}[\cdot]_{c_{\ell}^{2}m,\alpha}$, where
$[u]_{0,2}=\left\|\nabla u\right\|_{2}$ for all $u\in H^{1}(\mathbb{R}^{d})$.
Hence $L_{c_{\ell}^{2}m,\alpha}$ converges to $-\Delta$ in strong resolvent
sense.

####  6.2.4. Proof of estimate (2.16)

We start by the following preliminary estimate.

######  Lemma 6.10.

Let $\alpha\in(0,2)$, $d\geq 1$ and $m>0$. There exists $C_{d,\alpha}>0$ such
that

$R^{m}_{\lambda}(x)\leq
C_{d,\alpha}\begin{cases}\Big{(}1+\frac{me^{-\lambda/m}}{\lambda}\Big{)}\frac{1}{|x|^{d-\alpha}}&\alpha<d\\\
\\\
\log\left(1+\frac{1}{\lambda|x|}\right)+1+\frac{m^{1/\alpha}}{\lambda}e^{-\lambda/m}&\alpha=d=1\\\
\\\
|x|^{\alpha-1}+\lambda^{-\frac{\alpha-1}{\alpha}}+\frac{m^{1/\alpha}}{\lambda}e^{-\lambda/m}&\alpha>d=1\end{cases}$

holds for all $x\in\mathbb{R}^{d}$ and $\lambda>0$.

###### Proof.

As in the proof of Lemma 6.7, we have

$I_{1}^{m}\leq C_{d,\alpha}|x|^{-(d-\alpha)}.$

If $d>\alpha$, we obtain

$I_{2}^{m}\leq C_{d,\alpha}|x|^{-(d-\alpha)}.$

If $d=\alpha=1$, by making use of (6.2) it follows that

$\displaystyle I_{2}^{m}\leq
c_{1}\int_{\Phi_{1,1}^{m}(|x|)}^{\frac{1}{m}}t^{-1}e^{-\lambda t}dt$
$\displaystyle\leq c_{1}\int_{\Phi_{1,1}^{m}(|x|)}^{\infty}t^{-1}e^{-\lambda
t}dt=c_{1}\int_{\lambda\Phi_{1,1}^{m}(|x|)}^{\infty}t^{-1}e^{-t}dt$
$\displaystyle=c_{1}E_{1}(\lambda\Phi_{1,1}^{m}(|x|))\leq
c_{1}\log\left(1+\frac{1}{C_{d,\alpha}\lambda|x|}\right)\leq
C_{d,\alpha}\log\left(1+\frac{1}{\lambda|x|}\right).$

Finally, if $\alpha>d=1$, then

$I_{2}^{m}\leq
c_{1}\int_{\Phi_{1,\alpha}^{m}(|x|)}^{\frac{1}{m}}t^{-\frac{1}{\alpha}}e^{-\lambda
t}dt\leq c_{1}\int_{0}^{\infty}t^{-\frac{1}{\alpha}}e^{-\lambda
t}dt=c_{1}\lambda^{-\frac{\alpha-1}{\alpha}}\Gamma\left(\frac{\alpha-1}{\alpha}\right).$

Next we estimate $I_{3}^{m}$. It is necessary to do this only if
$|x|^{\alpha}>\frac{1}{m}$. In this case by (6.9) we get

$I_{3}^{m}\leq
c_{1}m^{\frac{d}{\alpha}-\frac{d}{2}}e^{-cm^{\frac{1}{\alpha}}|x|}\int_{\frac{1}{m}}^{m^{\frac{1-\alpha}{\alpha}}|x|}e^{-\lambda
t}t^{-\frac{d}{2}}dt\leq\frac{c_{1}m^{\frac{d}{\alpha}}}{\lambda}e^{-cm^{\frac{1}{\alpha}}|x|-\frac{\lambda}{m}}.$
(6.11)

It is clear that if $\alpha\geq d=1$

$I_{3}^{m}\leq
C_{1,\alpha}\frac{m^{\frac{1}{\alpha}}}{\lambda}e^{-\frac{\lambda}{m}}.$

For $\alpha<d$, by (6.3) we have

$I_{3}^{m}\leq\frac{c_{1}m}{\lambda
c^{d-\alpha}|x|^{d-\alpha}}\left(cm^{\frac{1}{\alpha}}|x|\right)^{d-\alpha}e^{-cm^{\frac{1}{\alpha}}|x|}e^{-\frac{\lambda}{m}}\leq\frac{c_{1}m(d-\alpha)^{d-\alpha}}{\lambda
c^{d-\alpha}e^{d-\alpha}|x|^{d-\alpha}}e^{-\frac{\lambda}{m}}.$

Finally, for $I_{4}^{m}$ we have two cases. If $|x|^{\alpha}\leq\frac{1}{m}$,
then for $d>\alpha$ we obtain

$\displaystyle I_{4}^{m}$ $\displaystyle\leq
c_{1}m^{\frac{d}{\alpha}-\frac{d}{2}}\int_{\frac{1}{m}}^{\infty}e^{-\lambda
t}t^{-\frac{d}{2}}e^{-cm^{\frac{2}{\alpha}-1}|x|^{2}/t}dt$
$\displaystyle\leq\frac{c_{1}m^{1-\frac{\alpha}{2}}}{c^{\frac{d-\alpha}{2}}|x|^{d-\alpha}}\int_{\frac{1}{m}}^{\infty}e^{-\lambda
t}t^{-\frac{\alpha}{2}}\left(cm^{\frac{2}{\alpha}-1}|x|^{2}/t\right)^{\frac{d-\alpha}{2}}e^{-cm^{\frac{2}{\alpha}-1}|x|^{2}/t}dt$
$\displaystyle\leq\frac{c_{1}m^{1-\frac{\alpha}{2}}}{e^{\frac{d-\alpha}{2}}c^{\frac{d-\alpha}{2}}|x|^{d-\alpha}}\left(\frac{d-\alpha}{2}\right)^{\frac{d-\alpha}{2}}\int_{\frac{1}{m}}^{\infty}e^{-\lambda
t}t^{-\frac{\alpha}{2}}dt\leq\frac{c_{2}me^{-\frac{\lambda}{m}}}{e^{\frac{d-\alpha}{2}}c^{\frac{d-\alpha}{2}}\lambda|x|^{d-\alpha}}\left(\frac{d-\alpha}{2}\right)^{\frac{d-\alpha}{2}}.$

For $\alpha\geq d=1$ we estimate like

$\displaystyle I_{4}^{m}\leq
c_{1}m^{\frac{1}{\alpha}-\frac{1}{2}}\int_{\frac{1}{m}}^{\infty}e^{-\lambda
t}t^{-\frac{1}{2}}e^{-cm^{\frac{2}{\alpha}-1}|x|^{2}/t}dt\leq
c_{1}\frac{m^{\frac{1}{\alpha}}}{\lambda}e^{-\frac{\lambda}{m}}.$

In case $|x|^{\alpha}>\frac{1}{m}$, we obtain

$\displaystyle I_{4}^{m}$ $\displaystyle\leq
c_{1}m^{\frac{d}{\alpha}-\frac{d}{2}}\int_{m^{\frac{1-\alpha}{\alpha}}|x|}^{\infty}e^{-\lambda
t}t^{-\frac{d}{2}}e^{-cm^{\frac{2}{\alpha}-1}|x|^{2}/t}dt\leq
c_{1}m^{\frac{d}{\alpha}-\frac{d}{2}}\int_{\frac{1}{m}}^{\infty}e^{-\lambda
t}t^{-\frac{d}{2}}e^{-cm^{\frac{2}{\alpha}-1}|x|^{2}/t}dt$

and then we proceed as in the previous case. ∎

The following estimate is sharper for large values of $|x|$.

######  Lemma 6.11.

Let $\alpha\in(0,2)$, $d\geq 1$ and $m>0$. Then there exists $C_{d,\alpha}>0$
such that for all $x\in\mathbb{R}^{d}\setminus\\{0\\}$ and $\lambda>0$ we have

$R^{m}_{\lambda}(x)\leq
C_{d,\alpha}\left(\frac{1}{\lambda^{2}}+\frac{e^{-\frac{\lambda}{m}}}{m\lambda}+\frac{m^{\frac{\alpha}{2}-1}}{\lambda^{\frac{\alpha}{2}+1}}\Gamma\left(\frac{\alpha}{2}+1;\frac{\lambda}{m}\right)\right)\frac{1}{|x|^{d+\alpha}}.$

###### Proof.

We estimate $I^{m}_{j}$, $j=1,2,3,4$, in the proof of Lemma 6.10 in different
ways. First consider $I_{1}^{m}$. By (6.6) we get

$I_{1}^{m}\leq
c_{1}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)\int_{0}^{\min\left\\{\Phi_{d,\alpha}^{m}(|x|),\frac{1}{m}\right\\}}te^{-\lambda
t}dt\leq
c_{1}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)\int_{0}^{\infty}te^{-\lambda
t}dt\leq\frac{C_{d,\alpha}}{\lambda^{2}|x|^{d+\alpha}}.$

Next consider $I_{2}^{m}$. We need to obtain an estimate only for the case
$\Phi_{d,\alpha}^{m}(|x|)<\frac{1}{m}$. We have

$\displaystyle I_{2}^{m}$ $\displaystyle\leq
c_{1}\int_{\Phi_{d,\alpha}^{m}(|x|)}^{\frac{1}{m}}t^{-\frac{d}{\alpha}}e^{-\lambda
t}dt\leq\frac{2c_{1}}{\lambda
e}\int_{\Phi_{d,\alpha}^{m}(|x|)}^{\infty}t^{-\frac{d}{\alpha}-1}e^{-\frac{\lambda
t}{2}}dt\leq\frac{2\alpha c_{1}}{d\lambda
e}e^{-\frac{\lambda\Phi_{d,\alpha}^{m}(|x|)}{2}}(\Phi_{d,\alpha}^{m}(|x|))^{-\frac{d}{\alpha}}$
$\displaystyle=\frac{4\alpha
c_{1}}{d\lambda^{2}e}\left(\frac{\lambda\Phi_{d,\alpha}^{m}(|x|)}{2}\right)e^{-\frac{\lambda\Phi_{d,\alpha}^{m}(|x|)}{2}}(\Phi_{d,\alpha}^{m}(|x|))^{-\frac{d}{\alpha}-1}\leq\frac{4\alpha
c_{1}}{d\lambda^{2}e^{2}}(\Phi_{d,\alpha}^{m}(|x|))^{-\frac{d+\alpha}{\alpha}}\leq\frac{C_{d,\alpha}}{\lambda^{2}|x|^{d+\alpha}},$

where we also used (6.8). Next we need to control $I_{3}^{m}$. We only need to
consider the case $|x|^{\alpha}>\frac{1}{m}$. Starting from (6.11) and using
(6.3) we obtain

$\displaystyle
I_{3}^{m}\leq\frac{c_{1}m^{\frac{d}{\alpha}}}{\lambda}e^{-cm^{\frac{1}{\alpha}}|x|-\frac{\lambda}{m}}$
$\displaystyle=\frac{c_{1}}{\lambda
c^{d+\alpha}m|x|^{d+\alpha}}\left(cm^{\frac{1}{\alpha}}|x|\right)^{d+\alpha}e^{-cm^{\frac{1}{\alpha}}|x|-\frac{\lambda}{m}}$
$\displaystyle\leq\frac{(d+\alpha)^{d+\alpha}c_{1}}{\lambda
e^{d+\alpha}c^{d+\alpha}m|x|^{d+\alpha}}e^{-\frac{\lambda}{m}}.$

Finally, we consider $I_{4}^{m}$. If $|x|^{\alpha}\leq\frac{1}{m}$, we have

$\displaystyle I_{4}^{m}$ $\displaystyle\leq
c_{1}m^{\frac{d}{\alpha}-\frac{d}{2}}\int_{\frac{1}{m}}^{\infty}e^{-\lambda
t}t^{-\frac{d}{2}}e^{-cm^{\frac{2}{\alpha}-1}|x|^{2}/t}dt$
$\displaystyle\leq\frac{c_{1}m^{\frac{\alpha}{2}-1}}{c^{\frac{d+\alpha}{2}}|x|^{d+\alpha}}\int_{\frac{1}{m}}^{\infty}e^{-\lambda
t}t^{\frac{\alpha}{2}}\left(cm^{\frac{2}{\alpha}-1}|x|^{2}/t\right)^{\frac{d+\alpha}{2}}e^{-cm^{\frac{2}{\alpha}-1}|x|^{2}/t}dt$
$\displaystyle\leq\frac{c_{1}m^{\frac{\alpha}{2}-1}}{e^{\frac{d+\alpha}{2}}c^{\frac{d+\alpha}{2}}|x|^{d+\alpha}}\left(\frac{d+\alpha}{2}\right)^{\frac{d+\alpha}{2}}\int_{\frac{1}{m}}^{\infty}e^{-\lambda
t}t^{\frac{\alpha}{2}}dt$
$\displaystyle\leq\frac{c_{1}m^{\frac{\alpha}{2}-1}}{e^{\frac{d-\alpha}{2}}c^{\frac{d-\alpha}{2}}\lambda^{\frac{\alpha}{2}+1}|x|^{d+\alpha}}\left(\frac{d+\alpha}{2}\right)^{\frac{d+\alpha}{2}}\Gamma\left(\frac{\alpha}{2}+1;\frac{\lambda}{m}\right),$

where

$\Gamma\left(a;z\right)=\int_{z}^{\infty}t^{a-1}e^{-t}dt,\quad
a\in\mathbb{R},\,z\geq 0,$

is the upper incomplete Gamma function. If $|x|^{\alpha}>\frac{1}{m}$, we have
instead

$\displaystyle I_{4}^{m}$ $\displaystyle\leq
c_{1}m^{\frac{d}{\alpha}-\frac{d}{2}}\int_{m^{\frac{1-\alpha}{\alpha}}|x|}^{\infty}e^{-\lambda
t}t^{-\frac{d}{2}}e^{-cm^{\frac{2}{\alpha}-1}|x|^{2}/t}dt\leq
c_{1}m^{\frac{d}{\alpha}-\frac{d}{2}}\int_{\frac{1}{m}}^{\infty}e^{-\lambda
t}t^{-\frac{d}{2}}e^{-cm^{\frac{2}{\alpha}-1}|x|^{2}/t}dt$

and we proceed as in the previous case. ∎

######  Remark 6.12.

Since

$\Gamma\left(a,z\right)\sim z^{a-1}e^{-z},\quad z\to\infty,$

on taking $m\to 0$ with $\lambda>0$ fixed it follows that

$\frac{m^{\frac{\alpha}{2}}-1}{\lambda^{\frac{\alpha}{2}+1}}\Gamma\left(\frac{\alpha}{2}+1;\frac{\lambda}{m}\right)\sim\frac{e^{-\frac{\lambda}{m}}}{m\lambda}.$

The same holds for $\lambda\to\infty$ if $m$ is fixed.

Combining Lemmas 6.10 and 6.11´we get the following result.

######  Proposition 6.13.

Let $\alpha\in(0,2)$, $d\geq 1$. Then for all $\lambda>0$ and $m>0$ there
exists a constant $C_{d,\alpha,\lambda,m}$ such that

$R_{\lambda}^{m}(x)\leq C_{d,\alpha,\lambda,m}H^{0}_{d,\alpha}(|x|),$

for all $x\in\mathbb{R}^{d}\setminus\\{0\\}$, where $H^{0}_{d,\alpha}$ is
defined in (2.17).

This shows (2.16) for $m>0$.

####  6.2.5. Proof of estimate (2.19)

To provide the upper bound we improve the bound given in Lemma 6.11 for large
values of $\lambda$.

######  Lemma 6.14.

Let $\alpha\in(0,2)$, $d\geq 1$ and $m>0$. Then there exists a constant
$C_{d,\alpha}>0$ such that for all $\lambda>m$ and
$x\in\mathbb{R}^{d}\setminus\\{0\\}$

$R^{m}_{\lambda}(x)\leq
C_{d,\alpha}\frac{|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)}{(\lambda-m)^{2}}.$

###### Proof.

By [8, eq. (2.16)] there exists a constant $c_{1}$ dependent on $\alpha$ and
$d$ such that for all $t>0$ and $x\in\mathbb{R}^{d}\setminus\\{0\\}$

$p_{t}(x)\leq c_{1}te^{mt}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)$

If $\lambda>m$, we have

$R^{m}_{\lambda}(x)\leq
c_{1}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)\int_{0}^{\infty}te^{-(\lambda-m)t}dt\leq
c_{1}\frac{|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)}{(\lambda-m)^{2}}.$

∎

Combining the previous estimate with Lemma 6.10 gives the following result.

######  Corollary 6.15.

Let $\alpha\in(0,2)$, $d\geq 1$. Then for all $m>0$ and $\lambda>m$, there
exists a constant $C_{d,\alpha,\lambda,m}$ such that for all
$x\in\mathbb{R}^{d}\setminus\\{0\\}$

$R_{\lambda}^{m}(x)\leq C_{d,\alpha,\lambda,m}H^{m}_{d,\alpha}(|x|)$

where $H^{m}_{d,\alpha}$ is defined in (2.20)

######  Remark 6.16.

We note that (6.5) gives $\lim_{m\downarrow
0}H^{m}_{d,\alpha}(r)=H^{0}_{d,\alpha}(r)$.

This proves the upper bound in (2.19). Finally, we prove the lower bound in
(2.19), together with (2.21).

######  Lemma 6.17.

Let $\alpha\in(0,2)$ and $d\geq 1$. Then for all $\lambda>0$ and $m>0$ there
exists a constant $C_{d,\alpha,\lambda,m}>0$ such that

$R_{\lambda}^{m}(x)\geq C_{d,\alpha,\lambda,m}H^{m}_{d,\alpha}(|x|),$

for all $x\in\mathbb{R}^{d}\setminus\\{0\\}$, where $H^{m}_{d,\alpha}$ is
defined in (2.20).

###### Proof.

Throughout this proof we will use the lower bounds provided in (6.10). First
we give an estimate for $|x|>1$. We need to distinguish among two cases. If
$\Phi_{d,\alpha}^{m}(|x|)\geq\frac{1}{m}$, we have

$\displaystyle R_{\lambda}^{m}(x)$
$\displaystyle\geq\int_{0}^{\frac{1}{m}}e^{-\lambda t}p_{t}^{m}(x)dt\geq
c_{2}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)\int_{0}^{\frac{1}{m}}te^{-\lambda
t}dt$
$\displaystyle=\frac{c_{2}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)}{\lambda^{2}}\left(1-e^{-\frac{\lambda}{m}}-\frac{\lambda}{m}e^{-\frac{\lambda}{m}}\right)\geq\frac{c_{2}}{2}m^{-2}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)e^{-\lambda/m}.$

If $\Phi_{d,\alpha}^{m}(|x|)\leq\frac{1}{m}$, we have

$\displaystyle R_{\lambda}^{m}(x)$
$\displaystyle\geq\int_{0}^{\Phi_{d,\alpha}^{m}(|x|)}e^{-\lambda
t}p_{t}^{m}(x)dt\geq
c_{2}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)\int_{0}^{\Phi_{d,\alpha}^{m}(|x|)}te^{-\lambda
t}dt$ $\displaystyle\geq
c_{2}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)\int_{0}^{\Phi_{d,\alpha}^{m}(1)}te^{-\lambda
t}dt$
$\displaystyle\geq\frac{c_{2}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)}{\lambda^{2}}\left(1-e^{-\lambda\Phi_{d,\alpha}^{m}(1)}-\lambda\Phi_{d,\alpha}^{m}(1)e^{-\lambda\Phi_{d,\alpha}^{m}(1)}\right),$

where we used that $\Phi_{d,\alpha}^{m}$ is strictly increasing. Next let
$|x|\leq 1$ and consider the following cases. If $d>\alpha$ and
$\Phi_{d,\alpha}^{m}(|x|)\leq\frac{1}{m}$, we get

$\displaystyle R_{\lambda}^{m}(x)$
$\displaystyle\geq\int_{0}^{\Phi_{d,\alpha}^{m}(|x|)}e^{-\lambda
t}p_{t}^{m}(x)dt\geq
c_{2}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)\int_{0}^{\Phi_{d,\alpha}^{m}(|x|)}te^{-\lambda
t}dt$
$\displaystyle=\frac{c_{2}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)}{\lambda^{2}}e^{-\lambda\Phi_{d,\alpha}^{m}(|x|)}\left(e^{\lambda\Phi_{d,\alpha}^{m}(|x|)}-1-\lambda\Phi_{d,\alpha}^{m}(|x|)\right)$
$\displaystyle\geq
c_{2}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)(\Phi_{d,\alpha}^{m}(|x|))^{2}e^{-\lambda\Phi_{d,\alpha}^{m}(|x|)}.$

Now observe that for $|x|\leq 1$ the second asymptotic relation in (2.3)
implies that there exists a constant $C_{d,\alpha}$ such that
$\kappa_{m}(|x|)\geq C_{d,\alpha}|x|^{-\frac{d+\alpha}{2}}$. Hence, by also
using (6.8) and the fact that $\Phi_{d,\alpha}^{m}$ is increasing, we obtain

$\displaystyle R_{\lambda}^{m}(x)\geq
C_{d,\alpha}|x|^{-(d-\alpha)}e^{-\lambda\Phi_{d,\alpha}^{m}(1)}.$

Next, if $d>\alpha$ and $\Phi_{d,\alpha}^{m}(|x|)>\frac{1}{m}$, then we have

$\displaystyle R_{\lambda}^{m}(x)$
$\displaystyle\geq\int_{0}^{\frac{1}{m}}e^{-\lambda t}p_{t}^{m}(x)dt\geq
c_{2}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)\int_{0}^{\frac{1}{m}}te^{-\lambda
t}dt$
$\displaystyle=\frac{c_{2}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)}{\lambda^{2}}e^{-\frac{\lambda}{m}}\left(e^{\frac{\lambda}{m}}-1-\frac{\lambda}{m}\right)$
$\displaystyle\geq\frac{c_{2}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)}{m^{2}}e^{-\frac{\lambda}{m}}\geq\frac{c_{2}|x|^{-\frac{d+\alpha}{2}}\kappa_{m}(|x|)(\Phi_{d,\alpha}^{m}(|x|))^{2}}{(\Phi_{d,\alpha}^{m}(1))^{2}m^{2}}e^{-\frac{\lambda}{m}}\geq
C_{d,\alpha}\frac{|x|^{-(d-\alpha)}}{(\Phi_{d,\alpha}^{m}(|1|))^{2}m^{2}}e^{-\frac{\lambda}{m}}.$

Next assume that $d=\alpha=1$ and let $\Phi_{1,1}^{m}(|x|)\leq\frac{1}{m}$.
Then we start from

$\displaystyle R_{\lambda}^{m}(x)\geq I_{2}^{m}+I_{3}^{m}+I_{4}^{m}$

and we first search for some lower bounds of $I_{3}^{m}$ and $I_{4}^{m}$.
Again, we have to distinguish two cases. If $|x|\leq\frac{1}{m}$, then
$I_{3}^{m}=0$ and we have

$I_{4}^{m}\geq c_{2}m^{\frac{1}{2}}\int_{\frac{1}{m}}^{\infty}e^{-\lambda
t}t^{-\frac{1}{2}}e^{-cm|x|^{2}/t}dt.$

Now observe that, for fixed $|x|$, the function
$f(r)=r^{-\frac{1}{2}}e^{-cm|x|^{2}r}$ is decreasing, hence, for
$t>\frac{1}{m}$ and $|x|\leq 1$, we have

$t^{-\frac{1}{2}}e^{-cm|x|^{2}/t}\geq
m^{-\frac{1}{2}}e^{-cm^{2}|x|^{2}}t^{-1}\geq
m^{-\frac{1}{2}}e^{-cm^{2}}t^{-1}.$ (6.12)

Thus

$I_{4}^{m}\geq c_{2}e^{-cm^{2}}\int_{\frac{1}{m}}^{\infty}t^{-1}e^{-\lambda
t}dt.$

If $|x|>\frac{1}{m}$, we have

$I_{3}^{m}\geq
c_{2}m^{\frac{1}{2}}e^{-cm|x|}\int_{\frac{1}{m}}^{|x|}t^{-\frac{1}{2}}e^{-\lambda
t}dt\geq e^{-cm}\int_{\frac{1}{m}}^{|x|}t^{-1}e^{-\lambda t}dt.$

Furthermore

$I_{4}^{m}\geq c_{2}m^{\frac{1}{2}}\int_{|x|}^{\infty}e^{-\lambda
t}t^{-\frac{1}{2}}e^{-cm^{\frac{2}{\alpha}-1}|x|^{2}/t}dt\geq
c_{2}e^{-cm^{2}}\int_{|x|}^{\infty}t^{-1}e^{-\lambda t}dt,$

where we used again (6.12), since $t>|x|>\frac{1}{m}$ and $|x|\leq 1$. In
general, we have

$I_{3}^{m}+I_{4}^{m}\geq C_{1,1,m}\int_{\frac{1}{m}}^{\infty}t^{-1}e^{-\lambda
t}dt$

and then

$\displaystyle R_{\lambda}^{m}(x)\geq I_{2}^{m}+I_{3}^{m}+I_{4}^{m}$
$\displaystyle\geq
C_{1,1,m}\int_{\Phi_{1,1}^{m}(|x|)}^{\infty}t^{-1}e^{-\lambda
t}dt=C_{1,1,m}E_{1}(\lambda\Phi_{1,1}^{m}(|x|))$
$\displaystyle\geq\frac{C_{1,1,m}e^{-\lambda\Phi_{1,1}^{m}(|x|)}}{2}\log\left(1+\frac{2}{\lambda\Phi_{1,1}^{m}(|x|)}\right),$

where we used (6.2). Since $|x|\leq 1$, we can use (6.7) to finally obtain

$\displaystyle R_{\lambda}^{m}(x)\geq
C_{1,1,m,\lambda}\log\left(1+\frac{1}{|x|}\right).$

Now consider the case $d=\alpha=1$ and $\Phi^{m}_{1,1}(|x|)>\frac{1}{m}$.
Arguing as before, we have

$\displaystyle R_{\lambda}^{m}(x)$ $\displaystyle\geq I_{3}^{m}+I_{4}^{m}\geq
C_{1,1,m}\int_{\frac{1}{m}}^{\infty}t^{-1}e^{-\lambda t}dt\geq
C_{1,1,m}\int_{\Phi_{1,1}^{m}(|x|)}^{\infty}t^{-1}e^{-\lambda t}dt$
$\displaystyle=C_{1,1,m}E_{1}(\lambda\Phi_{1,1}^{m}(|x|))\geq\frac{C_{1,1,m}e^{-\lambda\Phi_{1,1}^{m}(|x|)}}{2}\log\left(1+\frac{2}{\lambda\Phi_{1,1}^{m}(|x|)}\right)\geq
C_{1,1,m,\lambda}\log\left(1+\frac{1}{|x|}\right).$

Finally, let $\alpha>d=1$ and make the split

$R_{\lambda}^{m}(x)\geq I_{3}^{m}+I_{4}^{m}.$

If $|x|\leq\frac{1}{m}$, then $I_{3}^{m}=0$ and

$I_{4}^{m}\geq
c_{2}m^{\frac{1}{\alpha}-\frac{1}{2}}\int_{\frac{1}{m}}^{\infty}e^{-\lambda
t}t^{-\frac{1}{2}}e^{-cm^{\frac{2}{\alpha}-1}|x|^{2}/t}dt.$

Since
$f_{\alpha}(r)=r^{\frac{1}{2}-\frac{1}{\alpha}}e^{-cm^{\frac{2}{\alpha}-1}|x|^{2}r}$
is decreasing, for $t>\frac{1}{m}$ and $|x|\leq 1$ we get

$t^{-\frac{1}{2}}e^{-cm^{\frac{2}{\alpha}-1}|x|^{2}/t}dt\geq
m^{\frac{1}{2}-\frac{1}{\alpha}}e^{-cm^{\frac{2}{\alpha}}}t^{-\frac{1}{\alpha}}$
(6.13)

and then

$I_{4}^{m}\geq
c_{2}e^{-cm^{\frac{2}{\alpha}}}\int_{\frac{1}{m}}^{\infty}e^{-\lambda
t}t^{-\frac{1}{\alpha}}dt.$

If $|x|>\frac{1}{m}$, we have

$I_{3}^{m}\geq
c_{2}m^{\frac{1}{\alpha}-\frac{1}{2}}e^{-cm^{\frac{1}{\alpha}}|x|}\int_{\frac{1}{m}}^{m^{\frac{1-\alpha}{\alpha}}|x|}t^{-\frac{1}{2}}e^{-\lambda
t}dt\geq
e^{-cm^{\frac{1}{\alpha}}}\int_{\frac{1}{m}}^{m^{\frac{1-\alpha}{\alpha}}|x|}t^{-\frac{1}{\alpha}}e^{-\lambda
t}dt$

and

$I_{4}^{m}\geq
c_{2}e^{-cm^{\frac{2}{\alpha}}}\int_{m^{\frac{1-\alpha}{\alpha}}|x|}^{\infty}e^{-\lambda
t}t^{-\frac{1}{\alpha}}dt,$

by (6.13) since $t>m^{\frac{1-\alpha}{\alpha}}|x|>\frac{1}{m}$. Thus we
finally obtain

$R_{\lambda}^{m}(x)\geq I_{3}^{m}+I_{4}^{m}\geq
C_{1,\alpha,\lambda,m}\int_{\frac{1}{m}}^{\infty}e^{-\lambda
t}t^{-\frac{1}{\alpha}}dt=C_{1,\alpha,\lambda,m}\Gamma\left(1-\frac{1}{\alpha},\frac{\lambda}{m}\right).$

∎

## Acknowledgments

G.A. is supported by the PRIN 2022XZSAFN Project “Anomalous Phenomena on
Regular and Irregular Domains: Approximating Complexity for the Applied
Sciences” CUP$\\_$E53D23005970006 and by the GNAMPA-INdAM group. A.I. thanks
JSPS KAKENHI Grant Numbers 20K03625, 21K03279 and 21KK0245. J.L. thanks the
hospitality of Tokyo University of Science on a research visit during which
part of this paper has been written.

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