By Theorem 4.1 in [MR3563609], it follows that $w$ vanishes identically, and
this proves the desired uniqueness result. ∎

The boundary behaviour of the Mittag-Leffler function for different values of
the fractional parameter $\alpha$ is depicted in Figure 3.1. In light of
(3.1), we notice in particular that, near $z=0$,

$E_{\alpha,\beta}\left(z\right)=\frac{1}{\Gamma\left(\beta\right)}+\frac{z}{\Gamma\left(\alpha+\beta\right)}+O(z^{2})$

and therefore, near $t=a$,

$E_{\alpha,1}\left(\lambda\left(t-a\right)^{\alpha}\right)=1+\frac{\lambda\left(t-a\right)^{\alpha}}{\Gamma\left(\alpha+1\right)}+O\big{(}\lambda^{2}\left(t-a\right)^{2\alpha}\big{)}.$

Figure 3.1: Behaviour of the Mittag-Leffler function
$E_{\alpha,1}\left(t^{\alpha}\right)$ near the origin for
$\alpha=\frac{1}{100}$, $\alpha=\frac{1}{20}$, $\alpha=\frac{1}{3}$,
$\alpha=\frac{2}{3}$ $\alpha=\frac{3}{2}$ and $\alpha=\frac{11}{2}$.

###  3.2 Sharp boundary behaviour for the time-fractional harmonic functions

In this section, we detect the optimal boundary behaviour of time-fractional
harmonic functions and of their derivatives. The result that we need for our
purposes is the following:

######  Lemma 3.3.

Let $\alpha\in(0,+\infty)\setminus\mathbb{N}$. There exists a function
$\psi:\mathbb{R}\to\mathbb{R}$ such that $\psi\in C^{\infty}((1,+\infty))$ and

$\displaystyle D^{\alpha}_{0}\psi(t)=0\qquad{\mbox{for all }}t\in(1,+\infty),$
(3.8) and $\displaystyle\lim_{\epsilon\searrow
0}\epsilon^{\ell-\alpha}\partial^{\ell}\psi(1+\epsilon
t)=\kappa_{\alpha,\ell}\,t^{\alpha-\ell},\qquad{\mbox{for all
}}\ell\in\mathbb{N},$ (3.9)

for some $\kappa_{\alpha,\ell}\in\mathbb{R}\setminus\\{0\\}$, where (3.9) is
taken in the sense of distribution for $t\in(0,+\infty)$.

###### Proof.

We use Lemma 2.5 in [CDV18], according to which (see in particular formula
(2.16) in [CDV18]) the claim in (3.8) holds true. Furthermore (see formulas
(2.19) and (2.20) in [CDV18]), we can write that, for all $t>1$,

$\psi(t)=-\frac{1}{\Gamma(\alpha)\Gamma([\alpha]+1-\alpha)}\iint_{[1,t]\times[0,3/4]}\partial^{[\alpha]+1}\psi_{0}(\sigma)\,(\tau-\sigma)^{[\alpha]-\alpha}\,(t-\tau)^{\alpha-1}\,d\tau\,d\sigma,$
(3.10)

for a suitable $\psi_{0}\in C^{[\alpha]+1}([0,1])$.

In addition, by Lemma 2.6 in [CDV18], we can write that

$\lim_{\epsilon\searrow 0}\epsilon^{-\alpha}\psi(1+\epsilon)=\kappa,$ (3.11)

for some $\kappa\neq 0$. Now we set

$(0,+\infty)\ni t\mapsto
f_{\epsilon}(t):=\epsilon^{\ell-\alpha}\partial^{\ell}\psi(1+\epsilon t).$

We observe that, for any $\varphi\in C^{\infty}_{0}((0,+\infty))$,

$\begin{split}&\int_{0}^{+\infty}f_{\epsilon}(t)\,\varphi(t)\,dt=\epsilon^{\ell-\alpha}\int_{0}^{+\infty}\partial^{\ell}\psi(1+\epsilon
t)\varphi(t)\,dt\\\
&=\epsilon^{-\alpha}\int_{0}^{+\infty}\frac{d^{\ell}}{dt^{\ell}}\big{(}\psi(1+\epsilon
t)\big{)}\varphi(t)\,dt=(-1)^{\ell}\,\epsilon^{-\alpha}\int_{0}^{+\infty}\psi(1+\epsilon
t)\,\partial^{\ell}\varphi(t)\,dt.\end{split}$ (3.12)

Also, in view of (3.10),

$\displaystyle\epsilon^{-\alpha}|\psi(1+\epsilon t)|$ $\displaystyle=$
$\displaystyle\left|\frac{\epsilon^{-\alpha}}{\Gamma(\alpha)\Gamma([\alpha]+1-\alpha)}\iint_{[1,1+\epsilon
t]\times[0,3/4]}\partial^{[\alpha]+1}\psi_{0}(\sigma)\,(\tau-\sigma)^{[\alpha]-\alpha}\,(1+\epsilon
t-\tau)^{\alpha-1}\,d\tau\,d\sigma\right|$ $\displaystyle\leq$ $\displaystyle
C\,\epsilon^{-\alpha}\,\int_{[1,1+\epsilon t]}(1+\epsilon
t-\tau)^{\alpha-1}\,d\tau$ $\displaystyle=$ $\displaystyle Ct^{\alpha},$

which is locally bounded in $t$, where $C>0$ here above may vary from line to
line.

As a consequence, we can pass to the limit in (3.12) and obtain that

$\lim_{\epsilon\searrow
0}\int_{0}^{+\infty}f_{\epsilon}(t)\,\varphi(t)\,dt=(-1)^{\ell}\,\int_{0}^{+\infty}\lim_{\epsilon\searrow
0}\epsilon^{-\alpha}\psi(1+\epsilon t)\,\partial^{\ell}\varphi(t)\,dt.$

This and (3.11) give that

$\lim_{\epsilon\searrow
0}\int_{0}^{+\infty}f_{\epsilon}(t)\,\varphi(t)\,dt=(-1)^{\ell}\,\kappa\,\int_{0}^{+\infty}t^{\alpha}\,\partial^{\ell}\varphi(t)\,dt=\kappa\,\alpha\dots(\alpha-\ell+1)\int_{0}^{+\infty}t^{\alpha-\ell}\,\varphi(t)\,dt,$

which establishes (3.9). ∎

##  Chapter 4 Boundary behaviour of solutions of space-fractional equations

In this chapter, we give precise asymptotics for the boundary behaviour of
solutions of space-fractional equations. The cases of the eigenfunctions and
of the Dirichlet problem with vanishing forcing term will be studied in
detail. To this end, we will also exploit useful representation formulas of
the solutions in terms of suitable Green functions.

###  4.1 Green representation formulas and solution of $(-\Delta)^{s}u=f$ in
$B_{1}$ with homogeneous Dirichlet datum

Our goal is to provide some representation results on the solution of
$(-\Delta)^{s}u=f$ in a ball, with $u=0$ outside this ball, for all $s>0$. Our
approach is an extension of the Green formula methods introduced in
[MR3673669] and [AJS3]: differently from the previous literature, we are not
assuming here that $f$ is regular in the whole of the ball, but merely that it
is Hölder continuous near the boundary and sufficiently integrable inside.
Given the type of singularity of the Green function, these assumptions are
sufficient to obtain meaningful representations, which in turn will be useful
to deal with the eigenfunction problem in the subsequent Chapter 4.2.

####  4.1.1 Solving $(-\Delta)^{s}u=f$ in $B_{1}$ for discontinuous $f$
vanishing near $\partial B_{1}$

Now, we want to extend the representation results of [MR3673669] and [AJS3] to
the case in which the right hand side is not Hölder continuous, but merely in
a Lebesgue space, but it has the additional property of vanishing near the
boundary of the domain. To this end, fixed $s>0$, we consider the polyharmonic
Green function in $B_{1}\subset{\mathbb{R}}^{n}$, given, for every $x\neq
y\in\mathbb{R}^{n}$, by

$\begin{split}&\mathcal{G}_{s}\left(x,y\right):=\frac{k(n,s)}{\left|x-y\right|^{n-2s}}\,\int_{0}^{r_{0}\left(x,y\right)}\frac{\eta^{s-1}}{\left(\eta+1\right)^{\frac{n}{2}}}\,d\eta,\\\
{\mbox{where
}}\quad&r_{0}\left(x,y\right):=\frac{\left(1-\left|x\right|^{2}\right)_{+}\left(1-\left|y\right|^{2}\right)_{+}}{\left|x-y\right|^{2}},\\\
{\mbox{with
}}\quad&k(n,s):=\frac{\Gamma\left(\frac{n}{2}\right)}{\pi^{\frac{n}{2}}\,4^{s}\Gamma^{2}\left(s\right)}.\end{split}$
(4.1)

Given $x\in B_{1}$, we also set

$d(x):=1-|x|.$ (4.2)

In this setting, we have:

######  Proposition 4.1.

Let $r\in(0,1)$ and $f\in L^{2}(B_{1})$, with $f=0$ in
${\mathbb{R}}^{n}\setminus B_{r}$. Let

$u(x):=\begin{cases}\displaystyle\int_{B_{1}}\mathcal{G}_{s}\left(x,y\right)\,f(y)\,dy&{\mbox{
if }}x\in B_{1},\\\ 0&{\mbox{ if }}x\in{\mathbb{R}}^{n}\setminus
B_{1}.\end{cases}$ (4.3)

Then:

$u\in L^{1}(B_{1}),{\mbox{ and }}\|u\|_{L^{1}(B_{1})}\leq
C\,\|f\|_{L^{1}(B_{1})},$ (4.4) ${\mbox{for every $R\in(r,1)$, }}\sup_{x\in
B_{1}\setminus B_{R}}d^{-s}(x)\,|u(x)|\leq C_{R}\,\|f\|_{L^{1}(B_{1})},$ (4.5)
${\mbox{$u$ satisfies }}(-\Delta)^{s}u=f{\mbox{ in }}B_{1}{\mbox{ in the sense
of distributions,}}$ (4.6)

and

$u\in W^{2s,2}_{loc}(B_{1}).$ (4.7)

Here above, $C>0$ is a constant depending on $n$, $s$ and $r$, $C_{R}>0$ is a
constant depending on $n$, $s$, $r$ and $R$ and $C_{\rho}>0$ is a constant
depending on $n$, $s$, $r$ and $\rho$.

When $f\in C^{\alpha}(B_{1})$ for some $\alpha\in(0,1)$, Proposition 4.1 boils
down to the main results of [MR3673669] and [AJS3].

###### Proof of Proposition 4.1.

We recall the following useful estimate, see Lemma 3.3 in [AJS3]: for any
$\epsilon\in\left(0,\,\min\\{n,s\\}\right)$, and any $\bar{R}$, $\bar{r}>0$,

$\frac{1}{\bar{R}^{n-2s}}\,\int_{0}^{\bar{r}/\bar{R}^{2}}\frac{\eta^{s-1}}{\left(\eta+1\right)^{\frac{n}{2}}}\,d\eta\leq\frac{2}{s}\;\frac{\bar{r}^{s-(\epsilon/2)}}{\bar{R}^{n-\epsilon}},$

and so, by (4.1) and (4.2), for every $x$, $y\in B_{1}$,

$\mathcal{G}_{s}\left(x,y\right)\leq\frac{C\,d^{s-(\epsilon/2)}(x)\,d^{s-(\epsilon/2)}(y)}{|x-y|^{n-\epsilon}}$

for some $C>0$. Hence, recalling (4.3),

$\displaystyle\int_{B_{1}}|u(x)|\,dx$ $\displaystyle\leq$
$\displaystyle\int_{B_{1}}\left(\int_{B_{1}}\mathcal{G}_{s}\left(x,y\right)\,|f(y)|\,dy\right)\,dx$
$\displaystyle\leq$ $\displaystyle
C\int_{B_{1}}\left(\int_{B_{1}}\frac{|f(y)|}{|x-y|^{n-\epsilon}}\,dy\right)\,dx$
$\displaystyle=$ $\displaystyle
C\int_{B_{1}}\left(\int_{B_{1}}\frac{|f(y)|}{|x-y|^{n-\epsilon}}\,dx\right)\,dy$
$\displaystyle=$ $\displaystyle C\int_{B_{1}}|f(y)|\,dy,$

up to renaming $C>0$ line after line, and this proves (4.4).

Now, if $x\in B_{1}\setminus B_{R}$ and $y\in B_{r}$, with $0<r<R<1$, we have
that

$|x-y|\geq|x|-|y|\geq R-r$

and accordingly

$r_{0}\left(x,y\right)\leq\frac{2d(x)}{(R-r)^{2}},$

which in turn implies that

$\mathcal{G}_{s}\left(x,y\right)\leq\frac{k(n,s)}{\left|x-y\right|^{n-2s}}\,\int_{0}^{{2d(x)}/{(R-r)^{2}}}\frac{\eta^{s-1}}{\left(\eta+1\right)^{\frac{n}{2}}}\,d\eta,\leq{C\,d^{s}(x)},$

for some $C>0$. As a consequence, since $f$ vanishes outside $B_{r}$, we see
that, for any $x\in B_{1}\setminus B_{R}$,

$\displaystyle|u(x)|\leq\int_{B_{r}}\mathcal{G}_{s}\left(x,y\right)\,|f(y)|\,dy\leq
C\,d^{s}(x)\int_{B_{r}}|f(y)|\,dy,$

which proves (4.5).

Now, we fix $\hat{r}\in(r,1)$ and consider a mollification of $f$, that we
denote by $f_{j}\in C^{\infty}_{0}(B_{\hat{r}})$, with $f_{j}\to f$ in
$L^{2}(B_{1})$ as $j\to+\infty$. We also write $\mathcal{G}_{s}*f$ as a short
notation for the right hand side of (4.3). Then, by [MR3673669] and [AJS3], we
know that $u_{j}:=\mathcal{G}_{s}*f_{j}$ is a (locally smooth, hence
distributional) solution of $(-\Delta)^{s}u_{j}=f_{j}$. Furthermore, if we set
$\tilde{u}_{j}:=u_{j}-u$ and $\tilde{f}_{j}:=f_{j}-f$ we have that

$\tilde{u}_{j}=\mathcal{G}_{s}*(f_{j}-f)=\mathcal{G}_{s}*\tilde{f}_{j},$

and therefore, by (4.4),

$\|\tilde{u}_{j}\|_{L^{1}(B_{1})}\leq C\,\|\tilde{f}_{j}\|_{L^{1}(B_{1})},$

which is infinitesimal as $j\to+\infty$. This says that $u_{j}\to u$ in
$L^{1}(B_{1})$ as $j\to+\infty$, and consequently, for any $\varphi\in
C^{\infty}_{0}(B_{1})$,

$\displaystyle\int_{B_{1}}u(x)\,(-\Delta)^{s}\varphi(x)\,dx=\lim_{j\to+\infty}\int_{B_{1}}u_{j}(x)\,(-\Delta)^{s}\varphi(x)\,dx$
$\displaystyle\qquad=\lim_{j\to+\infty}\int_{B_{1}}f_{j}(x)\,\varphi(x)\,dx=\int_{B_{1}}f(x)\,\varphi(x)\,dx,$

thus completing the proof of (4.6).

Now, to prove (4.7), we can suppose that
$s\in(0,+\infty)\setminus{\mathbb{N}}$, since the case of integer $s$ is
classical, see e.g. [MR1814364]. First of all, we claim that

(4.7) holds true for every $s\in(0,1)$. (4.8)

For this, we first claim that if $g\in C^{\infty}(B_{1})$ and $v$ is a
(locally smooth) solution of $(-\Delta)^{s}v=g$ in $B_{1}$, with $v=0$ outside
$B_{1}$, then $v\in W^{2s,2}_{loc}(B_{1})$, and, for any $\rho\in(0,1)$,

$\|v\|_{W^{2s,2}(B_{\rho})}\leq C_{\rho}\,\|g\|_{L^{2}(B_{1})}.$ (4.9)

This claim can be seen as a localization of Lemma 3.1 of [MR2863859], or a
quantification of the last claim in Theorem 1.3 of [MR3641649]. To prove
(4.9), we let $R_{-}<R_{+}\in(\rho,1)$, and consider $\eta\in
C^{\infty}_{0}(B_{R_{+}})$ with $\eta=1$ in $B_{R_{-}}$. We let
$v^{*}:=v\eta$, and we recall formulas (3.2), (3.3) and (A.5) in [MR3641649],
according to which

$\begin{split}&(-\Delta)^{s}v^{*}-\eta(-\Delta)^{s}v=g^{*}\quad{\mbox{ in
}}{\mathbb{R}}^{n},\\\ {\mbox{with
}}\quad&\|g^{*}\|_{L^{2}({\mathbb{R}}^{n})}\leq
C\,\|v\|_{W^{s,2}({\mathbb{R}}^{n})},\end{split}$

for some $C>0$.

Moreover, using a notation taken from [MR3641649] we denote by
$W^{s,2}_{0}(\overline{B_{1}})$ the space of functions in
$W^{s,2}({\mathbb{R}}^{n})$ vanishing outside $B_{1}$ and we consider the dual
space $W^{-s,2}_{0}(\overline{B_{1}})$. We remark that if $h\in L^{2}(B_{1})$
we can naturally identify $h$ as an element of
$W^{-s,2}_{0}(\overline{B_{1}})$ by considering the action of $h$ on any
$\varphi\in W^{s,2}_{0}(\overline{B_{1}})$ as defined by

$\int_{B_{1}}h(x)\,\varphi(x)\,dx.$

With respect to this, we have that

$\|h\|_{W^{-s,2}_{0}(\overline{B_{1}})}=\sup_{{\varphi\in
W^{s,2}_{0}(\overline{B_{1}})}\atop{\|\varphi\|_{W^{s,2}_{0}(\overline{B_{1}})}=1}}\int_{B_{1}}h(x)\,\varphi(x)\,dx\leq\|h\|_{L^{2}(B_{1})}.$
(4.10)

We notice also that

$\|v\|_{W^{s,2}({\mathbb{R}}^{n})}\leq C\,\|g\|_{W^{-s,2}(\overline{B_{1}})},$

in light of Proposition 2.1 of [MR3641649]. This and (4.10) give that

$\|v\|_{W^{s,2}({\mathbb{R}}^{n})}\leq C\,\|g\|_{L^{2}(B_{1})}.$

Then, by Lemma 3.1 of [MR2863859] (see in particular formula (3.2) there,
applied here with $\lambda:=0$), we obtain that

$\begin{split}\|v^{*}\|_{W^{2s,2}({\mathbb{R}}^{n})}\,&\leq
C\,\|\eta(-\Delta)^{s}v+g^{*}\|_{L^{2}({\mathbb{R}}^{n})}\\\ &\leq
C\,\big{(}\|(-\Delta)^{s}v\|_{L^{2}(B_{R_{+}})}+\|g^{*}\|_{L^{2}({\mathbb{R}}^{n})}\big{)}\\\
&=C\,\big{(}\|g\|_{L^{2}(B_{R_{+}})}+\|g^{*}\|_{L^{2}({\mathbb{R}}^{n})}\big{)}\\\
&\leq
C\,\big{(}\|g\|_{L^{2}(B_{1})}+\|v\|_{W^{s,2}({\mathbb{R}}^{n})}\big{)}\\\
&\leq C\,\|g\|_{L^{2}(B_{1})},\end{split}$ (4.11)

up to renaming $C>0$ step by step. On the other hand, since $v^{*}=v$ in
$B_{\rho}$,

$\|v\|_{W^{2s,2}(B_{\rho})}=\|v^{*}\|_{W^{2s,2}(B_{\rho})}\leq\|v^{*}\|_{W^{2s,2}({\mathbb{R}}^{n})}.$

From this and (4.11), we obtain (4.9), as desired.

Now, we let $f_{j}$, $\tilde{f}_{j}$, $u_{j}$ and $\tilde{u}_{j}$ as above and
make use of (4.9) to write

$\begin{split}&\|u_{j}\|_{W^{2s,2}(B_{\rho})}\leq
C_{\rho}\,\|f_{j}\|_{L^{2}(B_{1})}\\\ {\mbox{and
}}\quad&\|\tilde{u}_{j}\|_{W^{2s,2}(B_{\rho})}\leq
C_{\rho}\,\|\tilde{f}_{j}\|_{L^{2}(B_{1})}.\end{split}$ (4.12)

As a consequence, taking the limit as $j\to+\infty$ we obtain that

$\|u\|_{W^{2s,2}(B_{\rho})}\leq C_{\rho}\,\|f\|_{L^{2}(B_{1})},$

that is (4.7) in this case, namely the claim in (4.8).

Now, to prove (4.7), we argue by induction on the integer part of $s$. When
the integer part of $s$ is zero, the basis of the induction is warranted by
(4.8). Then, to perform the inductive step, given
$s\in(0,+\infty)\setminus{\mathbb{N}}$, we suppose that (4.7) holds true for
$s-1$, namely

$\mathcal{G}_{s-1}*f\in W^{2s-2,2}_{loc}(B_{1}).$ (4.13)

Then, following [AJS3], it is convenient to introduce the notation

$[x,y]:=\sqrt{|x|^{2}|y|^{2}-2x\cdot y+1}$

and consider the auxiliary kernel given, for every $x\neq y\in B_{1}$, by

$P_{s-1}(x,y):=\frac{(1-|x|^{2})^{s-2}_{+}(1-|y|^{2})^{s-1}_{+}(1-|x|^{2}|y|^{2})}{[x,y]^{n}}.$
(4.14)

We point out that if $x\in B_{r}$ with $r\in(0,1)$, then

$[x,y]^{2}\geq|x|^{2}|y|^{2}-2|x|\,|y|+1=(1-|x|\,|y|)^{2}\geq(1-r)^{2}>0.$
(4.15)

Consequently, since $f$ is supported in $B_{r}$,

$P_{s-1}*f\in C^{\infty}({\mathbb{R}}^{n}).$ (4.16)

Then, we recall that

$-\Delta_{x}\mathcal{G}_{s}(x,y)=\mathcal{G}_{s-1}(x,y)-CP_{s-1}(x,y),$ (4.17)

for some $C\in{\mathbb{R}}$, see Lemma 3.1 in [AJS3].

As a consequence, in view of (4.13), (4.16), (4.17), we conclude that

$-\Delta(\mathcal{G}_{s}*f)=(-\Delta_{x}\mathcal{G}_{s})*f\in
W^{2s-2,2}_{loc}(B_{1}).$

This and the classical elliptic regularity theory (see e.g. [MR1814364]) give
that $\mathcal{G}_{s}*f\in W^{2s,2}_{loc}(B_{1})$, which completes the
inductive proof and establishes (4.7). ∎

####  4.1.2 Solving $(-\Delta)^{s}u=f$ in $B_{1}$ for $f$ Hölder continuous
near $\partial B_{1}$

Our goal is now to extend the representation results of [MR3673669] and [AJS3]
to the case in which the right hand side is not Hölder continuous in the whole
of the ball, but merely in a neighborhood of the boundary. This result is
obtained here by superposing the results in [MR3673669] and [AJS3] with
Proposition 4.1 here, taking advantage of the linear structure of the problem.

######  Proposition 4.2.

Let $f\in L^{2}(B_{1})$. Let $\alpha$, $r\in(0,1)$ and assume that

$f\in C^{\alpha}(B_{1}\setminus B_{r}).$ (4.18)

In the notation of (4.1), let

$u(x):=\begin{cases}\displaystyle\int_{B_{1}}\mathcal{G}_{s}\left(x,y\right)\,f(y)\,dy&{\mbox{
if }}x\in B_{1},\\\ 0&{\mbox{ if }}x\in{\mathbb{R}}^{n}\setminus
B_{1}.\end{cases}$ (4.19)

Then, in the notation of (4.2), we have that:

${\mbox{for every $R\in(r,1)$, }}\sup_{x\in B_{1}\setminus
B_{R}}d^{-s}(x)\,|u(x)|\leq
C_{R}\,\big{(}\|f\|_{L^{1}(B_{1})}+\|f\|_{L^{\infty}(B_{1}\setminus
B_{r})}\big{)},$ (4.20) ${\mbox{$u$ satisfies }}(-\Delta)^{s}u=f{\mbox{ in
}}B_{1}{\mbox{ in the sense of distributions,}}$ (4.21)

and

$u\in W^{2s,2}_{loc}(B_{1}).$ (4.22)

Here above, $C>0$ is a constant depending on $n$, $s$ and $r$, $C_{R}>0$ is a
constant depending on $n$, $s$, $r$ and $R$ and $C_{\rho}>0$ is a constant
depending on $n$, $s$, $r$ and $\rho$.

###### Proof.

We take $r_{1}\in(r,1)$ and $\eta\in C^{\infty}_{0}(B_{r_{1}})$ with $\eta=1$
in $B_{r}$. Let also

$f_{1}:=f\eta\qquad{\mbox{and}}\qquad f_{2}:=f-f_{1}.$

We observe that $f_{1}\in L^{2}(B_{1})$, and that $f_{1}=0$ outside
$B_{r_{1}}$. Therefore, we are in the position of applying Proposition 4.1 and
find a function $u_{1}$ (obtained by convolving $\mathcal{G}_{s}$ against
$f_{1}$) such that

$\displaystyle{\mbox{for every $R\in(r_{1},1)$, }}\sup_{x\in B_{1}\setminus
B_{R}}d^{-s}(x)\,|u_{1}(x)|\leq C_{R}\,\|f_{1}\|_{L^{1}(B_{1})},$ (4.24)
$\displaystyle{\mbox{$u_{1}$ satisfies }}(-\Delta)^{s}u_{1}=f_{1}{\mbox{ in
}}B_{1}{\mbox{ in the sense of distributions,}}$ and $\displaystyle u_{1}\in
W^{2s,2}_{loc}(B_{1}).$ (4.25)

On the other hand, we have that $f_{2}=f(1-\eta)$ vanishes outside
$B_{1}\setminus B_{r}$ and it is Hölder continuous. Accordingly, we can apply
Theorem 1.1 of [AJS3] and find a function $u_{2}$ (obtained by convolving
$\mathcal{G}_{s}$ against $f_{2}$) such that

$\displaystyle{\mbox{for every $R\in(r_{1},1)$, }}\sup_{x\in B_{1}\setminus
B_{R}}d^{-s}(x)\,|u_{2}(x)|\leq C_{R}\,\|f_{2}\|_{L^{\infty}(B_{1})},$ (4.27)
$\displaystyle{\mbox{$u_{2}$ satisfies }}(-\Delta)^{s}u_{2}=f_{2}{\mbox{ in
}}B_{1}{\mbox{ in the sense of distributions,}}$ and $\displaystyle u_{2}\in
C^{2s+\alpha}_{loc}(B_{1}).$ (4.28)

Then, $f=f_{1}+f_{2}$, and thus, in view of (4.19), we have that
$u=u_{1}+u_{2}$. Also, $u$ satisfies (4.20), thanks to (4.24) and (4.27),
(4.21), thanks to (4.24) and (4.27), and (4.22), thanks to (4.25) and (4.28).
∎

###  4.2 Existence and regularity for the first eigenfunction of the higher
order fractional Laplacian

The goal of these pages is to study the boundary behaviour of the first
Dirichlet eigenfunction of higher order fractional equations.

For this, writing $s=m+\sigma$, with $m\in{\mathbb{N}}$ and $\sigma\in(0,1)$,
we define the energy space

$H_{0}^{s}\left(B_{1}\right):=\left\\{u\in
H^{s}\left(\mathbb{R}^{n}\right);\;u=0\;\text{in}\;\mathbb{R}^{n}\setminus
B_{1}\right\\},$ (4.29)

endowed with the Hilbert norm

$\left\|u\right\|_{H_{0}^{s}\left(B_{1}\right)}:=\left(\sum_{\left|\alpha\right|\leq
m}{\left\|\partial^{\alpha}u\right\|^{2}_{L^{2}\left(B_{1}\right)}}+\mathcal{E}_{s}\left(u,u\right)\right)^{\frac{1}{2}},$
(4.30)

where

$\mathcal{E}_{s}\left(u,v\right)=\int_{\mathbb{R}^{n}}{\left|\xi\right|^{2s}\mathcal{F}u\left(\xi\right)\overline{\mathcal{F}v\left(\xi\right)}\,d\xi},$
(4.31)

being $\mathcal{F}$ the Fourier transform and using the notation
$\overline{z}$ to denote the complex conjugated of a complex number $z$.

In this setting, we consider $u\in H^{s}_{0}(B_{1})$ to be such that

$\begin{cases}\left(-\Delta\right)^{s}u=\lambda_{1}u&\quad\text{ in }B_{1},\\\
u=0&\quad\text{ in }\mathbb{R}^{n}\setminus\overline{B_{1}},\end{cases}$
(4.32)

for every $s>0$, with $\lambda_{1}$ as small as possible.

The existence of solutions of (4.32) is ensured via variational techniques, as
stated in the following result:

######  Lemma 4.3.

The functional $\mathcal{E}_{s}\left(u,u\right)$ attains its minimum
$\lambda_{1}$ on the functions in $H^{s}_{0}(B_{1})$ with unit norm in
$L^{2}(B_{1})$.

The minimizer satisfies (4.32).

In addition, $\lambda_{1}>0$.

###### Proof.

The proof is based on the direct method in the calculus of variations. We
provide some details for completeness. Let $s=m+\sigma$, with $m\in\mathbb{N}$
and $\sigma\in(0,1)$. Let us consider a minimizing sequence $u_{j}\in
H^{s}_{0}(B_{1})\subseteq H^{m}({\mathbb{R}}^{n})$ such that
$\|u_{j}\|_{L^{2}(B_{1})}=1$ and

$\lim_{j\to+\infty}\mathcal{E}_{s}\left(u_{j},u_{j}\right)=\inf_{{u\in
H^{s}_{0}(B_{1})}\atop{\|u\|_{L^{2}(B_{1})}=1}}\mathcal{E}_{s}\left(u,u\right).$

In particular, we have that $u_{j}$ is bounded in $H^{s}_{0}(B_{1})$ uniformly
in $j$, so, up to a subsequence, it converges to some $u_{\star}$ weakly in
$H^{s}_{0}(B_{1})$ and strongly in $L^{2}(B_{1})$ as $j\to+\infty$.

The weak lower semicontinuity of the seminorm
$\mathcal{E}_{s}\left(\cdot,\cdot\right)$ then implies that $u_{\star}$ is the
desired minimizer.

Then, given $\phi\in C^{\infty}_{0}(B_{1})$, we have that

$\mathcal{E}_{s}\left(u_{\star}+\epsilon\phi,u_{\star}+\epsilon\phi\right)\geq\mathcal{E}_{s}\left(u_{\star},u_{\star}\right),$

for every $\epsilon\in{\mathbb{R}}$, and this gives that (4.32) is satisfied
in the sense of distributions, and also in the classical sense by the elliptic
regularity theory.

Finally, we have that $\mathcal{E}_{s}\left(u_{\star},u_{\star}\right)>0$,
since $u_{\star}$ (and thus ${\mathcal{F}}u_{\star}$) does not vanish
identically. Consequently,

$\lambda_{1}=\frac{\mathcal{E}_{s}\left(u_{\star},u_{\star}\right)}{\|u_{\star}\|_{L^{2}(B_{1})}^{2}}=\mathcal{E}_{s}\left(u_{\star},u_{\star}\right)>0,$

as desired. ∎

Our goal is now to apply Proposition 4.2 to solutions of (4.32), taking
$f:=\lambda u$. To this end, we have to check that condition (4.18) is
satisfied, namely that solutions of (4.32) are Hölder continuous in
$B_{1}\setminus B_{r}$, for any $0<r<1$.

To this aim, we prove that polyharmonic operators of any order $s>0$ always
admit a first eigenfunction in the ball which does not change sign and which
is radially symmetric. For this, we start discussing the sign property:

######  Lemma 4.4.

There exists a nontrivial solution of (4.32) that does not change sign.

###### Proof.

We exploit a method explained in detail in Section 3.1 of [MR2667016]. As a
matter of fact, when $s\in{\mathbb{N}}$, the desired result is exactly Theorem
3.7 in [MR2667016].

Let $u$ be as in Lemma 4.3. If either $u\geq 0$ or $u\leq 0$, then the desired
result is proved. Hence, we argue by contradiction, assuming that $u$ attains
strictly positive and strictly negative values. We define

${\mathcal{K}}:=\\{w:{\mathbb{R}}^{n}\to{\mathbb{R}}{\mbox{ s.t.
$\mathcal{E}_{s}\left(w,w\right)<+\infty$, and $w\geq 0$ in $B_{1}$}}\\}.$

Also, we set

${\mathcal{K}}^{\star}:=\\{w\in H^{s}_{0}(B_{1}){\mbox{ s.t.
}}\mathcal{E}_{s}\left(w,v\right)\leq 0{\mbox{ for all
}}v\in{\mathcal{K}}\\}.$

We claim that

${\mbox{if\leavevmode\nobreak\ $w\in{\mathcal{K}}^{\star}$, then $w\leq 0$}}.$
(4.33)

To prove this, we recall the notation in (4.1), take $\phi\in
C^{\infty}_{0}(B_{1})\cap{\mathcal{K}}$, and let

$v_{\phi}(x):=\begin{cases}\displaystyle\int_{B_{1}}\mathcal{G}_{s}\left(x,y\right)\,\phi(y)\,dy&{\mbox{
if }}x\in B_{1},\\\ 0&{\mbox{ if }}x\in{\mathbb{R}}^{n}\setminus
B_{1}.\end{cases}$

Then $v_{\phi}\in{\mathcal{K}}$ and it satisfies
$\left(-\Delta\right)^{s}v_{\phi}=\phi$ in $B_{1}$, thanks to [MR3673669] or
[AJS3].

Consequently, we can write, for every $x\in B_{1}$,

$\phi(x)={\mathcal{F}}^{-1}(|\xi|^{2s}{\mathcal{F}}v_{\phi})(x).$

Hence, for every $w\in{\mathcal{K}}^{\star}$,

$\displaystyle 0$ $\displaystyle\geq$
$\displaystyle\mathcal{E}_{s}\left(w,v_{\phi}\right)$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{n}}\left|\xi\right|^{2s}\mathcal{F}v_{\phi}\left(\xi\right)\overline{\mathcal{F}w\left(\xi\right)}\,d\xi$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{n}}{\mathcal{F}}^{-1}(\left|\xi\right|^{2s}\mathcal{F}v_{\phi})(x)\,w\left(x\right)\,dx$
$\displaystyle=$
$\displaystyle\int_{B_{1}}{\mathcal{F}}^{-1}(\left|\xi\right|^{2s}\mathcal{F}v_{\phi})(x)\,w\left(x\right)\,dx$
$\displaystyle=$ $\displaystyle\int_{B_{1}}\phi(x)\,w\left(x\right)\,dx.$

Since $\phi$ is arbitrary and nonnegative, this gives that $w\leq 0$, and this
establishes (4.33).

Furthermore, by Theorem 3.4 in [MR2667016], we can write

$u=u_{1}+u_{2},$

with $u_{1}\in{\mathcal{K}}\setminus\\{0\\}$,
$u_{2}\in{\mathcal{K}}^{\star}\setminus\\{0\\}$, and
$\mathcal{E}_{s}\left(u_{1},u_{2}\right)=0$.

We observe that

$\mathcal{E}_{s}\left(u_{1}-u_{2},u_{1}-u_{2}\right)=\mathcal{E}_{s}\left(u_{1},u_{1}\right)+\mathcal{E}_{s}\left(u_{2},u_{2}\right)+2\mathcal{E}_{s}\left(u_{1},u_{2}\right)=\mathcal{E}_{s}\left(u_{1},u_{1}\right)+\mathcal{E}_{s}\left(u_{2},u_{2}\right).$

In the same way,

$\mathcal{E}_{s}\left(u,u\right)=\mathcal{E}_{s}\left(u_{1}+u_{2},u_{1}+u_{2}\right)=\mathcal{E}_{s}\left(u_{1},u_{1}\right)+\mathcal{E}_{s}\left(u_{2},u_{2}\right),$

and therefore

$\mathcal{E}_{s}\left(u_{1}-u_{2},u_{1}-u_{2}\right)=\mathcal{E}_{s}\left(u,u\right).$
(4.34)

On the other hand,

$\displaystyle\|u_{1}-u_{2}\|_{L^{2}(B_{1})}^{2}-\|u\|_{L^{2}(B_{1})}^{2}$
$\displaystyle=$
$\displaystyle\|u_{1}-u_{2}\|_{L^{2}(B_{1})}^{2}-\|u_{1}+u_{2}\|_{L^{2}(B_{1})}^{2}$
$\displaystyle=$ $\displaystyle-4\int_{B_{1}}u_{1}(x)\,u_{2}(x)\,dx.$

As a consequence, since $u_{2}\leq 0$ in view of (4.33), we conclude that

$\|u_{1}-u_{2}\|_{L^{2}(B_{1})}^{2}-\|u\|_{L^{2}(B_{1})}^{2}\geq 0.$

This and (4.34) say that the function $u_{1}-u_{2}$ is also a minimizer for
the variational problem in Lemma 4.3. Since now $u_{1}-u_{2}\geq 0$, the
desired result follows. ∎

Now, we define the spherical mean of a function $v$ by

$v_{\sharp}(x):=\frac{1}{\left|\mathbb{S}^{n-1}\right|}\int_{\mathbb{S}^{n-1}}v({\mathcal{R}}_{\omega}\,x)\,d{\mathcal{H}}^{n-1}(\omega)$

where ${\mathcal{R}}_{\omega}$ is the rotation corresponding to the solid
angle $\omega\in{\mathbb{S}^{n-1}}$, ${\mathcal{H}}^{n-1}$ is the standard
Hausdorff measure, and
$\left|\mathbb{S}^{n-1}\right|={\mathcal{H}}^{n-1}(\mathbb{S}^{n-1})$. Notice
that $v_{\sharp}(x)=v_{\sharp}({\mathcal{R}}_{\varpi}x)$ for any
$\varpi\in\mathbb{S}^{n-1}$, that is $v_{\sharp}$ is rotationally invariant.

Then, we have:

######  Lemma 4.5.

Any positive power of the Laplacian commutes with the spherical mean, that is

$\big{(}(-\Delta)^{s}v\big{)}_{\sharp}(x)=(-\Delta)^{s}v_{\sharp}(x).$

###### Proof.

By density, we prove the claim for a function $v$ in the Schwartz space of
smooth and rapidly decreasing functions. In this setting, writing
${\mathcal{R}}_{\omega}^{T}$ to denote the transpose of the rotation
${\mathcal{R}}_{\omega}$, and changing variable
$\eta:={\mathcal{R}}_{\omega}^{T}\,\xi$, we have that

$\begin{split}(-\Delta)^{s}v({\mathcal{R}}_{\omega}\,x)\,&=\int_{{\mathbb{R}}^{n}}|\xi|^{2s}{\mathcal{F}}v(\xi)\,e^{2\pi
i{\mathcal{R}}_{\omega}\,x\cdot\xi}\,d\xi\\\
&=\int_{{\mathbb{R}}^{n}}|\xi|^{2s}{\mathcal{F}}v(\xi)\,e^{2\pi
ix\cdot{\mathcal{R}}_{\omega}^{T}\,\xi}\,d\xi\\\
&=\int_{{\mathbb{R}}^{n}}|\eta|^{2s}{\mathcal{F}}v({\mathcal{R}}_{\omega}\,\eta)\,e^{2\pi
ix\cdot\eta}\,d\eta.\end{split}$ (4.35)

On the other hand, using the substitution $y:={\mathcal{R}}_{\omega}^{T}\,x$,

$\displaystyle{\mathcal{F}}v({\mathcal{R}}_{\omega}\,\eta)$ $\displaystyle=$
$\displaystyle\int_{{\mathbb{R}}^{n}}v(x)\,e^{-2\pi
ix\cdot{\mathcal{R}}_{\omega}\,\eta}\,dx$ $\displaystyle=$
$\displaystyle\int_{{\mathbb{R}}^{n}}v(x)\,e^{-2\pi
i{\mathcal{R}}_{\omega}^{T}\,x\cdot\eta}\,dx$ $\displaystyle=$
$\displaystyle\int_{{\mathbb{R}}^{n}}v({\mathcal{R}}_{\omega}\,y)\,e^{-2\pi
iy\cdot\eta}\,dy,$

and therefore, recalling (4.35),

$(-\Delta)^{s}v({\mathcal{R}}_{\omega}\,x)=\iint_{{\mathbb{R}}^{n}\times{\mathbb{R}}^{n}}|\eta|^{2s}v({\mathcal{R}}_{\omega}\,y)\,e^{2\pi
i(x-y)\cdot\eta}\,dy\,d\eta.$

As a consequence,

$\displaystyle\big{(}(-\Delta)^{s}v\big{)}_{\sharp}(x)$ $\displaystyle=$
$\displaystyle\frac{1}{\left|\mathbb{S}^{n-1}\right|}\int_{\mathbb{S}^{n-1}}(-\Delta)^{s}v({\mathcal{R}}_{\omega}\,x)\,d{\mathcal{H}}^{n-1}(\omega)$
$\displaystyle=$
$\displaystyle\frac{1}{\left|\mathbb{S}^{n-1}\right|}\iiint_{\mathbb{S}^{n-1}\times{\mathbb{R}}^{n}\times{\mathbb{R}}^{n}}|\eta|^{2s}v({\mathcal{R}}_{\omega}\,y)\,e^{2\pi
i(x-y)\cdot\eta}\,d{\mathcal{H}}^{n-1}(\omega)\,dy\,d\eta$ $\displaystyle=$
$\displaystyle\iint_{{\mathbb{R}}^{n}\times{\mathbb{R}}^{n}}|\eta|^{2s}v_{\sharp}(y)\,e^{2\pi
i(x-y)\cdot\eta}\,dy\,d\eta$ $\displaystyle=$
$\displaystyle\int_{{\mathbb{R}}^{n}}|\eta|^{2s}{\mathcal{F}}(v_{\sharp})(\eta)\,e^{2\pi
ix\cdot\eta}\,d\eta$ $\displaystyle=$
$\displaystyle(-\Delta)^{s}v_{\sharp}(x),$

as desired. ∎

It is also useful to observe that the spherical mean is compatible with the
energy bounds. In particular we have the following observation:

######  Lemma 4.6.

We have that

$\mathcal{E}_{s}\left(v_{\sharp},v_{\sharp}\right)\leq\mathcal{E}_{s}\left(v,v\right).$
(4.36)

Moreover,

if $v\in H^{s}_{0}(B_{1})$, then so does $v_{\sharp}$. (4.37)

###### Proof.

We see that

$\displaystyle{\mathcal{F}}(v_{\sharp})(\xi)$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{n}}v_{\sharp}(x)\,e^{-2\pi ix\cdot\xi}\,dx$
$\displaystyle=$
$\displaystyle\frac{1}{\left|\mathbb{S}^{n-1}\right|}\iint_{\mathbb{S}^{n-1}\times{\mathbb{R}^{n}}}v({\mathcal{R}}_{\omega}\,x)\,e^{-2\pi
ix\cdot\xi}\,d{\mathcal{H}}^{n-1}(\omega)\,dx$

and therefore, taking the complex conjugated,

$\overline{{\mathcal{F}}(v_{\sharp})(\xi)}=\frac{1}{\left|\mathbb{S}^{n-1}\right|}\iint_{\mathbb{S}^{n-1}\times{\mathbb{R}^{n}}}v({\mathcal{R}}_{\omega}\,x)\,e^{2\pi
ix\cdot\xi}\,d{\mathcal{H}}^{n-1}(\omega)\,dx.$

Hence, by (4.31), and exploiting the changes of variables
$y:={\mathcal{R}}_{\omega}\,x$ and
$\tilde{y}:={\mathcal{R}}_{\tilde{\omega}}\,\tilde{x}$,

$\displaystyle\mathcal{E}_{s}\left(v_{\sharp},v_{\sharp}\right)$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{n}}\left|\xi\right|^{2s}\mathcal{F}(v_{\sharp})\left(\xi\right)\overline{\mathcal{F}(v_{\sharp})\left(\xi\right)}\,d\xi$
$\displaystyle=$
$\displaystyle\frac{1}{\left|\mathbb{S}^{n-1}\right|^{2}}\iiiint\\!\\!\\!\int_{\mathbb{S}^{n-1}\times\mathbb{S}^{n-1}\times{\mathbb{R}^{n}}\times{\mathbb{R}^{n}}\times{\mathbb{R}^{n}}}|\xi|^{2s}v({\mathcal{R}}_{\omega}\,x)\,v({\mathcal{R}}_{\tilde{\omega}}\,\tilde{x})\,e^{2\pi
i(\tilde{x}-x)\cdot\xi}\,d{\mathcal{H}}^{n-1}(\omega)\,d{\mathcal{H}}^{n-1}(\tilde{\omega})\,dx\,d\tilde{x}\,d\xi$
$\displaystyle=$
$\displaystyle\frac{1}{\left|\mathbb{S}^{n-1}\right|^{2}}\iiiint\\!\\!\\!\int_{\mathbb{S}^{n-1}\times\mathbb{S}^{n-1}\times{\mathbb{R}^{n}}\times{\mathbb{R}^{n}}\times{\mathbb{R}^{n}}}|\xi|^{2s}v(y)\,v(\tilde{y})\,e^{2\pi
i\tilde{y}\cdot{\mathcal{R}}_{\tilde{\omega}}\,\xi}e^{-2\pi
iy\cdot{\mathcal{R}}_{\omega}\,\xi}\,d{\mathcal{H}}^{n-1}(\omega)\,d{\mathcal{H}}^{n-1}(\tilde{\omega})\,dy\,d\tilde{y}\,d\xi$
$\displaystyle=$
$\displaystyle\frac{1}{\left|\mathbb{S}^{n-1}\right|^{2}}\iiint_{\mathbb{S}^{n-1}\times\mathbb{S}^{n-1}\times{\mathbb{R}^{n}}}|\xi|^{2s}\,{\mathcal{F}}v({\mathcal{R}}_{\omega}\,\xi)\,\overline{{\mathcal{F}}v({\mathcal{R}}_{\tilde{\omega}}\,\xi)}\,\,d{\mathcal{H}}^{n-1}(\omega)\,d{\mathcal{H}}^{n-1}(\tilde{\omega})\,d\xi.$

Consequently, using the Cauchy-Schwarz Inequality, and the substitutions
$\eta:={\mathcal{R}}_{\omega}\,\xi$ and
$\tilde{\eta}:={\mathcal{R}}_{\tilde{\omega}}\,\xi$,

$\displaystyle\mathcal{E}_{s}\left(v_{\sharp},v_{\sharp}\right)$
$\displaystyle\leq$
$\displaystyle\frac{1}{\left|\mathbb{S}^{n-1}\right|^{2}}\iiint_{\mathbb{S}^{n-1}\times\mathbb{S}^{n-1}\times{\mathbb{R}^{n}}}|\xi|^{2s}\,\big{|}{\mathcal{F}}v({\mathcal{R}}_{\omega}\,\xi)\big{|}\,\big{|}{\mathcal{F}}v({\mathcal{R}}_{\tilde{\omega}}\,\xi)\big{|}\,\,d{\mathcal{H}}^{n-1}(\omega)\,d{\mathcal{H}}^{n-1}(\tilde{\omega})\,d\xi$
$\displaystyle\leq$
$\displaystyle\frac{1}{\left|\mathbb{S}^{n-1}\right|^{2}}\left(\iiint_{\mathbb{S}^{n-1}\times\mathbb{S}^{n-1}\times{\mathbb{R}^{n}}}|\xi|^{2s}\,\big{|}{\mathcal{F}}v({\mathcal{R}}_{\omega}\,\xi)\big{|}^{2}\,d{\mathcal{H}}^{n-1}(\omega)\,d{\mathcal{H}}^{n-1}(\tilde{\omega})\,d\xi\right)^{\frac{1}{2}}$
$\displaystyle\qquad\cdot\left(\iiint_{\mathbb{S}^{n-1}\times\mathbb{S}^{n-1}\times{\mathbb{R}^{n}}}|\xi|^{2s}\,\big{|}{\mathcal{F}}v({\mathcal{R}}_{\tilde{\omega}}\,\xi)\big{|}^{2}\,d{\mathcal{H}}^{n-1}(\omega)\,d{\mathcal{H}}^{n-1}(\tilde{\omega})\,d\xi\right)^{\frac{1}{2}}$
$\displaystyle=$
$\displaystyle\frac{1}{\left|\mathbb{S}^{n-1}\right|^{2}}\left(\iiint_{\mathbb{S}^{n-1}\times\mathbb{S}^{n-1}\times{\mathbb{R}^{n}}}|\eta|^{2s}\,\big{|}{\mathcal{F}}v(\eta)\big{|}^{2}\,d{\mathcal{H}}^{n-1}(\omega)\,d{\mathcal{H}}^{n-1}(\tilde{\omega})\,d\eta\right)^{\frac{1}{2}}$
$\displaystyle\qquad\cdot\left(\iiint_{\mathbb{S}^{n-1}\times\mathbb{S}^{n-1}\times{\mathbb{R}^{n}}}|\tilde{\eta}|^{2s}\,\big{|}{\mathcal{F}}v(\tilde{\eta})\big{|}^{2}\,d{\mathcal{H}}^{n-1}(\omega)\,d{\mathcal{H}}^{n-1}(\tilde{\omega})\,d\tilde{\eta}\right)^{\frac{1}{2}}$
$\displaystyle=$
$\displaystyle\left(\int_{{\mathbb{R}^{n}}}|\eta|^{2s}\,\big{|}{\mathcal{F}}v(\eta)\big{|}^{2}\,d\eta\right)^{\frac{1}{2}}\left(\int_{\mathbb{R}^{n}}|\tilde{\eta}|^{2s}\,\big{|}{\mathcal{F}}v(\tilde{\eta})\big{|}^{2}\,d\tilde{\eta}\right)^{\frac{1}{2}}$
$\displaystyle=$ $\displaystyle\mathcal{E}_{s}\left(v,v\right).$

This proves (4.36).

Now, we prove (4.37). For this, we observe that

$\frac{\partial^{\ell}v_{\sharp}}{\partial x_{j_{1}}\dots\partial
x_{j_{\ell}}}(x)=\frac{1}{\left|\mathbb{S}^{n-1}\right|}\sum_{k_{1},\dots,k_{\ell}=1}^{n}\int_{\mathbb{S}^{n-1}}\frac{\partial^{\ell}v}{\partial
x_{k_{1}}\dots\partial
x_{k_{\ell}}}({\mathcal{R}}_{\omega}\,x)\;{\mathcal{R}}_{\omega}^{k_{1}j_{1}}\dots\mathcal{R}_{\omega}^{k_{\ell}j_{\ell}}\;d{\mathcal{H}}^{n-1}(\omega),$

for every $\ell\in{\mathbb{N}}$ and $j_{1},\dots,j_{\ell}\in\\{1,\dots,n\\}$,
where ${\mathcal{R}}_{\omega}^{jk}$ denotes the $(j,k)$ component of the
matrix ${\mathcal{R}}_{\omega}$. In particular,

$\left|\frac{\partial^{\ell}v_{\sharp}}{\partial x_{j_{1}}\dots\partial
x_{j_{\ell}}}(x)\right|\leq
C\,\sum_{k_{1},\dots,k_{\ell}=1}^{n}\int_{\mathbb{S}^{n-1}}\left|\frac{\partial^{\ell}v}{\partial
x_{k_{1}}\dots\partial
x_{k_{\ell}}}({\mathcal{R}}_{\omega}\,x)\right|\;d{\mathcal{H}}^{n-1}(\omega),$

for some $C>0$ only depending on $n$ and $\ell$, and hence

$\displaystyle\left\|\frac{\partial^{\ell}v_{\sharp}}{\partial
x_{j_{1}}\dots\partial x_{j_{\ell}}}(x)\right\|_{L^{2}(B_{1})}^{2}$
$\displaystyle\leq$ $\displaystyle
C\,\sum_{k_{1},\dots,k_{\ell}=1}^{n}\iint_{\mathbb{S}^{n-1}\times
B_{1}}\left|\frac{\partial^{\ell}v}{\partial x_{k_{1}}\dots\partial
x_{k_{\ell}}}({\mathcal{R}}_{\omega}\,x)\right|^{2}\;d{\mathcal{H}}^{n-1}(\omega)\,dx$
$\displaystyle=$ $\displaystyle
C\,\sum_{k_{1},\dots,k_{\ell}=1}^{n}\iint_{\mathbb{S}^{n-1}\times
B_{1}}\left|\frac{\partial^{\ell}v}{\partial x_{k_{1}}\dots\partial
x_{k_{\ell}}}(y)\right|^{2}\;d{\mathcal{H}}^{n-1}(\omega)\,dy$
$\displaystyle=$ $\displaystyle
C\,\sum_{k_{1},\dots,k_{\ell}=1}^{n}\left\|\frac{\partial^{\ell}v}{\partial
x_{k_{1}}\dots\partial x_{k_{\ell}}}\right\|_{L^{2}(B_{1})}^{2},$

up to renaming $C$.

This, together with (4.30) and (4.36), gives (4.37), as desired. ∎

With this preliminary work, we can now find a nontrivial, nonnegative and
radial solution of (4.32).

######  Proposition 4.7.

There exists a solution of (4.32) in $H^{s}_{0}(B_{1})$ which is radial,
nonnegative and with unit norm in $L^{2}(B_{1})$.

###### Proof.

Let $u\geq 0$ be a nontrivial solution of (4.32), whose existence is warranted
by Lemma 4.4.

Then, we have that $u_{\sharp}\geq 0$. Moreover,

$\displaystyle\int_{B_{1}}u_{\sharp}(x)\,dx=\frac{1}{\left|\mathbb{S}^{n-1}\right|}\iint_{\mathbb{S}^{n-1}\times
B_{1}}u({\mathcal{R}}_{\omega}\,x)\,d{\mathcal{H}}^{n-1}(\omega)\,dx$
$\displaystyle\qquad=\frac{1}{\left|\mathbb{S}^{n-1}\right|}\iint_{\mathbb{S}^{n-1}\times
B_{1}}u(y)\,d{\mathcal{H}}^{n-1}(\omega)\,dy=\int_{B_{1}}u(y)\,dy>0,$

and therefore $u_{\sharp}$ does not vanish identically.

As a consequence, we can define

$u_{\star}:=\frac{u_{\sharp}}{\|u_{\sharp}\|_{L^{2}(B_{1})}}.$

We know that $u_{\star}\in H^{s}_{0}(B_{1})$, due to (4.37). Moreover, in view
of Lemma 4.5,

$(-\Delta)^{s}u_{\star}=\frac{(-\Delta)^{s}u_{\sharp}}{\|u_{\sharp}\|_{L^{2}(B_{1})}}=\frac{\big{(}(-\Delta)^{s}u\big{)}_{\sharp}}{\|u_{\sharp}\|_{L^{2}(B_{1})}}=\frac{\lambda_{1}\,u_{\sharp}}{\|u_{\sharp}\|_{L^{2}(B_{1})}}=\lambda_{1}\,u_{\star},$

which gives the desired result. ∎

Now, we are in the position of proving the following result.

######  Lemma 4.8.

Let $s\geq 1$ and $r\in(0,1)$. If $u\in H_{0}^{s}\left(B_{1}\right)$ and $u$
is radial, then $u\in C^{\alpha}\left({\mathbb{R}}^{n}\setminus B_{r}\right)$
for any $\alpha\in\left[0,\frac{1}{2}\right]$.

###### Proof.

We write

$u\left(x\right)=u_{0}\left(\left|x\right|\right),\qquad\mbox{for some
}\;u_{0}:[0,+\infty)\rightarrow\mathbb{R}$ (4.38)

and we observe that $u\in H_{0}^{s}\left(B_{1}\right)\subset
H^{1}\left({\mathbb{R}}^{n}\right)$.

Accordingly, for any $0<r<1$, we have

$\infty>\int_{{\mathbb{R}}^{n}\setminus
B_{r}}|u(x)|^{2}\,dx=\int_{r}^{+\infty}|u_{0}(\rho)|^{2}\rho^{n-1}\,d\rho\geq
r^{n-1}\int_{r}^{+\infty}|u_{0}(\rho)|^{2}\,d\rho$ (4.39)

and

$\infty>\int_{{\mathbb{R}}^{n}\setminus B_{r}}{|\nabla
u(x)|^{2}\,dx}=\int_{r}^{+\infty}{\left|u_{0}^{\prime}(\rho)\right|^{2}\rho^{n-1}\,d\rho}\geq
r^{n-1}\int_{r}^{+\infty}{\left|u_{0}^{\prime}(\rho)\right|^{2}\,d\rho}.$
(4.40)

Thanks to (4.39) and (4.40) we have that $u_{0}\in
H^{1}\left(\left(r,+\infty\right)\right)$, with $u_{0}=0$ in $[1,+\infty)$.

Then, from the Morrey Embedding Theorem, it follows that $u_{0}\in
C^{\alpha}\left(\left(r,+\infty\right)\right)$ for any
$\alpha\in\left[0,\frac{1}{2}\right]$, which leads to the desired result. ∎

######  Corollary 4.9.

Let $s\in(0,+\infty)$. There exists a radial, nonnegative and nontrivial
solution of (4.32) which belongs to $H^{s}_{0}(B_{1})\cap
C^{\alpha}({\mathbb{R}}^{n}\setminus B_{1/2})$, for some $\alpha\in(0,1)$.

###### Proof.

If $s\in(0,1)$, the desired claim follows from Corollary 8 in [DSV1].

If instead $s\geq 1$, we obtain the desired result as a consequence of
Proposition 4.7 and Lemma 4.8. ∎

###  4.3 Boundary asymptotics of the first eigenfunctions of $(-\Delta)^{s}$

In Lemma 4 of [DSV1], some precise asymptotics at the boundary for the first
Dirichlet eigenfunction of $(-\Delta)^{s}$ have been established in the range
$s\in(0,1)$.

Here, we obtain a related expansion in the range $s>0$ for the eigenfunction
provided in Corollary 4.9. The result that we obtain is the following:

######  Proposition 4.10.

There exists a nontrivial solution $\phi_{*}$ of (4.32) which belongs to
$H^{s}_{0}(B_{1})\cap C^{\alpha}({\mathbb{R}}^{n}\setminus B_{1/2})$, for some
$\alpha\in(0,1)$, and such that, for every $e\in\partial B_{1}$ and
$\beta=(\beta_{1},\dots,\beta_{n})\in{\mathbb{N}}^{n}$,

$\lim_{\epsilon\searrow
0}\epsilon^{\left|\beta\right|-s}\partial^{\beta}\phi_{*}\left(e+\epsilon
X\right)=\left(-1\right)^{\left|\beta\right|}k_{*}\,s\left(s-1\right)\ldots\left(s-\left|\beta\right|+1\right)e_{1}^{\beta_{1}}\ldots
e_{n}^{\beta_{n}}\left(-e\cdot X\right)_{+}^{s-\left|\beta\right|},$

in the sense of distribution, with $|\beta|:=\beta_{1}+\dots+\beta_{n}$ and
$k_{*}>0$.

The proof of Proposition 4.10 relies on Proposition 4.2 and some auxiliary
computations on the Green function in (4.1). We start with the following
result:

######  Lemma 4.11.

Let $0<r<1$, $e\in\partial B_{1}$, $s>0$, $f\in
C^{\alpha}(\mathbb{R}^{n}\setminus B_{r})\cap L^{2}(\mathbb{R}^{n})$ for some
$\alpha\in(0,1)$, and $f=0$ outside $B_{1}$. Then the integral

$\int_{B_{1}}f(z)\frac{(1-|z|^{2})^{s}}{s|z-e|^{n}}\,dz$ (4.41)

is finite.

###### Proof.

We denote by $I$ the integral in (4.41). We let

$I_{1}:=\int_{B_{1}\setminus
B_{r}}f(z)\frac{(1-|z|^{2})^{s}}{s|z-e|^{n}}\,dz\qquad{\mbox{and}}\qquad
I_{2}:=\int_{B_{r}}f(z)\frac{(1-|z|^{2})^{s}}{s|z-e|^{n}}\,dz.$

Then, we have that

$I=I_{1}+I_{2}.$ (4.42)

Now, if $z\in B_{1}\setminus B_{r}$, we have that

$f(z)\leq|f(z)-f(e)|\leq C|z-e|^{\alpha},$

therefore

$I_{1}\leq\int_{B_{1}\setminus
B_{r}}\frac{(1-|z|^{2})^{s}}{s|z-e|^{n-\alpha}}\,dz<\infty.$ (4.43)

If instead $z\in B_{r}$,

$|z-e|\geq 1-r>0,$

and consequently

$I_{2}\leq\frac{1}{s\,(1-r)^{n}}\int_{B_{r}}f(z)\,dz<\infty.$ (4.44)

The desired result follows from (4.42), (4.43) and (4.44). ∎

Next result gives a precise boundary behaviour of the Green function for any
$s>0$ (the case in which $s\in(0,1)$ and $f\in C^{\alpha}(\mathbb{R}^{n})$ was
considered in Lemma 6 of [DSV1], and in fact the proof presented here also
simplifies the one in Lemma 6 of [DSV1] for the setting considered there).

######  Lemma 4.12.

Let $e$, $\omega\in\partial B_{1}$, $\epsilon_{0}>0$ and $r\in(0,1)$. Assume
that

$e+\epsilon\omega\in B_{1},$ (4.45)

for any $\epsilon\in(0,\epsilon_{0}]$. Let $f\in
C^{\alpha}(\mathbb{R}^{n}\setminus B_{r})\cap L^{2}(\mathbb{R}^{n})$ for some
$\alpha\in(0,1)$, with $f=0$ outside $B_{1}$.
Then

$\lim_{\epsilon\searrow
0}\epsilon^{-s}\int_{B_{1}}f(z)\mathcal{G}_{s}(e+\epsilon\omega,z)\,dz=k(n,s)\,\int_{B_{1}}f(z)\frac{(-2e\cdot\omega)^{s}(1-|z|^{2})^{s}}{s|z-e|^{n}}\,dz,$
(4.46)

for a suitable normalizing constant $k(n,s)>0$.

###### Proof.

In light of (4.45), we have that

$1>|e+\epsilon\omega|^{2}=1+\epsilon^{2}+2\epsilon e\cdot\omega,$

and therefore

$-e\cdot\omega>\frac{\epsilon}{2}>0.$ (4.47)

Moreover, if $r_{0}$ is as given in (4.1), we have that, for all $z\in B_{1}$,

$r_{0}(e+\epsilon\omega,z)=\frac{\epsilon(-\epsilon-2e\cdot\omega)(1-|z|^{2})}{|z-e-\epsilon\omega|^{2}}\leq\frac{3\epsilon}{|z-e-\epsilon\omega|^{2}}.$
(4.48)

Also, a Taylor series representation allows us to write, for any $t\in(-1,1)$,

$\frac{t^{s-1}}{(t+1)^{\frac{n}{2}}}=\sum_{k=0}^{\infty}\binom{-n/2}{k}t^{k+s-1}.$
(4.49)

We also notice that

$\begin{split}&\left|\binom{-n/2}{k}\right|=\left|\frac{-\frac{n}{2}\left(-\frac{n}{2}-1\right)\,...\,\left(-\frac{n}{2}-k+1\right)}{k!}\right|=\frac{\frac{n}{2}\left(\frac{n}{2}+1\right)\,...\,\left(\frac{n}{2}+(k-1)\right)}{k!}\\\
&\qquad\leq\frac{n\left(n+1\right)\,...\,\left(n+(k-1)\right)}{k!}\leq\frac{\left(n+(k-1)\right)!}{k!}=(k+1)\,...\,\left(n+(k-1)\right)\\\
&\qquad\leq(n+k+1)^{n+1}.\end{split}$ (4.50)

This and the Root Test give that the series in (4.49) is uniformly convergent
on compact sets in $(-1,1)$.

As a consequence, if we set

$r_{1}(x,z):=\min\left\\{\frac{1}{2},r_{0}(x,z)\right\\},$ (4.51)

we can switch integration and summation signs and obtain that

$\int_{0}^{r_{1}(x,z)}\frac{t^{s-1}}{(t+1)^{\frac{n}{2}}}\,dt=\sum_{k=0}^{\infty}c_{k}(r_{1}(x,z))^{k+s},$
(4.52)

where

$c_{k}:=\frac{1}{k+s}\binom{-n/2}{k}.$

Once again, the bound in (4.50), together with (4.51), give that the series in
(4.52) is convergent.

Now, we omit for simplicity the normalizing constant $k(n,s)$ in the
definition of the Green function in (4.1), and we define

$\mathcal{G}(x,z):=|z-x|^{2s-n}\sum_{k=0}^{\infty}c_{k}(r_{1}(x,z))^{k+s}$
(4.53)

and

$g(x,z):=|z-x|^{2s-n}\int_{r_{1}(x,z)}^{r_{0}(x,z)}\frac{t^{s-1}}{(t+1)^{\frac{n}{2}}}\,dt.$

Using (4.1) and (4.52), and dropping dimensional constants for the sake of
shortness, we can write

$\mathcal{G}_{s}(x,z)=\mathcal{G}(x,z)+g(x,z).$ (4.54)

Now, we show that

$g(x,z)\leq\begin{cases}C\chi(r,z)\,|z-x|^{2s-n}&\quad\text{if}\quad n>2s,\\\
C\chi(r,z)\,\log r_{0}(x,z)&\quad\text{if}\quad n=2s,\\\
C\chi(r,z)\,|z-x|^{2s-n}(r_{0}(x,z))^{s-\frac{n}{2}}&\quad\text{if}\quad
n<2s,\end{cases}$ (4.55)

where $\chi(r,z)=1$ if $r_{0}(x,z)>\frac{1}{2}$ and $\chi(r,z)=0$ if
$r_{0}(x,z)\leq\frac{1}{2}$. To check this, we notice that if
$r_{0}(x,z)\leq\frac{1}{2}$ we have that $r_{1}(x,z)=r_{0}(x,z)$, due to
(4.51), and therefore $g(x,z)=0$.

On the other hand, if $r_{0}(x,z)>\frac{1}{2}$, we deduce from (4.51) that
$r_{1}(x,z)=\frac{1}{2}$, and consequently

$g(x,z)\leq|z-x|^{2s-n}\int_{1/2}^{r_{0}(x,z)}t^{s-\frac{n}{2}-1}dt\leq\begin{cases}C|z-x|^{2s-n}&\quad\text{if}\quad
n>2s,\\\ C\log r_{0}(x,z)&\quad\text{if}\quad n=2s,\\\
C|z-x|^{2s-n}(r_{0}(x,z))^{s-\frac{n}{2}}&\quad\text{if}\quad
n<2s,\end{cases}$

for some constant $C>0$. This completes the proof of (4.55).

Now, we exploit the bound in (4.55) when $x=e+\epsilon\omega$. For this, we
notice that if $r_{0}(e+\epsilon\omega,z)>\frac{1}{2}$, recalling (4.48), we
find that

$|z-e-\epsilon\omega|^{2}\leq 6\epsilon<9\epsilon,$ (4.56)

and therefore $z\in B_{3\sqrt{\epsilon}}(e+\epsilon\omega)$.

Hence, using (4.55),

$\begin{split}&\left|\int_{B_{1}}f(z)g(e+\epsilon\omega,z)dz\right|\leq\int_{B_{3\sqrt{\epsilon}}(e+\epsilon\omega)}|f(z)||g(e+\epsilon\omega,z)|dz\\\
&\leq\begin{cases}C\displaystyle\int_{B_{3\sqrt{\epsilon}}(e+\epsilon\omega)}|f(z)||z-e-\epsilon\omega|^{2s-n}dz&\quad\text{if}\quad
n>2s,\\\ C\displaystyle\int_{B_{3\sqrt{\epsilon}}(e+\epsilon\omega)}|f(z)|\log
r_{0}(e+\epsilon\omega,z)dz&\quad\text{if}\quad n=2s,\\\
C\displaystyle\int_{B_{3\sqrt{\epsilon}}(e+\epsilon\omega)}|f(z)||z-e-\epsilon\omega|^{2s-n}(r_{0}(e+\epsilon\omega,z))^{s-\frac{n}{2}}dz&\quad\text{if}\quad
n<2s.\end{cases}\end{split}$ (4.57)

Now, if $z\in B_{3\sqrt{\epsilon}}(e+\epsilon\omega)$, then

$|z-e|\leq|z-e-\epsilon\omega|+|\epsilon\omega|\leq
3\sqrt{\epsilon}+\epsilon<4\sqrt{\epsilon}.$ (4.58)

Furthermore, for a given $r\in(0,1)$, we have that
$B_{3\sqrt{\epsilon}}(e+\epsilon\omega)\subseteq{\mathbb{R}}^{n}\setminus
B_{r}$, provided that $\epsilon$ is sufficiently small.

Hence, if $z\in B_{3\sqrt{\epsilon}}(e+\epsilon\omega)$, we can exploit the
regularity of $f$ and deduce that

$|f(z)|=|f(z)-f(e)|\leq C|z-e|^{\alpha}.$

This and (4.58) lead to

$|f(z)|\leq C\epsilon^{\frac{\alpha}{2}},$ (4.59)

for every $z\in B_{3\sqrt{\epsilon}}(e+\epsilon\omega)$.

Thanks to (4.48), (4.57) and (4.59), we have that

$\begin{split}\left|\int_{B_{1}}f(z)g(e+\epsilon\omega,z)dz\right|&\leq\begin{cases}C\epsilon^{\frac{\alpha}{2}}\displaystyle\int_{B_{3\sqrt{\epsilon}}(e+\epsilon\omega)}|z-e-\epsilon\omega|^{2s-n}dz&\quad\text{if}\quad
n>2s,\\\
C\epsilon^{\frac{\alpha}{2}}\displaystyle\int_{B_{3\sqrt{\epsilon}}(e+\epsilon\omega)}\log\frac{3\epsilon}{|z-e-\epsilon\omega|^{2}}dz&\quad\text{if}\quad
n=2s,\\\
C\epsilon^{\frac{\alpha}{2}+s-\frac{n}{2}}\displaystyle\int_{B_{3\sqrt{\epsilon}}(e+\epsilon\omega)}dz&\quad\text{if}\quad
n<2s\end{cases}\\\ &\leq C\epsilon^{\frac{\alpha}{2}+s},\end{split}$

up to renaming $C$.

This and (4.54) give that

$\int_{B_{1}}f(z)\mathcal{G}_{s}(e+\epsilon\omega,z)dz=\int_{B_{1}}f(z)\mathcal{G}(e+\epsilon\omega,z)dz+o(\epsilon^{s}).$
(4.60)

Now, we consider the series in (4.53), and we split the contribution coming
from the index $k=0$ from the ones coming from the indices $k>0$, namely we
write

$\begin{split}&\mathcal{G}(x,z)=\mathcal{G}_{0}(x,z)+\mathcal{G}_{1}(x,z),\\\
\quad\text{with}\quad&\mathcal{G}_{0}(x,z):=\frac{|z-x|^{2s-n}}{s}(r_{1}(x,z))^{s}\\\
\quad\text{and}\quad&\mathcal{G}_{1}(x,z):=|z-x|^{2s-n}\sum_{k=1}^{+\infty}c_{k}\,(r_{1}(x,z))^{k+s}.\end{split}$
(4.61)

Firstly, we consider the contribution given by the term $\mathcal{G}_{1}$.
Thanks to (4.51) and (4.59), we have that

$\begin{split}&\left|\int_{B_{1}\cap
B_{3\sqrt{\epsilon}}(e+\epsilon\omega)}f(z)\mathcal{G}_{1}(e+\epsilon\omega,z)dz\right|\leq\int_{B_{3\sqrt{\epsilon}}(e+\epsilon\omega)}|f(z)|\mathcal{G}_{1}(e+\epsilon\omega,z)dz\\\
&\quad\quad\leq
C\epsilon^{\frac{\alpha}{2}}\int_{B_{3\sqrt{\epsilon}}(e+\epsilon\omega)}|z-e-\epsilon\omega|^{2s-n}\sum_{k=1}^{+\infty}|c_{k}|\,(r_{1}(e+\epsilon\omega,z))^{k+s}dz\\\
&\quad\quad\leq
C\epsilon^{\frac{\alpha}{2}}\int_{B_{3\sqrt{\epsilon}}(e+\epsilon\omega)}|z-e-\epsilon\omega|^{2s-n}\sum_{k=1}^{+\infty}|c_{k}|\,\left(\frac{1}{2}\right)^{k+s}dz\\\
&\quad\quad\leq
C\epsilon^{\frac{\alpha}{2}}\int_{B_{3\sqrt{\epsilon}}(e+\epsilon\omega)}|z-e-\epsilon\omega|^{2s-n}dz\\\
&\quad\quad\leq C\epsilon^{\frac{\alpha}{2}+s},\end{split}$ (4.62)

up to renaming the constant $C$ step by step.

On the other hand, for every $z\in{\mathbb{R}}^{n}$,

$|z|=|e+\epsilon\omega+z-e-\epsilon\omega|\geq|e+\epsilon\omega|-|z-e-\epsilon\omega|\geq
1-\epsilon-|z-e-\epsilon\omega|.$

Therefore, for every $z\in B_{1}\setminus\left(B_{r}\cup
B_{3\sqrt{\epsilon}}(e+\epsilon\omega)\right)$, we can take
$e_{*}:=\frac{z}{|z|}$ and obtain that

$\begin{split}&|f(z)|=|f(z)-f(e_{*})|\leq
C|z-e_{*}|^{\alpha}=C(1-|z|)^{\alpha}\\\ &\qquad\leq
C(\epsilon+|z-e-\epsilon\omega|)^{\alpha}\leq
C|z-e-\epsilon\omega|^{\alpha},\end{split}$ (4.63)

up to renaming $C>0$.

Also, using (4.48), we see that, for any $k>0$,

$\begin{split}&(r_{0}(e+\epsilon\omega,z))^{s+\frac{\alpha}{4}}\left(\frac{1}{2}\right)^{k-\frac{\alpha}{4}}\leq\frac{C\epsilon^{s+\frac{\alpha}{4}}}{2^{k}|z-e-\epsilon\omega|^{2s+\frac{\alpha}{2}}}.\end{split}$
(4.64)

This, (4.51) and (4.63) give that if $z\in B_{1}\setminus\left(B_{r}\cup
B_{3\sqrt{\epsilon}}(e+\epsilon\omega)\right)$, then

$\begin{split}|f(z)\mathcal{G}_{1}(e+\epsilon\omega,z)|\,&\leq
C|z-e-\epsilon\omega|^{\alpha+2s-n}\sum_{k=1}^{+\infty}|c_{k}|\,(r_{1}(e+\epsilon\omega,z))^{k+s}\\\
&=C|z-e-\epsilon\omega|^{\alpha+2s-n}\sum_{k=1}^{+\infty}|c_{k}|\,(r_{1}(e+\epsilon\omega,z))^{s+\frac{\alpha}{4}}(r_{1}(e+\epsilon\omega,z))^{k-\frac{\alpha}{4}}\\\
&\leq
C|z-e-\epsilon\omega|^{\alpha+2s-n}\sum_{k=1}^{+\infty}|c_{k}|\,(r_{0}(e+\epsilon\omega,z))^{s+\frac{\alpha}{4}}\left(\frac{1}{2}\right)^{k-\frac{\alpha}{4}}\\\
&\leq
C\epsilon^{s+\frac{\alpha}{4}}|z-e-\epsilon\omega|^{\frac{\alpha}{2}-n}\sum_{k=1}^{+\infty}\frac{|c_{k}|}{2^{k}},\end{split}$

where the latter series is absolutely convergent thanks to (4.50).

This implies that, if we set $E:=B_{1}\setminus\left(B_{r}\cup
B_{3\sqrt{\epsilon}}(e+\epsilon\omega)\right)$, it holds that

$\begin{split}&\left|\int_{E}f(z)\mathcal{G}_{1}(e+\epsilon\omega,z)dz\right|\leq
C\epsilon^{s+\frac{\alpha}{4}}\int_{E}|z-e-\epsilon\omega|^{\frac{\alpha}{2}-n}dz\\\
\quad\quad&\qquad\qquad\leq
C\epsilon^{s+\frac{\alpha}{4}}\int_{B_{1}}|z-e-\epsilon\omega|^{\frac{\alpha}{2}-n}dz\leq
C\epsilon^{s+\frac{\alpha}{4}}\int_{B_{3}}|z|^{\frac{\alpha}{2}-n}dz\leq
C\epsilon^{s+\frac{\alpha}{4}}.\end{split}$ (4.65)

Moreover, if $z\in B_{r}$, we have that

$|e+\epsilon\omega-z|\geq 1-\epsilon-r,$

and therefore, recalling (4.64),

$\begin{split}\sup_{z\in
B_{r}}|\mathcal{G}_{1}(e+\epsilon\omega,z)|\,&\leq|z-e-\epsilon\omega|^{2s-n}\sum_{k=1}^{+\infty}|c_{k}|\,\big{(}r_{1}(e+\epsilon\omega,z)\big{)}^{s+\frac{\alpha}{4}}\big{(}r_{1}(e+\epsilon\omega,z)\big{)}^{k-\frac{\alpha}{4}}\\\
&\leq|z-e-\epsilon\omega|^{2s-n}\sum_{k=1}^{+\infty}|c_{k}|\,\big{(}r_{0}(e+\epsilon\omega,z)\big{)}^{s+\frac{\alpha}{4}}\left(\frac{1}{2}\right)^{k-\frac{\alpha}{4}}\\\
&\leq
C\,|z-e-\epsilon\omega|^{-n-\frac{\alpha}{2}}\sum_{k=1}^{+\infty}\frac{|c_{k}|}{2^{k}}\\\
&\leq
C(1-\epsilon-r)^{-n-\frac{\alpha}{2}}\,\epsilon^{s+\frac{\alpha}{4}},\end{split}$

up to renaming $C$.

As a consequence, we find that

$\begin{split}\left|\int_{B_{r}}f(z)\mathcal{G}_{1}(e+\epsilon\omega,z)dz\right|&\leq\sup_{z\in
B_{r}}|\mathcal{G}_{1}(e+\epsilon\omega,z)|\,\left\|f\right\|_{L^{1}(B_{r})}\\\
&\leq\left\|f\right\|_{L^{1}(B_{r})}(1-\epsilon-r)^{-n-\frac{\alpha}{2}}\epsilon^{s+\frac{\alpha}{4}}\\\
&\leq\left\|f\right\|_{L^{1}(B_{r})}2^{n+\frac{\alpha}{2}}(1-r)^{-n-\frac{\alpha}{2}}\epsilon^{s+\frac{\alpha}{4}}\\\
&=C\epsilon^{s+\frac{\alpha}{4}},\end{split}$ (4.66)

as long as $\epsilon$ is suitably small with respect to $r$, and $C$ is a
positive constant which depends on $\|f\|_{L^{1}(B_{r})}$, $r$, $n$ and
$\alpha$.

Then, by (4.62), (4.65) and (4.66) we conclude that

$\int_{B_{1}}f(z)\mathcal{G}_{1}(e+\epsilon\omega,z)dz=o(\epsilon^{s}).$
(4.67)

Inserting this information into (4.60), and recalling (4.61), we obtain

$\int_{B_{1}}f(z)\mathcal{G}_{s}(e+\epsilon\omega,z)dz=\int_{B_{1}}f(z)\mathcal{G}_{0}(e+\epsilon\omega,z)dz+o(\epsilon^{s}).$
(4.68)

Now, we define

$\mathcal{D}_{1}:=\left\\{z\in B_{1}\quad\text{s.t.}\quad
r_{0}(e+\epsilon\omega,z)>1/2\right\\}$

and

$\mathcal{D}_{2}:=\left\\{z\in B_{1}\quad\text{s.t.}\quad
r_{0}(e+\epsilon\omega,z)\leq 1/2\right\\}.$

If $z\in\mathcal{D}_{1}$, then $z\in B_{1}\setminus B_{r}$, thanks to (4.56),
and hence we can use (4.57) and (4.59) and write

$|f(z)\mathcal{G}_{0}(e+\epsilon\omega,z)|\leq
C\epsilon^{\frac{\alpha}{2}}|z-e-\epsilon\omega|^{2s-n}.$

Then, recalling again (4.57),

$\left|\int_{\mathcal{D}_{1}}f(z)\mathcal{G}_{1}(e+\epsilon\omega,z)dz\right|\leq
C\epsilon^{\frac{\alpha}{2}}\int_{B_{3\sqrt{\epsilon}}(e+\epsilon\omega)}|z-e-\epsilon\omega|^{2s-n}dz=C\epsilon^{\frac{\alpha}{2}+s},$
(4.69)

up to renaming the constant $C>0$. This information and (4.68) give that

$\int_{B_{1}}f(z)\mathcal{G}_{s}(e+\epsilon\omega,z)dz=\int_{\mathcal{D}_{2}}f(z)\mathcal{G}_{0}(e+\epsilon\omega,z)dz+o(\epsilon^{s}).$

Now, by (4.48) and (4.51), if $z\in\mathcal{D}_{2}$,

$\mathcal{G}_{0}(e+\epsilon\omega,z)=\frac{|z-e-\epsilon\omega|^{2s-n}}{s}(r_{0}(e+\epsilon\omega))^{s}=\frac{\epsilon^{s}(-\epsilon-2e\cdot\omega)^{s}(1-|z|^{2})^{s}}{s|z-e-\epsilon\omega|^{n}}.$

Hence, we have

$\begin{split}&\lim_{\epsilon\searrow
0}\epsilon^{-s}\int_{B_{1}}f(z)\mathcal{G}_{s}(e+\epsilon\omega,z)dz\\\
=\;&\lim_{\epsilon\searrow
0}\epsilon^{-s}\int_{\mathcal{D}_{2}}f(z)\mathcal{G}_{0}(e+\epsilon\omega,z)dz\\\
=\;&\lim_{\epsilon\searrow
0}\int_{\left\\{2\epsilon(-\epsilon-2e\cdot\omega)(1-|z|^{2})\leq|z-e-\epsilon\omega|^{2}\right\\}}f(z)\frac{(-\epsilon-2e\cdot\omega)^{s}(1-|z|^{2})^{s}}{s|z-e-\epsilon\omega|^{n}}dz.\end{split}$
(4.70)

Now we set

$F_{\epsilon}(z):=\begin{cases}f(z)\displaystyle\frac{(-\epsilon-2e\cdot\omega)^{s}(1-|z|^{2})^{s}}{s|z-e-\epsilon\omega|^{n}}&\quad\text{if}\quad
2\epsilon(-\epsilon-2e\cdot\omega)(1-|z|^{2})\leq|z-e-\epsilon\omega|^{2},\\\
0&\quad\text{otherwise},\end{cases}$ (4.71)

and we prove that for any $\eta>0$ there exists $\delta>0$ independent of
$\epsilon$ such that, for any $E\subset\mathbb{R}^{n}$ with $|E|\leq\delta$,
we have

$\int_{B_{1}\cap E}|F_{\epsilon}(z)|dz\leq\eta.$ (4.72)

To this aim, given $\eta$ and $E$ as above, we define

$\rho:=\min\left\\{\epsilon(-\epsilon-2e\cdot\omega),\,\sqrt{{2\epsilon(-\epsilon-2e\cdot\omega)}(1-r)},\,\left(\frac{2^{s+\alpha}s^{2}\,\epsilon^{s+\alpha}\,(-\epsilon-2e\cdot\omega)^{\alpha}\eta}{3^{2s}\,\|f\|_{C^{\alpha}(B_{1}\setminus
B_{r})}\,|\partial B_{1}|}\right)^{\frac{1}{2\alpha}}\right\\}.$ (4.73)

We stress that the above definition is well-posed, thanks to (4.47). In
addition, using the integrability of $f$, we take $\delta>0$ such that if
$A\subseteq B_{1}$ and $|A|\leq\delta$ then

$\int_{A}|f(x)|\,dx\leq\frac{s\rho^{n}\eta}{2\cdot 3^{s}}.$ (4.74)

We set

$E_{1}:=E\cap B_{\rho}(e+\epsilon\omega)\qquad{\mbox{and}}\qquad
E_{2}:=E\setminus B_{\rho}(e+\epsilon\omega).$ (4.75)

From (4.71), we see that

$|F_{\epsilon}(z)|\leq\frac{|f(z)|\,\chi_{\star}(z)}{2^{s}s\,\epsilon^{s}|z-e-\epsilon\omega|^{n-2s}},$

where

$\chi_{\star}(z):=\begin{cases}1&\quad\text{if}\quad
2\epsilon(-\epsilon-2e\cdot\omega)(1-|z|^{2})\leq|z-e-\epsilon\omega|^{2},\\\
0&\quad\text{otherwise},\end{cases}$

and therefore

$\int_{B_{1}\cap E_{1}}|F_{\epsilon}(z)|\,dz\leq\int_{B_{1}\cap
E_{1}}\frac{|f(z)|\,\chi_{\star}(z)}{2^{s}s\,\epsilon^{s}|z-e-\epsilon\omega|^{n-2s}}\,dz.$
(4.76)

Now, for every $z\in B_{1}\cap E_{1}\subseteq B_{\rho}(e+\epsilon\omega)$ for
which $\chi_{\star}(z)\neq 0$, we have that

$2\epsilon(-\epsilon-2e\cdot\omega)(1-|z|^{2})\leq|z-e-\epsilon\omega|^{2}\leq\rho^{2},$

and hence

$|z|\geq\sqrt{1-\frac{\rho^{2}}{2\epsilon(-\epsilon-2e\cdot\omega)}}\geq
1-\frac{\rho^{2}}{2\epsilon(-\epsilon-2e\cdot\omega)},$

which in turn gives that $|z|\geq r$, recall (4.73).

From this and (4.76) we deduce that

$\begin{split}&\int_{B_{1}\cap
E_{1}}|F_{\epsilon}(z)|\,dz\leq\int_{1-\frac{\rho^{2}}{2\epsilon(-\epsilon-2e\cdot\omega)}\leq|z|<1}\frac{\|f\|_{C^{\alpha}(B_{1}\setminus
B_{r})}\,(1-|z|)^{\alpha}}{2^{s}s\,\epsilon^{s}|z-e-\epsilon\omega|^{n-2s}}\,dz\\\
&\qquad\leq\frac{\|f\|_{C^{\alpha}(B_{1}\setminus
B_{r})}}{2^{s}s\,\epsilon^{s}}\,\left(\frac{\rho^{2}}{2\epsilon(-\epsilon-2e\cdot\omega)}\right)^{\alpha}\int_{1-\frac{\rho^{2}}{2\epsilon(-\epsilon-2e\cdot\omega)}\leq|z|<1}\frac{dz}{|z-e-\epsilon\omega|^{n-2s}}\\\
&\qquad\leq\frac{\|f\|_{C^{\alpha}(B_{1}\setminus
B_{r})}}{2^{s}s\,\epsilon^{s}}\,\left(\frac{\rho^{2}}{2\epsilon(-\epsilon-2e\cdot\omega)}\right)^{\alpha}\int_{B_{3}}\frac{dx}{|x|^{n-2s}}\\\
&\qquad=\frac{3^{2s}\,\|f\|_{C^{\alpha}(B_{1}\setminus B_{r})}\,|\partial
B_{1}|}{2^{s+\alpha+1}s^{2}\,\epsilon^{s+\alpha}\,(-\epsilon-2e\cdot\omega)^{\alpha}}\,\;\rho^{2\alpha}\\\
&\qquad\leq\frac{\eta}{2},\end{split}$ (4.77)

where (4.73) has been exploited in the last inequality.

We also point out that, by (4.71), (4.74) and (4.75),

$\displaystyle\int_{B_{1}\cap E_{2}}|F_{\epsilon}(z)|\,dz$ $\displaystyle\leq$
$\displaystyle\int_{(B_{1}\setminus B_{\rho}(e+\epsilon\omega))\cap
E}|f(z)|\,\frac{(-\epsilon-2e\cdot\omega)^{s}(1-|z|^{2})^{s}}{s|z-e-\epsilon\omega|^{n}}\,dz$
$\displaystyle\leq$ $\displaystyle\frac{3^{s}}{s\rho^{n}}\int_{B_{1}\cap
E}|f(z)|\,dz$ $\displaystyle\leq$ $\displaystyle\frac{\eta}{2}.$

This, (4.75) and (4.77) give (4.72), as desired.

Notice also that the sequence $F_{\epsilon}(z)$ converges pointwise to the
function

$F(z):=f(z)\frac{(-2e\cdot\omega)^{s}(1-|z|^{2})^{s}}{s|z-e|^{n}}.$

Hence (4.70), (4.72) and the Vitali Convergence Theorem allow us to conclude
that

$\begin{split}\lim_{\epsilon\searrow
0}\int_{B_{1}}f(z)\mathcal{G}_{s}(e+\epsilon\omega,z)dz&=\lim_{\epsilon\searrow
0}\int_{B_{1}}F_{\epsilon}(z)dz\\\
&=\int_{B_{1}}f(z)\frac{(-2e\cdot\omega)^{s}(1-|z|^{2})^{s}}{s|z-e|^{n}}dz,\end{split}$
(4.78)

which establishes the claim of Lemma 4.12 (notice that the finiteness of the
latter quantity in (4.78) follows from (4.11)). ∎

With this preliminary work, we can now establish the boundary behaviour of
solutions which is needed in our setting. As a matter of fact, from Lemma 4.12
we immediately deduce that:

######  Corollary 4.13.

Let $e$, $\omega\in\partial B_{1}$, $\epsilon_{0}>0$ and $r\in(0,1)$.

Assume that $e+\epsilon\omega\in B_{1}$, for any
$\epsilon\in(0,\epsilon_{0}]$. Let $f\in C^{\alpha}(\mathbb{R}^{n}\setminus
B_{r})\cap L^{2}(\mathbb{R}^{n})$ for some $\alpha\in(0,1)$, with $f=0$
outside $B_{1}$.

Let $u$ be as in (4.19). Then,

$\lim_{\epsilon\searrow
0}\epsilon^{-s}u(e+\epsilon\omega)=k(n,s)(-2e\cdot\omega)^{s}\int_{B_{1}}f(z)\frac{(1-|z|^{2})^{s}}{s|z-e|^{n}}dz,$

where $k(n,s)$ denotes a positive normalizing constant.

Now we apply the previous results to detect the boundary growth of a suitable
first eigenfunction. For our purposes, the statement that we need is the
following:

######  Corollary 4.14.

There exists a nontrivial solution $\phi_{*}$ of (4.32) which belongs to
$H^{s}_{0}(B_{1})\cap C^{\alpha}({\mathbb{R}}^{n}\setminus B_{1/2})$, for some
$\alpha\in(0,1)$, and such that, for every $e\in\partial B_{1}$,

$\lim_{\epsilon\searrow
0}\epsilon^{-s}\phi_{*}(e+\epsilon\omega)=k_{*}\,(-e\cdot\omega)^{s}_{+},$
(4.79)

for a suitable constant $k_{*}>0$.

Furthermore, for every $R\in(r,1)$, there exists $C_{R}>0$ such that

$\sup_{x\in B_{1}\setminus B_{R}}d^{-s}(x)\,|\phi_{*}(x)|\leq C_{R}.$ (4.80)

###### Proof.

Let $\alpha\in(0,1)$ and $\phi\in H^{s}_{0}(B_{1})\cap
C^{\alpha}({\mathbb{R}}^{n}\setminus B_{1/2})$ be the nonnegative and
nontrivial solution of (4.32), as given in Corollary 4.9.

In the spirit of (4.19), we define

$\phi_{*}(x):=\begin{cases}\displaystyle\lambda_{1}\int_{B_{1}}\mathcal{G}_{s}\left(x,y\right)\,\phi(y)\,dy&{\mbox{
if }}x\in B_{1},\\\ 0&{\mbox{ if }}x\in{\mathbb{R}}^{n}\setminus
B_{1}.\end{cases}$

We stress that we can use Proposition 4.2 in this context, with
$f:=\lambda_{1}\phi$, since condition (4.18) is satisfied in this case.

Then, from (4.20) and (4.22), we know that $\phi_{*}\in H^{s}_{0}(B_{1})$ and,
from (4.21),

$(-\Delta)^{s}\phi_{*}=\lambda_{1}\,\phi{\mbox{ in }}B_{1}.$

In particular, we have that $(-\Delta)^{s}(\phi-\phi_{*})=0$ in $B_{1}$, and
$\phi-\phi_{*}\in H^{s}_{0}(B_{1})$, which give that $\phi-\phi_{*}$ vanishes
identically. Hence, we can write that $\phi=\phi_{*}$, and thus $\phi_{*}$ is
a solution of (4.32).

Now, we check (4.79). For this, we distinguish two cases. If $e\cdot\omega\geq
0$, we have that

$|e+\epsilon\omega|^{2}=1+2\epsilon e\cdot\omega+\epsilon^{2}>1,$

for all $\epsilon>0$. Then, in this case
$e+\epsilon\omega\in{\mathbb{R}}^{n}\setminus B_{1}$, and therefore
$\phi_{*}(e+\epsilon\omega)=0$. This gives that, in this case,

$\lim_{\epsilon\searrow 0}\epsilon^{-s}\phi_{*}(e+\epsilon\omega)=0.$ (4.81)

If instead $e\cdot\omega<0$, we see that

$|e+\epsilon\omega|^{2}=1+2\epsilon e\cdot\omega+\epsilon^{2}<1,$

for all $\epsilon>0$ sufficiently small. Hence, we can exploit Corollary 4.13
and find that

$\lim_{\epsilon\searrow
0}\epsilon^{-s}\phi_{*}(e+\epsilon\omega)=\lambda_{1}\,k(n,s)(-2e\cdot\omega)^{s}\int_{B_{1}}\phi(z)\frac{(1-|z|^{2})^{s}}{s|z-e|^{n}}\,dz,$
(4.82)

with $k(n,s)>0$. Then, we define

$k_{*}:=2^{s}\,k(n,s)\int_{B_{1}}\phi(z)\frac{(1-|z|^{2})^{s}}{s|z-e|^{n}}\,dz.$

We observe that $k_{*}$ is positive by construction, with $k(n,s)>0$. Also, in
light of Lemma 4.11, we know that $k_{*}$ is finite. Hence, from (4.81) and
(4.82) we obtain (4.79), as desired.

It only remains to check (4.80). For this, we use (4.21), and we see that, for
every $R\in(r,1)$,

$\sup_{x\in B_{1}\setminus B_{R}}d^{-s}(x)\,|\phi_{*}(x)|\leq
C_{R}\,\lambda_{1}\,\big{(}\|\phi\|_{L^{1}(B_{1})}+\|\phi\|_{L^{\infty}(B_{1}\setminus
B_{r})}\big{)},$

and this gives (4.80) up to renaming $C_{R}$. ∎

Now, we can complete the proof of Proposition 4.10, by arguing as follows.

###### Proof of Proposition 4.10.

Let $\psi$ be a test function in $C^{\infty}_{0}(\mathbb{R}^{n})$. Let also
$R:=\frac{r+1}{2}\in(r,1)$ and

$g_{\epsilon}(X):=\epsilon^{-s}\phi_{*}(e+\epsilon X)\partial^{\beta}\psi(X).$

We claim that

$\sup_{{X\in{\mathbb{R}}^{n}}}|g_{\epsilon}(X)|\leq C,$ (4.83)

for some $C>0$ independent of $\epsilon$. To prove this, we distinguish three
cases. If $e+\epsilon X\in{\mathbb{R}}^{n}\setminus B_{1}$, we have that
$\phi_{*}(e+\epsilon X)=0$ and thus $g_{\epsilon}(X)=0$. If instead
$e+\epsilon X\in B_{R}$, we observe that

$R>|e+\epsilon X|\geq 1-\epsilon|X|,$

and therefore $|X|\geq\frac{1-R}{\epsilon}$. In particular, in this case $X$
falls outside the support of $\psi$, as long as $\epsilon>0$ is sufficiently
small, and consequently $\partial^{\beta}\psi(X)=0$ and $g_{\epsilon}(X)=0$.

Hence, to complete the proof of (4.83), we are only left with the case in
which $e+\epsilon X\in B_{1}\setminus B_{R}$. In this situation, we make use
of (4.80) and we find that

$\displaystyle|\phi_{*}(e+\epsilon X)|\leq C\,d^{s}(e+\epsilon
X)=C\,(1-|e+\epsilon X|)^{s}$ $\displaystyle\qquad\leq C\,(1-|e+\epsilon
X|)^{s}(1+|e+\epsilon X|)^{s}=C\,(1-|e+\epsilon X|^{2})^{s}$
$\displaystyle\qquad=C\,\epsilon^{s}(-2e\cdot X-\epsilon|X|^{2})^{s}\leq
C\epsilon^{s},$

for some $C>0$ possibly varying from line to line, and this completes the
proof of (4.83).

Now, from (4.83) and the Dominated Convergence Theorem, we obtain that

$\lim_{\epsilon\searrow
0}\int_{\mathbb{R}^{n}}\epsilon^{-s}\phi_{*}(e+\epsilon
X)\partial^{\beta}\psi(X)dX=\int_{\mathbb{R}^{n}}\lim_{\epsilon\searrow
0}\epsilon^{-s}\phi_{*}(e+\epsilon X)\partial^{\beta}\psi(X)dX.$ (4.84)

On the other hand, by Corollary 4.14, used here with $\omega:=\frac{X}{|X|}$,
we know that

$\displaystyle\lim_{\epsilon\searrow 0}\epsilon^{-s}\phi_{*}(e+\epsilon
X)=\lim_{\epsilon\searrow
0}\epsilon^{-s}\phi_{*}(e+\epsilon|X|\omega)=|X|^{s}\lim_{\epsilon\searrow
0}\epsilon^{-s}\phi_{*}(e+\epsilon\omega)$
$\displaystyle\qquad=k_{*}\,|X|^{s}\,(-e\cdot\omega)^{s}_{+}=k_{*}\,(-e\cdot
X)^{s}_{+}.$

Substituting this into (4.84), we thus find that

$\lim_{\epsilon\searrow
0}\int_{\mathbb{R}^{n}}\epsilon^{-s}\phi_{*}(e+\epsilon
X)\partial^{\beta}\psi(X)dX=k_{*}\,\int_{\mathbb{R}^{n}}(-e\cdot
X)^{s}_{+}\partial^{\beta}\psi(X)dX.$

As a consequence, integrating by parts twice,

$\begin{split}&\lim_{\epsilon\searrow
0}\epsilon^{|\beta|-s}\int_{\mathbb{R}^{n}}\partial^{\beta}\phi_{*}(e+\epsilon
X)\psi(X)dX=\lim_{\epsilon\searrow
0}\int_{\mathbb{R}^{n}}\partial^{\beta}\Big{(}\epsilon^{-s}\phi_{*}(e+\epsilon
X)\Big{)}\psi(X)dX\\\ &\qquad=(-1)^{|\beta|}\lim_{\epsilon\searrow
0}\int_{\mathbb{R}^{n}}\epsilon^{-s}\phi_{*}(e+\epsilon
X)\partial^{\beta}\psi(X)dX\\\
&\qquad=(-1)^{|\beta|}\,k_{*}\,\int_{\mathbb{R}^{n}}(-e\cdot
X)^{s}_{+}\partial^{\beta}\psi(X)dX\\\
&\qquad=k_{*}\,\int_{\mathbb{R}^{n}}\partial^{\beta}(-e\cdot
X)^{s}_{+}\psi(X)dX\\\
&\qquad=(-1)^{|\beta|}\,k_{*}\,s(s-1)\ldots(s-|\beta|+1)e_{1}^{\beta_{1}}\ldots
e_{n}^{\beta_{n}}\int_{\mathbb{R}^{n}}(-e\cdot
X)^{s-|\beta|}_{+}\psi(X)dX.\end{split}$

Since the test function $\psi$ is arbitrary, the claim in Proposition 4.10 is
proved. ∎

###  4.4 Boundary behaviour of $s$-harmonic functions

In this section we analyze the asymptotic behaviour of $s$-harmonic functions,
with a “spherical bump function” as exterior Dirichlet datum.

The result needed for our purpose is the following:

######  Lemma 4.15.

Let $s>0$. Let $m\in\mathbb{N}_{0}$ and $\sigma\in(0,1)$ such that
$s=m+\sigma$.

Then, there exists

$\psi\in H^{s}(\mathbb{R}^{n})\cap C^{s}_{0}(\mathbb{R}^{n})$ such that
$(-\Delta)^{s}\psi=0$ in $B_{1}$, (4.85)

and, for every $x\in\partial B_{1-\epsilon}$,

$\psi(x)=k\,\epsilon^{s}+o(\epsilon^{s}),$ (4.86)

as $\epsilon\searrow 0$, for some $k>0$.

###### Proof.

Let $\overline{\psi}\in C^{\infty}(\mathbb{R},\,[0,1])$ such that
$\overline{\psi}=0$ in $\mathbb{R}\setminus(2,3)$ and $\overline{\psi}>0$ in
$(2,3)$. Let $\psi_{0}(x):=(-1)^{m}\overline{\psi}(|x|)$. We recall the
Poisson kernel

$\Gamma_{s}(x,y):=(-1)^{m}\frac{\gamma_{n,\sigma}}{|x-y|^{n}}\frac{(1-|x|^{2})^{s}_{+}}{(|y|^{2}-1)^{s}},$

for $x\in\mathbb{R}^{n}$, $y\in\mathbb{R}^{n}\setminus\overline{B_{1}}$, and a
suitable normalization constant $\gamma_{n,\sigma}>0$ (see formulas (1.10) and
(1.30) in [ABX]). We define

$\psi(x):=\displaystyle\int_{{\mathbb{R}}^{n}\setminus
B_{1}}\Gamma_{s}(x,y)\,\psi_{0}(y)\,dy+\psi_{0}(x).$

Notice that $\psi_{0}=0$ in $B_{3/2}$ and therefore we can exploit Theorem in
[ABX] and obtain that (4.85) is satisfied (notice also that $\psi=\psi_{0}$
outside $B_{1}$, hence $\psi$ is compactly supported).

Furthermore, to prove (4.86) we borrow some ideas from Lemma 2.2 in
[MR3626547] and we see that, for any $x\in\partial B_{1-\epsilon}$,

$\begin{split}\psi(x)&=c(-1)^{m}\int_{\mathbb{R}^{n}\setminus
B_{1}}\frac{\psi_{0}(y)(1-|x|^{2})^{s}}{(|y|^{2}-1)^{s}|x-y|^{n}}dy+\psi_{0}(x)\\\
&=c(-1)^{m}\int_{\mathbb{R}^{n}\setminus
B_{1}}\frac{\psi_{0}(y)(1-|x|^{2})^{s}}{(|y|^{2}-1)^{s}|x-y|^{n}}dy\\\
&=c\,(1-|x|^{2})^{s}\int_{2}^{3}\left[\int_{\mathbb{S}^{n-1}}\frac{\rho^{n-1}\overline{\psi}(\rho)}{(\rho^{2}-1)^{s}|x-\rho\omega|^{n}}d\omega\right]d\rho\\\
&=c\,(2\epsilon-\epsilon^{2})^{s}\int_{2}^{3}\left[\int_{\mathbb{S}^{n-1}}\frac{\rho^{n-1}\overline{\psi}(\rho)}{(\rho^{2}-1)^{s}|(1-\epsilon)e_{1}-\rho\omega|^{n}}d\omega\right]d\rho\\\
&=2^{s}c\,\epsilon^{s}\int_{2}^{3}\left[\int_{\mathbb{S}^{n-1}}\frac{\rho^{n-1}\overline{\psi}(\rho)}{(\rho^{2}-1)^{s}|e_{1}-\rho\omega|^{n}}d\omega\right]d\rho+o(\epsilon^{s})\\\
&=c\epsilon^{s}+o(\epsilon^{s}),\end{split}$

where $c>0$ is a constant possibly varying from line to line, and this
establishes (4.86). ∎

######  Remark 4.16.

As in Proposition 4.10, one can extend (4.86) to higher derivatives (in the
distributional sense), obtaining, for any $e\in\partial B_{1}$ and
$\beta\in\mathbb{N}^{n}$

$\lim_{\epsilon\searrow 0}\epsilon^{|\beta|-s}\partial^{\beta}\psi(e+\epsilon
X)=k_{\beta}\,e_{1}^{\beta_{1}}\dots e_{n}^{\beta_{n}}(-e\cdot
X)_{+}^{s-|\beta|},$

for some $\kappa_{\beta}\neq 0$.

Using Lemma 4.15, in the spirit of [MR3626547], we can construct a sequence of
$s$-harmonic functions approaching $(x\cdot e)^{s}_{+}$ for a fixed unit
vector $e$, by using a blow-up argument. Namely, we prove the following:

######  Corollary 4.17.

Let $e\in\partial B_{1}$. There exists a sequence $v_{e,j}\in
H^{s}(\mathbb{R}^{n})\cap C^{s}(\mathbb{R}^{n})$ such that
$(-\Delta)^{s}v_{e,j}=0$ in $B_{1}(e)$, $v_{e,j}=0$ in
$\mathbb{R}^{n}\setminus B_{4j}(e)$, and

$v_{e,j}\to\kappa(x\cdot e)^{s}_{+}\quad\mbox{in}\quad L^{1}(B_{1}(e)),$

as $j\to+\infty$, for some $\kappa>0$.

###### Proof.

Let $\psi$ be as in Lemma 4.15 and define

$v_{e,j}(x):=j^{s}\psi\left(\frac{x}{j}-e\right).$

The $s$-harmonicity and the property of being compactly supported follow by
the ones of $\psi$. We now prove the convergence. To this aim, given $x\in
B_{1}(e)$, we write $p_{j}:=\frac{x}{j}-e$ and $\epsilon_{j}:=1-|p_{j}|$.
Recall that since $x\in B_{1}(e)$, then $|x-e|^{2}<1$, which implies that
$|x|^{2}<2x\cdot e$ and $x\cdot e>0$ for any $x\in B_{1}(e)$.
As a consequence

$|p_{j}|^{2}=\left|\frac{x}{j}-e\right|^{2}=\frac{|x|^{2}}{j^{2}}+1-2\frac{x}{j}\cdot
e=1-\frac{2}{j}(x\cdot e)_{+}+o\left(\frac{1}{j}\right)(x\cdot e)^{2}_{+},$

and so

$\epsilon_{j}=\frac{(1+o(1))}{j}(x\cdot e)_{+}.$

Therefore, using (4.86),

$\begin{split}v_{e,j}(x)&=j^{s}\psi(p_{j})\\\
&=j^{s}\kappa(\epsilon_{j}^{s}+o(\epsilon^{s}_{j}))\\\
&=j^{s}\left(\frac{\kappa}{j^{s}}(x\cdot
e)_{+}^{s}+o\left(\frac{1}{j^{s}}\right)\right)\\\ &=\kappa(x\cdot
e)^{s}_{+}+o(1).\end{split}$

Integrating over $B_{1}(e)$, we obtain the desired $L^{1}$-convergence. ∎

Now, we show that, as in the case $s\in(0,1)$ proved in Theorem 3.1 of
[MR3626547], we can find an $s$-harmonic function with an arbitrarily large
number of derivatives prescribed at some point.

######  Proposition 4.18.

For any $\beta\in\mathbb{N}^{n}$, there exist $p\in\mathbb{R}^{n}$, $R>r>0$,
and $v\in H^{s}(\mathbb{R}^{n})\cap C^{s}(\mathbb{R}^{n})$ such that

$\begin{cases}(-\Delta)^{s}v=0&\quad\text{in}\quad B_{r}(p),\\\
v=0&\quad\text{in}\quad\mathbb{R}^{n}\setminus B_{R}(p),\end{cases}$ (4.87)
$D^{\alpha}v(p)=0\quad{\mbox{ for any }}\;\alpha\in\mathbb{N}^{n}\quad{\mbox{
with }}\;|\alpha|\leq|\beta|-1,$ $D^{\alpha}v(p)=0\quad{\mbox{ for any
}}\;\alpha\in\mathbb{N}^{n}\quad{\mbox{ with }}\;|\alpha|=|\beta|\quad{\mbox{
and }}\;\alpha\neq\beta$

and

$D^{\beta}v(p)=1.$

###### Proof.

Let $\mathcal{Z}$ be the set of all pairs
$(v,x)\in\left(H^{s}(\mathbb{R}^{n})\cap C^{s}(\mathbb{R}^{n})\right)\times
B_{r}(p)$ that satisfy (4.87) for some $R>r>0$ and $p\in\mathbb{R}^{n}$.

To each pair $(v,x)\in\mathcal{Z}$ we associate the vector
$\left(D^{\alpha}v(x)\right)_{|\alpha|\leq|\beta|}\in\mathbb{R}^{K^{\prime}}$,
for some $K^{\prime}=K^{\prime}_{|\beta|}$ and consider $\mathcal{V}$ to be
the vector space spanned by this construction, namely we set

$\mathcal{V}:=\Big{\\{}\left(D^{\alpha}v(x)\right)_{|\alpha|\leq|\beta|},\quad{\mbox{
with }}\;(v,x)\in\mathcal{Z}\Big{\\}}.$

We claim that

$\mathcal{V}=\mathbb{R}^{K^{\prime}}.$ (4.88)

To check this, we suppose by contradiction that $\mathcal{V}$ lies in a proper
subspace of $\mathbb{R}^{K^{\prime}}$. Then, $\mathcal{V}$ must lie in a
hyperplane, hence there exists

$c=(c_{\alpha})_{|\alpha|\leq|\beta|}\in\mathbb{R}^{K^{\prime}}\setminus\left\\{0\right\\}$
(4.89)

which is orthogonal to any vector
$\left(D^{\alpha}v(x)\right)_{|\alpha|\leq|\beta|}$ with
$(v,x)\in\mathcal{Z}$, that is

$\sum_{|\alpha|\leq|\beta|}c_{\alpha}D^{\alpha}v(x)=0.$ (4.90)

We notice that the pair $(v_{e,j},x)$, with $v_{j}$ as in Corollary 4.17,
$e\in\partial B_{1}$ and $x\in B_{1}(e)$, belongs to $\mathcal{Z}$.
Consequently, fixed $\xi\in\mathbb{R}^{n}\setminus B_{1/2}$ and set
$e:=\frac{\xi}{|\xi|}$, we have that (4.90) holds true when $v:=v_{e,j}$ and
$x\in B_{1}(e)$, namely

$\sum_{|\alpha|\leq|\beta|}c_{\alpha}D^{\alpha}v(x)=0.$

Let now $\varphi\in C_{0}^{\infty}(B_{1}(e))$. Integrating by parts, by
Corollary 4.17 and the Dominated Convergence Theorem, we have that

$\displaystyle
0=\lim_{j\to+\infty}\int_{\mathbb{R}^{n}}\sum_{|\alpha|\leq|\beta|}c_{\alpha}D^{\alpha}v_{e,j}(x)\varphi(x)\,dx=\lim_{j\to+\infty}\int_{\mathbb{R}^{n}}\sum_{|\alpha|\leq|\beta|}(-1)^{|\alpha|}c_{\alpha}v_{e,j}(x)D^{\alpha}\varphi(x)\,dx$
$\displaystyle\qquad=\kappa\int_{\mathbb{R}^{n}}\sum_{|\alpha|\leq|\beta|}(-1)^{|\alpha|}c_{\alpha}(x\cdot
e)^{s}_{+}D^{\alpha}\varphi(x)\,dx=\kappa\int_{\mathbb{R}^{n}}\sum_{|\alpha|\leq|\beta|}c_{\alpha}D^{\alpha}(x\cdot
e)^{s}_{+}\varphi(x)\,dx.$

This gives that, for every $x\in B_{1}(e)$,

$\sum_{|\alpha|\leq|\beta|}c_{\alpha}D^{\alpha}(x\cdot e)^{s}_{+}=0.$

Moreover, for every $x\in B_{1}(e)$,

$D^{\alpha}(x\cdot e)^{s}_{+}=s(s-1)\ldots(s-|\alpha|+1)(x\cdot
e)^{s-|\alpha|}_{+}e_{1}^{\alpha_{1}}\ldots e_{n}^{\alpha_{n}}.$

In particular, for $x=\frac{e}{|\xi|}\in B_{1}(e)$,

$D^{\alpha}(x\cdot
e)^{s}_{+}\big{|}_{|_{x=e/{|\xi|}}}=s(s-1)\ldots(s-|\alpha|+1)|\xi|^{-s}\xi_{1}^{\alpha_{1}}\ldots\xi_{n}^{\alpha_{n}}.$

And, using the usual multi-index notation, we write

$\sum_{|\alpha|\leq|\beta|}c_{\alpha}s(s-1)\ldots(s-|\alpha|+1)\xi^{\alpha}=0,$
(4.91)

for any $\xi\in\mathbb{R}^{n}\setminus B_{1/2}$. The identity (4.91) describes
a polynomial in $\xi$ which vanishes for any $\xi$ in an open subset of
$\mathbb{R}^{n}$. As a result, the Identity Principle for polynomials leads to

$c_{\alpha}s(s-1)\ldots(s-|\alpha|+1)=0,$

for all $|\alpha|\leq|\beta|$.

Consequently, since $s\in\mathbb{R}\setminus\mathbb{N}$, the product
$s(s-1)\ldots(s-|\alpha|+1)$ never vanishes, and so the coefficients
$c_{\alpha}$ are forced to be null for any $|\alpha|\leq|\beta|$. This is in
contradiction with (4.89), and therefore the proof of (4.88) is complete.

From this, the desired claim in Proposition 4.18 plainly follows. ∎

##  Chapter 5 Proof of the main result

This chapter is devoted to the proof of the main result in Theorem 2.1. This
will be accomplished by an auxiliary result of purely nonlocal type which will
allow us to prescribe an arbitrarily large number of derivatives at a point
for the solution of a fractional equation.

###  5.1 A result which implies Theorem 2.1

We will use the notation

$\Lambda_{-\infty}:=\Lambda_{(-\infty,\dots,-\infty)},$ (5.1)

that is we exploit (2.8) with $a_{1}:=\dots:=a_{l}:=-\infty$. This section
presents the following statement:

######  Theorem 5.1.

Suppose that

$\begin{split}&{\mbox{either there exists\leavevmode\nobreak\
$i\in\\{1,\dots,M\\}$ such that\leavevmode\nobreak\
${\mbox{\large{\wedn{b}}}}_{i}\neq 0$ and\leavevmode\nobreak\
$s_{i}\not\in{\mathbb{N}}$,}}\\\ &{\mbox{or there exists\leavevmode\nobreak\
$i\in\\{1,\dots,l\\}$ such that\leavevmode\nobreak\
${\mbox{\large{\wedn{c}}}}\,_{i}\neq 0$ and
$\alpha_{i}\not\in{\mathbb{N}}$.}}\end{split}$

Let $\ell\in\mathbb{N}$, $f:\mathbb{R}^{N}\rightarrow\mathbb{R}$, with $f\in
C^{\ell}\big{(}\overline{B_{1}^{N}}\big{)}$. Fixed $\epsilon>0$, there exist

$\begin{split}&u=u_{\epsilon}\in C^{\infty}\left(B_{1}^{N}\right)\cap
C\left(\mathbb{R}^{N}\right),\\\
&a=(a_{1},\dots,a_{l})=(a_{1,\epsilon},\dots,a_{l,\epsilon})\in(-\infty,0)^{l},\\\
{\mbox{and }}\quad&R=R_{\epsilon}>1\end{split}$

such that:

  * •

for every $h\in\\{1,\dots,l\\}$ and
$(x,y,t_{1},\dots,t_{h-1},t_{h+1},\dots,t_{l})$

the map ${\mathbb{R}}\ni t_{h}\mapsto u(x,y,t)$ belongs to
$C^{k_{h},\alpha_{h}}_{-\infty}$, (5.2)

in the notation of formula (1.4) of [CDV18],

  * •

it holds that

$\left\\{\begin{matrix}\Lambda_{-\infty}u=0&\mbox{ in
}\;B_{1}^{N-l}\times(-1,+\infty)^{l},\\\ u(x,y,t)=0&\mbox{ if }\;|(x,y)|\geq
R,\end{matrix}\right.$ (5.3)
$\partial^{k_{h}}_{t_{h}}u(x,y,t)=0\qquad{\mbox{if
}}t_{h}\in(-\infty,a_{h}),\qquad{\mbox{for all }}h\in\\{1,\dots,l\\},$ (5.4)

and

$\left\|u-f\right\|_{C^{\ell}(B_{1}^{N})}<\epsilon.$ (5.5)

The proof of Theorem 5.1 will basically occupy the rest of this work, and this
will lead us to the completion of the proof of Theorem 2.1. Indeed, we have
that:

######  Lemma 5.2.

If the statement of Theorem 5.1 holds true, then the statement in Theorem 2.1
holds true.

###### Proof.

Assume that the claims in Theorem 5.1 are satisfied. Then, by (5.2) and (5.4),
we are in the position of exploting Lemma A.1 in [CDV18] and conclude that, in
$B_{1}^{N-l}\times(-1,+\infty)^{l}$,

$D^{\alpha_{h}}_{t_{h},a_{h}}u=D^{\alpha_{h}}_{t_{h},-\infty}u,$

for every $h\in\\{1,\dots,l\\}$. This and (5.3) give that

$\Lambda_{a}u=\Lambda_{-\infty}u=0\qquad\mbox{ in
}\;B_{1}^{N-l}\times(-1,+\infty)^{l}.$ (5.6)

We also define

$\underline{a}:=\min_{h\in\\{1,\dots,l\\}}a_{h}$

and take $\tau\in C^{\infty}_{0}([-\underline{a}-2,3])$ with $\tau=1$ in
$[-\underline{a}-1,1]$. Let

$U(x,y,t):=u(x,y,t)\,\tau(t_{1})\dots\tau(t_{l}).$ (5.7)

Our goal is to prove that $U$ satisfies the theses of Theorem 2.1. To this
end, we observe that $u=U$ in $B^{N}_{1}$, therefore (2.12) for $U$ plainly
follows from (5.5).

In addition, from (2.6), we see that $D^{\alpha_{h}}_{t_{h},a_{h}}$ at a point
$t_{h}\in(-1,1)$ only depends on the values of the function between $a_{h}$
and $1$. Since the cutoffs in (5.7) do not alter these values, we see that
$D^{\alpha_{h}}_{t_{h},a_{h}}U=D^{\alpha_{h}}_{t_{h},a_{h}}u$ in $B_{1}^{N}$,
and accordingly $\Lambda_{a}U=\Lambda_{a}u$ in $B_{1}^{N}$. This and (5.6) say
that

$\Lambda_{a}U=0\qquad{\mbox{in }}B_{1}^{N}.$ (5.8)

Also, since $u$ in Theorem 5.1 is compactly supported in the variable $(x,y)$,
we see from (5.7) that $U$ is compactly supported in the variables $(x,y,t)$.
This and (5.8) give that (2.11) is satisfied by $U$ (up to renaming $R$). ∎

###  5.2 A pivotal span result towards the proof of Theorem 5.1

In what follows, we let $\Lambda_{-\infty}$ be as in (5.1), we recall the
setting in (2.1), and we use the following multi-indices notations:

$\begin{split}&\iota=\left(i,I,\mathfrak{I}\right)=\left(i_{1},\ldots,i_{n},I_{1},\ldots,I_{M},\mathfrak{I}_{1},\ldots,\mathfrak{I}_{l}\right)\in\mathbb{N}^{N}\\\
{\mbox{and
}}&\partial^{\iota}w=\partial^{i_{1}}_{x_{1}}\ldots\partial^{i_{n}}_{x_{n}}\partial^{I_{1}}_{y_{1}}\ldots\partial^{I_{M}}_{y_{M}}\partial^{\mathfrak{I}_{1}}_{t_{1}}\ldots\partial^{\mathfrak{I}_{l}}_{t_{l}}w.\end{split}$
(5.9)

Inspired by Lemma 5 of [DSV1], we consider the span of the derivatives of
functions in $\ker\Lambda_{-\infty}$, with derivatives up to a fixed order
$K\in{\mathbb{N}}$. We want to prove that the derivatives of such functions
span a maximal vectorial space.

For this, we denote by $\partial^{K}w(0)$ the vector with entries given, in
some prescribed order, by $\partial^{\iota}w(0)$ with $\left|\iota\right|\leq
K$.

We notice that

$\partial^{K}w(0)\in\mathbb{R}^{K^{\prime}}$ for some
$K^{\prime}\in{\mathbb{N}}$, (5.10)

with $K^{\prime}$ depending on $K$.

Now, we adopt the notation in formula (1.4) of [CDV18], and we denote by
$\mathcal{A}$ the set of all functions $w=w(x,y,t)$ such that for all
$h\in\\{1,\ldots,l\\}$ and all
$(x,y,t_{1},\ldots,t_{h-1},t_{h+1},\ldots,t_{l})\in\mathbb{R}^{N-1}$, the map
${\mathbb{R}}\ni t_{h}\mapsto w(x,y,t)$ belongs to
$C^{\infty}((a_{h},+\infty))\cap C^{k_{h},\alpha_{h}}_{-\infty}$, and (5.4)
holds true for some $a_{h}\in(-2,0)$.

We also set

$\begin{split}\mathcal{H}:=\Big{\\{}w\in C(\mathbb{R}^{N})\cap
C_{0}(\mathbb{R}^{N-l})\cap C^{\infty}(\mathcal{N})\cap\mathcal{A},\text{ for
some neighborhood $\mathcal{N}$ of the origin, }\\\ \text{ such that
}\Lambda_{-\infty}w=0\text{ in }\mathcal{N}\Big{\\}}\end{split}$

and, for any $w\in\mathcal{H}$, let $\mathcal{V}_{K}$ be the vector space
spanned by the vector $\partial^{K}w(0)$.

By (5.10), we know that $\mathcal{V}_{K}\subseteq\mathbb{R}^{K^{\prime}}$. In
fact, we show that equality holds in this inclusion, as stated in the
following111Notice that results analogous to Lemma 5.3 cannot hold for
solutions of local operators: for instance, pure second derivatives of
harmonic functions have to satisfy a linear equation, so they are forced to
lie in a proper subspace. In this sense, results such as Lemma 5.3 here reveal
a truly nonlocal phenomenon. result:

######  Lemma 5.3.

It holds that $\mathcal{V}_{K}=\mathbb{R}^{K^{\prime}}$.

The proof of Lemma 5.3 is by contradiction. Namely, if $\mathcal{V}_{K}$ does
not exhaust the whole of $\mathbb{R}^{K^{\prime}}$ there exists

$\theta\in\partial B_{1}^{K^{\prime}}$ (5.11)

such that

$\mathcal{V}_{K}\subseteq\left\\{\zeta\in\mathbb{R}^{K^{\prime}}\text{ s.t.
}\theta\cdot\zeta=0\right\\}.$ (5.12)

In coordinates, recalling (5.9), we write $\theta$ as
$\theta_{\iota}=\theta_{i,I,\mathfrak{I}}$, with
$i\in\mathbb{N}^{p_{1}+\dots+p_{n}}$, $I\in\mathbb{N}^{m_{1}+\dots+m_{M}}$ and
$\mathfrak{I}\in\mathbb{N}^{l}$. We consider

$\begin{split}&{\mbox{a multi-index
$\overline{I}\in\mathbb{N}^{m_{1}+\dots+m_{M}}$ such that it
maximizes\leavevmode\nobreak\ $|I|$}}\\\ &{\mbox{among all the multi-
indexes\leavevmode\nobreak\ $(i,I,\mathfrak{I})$ for which\leavevmode\nobreak\
$\left|i\right|+\left|I\right|+|\mathfrak{I}|\leq K$}}\\\
&{\mbox{and\leavevmode\nobreak\ $\theta_{i,I,\mathfrak{I}}\neq 0$ for
some\leavevmode\nobreak\ $(i,\mathfrak{I})$.}}\end{split}$ (5.13)

Some comments on the setting in (5.13). We stress that, by (5.11), the set
$\mathcal{S}$ of indexes $I$ for which there exist indexes $(i,\mathfrak{I})$
such that $|i|+|I|+|\mathfrak{I}|\leq K$ and $\theta_{i,I,\mathfrak{I}}\neq 0$
is not empty. Therefore, since ${\mathcal{S}}$ is a finite set, we can take

$S:=\sup_{I\in{\mathcal{S}}}|I|=\max_{I\in{\mathcal{S}}}|I|\in{\mathbb{N}}\cap[0,K].$

Hence, we consider a multi-index $\overline{I}$ for which $|\overline{I}|=S$
to obtain the setting in (5.13). By construction, we have that

  * •

$|i|+|\overline{I}|+|\mathfrak{I}|\leq K$,

  * •

if $|I|>|\overline{I}|$, then $\theta_{i,I,\mathfrak{I}}=0$,

  * •

and there exist multi-indexes $i$ and $\mathfrak{I}$ such that
$\theta_{i,\overline{I},\mathfrak{I}}\neq 0$.

As a variation of the setting in (5.13), we can also consider

$\begin{split}&{\mbox{a multi-index $\overline{\mathfrak{I}}\in\mathbb{N}^{l}$
such that it maximizes\leavevmode\nobreak\ $|\mathfrak{I}|$}}\\\ &{\mbox{among
all the multi-indexes\leavevmode\nobreak\ $(i,I,\mathfrak{I})$ for
which\leavevmode\nobreak\ $\left|i\right|+\left|I\right|+|\mathfrak{I}|\leq
K$}}\\\ &{\mbox{and\leavevmode\nobreak\ $\theta_{i,I,\mathfrak{I}}\neq 0$ for
some\leavevmode\nobreak\ $(i,I)$.}}\end{split}$ (5.14)

In the setting of (5.13) and (5.14), we claim that there exists an open set of
$\mathbb{R}^{p_{1}+\ldots+p_{n}}\times\mathbb{R}^{m_{1}+\ldots+m_{M}}\times\mathbb{R}^{l}$
such that for every
$(\underline{\mbox{\large{\wedn{x}}}}\,,\underline{\mbox{\large{\wedn{y}}}}\,,\underline{\mbox{\large{\wedn{t}}}}\,)$
in such open set we have that

$\begin{split}{\mbox{either }}\qquad&0=\sum_{{|i|+|I|+|\mathfrak{I}|\leq
K}\atop{|I|=|\overline{I}|}}C_{i,I,\mathfrak{I}}\;\theta_{i,I,\mathfrak{I}}\;\underline{\mbox{\large{\wedn{x}}}}\,^{i}\underline{\mbox{\large{\wedn{y}}}}\,^{{I}}\underline{\mbox{\large{\wedn{t}}}}\,^{\mathfrak{I}},\qquad{\mbox{
with }}\qquad C_{i,I,\mathfrak{I}}\neq 0,\\\ {\mbox{or
}}\qquad&0=\sum_{{|i|+|I|+|\mathfrak{I}|\leq
K}\atop{|\mathfrak{I}|=|\overline{\mathfrak{I}}|}}C_{i,I,\mathfrak{I}}\;\theta_{i,I,\mathfrak{I}}\;\underline{\mbox{\large{\wedn{x}}}}\,^{i}\underline{\mbox{\large{\wedn{y}}}}\,^{{I}}\underline{\mbox{\large{\wedn{t}}}}\,^{\mathfrak{I}},\qquad{\mbox{
with }}\qquad C_{i,I,\mathfrak{I}}\neq 0.\end{split}$ (5.15)

In our framework, the claim in (5.15) will be pivotal towards the completion
of the proof of Lemma 5.3. Indeed, let us suppose for the moment that (5.15)
is established and let us complete the proof of Lemma 5.3 by arguing as
follows.

Formula (5.15) says that $\theta\cdot\partial^{K}w(0)$ is a polynomial which
vanishes for any triple
$(\underline{\mbox{\large{\wedn{x}}}}\,,\underline{\mbox{\large{\wedn{y}}}}\,,\underline{\mbox{\large{\wedn{t}}}}\,)$
in an open subset of
$\mathbb{R}^{p_{1}+\ldots+p_{n}}\times\mathbb{R}^{m_{1}+\ldots+m_{M}}\times\mathbb{R}^{l}$.
Hence, using the identity principle of polynomials, we have that each
$C_{i,I,\mathfrak{I}}\;\theta_{i,I,\mathfrak{I}}$ is equal to zero whenever
$|i|+|I|+|\mathfrak{I}|\leq K$ and either $|I|=|\overline{I}|$ (if the first
identity in (5.15) holds true) or $|\mathfrak{I}|=|\overline{\mathfrak{I}}|$
(if the second identity in (5.15) holds true). Then, since
$C_{i,I,\mathfrak{I}}\neq 0$, we conclude that each
$\theta_{i,I,\mathfrak{I}}$ is zero as long as either $|I|=|\overline{I}|$ (in
the first case) or $|\mathfrak{I}|=|\overline{\mathfrak{I}}|$ (in the second
case), but this contradicts either the definition of $\overline{I}$ in (5.13)
(in the first case) or the definition of $\overline{\mathfrak{I}}$ in (5.14)
(in the second case). This would therefore complete the proof of Lemma 5.3.

In view of the discussion above, it remains to prove (5.15). To this end, we
distinguish the following four cases:

  1. 1.

there exist $i\in\\{1,\dots,n\\}$ and $j\in\\{1,\dots,M\\}$ such that
${\mbox{\large{\wedn{a}}}}_{i}\neq 0$ and ${\mbox{\large{\wedn{b}}}}_{j}\neq
0$,

  2. 2.

there exist $i\in\\{1,\dots,n\\}$ and $h\in\\{1,\dots,l\\}$ such that
${\mbox{\large{\wedn{a}}}}_{i}\neq 0$ and ${\mbox{\large{\wedn{c}}}}\,_{h}\neq
0$,

  3. 3.

we have that
${\mbox{\large{\wedn{a}}}}_{1}=\dots={\mbox{\large{\wedn{a}}}}_{n}=0$, and
there exists $j\in\\{1,\dots,M\\}$ such that
${\mbox{\large{\wedn{b}}}}_{j}\neq 0$,

  4. 4.

we have that
${\mbox{\large{\wedn{a}}}}_{1}=\dots={\mbox{\large{\wedn{a}}}}_{n}=0$, and
there exists $h\in\\{1,\dots,l\\}$ such that
${\mbox{\large{\wedn{c}}}}\,_{h}\neq 0$.

Notice that cases 1 and 3 deal with the case in which space-fractional
diffusion is present (and in case 1 one also has classical derivatives, while
in case 3 the classical derivatives are absent).

Similarly, cases 2 and 4 deal with the case in which time-fractional diffusion
is present (and in case 2 one also has classical derivatives, while in case 4
the classical derivatives are absent).

Of course, the case in which both space- and time-fractional diffusion occur
is already comprised by the previous cases (namely, it is comprised in both
cases 1 and 2 if classical derivatives are also present, and in both cases 3
and 4 if classical derivatives are absent).

###### Proof of (5.15), case 1.

For any $j\in\left\\{1,\ldots,M\right\\}$ we denote by
$\tilde{\phi}_{\star,j}$ the first eigenfunction for
$(-\Delta)^{s_{j}}_{y_{j}}$ vanishing outside $B_{1}^{m_{j}}$ given in
Corollary 4.9. We normalize it such that
$\|\tilde{\phi}_{\star,j}\|_{L^{2}(\mathbb{R}^{m_{j}})}=1$, and we write
$\lambda_{\star,j}\in(0,+\infty)$ to indicate the corresponding first
eigenvalue (which now depends on $s_{j}$), namely we write

$\begin{cases}(-\Delta)^{s_{j}}_{y_{j}}\tilde{\phi}_{\star,j}=\lambda_{\star,j}\tilde{\phi}_{\star,j}&\quad\text{in}\,B_{1}^{m_{j}},\\\
\tilde{\phi}_{\star,j}=0&\quad\text{in}\,\mathbb{R}^{m_{j}}\setminus\overline{B_{1}^{m_{j}}}.\end{cases}$
(5.16)

Up to reordering the variables and/or taking the operators to the other side
of the equation, given the assumptions of case 1, we can suppose that

${\mbox{\large{\wedn{a}}}}_{1}\neq 0$ (5.17)

and

${\mbox{${\mbox{\large{\wedn{b}}}}_{M}>0$}}.$ (5.18)

In view of (5.17), we can define

$R:=\left(\frac{1}{|{\mbox{\large{\wedn{a}}}}_{1}|}\displaystyle\left(\sum_{j=1}^{M-1}{|{\mbox{\large{\wedn{b}}}}_{j}|\lambda_{\star,j}}+\sum_{h=1}^{l}|{\mbox{\large{\wedn{c}}}}\,_{h}|\right)\right)^{1/|r_{1}|}.$
(5.19)

Now, we fix two sets of free parameters

$\underline{\mbox{\large{\wedn{x}}}}\,_{1}\in(R+1,R+2)^{p_{1}},\ldots,\underline{\mbox{\large{\wedn{x}}}}\,_{n}\in(R+1,R+2)^{p_{n}},$
(5.20)

and

$\underline{\mbox{\large{\wedn{t}}}}\,_{\star,1}\in\left(\frac{1}{2},1\right),\dots,\underline{\mbox{\large{\wedn{t}}}}\,_{\star,l}\in\left(\frac{1}{2},1\right).$
(5.21)

We also set

$\lambda_{j}:=\lambda_{\star,j}$ for $j\in\left\\{1,\ldots,M-1\right\\}$,
(5.22)

where $\lambda_{\star,j}$ is defined as in (5.16), and

$\lambda_{M}\,:=\,\frac{1}{{\mbox{\large{\wedn{b}}}}_{M}}\left(\sum_{j=1}^{n}{\left|{\mbox{\large{\wedn{a}}}}_{j}\right|\underline{\mbox{\large{\wedn{x}}}}\,_{j}^{r_{j}}}-\sum_{j=1}^{M-1}{{\mbox{\large{\wedn{b}}}}_{j}\lambda_{j}}-\sum_{h=1}^{l}{\mbox{\large{\wedn{c}}}}\,_{h}\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}\right).$
(5.23)

Notice that this definition is well-posed, thanks to (5.18). In addition, from
(5.20), we can write
$\underline{\mbox{\large{\wedn{x}}}}\,_{j}=(\underline{\mbox{\large{\wedn{x}}}}\,_{j1},\dots,\underline{\mbox{\large{\wedn{x}}}}\,_{jp_{j}})$,
and we know that $\underline{\mbox{\large{\wedn{x}}}}\,_{j\ell}>R+1$ for any
$j\in\\{1,\dots,n\\}$ and any $\ell\in\\{1,\dots,p_{j}\\}$. Therefore,

$\underline{\mbox{\large{\wedn{x}}}}\,_{j}^{r_{j}}=\underline{\mbox{\large{\wedn{x}}}}\,_{j1}^{r_{j1}}\dots\underline{\mbox{\large{\wedn{x}}}}\,_{jp_{j}}^{r_{jp_{j}}}\geq
0.$ (5.24)

From this, (5.19) and (5.21), we deduce that

$\displaystyle\sum_{j=1}^{n}{\left|{\mbox{\large{\wedn{a}}}}_{j}\right|\underline{\mbox{\large{\wedn{x}}}}\,_{j}^{r_{j}}}\geq\left|{\mbox{\large{\wedn{a}}}}_{1}\right|\underline{\mbox{\large{\wedn{x}}}}\,_{1}^{r_{1}}\geq\left|{\mbox{\large{\wedn{a}}}}_{1}\right|(R+1)^{|r_{1}|}>\left|{\mbox{\large{\wedn{a}}}}_{1}\right|R^{|r_{1}|}$
$\displaystyle\qquad=\sum_{j=1}^{M-1}{|{\mbox{\large{\wedn{b}}}}_{j}|\lambda_{j}}+\sum_{h=1}^{l}|{\mbox{\large{\wedn{c}}}}\,_{h}|\geq\sum_{j=1}^{M-1}{{\mbox{\large{\wedn{b}}}}_{j}\lambda_{j}}+\sum_{h=1}^{l}{\mbox{\large{\wedn{c}}}}\,_{h}\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h},$

and consequently, by (5.23),

$\lambda_{M}>0.$ (5.25)

We also set

$\omega_{j}:=\begin{cases}1&\quad\text{if }\,j=1,\dots,M-1,\\\
\displaystyle\frac{\lambda_{\star,M}^{1/2s_{M}}}{\lambda_{M}^{1/2s_{M}}}&\quad\text{if
}\,j=M.\end{cases}$ (5.26)

Notice that this definition is well-posed, thanks to (5.25). In addition, by
(5.16), we have that, for any $j\in\\{1,\dots,M\\}$, the functions

$\phi_{j}\left(y_{j}\right):=\tilde{\phi}_{\star,j}\left(\frac{y_{j}}{\omega_{j}}\right)$
(5.27)

are eigenfunctions of $(-\Delta)^{s_{j}}_{y_{j}}$ in $B_{\omega_{j}}^{m_{j}}$
with external homogenous Dirichlet boundary condition, and eigenvalues
$\lambda_{j}$: namely, we can rewrite (5.16) as

$\begin{cases}(-\Delta)^{s_{j}}_{y_{j}}{\phi}_{j}=\lambda_{j}{\phi}_{j}&\quad\text{in}\,B_{\omega_{j}}^{m_{j}},\\\
{\phi}_{j}=0&\quad\text{in}\,\mathbb{R}^{m_{j}}\setminus\overline{B_{\omega_{j}}^{m_{j}}}.\end{cases}$
(5.28)

Now, we define

$\psi_{\star,h}(t_{h}):=E_{\alpha_{h},1}(t_{h}^{\alpha_{h}}),$ (5.29)

where $E_{\alpha_{h},1}$ denotes the Mittag-Leffler function with parameters
$\alpha:=\alpha_{h}$ and $\beta:=1$ as defined in (3.1).

Moreover, we consider $a_{h}\in(-2,0)$, for every $h=1,\dots,l$, to be chosen
appropriately in what follows (the precise choice will be performed in
(5.48)), and, recalling (5.21), we let

$\underline{\mbox{\large{\wedn{t}}}}\,_{h}:=\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}^{1/{\alpha_{h}}},$
(5.30)

and we define

$\psi_{h}(t_{h}):=\psi_{\star,h}\big{(}\underline{\mbox{\large{\wedn{t}}}}\,_{h}(t_{h}-a_{h})\big{)}=E_{\alpha_{h},1}\big{(}\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}(t_{h}-a_{h})^{\alpha_{h}}\big{)}.$
(5.31)

We point out that, thanks to Lemma 3.2, the function in (5.31), solves

$\begin{cases}D^{\alpha_{h}}_{t_{h},a_{h}}\psi_{h}(t_{h})=\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}\psi_{h}(t_{h})&\quad\text{in
}\,(a_{h},+\infty),\\\ \psi_{h}(a_{h})=1,\\\
\partial^{m}_{t_{h}}\psi_{h}(a_{h})=0&\quad\text{for every
}\,m\in\\{1,\dots,[\alpha_{h}]\\}.\end{cases}$ (5.32)

Moreover, for any $h\in\\{1,\ldots,l\\}$, we define

$\psi^{\star}_{h}(t_{h}):=\begin{cases}\psi_{h}(t_{h})\qquad\text{ if
}\,t_{h}\in[a_{h},+\infty)\\\ 1\qquad\qquad\text{ if
}\,t_{h}\in(-\infty,a_{h}).\end{cases}$ (5.33)

Thanks to (5.32) and Lemma A.3 in [CDV18] applied here with $b:=a_{h}$,
$a:=-\infty$, $u:=\psi_{h}$, $u_{\star}:=\psi^{\star}_{h}$, we have that
$\psi^{\star}_{h}\in C^{k_{h},\alpha_{h}}_{-\infty}$, and

$D^{\alpha_{h}}_{t_{h},-\infty}\psi_{h}^{\star}(t_{h})=D^{\alpha_{h}}_{t_{h},a_{h}}\psi_{h}(t_{h})=\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}\psi_{h}(t_{h})=\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}\psi_{h}^{\star}(t_{h})\,\text{
in every interval }\,I\Subset(a_{h},+\infty).$ (5.34)

We observe that the setting in (5.33) is compatible with the ones in (5.2) and
(5.4) .

From (3.1) and (5.31), we see that

$\psi_{h}(t_{h})=\sum_{j=0}^{+\infty}{\frac{\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}^{j}\,(t_{h}-a_{h})^{\alpha_{h}j}}{\Gamma\left(\alpha_{h}j+1\right)}}.$

Consequently, for every ${\mathfrak{I}_{h}}\in\mathbb{N}$, we have that

$\partial^{\mathfrak{I}_{h}}_{t_{h}}\psi_{h}(t_{h})=\sum_{j=0}^{+\infty}{\frac{\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}^{j}\,\alpha_{h}j(\alpha_{h}j-1)\dots(\alpha_{h}j-\mathfrak{I}_{h}+1)(t_{h}-a_{h})^{\alpha_{h}j-\mathfrak{I}_{h}}}{\Gamma\left(\alpha_{h}j+1\right)}}.$
(5.35)

Now, we define, for any $i\in\left\\{1,\ldots,n\right\\}$,

$\overline{{\mbox{\large{\wedn{a}}}}}_{i}:=\begin{cases}\displaystyle\frac{{\mbox{\large{\wedn{a}}}}_{i}}{\left|{\mbox{\large{\wedn{a}}}}_{i}\right|}\quad\text{
if }{\mbox{\large{\wedn{a}}}}_{i}\neq 0,\\\ 1\quad\text{ if
}{\mbox{\large{\wedn{a}}}}_{i}=0.\end{cases}$

We notice that

$\overline{{\mbox{\large{\wedn{a}}}}}_{i}\neq 0$ for all
$i\in\left\\{1,\ldots,n\right\\}$, (5.36)

and

${{\mbox{\large{\wedn{a}}}}}_{i}\overline{{\mbox{\large{\wedn{a}}}}}_{i}=|{{\mbox{\large{\wedn{a}}}}}_{i}|.$
(5.37)

Now, for each $i\in\\{1,\dots,n\\}$, we consider the multi-index
$r_{i}=(r_{i1},\dots,r_{ip_{i}})\in\mathbb{N}^{p_{i}}$. This multi-index acts
on $\mathbb{R}^{p_{i}}$, whose variables are denoted by
$x_{i}=(x_{i1},\dots,x_{ip_{i}})\in\mathbb{R}^{p_{i}}$. We let
$\overline{v}_{i1}$ be the solution of the Cauchy problem

$\begin{cases}\partial^{r_{i1}}_{x_{i1}}\overline{v}_{i1}=-\overline{{\mbox{\large{\wedn{a}}}}}_{i}\overline{v}_{i1}\\\
\partial^{\beta_{1}}_{x_{i1}}\overline{v}_{i1}\left(0\right)=1\quad\text{ for
every }\beta_{1}\leq r_{i1}-1.\end{cases}$ (5.38)

We notice that the solution of the Cauchy problem in (5.38) exists at least in
a neighborhood of the origin of the form $[-\rho_{i1},\rho_{i1}]$ for a
suitable $\rho_{i1}>0$.

Moreover, if $p_{i}\geq 2$, for any $\ell\in\\{2,\dots,p_{i}\\}$, we consider
the solution of the following Cauchy problem:

$\begin{cases}\partial^{r_{i\ell}}_{x_{i\ell}}\overline{v}_{i\ell}=\overline{v}_{i\ell}\\\
\partial^{\beta_{\ell}}_{x_{i\ell}}\overline{v}_{i\ell}\left(0\right)=1\quad\text{
for every }\beta_{\ell}\leq r_{i\ell}-1.\end{cases}$ (5.39)

As above, these solutions are well-defined at least in a neighborhood of the
origin of the form $[-\rho_{i\ell},\rho_{i\ell}]$, for a suitable
$\rho_{i\ell}>0$.

Then, we define

$\overline{\rho}_{i}:=\min\\{\rho_{i1},\dots,\rho_{ip_{i}}\\}=\min_{\ell\in\\{1,\dots,p_{i}\\}}\rho_{i\ell}.$

In this way, for every $x_{i}=(x_{i1},\dots,x_{ip_{i}})\in
B^{p_{i}}_{\overline{\rho}_{i}}$, we set

$\overline{v}_{i}(x_{i}):=\overline{v}_{i1}(x_{i1})\ldots\overline{v}_{i{p_{i}}}(x_{ip_{i}}).$
(5.40)

By (5.38) and (5.39), we have that

$\begin{cases}\partial^{r_{i}}_{x_{i}}\overline{v}_{i}=-\overline{{\mbox{\large{\wedn{a}}}}}_{i}\overline{v}_{i}\\\
\\\
\partial^{\beta}_{x_{i}}\overline{v}_{i}\left(0\right)=1\quad\begin{matrix}&\text{
for every $\beta=(\beta_{1},\dots\beta_{p_{i}})\in{\mathbb{N}}^{p_{i}}$}\\\
&\text{ such that\leavevmode\nobreak\ $\beta_{\ell}\leq r_{i\ell}-1$ for
each\leavevmode\nobreak\
$\ell\in\\{1,\dots,p_{i}\\}$.}\end{matrix}\end{cases}$ (5.41)

Now, we define

$\rho:=\min\\{\overline{\rho}_{1},\dots\overline{\rho}_{n}\\}=\min_{i\in\\{1,\dots,n\\}}\overline{\rho}_{i}.$

We take

$\overline{\tau}\in
C_{0}^{\infty}\left(B^{p_{1}+\ldots+p_{n}}_{\rho/(R+2)}\right),$

with $\overline{\tau}=1$ in $B^{p_{1}+\ldots+p_{n}}_{\rho/(2(R+2))}$, and, for
every
$x=(x_{1},\dots,x_{n})\in{\mathbb{R}}^{p_{1}}\times\dots\times{\mathbb{R}}^{p_{n}}$,
we set

$\tau_{1}\left(x_{1},\ldots,x_{n}\right):=\overline{\tau}\left(\underline{\mbox{\large{\wedn{x}}}}\,_{1}\otimes
x_{1},\ldots,\underline{\mbox{\large{\wedn{x}}}}\,_{n}\otimes x_{n}\right).$
(5.42)

We recall that the free parameters
$\underline{\mbox{\large{\wedn{x}}}}\,_{1},\dots,\underline{\mbox{\large{\wedn{x}}}}\,_{n}$
have been introduced in (5.20), and we have used here the notation

$\underline{\mbox{\large{\wedn{x}}}}\,_{i}\otimes
x_{i}=(\underline{\mbox{\large{\wedn{x}}}}\,_{i1},\dots,\underline{\mbox{\large{\wedn{x}}}}\,_{ip_{i}})\otimes(x_{i1},\dots,x_{ip_{i}}):=(\underline{\mbox{\large{\wedn{x}}}}\,_{i1}x_{i1},\dots,\underline{\mbox{\large{\wedn{x}}}}\,_{ip_{i}}x_{ip_{i}})\in{\mathbb{R}}^{p_{i}},$

for every $i\in\\{1,\dots,n\\}$.

We also set, for any $i\in\\{1,\dots,n\\}$,

$v_{i}\left(x_{i}\right):=\overline{v}_{i}\left(\underline{\mbox{\large{\wedn{x}}}}\,_{i}\otimes
x_{i}\right).$ (5.43)

We point out that if $x_{i}\in B^{p_{i}}_{\overline{\rho}_{i}/(R+2)}$ we have
that

$|\underline{\mbox{\large{\wedn{x}}}}\,_{i}\otimes
x_{i}|^{2}=\sum_{\ell=1}^{p_{i}}(\underline{\mbox{\large{\wedn{x}}}}\,_{i\ell}x_{i\ell})^{2}\leq(R+2)^{2}\sum_{\ell=1}^{p_{i}}x_{i\ell}^{2}<\overline{\rho}_{i}^{2},$

thanks to (5.20), and therefore the setting in (5.43) is well-defined for
every $x_{i}\in B^{p_{i}}_{\overline{\rho}_{i}/(R+2)}$.

Recalling (5.41) and (5.43), we see that, for any $i\in\\{1,\dots,n\\}$,

$\partial_{x_{i}}^{r_{i}}v_{i}(x_{i})=\underline{\mbox{\large{\wedn{x}}}}\,_{i}^{r_{i}}\partial_{x_{i}}^{r_{i}}\overline{v}_{i}\left(\underline{\mbox{\large{\wedn{x}}}}\,_{i}\otimes
x_{i}\right)=-\overline{\mbox{\large{\wedn{a}}}}_{i}\underline{\mbox{\large{\wedn{x}}}}\,_{i}^{r_{i}}\overline{v}_{i}\left(\underline{\mbox{\large{\wedn{x}}}}\,_{i}\otimes
x_{i}\right)=-\overline{\mbox{\large{\wedn{a}}}}_{i}\underline{\mbox{\large{\wedn{x}}}}\,_{i}^{r_{i}}{v}_{i}(x_{i}).$
(5.44)

We take $e_{1},\ldots,e_{M}$, with

$e_{j}\in\partial B_{\omega_{j}}^{m_{j}},$ (5.45)

and we introduce an additional set of free parameters $Y_{1},\ldots,Y_{M}$
with

$Y_{j}\in\mathbb{R}^{m_{j}}\qquad{\mbox{ and }}\qquad e_{j}\cdot Y_{j}<0.$
(5.46)

We let $\epsilon>0$, to be taken small possibly depending on the free
parameters $e_{j}$, $Y_{j}$ and $\underline{\mbox{\large{\wedn{t}}}}\,_{h}$,
and we define

$\begin{split}w\left(x,y,t\right):=&\tau_{1}\left(x\right)v_{1}\left(x_{1}\right)\cdot\ldots\cdot
v_{n}\left(x_{n}\right)\phi_{1}\left(y_{1}+e_{1}+\epsilon
Y_{1}\right)\cdot\ldots\cdot\phi_{M}\left(y_{M}+e_{M}+\epsilon Y_{M}\right)\\\
&\times\psi^{\star}_{1}(t_{1})\cdot\ldots\cdot\psi^{\star}_{l}(t_{l}),\end{split}$
(5.47)

where the setting in (5.27), (5.33), (5.42) and (5.43) has been exploited.

We also notice that $w\in C\left(\mathbb{R}^{N}\right)\cap
C_{0}\left(\mathbb{R}^{N-l}\right)\cap\mathcal{A}$. Moreover, if

$a=(a_{1},\dots,a_{l}):=\left(-\frac{\epsilon}{\underline{\mbox{\large{\wedn{t}}}}\,_{1}},\dots,-\frac{\epsilon}{\underline{\mbox{\large{\wedn{t}}}}\,_{l}}\right)\in(-\infty,0)^{l}$
(5.48)

and $(x,y)$ is sufficiently close to the origin and
$t\in(a_{1},+\infty)\times\dots\times(a_{l},+\infty)$, we have that

$\displaystyle\Lambda_{-\infty}w\left(x,y,t\right)$ $\displaystyle=$
$\displaystyle\left(\sum_{i=1}^{n}{\mbox{\large{\wedn{a}}}}_{i}\partial^{r_{i}}_{x_{i}}+\sum_{j=1}^{M}{\mbox{\large{\wedn{b}}}}_{j}(-\Delta)^{s_{j}}_{y_{j}}+\sum_{h=1}^{l}{\mbox{\large{\wedn{c}}}}\,_{h}D^{\alpha_{h}}_{t_{h},-\infty}\right)w\left(x,y,t\right)$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{n}{\mbox{\large{\wedn{a}}}}_{i}v_{1}\left(x_{1}\right)\ldots
v_{i-1}\left(x_{i-1}\right)\partial^{r_{i}}_{x_{i}}v_{i}\left(x_{i}\right)v_{i+1}\left(x_{i+1}\right)\ldots
v_{n}\left(x_{n}\right)$
$\displaystyle\qquad\times\phi_{1}\left(y_{1}+e_{1}+\epsilon
Y_{1}\right)\ldots\phi_{M}\left(y_{M}+e_{M}+\epsilon
Y_{M}\right)\psi^{\star}_{1}\left(t_{1}\right)\ldots\psi^{\star}_{l}\left(t_{l}\right)$
$\displaystyle+\sum_{j=1}^{M}{\mbox{\large{\wedn{b}}}}_{j}v_{1}\left(x_{1}\right)\ldots
v_{n}\left(x_{n}\right)\phi_{1}\left(y_{1}+e_{1}+\epsilon
Y_{1}\right)\ldots\phi_{j-1}\left(y_{j-1}+e_{j-1}+\epsilon Y_{j-1}\right)$
$\displaystyle\qquad\times(-\Delta)^{s_{j}}_{y_{j}}\phi_{j}\left(y_{j}+e_{j}+\epsilon
Y_{j}\right)\phi_{j+1}\left(y_{j+1}+e_{j+1}+\epsilon
Y_{j+1}\right)\ldots\phi_{M}\left(y_{M}+e_{M}+\epsilon Y_{M}\right)$
$\displaystyle\qquad\times\psi^{\star}_{1}\left(t_{1}\right)\ldots\psi^{\star}_{l}\left(t_{l}\right)$
$\displaystyle+\sum_{h=1}^{l}{\mbox{\large{\wedn{c}}}}\,_{h}v_{1}\left(x_{1}\right)\ldots
v_{n}\left(x_{n}\right)\phi_{1}\left(y_{1}+e_{1}+\epsilon
Y_{1}\right)\ldots\phi_{M}\left(y_{M}+e_{M}+\epsilon
Y_{M}\right)\psi^{\star}_{1}\left(t_{1}\right)\ldots\psi^{\star}_{h-1}\left(t_{h-1}\right)$
$\displaystyle\qquad\times
D^{\alpha_{h}}_{t_{h},-\infty}\psi^{\star}_{h}\left(t_{h}\right)\psi^{\star}_{h+1}(t_{h+1})\ldots\psi^{\star}_{l}\left(t_{l}\right)$
$\displaystyle=$
$\displaystyle-\sum_{i=1}^{n}{\mbox{\large{\wedn{a}}}}_{i}\overline{\mbox{\large{\wedn{a}}}}_{i}\underline{\mbox{\large{\wedn{x}}}}\,_{i}^{r_{i}}v_{1}\left(x_{1}\right)\ldots
v_{n}\left(x_{n}\right)\phi_{1}\left(y_{1}+e_{1}+\epsilon
Y_{1}\right)\ldots\phi_{M}\left(y_{M}+e_{M}+\epsilon
Y_{M}\right)\psi^{\star}_{1}(t_{1})\ldots\psi^{\star}_{l}(t_{l})$
$\displaystyle+\sum_{j=1}^{M}{\mbox{\large{\wedn{b}}}}_{j}\lambda_{j}v_{1}\left(x_{1}\right)\ldots
v_{n}\left(x_{n}\right)\phi_{1}\left(y_{1}+e_{1}+\epsilon
Y_{1}\right)\ldots\phi_{M}\left(y_{M}+e_{M}+\epsilon
Y_{M}\right)\psi^{\star}_{1}(t_{1})\ldots\psi^{\star}_{l}(t_{l})$
$\displaystyle+\sum_{h=1}^{l}{\mbox{\large{\wedn{c}}}}\,_{h}\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}v_{1}\left(x_{1}\right)\ldots
v_{n}\left(x_{n}\right)\phi_{1}\left(y_{1}+e_{1}+\epsilon
Y_{1}\right)\ldots\phi_{M}\left(y_{M}+e_{M}+\epsilon
Y_{M}\right)\psi^{\star}_{1}(t_{1})\ldots\psi^{\star}_{l}(t_{l})$
$\displaystyle=$
$\displaystyle\left(-\sum_{i=1}^{n}{\mbox{\large{\wedn{a}}}}_{i}\overline{\mbox{\large{\wedn{a}}}}_{i}\underline{\mbox{\large{\wedn{x}}}}\,_{i}^{r_{i}}+\sum_{j=1}^{M}{\mbox{\large{\wedn{b}}}}_{j}\lambda_{j}+\sum_{h=1}^{l}{\mbox{\large{\wedn{c}}}}\,_{h}\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}\right)w(x,y,t),$

thanks to (5.28), (5.34) and (5.44) .

Consequently, making use of (5.22), (5.23) and (5.37), if $(x,y)$ lies near
the origin and $t\in(a_{1},+\infty)\times\dots\times(a_{l},+\infty)$, we have
that

$\displaystyle\Lambda_{-\infty}w\left(x,y,t\right)=\left(-\sum_{i=1}^{n}|{\mbox{\large{\wedn{a}}}}_{i}|\underline{\mbox{\large{\wedn{x}}}}\,_{i}^{r_{i}}+\sum_{j=1}^{M-1}{\mbox{\large{\wedn{b}}}}_{j}\lambda_{j}+{\mbox{\large{\wedn{b}}}}_{M}\lambda_{M}+\sum_{h=1}^{l}{\mbox{\large{\wedn{c}}}}\,_{h}\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}\right)w(x,y,t)$
$\displaystyle\qquad=\left(-\sum_{i=1}^{n}|{\mbox{\large{\wedn{a}}}}_{i}|\underline{\mbox{\large{\wedn{x}}}}\,_{i}^{r_{i}}+\sum_{j=1}^{M-1}{\mbox{\large{\wedn{b}}}}_{j}\lambda_{\star,j}+{\mbox{\large{\wedn{b}}}}_{M}\lambda_{M}+\sum_{h=1}^{l}{\mbox{\large{\wedn{c}}}}\,_{h}\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}\right)w(x,y,t)=0.$

This says that $w\in\mathcal{H}$. Thus, in light of (5.12) we have that

$0=\theta\cdot\partial^{K}w\left(0\right)=\sum_{\left|\iota\right|\leq
K}{\theta_{\iota}\partial^{\iota}w\left(0\right)}=\sum_{|i|+|I|+|\mathfrak{I}|\leq
K}\theta_{i,I,\mathfrak{I}}\,\partial_{x}^{i}\partial_{y}^{I}\partial_{t}^{\mathfrak{I}}w\left(0\right).$
(5.49)

Now, we recall (5.40) and we claim that, for any $j\in\\{1,\dots,n\\}$, any
$\ell\in\\{1,\dots,p_{j}\\}$ and any $i_{j\ell}\in\mathbb{N}$, we have that

$\partial^{i_{j\ell}}_{x_{j\ell}}\overline{v}_{j\ell}(0)\neq 0.$ (5.50)

We prove it by induction over $i_{j\ell}$. Indeed, if
$i_{j\ell}\in\left\\{0,\ldots,r_{j\ell}-1\right\\}$, then the initial
condition in (5.38) (if $\ell=1$) or (5.39) (if $\ell\geq 2$) gives that
$\partial^{i_{j\ell}}_{x_{i\ell}}\overline{v}_{i\ell}\left(0\right)=1$, and so
(5.50) is true in this case.

To perform the inductive step, let us now suppose that the claim in (5.50)
still holds for all $i_{j\ell}\in\left\\{0,\ldots,i_{0}\right\\}$ for some
$i_{0}$ such that $i_{0}\geq r_{j\ell}-1$. Then, using the equation in (5.38)
(if $\ell=1$) or in (5.39) (if $\ell\geq 2$), we have that

$\partial^{i_{0}+1}_{x_{j\ell}}\overline{v}_{j}=\partial^{i_{0}+1-r_{j\ell}}_{x_{j\ell}}\partial^{r_{j\ell}}_{x_{j\ell}}\overline{v}_{j}=-\tilde{a}_{j}\partial^{i_{0}+1-r_{j\ell}}_{x_{j\ell}}\overline{v}_{j},$
(5.51)

with

$\tilde{a}_{j}:=\begin{cases}\overline{a}_{j}&{\mbox{ if }}\ell=1,\\\
-1&{\mbox{ if }}\ell\geq 2.\end{cases}$

Notice that $\tilde{a}_{j}\neq 0$, in view of (5.36), and
$\partial^{i_{0}+1-r_{j\ell}}_{x_{j\ell}}\overline{v}_{j}\left(0\right)\neq
0$, by the inductive assumption. These considerations and (5.51) give that
$\partial^{i_{0}+1}_{x_{j\ell}}\overline{v}_{j}\left(0\right)\neq 0$, and this
proves (5.50).

Now, using (5.40) and (5.50) we have that, for any $j\in\\{1,\dots,n\\}$ and
any $i_{j}\in\mathbb{N}^{p_{j}}$,

$\partial^{i_{j}}_{x_{j}}\overline{v}_{j}(0)\neq 0.$

This, (5.20) and the computation in (5.44) give that, for any
$j\in\\{1,\dots,n\\}$ and any $i_{j}\in\mathbb{N}^{p_{j}}$,

$\partial^{i_{j}}_{x_{j}}{v}_{j}(0)=\underline{\mbox{\large{\wedn{x}}}}\,_{j}^{i_{j}}\partial^{i_{j}}_{x_{j}}\overline{v}_{j}(0)\neq
0.$ (5.52)

We also notice that, in light of (5.33), (5.47) and (5.49),

$\begin{split}&0=\sum_{|i|+|I|+|\mathfrak{I}|\leq
K}\theta_{i,I,\mathfrak{I}}\,\partial^{i_{1}}_{x_{1}}{v}_{1}(0)\ldots\partial^{i_{n}}_{x_{n}}{v}_{n}(0)\,\partial_{y_{1}}^{I_{1}}\phi_{1}\left(e_{1}+\epsilon
Y_{1}\right)\ldots\partial_{y_{M}}^{I_{M}}\phi_{M}\left(e_{M}+\epsilon
Y_{M}\right)\\\
&\qquad\qquad\qquad\times\partial^{\mathfrak{I}_{1}}_{t_{1}}\psi_{1}(0)\ldots\partial^{\mathfrak{I}_{l}}\psi_{l}(0).\end{split}$
(5.53)

Now, by (5.27) and Proposition 4.10 (applied to $s:=s_{j}$, $\beta:=I_{j}$,
$e:=\frac{e_{j}}{\omega_{j}}\in\partial B_{1}^{m_{j}}$, due to (5.45), and
$X:=\frac{Y_{j}}{\omega_{j}}$), we see that, for any $j=1,\ldots,M$,

$\begin{split}\omega_{j}^{|I_{j}|}\lim_{\epsilon\searrow
0}\epsilon^{|I_{j}|-s_{j}}\partial_{y_{j}}^{I_{j}}\phi_{j}\left(e_{j}+\epsilon
Y_{j}\right)\;=\;&\lim_{\epsilon\searrow
0}\epsilon^{|I_{j}|-s_{j}}\partial_{y_{j}}^{I_{j}}\tilde{\phi}_{\star,j}\left(\frac{e_{j}+\epsilon
Y_{j}}{\omega_{j}}\right)\\\
=\;&\kappa_{j}\frac{e_{j}^{I_{j}}}{\omega_{j}^{|I_{j}|}}\left(-\frac{e_{j}}{\omega_{j}}\cdot\frac{Y_{j}}{\omega_{j}}\right)_{+}^{s_{j}-|I_{j}|},\end{split}$
(5.54)

with $\kappa_{j}\neq 0$, in the sense of distributions (in the coordinates
$Y_{j}$).

Moreover, using (5.35) and (5.48), it follows that

$\displaystyle\partial^{\mathfrak{I}_{h}}_{t_{h}}\psi_{h}(0)$ $\displaystyle=$
$\displaystyle\sum_{j=0}^{+\infty}{\frac{\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}^{j}\,\alpha_{h}j(\alpha_{h}j-1)\dots(\alpha_{h}j-\mathfrak{I}_{h}+1)(0-a_{h})^{\alpha_{h}j-\mathfrak{I}_{h}}}{\Gamma\left(\alpha_{h}j+1\right)}}$
$\displaystyle=$
$\displaystyle\sum_{j=0}^{+\infty}{\frac{\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}^{j}\,\alpha_{h}j(\alpha_{h}j-1)\dots(\alpha_{h}j-\mathfrak{I}_{h}+1)\,{\epsilon}^{\alpha_{h}j-\mathfrak{I}_{h}}}{\Gamma\left(\alpha_{h}j+1\right)\;{\underline{\mbox{\large{\wedn{t}}}}\,_{h}}^{\alpha_{h}j-\mathfrak{I}_{h}}}}$
$\displaystyle=$
$\displaystyle\sum_{j=1}^{+\infty}{\frac{\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}^{j}\,\alpha_{h}j(\alpha_{h}j-1)\dots(\alpha_{h}j-\mathfrak{I}_{h}+1)\,{\epsilon}^{\alpha_{h}j-\mathfrak{I}_{h}}}{\Gamma\left(\alpha_{h}j+1\right)\;{\underline{\mbox{\large{\wedn{t}}}}\,_{h}}^{\alpha_{h}j-\mathfrak{I}_{h}}}}.$

Accordingly, recalling (5.30), we find that

$\begin{split}&\lim_{\epsilon\searrow
0}\epsilon^{\mathfrak{I}_{h}-\alpha_{h}}\partial^{\mathfrak{I}_{h}}_{t_{h}}\psi_{h}(0)=\lim_{\epsilon\searrow
0}\sum_{j=1}^{+\infty}{\frac{\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}^{j}\,\alpha_{h}j(\alpha_{h}j-1)\dots(\alpha_{h}j-\mathfrak{I}_{h}+1)\,{\epsilon}^{\alpha_{h}(j-1)}}{\Gamma\left(\alpha_{h}j+1\right)\;{\underline{\mbox{\large{\wedn{t}}}}\,_{h}}^{\alpha_{h}j-\mathfrak{I}_{h}}}}\\\
&\qquad={\frac{\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}\,\alpha_{h}(\alpha_{h}-1)\dots(\alpha_{h}-\mathfrak{I}_{h}+1)}{\Gamma\left(\alpha_{h}+1\right)\;\underline{\mbox{\large{\wedn{t}}}}\,_{h}^{\alpha_{h}-\mathfrak{I}_{h}}}}={\frac{\underline{\mbox{\large{\wedn{t}}}}\,_{h}^{\mathfrak{I}_{h}}\,\alpha_{h}(\alpha_{h}-1)\dots(\alpha_{h}-\mathfrak{I}_{h}+1)}{\Gamma\left(\alpha_{h}+1\right)}}.\end{split}$
(5.55)

Also, recalling (5.13), we can write (5.53) as

$\begin{split}&0=\sum_{{|i|+|I|+|\mathfrak{I}|\leq
K}\atop{|I|\leq|\overline{I}|}}\theta_{i,I,\mathfrak{I}}\,\partial^{i_{1}}_{x_{1}}{v}_{1}(0)\ldots\partial^{i_{n}}_{x_{n}}{v}_{n}(0)\,\partial_{y_{1}}^{I_{1}}\phi_{1}\left(e_{1}+\epsilon
Y_{1}\right)\ldots\partial_{y_{M}}^{I_{M}}\phi_{M}\left(e_{M}+\epsilon
Y_{M}\right)\\\
&\qquad\qquad\times\partial^{\mathfrak{I}_{1}}_{t_{1}}\psi_{1}(0)\ldots\partial^{\mathfrak{I}_{l}}_{t_{l}}\psi_{l}(0).\end{split}$
(5.56)

Moreover, we define

$\Xi:=\left|\overline{I}\right|-\sum_{j=1}^{M}{s_{j}}+|\mathfrak{I}|-\sum_{h=1}^{l}{\alpha_{h}}.$

Then, we multiply (5.56) by $\epsilon^{\Xi}\in(0,+\infty)$, and we send
$\epsilon$ to zero. In this way, we obtain from (5.54), (5.55) and (5.56) that

$\displaystyle 0$ $\displaystyle=$ $\displaystyle\lim_{\epsilon\searrow
0}\epsilon^{\Xi}\sum_{{|i|+|I|+|\mathfrak{I}|\leq
K}\atop{|I|\leq|\overline{I}|}}\theta_{i,I,\mathfrak{I}}\,\partial^{i_{1}}_{x_{1}}{v}_{1}(0)\ldots\partial^{i_{n}}_{x_{n}}{v}_{n}(0)\,\partial_{y_{1}}^{I_{1}}\phi_{1}\left(e_{1}+\epsilon
Y_{1}\right)\ldots\partial_{y_{M}}^{I_{M}}\phi_{M}\left(e_{M}+\epsilon
Y_{M}\right)$
$\displaystyle\qquad\times\partial^{\mathfrak{I}_{1}}_{t_{1}}\psi_{1}(0)\ldots\partial^{\mathfrak{I}_{l}}_{t_{l}}\psi_{l}(0)$
$\displaystyle=$ $\displaystyle\lim_{\epsilon\searrow
0}\sum_{{|i|+|I|+|\mathfrak{I}|\leq
K}\atop{|I|\leq|\overline{I}|}}\epsilon^{|\overline{I}|-|I|}\theta_{i,I,\mathfrak{I}}\,\partial^{i_{1}}_{x_{1}}{v}_{1}(0)\ldots\partial^{i_{n}}_{x_{n}}{v}_{n}(0)$
$\displaystyle\qquad\times\epsilon^{|I_{1}|-s_{1}}\partial_{y_{1}}^{I_{1}}\phi_{1}\left(e_{1}+\epsilon
Y_{1}\right)\ldots\epsilon^{|I_{M}|-s_{M}}\partial_{y_{M}}^{I_{M}}\phi_{M}\left(e_{M}+\epsilon
Y_{M}\right)$
$\displaystyle\qquad\times\epsilon^{\mathfrak{I}_{1}-\alpha_{1}}\partial^{\mathfrak{I}_{1}}_{t_{1}}\psi_{1}(0)\ldots\epsilon^{\mathfrak{I}_{l}-\alpha_{l}}\partial^{\mathfrak{I}_{l}}_{t_{l}}\psi_{l}(0)$
$\displaystyle=$ $\displaystyle\sum_{{|i|+|I|+|\mathfrak{I}|\leq
K}\atop{|I|=|\overline{I}|}}\tilde{C}_{i,I,\mathfrak{I}}\,\theta_{i,I,\mathfrak{I}}\,\partial^{i_{1}}_{x_{1}}{v}_{1}(0)\ldots\partial^{i_{n}}_{x_{n}}{v}_{n}(0)$
$\displaystyle\qquad\times e_{1}^{I_{1}}\ldots
e_{M}^{I_{M}}\,\left(-e_{1}\cdot
Y_{1}\right)_{+}^{s_{1}-|I_{1}|}\ldots\left(-e_{M}\cdot
Y_{M}\right)_{+}^{s_{M}-|I_{M}|}\underline{\mbox{\large{\wedn{t}}}}\,_{1}^{\mathfrak{I}_{1}}\ldots\underline{\mbox{\large{\wedn{t}}}}\,_{l}^{\mathfrak{I}_{l}},$

for a suitable $\tilde{C}_{i,I,\mathfrak{I}}\neq 0$ (strictly speaking, the
above identity holds in the sense of distribution with respect to the
coordinates $Y$ and $\underline{\mbox{\large{\wedn{t}}}}\,$, but since the
left hand side vanishes, we can consider it also a pointwise identity).

Hence, recalling (5.52),

$\begin{split}0&\,=\,\sum_{{|i|+|I|+|\mathfrak{I}|\leq
K}\atop{|I|=|\overline{I}|}}C_{i,I,\mathfrak{I}}\,\theta_{i_{1},\dots,i_{n},I_{1},\dots,I_{M},\mathfrak{I}_{1},\dots,\mathfrak{I}_{l}}\;\underline{\mbox{\large{\wedn{x}}}}\,_{1}^{i_{1}}\ldots\underline{\mbox{\large{\wedn{x}}}}\,_{n}^{i_{n}}\\\
&\qquad\qquad\times e_{1}^{I_{1}}\ldots e_{M}^{I_{M}}\,\left(-e_{1}\cdot
Y_{1}\right)_{+}^{s_{1}-|I_{1}|}\ldots\left(-e_{M}\cdot
Y_{M}\right)_{+}^{s_{M}-|I_{M}|}\underline{\mbox{\large{\wedn{t}}}}\,_{1}^{\mathfrak{I}_{1}}\ldots\underline{\mbox{\large{\wedn{t}}}}\,_{l}^{\mathfrak{I}_{l}}\\\
&\,=\,\left(-e_{1}\cdot Y_{1}\right)_{+}^{s_{1}}\ldots\left(-e_{M}\cdot
Y_{M}\right)_{+}^{s_{M}}\\\
&\qquad\qquad\times\sum_{{|i|+|I|+|\mathfrak{I}|\leq
K}\atop{|I|=|\overline{I}|}}C_{i,I,\mathfrak{I}}\,\theta_{i,I,\mathfrak{I}}\;\underline{\mbox{\large{\wedn{x}}}}\,^{i}\,e^{I}\,\left(-e_{1}\cdot
Y_{1}\right)_{+}^{-|I_{1}|}\ldots\left(-e_{M}\cdot
Y_{M}\right)_{+}^{-|I_{M}|}\underline{\mbox{\large{\wedn{t}}}}\,^{\mathfrak{I}},\end{split}$
(5.57)

for a suitable $C_{i,I,\mathfrak{I}}\neq 0$.

We observe that the equality in (5.57) is valid for any choice of the free
parameters
$(\underline{\mbox{\large{\wedn{x}}}}\,,Y,\underline{\mbox{\large{\wedn{t}}}}\,)$
in an open subset of
$\mathbb{R}^{p_{1}+\dots+p_{n}}\times\mathbb{R}^{m_{1}+\dots+m_{M}}\times\mathbb{R}^{l}$,
as prescribed in (5.20), (5.21) and (5.46).

Now, we take new free parameters,
$\underline{\mbox{\large{\wedn{y}}}}\,_{1},\ldots,\underline{\mbox{\large{\wedn{y}}}}\,_{M}$
with
$\underline{\mbox{\large{\wedn{y}}}}\,_{j}\in\mathbb{R}^{m_{j}}\setminus\\{0\\}$,
and we define

$e_{j}:=\frac{\omega_{j}\underline{\mbox{\large{\wedn{y}}}}\,_{j}}{|\underline{\mbox{\large{\wedn{y}}}}\,_{j}|}\quad{\mbox{
and }}\quad
Y_{j}:=-\frac{\underline{\mbox{\large{\wedn{y}}}}\,_{j}}{|\underline{\mbox{\large{\wedn{y}}}}\,_{j}|^{2}}.$
(5.58)

We stress that the setting in (5.58) is compatible with that in (5.46), since

$e_{j}\cdot
Y_{j}=-\frac{\omega_{j}\underline{\mbox{\large{\wedn{y}}}}\,_{j}}{|\underline{\mbox{\large{\wedn{y}}}}\,_{j}|}\cdot\frac{\underline{\mbox{\large{\wedn{y}}}}\,_{j}}{|\underline{\mbox{\large{\wedn{y}}}}\,_{j}|^{2}}=-\frac{\omega_{j}}{|\underline{\mbox{\large{\wedn{y}}}}\,_{j}|}<0,$

thanks to (5.26). We also notice that, for all $j\in\\{1,\dots,M\\}$,

$e_{j}^{I_{j}}\left(-e_{j}\cdot
Y_{j}\right)_{+}^{-|I_{j}|}=\frac{\omega_{j}^{|I_{j}|}\underline{\mbox{\large{\wedn{y}}}}\,_{j}^{I_{j}}}{|\underline{\mbox{\large{\wedn{y}}}}\,_{j}|^{|I_{j}|}}\,\frac{|\underline{\mbox{\large{\wedn{y}}}}\,_{j}|^{|I_{j}|}}{\omega_{j}^{|I_{j}|}}=\underline{\mbox{\large{\wedn{y}}}}\,_{j}^{I_{j}},$

and hence

$e^{I}\,\left(-e_{1}\cdot Y_{1}\right)_{+}^{-|I_{1}|}\ldots\left(-e_{M}\cdot
Y_{M}\right)_{+}^{-|I_{M}|}=\underline{\mbox{\large{\wedn{y}}}}\,^{I}.$

Plugging this into formula (5.57), we obtain the first identity in (5.15), as
desired. Hence, the proof of (5.15) in case 1 is complete. ∎

###### Proof of (5.15), case 2.

Thanks to the assumptions given in case 2, we can suppose that formula (5.17)
still holds, and also that

${\mbox{${\mbox{\large{\wedn{c}}}}\,_{l}>0$}}.$ (5.59)

In addition, for any $j\in\\{1,\ldots,M\\}$, we consider $\lambda_{j}$ and
$\phi_{j}$ as in (5.28).

Then, we define

$R:=\left(\frac{1}{|{\mbox{\large{\wedn{a}}}}_{1}|}\displaystyle\left(\sum_{h=1}^{l-1}|{\mbox{\large{\wedn{c}}}}\,_{h}|+\sum_{j=1}^{M}|{\mbox{\large{\wedn{b}}}}_{j}|\lambda_{j}\right)\right)^{1/|r_{1}|}.$
(5.60)

We notice that, in light of (5.17), the setting in (5.60) is well-defined.

Now, we fix two sets of free parameters
$\underline{\mbox{\large{\wedn{x}}}}\,_{1},\ldots,\underline{\mbox{\large{\wedn{x}}}}\,_{n}$
as in (5.20) and
$\,\underline{\mbox{\large{\wedn{t}}}}\,_{\star,1},\dots,\underline{\mbox{\large{\wedn{t}}}}\,_{\star,l}$
as in (5.21), here taken with $R$ as in (5.60). Moreover, we define

$\lambda\,:=\,\frac{1}{{\mbox{\large{\wedn{c}}}}\,_{l}\,\underline{\mbox{\large{\wedn{t}}}}\,_{\star,l}}\left(\sum_{j=1}^{n}{\left|{\mbox{\large{\wedn{a}}}}_{j}\right|\underline{\mbox{\large{\wedn{x}}}}\,_{j}^{r_{j}}}-\sum_{j=1}^{M}{\mbox{\large{\wedn{b}}}}_{j}\lambda_{j}-\sum_{h=1}^{l-1}{\mbox{\large{\wedn{c}}}}\,_{h}\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}\right).$
(5.61)

We notice that (5.61) is well-defined, thanks to (5.21) and (5.59).
Furthermore, recalling (5.20), (5.24) and (5.60), we find that

$\begin{split}\sum_{i=1}^{n}&|{\mbox{\large{\wedn{a}}}}_{i}|\underline{\mbox{\large{\wedn{x}}}}\,_{i}^{r_{i}}\geq|{\mbox{\large{\wedn{a}}}}_{1}|\underline{\mbox{\large{\wedn{x}}}}\,_{1}^{r_{1}}>|{\mbox{\large{\wedn{a}}}}_{1}|(R+1)^{|r_{1}|}>|{\mbox{\large{\wedn{a}}}}_{1}|R^{|r_{1}|}\\\
&=\sum_{h=1}^{l-1}|{\mbox{\large{\wedn{c}}}}\,_{h}|+\sum_{j=1}^{M}|{\mbox{\large{\wedn{b}}}}_{j}|\lambda_{j}\geq\sum_{h=1}^{l-1}{\mbox{\large{\wedn{c}}}}\,_{h}\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}+\sum_{j=1}^{M}{\mbox{\large{\wedn{b}}}}_{j}\lambda_{j}.\end{split}$

Consequently, by (5.61),

$\lambda>0.$ (5.62)

Hence, we can define

$\overline{\lambda}:=\lambda^{1/{\alpha_{l}}}.$ (5.63)

Moreover, we consider $a_{h}\in(-2,0)$, for every $h\in\\{1,\dots,l\\}$, to be
chosen appropriately in what follows (the exact choice will be performed in
(5.70)), and, using the notation in (5.29) and (5.30), we define

$\psi_{h}(t_{h}):=\psi_{\star,h}\big{(}\underline{\mbox{\large{\wedn{t}}}}\,_{h}(t_{h}-a_{h})\big{)}=E_{\alpha_{h},1}\big{(}\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}(t_{h}-a_{h})^{\alpha_{h}}\big{)}\quad\text{if
}\,h\in\\{1,\dots,l-1\\}$ (5.64)

and

$\psi_{l}(t_{l}):=\psi_{\star,l}\big{(}\overline{\lambda}\,\underline{\mbox{\large{\wedn{t}}}}\,_{l}(t_{l}-a_{l})\big{)}=E_{\alpha_{l},1}\big{(}\lambda\,\underline{\mbox{\large{\wedn{t}}}}\,_{\star,l}(t_{l}-a_{l})^{\alpha_{l}}\big{)}.$
(5.65)

We recall that, thanks to Lemma 3.2, the function in (5.64) solves (5.32) and
satisfies (5.35) for any $h\in\\{1,\dots,l-1\\}$, while the function in (5.65)
solves

$\begin{cases}D^{\alpha_{l}}_{t_{l},a_{l}}\psi_{l}(t_{l})=\lambda\,\underline{\mbox{\large{\wedn{t}}}}\,_{\star,l}\psi_{l}(t_{l})&\quad\text{in
}\,(a_{l},+\infty),\\\ \psi_{l}(a_{l})=1,\\\
\partial^{m}_{t_{l}}\psi_{l}(a_{l})=0&\quad\text{for every
}\,m\in\\{1,\dots,[\alpha_{l}]\\}.\end{cases}$ (5.66)

As in (5.33), we extend the functions $\psi_{h}$ constantly in
$(-\infty,a_{h})$, calling $\psi^{\star}_{h}$ this extended function. In this
way, Lemma A.3 in [CDV18] translates (5.66) into

$D^{\alpha_{h}}_{t_{h},-\infty}\psi_{h}^{\star}(t_{h})=\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}\psi_{h}(t_{h})=\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}\psi_{h}^{\star}(t_{h})\,\text{
in every interval }\,I\Subset(a_{h},+\infty).$ (5.67)

Now, we let $\epsilon>0$, to be taken small possibly depending on the free
parameters, and we exploit the functions defined in (5.42) and (5.43),
provided that one replaces the positive constant $R$ defined in (5.19) with
the one in (5.60), when necessary.

With this idea in mind, for any $j\in\\{1,\ldots,M\\}$, we let222Comparing
(5.68) with (5.45), we observe that (5.45) reduces to (5.68) with the choice
$\omega_{j}:=1$.

$e_{j}\in\partial B^{m_{j}}_{1},$ (5.68)

and we define

$\begin{split}w\left(x,y,t\right):=&\tau_{1}\left(x\right)v_{1}\left(x_{1}\right)\cdot\ldots\cdot
v_{n}\left(x_{n}\right)\phi_{1}\left(y_{1}+e_{1}+\epsilon
Y_{1}\right)\cdot\ldots\cdot\phi_{M}\left(y_{M}+e_{M}+\epsilon Y_{M}\right)\\\
&\times\psi^{\star}_{1}(t_{1})\cdot\ldots\cdot\psi^{\star}_{l}(t_{l}),\end{split}$
(5.69)

where the setting in (5.28), (5.42), (5.43), (5.46), (5.64) and (5.65) has
been exploited.

We also notice that $w\in C\left(\mathbb{R}^{N}\right)\cap
C_{0}(\mathbb{R}^{N-l})\cap\mathcal{A}$. Moreover, if

$a=(a_{1},\dots,a_{l}):=\left(-\frac{\epsilon}{\underline{\mbox{\large{\wedn{t}}}}\,_{1}},\dots,-\frac{\epsilon}{\,\underline{\mbox{\large{\wedn{t}}}}\,_{l}}\right)\in(-\infty,0)^{l}$
(5.70)

and $(x,y)$ is sufficiently close to the origin and
$t\in(a_{1},+\infty)\times\dots\times(a_{l},+\infty)$, we have that

$\displaystyle\Lambda_{-\infty}w\left(x,y,t\right)$ $\displaystyle=$
$\displaystyle\left(\sum_{i=1}^{n}{\mbox{\large{\wedn{a}}}}_{i}\partial^{r_{i}}_{x_{i}}+\sum_{j=1}^{M}{\mbox{\large{\wedn{b}}}}_{j}(-\Delta)^{s_{j}}_{y_{j}}+\sum_{h=1}^{l}{\mbox{\large{\wedn{c}}}}\,_{h}D^{\alpha_{h}}_{t_{h},-\infty}\right)w\left(x,y,t\right)$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{n}{\mbox{\large{\wedn{a}}}}_{i}v_{1}\left(x_{1}\right)\ldots
v_{i-1}\left(x_{i-1}\right)\partial^{r_{i}}_{x_{i}}v_{i}\left(x_{i}\right)v_{i+1}\left(x_{i+1}\right)\ldots
v_{n}\left(x_{n}\right)$
$\displaystyle\qquad\times\phi_{1}\left(y_{1}+e_{1}+\epsilon
Y_{1}\right)\ldots\phi_{M}\left(y_{M}+e_{M}+\epsilon
Y_{M}\right)\psi^{\star}_{1}\left(t_{1}\right)\ldots\psi^{\star}_{l-1}\left(t_{l-1}\right)\psi^{\star}_{l}\left(t_{l}\right)$
$\displaystyle+\sum_{j=1}^{M}{\mbox{\large{\wedn{b}}}}_{j}v_{1}\left(x_{1}\right)\ldots
v_{n}\left(x_{n}\right)\phi_{1}\left(y_{1}+e_{1}+\epsilon
Y_{1}\right)\ldots\phi_{j-1}\left(y_{j-1}+e_{j-1}+\epsilon Y_{j-1}\right)$
$\displaystyle\qquad\times(-\Delta)^{s_{j}}_{y_{j}}\phi_{j}\left(y_{j}+e_{j}+\epsilon
Y_{j}\right)\phi_{j+1}\left(y_{j+1}+e_{j+1}+\epsilon
Y_{j+1}\right)\ldots\phi_{M}\left(y_{M}+e_{M}+\epsilon Y_{M}\right)$
$\displaystyle\qquad\times\psi^{\star}_{1}\left(t_{1}\right)\ldots\psi^{\star}_{l-1}\left(t_{l-1}\right)\psi^{\star}_{l}\left(t_{l}\right)$
$\displaystyle+\sum_{h=1}^{l}{\mbox{\large{\wedn{c}}}}\,_{h}v_{1}\left(x_{1}\right)\ldots
v_{n}\left(x_{n}\right)\phi_{1}\left(y_{1}+e_{1}+\epsilon
Y_{1}\right)\ldots\phi_{M}\left(y_{M}+e_{M}+\epsilon
Y_{M}\right)\psi^{\star}_{1}\left(t_{1}\right)\ldots\psi^{\star}_{h-1}\left(t_{h-1}\right)$
$\displaystyle\qquad\times
D^{\alpha_{h}}_{t_{h},-\infty}\psi^{\star}_{h}\left(t_{h}\right)\psi^{\star}_{h+1}(t_{h+1})\ldots\psi^{\star}_{l-1}\left(t_{l-1}\right)\psi^{\star}_{l}\left(t_{l}\right)$
$\displaystyle=$
$\displaystyle-\sum_{i=1}^{n}{\mbox{\large{\wedn{a}}}}_{i}\overline{\mbox{\large{\wedn{a}}}}_{i}\underline{\mbox{\large{\wedn{x}}}}\,_{i}^{r_{i}}v_{1}\left(x_{1}\right)\ldots
v_{n}\left(x_{n}\right)\phi_{1}\left(y_{1}+e_{1}+\epsilon
Y_{1}\right)\ldots\phi_{M}\left(y_{M}+e_{M}+\epsilon Y_{M}\right)$
$\displaystyle\qquad\times\psi^{\star}_{1}(t_{1})\ldots\psi^{\star}_{l-1}(t_{l-1})\psi^{\star}_{l}(t_{l})$
$\displaystyle+\sum_{j=1}^{M}{\mbox{\large{\wedn{b}}}}_{j}\lambda_{j}v_{1}\left(x_{1}\right)\ldots
v_{n}\left(x_{n}\right)\phi_{1}\left(y_{1}+e_{1}+\epsilon
Y_{1}\right)\ldots\phi_{M}\left(y_{M}+e_{M}+\epsilon Y_{M}\right)$
$\displaystyle\qquad\times\psi^{\star}_{1}(t_{1})\ldots\psi^{\star}_{l-1}\left(t_{l-1}\right)\psi^{\star}_{l}(t_{l})$
$\displaystyle+\sum_{h=1}^{l-1}{\mbox{\large{\wedn{c}}}}\,_{h}\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}v_{1}\left(x_{1}\right)\ldots
v_{n}\left(x_{n}\right)\phi_{1}\left(y_{1}+e_{1}+\epsilon
Y_{1}\right)\ldots\phi_{M}\left(y_{M}+e_{M}+\epsilon Y_{M}\right)$
$\displaystyle\qquad\times\psi^{\star}_{1}(t_{1})\ldots\psi^{\star}_{l-1}(t_{l-1})\psi^{\star}_{l}(t_{l})$
$\displaystyle+{\mbox{\large{\wedn{c}}}}\,_{l}\lambda\underline{\mbox{\large{\wedn{t}}}}\,_{\star,l}v_{1}\left(x_{1}\right)\ldots
v_{n}\left(x_{n}\right)\phi_{1}\left(y_{1}+e_{1}+\epsilon
Y_{1}\right)\ldots\phi_{M}\left(y_{M}+e_{M}+\epsilon Y_{M}\right)$
$\displaystyle\qquad\times\psi^{\star}_{1}(t_{1})\ldots\psi^{\star}_{l-1}(t_{l-1})\psi^{\star}_{l}(t_{l})$
$\displaystyle=$
$\displaystyle\left(-\sum_{i=1}^{n}{\mbox{\large{\wedn{a}}}}_{i}\overline{\mbox{\large{\wedn{a}}}}_{i}\underline{\mbox{\large{\wedn{x}}}}\,_{i}^{r_{i}}+\sum_{j=1}^{M}{\mbox{\large{\wedn{b}}}}_{j}\lambda_{j}+\sum_{h=1}^{l-1}{\mbox{\large{\wedn{c}}}}\,_{h}\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}+{\mbox{\large{\wedn{c}}}}\,_{l}\lambda\underline{\mbox{\large{\wedn{t}}}}\,_{\star,l}\right)w(x,y,t),$

thanks to (5.28), (5.32), (5.44) and (5.67).

Consequently, making use of (5.37) and (5.61), when $(x,y)$ is near the origin
and $t\in(a_{1},+\infty)\times\dots\times(a_{l},+\infty)$, we have that

$\Lambda_{-\infty}w\left(x,y,t\right)=\left(-\sum_{i=1}^{n}|{\mbox{\large{\wedn{a}}}}_{i}|\underline{\mbox{\large{\wedn{x}}}}\,_{i}^{r_{i}}+\sum_{j=1}^{M}{\mbox{\large{\wedn{b}}}}_{j}\lambda_{j}+\sum_{h=1}^{l-1}{\mbox{\large{\wedn{c}}}}\,_{h}\underline{\mbox{\large{\wedn{t}}}}\,_{\star,h}+\lambda{\mbox{\large{\wedn{c}}}}\,_{l}\underline{\mbox{\large{\wedn{t}}}}\,_{\star,l}\right)w(x,y,t)=0.$

This says that $w\in\mathcal{H}$. Thus, in light of (5.12) we have that

$0=\theta\cdot\partial^{K}w\left(0\right)=\sum_{\left|\iota\right|\leq
K}{\theta_{\iota}\partial^{\iota}w\left(0\right)}=\sum_{|i|+|I|+|\mathfrak{I}|\leq
K}\theta_{i,I,\mathfrak{I}}\,\partial_{x}^{i}\partial_{y}^{I}\partial_{t}^{\mathfrak{I}}w\left(0\right).$

Hence, in view of (5.52) and (5.69),

$\begin{split}0\,&=\,\sum_{|i|+|I|+|\mathfrak{I}|\leq
K}\theta_{i,I,\mathfrak{I}}\,\partial^{i_{1}}_{x_{1}}{v}_{1}(0)\ldots\partial^{i_{n}}_{x_{n}}{v}_{n}(0)\\\
&\qquad\times\partial^{I_{1}}_{y_{1}}{\phi}_{1}(e_{1}+\epsilon
Y_{1})\ldots\partial^{I_{M}}_{y_{M}}{\phi}_{M}(e_{M}+\epsilon
Y_{M})\,\partial_{t_{1}}^{\mathfrak{I}_{1}}\psi_{1}(0)\ldots\partial_{t_{l}}^{\mathfrak{I}_{l}}\psi_{l}(0)\\\
&=\,\sum_{|i|+|I|+|\mathfrak{I}|\leq
K}\theta_{i,I,\mathfrak{I}}\,\underline{\mbox{\large{\wedn{x}}}}\,_{1}^{r_{1}}\ldots\underline{\mbox{\large{\wedn{x}}}}\,_{n}^{r_{n}}\,\partial^{i_{1}}_{x_{1}}\overline{v}_{1}(0)\ldots\partial^{i_{n}}_{x_{n}}\overline{v}_{n}(0)\\\
&\qquad\times\partial^{I_{1}}_{y_{1}}{\phi}_{1}(e_{1}+\epsilon
Y_{1})\ldots\partial^{I_{M}}_{y_{M}}{\phi}_{M}(e_{M}+\epsilon
Y_{M})\,\partial_{t_{1}}^{\mathfrak{I}_{1}}\psi_{1}(0)\ldots\partial_{t_{l}}^{\mathfrak{I}_{l}}\psi_{l}(0).\end{split}$
(5.71)

Moreover, using (3.1), (5.65) and (5.70), it follows that

$\displaystyle\partial^{\mathfrak{I}_{l}}_{t_{l}}\psi_{l}(0)$ $\displaystyle=$
$\displaystyle\sum_{j=0}^{+\infty}{\frac{\lambda^{j}\,\underline{\mbox{\large{\wedn{t}}}}\,_{\star,l}^{j}\,\alpha_{l}j(\alpha_{l}j-1)\dots(\alpha_{l}j-\mathfrak{I}_{l}+1)(0-a_{l})^{\alpha_{l}j-\mathfrak{I}_{l}}}{\Gamma\left(\alpha_{l}j+1\right)}}$
$\displaystyle=$
$\displaystyle\sum_{j=0}^{+\infty}{\frac{\lambda^{j}\,\underline{\mbox{\large{\wedn{t}}}}\,_{\star,l}^{j}\,\alpha_{l}j(\alpha_{l}j-1)\dots(\alpha_{l}j-\mathfrak{I}_{l}+1)\,{\epsilon}^{\alpha_{l}j-\mathfrak{I}_{l}}}{\Gamma\left(\alpha_{l}j+1\right)\;{\underline{\mbox{\large{\wedn{t}}}}\,_{l}}^{\alpha_{l}j-\mathfrak{I}_{l}}}}$
$\displaystyle=$
$\displaystyle\sum_{j=1}^{+\infty}{\frac{\lambda^{j}\,\underline{\mbox{\large{\wedn{t}}}}\,_{\star,l}^{j}\,\alpha_{l}j(\alpha_{l}j-1)\dots(\alpha_{l}j-\mathfrak{I}_{l}+1)\,{\epsilon}^{\alpha_{l}j-\mathfrak{I}_{l}}}{\Gamma\left(\alpha_{l}j+1\right)\;{\underline{\mbox{\large{\wedn{t}}}}\,_{l}}^{\alpha_{l}j-\mathfrak{I}_{l}}}}.$

Accordingly, by (5.30), we find that

$\begin{split}&\lim_{\epsilon\searrow
0}\epsilon^{\mathfrak{I}_{l}-\alpha_{l}}\partial^{\mathfrak{I}_{l}}_{t_{l}}\psi_{l}(0)=\lim_{\epsilon\searrow
0}\sum_{j=1}^{+\infty}{\frac{\lambda^{j}\,\underline{\mbox{\large{\wedn{t}}}}\,_{\star,l}^{j}\,\alpha_{l}j(\alpha_{l}j-1)\dots(\alpha_{l}j-\mathfrak{I}_{l}+1)\,{\epsilon}^{\alpha_{l}(j-1)}}{\Gamma\left(\alpha_{l}j+1\right)\;{\underline{\mbox{\large{\wedn{t}}}}\,_{l}}^{\alpha_{l}j-\mathfrak{I}_{l}}}}\\\
&\qquad={\frac{\lambda\,\underline{\mbox{\large{\wedn{t}}}}\,_{\star,l}\,\alpha_{l}(\alpha_{l}-1)\dots(\alpha_{l}-\mathfrak{I}_{l}+1)}{\Gamma\left(\alpha_{l}+1\right)\;\underline{\mbox{\large{\wedn{t}}}}\,_{l}^{\alpha_{l}-\mathfrak{I}_{l}}}}={\frac{\lambda\,\underline{\mbox{\large{\wedn{t}}}}\,_{l}^{\mathfrak{I}_{l}}\,\alpha_{l}(\alpha_{l}-1)\dots(\alpha_{l}-\mathfrak{I}_{l}+1)}{\Gamma\left(\alpha_{l}+1\right)}}.\end{split}$
(5.72)

Hence, recalling (5.14), we can write (5.71) as

$\begin{split}&0=\sum_{{|i|+|I|+|\mathfrak{I}|\leq
K}\atop{|\mathfrak{I}|\leq|\overline{\mathfrak{I}}|}}\theta_{i,I,\mathfrak{I}}\,\underline{\mbox{\large{\wedn{x}}}}\,_{1}^{r_{1}}\ldots\underline{\mbox{\large{\wedn{x}}}}\,_{n}^{r_{n}}\,\partial^{i_{1}}_{x_{1}}\overline{v}_{1}(0)\ldots\partial^{i_{n}}_{x_{n}}\overline{v}_{n}(0)\\\
&\qquad\qquad\times\partial^{I_{1}}_{y_{1}}{\phi}_{1}(e_{1}+\epsilon
Y_{1})\ldots\partial^{I_{M}}_{y_{M}}{\phi}_{M}(e_{M}+\epsilon
Y_{M})\partial^{\mathfrak{I}_{1}}_{t_{1}}\psi_{1}(0)\ldots\partial^{\mathfrak{I}_{l}}_{t_{l}}\psi_{l}(0).\end{split}$
(5.73)

Moreover, we define

$\Xi:=|\overline{\mathfrak{I}}|-\sum_{h=1}^{l}{\alpha_{h}}+\left|I\right|-\sum_{j=1}^{M}{s_{j}}.$

Then, we multiply (5.73) by $\epsilon^{\Xi}\in(0,+\infty)$, and we send
$\epsilon$ to zero. In this way, we obtain from (5.55), used here for
$h\in\\{1,\ldots,l-1\\}$, (5.72) and (5.73) that

$\displaystyle 0$ $\displaystyle=$ $\displaystyle\lim_{\epsilon\searrow
0}\epsilon^{\Xi}\sum_{{|i|+|I|+|\mathfrak{I}|\leq