Appendix A: Boundary Behavior of Solutions of Fractional Laplace Equations and Proof of Lemma 4
In this section, we detect the exact boundary behavior of solutions of fractional Laplace equations in a ball with Dirichlet data, in order to prove Lemma 4. For estimates in general domains, see e.g., [6] and the references therein. Let us remark that, in our context, we do not only obtain bounds from above and below, but also a precise asymptotics in the limits which approach the boundary.
In order to obtain our bounds, we make use of the fractional Green function, whose setting goes as follows: Given \(s\in (0,1)\) and x, \(z\in B_1\), we consider the function
$$\begin{aligned} G(x,z):=|z-x|^{2s-n}\int _0^{r_0(x,z)} \frac{t^{s-1}\,\mathrm{d}t}{ (t+1)^{\frac{n}{2}}}, \end{aligned}$$
(46)
withFootnote 3
$$\begin{aligned} r_0(x,z):=\frac{(1-|x|^2)\,(1-|z|^2)}{|z-x|^2}. \end{aligned}$$
(47)
Up to normalization factors, the function G plays the role of a Green function in the fractional setting, as discussed for instance in [1] and in the references therein.
If x lies in an \(\varepsilon \)-neighborhood of \(\partial B_1\), then G is of order \(\varepsilon ^s\), as stated precisely in the next result:
Lemma 6
Let \(e\in \partial B_1\), \(\varepsilon _o>0\) and \(\omega \in \partial B_1\). Assume that \( e+\varepsilon \omega \in B_1\) for all \(\varepsilon \in (0,\varepsilon _o]\). Let \(f\in C^\alpha ({\mathbb {R}}^{d})\) for some \(\alpha \in (0,1)\), with \(f=0\) outside \(B_1\).
Then
$$\begin{aligned} \lim _{\varepsilon \searrow 0}\varepsilon ^{-s}\int _{B_1} f(z)\,G(e+\varepsilon \omega ,z)\,\mathrm{d}z = \int _{B_1} f(z)\, \frac{(-2\,e\cdot \omega )^s\,(1-|z|^2)^s }{ s\,|z-e|^n } \,\mathrm{d}z.\end{aligned}$$
(48)
The rather technical proof of Lemma 6 is postponed to Appendix B, for the facility of the reader. Here, we deduce from Lemma 6 the boundary estimates needed to the proof of our main result:
Proposition 7
Let \(e\in \partial B_1\), \(\varepsilon _o>0\) and \(\omega \in \partial B_1\). Assume that \( e+\varepsilon \omega \in B_1\) for all \(\varepsilon \in (0,\varepsilon _o]\). Let \(f\in C^\alpha ({\mathbb {R}}^{n})\) for some \(\alpha \in (0,1)\), with \(f=0\) outside \(B_1\).
Let u be a weak solution of
$$\begin{aligned} \left\{ \begin{array}{ll} (-\Delta )^s u =f &{}\quad {\text{ in } } B_1,\\ u=0 &{}\quad {\text{ in } }{\mathbb {R}}^n\setminus B_1. \end{array}\right. \end{aligned}$$
Then
$$\begin{aligned} \lim _{\varepsilon \searrow 0}\varepsilon ^{-s}\,u(e+\varepsilon \omega ) = \kappa (n,s)\,(-2\,e\cdot \omega )^s\,\int _{B_1} f(z)\, \frac{(1-|z|^2)^s }{ s\,|z-e|^n } \,\mathrm{d}z,\ \end{aligned}$$
where
$$\begin{aligned} \kappa (n,s):= \frac{\Gamma \left( \frac{n}{2}\right) }{4^s\,\pi ^{\frac{n}{2}}\,\Gamma ^2(s)}, \end{aligned}$$
being \(\Gamma \) the Euler’s \(\Gamma \)-function.
Proof
We know from Theorems 1 and 2 in [8] that u is actually continuous in \({\mathbb {R}}^n\) and it is a viscosity solution of the equation. Also, by the fractional Green Representation Theorem (see e.g., Theorem 3.2 in [1] and the references therein), we have that
$$\begin{aligned} u(e+\varepsilon \omega )= \kappa (n,s)\, \int _{B_1} f(z)\,G(e+\varepsilon \omega ,z)\,\mathrm{d}z, \end{aligned}$$
with G as in (46). Hence, the desired result follows from (48). \(\square \)
As a simple consequence, we can characterize the boundary behavior of the first eigenfunction for the fractional Laplacian with Dirichlet data (see e.g., Appendix A in [7] for a discussion on fractional eigenvalues).
Corollary 8
Let \(e\in \partial B_1\). Let \(\phi _\star \) be the first eigenfunction for \((-\Delta )^s\), normalized to be positive and such that \(\Vert \phi _\star \Vert _{L^2(B_1)}=1\), and let \(\lambda _\star >0\) be the corresponding eigenvalue. Then,
$$\begin{aligned} \Vert \phi _\star \Vert _{C^s({\mathbb {R}}^n)}\leqslant C, \end{aligned}$$
(49)
for some \(C>0\) depending only on n and s, and
$$\begin{aligned} \lim _{\varepsilon \searrow 0}\varepsilon ^{-s}\,\phi _\star (e+\varepsilon \omega ) = \kappa _\star \,\lambda _\star \,(-\,e\cdot \omega )^s_+, \end{aligned}$$
(50)
where \(\kappa _\star \) is as in (12).
Proof
The idea is that, since \((-\Delta )^s \phi _\star =\lambda _\star \,\phi _\star \), we can use Proposition 7 and get the desired result. More precisely, we have that \(\phi _\star \) is \(C^s(B_1)\) (see the proof of Corollary 8 in [8] to obtain the continuity and then Proposition 1.1 in [6] to get the Hölder estimate in (49)).
Notice that, by (49), we have that the quantity \(\kappa _\star \) defined in (12) is finite, while the positivity of \(\phi _\star \) implies that \(\kappa _\star >0\).
Also, the Hölder estimate in (49) allows to use Proposition 7 with \(f:=\lambda _\star \,\phi _\star \). Accordingly, for any \(\omega \in \partial B_1\) for which there exists \(\varepsilon _o>0\) such that \( e+\varepsilon \omega \in B_1\) for all \(\varepsilon \in (0,\varepsilon _o]\), we have that
$$\begin{aligned} \lim _{\varepsilon \searrow 0}\varepsilon ^{-s}\,\phi _\star (e+\varepsilon \omega ) = \kappa _\star \,\lambda _\star \,(-\,e\cdot \omega )^s. \end{aligned}$$
(51)
Now, we distinguish two cases: if \(e\cdot \omega <0\), then
$$\begin{aligned} |e+\varepsilon \omega |^2 =1+2\varepsilon e\cdot \omega +\varepsilon ^2<1 \end{aligned}$$
for small \(\varepsilon \), and so \(e+\varepsilon \omega \in B_1\) for small \(\varepsilon \), and hence (50) follows from (51).
If instead \(e\cdot \omega \geqslant 0\), then \(e+\varepsilon \omega \in {\mathbb {R}}^n\setminus B_1\), thus \(\phi _\star (e+\varepsilon \omega )=0\), which obviously implies (50) in this case. \(\square \)
From Corollary 8, we can now complete the proof of Lemma 4, by arguing as follows:
Proof of Lemma 4
Let \(\psi \in C^\infty _0({\mathbb {R}}^n)\). We write \(X=\rho \omega \), with \(\rho \geqslant 0\) and \(\omega \in S^{n-1}\). Notice that \((\varepsilon \rho )^{-s}\,|\phi _\star (e+\varepsilon \rho \omega )|\leqslant C\), thanks to (49).
So, we use (51) and the Dominated Convergence Theorem to see that
$$\begin{aligned}&\lim _{\varepsilon \searrow 0} \varepsilon ^{|\alpha |-s} \int _{{\mathbb {R}}^n} \partial ^\alpha \phi _\star (e+\varepsilon X)\, \psi (X)\,\mathrm{d}X\\&\qquad = \lim _{\varepsilon \searrow 0} \int _{{\mathbb {R}}^n} \partial ^\alpha _X \big ( \varepsilon ^{-s}\phi _\star (e+\varepsilon X)\big )\, \psi (X)\,\mathrm{d}X\\&\qquad = (-1)^{|\alpha |} \lim _{\varepsilon \searrow 0} \int _{{\mathbb {R}}^n} \varepsilon ^{-s}\phi _\star (e+\varepsilon X)\, \partial ^\alpha \psi (X)\,\mathrm{d}X\\&\qquad = (-1)^{|\alpha |} \lim _{\varepsilon \searrow 0} \int _{0}^{+\infty }\,\mathrm{d}\rho \,\int _{S^{n-1}}\,\mathrm{d}\omega \, \rho ^{n-1} \rho ^s \,(\varepsilon \rho )^{-s}\phi _\star (e+\varepsilon \rho \omega )\, \partial ^\alpha \psi (\rho \omega )\\&\qquad = (-1)^{|\alpha |} \,\kappa _\star \,\lambda _\star \, \int _{0}^{+\infty }\,\mathrm{d}\rho \,\int _{S^{n-1}}\,\mathrm{d}\omega \, \rho ^{n-1} \rho ^s \,(-e\cdot \omega )_+^s\, \partial ^\alpha \psi (\rho \omega )\\&\qquad = (-1)^{|\alpha |} \,\kappa _\star \,\lambda _\star \, \int _{0}^{+\infty }\,\mathrm{d}\rho \,\int _{S^{n-1}}\,\mathrm{d}\omega \, \rho ^{n-1} (-e\cdot \rho \omega )^s_+\, \partial ^\alpha \psi (\rho \omega )\\&\qquad = (-1)^{|\alpha |} \,\kappa _\star \,\lambda _\star \, \int _{{\mathbb {R}}^n} (-e\cdot X)^s_+\, \partial ^\alpha \psi (X)\,\mathrm{d}X\\&\qquad = \kappa _\star \,\lambda _\star \, \int _{{\mathbb {R}}^n} \partial ^\alpha _X (-e\cdot X)^s_+\, \psi (X)\,\mathrm{d}X\\&\qquad = (-1)^{|\alpha |}\,\kappa _\star \,\lambda _\star \,s\,(s{-}1)\dots (s{-}|\alpha |{+}1)\, e_1^{\alpha _1}\dots e_n^{\alpha _n}\,\int _{{\mathbb {R}}^n} (-e\cdot X)_+^{s-|\alpha |}\, \, \psi (X)\,\mathrm{d}X, \end{aligned}$$
and this gives the desired result, since \(\psi \) is an arbitrary test function. \(\square \)
Appendix B—Green Function Computations
Now, we present the proof of Lemma 6. We recall that such result gives some precise asymptotics on the boundary behavior of the Green function of the fractional Laplacian, which in turn have been exploited in Appendix A to obtain precise boundary information on the solutions of fractional Laplace equations.
Proof of Lemma 6
We remark that the condition \(e+\varepsilon \omega \in B_1\) for all \(\varepsilon \in (0,\varepsilon _o]\) says that
$$\begin{aligned} 1> |e+\varepsilon \omega |^2 = 1 +\varepsilon ^2 +2\varepsilon e\cdot \omega \end{aligned}$$
and in particular
$$\begin{aligned} -e\cdot \omega> \frac{\varepsilon }{2}>0. \end{aligned}$$
(52)
From (47), we have that
$$\begin{aligned} r_0(e+\varepsilon \omega ,z)= \frac{\varepsilon \,(-\varepsilon -2\,e\cdot \omega )\,(1-|z|^2)}{|z-e-\varepsilon \omega |^2}. \end{aligned}$$
(53)
In particular,
$$\begin{aligned} r_0(e+\varepsilon \omega ,z)\leqslant \frac{3\varepsilon }{|z-e+\varepsilon \omega |^2}. \end{aligned}$$
(54)
Moreover, using a Taylor binomial series,
$$\begin{aligned} (t+1)^{-\frac{n}{2}} =\sum _{k=0}^{+\infty } \left( \begin{array}{c} {-n/2}\\ {k} \end{array}\right) \,t^k \end{aligned}$$
and therefore
$$\begin{aligned} \frac{t^{s-1}}{ (t+1)^{\frac{n}{2}}}=\sum _{k=0}^{+\infty } \left( \begin{array}{c} {-n/2}\\ {k} \end{array}\right) \,t^{k+s-1}. \end{aligned}$$
(55)
Since the bounds on the binomial coefficients yield that
$$\begin{aligned} \left| \left( \begin{array}{c} {-n/2}\\ {k} \end{array}\right) \right| \leqslant C\,k^{\frac{n}{2}}, \end{aligned}$$
(56)
it follows from the root test that the series in (55) is uniformly convergent for any t in a compact subset of \((-1,1)\). In particular, if we set
$$\begin{aligned} r_1(x,z):= \min \left\{ r_0(x,z),\,\frac{1}{2}\right\} ,\end{aligned}$$
(57)
we can exchange the integration and summation signs and find that
$$\begin{aligned} \int _0^{r_1(x,z)} \frac{t^{s-1}\,\mathrm{d}t}{ (t+1)^{\frac{n}{2}}} = \sum _{k=0}^{+\infty } c_k\,\big ( r_1(x,z)\big )^{k+s}, \end{aligned}$$
with
$$\begin{aligned} c_k:= \frac{1}{k+s}\left( \begin{array}{c} {-n/2}\\ {k} \end{array}\right) . \end{aligned}$$
Therefore, we have
$$\begin{aligned} G(x,z)={{\mathcal {G}}}(x,z)+g(x,z), \end{aligned}$$
(58)
with
$$\begin{aligned}&{{\mathcal {G}}}(x,z):= |z-x|^{2s-n}\,\sum _{k=0}^{+\infty } c_k\,\big ( r_1(x,z)\big )^{k+s} \\ {\text{ and } }&g(x,z):= |z-x|^{2s-n}\int _{r_1(x,z)}^{r_0(x,z)} \frac{t^{s-1}\,\mathrm{d}t}{ (t+1)^{\frac{n}{2}}}. \end{aligned}$$
Notice that \(g(x,z)=0\) if \(r_0(x,z)\leqslant 1/2\). Also, if \(r_0(x,z)> 1/2\),
$$\begin{aligned} 0\leqslant g(x,z)\leqslant & {} |z-x|^{2s-n}\int _{1/2}^{r_0(x,z)} \frac{t^{s-1}\,\mathrm{d}t}{ t^{\frac{n}{2}}} \\\leqslant & {} \left\{ \begin{matrix} C\, |z-x|^{2s-n} &{} { \text{ if } }\quad n>2s,\\ C\,\log r_0(x,z) &{} { \text{ if } }\quad n=2s, \\ C \,|z-x|^{2s-n} \,\big ( r_0(x,z)\big )^{s-\frac{n}{2}} &{} { \text{ if } }\quad n<2s, \end{matrix}\right. \end{aligned}$$
for some \(C>0\). Now we compute this expression in \(x:=e+\varepsilon \omega \). Notice that the condition \(r_0(e+\varepsilon \omega ,z)=r_0(x,z)> 1/2\), combined with (54), says that
$$\begin{aligned} |z-e-\varepsilon \omega |^2 \leqslant 9\varepsilon . \end{aligned}$$
(59)
As a consequence
$$\begin{aligned} \begin{aligned}&\left| \int _{B_1} f(z)\,g(e+\varepsilon \omega ,z)\,\mathrm{d}z\right| \\&\quad \leqslant \int _{ B_{3\sqrt{\varepsilon }}(e+\varepsilon \omega )} |f(z)|\,|g(e+\varepsilon \omega ,z)|\,\mathrm{d}z\\&\quad \leqslant \left\{ \begin{matrix} C\, \int _{ B_{3\sqrt{\varepsilon }}(e+\varepsilon \omega )} |f(z)|\,|z-e+\varepsilon \omega |^{2s-n} \,\mathrm{d}z &{} { \text{ if } }\quad n>2s,\\ C\,\int _{ B_{3\sqrt{\varepsilon }}(e+\varepsilon \omega )} |f(z)|\,\log r_0(e+\varepsilon \omega ,z) \,\mathrm{d}z&{} { \text{ if } }\quad n=2s, \\ C \,\int _{ B_{3\sqrt{\varepsilon }}(e+\varepsilon \omega )} |f(z)|\,|z-e+\varepsilon \omega |^{2s-n} \,\big ( r_0(e+\varepsilon \omega ,z)\big )^{s-\frac{n}{2}} \,\mathrm{d}z&{} { \text{ if } }\quad n<2s. \end{matrix}\right. \qquad \end{aligned} \end{aligned}$$
(60)
Now, if \(z\in B_{3\sqrt{\varepsilon }}(e+\varepsilon \omega )\), then \(|z-e|\leqslant 4\sqrt{\varepsilon }\) and so
$$\begin{aligned} |f(z)|\leqslant C\varepsilon ^{\frac{\alpha }{2}}, \end{aligned}$$
(61)
with \(C>0\) depending on f. Hence recalling (54), after renaming \(C>0\), we deduce from (60) that
$$\begin{aligned} \left| \int _{B_1} f(z)\,g(e+\varepsilon \omega ,z)\,\mathrm{d}z\right|\leqslant & {} \left\{ \begin{matrix} C\varepsilon ^{\frac{\alpha }{2}}\, \int _{ B_{3\sqrt{\varepsilon }}(e+\varepsilon \omega )} |z-e+\varepsilon \omega |^{2s-n} \,\mathrm{d}z &{} { \text{ if } }\quad n>2s,\\ C\varepsilon ^{\frac{\alpha }{2}}\,\int _{ B_{3\sqrt{\varepsilon }}(e+\varepsilon \omega )} \log \frac{3\varepsilon }{|z-e+\varepsilon \omega |^2} \,\mathrm{d}z&{} { \text{ if } }\quad n=2s, \\ C \varepsilon ^{\frac{\alpha }{2}+s-\frac{n}{2}}\,\int _{ B_{3\sqrt{\varepsilon }}(e+\varepsilon \omega )} 1\,\mathrm{d}z&{} { \text{ if } }\quad n<2s, \end{matrix}\right. \\\leqslant & {} C\varepsilon ^{\frac{\alpha }{2}+s}. \end{aligned}$$
This and (58) give that
$$\begin{aligned} \int _{B_1} f(z)\,G(e+\varepsilon \omega ,z)\,\mathrm{d}z= \int _{B_1} f(z)\,{{\mathcal {G}}}(e+\varepsilon \omega ,z)\,\mathrm{d}z+o(\varepsilon ^s). \end{aligned}$$
(62)
Now we consider the series defining \({{\mathcal {G}}}\) and we split the contribution coming from the index \(k=0\) from the ones coming from the indices \(k\geqslant 1\), namely we write
$$\begin{aligned} \begin{aligned}&{{\mathcal {G}}}(x,z)= {{\mathcal {G}}}_0(x,z)+{{\mathcal {G}}}_1(x,z)\\ {\text{ with } }\quad&{{\mathcal {G}}}_0(x,z):= \frac{|z-x|^{2s-n}}{s}\,\big ( r_1(x,z)\big )^{s} \\ {\text{ and } }\quad&{{\mathcal {G}}}_1(x,z):= |z-x|^{2s-n}\,\sum _{k=1}^{+\infty } c_k \,\big ( r_1(x,z)\big )^{k+s}. \end{aligned}\end{aligned}$$
(63)
So, we use (57) and (61) and obtain that
$$\begin{aligned} \begin{aligned}&\left| \int _{B_1\cap B_{3\sqrt{\varepsilon }}(e+\varepsilon \omega ) } f(z)\,{{\mathcal {G}}}_1(e+\varepsilon \omega ,z)\,\mathrm{d}z \right| \\&\qquad \leqslant \int _{ B_{3\sqrt{\varepsilon }}(e+\varepsilon \omega ) } |f(z)|\,{{\mathcal {G}}}_1(e+\varepsilon \omega ,z)\,\mathrm{d}z\\&\qquad \leqslant C\varepsilon ^{\frac{\alpha }{2}} \int _{ B_{3\sqrt{\varepsilon }}(e+\varepsilon \omega ) } |z-e-\varepsilon \omega |^{2s-n}\,\sum _{k=1}^{+\infty } |c_k|\,\big ( r_1(e+\varepsilon \omega ,z)\big )^{k+s} \,\mathrm{d}z \\&\qquad \leqslant C\varepsilon ^{\frac{\alpha }{2}} \int _{ B_{3\sqrt{\varepsilon }}(e+\varepsilon \omega ) } |z-e-\varepsilon \omega |^{2s-n}\,\sum _{k=1}^{+\infty } |c_k| \,\big ( 1/2\big )^{k+s} \,\mathrm{d}z \\&\qquad \leqslant C\varepsilon ^{\frac{\alpha }{2}} \int _{ B_{3\sqrt{\varepsilon }}(e+\varepsilon \omega ) } |z-e-\varepsilon \omega |^{2s-n} \,\mathrm{d}z\\&\qquad \leqslant C\varepsilon ^{\frac{\alpha }{2}+s}, \end{aligned} \end{aligned}$$
(64)
up to renaming \(C>0\). On the other hand,
$$\begin{aligned} |z|= |e+\varepsilon \omega +z-e-\varepsilon \omega | \geqslant |e+\varepsilon \omega | -|z-e-\varepsilon \omega |\geqslant 1-\varepsilon -|z-e-\varepsilon \omega | \end{aligned}$$
and therefore
$$\begin{aligned} |f(z)|\leqslant C\,\big ( 1-|z|\big )^\alpha \leqslant C\,\big ( \varepsilon +|z-e-\varepsilon \omega |\big )^\alpha . \end{aligned}$$
In particular, if \(|z-e-\varepsilon \omega |>3\sqrt{\varepsilon }\), then
$$\begin{aligned} |f(z)|\leqslant C\,|z-e-\varepsilon \omega |^\alpha . \end{aligned}$$
(65)
Also, using (54) and (57), for any \(k\geqslant 1\)
$$\begin{aligned} \begin{aligned} \big ( r_1(e+\varepsilon \omega ,z)\big )^{k+s}&= \big (r_1(e+\varepsilon \omega ,z)\big )^{s+\frac{\alpha }{4}}\,\big (r_1(e+\varepsilon \omega ,z)\big )^{k-\frac{\alpha }{4}} \\&\leqslant \big ( r_0(e+\varepsilon \omega ,z)\big )^{s+\frac{\alpha }{4}} \left( \frac{1}{2}\right) ^{k-\frac{\alpha }{4}} \\&\leqslant \frac{C\,\varepsilon ^{s+\frac{\alpha }{4}}}{2^k\, |z-e-\varepsilon \omega |^{2s+\frac{\alpha }{2}} }.\end{aligned} \end{aligned}$$
This and (65) give that, if \(z\in B_1\setminus B_{3\sqrt{\varepsilon }}(e+\varepsilon \omega )\), then
$$\begin{aligned}&\left| f(z)\,{{\mathcal {G}}}_1(e+\varepsilon \omega ,z)\right| \\&\qquad \leqslant C\,|z-e-\varepsilon \omega |^{\alpha +2s-n}\,\sum _{k=1}^{+\infty } |c_k|\,\big ( r_1(e+\varepsilon \omega ,z)\big )^{k+s} \\&\qquad \leqslant C\varepsilon ^{s+\frac{\alpha }{4}}\,|z-e-\varepsilon \omega |^{\frac{\alpha }{2}-n}\,\sum _{k=1}^{+\infty } \frac{|c_k|}{2^k}, \end{aligned}$$
and the latter series is convergent, thanks to (56). This implies that
$$\begin{aligned} \left| \int _{B_1\setminus B_{3\sqrt{\varepsilon }}(e+\varepsilon \omega ) } f(z)\,{{\mathcal {G}}}_1(e+\varepsilon \omega ,z)\,\mathrm{d}z \right|\leqslant & {} C\varepsilon ^{s+\frac{\alpha }{4}}\,\int _{B_1\setminus B_{3\sqrt{\varepsilon }}(e+\varepsilon \omega ) } |z-e-\varepsilon \omega |^{\frac{\alpha }{2}-n}\,\mathrm{d}z\\\leqslant & {} C\varepsilon ^{s+\frac{\alpha }{4}}\,\int _{B_1} |z-e-\varepsilon \omega |^{\frac{\alpha }{2}-n}\,\mathrm{d}z\\\leqslant & {} C\varepsilon ^{s+\frac{\alpha }{4}}. \end{aligned}$$
By this and (64), we conclude that
$$\begin{aligned} \int _{B_1} f(z)\,{{\mathcal {G}}}_1(e+\varepsilon \omega ,z)\,\mathrm{d}z= o(\varepsilon ^s). \end{aligned}$$
Hence, we insert this information into (62) and, recalling (63), we obtain
$$\begin{aligned} \int _{B_1} f(z)\,G(e+\varepsilon \omega ,z)\,\mathrm{d}z= \int _{B_1} f(z)\,{{\mathcal {G}}}_0(e+\varepsilon \omega ,z)\,\mathrm{d}z+o(\varepsilon ^s). \end{aligned}$$
(66)
Now we define
$$\begin{aligned}&{{\mathcal {D}}}_1:= \{ z\in B_1 { \text{ s.t. } } r_0(e+\varepsilon \omega ,z)>1/2\}\\ { \text{ and } }&{{\mathcal {D}}}_2:= \{ z\in B_1 { \text{ s.t. } } r_0(e+\varepsilon \omega ,z)\leqslant 1/2\}. \end{aligned}$$
If \(z\in {{\mathcal {D}}}_1\), then (59) holds true, and so we can use (61), to find that
$$\begin{aligned} \big | f(z)\,{{\mathcal {G}}}_0(e+\varepsilon \omega ,z)\big |\leqslant C\varepsilon ^{\frac{\alpha }{2}}|z-e+\varepsilon \omega |^{2s-n}. \end{aligned}$$
Consequently, recalling (59),
$$\begin{aligned}&\left| \int _{{{\mathcal {D}}}_1} f(z)\,{{\mathcal {G}}}_0(e+\varepsilon \omega ,z)\,\mathrm{d}z\right| \leqslant C\varepsilon ^{\frac{\alpha }{2}}\, \int _{B_{3\sqrt{\varepsilon }}(e+\varepsilon \omega )} |z-e+\varepsilon \omega |^{2s-n}\,\mathrm{d}z=C\varepsilon ^{\frac{\alpha }{2}+s}, \end{aligned}$$
up to renaming \(C>0\) once again. In this way, formula (66) reduces to
$$\begin{aligned} \int _{B_1} f(z)\,G(e+\varepsilon \omega ,z)\,\mathrm{d}z= \int _{{{\mathcal {D}}}_2} f(z)\,{{\mathcal {G}}}_0(e+\varepsilon \omega ,z)\,\mathrm{d}z+o(\varepsilon ^s). \end{aligned}$$
(67)
Now, by (57) and (53), if \(z\in {{{\mathcal {D}}}_2}\),
$$\begin{aligned} {{\mathcal {G}}}_0(e+\varepsilon \omega ,z) = \frac{|z-e-\varepsilon \omega |^{2s-n}}{s}\,\big ( r_0(e+\varepsilon \omega ,z)\big )^{s} = \frac{ \varepsilon ^s\,(-\varepsilon -2\,e\cdot \omega )^s\,(1-|z|^2)^s }{ s\,|z-e-\varepsilon \omega |^n }. \end{aligned}$$
Hence, (67) gives that
$$\begin{aligned} \begin{aligned}&\lim _{\varepsilon \searrow 0}\varepsilon ^{-s}\int _{B_1} f(z) \,G(e+\varepsilon \omega ,z)\,\mathrm{d}z\\&\quad =\,\lim _{\varepsilon \searrow 0} \int _{ \{ {2\varepsilon \,(-\varepsilon -2\,e\cdot \omega )\,(1-|z|^2)} \leqslant {|z-e-\varepsilon \omega |^2} \} } f(z)\, \frac{(-\varepsilon -2\,e\cdot \omega )^s\,(1-|z|^2)^s }{ s\,|z-e-\varepsilon \omega |^n }\,\mathrm{d}z.\end{aligned} \end{aligned}$$
(68)
Now, we show the following uniform integrability condition: we set
$$\begin{aligned} F_\varepsilon (z):= \left\{ \begin{array}{ll} f(z)\, \displaystyle \frac{(-\varepsilon -2\,e\cdot \omega )^s\,(1-|z|^2)^s }{ s\,|z-e-\varepsilon \omega |^n } &{} { \text{ if } } {2\varepsilon \,(-\varepsilon -2\,e\cdot \omega )\,(1-|z|^2)} \leqslant {|z-e-\varepsilon \omega |^2}, \\ \,&{}\, \\ 0 &{} { \text{ otherwise }}, \end{array}\right. \end{aligned}$$
and we prove that for any \(\eta >0\) there exists \(\delta >0\) (depending on \(\eta \), e and \(\omega \), but independent of \(\varepsilon \)) such that, for any \(E\subset {\mathbb {R}}^{d}\) with \(|E|\leqslant \delta \), we have
$$\begin{aligned} \int _{B_1\cap E} \big | F_\varepsilon (z)\big |\,\mathrm{d}z\leqslant \eta . \end{aligned}$$
(69)
To this aim, we take E as above and
$$\begin{aligned} \rho := c_\star \,\varepsilon , \end{aligned}$$
with \(c_\star \in \left( 0, \frac{1}{10}\right) \) to be conveniently chosen in the sequel (also in dependence of \(\omega \) and e), and we set \(E_1:=E\cap B_{\rho }(e+\varepsilon \omega )\), \(E_2:=E\setminus E_1\).
We claim that
$$\begin{aligned} {E_1 \text{ is } \text{ empty. }} \end{aligned}$$
(70)
For this, we argue by contradiction: if there existed \(z\in E_1\), then
$$\begin{aligned}&{\varepsilon \,(-\,e\cdot \omega )\,(1-|z|^2)} \leqslant {2\varepsilon \,(-\varepsilon -2\,e\cdot \omega )\,(1-|z|^2)} \leqslant {|z-e-\varepsilon \omega |^2}\leqslant \rho ^2, \end{aligned}$$
if \(\varepsilon \) is small enough in dependence of the fixed e and \(\omega \) (recall (52)), and thus
$$\begin{aligned} 1-|z|^2 \leqslant \frac{C\rho ^2}{\varepsilon }, \end{aligned}$$
(71)
with \(C>0\) also depending on e and \(\omega \). On the other hand, we have that \(E_1\subseteq B_{\rho }(e+\varepsilon \omega )\), and therefore
$$\begin{aligned} |z|\leqslant |e+\varepsilon \omega | +|z-e-\varepsilon \omega |\leqslant \sqrt{1+\varepsilon ^2+2\varepsilon e\cdot \omega }+\rho \leqslant 1 -\frac{-\varepsilon e\cdot \omega }{10}+C\varepsilon ^2+\rho , \end{aligned}$$
and then
$$\begin{aligned} |z|^2 \leqslant 1 -\frac{\varepsilon e\cdot \omega }{5}+C\varepsilon ^2+\rho ^2. \end{aligned}$$
This is a contradiction with (71) if \(c_\star \) is appropriately small and thus (70) is proved.
So, from now on, \(c_\star \) is fixed suitably small. We observe that if \(z\in E_2\) then
$$\begin{aligned} |z-e-\varepsilon \omega | \geqslant \rho = c_\star \varepsilon , \end{aligned}$$
and consequently
$$\begin{aligned} \int _{B_1\cap E_2} \big | F_\varepsilon (z)\big |\,\mathrm{d}z\leqslant \int _{\begin{array}{c} {B_1\cap E}\\ { \{ |z-e-\varepsilon \omega | \geqslant c_\star \varepsilon \}} \end{array}} \frac{C\,(1-|z|)^{s+\alpha } }{ |z-e-\varepsilon \omega |^n }\,\mathrm{d}z. \end{aligned}$$
(72)
Now, we distinguish two cases, either \(\delta \leqslant \varepsilon ^{2n}\) or \(\delta >\varepsilon ^{2n}\). If \(\delta \leqslant \varepsilon ^{2n}\), we use (72) to get that
$$\begin{aligned} \int _{B_1\cap E_2} \big | F_\varepsilon (z)\big |\,\mathrm{d}z\leqslant \int _{ \begin{array}{c} {B_1\cap E}\\ { \{ |z-e-\varepsilon \omega | \geqslant c_\star \varepsilon \}} \end{array}} \frac{C }{ \varepsilon ^n }\,\mathrm{d}z\leqslant \frac{C \delta }{ \varepsilon ^n } \leqslant C\,\sqrt{\delta }. \end{aligned}$$
(73)
If instead
$$\begin{aligned} \delta >\varepsilon ^{2n}, \end{aligned}$$
(74)
we observe that
$$\begin{aligned} |z-e-\varepsilon \omega |\geqslant 1-|z|-\varepsilon , \end{aligned}$$
and we deduce from (72) that
$$\begin{aligned} \begin{aligned} \int _{B_1\cap E_2} \big | F_\varepsilon (z)\big |\,\mathrm{d}z&\leqslant \int _{ \begin{array}{c} {B_1\cap E}\\ { \{ |z-e-\varepsilon \omega | \geqslant c_\star \varepsilon \}} \end{array}} \frac{C\,(|z-e-\varepsilon \omega |+\varepsilon )^{s+\alpha } }{ |z-e-\varepsilon \omega |^n }\,\mathrm{d}z\\&\leqslant C\,\int _{ \begin{array}{c} {B_1\cap E}\\ { \{ |z-e-\varepsilon \omega | \geqslant c_\star \varepsilon \}} \end{array}} \frac{|z-e-\varepsilon \omega |^{s+\alpha } }{ |z-e-\varepsilon \omega |^n }\,\mathrm{d}z\\&\quad + C\,\int _{\begin{array}{c} {B_1\cap E}\\ { \{ |z-e-\varepsilon \omega | \geqslant c_\star \varepsilon \}} \end{array}} \frac{\varepsilon ^{s+\alpha } }{ |z-e-\varepsilon \omega |^n }\,\mathrm{d}z\\ {}&=:I_1+I_2. \end{aligned}\end{aligned}$$
(75)
To estimate \(I_1\), we split it into
$$\begin{aligned}&I_{1,1}:= C\,\int _{\begin{array}{c} {B_1\cap E}\\ { \{ c_\star \varepsilon \leqslant |z-e-\varepsilon \omega | \leqslant \delta ^{1/{2n}} \}} \end{array}} |z-e-\varepsilon \omega |^{s+\alpha -n} \,\mathrm{d}z\\ {\text{ and } }&I_{1,2}:= C\,\int _{\begin{array}{c} {B_1\cap E}\\ { \{ |z-e-\varepsilon \omega | >\delta ^{1/{2n}}\}} \end{array}} |z-e-\varepsilon \omega |^{s+\alpha -n} \,\mathrm{d}z. \end{aligned}$$
Using polar coordinates, we find that
$$\begin{aligned} I_{1,1}\leqslant C\, \int _{c_\star \varepsilon }^{ \delta ^{1/{2n}} } t^{n-1}\,t^{s+\alpha -n} \,\mathrm{d}t \leqslant C\, \left[ \left( \delta ^{\frac{1}{2n}}\right) ^{s+\alpha }- \left( c_\star \varepsilon \right) ^{s+\alpha }\right] \leqslant C\,\delta ^{\frac{s+\alpha }{2n}}. \end{aligned}$$
(76)
In addition,
$$\begin{aligned} I_{1,2}\leqslant C\,\int _{ \begin{array}{c} {B_1\cap E}\\ { \{ |z-e-\varepsilon \omega | >\delta ^{1/{2n}}\}} \end{array}} |z-e-\varepsilon \omega |^{s-n}\,\mathrm{d}z \leqslant C\,\int _E \delta ^{{\frac{s-n}{2n}} }\,\mathrm{d}z \leqslant C\delta ^{1+{\frac{s-n}{2n}}}= C\delta ^{{\frac{s+n}{2n}}}. \end{aligned}$$
This and (76) say that
$$\begin{aligned} I_{1}\leqslant C\,\delta ^{\frac{s+\alpha }{2n}}+C\delta ^{{\frac{s+n}{2n}}}. \end{aligned}$$
(77)
Moreover,
$$\begin{aligned} I_2 \leqslant C\varepsilon ^{s+\alpha }\,\int _{c_\star \varepsilon }^2\frac{t^{n-1}}{t^N}\,\mathrm{d}t\leqslant C\varepsilon ^{s+\alpha }\,|\log \varepsilon |\leqslant C\varepsilon ^s\leqslant C\delta ^{\frac{s}{2n}}, \end{aligned}$$
thanks to (74). Hence, using this and (77), and recalling (75), we obtain that
$$\begin{aligned} \int _{B_1\cap E_2} \big | F_\varepsilon (z)\big |\,\mathrm{d}z\leqslant I_1+I_2\leqslant C\,\delta ^{\frac{s+\alpha }{2n}}+C\delta ^{{\frac{s+n}{2n}}}+ C\delta ^{\frac{s}{2n}}. \end{aligned}$$
We now combining this estimate, which is coming from the case in (74), with (73), which was coming from the complementary case, and we see that, in any case,
$$\begin{aligned} \int _{B_1\cap E_2} \big | F_\varepsilon (z)\big |\,\mathrm{d}z\leqslant C\,\delta ^\kappa , \end{aligned}$$
for some \(\kappa >0\). From this and (70), we obtain that
$$\begin{aligned} \int _{B_1\cap E} \big | F_\varepsilon (z)\big |\,\mathrm{d}z\leqslant C\,\delta ^\kappa , \end{aligned}$$
Then, choosing \(\delta \) suitably small with respect to \(\eta \), we establish (69), as desired.
Notice also that \(F_\varepsilon \) converges pointwise to \( f(z)\, \frac{(-2\,e\cdot \omega )^s\,(1-|z|^2)^s }{ s\,|z-e|^n }\). Hence, using (68), (69) and the Vitali Convergence Theorem, we conclude that
$$\begin{aligned} \lim _{\varepsilon \searrow 0}\varepsilon ^{-s}\int _{B_1} f(z)\,G(e+\varepsilon \omega ,z)\,\mathrm{d}z= & {} \lim _{\varepsilon \searrow 0} \int _{B_1} F_\varepsilon (z)\,\mathrm{d}z\\= & {} \int _{B_1} f(z)\, \frac{(-2\,e\cdot \omega )^s\,(1-|z|^2)^s }{ s\,|z-e|^n }\,\mathrm{d}z, \end{aligned}$$
which establishes (48). \(\square \)
