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Research article Special Issues

The fractional Malmheden theorem

  • We provide a fractional counterpart of the classical results by Schwarz and Malmheden on harmonic functions. From that we obtain a representation formula for s-harmonic functions as a linear superposition of weighted classical harmonic functions which also entails a new proof of the fractional Harnack inequality. This proof also leads to optimal constants for the fractional Harnack inequality in the ball.

    Citation: Serena Dipierro, Giovanni Giacomin, Enrico Valdinoci. The fractional Malmheden theorem[J]. Mathematics in Engineering, 2023, 5(2): 1-28. doi: 10.3934/mine.2023024

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  • We provide a fractional counterpart of the classical results by Schwarz and Malmheden on harmonic functions. From that we obtain a representation formula for s-harmonic functions as a linear superposition of weighted classical harmonic functions which also entails a new proof of the fractional Harnack inequality. This proof also leads to optimal constants for the fractional Harnack inequality in the ball.



    In 1934, Harry William Malmheden [13] proved a simple algorithm to compute the value of a harmonic function at a point of B1, knowing its value on the boundary.

    The Malmheden theorem makes use of two fundamental geometric ingredients:

    1) the notion of affine interpolation between the values of a given function at two different points of the space,

    2) the projections of a point inside a ball to the boundary in a given direction.

    Hence, to state the Malmheden theorem explicitly, we now formalize these two notions into a precise mathematical setting. We start by introducing a notation for the affine interpolation between the values of some given function. That is, given a set KRn, a function f:KR, two distinct points a, bK, and a point x on the segment L joining a and b, we define La,bf(x) as the affine function on L such that La,bf(a)=f(a) and La,bf(b)=f(b).

    Of course, one can write this affine function explicitly by using the analytic expression

    La,bf(x)=(xa)e|ba|f(b)+(bx)e|ba|f(a),where e:=ba|ba|. (1.1)

    One can call La,bf(x) the "affine function of f with extrema a and b evaluated at the point x".

    Now we discuss the notation regarding the projections of a point inside a ball to the boundary of the ball in a given direction. For this, given a point xB1 and a direction eB1 we consider the intersections Qx+(e) and Qx(e) of B1 with the straight line passing through xB1 with direction e, with the convention that Qx+(e)Qx(e) has the same orientation of e, see Figure 1.

    Figure 1.  The projections Qx±(e) used in the Malmheden theorem.

    Clearly, from the analytic point of view, one can write explicitly these projections in the form

    Qx+(e):=x+rx+(e)eand Qx(e):=x+rx(e)e, (1.2)

    where

    rx+(e)=xe+(xe)2|x|2+1and rx(e)=xe(xe)2|x|2+1. (1.3)

    We note from Eqs (1.2) and (1.3) that Qx±(e) are continuous functions in (x,e)B1×B1. Moreover,

    limx0Qx±(e)=±e (1.4)

    for each eB1. This tells us that the maps Qx± simply reduce to ±idB1 when x=0.

    Given a boundary datum f:B1R, the core of the Malmheden Theorem is thus to consider, for every point xB1 and every direction eB1, the affine function of f with extrema Qx(e) and Qx+(e), namely the function

    LQx(e),Qx+(e)f(x) (1.5)

    and then to average in all directions e.

    The remarkable result by Malmheden is that this averaging procedure of linear interpolations produces precisely the solution of the classical Dirichlet problem in B1 with boundary datum f, according to the following classical statement (see [13]):

    Theorem 1.1 (Malmheden theorem). Let n2 and f:B1R be continuous. Then

    uf(x):=B1LQx(e),Qx+(e)f(x)dHn1e (1.6)

    is the harmonic function in B1 with boundary datum f.

    As usual, here above and in the rest of this paper, we denoted by Hn1 the (n1)-Hausdorff measure (hence, the integral on the right hand side of (1.6) is simply the spherical integral along B1; we kept the explicit notation with the Hausdorff measure to have a typographical evidence of the surface integrals, to be distinguished by the classical volume ones).

    We remark that Theorem 1.1 contains the Mean Value Theorem for harmonic functions as a particular case: indeed, in light of (1.1) and (1.4), if we take x:=0 then (1.6) reduces to

    uf(0)=B1Le,ef(0)dHn1e=B1(f(e)2+f(e)2)dHn1e=B1f(e)dHn1e, (1.7)

    which is the content of the Mean Value Theorem.

    We also stress that an elegant result such as Theorem 1.1 is specific for balls and cannot be extended in general to other domains, as pointed out in [1].

    Interestingly, Theorem 1.1 contains as a particular case a classical result due to Hermann Amandus Schwarz [18] about the Dirichlet problem in the plane and related to conformal mappings in the complex framework.

    To state Schwarz result it is convenient to introduce the reflection of a point ωB1 through a point xB1, see Figure 2. More precisely, given xB1 and ωB1 we define

    Qx(ω):=ω2(xω)ω|xω|2(xω). (1.8)
    Figure 2.  The reflection Qx(ω) used in the Schwarz theorem.

    Comparing with (1.2), one sees that if e:=ωx|ωx| then Qx+(e)=ω and Qx(e)=Qx(ω).

    In this setting, the result by Schwarz is that the average of the boundary datum composed with the above reflection returns the solution of the Dirichlet problem in the ball. More explicitly:

    Theorem 1.2 (Schwarz theorem). Let n=2 and f:B1R be continuous. Then

    uf(x):=B1f(Qx(e))dHn1e (1.9)

    is the harmonic function in B1 with external datum f.

    Theorem 1.2 can be proved in several ways using either complex or real analysis (see e.g., [9,15,17]), but it is also a direct consequence of Theorem 1.1, see e.g., [8] for a detailed presentation of this classical argument.

    Example 1.3. A very neat application of Theorem 1.2 (see e.g., [16]) consists in the determination of the stationary temperature u at a point x in a plate (say B1) when the temperature along the boundary of the plate is 1 along some arc Σ and 0 outside. In this case, the reflection in (1.8) sends Σ into an arc Σ (the symmetric of Σ through x, see Figure 3) and it therefore follows from Theorem 1.2 that

    u(x)=|Σ|2π,
    Figure 3.  The geometric construction to detect the temperature of a plate at the point x.

    where |Σ| is the length of the arc Σ, thus providing an elementary geometric construction to solve a problem of physical relevance.

    The objective of this paper is to obtain a fractional counterpart for the Malmheden and Schwarz theorems.

    We will thus replace the notion of harmonic functions in B1 with that of s-harmonic functions, namely functions whose fractional Laplacian vanishes in B1, that is, for all xB1,

    Rnu(x)u(y)|xy|n+2sdy=0, (1.10)

    where the integral above is intended in the principal value sense. Here above and throughout the paper the fractional parameter s(0,1).

    Rather than a boundary value along B1, as usual in the nonlocal setting, we complement (1.10) with an external condition of the type u=f in RnB1.

    We recall that in general s-harmonic functions behave way more wildly that their classical counterparts, see e.g., [7]. Therefore, in principle one cannot easily expect that a "simple formulation" such as the one in Theorems 1.1 and 1.2 accounts for all the complex situations arising in the fractional setting.

    However, we will prove that a counterpart of Theorems 1.1 and 1.2 carries over to the case of the fractional Laplacian, considering the following structural modifications:

    1) the classical spherical averages are replaced by suitable weighted averages on spheres of radius larger than 1,

    2) the geometric transformations in (1.2) and (1.8) are scaled in dependence of the radius of each of these spheres.

    To clarify these points, and thus reconsider (1.5) in a nonlocal setting, given ρ>1 and f:RnB1R, for all xB1 we define

    fρ(x):=f(ρx). (1.11)

    Hence, in the notation of (1.1), we define

    Lf,e,ρ(x):=LQx/ρ(e),Qx/ρ+(e)fρ(xρ). (1.12)

    Notice that when ρ=1 the above setting reduces to (1.5), otherwise one is considering here a similar framework but for a rescaled version of the function f and rescaled points.

    To detect the long-range effect of the fractional Laplacian, it is also useful to consider the kernel

    B1×(1,+)(x,ρ)E(x,ρ):=c(n,s)ρ(1|x|2)s(ρ21)s(ρ2|x|2), (1.13)

    where

    c(n,s):=Γ(n2)sin(πs)πn2+1. (1.14)

    For our purposes, the kernel E will play the role of a suitable spherical average* of a fractional Poisson kernel and the constant c(n,s) is merely needed for normalization purposes.

    *More precisely, the intuition behind E is \label{INTUK} that it satisfies, for each xB1 and ρ(1,+),

    E(x,ρ)=ρn1BρP(x,y)dHn1y,

    where P is the fractional Poisson kernel.

    This relation can be obtained as a consequence of our fractional Malmheden theorem. Though our approach does not pass explicitly through this identity, for the sake of completeness we provide an independent proof in Appendix A.1.

    We also define the space

    L1s(RnB1):={f:RnR measurable :RnB1|f(x)||x|n+2sdx<}. (1.15)

    With this, we can state the main result of this paper as follows:

    Theorem 1.4 (fractional Malmheden theorem). Let n2, s(0,1), R>1and fL(BRB1)L1s(RnB1).

    Then, the unique solution (up to a zero measure subset of RnB1) to the problem

    {(Δ)su=0inB1,u=finRnB1 (1.16)

    can be written, for each xB1, as

    u(s)f(x):=1(B1E(x,ρ)Lf,e,ρ(x)dHn1e)dρ. (1.17)

    As a fractional counterpart of the observation in (1.7), we point out that Theorem 1.4 entails as a straightforward consequence the Mean Value Formula for s-harmonis functions. Indeed, by the changes of variable e:=ω/|ω| and y:=ρω/|ω|,

    u(s)f(0)=1(B1E(0,ρ)Lf,e,ρ(0)dHn1e)dρ=c(n,s)1(B11ρ(ρ21)sLe,efρ(0)dHn1e)dρ=c(n,s)1(B11ρ(ρ21)s(f(ρe)2+f(ρe)2)dHn1e)dρ=c(n,s)1(Bρ1ρn(ρ21)s(f(ω)2+f(ω)2)dHn1ω)dρ=c(n,s)RnB11|y|n(|y|21)s(f(y)2+f(y)2)dy=c(n,s)RnB1f(y)|y|n(|y|21)sdy,

    which is the Mean Value Formula for s-harmonis functions, see e.g., formula (1.3) in [4].

    We consider Theorem 1.4 as the natural fractional counterpart of Theorem 1.1 and we mention that indeed one can "recover" Theorem 1.1 in the limit as s1, according to the following result:

    Proposition 1.5. Let n2, s0(0,1), R>1and fC(BRB1)L1s(RnB1) for each s(s0,1]. Then, for each xB1, it holds that

    lims1u(s)f(x)=uf(x),

    where u(s)f and uf are defined in (1.17) and (1.6), respectively.

    As a straightforward consequence of the classical Malmheden theorem (Theorem 1.1) and its fractional formulation theorem (Theorem 1.4), we deduce the following result.

    Theorem 1.6 (An s-harmonic function is the superposition of classical harmonic functions). Let n2, s(0,1) and fC(RnB1)L1s(RnB1). For each ρ>1 we define ufρ as the unique solution to the Dirichlet problem

    {Δu=0inB1,u|B1=fρ|B1, (1.18)

    where fρ is defined in (1.11).

    Then the unique solution u(s)f to (1.16) can be written as

    u(s)f(x)=|B1|1E(x,ρ)ufρ(xρ)dρ. (1.19)

    The interest of Theorem 1.6 is that it allows us to write an s-harmonic function in B1 as a weighted integral of classical harmonic functions, where the weight coincide with E(x,ρ). Besides being interesting in itself, this result is very useful to deduce properties of s-harmonic functions, as the Harnack inequality (see Section 4), starting from their local counterpart.

    As a matter of fact, as a consequence of Theorem 1.6 one obtains a new proof of the Harnack inequality for s-harmonic functions in B1. The result goes as follows:

    Theorem 1.7 (Harnack inequality). Let n2, s(0,1), R>1 and u be non negative, s-harmonic in B1 and such that uL(BRB1)L1s(RnB1).

    Then, for each r(0,1) and xBr,

    (1r2)s(1+r)nu(0)u(x)(1r2)s(1r)nu(0). (1.20)

    The constants in (1.20) are optimal, and for s1 they converge to the optimal constants of the classical Harnack inequality in Br for harmonic functions in B1.

    For different proofs of the fractional Harnack inequality see [6,11,12] and the references therein.

    Another consequence of Theorem 1.4 is the fractional version of Schwarz result:

    Theorem 1.8 (fractional Schwarz theorem). Let n=2, s(0,1), R>1and fL(BRB1)L1s(R2B1).Then, the unique solution (up to a zero measure subset of R2B1) to the problem

    {(Δ)su=0inB1,u=finR2B1

    can be written, for each xB1, as

    u(s)f(x):=1(B1E(x,ρ)fρ(Qx/ρ(e))dH1e)dρ. (1.21)

    This is a fractional counterpart of Theorem 1.2, in the sense of Proposition 1.9 below. Proposition 1.9 is a straightforward consequence of Theorems 1.2 and 1.8 and Proposition 1.5.

    Proposition 1.9. Let n=2, s0(0,1), R>1and fC(BRB1)L1s(R2B1) for each s(s0,1]. Then, for each xB1, it holds that

    lims1u(s)f(x)=uf(x)

    where u(s)f, uf are defined in (1.21) and (1.9), respectively.

    Remark 1.10. It is worth pointing out that from Theorem 1.4 we can evince the identity

    1E(x,ρ)dρ=1|B1|, (1.22)

    for each xB1. Indeed, if we consider as external data f=1 in RnB1, then the unique solution to the problem (1.16) is u=1 in Rn. Therefore, according to (1.17) and the fact that in this case the linear interpolation L1,e,ρ(x)=1 for each xB1, we obtain that

    1=1(B1E(x,ρ)L1,e,ρ(x)dHn1e)dρ=1|B1|E(x,ρ)dρ,

    which gives (1.22).

    As an application of Theorem 1.8, we have:

    Example 1.11. Let n=2 and take an arc ΣB1. Consider the function defined on R2B1 as

    ˜χΣ(y):={1ify|y|Σ,0ify|y|B1Σ. (1.23)

    It is clear that ˜χΣ is positively homogeneous of degree zero, and furthermore ˜χΣL(R2B1)L1s(R2B1). Then by Theorem 1.8 we get that for each xB1

    u(s)˜χΣ(x)=1E(x,ρ)|Σx/ρ|dρ (1.24)

    where Σx/ρ is the projected arc of Σ on B1 through the focal point x/ρ, as constructed in Example 1.3. We denoted with |Σx/ρ| its length.

    This gives a simple geometrical procedure to compute the solution of

    {(Δ)su=0inB1,u=˜χΣinR2B1

    at a point x of the two dimensional disc when the non local boundary condition is given by (1.23). Note that as ρ is getting larger, the measure of Σx/ρ reaches the one of Σ, or more precisely

    limρ|Σx/ρ|=|Σ|.

    If x=0, formula (1.24) boils down to

    u(s)˜χΣ(0)=c(n,s)|Σ|11ρ(ρ21)sdρ=|Σ|2π, (1.25)

    where we have applied identity (1.22). This example can be seen as the fractional counterpart of Example 1.3.

    This paper is organized as follows. In Section 2 we give some preliminary results on the s-harmonic function written as a convolution with the fractional Poisson kernel.

    Section 3 is devoted to the proofs of the fractional Malmheden and Schwarz results, that is Theorems 1.4 and 1.8, and of the convergence result in Proposition 1.5.

    In Section 4 we use these results to provide a proof of the well-known Harnack inequality for s-harmonic functions under some regularity assumptions on the external datum f:RnB1R, that is we prove Theorem 1.7.

    In this section, we revisit the well-established result according to which fractional harmonic functions can be represented as an integral of the datum outside the domain against a suitable Poisson kernel. For completeness, we extend this result to the case in which the datum is not necessarily continuous, so to be able to present the results of this paper in a suitable generality. Notice that the extension to functions that are not necessarily continuous is also useful for us to comprise situations as in Example 1.11.

    The framework that we consider is the following. For n2 and s(0,1), we consider the space L1s(RnB1) as defined in (1.15). Given fL1s(RnB1), we denote the norm on L1s(RnB1) by

    fL1s(RnB1):=RnB1|f(x)||x|n+2sdx.

    Furthermore, we define the following fractional Poisson kernel in the unit ball

    P(x,y):=c(n,s)(1|x|2|y|21)s1|xy|n (2.1)

    for xB1 and yRnB1, and c(n,s) is the normalizing constant in (1.14).

    As customary, the role of the constant c(n,s) is to normalize the Poisson kernel, namely we have that

    RnB1P(x,y)dy=1, (2.2)

    see e.g., formula (1.14) and Lemma A.5 in [3].

    We also remark that

    P(,y)C(B1)

    and, for every ρ(0,1), αNn and yRnB1,

    supxBρ|DαxP(x,y)|Cρ(|y|1)s|y|n+s+|α|, (2.3)

    where Cρ>0 depends only on ρ, n and s and, as usual, we have denoted the length of the multi-index α as |α|:=α1++αn.

    Then, we define

    u(s)f(x):={RnB1P(x,y)f(y)dyifxB1,f(x)ifxRnB1, (2.4)

    and we have the following result on the representation of s-harmonic functions:

    Theorem 2.1. Let n2, s(0,1) and fC(RnB1)L1s(RnB1). Then the function in (2.4) is the unique pointwise continuous solution to the problem

    {(Δ)su(s)f=0inB1,u(s)f=finRnB1. (2.5)

    For a proof of Theorem 2.1 see e.g. Theorem 2.10 in [3].

    We now generalize Theorem 2.1 by allowing external data that are not necessarily continuous:

    Proposition 2.2. Let n2, s(0,1), R>1 and fL(BRB1)L1s(RnB1). Then the function defined in (2.4) is the unique solution (up to a zero measure subset of RnB1) to the problem in (2.5).

    Proof. We argue by approximation, owing to Theorem 2.1. The gist is indeed to take a sequence of continuous functions fk approaching f as k+, use Theorem 2.1 and then pass to the limit. To implement this idea, one needs to take care of some regularity issues.

    The details of this technical argument go as follows. By (2.3), for each xB1 and multi-index α we have that

    DαxP(x,)f()L1(Rn¯B1).

    As a consequence, we obtain that u(s)f(x) in (2.4) is well defined and smooth inside B1.

    To complete the proof of Proposition 2.2, we need to show that u(s)f, as defined in (2.4), is the unique solution of (2.5). To do so, we start by checking that u(s)f is s-harmonic in B1. We consider a sequence {fk}kC(RnB1)L1s(RnB1), such that

    fkf in L1s(RnB1) as k+. (2.6)

    More specifically, we take fk:=(χBk˜f)η1k with k2, where ˜f is defined as

    ˜f(x):={f(x)ifxRnB1,0ifxB1,

    and η1k is a mollifier of radius 1k, while χBk is the characteristic function of Bk. We also let u(s)fk be the unique pointwise continuous solution to the problem (2.5), according to Theorem 2.1.

    Then we have that for each multi-index α

    Dαu(s)fkDαu(s)fLloc(B1)0 as k+. (2.7)

    Indeed for each multi-index α and gL(BRB1)L1s(RnB1) one finds that

    Dαug(x)=RnB1DαxP(x,y)g(y)dy

    for each xB1, and therefore, choosing R0(1,R), we see that, for every xB with BB1,

    |Dαu(s)fk(x)Dαu(s)f(x)|RnBR0|DαxP(x,y)||fk(y)f(y)|dy+BR0B1|DαxP(x,y)||fk(y)f(y)|dycRnBR0|fk(y)f(y)||y|n+2sdy+BR0B1|DαxP(x,y)||fk(y)f(y)|dy, (2.8)

    where c is a positive constant depending on α, R0, n, s and B. The first term in the third line in (2.8) converges to zero as k+, thanks to (2.6). We also observe that, if yBR0B1, then

    |fk(y)f(y)|2fL(BRB1)

    and therefore, by the dominated convergence theorem, we have that also the second term in the third line in (2.8) converges to zero as k+. These considerations prove (2.7).

    Furthermore note that if α=0, taking R0(1,R) and using also (2.2), we have that, for all xB1,

    |u(s)fk(x)|RnBR0|P(x,y)||fk(y)|dy+BR0B1|P(x,y)||fk(y)|dyCRnBR0|fk(y)||y|n+2sdy+BR0B1|P(x,y)||fk(y)|dyCfkL1s(RnB1)+fL(BRB1), (2.9)

    where C is a positive constant depending on R0, n and s. Now, we observe that the sequence fkL1s(RnB1) is uniformly bounded, thanks to (2.6). Accordingly, from (2.9) we see that

    u(s)fk is uniformly bounded in B1. (2.10)

    Now, if xB1, taking δ(0,1|x|), we have that

    (Δ)su(s)fk(x)(Δ)su(s)f(x)=Rnu(s)fk(x)u(s)fk(y)u(s)f(x)+u(s)f(y)|xy|n+2sdy=A+B+C+D+E+F, (2.11)

    where

    A:=RnB1u(s)fk(x)u(s)f(x)|xy|n+2sdy,B:=RnB1u(s)fk(y)u(s)f(y)|xy|n+2sdy,C:=Bδ(x)u(s)fk(x)u(s)fk(y)|xy|n+2sdy,D:=Bδ(x)u(s)f(y)u(s)f(x)|xy|n+2sdy,E:=B1Bδ(x)u(s)fk(x)u(s)f(x)|xy|n+2sdy and F:=B1Bδ(x)u(s)f(y)u(s)fk(y)|xy|n+2sdy.

    Notice that

    |A+E|RnBδ(x)|u(s)fk(x)u(s)f(x)||xy|n+2sdy|u(s)fk(x)u(s)f(x)|RnBδdz|z|n+2sCδ2s|u(s)fk(x)u(s)f(x)|,

    which converges to zero as k+, thanks to (2.7).

    Furthermore, we observe that if yRnB1 then

    |xy||y||x|=δ|y|+(1δ)|y||x|δ|y|+1δ|x|δ|y|,

    and thus

    |B|RnB1|fk(y)f(y)||xy|n+2sdy1δn+2sRnB1|fk(y)f(y)||y|n+2sdy

    which, in light of (2.6), converges to zero as k+.

    Moreover, from (2.7) and the dominated convergence theorem, we see that the quantity C+D converges to zero as k+.

    Finally, recalling (2.10) and making again use of the dominated convergence theorem, we have that F converges to zero as k+.

    These considerations and (2.11) give that (Δ)su(s)fk(x) converges to (Δ)su(s)f(x) as k+ for every xB1. Since, by Theorem 2.1, we know that (Δ)su(s)fk(x)=0 for each xB1, we conclude that (Δ)su(s)f(x)=0. This proves that u(s)f solves (2.5).

    It is only left to show the uniqueness statement. Suppose that there exists u1:RnR satisfying

    {(Δ)su1=0inB1,u1=finRnB1.

    Then both v:=u(s)fu1 and v=u1u(s)f are solutions to

    {(Δ)su=0inB1,u=0inRnB1,

    and therefore by the maximum principle for the fractional Laplacian (see e.g., Theorem 2.3.2. in [5]) we have that v=0 in B1, leading to uniqueness.

    In this section we provide the proofs of the fractional Malmheden and Schwarz results, as stated in Theorems 1.4 and 1.8, and of the convergence result in Proposition 1.5.

    We start with the main argument to prove Theorem 1.4. For this, we employ the following change of variable result (see Lemma 2.13.3 in [8] for the proof of it):

    Lemma 3.1. Let n2, xB1 and Qx± and rx±be defined as in (1.2) and (1.3), respectively. Then for each eB1 it holdsthat

    |detDQx±(e)|=(±rx±(e))n1|x|2rx±(e)xe,

    and for each continuous f:B1R we have that

    B1f(e)dHn1e=B1f(Qx±(e))(±rx±(e))n1|x|2rx±(e)xedHn1e.

    With this notation, Theorem 1.4 will be a consequence of the following statement.

    Theorem 3.2. Let n2, s(0,1), R>1and fL(BRB1)L1s(RnB1).Let u(s)f be as in (2.4).

    Then, for each xB1,

    u(s)f(x)=1(B1E(x,ρ)Lf,e,ρ(x)dHn1e)dρ, (3.1)

    where the notation in (1.12) and (1.13) has been used.

    Furthermore, if f is positively homogeneous of degree γ for some γ0, then we have that

    u(s)f(x)=1(B1ργE(x,ρ)Lf,e,1(xρ)dHn1e)dρ. (3.2)

    Proof. We first suppose that fC(RnB1)L1s(RnB1). Let xB1, then, using polar coordinates, from (2.1) and (2.4) we get that

    u(s)f(x)=RnB1P(x,y)f(y)dy=c(n,s)(1|x|2)sRnB11(|y|21)sf(y)|xy|ndy=c(n,s)(1|x|2)s11(ρ21)s(Bρf(ω)|xω|ndHn1ω)dρ=c(n,s)(1|x|2)s1ρn1(ρ21)s(B1f(ρe)|xρe|ndHn1e)dρ=c(n,s)(1|x|2)s11ρ(ρ21)s(B1f(ρe)|xρe|ndHn1e)dρ=c(n,s)(1|x|2)s1ρ(ρ2|x|2)(ρ21)s(B1f(ρe)|xρe|n(1|x|2ρ2)dHn1e)dρ=:I. (3.3)

    Hence, defining

    g(e):=f(ρe)|xρe|n(1|x|2ρ2)

    and applying Lemma 3.1 we obtain that

    Ic(n,s)(1|x|2)s=1ρ(ρ2|x|2)(ρ21)s(B1g(e)dHn1e)dρ=1ρ(ρ2|x|2)(ρ21)s(B1g(Qx/ρ(e))(rx/ρ(e))n1|x/ρ|2(x/ρe)rx/ρ(e)dHn1e)dρ=1ρ(ρ2|x|2)(ρ21)s(B1f(ρQx/ρ(e))|xρQx/ρ(e)|n(1|x/ρ|2)(rx/ρ(e))n1|x/ρ|2(x/ρe)rx/ρ(e)dHn1e)dρ. (3.4)

    From Eqs (1.2) and (1.3) we deduce that

    |xρQx/ρ(e)|=|rx/ρ(e)|=rx/ρ(e),

    and also (see formula (2.13.25) in [8])

    1|x/ρ|21|x/ρ|2(x/ρe)rx/ρ(e)=2rx/ρ+(e)rx/ρ+(e)rx/ρ(e)

    which, together with (3.4), gives that

    I=c(n,s)(1|x|2)s1ρ(ρ2|x|2)(ρ21)s(B12rx/ρ+(e)f(ρQx/ρ(e))rx/ρ+(e)rx/ρ(e)dHn1e)dρ. (3.5)

    We now observe that

    B12rx/ρ+(e)f(ρQx/ρ(e))rx/ρ+(e)rx/ρ(e)dHn1e=B1rx/ρ+(e)f(ρQx/ρ(e))rx/ρ+(e)rx/ρ(e)dHn1eB1rx/ρ(e)f(ρQx/ρ+(e)))rx/ρ+(e)rx/ρ(e)dHn1e=B1rx/ρ+(e)f(ρQx/ρ(e))rx/ρ(e)f(ρQx/ρ+(e))rx/ρ+(e)rx/ρ(e)dHn1e. (3.6)

    From (1.2), (1.3), (1.11) and (1.12) we also deduce that

    rx/ρ+(e)f(ρQx/ρ(e))rx/ρ(e)f(ρQx/ρ+(e))rx/ρ+(e)rx/ρ(e)=rx/ρ+(e)fρ(Qx/ρ(e))rx/ρ(e)fρ(Qx/ρ+(e))|Qx/ρ+(e)Qx/ρ(e)|=(Qx/ρ+(e)xρ)efρ(Qx/ρ(e))+(xρQx/ρ(e))efρ(Qx/ρ+(e))|Qx/ρ+(e)Qx/ρ(e)|=LQx/ρ(e),Qx/ρ+(e)fρ(xρ)=Lf,e,ρ(x).

    Plugging this information into (3.6) we find that

    B12rx/ρ+(e)f(ρQx/ρ(e))rx/ρ+(e)rx/ρ(e)dHn1e=B1Lf,e,ρ(x)dHn1e,

    which in turn, together with (3.5), gives that

    I=c(n,s)(1|x|2)s1ρ(ρ2|x|2)(ρ21)s(B1Lf,e,ρ(x)dHn1e)dρ.

    Thus, recalling (1.13), this and (3.3) establish the desired result in (3.1) under the additional assumption that f is continuous in RnB1.

    Now we remove the continuity assumption on f by an approximation argument. Given fL(BRB1)L1s(RnB1), we consider a sequence of functions {fk}kC(RnB1)L1s(RnB1) as in (2.6) and we let u(s)fk be the unique pointwise continuous solution to the problem (2.5), according to Theorem 2.1.

    By (2.7),

    u(s)fku(s)fLloc(B1)0. (3.7)

    Therefore we have that, for each xB1,

    u(s)f(x)=limk1(B1E(x,ρ)Lfk,e,ρ(x)dHn1e)dρ.

    From this, we claim that there exists a subsequence {fkj}j, such that

    u(s)f(x)=limj1(B1E(x,ρ)Lfkj,e,ρ(x)dHn1e)dρ=1(B1E(x,ρ)Lf,e,ρ(x)dHn1e)dρ. (3.8)

    For the convenience of the reader, the technical proof of (3.8) can be found in Appendix A.2.

    The claim in (3.8) establishes the desired result in (3.1) also for non continuous functions.

    To prove the claim in (3.2), we use (1.1) and (1.12) to see that

    Lf,e,ρ(x)=LQx/ρ(e),Qx/ρ+(e)fρ(xρ)=(xρQx/ρ(e))e|Qx/ρ+(e)Qx/ρ(e)|fρ(Qx/ρ+(e))+(Qx/ρ+(e)xρ)e|Qx/ρ+(e)Qx/ρ(e)|fρ(Qx/ρ(e))=(xρQx/ρ(e))e|Qx/ρ+(e)Qx/ρ(e)|f(ρQx/ρ+(e))+(Qx/ρ+(e)xρ)e|Qx/ρ+(e)Qx/ρ(e)|f(ρQx/ρ(e))=ργ[(xρQx/ρ(e))e|Qx/ρ+(e)Qx/ρ(e)|f(Qx/ρ+(e))+(Qx/ρ+(e)xρ)e|Qx/ρ+(e)Qx/ρ(e)|f(Qx/ρ(e))]=ργLQx/ρ(e),Qx/ρ+(e)f(xρ)=ργLf,e,1(xρ).

    The claim in (3.2) then follows from this and (3.1). This concludes the proof of Theorem 3.2.

    Proof of Theorem 1.4. From Proposition 2.2, we know that, under the hypotheses of Theorem 1.4, the function defined in (2.4) is the unique solution (up to a zero measure subset of RnB1) to the problem (1.16). Then, the desired result in Theorem 1.4 follows from this and Theorem 3.2.

    We now give the proof of the Schwarz result in Theorem 1.8.

    Proof of Theorem 1.8. We first suppose that fC(R2B1)L1s(R2B1). Under these assumptions we can apply Theorem 1.6 in dimension 2 and get

    u(s)f(x)=2π1E(x,ρ)˜ufρ(xρ)dρ,

    where ˜ufρ has been defined in the statement of Theorem 1.6. Therefore, when n=2 we can apply Theorem 1.2 to ˜ufρ and get

    ˜ufρ(xρ)=12πB1fρ(Qx/ρ(e))dHn1e,

    which leads to (1.21) in the case in which fC(R2B1).

    Suppose now that fL(BRB1)L1s(R2B1), and consider a sequence of functions {fk}kC(R2B1)L1s(R2B1) as in (2.6). We let u(s)fk be the unique pointwise continuous solution to the problem (2.5), according to Theorem 2.1. From the previous step, we have that, for each xB1,

    u(s)fk(x)=1E(x,ρ)(B1fk(ρQx/ρ(e))dHn1e)dρ.

    By this and (3.7), we have that, for each xB1,

    u(s)f(x)=limk1(B1E(x,ρ)fk(ρQx/ρ(e))dHn1e)dρ.

    From this, one sees that there exists a subsequence {fkj}j such that

    u(s)f(x)=limj1(B1E(x,ρ)fkj(ρQx/ρ(e))dHn1e)dρ=1(B1E(x,ρ)f(ρQx/ρ(e))dHn1e)dρ. (3.9)

    For the facility of the reader, a detailed proof of (3.9) is given in Appendix A.3.

    We observe that the proof of Theorem 1.8 is completed, thanks to (3.9).

    We now deal with the convergence result in Proposition 1.5.

    Proof of Proposition 1.5. Let fC(BRB1)L1s(RnB1), for each s(s0,1]. Furthermore let u(s)f and uf as in the statement of Proposition 1.5. Then, Theorem 1.1 implies that the following identity holds for each ρ(1,R)

    ufρ(xρ)=B1LQx/ρ(e),Qx/ρ+(e)fρ(xρ)dHn1e=B1Lf,e,ρ(x)dHn1e, (3.10)

    thanks to (1.12), where ufρ is the unique solution to the classical Dirichlet problem (1.18).

    In particular, we have that

    uf(xρ)=B1LQx/ρ(e),Qx/ρ+(e)f(xρ)dHn1e=B1Lf,e,1(xρ)dHn1e. (3.11)

    Now, using (1.17), (1.22) and (3.10) we obtain for each xB1 and R0(1,R) the following identity

    u(s)f(x)uf(x)=1|B1|E(x,ρ)(B1Lf,e,ρ(x)dHn1euf(x))dρ=R01|B1|E(x,ρ)(ufρ(xρ)uf(x))dρ+R0|B1|E(x,ρ)(B1Lf,e,ρ(x)dHn1euf(x))dρ. (3.12)

    By the continuity of uf and f, we have that for each δ>0 there exists some R0(1,R) such that for each ρ(1,R0)

    |uf(xρ)uf(x)|δ for all xB1 and fρfL(B1)δ. (3.13)

    Also, we point out that

    Lfρ,e,1(xρ)=LQx/ρ(e),Qx/ρ+(e)fρ(xρ)=Lf,e,ρ(x),

    thanks to (1.12).

    Therefore, from this, (3.10), (3.11), and (3.13), for all xB1, we deduce that, if ρ(1,R0),

    |ufρ(xρ)uf(x)||ufρ(xρ)uf(xρ)|+|uf(xρ)uf(x)|=|B1Lf,e,ρ(x)Lf,e,1(xρ)dHn1e|+|uf(xρ)uf(x)|=|B1Lfρf,e,1(xρ)dHn1e|+|uf(xρ)uf(x)|B1|Lfρf,e,1(xρ)|dHn1e+δ. (3.14)

    Now, if xB1 and ρ(1,R0), we see that

    |Qx/ρ+(e)Qx/ρ(e)|=2(xρe)2|x|2ρ2+121|x|2ρ2=2ρρ2|x|22R01|x|2 (3.15)

    and thus, according to (1.1),

    |Lfρf,e,1(xρ)|=|(xρQx/ρ(e))e|Qx/ρ+(e)Qx/ρ(e)|(fρf)(Qx/ρ+(e))+(Qx/ρ+(e)xρ)e|Qx/ρ+(e)Qx/ρ(e)|(fρf)(Qx/ρ(e))|4R021|x|2fρfL(B1)2R0δ1|x|2, (3.16)

    and therefore, plugging this information into (3.14), we obtain that, if xB1 and ρ(1,R0),

    |ufρ(xρ)uf(x)|(2R01|x|2+1)δ. (3.17)

    Furthermore, employing the change of variable e:=ω/|ω| and recalling (1.13),

    R0E(x,ρ)(B1Lf,e,ρ(x)dHn1e|B1|uf(x))dρ=R0E(x,ρ)ρn1(BρLf,ω/|ω|,|ω|(x)dHn1ω|B1|uf(x))dρ=c(n,s)(1|x|2)sRnBR0|y|2(|y|21)s(|y|2|x|2)|y|n(Lf,y/|y|,|y|(x)|B1|uf(x))dy. (3.18)

    We also deduce from (1.1) the following pointwise estimate

    |Lf,y/|y|,|y|(x)|=|(x|y|Qx/|y|(y|y|))y|y||Qx/|y|+(y|y|)Qx/|y|(y|y|)|f|y|(Qx/|y|(y|y|))+(Qx/|y|+(y|y|)x|y|)y|y||Qx/|y|+(y|y|)Qx/|y|(y|y|)|f|y|(Qx/|y|(y|y|))|11|x|2(|f|y|(Qx/|y|(y|y|))|+|f|y|(Qx/|y|+(y|y|))|). (3.19)

    We claim that

    the right hand side in (3.19) is L1s(RnB1) for each s(s0,1]. (3.20)

    Indeed, if we define the following function

    F±:RnB1RnB1y|y|Qx/|y|±(y/|y|), (3.21)

    we see that it is C1 and invertible. Note that, recalling also the limits in (1.4), one finds that

    det(DF1±(z)L(RnB1)C, (3.22)

    for some positive constant C, depending on x. Therefore by applying Theorem 2 in Section 3.3.3 of [10], we obtain that

    RnB1|f|y|(Qx/|y|±(y/|y|))||y|n+2sdy=RnB1|f|y|(Qx/|y|±(y/|y|))|||y|Qx/|y|±(y/|y|)|n+2sdy=RnB1|f(z)||z|n+2s|det(DF1±(z))|dzCRnB1|f(z)||z|n+2sdz. (3.23)

    This and the fact that fL1s(RnB1) for each s(s0,1] entail that (3.20) holds true.

    As a consequence of (3.19) and (3.20) we have that the integrals in (3.18) are finite and bounded in s.

    Using this information and (3.17), we deduce from (3.12) that for each δ>0 there exists some R0(1,R) such that for each ρ(1,R0) we have

    |u(s)f(x)uf(x)|R01|B1|E(x,ρ)|ufρ(xρ)uf(x)|dρ+|R0|B1|E(x,ρ)(B1Lf,e,ρ(x)dHn1euf(x))dρ|R01|B1|E(x,ρ)(2R01|x|2+1)δdρ+|R0|B1|E(x,ρ)(B1Lf,e,ρ(x)dHn1euf(x))dρ|C(x,R)δ+c(n,s)(1|x|2)sRnBR0|y|2(|y|21)s(|y|2|x|2)|y|n(Lf,y/|y|,|y|(x)|B1|uf(x))dyC(x,R)δ+c(n,s)(1|x|2)sRnBR0|y|2(|y|21)s(|y|2|x|2)|y|n×(11|x|2(|f|y|(Qx/|y|(y|y|))|+|f|y|(Qx/|y|+(y|y|))|)|B1|uf(x))dy

    where C(x,R0) depends only on x and R0.

    By taking the limit as s1, we see that

    lims1c(n,s)(1|x|2)sRnBR0|y|2(|y|21)s(|y|2|x|2)|y|n×(11|x|2(|f|y|(Qx/|y|(y|y|))|+|f|y|(Qx/|y|+(y|y|))|)|B1|uf(x))dy=0

    since c(n,s)0 for s1 by (1.14). As a consequence

    lims1|u(s)f(x)uf(x)|C(x,R)δ.

    This and the arbitrariness of δ give the desired claim in Proposition 1.5.

    In this section we provide a new proof of the Harnack inequality for s-harmonic functions as stated in Theorem 1.7. Our strategy is to use the Fractional Malmheden Theorem to show that this result can be directly inferred from the classical Harnack inequality for harmonic functions.

    Proof of Theorem 1.7. For convenience we call u|RnB1=f. Let us first assume that fC(RnB1)L1s(RnB1). Under this assumption, we can apply Theorem 1.6 and obtain that

    u(x)=|B1|1E(x,ρ)ufρ(xρ)dρ,

    for each xB1, where ufρ has been defined in the statement of Theorem 1.6. Therefore, we have that

    u(0)=c(n,s)|B1|1ufρ(0)ρ(ρ21)sdρ. (4.1)

    Now we fix r(0,1) and we consider xBr. Applying the Harnack inequality for classical harmonic functions to ufρ, we have that

    ufρ(0)(1+|x|/ρ)n11|x|/ρufρ(xρ).

    From this, (1.13) and (4.1) we obtain that

    u(0)c(n,s)|B1|11ρ(ρ21)s(1+|x|/ρ)n11|x|/ρufρ(xρ)dρ=|B1|1E(x,ρ)(ρ2|x|2)ρ2(1|x|2)s(ρ+|x|)n1ρn2(ρ|x|)ufρ(xρ)dρ=|B1|1E(x,ρ)(ρ+|x|)nρn(1|x|2)sufρ(xρ)dρ=|B1|1E(x,ρ)g(ρ,t)ufρ(xρ)dρ, (4.2)

    where for convenience we have called t:=|x| in the last line and defined

    g(ρ,t):=(ρ+t)nρn(1t2)s,

    with (ρ,t)[1,)×[0,r].

    Since g(ρ,t) is decreasing in ρ and increasing in t, we have that

    (1+r)n(1r2)s=sup(ρ,t)[1,)×[0,r]g(ρ,t).

    Therefore, from this, (1.19) and (4.2) we obtain that

    u(0)|B1|(1+r)n(1r2)s1E(x,ρ)ufρ(xρ)dρ=(1+r)n(1r2)su(x),

    which establishes the first inequality in (1.20).

    To prove the second inequality in (1.20), we make use of the Harnack inequality for harmonic functions, thus obtaining that

    ufρ(xρ)1+|x|/ρ(1|x|/ρ)n1ufρ(0).

    Using this and (1.13) into (4.1), we find that

    u(0)=c(n,s)|B1|1ufρ(0)ρ(ρ21)sdρc(n,s)|B1|11ρ(ρ21)s(1|x|/ρ)n11+|x|/ρufρ(xρ)dρ=c(n,s)|B1|1(ρ|x|)n1ρn1(ρ21)s(ρ+|x|)ufρ(xρ)dρ=|B1|1E(x,ρ)(ρ|x|)nρn(1|x|2)sufρ(xρ)dρ.

    Using again the notation t:=|x|, we define the following function

    g1(ρ,t):=(ρt)nρn(1t2)s, (4.3)

    with (ρ,t)[1,)×[0,r], and we see that

    u(0)|B1|1E(x,ρ)g1(ρ,t)ufρ(xρ)dρ. (4.4)

    Since g1 is increasing in ρ, we have that, for all (ρ,t)[1,)×[0,r],

    g1(ρ,t)g1(1,t)=(1t)n(1t2)s=(1t)ns(1+t)s=:g2(t).

    Notice also that g2 is decreasing, and therefore, for all (ρ,t)[1,)×[0,r],

    g1(ρ,t)g2(r)=(1r)ns(1+r)s=(1r)n(1r2)s.

    Plugging this information into (4.4) and recalling (1.19), we get

    u(0)|B1|(1r)n(1r2)s1E(x,ρ)ufρ(xρ)dρ(1r)n(1r2)su(x),

    which completes the proof of (1.20) under the additional continuity assumption on f.

    To deal with the general case, we perform an approximation argument, whose details go as follows. If fL(BRB1)L1s(RnB1), we take a sequence {fk}kC(RnB1)L1s(RnB1) as in (2.6). Then for u(s)fk the two-sided inequality in (1.20) holds true, thanks to the previous step. Also, by (3.7), we have the local uniform convergence

    u(s)fkuLloc(B1)0 as k+,

    and therefore we deduce the two sided inequality (1.20) also in this case.

    It is only left to prove that the constants provided in equation (1.20) are optimal. To show this let us fix some direction eB1, a constant ϵ>0 and the function

    fϵ(y):={0ifyRnBϵ((1+ϵ)e),(|y|21)sifyBϵ((1+ϵ)e).

    Then the function

    u(s)fϵ(x):=c(n,s)Bϵ((ϵ+1)e)(1|x|2)s|yx|ndy

    is s-harmonic in B1, as a consequence of Proposition 2.2. Therefore, if we fix x=re for r(0,1), we have that

    u(s)fϵ(0)u(s)fϵ(re)=Bϵ((ϵ+1)e)dy|y|nBϵ((ϵ+1)e)(1r2)s|y+re|ndy,

    and thus, by Lebesgue Differentiation Theorem, we conclude that

    limϵ0u(s)fϵ(0)u(s)fϵ(re)=(1+r)n(1r2)s.

    This proves that the constant on the left hand side inequality in (1.20) is optimal.

    Similarly, taking x=re, one sees that

    limϵ0u(s)fϵ(re)u(s)fϵ(0)=(1r2)s(1r)n,

    which shows that the constant on the right hand side inequality in (1.20) is also optimal. This concludes the proof of Theorem 1.7.

    SD and EV are members of AustMS. Supported by the Australian Laureate Fellowship FL190100081 "Minimal surfaces, free boundaries and partial differential equations" and by the Australian Research Council DECRA DE180100957 "PDEs, free boundaries and applications".

    The authors declare no conflict of interest.

    Here we give a direct proof of the identity pointed out in the footnote on page 5. Namely, we establish that, if xB1 and ρ>1,

    E(x,ρ)=ρn1BρP(x,y)dHn1y, (A.1)

    being P the fractional Poisson kernel in (2.1).

    The identity in (A.1) has its own interest and it can be deduced from our fractional Malmheden theorem, by taking a datum f concentrating along a given sphere Bρ. For the sake of completeness however, we provide here an independent proof, only based on elementary computations and standard integral formulas.

    More specifically, we aim at showing that

    B1dHn1ω|xρω|n=ρ2nρ2|x|2. (A.2)

    Indeed, once (A.2) is established, we deduce from it, (1.13) and (2.1) that

    BρP(x,y)dHn1y=c(n,s)(1|x|2)sBρdHn1y(|y|21)s|xy|n=c(n,s)(1|x|2)s|B1|ρn1B1ρn1dHn1ω(ρ21)s|xρω|n=c(n,s)(1|x|2)s(ρ21)sB1dHn1ω|xρω|n=c(n,s)ρ2n(1|x|2)s(ρ21)s(ρ2|x|2)=ρ1nE(x,ρ)

    and this would complete the proof of (A.1).

    Hence, we focus now on proving (A.2). To this end, we use spherical coordinates on B1, with θ, θ1, , θn3[0,π] and θn2[0,2π), see e.g., Eq (A.23) in [3] or pages 60-61 in [14], which correspond to ω=ω(θ) of the form

    {ω1=sinθsinθ1sinθn3sinθn2,ω2=sinθsinθ1sinθn3cosθn2,ω3=sinθsinθ1cosθn3,ωn=cosθ

    and produce a surface element of the form

    sinn2θsinn3θ1sin2θn4sinθn3dθdθ1dθn2.

    Also, up to a rotation, to prove (A.2) we can assume that x=(0,,0,|x|). In this way, we find that

    |xρω|2=|x|2+ρ22ρxω=|x|2+ρ22ρ|x|ωn=|x|2+ρ22ρ|x|cosθ,

    whence it follows that

    B1dHn1ω|xρω|n=θ,θ1,,θn3[0,π]θn2[0,2π)sinn2θsinn3θ1sin2θn4sinθn3dθdθ1dθn2(|x|2+ρ22ρ|x|cosθ)n2=2π(n3j=1π0sinnj2θjdθj)θ[0,π]sinn2θdθ(|x|2+ρ22ρ|x|cosθ)n2.

    Thus, we use the notation τ:=ρ|x|(1,+) and, by Proposition A.9 in [3], we deduce that

    B1dHn1ω|xρω|n=2π|x|n(n3k=1π0sinkϑdϑ)π0sinn2θdθ(τ2+12τcosθ)n2=2π|x|nτn2(τ21)(n3k=1π0sinkϑdϑ)π0sinn2αdα=2π|x|nτn2(τ21)(n2k=1π0sinkϑdϑ).

    This and Proposition A.10 in [3] yield that

    B1dHn1ω|xρω|n=2πn2|x|nτn2(τ21)Γ(n2)=|B1||x|nτn2(τ21)=|B1|ρ2nρ2|x|2.

    The proof of (A.2) is thereby complete.

    We recall (1.13) and we employ the change of variable e:=ω/|ω| to see that

    1c(n,s)(1|x|2)s1(B1E(x,ρ)Lfk,e,ρ(x)dHn1e)dρ=1ρ(ρ21)s(ρ2|x|2)(B1Lfk,e,ρ(x)dHn1e)dρ=1ρ2ρn(ρ21)s(ρ2|x|2)(BρLfk,ω/|ω|,|ω|(x)dHn1ω)dρ=RnB1|y|2|y|n(|y|21)s(|y|2|x|2)Lfk,y/|y|,|y|(x)dy. (A.3)

    It also follows from (2.6) that, for a.e. yRnB1,

    Lfk,y/|y|,|y|(x)Lf,y/|y|,|y|(x) as k+. (A.4)

    Now we take R0(1,R) and we deduce from (A.3) that

    1c(n,s)(1|x|2)s1(B1E(x,ρ)Lfk,e,ρ(x)dHn1e)dρ=BR0B1|y|2|y|n(|y|21)s(|y|2|x|2)Lfk,y/|y|,|y|(x)dy+RnBR0|y|2|y|n(|y|21)s(|y|2|x|2)Lfk,y/|y|,|y|(x)dy. (A.5)

    Recalling the computation in (3.16), for k large enough we have that

    Lfk,y/|y|,|y|(x)L(BR0B1)2R01|x|2fL(BRB1).

    Consequently, using this, (A.4) and the dominated convergence theorem,

    limk+BR0B1|y|2|y|n(|y|21)s(|y|2|x|2)Lfk,y/|y|,|y|(x)dy=BR0B1|y|2|y|n(|y|21)s(|y|2|x|2)Lf,y/|y|,|y|(x)dy. (A.6)

    Also, we claim that there exists a subsequence {fkj}j such that

    Lfkj,y/|y|,|y|(x)Lf,y/|y|,|y|(x)L1s(RnBR0)0 as j+. (A.7)

    To show (A.7), we recall (3.15) and we observe that, for every xB1 and yRnBR0,

    |Lfk,y/|y|,|y|(x)|=|(x|y|Qx/|y|(y|y|))y|y||Qx/|y|+(y|y|)Qx/|y|(y|y|)|fk(|y|Qx/|y|+(y|y|))+(Qx/|y|+(y|y|)x|y|)y|y||Qx/|y|+(y|y|)Qx/|y|(y|y|)|fk(Qx/|y|(y|y|))|R01|x|2[|fk(|y|Qx/|y|(y|y|))|+|fk(|y|Qx/|y|(y|y|))|]. (A.8)

    Moreover, by (2.6) there exists a subsequence {fkj}j and a function hL1s(RnB1) such that |fkj(y)|h(y) for a.e. yRnB1 (see for instance Theorem 4.9 in [2]). Therefore, using this information into (A.8), we have

    |Lfkj,y/|y|,|y|(x)|R01|x|2[h(|y|Qx/|y|(y|y|))+h(|y|Qx/|y|+(y|y|))] (A.9)

    for a.e. yRnBR0.

    Now we recall the map F± defined in (3.21), which is C1 and invertible, and therefore, by Theorem 2 in Section 3.3.3 of [10] and (3.22), we get that

    RnBR0h(|y|Qx/|y|±(y/|y|))|y|n+2sdy=RnBR0h(|y|Qx/|y|±(y/|y|))||y|Qx/|y|±(y/|y|)|n+2sdy=RnBR0h(z)|z|n+2s|det(DF1±(z))|dzCRnBR0h(z)|z|n+2sdz.

    Accordingly, we deduce that

    h(|y|Qx/|y|±(y/|y|))L1s(RnBR0).

    This, the bound in (A.9), the pointwise convergence in (A.4) and the dominated convergence theorem lead to (A.7), as desired.

    Hence, putting together (A.5)–(A.7), we obtain that

    uf(x)=1B1E(x,ρ)Lf,e,ρ(x)dHn1edρ

    for each xB1, which completes the proof of (3.8).

    The proof of (3.9) is similar to the one of (3.8). We provide here the details for the convenience of the reader.

    From (1.13) with n=2 and the change of variable e:=ω/|ω|,

    1c(2,s)(1|x|2)s1(B1E(x,ρ)fk(ρQx/ρ(e))dHn1e)dρ=1ρ(ρ2|x|2)(ρ21)s(B1fk(ρQx/ρ(e))dHn1e)dρ=11(ρ2|x|2)(ρ21)s(Bρfk(|ω|Qx/|ω|(ω/|ω|))dHn1ω)dρ=RnB11(|y|2|x|2)(|y|21)sfk(|y|Qx/|y|(y/|y|))dy (A.10)

    By (2.6), we have that, for a.e. yRnB1,

    fk(|y|Qx/|y|(y/|y|))f(|y|Qx/|y|(y/|y|)) as k+. (A.11)

    Now we take R0(1,R) and we get from (A.10) that

    1c(2,s)(1|x|2)s1(B1E(x,ρ)fk(ρQx/ρ(e))dHn1e)dρ=BR0B11(|y|2|x|2)(|y|21)sfk(|y|Qx/|y|(y/|y|))dy+RnBR01(|y|2|x|2)(|y|21)sfk(|y|Qx/|y|(y/|y|))dy (A.12)

    Notice that, for k large enough,

    fkL(BR0B1)fL(BRB1).

    As a consequence,

    limk+BR0B11(|y|2|x|2)(|y|21)sfk(|y|Qx/|y|(y/|y|))dy=BR0B11(|y|2|x|2)(|y|21)sf(|y|Qx/|y|(y/|y|))dy (A.13)

    Furthemore, recalling (2.6) (see also Theorem 4.9 in [2]) we deduce the existence of a subsequence {fkj}j and of a function hL1s(RnB1) such that

    |fkj(|y|Qx/|y|(y|y|))|h(|y|Qx/|y|(y/|y|)) (A.14)

    for a.e. yRnBR0.

    Furthermore, we claim that

    h(|y|Qx/|y|(y/|y|)) belongs to L1s(RnBR0). (A.15)

    Indeed, the function

    F:RnBR0RnBR0y|y|Qx/|y|(y/|y|)

    is C1 and invertible. Moreover, since

    lim|y|Qx/|y|(y/|y|)=idB1,

    we find that

    det(DF1(z))L(RnBR0)˜C,

    for some positive constant ˜C>0.

    From this and Theorem 2 in Section 3.3.3 of [10] we have that

    RnBR0h(|y|Qx/|y|(y/|y|))|y|2+2sdy=RnBR0h(|y|Qx/|y|(y/|y|))||y|Qx/|y|(y/|y|)|2+2sdy=RnBR0h(z)|z|2+2s|det(DF1(z))|dz˜CRnBR0h(z)|z|2+2sdz,

    which establishes (A.15).

    From (A.11), (A.14) and (A.15) and the Dominated Convergence Theorem, we deduce that

    fkj(|y|Qx/|y|(y|y|))f(|y|Qx/|y|(y|y|))L1s(RnBR0)0 as j+.

    Gathering together this, (A.12) and (A.13), we conclude that

    u(s)f(x)=1(B1E(x,ρ)f(ρQx/ρ(e))dHn1e)dρ.

    This finishes the proof of (3.9).



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