
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.
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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 K⊆Rn, a function f:K→R, two distinct points a, b∈K, 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)=(x−a)⋅e|b−a|f(b)+(b−x)⋅e|b−a|f(a),where e:=b−a|b−a|. | (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 x∈B1 and a direction e∈∂B1 we consider the intersections Qx+(e) and Qx−(e) of ∂B1 with the straight line passing through x∈B1 with direction e, with the convention that Qx+(e)−Qx−(e) has the same orientation of e, see Figure 1.
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)=−x⋅e+√(x⋅e)2−|x|2+1and rx−(e)=−x⋅e−√(x⋅e)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,
limx→0Qx±(e)=±e | (1.4) |
for each e∈∂B1. This tells us that the maps Qx± simply reduce to ±id∂B1 when x=0.
Given a boundary datum f:∂B1→R, the core of the Malmheden Theorem is thus to consider, for every point x∈B1 and every direction e∈∂B1, 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 n⩾2 and f:∂B1→R be continuous. Then
uf(x):=−∫∂B1LQx−(e),Qx+(e)f(x)dHn−1e | (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 Hn−1 the (n−1)-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)=−∫∂B1L−e,ef(0)dHn−1e=−∫∂B1(f(e)2+f(−e)2)dHn−1e=−∫∂B1f(e)dHn−1e, | (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 ω∈∂B1 through a point x∈B1, see Figure 2. More precisely, given x∈B1 and ω∈∂B1 we define
Qx(ω):=ω−2(x−ω)⋅ω|x−ω|2(x−ω). | (1.8) |
Comparing with (1.2), one sees that if e:=ω−x|ω−x| 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:∂B1→R be continuous. Then
uf(x):=−∫∂B1f(Qx(e))dHn−1e | (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, |
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 x∈B1,
∫Rnu(x)−u(y)|x−y|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 Rn∖B1.
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:Rn∖B1→R, for all x∈∂B1 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(ρ2−1)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 x∈B1 and ρ∈(1,+∞),
E(x,ρ)=ρn−1−∫∂BρP(x,y)dHn−1y, |
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(Rn∖B1):={f:Rn→R measurable :∫Rn∖B1|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 n⩾2, s∈(0,1), R>1and f∈L∞(BR∖B1)∩L1s(Rn∖B1).
Then, the unique solution (up to a zero measure subset of Rn∖B1) to the problem
{(−Δ)su=0inB1,u=finRn∖B1 | (1.16) |
can be written, for each x∈B1, as
u(s)f(x):=∫∞1(∫∂B1E(x,ρ)Lf,e,ρ(x)dHn−1e)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)dHn−1e)dρ=c(n,s)∫∞1(∫∂B11ρ(ρ2−1)sL−e,efρ(0)dHn−1e)dρ=c(n,s)∫∞1(∫∂B11ρ(ρ2−1)s(f(ρe)2+f(−ρe)2)dHn−1e)dρ=c(n,s)∫∞1(∫∂Bρ1ρn(ρ2−1)s(f(ω)2+f(−ω)2)dHn−1ω)dρ=c(n,s)∫Rn∖B11|y|n(|y|2−1)s(f(y)2+f(−y)2)dy=c(n,s)∫Rn∖B1f(y)|y|n(|y|2−1)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 s↗1, according to the following result:
Proposition 1.5. Let n⩾2, s0∈(0,1), R>1and f∈C(BR∖B1)∩L1s(Rn∖B1) for each s∈(s0,1]. Then, for each x∈B1, it holds that
lims↗1u(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 n⩾2, s∈(0,1) and f∈C(Rn∖B1)∩L1s(Rn∖B1). 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 n⩾2, s∈(0,1), R>1 and u be non negative, s-harmonic in B1 and such that u∈L∞(BR∖B1)∩L1s(Rn∖B1).
Then, for each r∈(0,1) and x∈Br,
(1−r2)s(1+r)nu(0)⩽u(x)⩽(1−r2)s(1−r)nu(0). | (1.20) |
The constants in (1.20) are optimal, and for s↗1 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 f∈L∞(BR∖B1)∩L1s(R2∖B1).Then, the unique solution (up to a zero measure subset of R2∖B1) to the problem
{(−Δ)su=0inB1,u=finR2∖B1 |
can be written, for each x∈B1, 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 f∈C(BR∖B1)∩L1s(R2∖B1) for each s∈(s0,1]. Then, for each x∈B1, it holds that
lims↗1u(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 x∈B1. Indeed, if we consider as external data f=1 in Rn∖B1, 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 x∈B1, we obtain that
1=∫∞1(∫∂B1E(x,ρ)L1,e,ρ(x)dHn−1e)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 Σ⊂∂B1. Consider the function defined on R2∖B1 as
˜χΣ(y):={1ify|y|∈Σ,0ify|y|∈∂B1∖Σ. | (1.23) |
It is clear that ˜χΣ is positively homogeneous of degree zero, and furthermore ˜χΣ∈L∞(R2∖B1)⊂L1s(R2∖B1). Then by Theorem 1.8 we get that for each x∈B1
u(s)˜χΣ(x)=∫∞1E(x,ρ)|Σ′x/ρ|dρ | (1.24) |
where Σ′x/ρ is the projected arc of Σ on ∂B1 through the focal point x/ρ, as constructed in Example 1.3. We denoted with |Σ′x/ρ| its length.
This gives a simple geometrical procedure to compute the solution of
{(−Δ)su=0inB1,u=˜χΣinR2∖B1 |
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 Σ′x/ρ reaches the one of Σ, or more precisely
limρ→∞|Σ′x/ρ|=|Σ|. |
If x=0, formula (1.24) boils down to
u(s)˜χΣ(0)=c(n,s)|Σ|∫∞11ρ(ρ2−1)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:Rn∖B1→R, 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 n⩾2 and s∈(0,1), we consider the space L1s(Rn∖B1) as defined in (1.15). Given f∈L1s(Rn∖B1), we denote the norm on L1s(Rn∖B1) by
‖f‖L1s(Rn∖B1):=∫Rn∖B1|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|2−1)s1|x−y|n | (2.1) |
for x∈B1 and y∈Rn∖B1, 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
∫Rn∖B1P(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), α∈Nn and y∈Rn∖B1,
supx∈Bρ|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):={∫Rn∖B1P(x,y)f(y)dyifx∈B1,f(x)ifx∈Rn∖B1, | (2.4) |
and we have the following result on the representation of s-harmonic functions:
Theorem 2.1. Let n⩾2, s∈(0,1) and f∈C(Rn∖B1)∩L1s(Rn∖B1). Then the function in (2.4) is the unique pointwise continuous solution to the problem
{(−Δ)su(s)f=0inB1,u(s)f=finRn∖B1. | (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 n⩾2, s∈(0,1), R>1 and f∈L∞(BR∖B1)∩L1s(Rn∖B1). Then the function defined in (2.4) is the unique solution (up to a zero measure subset of Rn∖B1) 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 x∈B1 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}k⊂C(Rn∖B1)∩L1s(Rn∖B1), such that
fk→f in L1s(Rn∖B1) as k→+∞. | (2.6) |
More specifically, we take fk:=(χBk˜f)∗η1k with k⩾2, where ˜f is defined as
˜f(x):={f(x)ifx∈Rn∖B1,0ifx∈B1, |
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)fk−Dαu(s)f‖L∞loc(B1)→0 as k→+∞. | (2.7) |
Indeed for each multi-index α and g∈L∞(BR∖B1)∩L1s(Rn∖B1) one finds that
Dαug(x)=∫Rn∖B1DαxP(x,y)g(y)dy |
for each x∈B1, and therefore, choosing R0∈(1,R), we see that, for every x∈B′ with B′⋐B1,
|Dαu(s)fk(x)−Dαu(s)f(x)|⩽∫Rn∖BR0|DαxP(x,y)||fk(y)−f(y)|dy+∫BR0∖B1|DαxP(x,y)||fk(y)−f(y)|dy⩽c∫Rn∖BR0|fk(y)−f(y)||y|n+2sdy+∫BR0∖B1|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 y∈BR0∖B1, then
|fk(y)−f(y)|⩽2‖f‖L∞(BR∖B1) |
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 x∈B1,
|u(s)fk(x)|⩽∫Rn∖BR0|P(x,y)||fk(y)|dy+∫BR0∖B1|P(x,y)||fk(y)|dy⩽C∫Rn∖BR0|fk(y)||y|n+2sdy+∫BR0∖B1|P(x,y)||fk(y)|dy⩽C‖fk‖L1s(Rn∖B1)+‖f‖L∞(BR∖B1), | (2.9) |
where C is a positive constant depending on R0, n and s. Now, we observe that the sequence ‖fk‖L1s(Rn∖B1) 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 x∈B1, 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)|x−y|n+2sdy=A+B+C+D+E+F, | (2.11) |
where
A:=∫Rn∖B1u(s)fk(x)−u(s)f(x)|x−y|n+2sdy,B:=∫Rn∖B1u(s)fk(y)−u(s)f(y)|x−y|n+2sdy,C:=∫Bδ(x)u(s)fk(x)−u(s)fk(y)|x−y|n+2sdy,D:=∫Bδ(x)u(s)f(y)−u(s)f(x)|x−y|n+2sdy,E:=∫B1∖Bδ(x)u(s)fk(x)−u(s)f(x)|x−y|n+2sdy and F:=∫B1∖Bδ(x)u(s)f(y)−u(s)fk(y)|x−y|n+2sdy. |
Notice that
|A+E|⩽∫Rn∖Bδ(x)|u(s)fk(x)−u(s)f(x)||x−y|n+2sdy⩽|u(s)fk(x)−u(s)f(x)|∫Rn∖Bδdz|z|n+2s⩽Cδ2s|u(s)fk(x)−u(s)f(x)|, |
which converges to zero as k→+∞, thanks to (2.7).
Furthermore, we observe that if y∈Rn∖B1 then
|x−y|⩾|y|−|x|=δ|y|+(1−δ)|y|−|x|⩾δ|y|+1−δ−|x|⩾δ|y|, |
and thus
|B|⩽∫Rn∖B1|fk(y)−f(y)||x−y|n+2sdy⩽1δn+2s∫Rn∖B1|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 x∈B1. Since, by Theorem 2.1, we know that (−Δ)su(s)fk(x)=0 for each x∈B1, 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:Rn→R satisfying
{(−Δ)su1=0inB1,u1=finRn∖B1. |
Then both v:=u(s)f−u1 and −v=u1−u(s)f are solutions to
{(−Δ)su=0inB1,u=0inRn∖B1, |
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 n⩾2, x∈B1 and Qx± and rx±be defined as in (1.2) and (1.3), respectively. Then for each e∈∂B1 it holdsthat
|detDQx±(e)|=(±rx±(e))n1−|x|2−rx±(e)x⋅e, |
and for each continuous f:∂B1→R we have that
∫∂B1f(e)dHn−1e=∫∂B1f(Qx±(e))(±rx±(e))n1−|x|2−rx±(e)x⋅edHn−1e. |
With this notation, Theorem 1.4 will be a consequence of the following statement.
Theorem 3.2. Let n⩾2, s∈(0,1), R>1and f∈L∞(BR∖B1)∩L1s(Rn∖B1).Let u(s)f be as in (2.4).
Then, for each x∈B1,
u(s)f(x)=∫∞1(∫∂B1E(x,ρ)Lf,e,ρ(x)dHn−1e)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 γ⩾0, then we have that
u(s)f(x)=∫∞1(∫∂B1ργE(x,ρ)Lf,e,1(xρ)dHn−1e)dρ. | (3.2) |
Proof. We first suppose that f∈C(Rn∖B1)∩L1s(Rn∖B1). Let x∈B1, then, using polar coordinates, from (2.1) and (2.4) we get that
u(s)f(x)=∫Rn∖B1P(x,y)f(y)dy=c(n,s)(1−|x|2)s∫Rn∖B11(|y|2−1)sf(y)|x−y|ndy=c(n,s)(1−|x|2)s∫∞11(ρ2−1)s(∫∂Bρf(ω)|x−ω|ndHn−1ω)dρ=c(n,s)(1−|x|2)s∫∞1ρn−1(ρ2−1)s(∫∂B1f(ρe)|x−ρe|ndHn−1e)dρ=c(n,s)(1−|x|2)s∫∞11ρ(ρ2−1)s(∫∂B1f(ρe)|xρ−e|ndHn−1e)dρ=c(n,s)(1−|x|2)s∫∞1ρ(ρ2−|x|2)(ρ2−1)s(∫∂B1f(ρe)|xρ−e|n(1−|x|2ρ2)dHn−1e)dρ=:I. | (3.3) |
Hence, defining
g(e):=f(ρe)|xρ−e|n(1−|x|2ρ2) |
and applying Lemma 3.1 we obtain that
Ic(n,s)(1−|x|2)s=∫∞1ρ(ρ2−|x|2)(ρ2−1)s(∫∂B1g(e)dHn−1e)dρ=∫∞1ρ(ρ2−|x|2)(ρ2−1)s(∫∂B1g(Qx/ρ−(e))(−rx/ρ−(e))n1−|x/ρ|2−(x/ρ⋅e)rx/ρ−(e)dHn−1e)dρ=∫∞1ρ(ρ2−|x|2)(ρ2−1)s(∫∂B1f(ρQx/ρ−(e))|xρ−Qx/ρ−(e)|n(1−|x/ρ|2)(−rx/ρ−(e))n1−|x/ρ|2−(x/ρ⋅e)rx/ρ−(e)dHn−1e)dρ. | (3.4) |
From Eqs (1.2) and (1.3) we deduce that
|xρ−Qx/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/ρ⋅e)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)s∫∞1ρ(ρ2−|x|2)(ρ2−1)s(∫∂B12rx/ρ+(e)f(ρQx/ρ−(e))rx/ρ+(e)−rx/ρ−(e)dHn−1e)dρ. | (3.5) |
We now observe that
∫∂B12rx/ρ+(e)f(ρQx/ρ−(e))rx/ρ+(e)−rx/ρ−(e)dHn−1e=∫∂B1rx/ρ+(e)f(ρQx/ρ−(e))rx/ρ+(e)−rx/ρ−(e)dHn−1e−∫∂B1rx/ρ−(e)f(ρQx/ρ+(e)))rx/ρ+(e)−rx/ρ−(e)dHn−1e=∫∂B1rx/ρ+(e)f(ρQx/ρ−(e))−rx/ρ−(e)f(ρQx/ρ+(e))rx/ρ+(e)−rx/ρ−(e)dHn−1e. | (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))+(xρ−Qx/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)dHn−1e=∫∂B1Lf,e,ρ(x)dHn−1e, |
which in turn, together with (3.5), gives that
I=c(n,s)(1−|x|2)s∫∞1ρ(ρ2−|x|2)(ρ2−1)s(∫∂B1Lf,e,ρ(x)dHn−1e)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 Rn∖B1.
Now we remove the continuity assumption on f by an approximation argument. Given f∈L∞(BR∖B1)∩L1s(Rn∖B1), we consider a sequence of functions {fk}k⊂C(Rn∖B1)∩L1s(Rn∖B1) 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)fk−u(s)f‖L∞loc(B1)→0. | (3.7) |
Therefore we have that, for each x∈B1,
u(s)f(x)=limk→∞∫∞1(∫∂B1E(x,ρ)Lfk,e,ρ(x)dHn−1e)dρ. |
From this, we claim that there exists a subsequence {fkj}j, such that
u(s)f(x)=limj→∞∫∞1(∫∂B1E(x,ρ)Lfkj,e,ρ(x)dHn−1e)dρ=∫∞1(∫∂B1E(x,ρ)Lf,e,ρ(x)dHn−1e)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)=(xρ−Qx/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))=(xρ−Qx/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[(xρ−Qx/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 Rn∖B1) 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 f∈C(R2∖B1)∩L1s(R2∖B1). Under these assumptions we can apply Theorem 1.6 in dimension 2 and get
u(s)f(x)=2π∫∞1E(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)=12π∫∂B1fρ(Qx/ρ(e))dHn−1e, |
which leads to (1.21) in the case in which f∈C(R2∖B1).
Suppose now that f∈L∞(BR∖B1)∩L1s(R2∖B1), and consider a sequence of functions {fk}k⊂C(R2∖B1)∩L1s(R2∖B1) 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 x∈B1,
u(s)fk(x)=∫∞1E(x,ρ)(∫∂B1fk(ρQx/ρ(e))dHn−1e)dρ. |
By this and (3.7), we have that, for each x∈B1,
u(s)f(x)=limk→∞∫∞1(∫∂B1E(x,ρ)fk(ρQx/ρ(e))dHn−1e)dρ. |
From this, one sees that there exists a subsequence {fkj}j such that
u(s)f(x)=limj→∞∫∞1(∫∂B1E(x,ρ)fkj(ρQx/ρ(e))dHn−1e)dρ=∫∞1(∫∂B1E(x,ρ)f(ρQx/ρ(e))dHn−1e)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 f∈C(BR∖B1)∩L1s(Rn∖B1), 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)dHn−1e=−∫∂B1Lf,e,ρ(x)dHn−1e, | (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)dHn−1e=−∫∂B1Lf,e,1(xρ)dHn−1e. | (3.11) |
Now, using (1.17), (1.22) and (3.10) we obtain for each x∈B1 and R0∈(1,R) the following identity
u(s)f(x)−uf(x)=∫∞1|∂B1|E(x,ρ)(−∫∂B1Lf,e,ρ(x)dHn−1e−uf(x))dρ=∫R01|∂B1|E(x,ρ)(ufρ(xρ)−uf(x))dρ+∫∞R0|∂B1|E(x,ρ)(−∫∂B1Lf,e,ρ(x)dHn−1e−uf(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 x∈B1 and ‖fρ−f‖L∞(∂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 x∈B1, 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)dHn−1e|+|uf(xρ)−uf(x)|=|−∫∂B1Lfρ−f,e,1(xρ)dHn−1e|+|uf(xρ)−uf(x)|⩽−∫∂B1|Lfρ−f,e,1(xρ)|dHn−1e+δ. | (3.14) |
Now, if x∈B1 and ρ∈(1,R0), we see that
|Qx/ρ+(e)−Qx/ρ−(e)|=2√(xρ⋅e)2−|x|2ρ2+1⩾2√1−|x|2ρ2=2ρ√ρ2−|x|2⩾2R0√1−|x|2 | (3.15) |
and thus, according to (1.1),
|Lfρ−f,e,1(xρ)|=|(xρ−Qx/ρ−(e))⋅e|Qx/ρ+(e)−Qx/ρ−(e)|(fρ−f)(Qx/ρ+(e))+(Qx/ρ+(e)−xρ)⋅e|Qx/ρ+(e)−Qx/ρ−(e)|(fρ−f)(Qx/ρ−(e))|⩽4R02√1−|x|2‖fρ−f‖L∞(∂B1)⩽2R0δ√1−|x|2, | (3.16) |
and therefore, plugging this information into (3.14), we obtain that, if x∈B1 and ρ∈(1,R0),
|ufρ(xρ)−uf(x)|⩽(2R0√1−|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)dHn−1e−|∂B1|uf(x))dρ=∫∞R0E(x,ρ)ρn−1(∫∂BρLf,ω/|ω|,|ω|(x)dHn−1ω−|∂B1|uf(x))dρ=c(n,s)(1−|x|2)s∫Rn∖BR0|y|2(|y|2−1)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|))|⩽1√1−|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(Rn∖B1) for each s∈(s0,1]. | (3.20) |
Indeed, if we define the following function
F±:Rn∖B1→Rn∖B1y↦|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(DF−1±(z)‖L∞(Rn∖B1)⩽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
∫Rn∖B1|f|y|(Qx/|y|±(y/|y|))||y|n+2sdy=∫Rn∖B1|f|y|(Qx/|y|±(y/|y|))|||y|Qx/|y|±(y/|y|)|n+2sdy=∫Rn∖B1|f(z)||z|n+2s|det(DF−1±(z))|dz⩽C∫Rn∖B1|f(z)||z|n+2sdz. | (3.23) |
This and the fact that f∈L1s(Rn∖B1) 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)dHn−1e−uf(x))dρ|⩽∫R01|∂B1|E(x,ρ)(2R0√1−|x|2+1)δdρ+|∫∞R0|∂B1|E(x,ρ)(−∫∂B1Lf,e,ρ(x)dHn−1e−uf(x))dρ|⩽C(x,R)δ+c(n,s)(1−|x|2)s∫Rn∖BR0|y|2(|y|2−1)s(|y|2−|x|2)|y|n(Lf,y/|y|,|y|(x)−|∂B1|uf(x))dy⩽C(x,R)δ+c(n,s)(1−|x|2)s∫Rn∖BR0|y|2(|y|2−1)s(|y|2−|x|2)|y|n×(1√1−|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 s↗1, we see that
lims↗1c(n,s)(1−|x|2)s∫Rn∖BR0|y|2(|y|2−1)s(|y|2−|x|2)|y|n×(1√1−|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 s↗1 by (1.14). As a consequence
lims↗1|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|Rn∖B1=f. Let us first assume that f∈C(Rn∖B1)∩L1s(Rn∖B1). Under this assumption, we can apply Theorem 1.6 and obtain that
u(x)=|∂B1|∫∞1E(x,ρ)ufρ(xρ)dρ, |
for each x∈B1, 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(ρ2−1)sdρ. | (4.1) |
Now we fix r∈(0,1) and we consider x∈Br. Applying the Harnack inequality for classical harmonic functions to ufρ, we have that
ufρ(0)⩽(1+|x|/ρ)n−11−|x|/ρufρ(xρ). |
From this, (1.13) and (4.1) we obtain that
u(0)⩽c(n,s)|∂B1|∫∞11ρ(ρ2−1)s(1+|x|/ρ)n−11−|x|/ρufρ(xρ)dρ=|∂B1|∫∞1E(x,ρ)(ρ2−|x|2)ρ2(1−|x|2)s(ρ+|x|)n−1ρn−2(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(1−t2)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(1−r2)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(1−r2)s∫∞1E(x,ρ)ufρ(xρ)dρ=(1+r)n(1−r2)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)n−1ufρ(0). |
Using this and (1.13) into (4.1), we find that
u(0)=c(n,s)|∂B1|∫∞1ufρ(0)ρ(ρ2−1)sdρ⩾c(n,s)|∂B1|∫∞11ρ(ρ2−1)s(1−|x|/ρ)n−11+|x|/ρufρ(xρ)dρ=c(n,s)|∂B1|∫∞1(ρ−|x|)n−1ρn−1(ρ2−1)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):=(ρ−t)nρn(1−t2)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)=(1−t)n(1−t2)s=(1−t)n−s(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)=(1−r)n−s(1+r)s=(1−r)n(1−r2)s. |
Plugging this information into (4.4) and recalling (1.19), we get
u(0)⩾|∂B1|(1−r)n(1−r2)s∫∞1E(x,ρ)ufρ(xρ)dρ⩾(1−r)n(1−r2)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 f∈L∞(BR∖B1)∩L1s(Rn∖B1), we take a sequence {fk}k⊂C(Rn∖B1)∩L1s(Rn∖B1) 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)fk−u‖L∞loc(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 e∈∂B1, a constant ϵ>0 and the function
fϵ(y):={0ify∈Rn∖Bϵ((1+ϵ)e),(|y|2−1)sify∈Bϵ((1+ϵ)e). |
Then the function
u(s)fϵ(x):=c(n,s)∫Bϵ((ϵ+1)e)(1−|x|2)s|y−x|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|n∫Bϵ((ϵ+1)e)(1−r2)s|y+re|ndy, |
and thus, by Lebesgue Differentiation Theorem, we conclude that
limϵ→0u(s)fϵ(0)u(s)fϵ(−re)=(1+r)n(1−r2)s. |
This proves that the constant on the left hand side inequality in (1.20) is optimal.
Similarly, taking x=re, one sees that
limϵ→0u(s)fϵ(re)u(s)fϵ(0)=(1−r2)s(1−r)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 x∈B1 and ρ>1,
E(x,ρ)=ρn−1−∫∂BρP(x,y)dHn−1y, | (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
−∫∂B1dHn−1ω|x−ρω|n=ρ2−nρ2−|x|2. | (A.2) |
Indeed, once (A.2) is established, we deduce from it, (1.13) and (2.1) that
−∫∂BρP(x,y)dHn−1y=c(n,s)(1−|x|2)s−∫∂BρdHn−1y(|y|2−1)s|x−y|n=c(n,s)(1−|x|2)s|∂B1|ρn−1∫∂B1ρn−1dHn−1ω(ρ2−1)s|x−ρω|n=c(n,s)(1−|x|2)s(ρ2−1)s−∫∂B1dHn−1ω|x−ρω|n=c(n,s)ρ2−n(1−|x|2)s(ρ2−1)s(ρ2−|x|2)=ρ1−nE(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, …, θn−3∈[0,π] and θn−2∈[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=sinθsinθ1…sinθn−3sinθn−2,ω2=sinθsinθ1…sinθn−3cosθn−2,ω3=sinθsinθ1…cosθn−3,⋮ωn=cosθ |
and produce a surface element of the form
sinn−2θsinn−3θ1…sin2θn−4sinθn−3dθdθ1…dθn−2. |
Also, up to a rotation, to prove (A.2) we can assume that x=(0,…,0,|x|). In this way, we find that
|x−ρω|2=|x|2+ρ2−2ρx⋅ω=|x|2+ρ2−2ρ|x|ωn=|x|2+ρ2−2ρ|x|cosθ, |
whence it follows that
∫∂B1dHn−1ω|x−ρω|n=∫θ,θ1,…,θn−3∈[0,π]θn−2∈[0,2π)sinn−2θsinn−3θ1…sin2θn−4sinθn−3dθdθ1…dθn−2(|x|2+ρ2−2ρ|x|cosθ)n2=2π(n−3∏j=1∫π0sinn−j−2θjdθj)∫θ∈[0,π]sinn−2θdθ(|x|2+ρ2−2ρ|x|cosθ)n2. |
Thus, we use the notation τ:=ρ|x|∈(1,+∞) and, by Proposition A.9 in [3], we deduce that
∫∂B1dHn−1ω|x−ρω|n=2π|x|n(n−3∏k=1∫π0sinkϑdϑ)∫π0sinn−2θdθ(τ2+1−2τcosθ)n2=2π|x|nτn−2(τ2−1)(n−3∏k=1∫π0sinkϑdϑ)∫π0sinn−2αdα=2π|x|nτn−2(τ2−1)(n−2∏k=1∫π0sinkϑdϑ). |
This and Proposition A.10 in [3] yield that
∫∂B1dHn−1ω|x−ρω|n=2πn2|x|nτn−2(τ2−1)Γ(n2)=|∂B1||x|nτn−2(τ2−1)=|∂B1|ρ2−nρ2−|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)s∫∞1(∫∂B1E(x,ρ)Lfk,e,ρ(x)dHn−1e)dρ=∫∞1ρ(ρ2−1)s(ρ2−|x|2)(∫∂B1Lfk,e,ρ(x)dHn−1e)dρ=∫∞1ρ2ρn(ρ2−1)s(ρ2−|x|2)(∫∂BρLfk,ω/|ω|,|ω|(x)dHn−1ω)dρ=∫Rn∖B1|y|2|y|n(|y|2−1)s(|y|2−|x|2)Lfk,y/|y|,|y|(x)dy. | (A.3) |
It also follows from (2.6) that, for a.e. y∈Rn∖B1,
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)s∫∞1(∫∂B1E(x,ρ)Lfk,e,ρ(x)dHn−1e)dρ=∫BR0∖B1|y|2|y|n(|y|2−1)s(|y|2−|x|2)Lfk,y/|y|,|y|(x)dy+∫Rn∖BR0|y|2|y|n(|y|2−1)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∞(BR0∖B1)⩽2R0√1−|x|2‖f‖L∞(BR∖B1). |
Consequently, using this, (A.4) and the dominated convergence theorem,
limk→+∞∫BR0∖B1|y|2|y|n(|y|2−1)s(|y|2−|x|2)Lfk,y/|y|,|y|(x)dy=∫BR0∖B1|y|2|y|n(|y|2−1)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(Rn∖BR0)→0 as j→+∞. | (A.7) |
To show (A.7), we recall (3.15) and we observe that, for every x∈B1 and y∈Rn∖BR0,
|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|))|⩽R0√1−|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 h∈L1s(Rn∖B1) such that |fkj(y)|⩽h(y) for a.e. y∈Rn∖B1 (see for instance Theorem 4.9 in [2]). Therefore, using this information into (A.8), we have
|Lfkj,y/|y|,|y|(x)|⩽R0√1−|x|2[h(|y|Qx/|y|−(y|y|))+h(|y|Qx/|y|+(y|y|))] | (A.9) |
for a.e. y∈Rn∖BR0.
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
∫Rn∖BR0h(|y|Qx/|y|±(y/|y|))|y|n+2sdy=∫Rn∖BR0h(|y|Qx/|y|±(y/|y|))||y|Qx/|y|±(y/|y|)|n+2sdy=∫Rn∖BR0h(z)|z|n+2s|det(DF−1±(z))|dz⩽C∫Rn∖BR0h(z)|z|n+2sdz. |
Accordingly, we deduce that
h(|y|Qx/|y|±(y/|y|))∈L1s(Rn∖BR0). |
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)=∫∞1∫∂B1E(x,ρ)Lf,e,ρ(x)dHn−1edρ |
for each x∈B1, 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)s∫∞1(∫∂B1E(x,ρ)fk(ρQx/ρ(e))dHn−1e)dρ=∫∞1ρ(ρ2−|x|2)(ρ2−1)s(∫∂B1fk(ρQx/ρ(e))dHn−1e)dρ=∫∞11(ρ2−|x|2)(ρ2−1)s(∫∂Bρfk(|ω|Qx/|ω|(ω/|ω|))dHn−1ω)dρ=∫Rn∖B11(|y|2−|x|2)(|y|2−1)sfk(|y|Qx/|y|(y/|y|))dy | (A.10) |
By (2.6), we have that, for a.e. y∈Rn∖B1,
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)s∫∞1(∫∂B1E(x,ρ)fk(ρQx/ρ(e))dHn−1e)dρ=∫BR0∖B11(|y|2−|x|2)(|y|2−1)sfk(|y|Qx/|y|(y/|y|))dy+∫Rn∖BR01(|y|2−|x|2)(|y|2−1)sfk(|y|Qx/|y|(y/|y|))dy | (A.12) |
Notice that, for k large enough,
‖fk‖L∞(BR0∖B1)⩽‖f‖L∞(BR∖B1). |
As a consequence,
limk→+∞∫BR0∖B11(|y|2−|x|2)(|y|2−1)sfk(|y|Qx/|y|(y/|y|))dy=∫BR0∖B11(|y|2−|x|2)(|y|2−1)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 h∈L1s(Rn∖B1) such that
|fkj(|y|Qx/|y|(y|y|))|⩽h(|y|Qx/|y|(y/|y|)) | (A.14) |
for a.e. y∈Rn∖BR0.
Furthermore, we claim that
h(|y|Qx/|y|(y/|y|)) belongs to L1s(Rn∖BR0). | (A.15) |
Indeed, the function
F:Rn∖BR0→Rn∖BR0y↦|y|Qx/|y|(y/|y|) |
is C1 and invertible. Moreover, since
lim|y|→∞Qx/|y|(y/|y|)=id∂B1, |
we find that
‖det(DF−1(z))‖L∞(Rn∖BR0)⩽˜C, |
for some positive constant ˜C>0.
From this and Theorem 2 in Section 3.3.3 of [10] we have that
∫Rn∖BR0h(|y|Qx/|y|(y/|y|))|y|2+2sdy=∫Rn∖BR0h(|y|Qx/|y|(y/|y|))||y|Qx/|y|(y/|y|)|2+2sdy=∫Rn∖BR0h(z)|z|2+2s|det(DF−1(z))|dz⩽˜C∫Rn∖BR0h(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(Rn∖BR0)→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))dHn−1e)dρ. |
This finishes the proof of (3.9).
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1. | Julie Clutterbuck, Jiakun Liu, Preface to the Special Issue: Nonlinear PDEs and geometric analysis – Dedicated to Neil Trudinger on the occasion of his 80th birthday, 2023, 5, 2640-3501, 1, 10.3934/mine.2023095 |