Existence of Solutions to the Nonlocal Dirichlet Problem

Remark 3.14. Since (3.6) holds for the family of kernels satisfying (3.12), M_L⁺

0u and M_L⁻₀u are continuous functions in Ωas long asu∈C^1,1[Ω] is bounded.

The following result will be important for Section 3.5.

Corollary 3.15. Let G be a compact set such that G ⊂ Ω and let α₀ ∈ (0,2). Let v∈C^1,1[Ω] be bounded. There existsc₁ ≥1 such that for every α∈(α₀,2) the following inequality holds:

sup

x∈G

M_α⁻v(x)

≤c₁(A+kvk_∞), whereA >0 is the constant in Definition 2.12.

Proof. Let α ∈ (α₀,2). Using Lemma 3.13, M_α⁻v is continuous in Ω. This implies the existence of a point x0 ∈ G such that |M_α⁻v(x)| ≤ |M_α⁻v(x0)| for every x ∈ G. Since v∈C^1,1[Ω], we can findA >0 such that

|v(x₀+y)−v(x₀)− ∇v(x₀)·y| ≤A|y|² for every y∈B_r, wherer= 1∧dist(G, ∂Ω). Recall thatA is independent ofx₀. Hence,

M_α⁻v(x₀)

≤(2−α)Λ Z

Br

|∆v(x₀;y)|

|y|^n+α dy+ (2−α)Λ Z

B^c_r

|∆v(x₀;y)|

|y|^n+α dy

≤2A(2−α)Λnω_n Zr

0

s^1−αds+ 4kvk_∞(2−α)Λnω_n Z∞

r

s^−1−αds

≤2AΛnω_n+ 8

α₀ kvk_∞Λnω_nr⁻² ≤ 8

α₀Λnω_nr⁻²(A+kvk_∞).

3.3 Existence of Solutions to the Nonlocal Dirichlet Problem 51

I0 is the value we obtain when applying I to the constant function that is equal to zero. Hence, M_α⁻v(x)≤ Iv(x) ≤M_α⁺v(x) for everyx ∈Rⁿ wheneverv is bounded and v∈C^1,1(x).

Fix any α₀∈(0,2)throughout the section. We prove the following existence result.

Theorem 3.16. Let α ∈ (α₀,2) and assume that the bounded domain Ω satisfies the exterior ball condition. Then there exists a unique viscosity solutionu∈C(Ω)∩L^∞(Rⁿ) of (3.28).

Existence of solutions under similar assertions has been established in [CLD12, BCF12].

We adapt their argumentation to our setting introduced in (3.12).

Remark 3.17.

(i) Ωsatisfies the exterior ball condition, if there isr0 >0 such that for everyω∈∂Ω there exists x∈Rⁿ\ΩwithB_r0(x)∩Ω ={ω}.

For example, ΩwithC¹-boundary satisfies the exterior ball condition.

(ii) The lower bound for αin Theorem 3.16 is important to construct suitable barriers on ∂Ω as explained below. These barriers will be independent ofα∈(α₀,2).

The proof of Theorem 3.16 is based on Perron’s method which originated in local theory to prove existence of solutions to the Laplace equation

∆u= 0 inΩ, u=g on∂Ω

by combining the explicit formula in the special caseΩ =B_Rand the maximum principle.

We refer to [GT01] for a general overview of this technique.

As main technical tool, we prove a comparison principle which will be crucial for the construction of the solutions to (3.28). The corresponding result in [CLD12] is [CLD12, Theorem 4.10].

Theorem 3.18. Let α ∈ (α₀,2) and let I be as in (3.29). Assume that the bounded functions v:Rⁿ→Rand w:Rⁿ→R satisfy

(i) Iv≥f and Iw≤f in Ω in the viscosity sense for some f ∈C(Ω)and (ii) v≤w inRⁿ\Ω.

Then v≤w in Ω.

The proof of the comparison principle in [CLD12, Theorem 4.10] is based on the con-struction of a suitable bump function (cf. [CLD12, Lemma 4.11]). [CLD12, Lemma 4.11]

has to be adapted to our setting.

Lemma 3.19. Let ϕ: Rⁿ → R, ϕ(x) = min 1,^|x|₄²

. There exists δ > 0 such that for every α∈(α0,2)

M_α⁻ϕ≥δ in B₁.

Proof. Note thatϕis regular enough to evaluateM_α⁻ϕclassically inB₁. Letx∈B₁. We claim that∆ϕ(x;y)≥0for everyy ∈Rⁿ. Indeed, if we assumex±y∈B2 orx±y6∈B2

then the claim is easily verified. For example, ifx±y∈B₂ then

∆ϕ(x;y) = |x+y|²

4 +|x−y|² 4 −|x|²

2 = |y|²

2 ≥0. (3.30)

If onlyx+y∈B₂, we use thatϕ(x)≤ ¹₄ and obtain

∆ϕ(x;y) =ϕ(x+y) + 1−2ϕ(x)≥ ¹₂.

The same holds true if onlyx−y ∈B2. So, ∆ϕ(x;y) ≥0 for every y∈Rⁿ. Using this fact and (3.30), we obtain for everyα∈(α₀,2):

M_α⁻ϕ(x)≥(2−α) Z

B1

λk _|^y_y|

∆ϕ(x;y)µ(dy)

= ²⁻₂^αλ Z

B1

k(_|y|^y )|y|^−n+2−α dy

≥ ²⁻₂^αλ

1

Z

0

rⁿ⁻¹ Z

I

r^−n+2−ασ(dy)dr

= ²⁻₂^αλσ(I) Z1

0

r^1−αdr= ¹₂λ|I|=:δ,

whereσ denotes the surface measure on Sⁿ⁻¹ and I ⊂ Sⁿ⁻¹ is as in Section 3.2. This completes the proof.

Proof of Theorem 3.18. We provide the proof of [CLD12, Theorem 4.10] for complete-ness. Defineg=v−w. Using Lemma 3.10,

M_α⁺g≥0 inΩ (3.31)

in the viscosity sense. We show that sup

Ω

g ≤ sup

Rⁿ\Ω

g=: N which will prove the theorem sinceg≤0inRⁿ\Ω.

Choose R ≥1 large enough such that Ω ⊂ B_R and define ψ(x) = ϕ(x/R), where ϕ is the bump function in Lemma 3.19. For everyx∈Ω,

M_α⁻ψ(x) = sup

K∈K0

Z

Rⁿ

∆ϕ _R^x;_R^y

K(y)dy=Rⁿ sup

K∈K0

Z

Rⁿ

∆ϕ _R^x;z

K(Rz)dz

=RⁿR^−n−αM_α⁻ϕ(x/R)≥R^−αδ,

3.3 Existence of Solutions to the Nonlocal Dirichlet Problem 53

whereδ >0 is as in Lemma 3.19. Fix anyε >0 and consider ψε(x) =N+ε(1−ψ(x)), which satisfies M_α⁺ψ_ε=−εM_α⁻ψ≤ −εR^−αδ <0 inΩ.

We claim that ψ_ε≥ginΩ. Assume this is not true. Therefore, inf_Ω(ψ_ε−g)<0which implies the existence of somed >0such thatψε+dtouchesgfrom above at somex∈Ω.

Because of (3.31) and Lemma 3.9, 0 ≤ M_α⁺(ψ_ε +d)(x) = M_α⁺ψ_ε(x). Contradiction.

Therefore,ψ_ε≥ginΩ which leads tosup_Ωg≤N by lettingε&0.

The next technical result deals with the construction of suitable barriers to attain the boundary data in (3.28).

Lemma 3.20. There exist constants γ > 0 and c₀ ≥ 1 such that the continuous and nonnegative function Φ :Rⁿ→R,

Φ(x) = min(1, c0((|x| −1)⁺)^γ) satisfies

M_α⁺Φ≤0 in Rⁿ\B1 for every α∈(α0,2). (3.32) Remark 3.21. Note thatΦin Lemma 3.20 satisfies Φ = 0inB₁ and Φ = 1inRⁿ\B₂. Proof. The proof can be obtained by direct computation similar to the approach in Section 3.5. However, we will avoid lengthy computations at this point by adapting [CS11, Lemma 1 and Corollary 31] to our setting. For ξ ∈ Sⁿ⁻¹ let I_ξ = I_ξ,% be as in Section 3.2, where % > 0 is fixed. For α ∈ (0,2) let K⁰ = K⁰(n, λ,Λ, α, %)¹ denote the class of all measurable symmetric kernels K satisfying the following condition: There existsξ ∈Sⁿ⁻¹ (which may depend on K) such that

(2−α)1I_ξ y

|y|

λ|y|^−n−α≤K(y)≤(2−α)Λ|y|^−n−α for every y∈Rⁿ\ {0}.

Note that K₀(α) ⊂ K⁰(α), where K₀(α) is as in Section 3.2. Let L⁰(α) denote the cor-responding class of all linear-integro differential operators of the form (3.4) with kernels K ∈ K⁰(α). We just stress the dependence on α here. So if we prove the assertion for M_L⁺0(α)Φ, then this implies the assertion forM_L⁺

0(α)Φ.

Forγ ∈(0,^α₂⁰)defineϕ_γ :Rⁿ→R,

ϕγ(x) = ((|x| −1)⁺)^γ.

Note that for every r ∈ (0,1), we can find an open set U ⊃ ∂B_1+r such that ϕ_γ|_U ∈ C²(U), which implies – together with the fact that (3.16) holds for ϕγ withα ∈(α0,2) arbitrary – thatM_L⁺0(α)ϕ_γ(x) is well-defined for everyα∈(α₀,2)and everyx∈∂B_1+r.

1We thank R. Schwab for proposing this class of kernels.

Claim 1: Choosingγ ∈(0,^α₂⁰) and r∈(0,1)sufficiently small, we obtain M_L⁺0(α)ϕ_γ(x)≤0 for every x∈∂B_1+r andα∈(α₀,2).

Since the operator M_L⁺0(α) is rotational invariant, it is sufficient to prove the claim for x₀ = (1 +r)e₁, wheree₁ denotes the first unit vector. We consider two steps:

Step 1: In this step, we choose the number r ∈(0,1)appropriately. Define a function l_N :Rⁿ→R,

l_N(x) =

(−N, |x| ≤1

log(|x| −1)∨ −N, |x|>1,

where N > 0 will be chosen below. Note that M_L⁺0(α)l_N(x₀) is well-defined for every r∈(0,1)if N ≥ −log₂^r, becausel_N|_U⁰ ∈C²(U⁰)for some open setU⁰ ⊃∂B_1+r, andl_N satisfies (3.16) for arbitraryα >0.

Choose anyr∈(0,1)andN =N(r)≥ −log^r₂. For ξ ∈Sⁿ⁻¹ define M_2,ξ⁺l_N(x₀) = lim

α%2M_α,ξ⁺ l_N(x₀), whereM_α,ξ⁺ l_N(x₀) = (2−α)R

Rⁿ(Λ∆l_N(x₀;y)⁺−λ1Iξ(_|^y_y|)∆l_N(x₀;y)⁻)µ(dy)(cf. (3.18)).

It is easy to see that M_L⁺0(α)l_N(x₀) = sup_ξ∈Sn−1M_α,ξ⁺ l_N(x₀). Using (3.22) withk =1I_ξ

in (3.20), we obtain

M_2,ξ⁺l_N(x₀)→ −∞ as|x₀| &1 (andN → ∞) uniformly inξ∈Sⁿ⁻¹, because

M_2,ξ⁺l_N(x0)≤ M⁺(D²l_N(x0),^eλ_n,Λ)e and M⁺(D²log(|x0| −1),^eλ_n,Λ)e → −∞as |x0| &1.

Recall thatM⁺ denotes the maximal Pucci operator and eλ= λc3,Λ = (Λ +e λ) Sⁿ⁻¹

withc₃= inf_ξ∈Sn−1

R

Iξs²₁σ(ds)>0, whereσ denotes the surface measure on Sⁿ⁻¹. So we may choose r ∈ (0,1) sufficiently small (depending on λ,Λ, n and %) and take any N ≥ −log^r₂ such that there exists α1 ∈ (α0,2) with M_L⁺0(α)l_N(x0) <−1 for every α ∈ (α₁,2]. By choosing N ≥ −log^r₂ sufficiently large, we can also make sure that M_L⁺0(α)l_N(x0)<−1for every α∈[α0, α1]:

Forξ∈Sⁿ⁻¹ setS_ξ={y∈Rⁿ: _|^y_y| ∈I_ξ∧x₀+y∈B₁}. After another possible decrease of r (depending on %), we can find ν > 0 independent of ξ such that |S_ξ| ≥ ν. Let α∈[α0, α1]and K∈ K⁰(α). We write

L l_N(x₀) = Z

Rⁿ

∆l_N(x₀;y)⁺K(y)dy− Z

Rⁿ

∆l_N(x₀;y)⁻K(y)dy

=:I₁+I₂

3.3 Existence of Solutions to the Nonlocal Dirichlet Problem 55

and estimate −I₂: According to the definition ofK⁰, there exists ξ∈Sⁿ⁻¹ such that

−I₂≥2(2−α)λ Z

Sξ

(−N −logr)⁻|y|^−n−α dy= 2(2−α)λ(N + logr) Z

Sξ

|y|^−n−α dy

≥2(2−α1)λ(N + logr) ν (2 +r)ⁿ⁺²,

where we note that y ∈ S_ξ implies r ≤ |y| ≤ 2 +r. Hence, I₂ → −∞ as N → ∞ uniformly in ξ. I₁ on the other side can be estimated from above by some number c=c(n,Λ, α₀)≥1 becausel_N|_U⁰ ∈C²(U⁰) and l_N satisfies (3.16) for arbitraryα >0.

Hence, we can chooseN ≥ −log^r₂ large enough (depending onn, λ,Λ, %, α0 andα1) such that M_L⁺0(α)l_N(x₀)<−1for every α∈[α₀, α₁]. The choice ofr and N is now complete.

Step 2: It remains to choose γ ∈(0,^α₂⁰). Let γ ∈ (0,^α₂⁰) be arbitrary for the moment and letN, r be as in the end of Step 1. Define v_γ:Rⁿ→R,

vγ(x) = ϕ_γ(x)−1

γ ∨ −N.

Then

γlim&0v_γ(x) = log(|x| −1)∨ −N =l_N(x)

for every|x|>1and this convergence holds locally uniformly. Note that ^aγ_γ⁻¹ −−−→^γ&0 loga for every a >0.

Assume that Claim 1 does not hold. Then there exists a sequence (γ_j)_j∈N,γ_j ∈(0,^α₂⁰), converging to zero and a sequence (αj)_j∈N, αj ∈ (α0,2), such that M_L⁺0(αj)ϕγj(x0) ≥0, and thus M_L⁺0(αj)v_γj ≥0 inΩ =B_2(1+r)\B_1+r. Take a subsequence(α_j) (which we do not relabel) such that α_j → α∈[α₀,2]asj → ∞. Using the fact thatv_γj → l_N locally uniformly as j → ∞, there exists j₀ ∈ Nwith the property that for every j ≥j₀ there exists a small number δj ∈R (converging to zero as j → ∞) such that l_N +δj touches v_γj from above at a pointx_j ∈B_r/2(x₀)∩Ω. Setr_j = ^r₂− |x_j−x₀|and note thatr_j → ^r₂ as j→ ∞. We can now proceed as in the proof of [BCF12, Theorem 4.6]: Assume first that α < 2. Hence, there existsj1 ≥j0 and a numberαe∈(α0,2) such thatαj ≤αe for every j ≥ j₁. Since M_L⁺0(αj)v_γj(x_j) = sup_ξ∈RnM_α⁺

j,ξv_γj(x_j) ≥ 0, we see that for every ε >0there exists a sequence (ξj) withξj ∈Sⁿ⁻¹ such that for every j≥j1

−ε≤(2−αj) Z

Rⁿ

Λ∆v_γj(x_j;y)⁺−λ1I_ξj(_|^y_y|)∆v_γj(x_j;y)⁻

|y|^n+αj dy

≤(2−α_j) Z

B_rj

Λ∆ηj(xj;y)⁺−λ1I_ξj(_|y|^y )∆ηj(xj;y)⁻

|y|^n+αj dy

+ (2−αj) Z

Rⁿ\B_rj

Λ∆v_γj(x_j;y)⁺−λ1I_ξj(_|^y_y|)∆v_γj(x_j;y)⁻

|y|^n+αj dy, (3.33)

where η_j = l_N +δ_j. Note that ∆η_j(x_j;y) = ∆l_N(x_j;y) for every y. Since Sⁿ⁻¹ is compact, we may take a subsequence(ξj) (which we again do not relabel) converging to some ξ⁰ ∈ Sⁿ⁻¹. The first integrand from above is bounded by the integrable function A|y|^2−eα−n for some A > 0 since l_N|_B

r/2(x0) ∈ C²(B_r/2(x₀)) and B_rj(x_j) ⊂ B_r/2(x₀).

Moreover, we can bound the second integrand by an integrable functiong since

v_γj(z) 1 +|z|^n+αj ≤

N+ sup

0≤γ≤α0

2

(|z|−1)^γ−1

γ 1_{|z|≥2}

1

1 +|z|^n+α0 + 1 1 +|z|ⁿ⁺²

≤

N+ _α²

0(|z| −1)^α0/21_{|z|≥2}

1

1 +|z|^n+α0 + 1 1 +|z|ⁿ⁺²

=:g(z) for every z ∈Rⁿ and every j ∈ N. Sinceξ_j → ξ⁰,r_j → ^r₂ and x_j → x₀ as j → ∞, the dominated convergence theorem implies

−ε≤M_α,ξ⁺ 0l_N(x0)≤M_L⁺0(α)l_N(x0).

Asεis arbitrary we concludeM_L⁺0(α)l_N(x₀)≥0.

If α = 2 we start again with (3.33) and argue in a similar way as in the discussion following (3.17) to obtain

−ε≤ lim

j→∞M_α⁺

j,ξ_jv_γj(x_j) =M_2,ξ⁺0l_N(x₀)≤ sup

ξ∈Sⁿ⁻¹M_2,ξ⁺l_N(x₀), which givessup_ξ∈Sn−1M_2,ξ⁺l_N(x₀)≥0since εis arbitrary.

But both cases (α <2andα= 2) contradict the result in Step 1. Claim 1 is now proved.

Claim 2: Letr, γ be as in Claim 1. If 1<|x|<1 +r, we obtain the same result as in Claim 1.

We prove Claim 2 by the same scaling argument as in the proof of [CLD12, Lemma 4.15].

Letxbe a point such that(1 +r)x= (1 +sr)x0, wheres∈(0,1)andx0∈∂B1+r. Hence,

|x|= 1 +sr. Defines₀ = ^1+sr_1+r and set

v(y) =s⁻₀^γϕ_γ(s₀y) = ((|y| −s₀⁻¹)⁺)^γ.

Note thatv(y)≤ϕ_γ(y) for every y∈Rⁿ. We translate v such that it remains belowϕ_γ but touches it in a whole ray passing throughx and x0. We still denote this translation v. Let α∈(α₀,2). Then

M_L⁺0(α)ϕ_γ(x) = sup

K∈K⁰(α)

Z

Rⁿ

(ϕ_γ(x+y) +ϕ_γ(x−y)−2ϕ_γ(x))K(y)dy

=s^γ₀ sup

K∈K⁰(α)

Z

Rⁿ

v(^x+y_s

0 ) +v(^x_s^−y

0 )−2v(_s^x

0)

K(y)dy

≤s^n+γ₀ sup

K∈K⁰(α)

Z

Rⁿ

ϕ_γ(_s^x

0 +z) +ϕ_γ(_s^x

0 −z)−2ϕ_γ(_s^x

0)

K(s₀z)dz

=s^n+γ₀ s^−n−α₀ M_L⁺0(α)ϕ_γ(_s^x

0) =s^γ−α₀ M_L⁺0(α)ϕ_γ(x₀)≤0,

3.3 Existence of Solutions to the Nonlocal Dirichlet Problem 57

where the last inequality holds because of Claim 1. Claim 2 is now proved.

We finish the proof by choosing c₀ = r^−γ in the definition of Φ, where r ∈ (0,1) and γ ∈(0,^α₂⁰) are as in Claim 1. Because of Claims 1 and 2, (3.32) holds forx∈B1+r\B1. Since Φ attains its global maximum at every point x∈Rⁿ\B_1+r, (3.32) also holds for x∈Rⁿ\B_1+r.

The proof of Theorem 3.16 now follows exactly as in [CLD12, Theorem 4.12]. We again provide the proof for completeness.

Proof of Theorem 3.16. We denote by C⁺(Ω)the set of all functions v:Rⁿ →R which are upper semicontinuous in Ω. Let α ∈ (α₀,2) and consider the operator I = I_α in (3.29). Let S be the set of all viscosity subsolutions of the equation Iv = 0 in Ω with boundary data smaller than g:

S ={v∈ C⁺(Ω)∩L^∞(Rⁿ) : Iv≥0inΩ in the viscosity sense andv≤ginRⁿ\Ω}.

S is non-empty because v =− kgk_∞ ∈ S. Now set u(x) = sup_v∈Sv(x) and define the upper semicontinuous envelope of uinΩ by

u(x) =







rlim&0 sup

u(ξ) : ξ ∈B_r(x)∩Ω , x∈Ω

u(x), x∈Rⁿ\Ω.

u is the smallest function w ∈ C⁺(Ω) such that w ≥ u in Rⁿ. Analogously, the lower semicontinuous envelope of u inΩis defined by

u(x) =







r&0lim inf

u(ξ) : ξ∈B_r(x)∩Ω , x∈Ω

u(x), x∈Rⁿ\Ω.

Note that [CLD12, Theorem 4.13 and Theorem 4.14] are applicable to our situation.

Using [CLD12, Theorem 4.13], we obtain Iu ≥ 0 in Ω in the viscosity sense which impliesu∈S and thereforeu=uby the definition of u. [CLD12, Theorem 4.14] implies thatIu≤0 inΩin the viscosity sense sinceu∈S is the biggest subsolution. Using the comparison principle (Theorem 3.18), the aforementioned result implies thatu≥uinΩ.

Therefore, u =u=u in Rⁿ which impliesu ∈C(Ω)∩L^∞(Rⁿ) and Iu= 0 in Ωin the viscosity sense.

We now prove that the boundary values are attained in a continuous way. We claim that for every ε > 0 and every x ∈ Rⁿ\Ω, we can find bounded barriers v :Rⁿ → R and w:Rⁿ→Rsuch that

(i) Iw≤0 and Iv≥0 inΩin the viscosity sense, (ii) w≥g and v≤g inRⁿ\Ω,

(iii) w(x)≤g(x) +εandv(x)≥g(x)−ε.

Assume for now that we already proved this claim. Note thatv∈S and – in combination with the comparison principle – we obtain from (i) and (ii) thatv≤u≤w inRⁿ. This proves thatu=g inRⁿ\Ωby lettingε&0in (iii). In particular, u∈C(Ω)∩L^∞(Rⁿ) andu satisfies (3.28).

We prove (i)-(iii) forw. Ifx∈Rⁿ\Ω, we define w(y) =

(kgk_∞, y6=x g(x), y=x.

wis lower semicontinuous inRⁿandIw≤0inΩin the viscosity sense. (ii) and (iii) are trivially satisfied.

Now assume thatx∈∂Ωand letε >0. SinceΩsatisfies the exterior ball condition, there existr₀∈(0,1)(independent ofx∈∂Ω) andξ ∈Sⁿ⁻¹such thatB_r0(x+r₀ξ)∩Ω ={x}.

Note thatdist(Ω, x+r₀ξ)≥r₀. Define w(y) = 2kgk_∞Φ

y−(x+rξ) r

+g(x) +ε,

where 0 < r < r0 and Φ is the function in Lemma 3.20. We will choose r ∈ (0, r0) sufficiently small such thatwsatisfies (i)-(iii).

(i) is satisfied (for each choice ofr ∈(0, r₀)) because everyxe∈Ωsatisfies the inequality

|xe−(x+rξ)|> r, which implies

Iw(x)e ≤M_α⁺w(ex) = 2kgk_∞r^−αM_α⁺Φ

ex−(x+rξ) r

≤0

by Lemma 3.20 and the ellipticity ofI with respect toL₀. (iii) is trivially satisfied since Φ = 0on Sⁿ⁻¹ by Remark 3.21. It remains to prove (ii). Since g is continuous on ∂Ω, we can chooseδ >0such that

|g(z)−g(x)| ≤ε whenever z∈B_δ(x)∩(Rⁿ\Ω). (3.34) Recall that g is only defined on Rⁿ \Ω. Choose r ∈ (0, r₀) small enough such that B_2r(x+rξ)⊂B_δ(x). Let y∈Rⁿ\Ω. We consider two cases:

• Assume that y ∈B_δ(x)∩(Rⁿ\Ω). Then w(y)≥g(x) +ε≥ g(y). Note that the first inequality holds becauseΦis a nonnegative function and the second inequality holds because of (3.34).

• Assume that y∈(Rⁿ\B_δ(x))∩(Rⁿ\Ω). Theny ∈(Rⁿ\B_2r(x+rξ))∩(Rⁿ\Ω) by the choice of r from above, which implies that ^y−(x+rξ)_r 6∈ B2. Therefore, by Remark 3.21, w(y)≥ kgk_∞≥g(y).

The choice ofv in the respective situations is now obvious.

It remains to prove uniqueness of the solutions to (3.28). Assume that there are two solutions u₁ and u₂ of (3.28). Since Iu₁ ≥ 0 and Iu₂ ≤ 0 in Ω in the viscosity sense and u₁ = u₂ = g in Rⁿ\Ω, we can apply Theorem 3.18 and obtain u₁ ≤ u₂ in Rⁿ. Interchanging the roles of u₁ and u₂ gives the reverse inequality. Hence, u₁ = u₂ in Rⁿ.

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