Optimal rearrangement problem and normalized obstacle problem in the fractional setting

Theorem 1.1

There exists a unique solution f̂ ∈ 𝓡_β of the minimization problem (1.1). For the function û = u_f̂ there exists a constant α > 0 such that

• 0 < û ≤ α in D,

• f̂ = χ_{û<α},

• û = α in {f̂ = 0}.

Moreover, the function U = α – û is the minimizer of the functional

among functions w ∈ H¹(D) with boundary values α on ∂D, and solves the normalized obstacle problem equation

Theorem 3.1

There exists a unique minimizer f̂ ∈ 𝓡̄_β ∖ 𝓡_β such that

for any f ∈ 𝓡̄_β. Moreover, for some α > 0 the function û = u_f̂ satisfies the following conditions

Remark 3.2

Observe that this result shows a remarkable difference with the local optimal rearrangement problem, since the optimal configuration f̂ for the fractional case is not a characteristic function. See Theorem 1.1.

Lemma 3.3

The set 𝓡̄_β ⊂ L^∞(D) is convex and

where for a convex set C, ext(C) denotes the extreme points of C.

Proof

The proof is standard and is omitted. For a more general result see [8, Lemma 3].□

Lemma 3.4

Let Φ_s be the functional defined in (3.1). Then Φ_s : 𝓡̄_β → ℝ is strictly convex and sequentially lower semi-continuous with respect to the weak^* topology. Moreover, there exists a unique minimizer f̂ of the functional Φ_s in 𝓡̄_β.

Proof

The strict convexity is a direct consequence of the linearity u_f1+f2 = u_f1 + u_f2 and the strict convexity of t ↦ t². Moreover, from (2.5), Hölder’s inequality and (2.1), we obtain

Therefore, f ↦ u_f is strongly continuous from L²(D) into H0s (D) and hence, Φ_s is strongly continuous from L²(D) into ℝ. Since Φ_s is convex, it follows that is sequentially weakly lower semicontinuous.

Finally, observe that if f_n ∈ 𝓡̄_β is such that f_n ⇀∗ f weakly^* in L^∞(D), then f_n ⇀ f weakly in L²(D), and so

To finish the proof just notice that the existence of a minimizer follows from Banach-Alaoglu’s theorem and the uniqueness of the minimizer from the strict convexity of Φ_s and the convexity of 𝓡̄_β.□

Proof of Theorem 3.1

The proof will be divided into a series of claims.

1. ∫Du^f^dx≤∫Du^fdx for any f∈R¯β.

   Let us take Ψ: L²(D) → ℝ̄ defined as

   Ψ(f)=Φs(f)+ξR¯β(f),

   where ξ_𝓡̄β(f) is the indicator function, i.e.

   ξR¯β(f)=0,if f∈R¯β,∞,if f∉R¯β..

   Observe that Ψ is strictly convex there. Moreover, it is easy to see that f̂ minimizes Ψ in L²(D). Thus

   0∈∂Ψ(f^),

   where

   ∂Ψ(f^)=g∈L2(D):Ψ(f)−Ψ(f^)≥∫Dg(f−f^)dx, for any f∈L2(D)

   is the sub-differential of Ψ at f̂.

   From (2.5) and (2.2) we get that

   ∂Φs(f^)={2u^}.

   Moreover

   ∂ξR¯β(f^)=g∈L2(D):ξR¯β(f)−ξR¯β(f^)≥∫Dg(f−f^)dx, for any f∈L2(D)=g∈L2(D):0≥∫Dg(f−f^)dx, for any f∈R¯β.

   Therefore the equation

   0∈∂Ψ(f^)=∂Φs(f^)+∂ξR¯β(f^),

   implies that

   −u^∈∂ξR¯β(f^)

   and thus the claim.

2. There exists a function f̃ = χ_E ∈ 𝓡_β such that

   ∫Du^f~dx≤∫Du^fdx

   for any f ∈ 𝓡̄_β.

   This follows from Claim 1, Lemma 3.3, and the fact that the minimum of the linear functional L(f) = ∫_D ûf dx on a bounded closed convex set 𝓡̄_β is attained in an extreme point f̃ = χ_E ∈ 𝓡_β.

3. There exists α > 0 such that

   {u^<α}⊂E⊂{u^≤α}.

   The proof is an immediate consequence of the bathtub principle for f̃. See [21, Theorem 1.14].

4. f^=1 in {u^<α}.

   The proof is again an immediate consequence of the bathtub principle for f̂.

5. {u^>α}⊂{f^=0}.

   Since f̂, f̃ ∈ 𝓡̄_β, we have that

   β=∫Df^dx=∫{u^<α}f^dx+∫{u^=α}f^dx+∫{u^>α}f^dx=∫Df~dx=∫{u^<α}f~dx+∫{u^=α}f~dx+∫{u^>α}f~dx

   Therefore, by Claims 3 and 4, we obtain that

   ∫{u^=α}f^dx+∫{u^>α}f^dx=∫{u^=α}f~dx. (3.2)

   On the other hand, by Claims 1 and 2, we get

   ∫Du^f^dx=∫Du^f~dx

   that together with (3.2) give us

   α∫{u^=α}f~dx=α∫{u^=α}f^dx+α∫{u^>α}f^dx≤∫{u^=α}u^f^dx+∫{u^>α}u^f^dx=∫{u^=α}u^f~dx=α∫{u^=α}f~dx,

   which implies

   α∫{u^>α}f^dx=∫{u^>α}u^f^dx,

   and thus the claim.

6. {u^>α}=∅.

   For β > α let us take ϕ(x) = (û(x) – β)⁺. Since supp ϕ = ω ⊂ {û > α}, claim 5 implies that

   0=2〈(−Δ)su^,ϕ〉=∬R2n(u^(x)−u^(y))(ϕ(x)−ϕ(y))|x−y|n+2sdxdy=∫ω∫ω(u^(x)−u^(y))2|x−y|n+2sdxdy⏟≥0+∫ω∫Rn∖ω(u^(x)−u^(y))(u^(x)−β)|x−y|n+2sdydx⏟≥0+∫Rn∖ω∫ω(u^(x)−u^(y))(β−u^(y))|x−y|n+2sdydx⏟≥0+∫Rn∖ω∫Rn∖ω(u^(x)−u^(y))(0−0)|x−y|n+2sdydx⏟=0.

   Thus, |ω| = |{û > β}| = 0 for any β > α. Moreover, since û ∈ Cloc0,δ (D) for some δ > 0, it is easy to see that the claim follows.

7. |{f^=0}|=0.

   Since (–Δ)^s û = f̂ ∈ L^∞(D) and f̂ ≥ 0 it is enough to check f̂ > 0 point-wise.

   Taken Claim 4 we need to check this only in the set {û = α}. But

   f^(x)=p.v.∫Rnu^(x)−u^(y)|x−y|n+2sdy=limϵ→0∫Rn∖Bϵ(x)u^(x)−u^(y)|x−y|n+2sdy>∫Rn∖Dα|x−y|n+2sdy>0.

   This proves the claim.

   The proof of Theorem 3.1 is complete.□

Proposition 4.1

Let u ∈ L²(ℝⁿ) be fixed. Then we have that

where the first limit is understood as a limit if u ∈ H¹(ℝⁿ) and as lim inf |u|s2 = ∞ if u ⧸ ∈ H¹(ℝⁿ) and the second limit is in the sense of distributions.

Proposition 4.2

Given a sequence s_k → 1 and {u_k}_k∈ℕ ⊂ L²(ℝⁿ) such that

then there exists a function u ∈ H¹(ℝⁿ) such that (up to a subsequence),

Theorem 4.3

Under the above notations, f̂_s ⇀∗ f̂ weakly^* in L^∞ as s → 1. Moreover we also obtain that

as s → 1.

Definition 4.4

Let X be a metric space and F_n, F : X → ℝ̄. We say that F_n Γ–converges to F, and is denoted by F_n →Γ F, is the following two inequalities hold true

• (lim inf–inequality) For any x ∈ X and any sequence {x_n}_n∈ℕ ⊂ X such that x_n → x in X, it holds that

  F(x)≤lim infn→∞Fn(xn).

• (lim sup–inequality) For any x ∈ X, there exists a sequence {y_n}_n∈ℕ ⊂ X such that y_n → x in X and

  F(x)≥lim supn→∞Fn(yn).

Theorem 4.5

Let X be a metric space and F_n, F: X → R̄ be functions such that F_n →Γ F. Moreover, assume that for each n ∈ ℕ, there exists x_n ∈ X such that

and that {x_n}_n∈ℕ ⊂ X is precompact. Then

and every accumulation point of the sequence {x_n}_n∈ℕ is a minimum point of F.

Theorem 4.6

Given 0 < s < 1, let f_s ∈ L²(Ω) be such that f_s ⇀ f weakly in L²(Ω) and let u_s ∈ H0s (Ω) and u ∈ H01 (Ω) be the solutions to

and

respectively.

Then u_s → u strongly in L²(Ω). Moreover,

Proof

Let F_s, F: L²(ℝⁿ) → ℝ̄ given by

and

Since ∫_Ω f_s v_s dx → ∫_Ω fv dx if v_s → v strongly in L²(ℝⁿ), from Propositions 4.1 and 4.2 we can conclude that F_s →Γ F as s → 1.

Now, observe that

The trivial estimate |u_s|_s ≤ ∥f_s∥₂ imply that, for any s_k → 1, the sequence {u_sk}_k∈ℕ ⊂ L²(ℝⁿ) is precompact. Then from Theorem 4.5 we obtain that u_s → u strongly in L²(Ω).

Finally,

This completes the proof.□

Proof of Theorem 4.3

Let f̂_s ∈ 𝓡̄_β be the optimal load for Φ_s. Observe that, for a subsequence, f̂_s ⇀∗ f weakly^* in L^∞(Ω) for some f ∈ 𝓡̄_β. Moreover, this convergence also holds weakly in L²(Ω).

From Theorem 4.6 we have that û_s → u_f strongly in L²(Ω) and using Theorem 4.2 we get

On the other hand, let f̂ ∈ 𝓡̄_β be the optimal load for Φ. Then, using the final part of Theorem 4.6, we obtain

The proof is complete.□

Theorem 5.1

Let f̂ ∈ 𝓡̄_β be the solution to the optimal fractional rearrangement problem and û : = u_f̂ ∈ H0s (D) be given by (2.4). Let α > 0 be the constant given in Theorem 3.1. Then the function Û : = α – û minimizes the functional

over the set H_α = {v ∈ Hlocs (ℝⁿ): v – α ∈ H0s (D)}. Moreover, Û verifies the inequalities

in the sense of distributions.

Finally, the minimizer of J in H_α is unique and is the unique solution to the inequality (5.2).

Proof

Let

and observe that, since 0 ≤ f̂ ≤ 1, for any v ∈ H_α it follows that J(v) ≥ I(v).

Next, observe that I(Û) = J(Û) and so the set of inequalities

imply the desired result.

Next, observe that the inequalities

are the Euler-Lagrange equation for the functional J based on the variation u_ε(x) = u(x) + εϕ(x), with ϕ ∈ Cc∞ (D).

Now, the uniqueness of minimizer for J is an immediate consequence of the strict convexity of J.

Assume that the function U satisfies the inequalities (5.2), but the unique minimizer of the convex functional J is the function V ≠ U.

Since J is strictly convex and J(V) < J(U) by taking U_ε = U + ε(V – U) we will obtain

Thus for ψ = V – U

where the last inequality follows from (5.2). This is a contradiction and the result follows.□

Remark 5.2

This result again shows an interesting difference between the classical obstacle problem and the fractional normalized version. Observe that in the positivity set, we still have −(−Δ)^s Û = 1, but in the zero set the function Û is not sharmonic (even if it is identically zero!). The free boundary condition on ∂{Û > 0} is given by the fact that (−Δ)^s Û is a function bounded by 0 and 1 across the free boundary.

Theorem 5.3

Let Û be solution of the normalized fractional obstacle problem given by Theorem 5.1. Then Û is a solution to

Moreover, problem (5.3) is equivalent to (5.2). Finally, U verifies (5.3) if and only if it is a minimizer of J in H_α, where J and H_α are given in Theorem 5.1.

Before we start with the proof, let us observe that for u ∈ H^s(ℝⁿ),

and hence (−Δ)^s u^± ∈ H^–s(ℝⁿ). On the other hand (−Δ)^s u⁺ is a distribution and the expression

makes in general no sense, unless (−Δ)^s u⁺ is a signed measure in D. Let us further observe that since

we need to search for solutions of (5.3) only among functions u, such that (−Δ)^s u ≤ 0 in D. This leads us to the introduction of fractional subharmonic functions in D, which form a convex subset of H^s(D)

Here the inequality (−Δ)^s u ≤ 0 should be understood in the sense of distributions.

The following lemma is essential for the equation (5.3) to make sense.

Lemma 5.4

Let u ∈ Hsubs (D). Then u⁺ ∈ Hsubs (D).

Proof

If u is smooth, then the fractional laplacian has pointwise values. In this case, we simply compute:

• For x ∈ {u ≤ 0},

  (−Δ)su+(x)=p.v.∫Rn−u+(y)|x−y|n+2sdy≤0.

• For x ∈ {u > 0},

  (−Δ)su+(x)=p.v.∫Rnu(x)−u+(y)|x−y|n+2sdy=p.v.∫Rnu(x)−u(y)|x−y|n+2sdy−p.v.∫Rnu−(y)|x−y|n+2sdy≤0.

For a general u ∈ H^s(ℝⁿ), we take {ρ_ε}_ε>0 a smooth family of approximations of the identity such that ρ_ε(z) = ρ_ε(−z) and define u_ε = u ∗ ρ_ε.

The result of the lemma will follow from the identity

for every ϕ ∈ Cc∞ (ℝⁿ).

Indeed, assuming (5.4), if (−Δ)^s u ≤ 0, then (−Δ)^s u_ε ≤ 0 for every ε > 0. Hence, from the smooth case we conclude that (−Δ)^s uε+ ≤ 0 and since uε+ → u⁺ in H^s the result is proved.

It remains to prove (5.4). For that purpose, it is useful to introduce the notation

the Hölder quotient of order s of u. Then, using that ρ_ε(−z) = ρ_ε(z), we observe that

The proof is now complete.□

Corollary 5.5

If u ∈ Hsubs (D) then min{−(−Δ)^s u⁺;1} ∈ L^∞(D).

Corollary 5.5 allows us to formulate the following normalized fractional obstacle problem:

For α > 0 solve

among functions U ∈ H^s(ℝⁿ), such that U = α in D^c and U ∈ Hsubs (D).

The weak formulation of the equation (5.5) is

for any ϕ ∈ H0s (D).

Now we are ready to prove Theorem 5.3.

Proof

Proof of Theorem 5.3. We only need to show that problems (5.2) and (5.5) are equivalent

For convenience let us break down the proof into several claims.

1. Assume that U ∈ Hlocs (ℝⁿ) is a solution of (5.5). Then $U ≥ 0.

   Observe first the following general fact:

   (U(x)−U(y))(U−(x)−U−(y))=(U+(x)−U+(y))(U−(x)−U−(y))−(U−(x)−U−(y))2.

   This simple identity implies that

   〈(−Δ)sU,U−〉=〈(−Δ)sU+,U−〉−|U−|s2. (5.6)

   Now, let us take ϕ = U⁻ as a test function in the weak formulation of (5.5). Then we obtain

   〈(−Δ)sU,U−〉=−∫D[χ{U≤0}min(−(−Δ)sU+;1)+χ{U>0}]U−dx=−∫Dmin{−(−Δ)sU+;1}U−dx=∫Dmax{(−Δ)2U+;−1}U−dx (5.7)

   But

   ∫Dmax{(−Δ)sU+;−1}U−dx≥〈(−Δ)sU+,U−〉. (5.8)

   Therefore, combining (5.6), (5.7) and (5.8), we arrive at

   |U−|s2≤0,

   and so the claim is proved.

2. (5.5) implies (5.2).

   It is immediate from Claim 1.

3. (5.2) implies U ≥ 0.

   The argument is similar to the one of Claim 6 in Theorem 3.1. Let U ∈ Hlocs (ℝⁿ) be a solution to (5.2). Take β < 0 and ϕ = (U − β)⁻, so ω = supp ϕ ⊂ {U < 0}. Then

   0=2〈(−Δ)sU,ϕ〉=∬R2n(U(x)−U(y))(ϕ(x)−ϕ(y))|x−y|n+2sdxdy=∫ω∫ω(U(x)−U(y))(U(y)−U(x))|x−y|n+2sdxdy⏟≤0+∫ω∫R∖ω(U(x)−U(y))(β−U(x))|x−y|n+2sdxdy⏟≤0+∫R∖ω∫ω(U(x)−U(y))(U(y)−β)|x−y|n+2sdxdy⏟≤0+∫R∖ω∫R∖ω(U(x)−U(y))(0−0)|x−y|n+2sdxdy⏟=0≤0.

   Thus, |ω| = |{U < β}| = 0 for any β < 0.

4. (5.2) implies (5.5).

   Can be verified directly.□