Nonlocal self-improving properties

Nonlocal self-improving properties Tuomo Kuusi (Aalto University) July 23rd, 2015 Frontiers of Mathematics And Applications IV UIMP 2015

Part 1: The local setting Self-improving properties:

Meyers estimate Theorem (Meyers) Let u be a local weak solution to div (a(x)du) = 0 in R n, where a( ) is measurable and satisfies ξ 2 Λ a(x)ξ, ξ and a(x) Λ. Then u W 1,2 loc = u W 1,2+δ loc for some δ > 0 depending only on n, Λ

The Gehring lemma This is based on a modification of the seminal result of Gehring: Theorem (Gehring) Let f L p loc (Rn ) be such that ( 1/p ( f dx) p B B ) 1/q f q dx for q < p and for every ball B. Then f L p+δ loc (Rn ) for some δ > 0 and ( 1/(p+δ) ( ) 1/q f dx) p+δ f q dx B B

Inhomogeneous case Theorem (Elcrat-Meyers, Giaquinta-Modica) Let u W 1,2 loc be a weak solution to where z 2 Λ div a(x, Du) = f L 2+δ 0, a(x, z), z and a(x, z) Λ z.

Inhomogeneous case Theorem (Elcrat-Meyers, Giaquinta-Modica) Let u W 1,2 loc be a weak solution to where Then z 2 Λ div a(x, Du) = f L 2+δ 0, a(x, z), z and a(x, z) Λ z. u W 1,2 = u W 1,2+δ loc for some δ (0, δ 0 ] depending only on n, Λ, δ 0.

Inhomogeneous case Moreover, the local estimate ( Du 2+δ dx B R holds for any ball B R. ) 1 ( ) 1 2+δ Du 2 2 dx B 2R ( + f 2+δ 0 dx B 2R ) 1 2+δ 0

The Gehring lemma with additional terms Theorem (Gehring-Giaquinta-Modica) Let f L p loc (Ω) be such that for q < p, then ( ) 1/p ( 1/q ( f p dx f dx) q + 1 2 B B B ( ) 1/(p+δ) ( f p+δ dx 1 2 B B ) 1/q f q dx ( + B ) 1/p g p dx ) 1/(p+δ) g p+δ dx

Caccioppoli inequalities imply higher integrability Theorem Let u W 1,2 (R n ) such that for every ball B B(x 0, r) R n Du 2 dx 1 1 2 B r 2 u(x) (u) B 2 dx B holds; then there exists δ > 0 such that u W 1,2+δ loc (R n )

Caccioppoli inequalities imply higher integrability The proof is very simple: Sobolev-Poincaré yields ( B/2 1/2 ( (n+2)/2n Du dx) 2 Du dx) 2n/(n+2), B and the assertion follows from an adaptation of the Gehring lemma.

Gradient oscillations What about the higher regularity? For solutions to we have a( ) is Dini = Du C 0 a( ) C 0,σ = Du C 0,σ a( ) N σ,q = Du N σ,2 div (a(x)du) = 0 Recall that a( ) N σ,q means that a(x + h) a(x) q dx h qσ Oscillations of coefficients influence the oscillations of the gradient

Merely measurable coefficients If the coefficients are merely measurable, there is no gradient oscillation control.

Merely measurable coefficients If the coefficients are merely measurable, there is no gradient oscillation control. Indeed, consider n = 1 and with (a(x)u x ) x = 0, u(0) = 0, u(1) = 1. Then the solution is given by u(x) = M x 0 0 < ν a(x) L. dt a(t), u x(x) = M [ 1 ] 1 a(x), M := dt. 0 a(t) i.e., no gradient differentiability is possible when coefficients are just measurable.

Integrodifferential equations Part 2: The nonlocal setting

Integrodifferential equations We consider E K (u, η) = f η dx R n for every test function η Cc (R n ), where E K (u, η) := [u(x) u(y)][η(x) η(y)]k(x, y) dx dy. R n R n The Kernel K(, ) is assumed to be symmetric and measurable satisfying growth bounds for some Λ 1. 1 Λ K(x, y) Λ x y n+2α x y n+2α

Integrodifferential equations Heuristically speaking, the nonlocal equation E K (u, η) = R f η dx n for all η Cc (R n ) is the weak formulation of L K u(x) = p.v. (u(x) u(y))k(x, y) dy = 1 R n 2 f (x). In the case K(x, y) = c n,α x y n 2α the operator is the fractional Laplacian, and the equation reduces to ( ) α u = f.

Integrodifferential equations Furthermore, in the case K(x, y) = x y n 2α we have u(x) u(y) η(x) η(y) dx dy E K (u, η) = R n R n x y α x y α x y n, which is the nonlocal analog of Du, Dη dx.

Integrodifferential equations - some regularity results Bass & Kassmann (Comm. PDE, TAMS 05) Caffarelli & Silvestre (Comm. PDE 07, lifting and localization) Kassmann (Calc. Var. 09, measurable coefficients) Caffarelli & Chan & Vasseur (JAMS 07, lifting and localization) Bjorland & Caffarelli & Figalli (Adv. Math. 09, p-laplacean type) Caffarelli & Silvestre (Ann. Math. 11, fully nonlinear theory) Da Lio & Rivière (Analysis & PDE 11, Adv. Math. 11, systems and half-harmonic maps) Di Castro & K. & Palatucci (JFA 14, Poincare 15, p-growth and related theory)

Fractional energies For α (0, 1) and p [1, ), define the seminorm Then ( [u] α,p (R n ) := R n Rn u(x) u(y) p ) 1/p x y n+αp dx dy W α,p (R n ) = {u L p (R n ) : u L p (R n ) + [u] α,p (R n )}. In the case p = 2 the abbreviation is H α (R n ) = W α,2 (R n ). The usual gradient can be obtained by letting α 1, but only after renormalisation by a factor depending on 1 α, see Bourgain & Brezis & Mironescu

Integrodifferential equations Energy solutions are initially considered in u H α (R n ), f L 2 (R n ) : E K (u, η) = f η dx, R n and the analogue of the Meyers property would be now u W α,2+δ, δ > 0 upon considering f L q (R n ) for some q > 2

A first result Theorem (Bass & Ren, JFA 13) Define the α-gradient (Rn u(y) u(x) 2 Γ(x) := dy x y n+2α ) 1/2 for the solution. Then Γ L 2+δ

A first result Theorem (Bass & Ren, JFA 13) Define the α-gradient (Rn u(y) u(x) 2 Γ(x) := dy x y n+2α ) 1/2 for the solution. Then Γ L 2+δ This implies, via a delicate yet by-now classical characterisation of Bessel potential spaces due to Strichartz and Stein, that for some δ > 0. u W α,2+δ

Self-improving property Theorem (K. & Mingione & Sire - Analysis & PDE 2015) Let u H α be a solution to E K (u, η) = f η for all η Cc. If f L 2+δ 0, then u W α+δ,2+δ for some δ > 0.

Self-improving property Theorem (K. & Mingione & Sire - Analysis & PDE 2015) Let u H α be a solution to E K (u, η) = f η for all η Cc. If f L 2+δ 0, then u W α+δ,2+δ for some δ > 0. In particular, Sobolev embedding W s,q W t,p for q > p and s n q = t n p gives u W α+δ,2+δ for some δ (0, δ)

The general case In the local case the most general equation that can be considered is div (A(x)Du) = div (B(x)g) + f. This corresponds to take in the right hand side all possible orders of differentiation, that in the integer case means taking orders zero and one.

The general case We therefore consider equations of the type E K (u, η) = E H (g, η) + f η dx η Cc (R n ), R n where the kernel H( ) satisfies H(x, y) A model case is obviously given by Λ x y n+2β ( ) α u = ( ) β g + f, where the analysis can be done via Fourier analysis.

Dimension analysis reveals the optimal assumptions Let us start with ( ) α u = f L p. C-Z theory gives u W 2α,p. Then we recall the embedding W 2α,p W α,2 provided the following interpolation scale relation holds: 2α n p = α n 2. This gives f L 2n n+2α f L 2n n+2α +δ 0

Dimension analysis reveals the optimal assumptions Continue with ( ) α u = ( ) β g In the case α = β we immediately see g W α,2 In the case 2β α then we formally invert the operators α u β α/2 g 2β α g L 2. Therefore we arrive at g W 2β α,2 g W 2β α+δ0,2.

Dimension analysis reveals the optimal assumptions In the case 2β < α no differentiability is needed on g Consider W 2β α,2 as the dual of W α 2β,2 and eventually observe the embedding But now ( L therefore we conclude with W α 2β,2 L 2n n 2(α 2β) 2n n 2(α 2β). ) = L 2n n+2(α 2β) ; g L 2n n+2(α 2β) g L 2n n+2(α 2β) +δ 0.

The Theorem Theorem (K. & Mingione & Sire) Under the optimal assumptions f L 2n 2+2α +δ 0 loc g W 2β α+δ 0,2 g L 2n n+2(α 2β) +δ 0 if 2β α if 2β < α, any H α -solution u to the equation E K (u, η) = E H (g, η) + f, η η C c is such that u W α+δ,2+δ loc (R n )

A first sketch of the proof Part 4: A fractional approach to Gehring lemma

Caccioppoli inequalities imply higher integrability - local case Theorem Let u W 1,2 (R n ) such that for every ball B B(x 0, r) R n Du 2 dx 1 r 2 u(x) (u) B 2 dx B/2 holds; then there exists δ > 0 such that B u W 1,2+δ loc (R n )

Caccioppoli inequalities imply higher integrability - nonlocal case Theorem (K. & Mingione & Sire) Let u H α (R n ) such that for every ball B B(x 0, r) R n B B u(x) u(y) 2 1 dx dy x y n+2α r 2α u(x) (u) B 2 dx B u(y) (u) B + dy u(x) (u) x 0 y n+2α B dx R n \B holds; then there exists δ > 0 such that u W α+δ,2+δ loc (R n ) B

Key observation u W 1,2 means that Du 2 is integrable w.r.t. a finite measure (i.e. the Lebesgue measure)

Key observation u W 1,2 means that Du 2 is integrable w.r.t. a finite measure (i.e. the Lebesgue measure) u W α,2 means that [ u(x) u(y) x y α is integrable w.r.t. an infinite set function, that is dx dy E x y n. E Therefore there are potentially more regularity properties to exploit in the above fractional difference quotient. ] 2

Key idea: Dual pairs To each u and ε < (0, 1 α) we associate a function U(x, y) := u(x) u(y) x y α+ε and a finite and doubling measure dx dy µ(e) := x y n 2ε. Note that they are in duality in the sense that E u W α,2 U L 2 (µ).

Strategy: higher integrability for U w.r.t. µ We translate the Caccioppoli inequality for u in a reverse Hölder inequality for U w.r.t. µ We prove a version of Gehring lemma for dual pairs (µ, U) The higher integrability of U turns into the higher differentiability of u All estimates heavily degenerate when α 1 or α 0

Higher integrability = higher differentiability Assume U L 2+δ loc, i.e., U 2+δ dµ = B B B B u(x) u(y) 2+δ dx dy <. x y n+(2+δ)α+εδ Rewrite it as follows: u(x) u(y) 2+δ dx dy <. x y n+(2+δ)[α+εδ/(2+δ)] B B But this means that u W α+εδ/(2+δ),2+δ loc (R n ), i.e., we have gained also differentiability!

Reverse inequality for dual pairs (µ, U) Proposition (K. & Mingione & Sire) For every σ (0, 1), the Caccioppoli inequality implies for the dual pair (µ, U) that ( 1/2 ( ) U dµ) 2 c 1/q B σε 1/q 1/2 U q dµ 2B + σ ε 1/q 1/2 2 ( k(α ε) k=1 holds, where B = B B and [ ) 2n q n + 2α, 2. 2 k B ) 1/q U q dµ

The Gehring lemma for dual pairs (µ, U) Theorem (K. & Mingione & Sire) Assume that for every σ (0, 1) the pair (µ, U) satisfies ( 1/2 ( U dµ) 2 c(σ) B +σ 2B ) 1/q U q dµ k=2 1/q 2 ( k(α ε) U dµ) q, 2 k B where q (1, 2) and for every choice of B = B B. Then U L 2+δ loc for some δ > 0 and ( 1/(2+δ) U dµ) 2+δ B k=1 ) 1/q 2 ( k(α ε) U q dµ 2 k B

Sketch Part 5: Brief sketch of the Gehring lemma for dual pairs

Step 1: Conclusion via a level set inequality We reduce to prove the Gehring inequality on level sets, that is U 2 dµ λ 2 q U q dµ holds provided B {U>λ} λ 0 := k=1 B {U>λ} 1/2 2 ( k(α ε) U dµ) 2 λ 2 k B then standard Cavalieri s principle yields the result.

Step 1: Conclusion via a level set inequality Indeed, denoting U m = min(u, m), m λ 0, and ν = U 2 dµ, we have m Um δ dν = δ λ δ 1 ν({u > λ}) dλ 0 m λ δ 0 U 2 dµ + δ λ δ 1 U 2 dµ dλ λ 0 {U>λ} m λ δ 0 U 2 dµ + cδ λ δ+1 q U q dµ dλ λ 0 {U>λ} λ δ 0 U 2 cδ dµ + Um δ+2 q U q dµ δ + 2 q λ δ 0 U 2 dµ + cδ Um δ dν 2 q

Step 1: Conclusion via a level set inequality By choosing δ small enough to get cδ 2 q 1 2 we obtain, after reabsorption and sending m, U 2+δ dµ 2λ δ 0 U 2 dµ cλ 2+δ 0.

Step 1: Conclusion via a level set inequality By choosing δ small enough to get cδ 2 q 1 2 we obtain, after reabsorption and sending m, U 2+δ dµ 2λ δ 0 U 2 dµ cλ 2+δ 0. Thus the goal is to obtain the level set inequality U 2 dµ λ 2 q U q dµ. B {U>λ} B {U>λ}

Step 2: Global Calderón-Zygmund covering We will apply C-Z decomposition at level Mλ in R 2n for M 1 to get a disjoint family of product of dyadic cubes {Q i } i, where Q = Q 1 Q 2 and Q 1 and Q 2 are dyadic cubes in R n with same size, such that ( ) 1/2 U 2 dµ Mλ Q i and We let U Mλ outside i U λ := {Q i } We denote by k(q) the generation of Q, i.e., 2 nk(q) = Q 1. Q i

Step 2: Classification of the cubes We call off-diagonal cubes Q = Q 1 Q 2 those whose distance from the diagonal is larger than their sidelength: 2 k(q)+3 dist(q) := dist(q 1, Q 2 ). The nearly diagonal cubes are then collected to { Uλ d := Q = Q 1 Q 2 U λ : dist(q 1, Q 2 ) < 2 3 k(q)} and we set U nd λ = {Q U λ : Q / U d λ }.

Idea: Step 2: Splitting of the analysis The diagonal cubes in Uλ d can be treated with the aid of an auxiliary diagonal cover. To treat these we will use the assumed diagonal reverse type Hölder inequality ( 1/2 ( U dµ) 2 c(σ) B +σ B ) 1/q U q dµ k=2 1/q 2 ( k(α ε) U dµ) q. 2 k B For the non-diagonal cubes in Uλ nd the fractional Poincaré inequality implies automatically a reverse type Hölder s inequality. Unfortunately this comes with error terms, whose analysis lead to heavy combinatorial arguments.

Step 3: Diagonal exit time argument and covering We consider the quantity ( 1/2 Ψ(x, ϱ) := U dµ) 2, B(x, ϱ) B(x, ϱ) B(x, ϱ). Notice that B(x,ϱ) Ψ(x, 1) λ 0 := k=1 ) 1/2 2 ( k(α ε) U 2 dµ. 2 k B 1 Therefore we can find a radius ϱ(x), such that ( 1/2 Ψ(x, ϱ(x)) U dµ) 2 λ whenever λ λ 0 and B(x,ϱ(x)) x {y R n : sup Ψ(y, ϱ) > λ}. ϱ>0

Step 3: Diagonal exit time argument and covering By Vitali s covering theorem we can now extract an at most countable covering B(x, 2 10 n ϱ(x)) B(x j, 10 n+1 ϱ(x j )) x j J D and moreover we have j U 2 dµ λ 2 B(x j,10 n+1 ϱ(x j )) j = λ 2 j µ(b(x j, ϱ(x j ))) µ(b j ).

Step 4: First summation formula We recall the diagonal reverse Hölder inequality, that is ( ) 1/2 ( U 2 dµ c(σ) U q dµ B j 2B j +σ The exit time argument gives k=2 ( λ U 2 dµ B j ) 1/q ) 1/q 2 ( k(α ε) U q dµ. 2 k B j ) 1/2 so that ( ) 1/q λ c(σ) U q dµ + cσλ. 2B j

Step 4: First summation formula We have, for small enough universal σ, that ( λ U q dµ 2B j ) 1/q and hence µ(b j ) 1 λ q 2B j U q dµ. Adjusting the constants properly µ(b j ) 1 λ q 2B j {U>λ} U q dµ, summation yields λ 2 j µ(b j ) λ 2 q {U>λ} U q dµ.

Step 4: First summation formula Therefore, we have where we recall that U 2 dµ U 2 dµ Q U d Q {U>Mλ} λ j 10 n+1 B j λ 2 µ(b j ) j J D λ 2 q {U>λ} U q dµ, U d λ := {Q = Q 1 Q 2 U λ : dist(q 1, Q 2 ) < 2 3 k(q)}.

Step 5: Off-diagonal cubes I What happens for the off-diagonal cubes?

Step 5: Off-diagonal cubes I What happens for the off-diagonal cubes? By the fractional Poincaré inequality we get for ( 1/2 ( U dµ) 2 2n n+2s Q Q q < 2. We denote here for Q = Q 1 Q 2. ) 1/q U q dµ ( ) α+ε ( 2 k(q) + dist(q) ( ) α+ε ( 2 k(q) + dist(q) P 1 Q P 2 Q P 1 Q = Q 1 Q 1 P 2 Q = Q 2 Q 2 ) 1/q U q dµ ) 1/q U q dµ

Step 5: Off-diagonal cubes I But with a priori nasty diagonal correction terms: ( 1/2 ( U dµ) 2 Q Q ) 1/q U q dµ ( ) α+ε ( 2 k(q) + dist(q) ( ) α+ε ( 2 k(q) + dist(q) P 1 Q P 2 Q ) 1/q U q dµ U q dµ) 1/q.

Step 5: Off-diagonal cubes I It follows that µ(q) 1 λ q Q {U>λ} + 1 λ q µ(q) µ(p 1 Q) + 1 λ q µ(q) µ(p 2 Q) U q dµ ( 2 k(q) ) q(α+ε) dist(q) ( ) q(α+ε) 2 k(q) dist(q) U q dµ P 1 Q {U>λ} U q dµ. P 2 Q {U>λ}

Step 6: Further classification of off-diagonal cubes We further classify M h λ := { Q U nd λ } : U q dµ λ q, h {1, 2}, P h Q and N h λ := { Q U nd λ } : U q dµ λ q, h {1, 2}, P h Q and finally set M λ := M 1 λ M2 λ and N λ := N 1 λ N 2 λ.

Step 7: Soft summation We then have a first, soft summation formula µ(q) 1 λ q U q dµ Q M {U>λ} λ after adjusting parameters suitably.

Step 7: Soft summation We then have a first, soft summation formula µ(q) 1 λ q U q dµ Q M {U>λ} λ after adjusting parameters suitably. We now need a similar summation formula for N λ := Nλ 1 N λ 2.

Step 8: Hard summation We consider those cubes that are not covered by the diagonal covering N λ,nd := Q N λ : Q 10B j j J D and prove the hard summation formula obtained with some heavy covering and combinatorics argument µ(q) 1 λ q U q dµ Q N {U>λ} λ,nd

Step 8: Sketch of the proof for hard summation Get a disjoint covering of the projections of the bad cubes N λ and call it PN λ = {H} and operate the first decomposition Nλ,nd h = Nλ,nd h (H), H PN λ with N h λ,nd (H) := {Q N λ,nd : P h Q H}, h {1, 2}. Then define } [Nλ,nd h (H)] i := {Q Nλ,nd h (H) : k(q) = i + k(h)

Step 8: Sketch of the proof for hard summation so that N h λ,nd (H) = i [N h λ,nd (H)] i Finally, we set [Nλ,nd h (H)] i,j := {Q [Nλ,nd h (H)] i : 2 j k(h) dist(q) < 2 j+1 k(h)} to obtain a disjoint decomposition Nλ,nd h = [Nλ,nd h (H)] i,j H PN λ i,j

Step 8: Sketch of the proof for hard summation Combinatorics then gives Q N h λ,nd (H) ( ) q(α+ε) µ(q) 2 k(q) U q dµ µ(p h Q) dist(q) P h Q {U>κλ} U q dµ H {U>κλ} for every H PN λ Then the hard summation formula follows by summing up on H PN λ, and recalling that PN λ is a disjoint covering

Step 9: Final summation Since now U λ = U d λ U nd λ and U nd λ = M λ N λ,d N λ,nd, we have actually obtained a summation estimate for all of these collections and hence conclude with the desired inequality: U 2 dµ λ 2 µ(b j ) + λ 2 µ(q) {U>λ} j J D Q M λ N λ,nd λ 2 µ(b j ) + λ 2 q U q dµ j J {U>λ} D λ 2 q U q dµ {U>λ}

Non-homogeneous equations Equations of the type E K (u, η) = E H (g, η) + f η dx R n η Cc (R n ), where H(x, y) Λ x y n+2β f L 2+δ 0 loc g W 2β α+δ 0,2

Non-homogeneous equations - β = α We further define G(x, y) := g(x) g(y) x y α+ε, F (x, y) := f (x) so that G L 2+δg loc (R 2n ; µ), F L 2+δ f loc (R 2n ; µ)

Non-homogeneous equations - β = α The reverse Hölder inequality provided by the Caccioppoli inequality becomes ( 1/2 ( ) U dµ) 2 c 1/q B σε 1/q 1/2 U q dµ 2B + σ ε 1/q 1/2 2 ( k(α ε) ( +c 2B + c b[µ(b)] θ ε 1/p 1/2 k=1 ) 1/2 F 2 dµ k=1 2 k B ) 1/q U q dµ 1/2 2 ( k(α ε) G dµ) 2. 2 k B

Non-homogeneous equations - β = α Then a non-homogeneous implementation of the previous argument gives ( 1/(2+δ) ) 1/2 U dµ) 2+δ c 2 ( k(α ε) U 2 dµ B 2 k B k=1 +cϱ α ε 0 ( +c +c 2B ( 2B ) 1/(2+δ0 ) F 2+δ 0 dµ G 2(1+δ 1) dµ 2 ( k(α ε) k=1 ) 1/[p(1+δ1 )] 2 k B ) 1/p G p dµ

Closing remarks The sketch above gives a very simplified overview of the proof, which is actually featuring many more technicalities

Closing remarks The sketch above gives a very simplified overview of the proof, which is actually featuring many more technicalities Our approach is based only on energy estimates and does not use the linearity of the equation. Therefore it can be easily adapted to the case of nonlinear integrodifferential equations, with possibly degenerate operators of the type E K (u, η) := Φ(u(x) u(y))[η(x) η(y)]k(x, y) dx dy R n R n with Φ(t)t t p 1 t R \ {0}, 1 < p <.

Thank you for your attention!