four definitions for the fractional laplacian
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Four definitions for the fractional Laplacian N. Accomazzo - PowerPoint PPT Presentation

Four definitions for the fractional Laplacian N. Accomazzo (UPV/EHU), S. Baena (UB), A. Becerra Tom e (US), J. Mart nez (BCAM), A. Rodr guez (UCM), I. Soler (UM) VIII Escuela-Taller de An alisis Funcional Basque Center for


  1. Therefore |�∇ 2 u ( x ) y , y � + o ( | x | 3 ) | ˆ = dy | y | n + 2 s | y |≤ 1 Using Cauchy-Schwarz inequality |∇ 2 u ( x ) || y | 2 + | o ( | x | 3 ) | |∇ 2 u ( x ) y || y | + | o ( | x | 3 ) | ˆ ˆ ≤ dy ≤ dy | y | n + 2 s | y | n + 2 s | y |≤ 1 | y |≤ 1 1 ˆ = C | y | n − 2 ( 1 − s ) dy = | y |≤ 1 Group 3 Fractional Laplacian VIII Escuela-Taller 9 / 40

  2. Therefore |�∇ 2 u ( x ) y , y � + o ( | x | 3 ) | ˆ = dy | y | n + 2 s | y |≤ 1 Using Cauchy-Schwarz inequality |∇ 2 u ( x ) || y | 2 + | o ( | x | 3 ) | |∇ 2 u ( x ) y || y | + | o ( | x | 3 ) | ˆ ˆ ≤ dy ≤ dy | y | n + 2 s | y | n + 2 s | y |≤ 1 | y |≤ 1 ˆ 1 1 1 ˆ ˆ = C | y | n − 2 ( 1 − s ) dy = r 1 − 2 s d σ ( ω ) dr < ∞ | y |≤ 1 0 S n − 1 Group 3 Fractional Laplacian VIII Escuela-Taller 9 / 40

  3. Therefore |�∇ 2 u ( x ) y , y � + o ( | x | 3 ) | ˆ = dy | y | n + 2 s | y |≤ 1 Using Cauchy-Schwarz inequality |∇ 2 u ( x ) || y | 2 + | o ( | x | 3 ) | |∇ 2 u ( x ) y || y | + | o ( | x | 3 ) | ˆ ˆ ≤ dy ≤ dy | y | n + 2 s | y | n + 2 s | y |≤ 1 | y |≤ 1 ˆ 1 1 1 ˆ ˆ = C | y | n − 2 ( 1 − s ) dy = r 1 − 2 s d σ ( ω ) dr < ∞ | y |≤ 1 0 S n − 1 Now, for (b): � � 2 u ( x ) − u ( x + y ) − u ( x − y ) 1 ˆ ˆ � � dy � ≤ 4 � u � L ∞ ( R n ) | y | n + 2 s dy < ∞ � � | y | n + 2 s � � | y |≥ 1 | y |≥ 1 � Group 3 Fractional Laplacian VIII Escuela-Taller 9 / 40

  4. Translations and dilations Let h ∈ R n and λ > 0, the translation and dilation operators are defined, respectively, by τ h f ( x ) = f ( x + h ) ; δ λ f ( x ) = f ( λ x ) for every f : R n → R and every x ∈ R n Group 3 Fractional Laplacian VIII Escuela-Taller 10 / 40

  5. Translations and dilations Let h ∈ R n and λ > 0, the translation and dilation operators are defined, respectively, by τ h f ( x ) = f ( x + h ) ; δ λ f ( x ) = f ( λ x ) for every f : R n → R and every x ∈ R n Proposition 1 Let u ∈ S ( R n ) , then for every h ∈ R n and λ > 0 we have ( − ∆ ) s ( τ h u ) = τ h (( − ∆ ) s u ) and ( − ∆ ) s ( δ λ u ) = λ 2 s δ λ (( − ∆ ) s u ) Group 3 Fractional Laplacian VIII Escuela-Taller 10 / 40

  6. Translations and dilations Let h ∈ R n and λ > 0, the translation and dilation operators are defined, respectively, by τ h f ( x ) = f ( x + h ) ; δ λ f ( x ) = f ( λ x ) for every f : R n → R and every x ∈ R n Proposition 1 Let u ∈ S ( R n ) , then for every h ∈ R n and λ > 0 we have ( − ∆ ) s ( τ h u ) = τ h (( − ∆ ) s u ) and ( − ∆ ) s ( δ λ u ) = λ 2 s δ λ (( − ∆ ) s u ) In particular, ( − ∆ ) s is a homogeneous operator of order 2 s . Group 3 Fractional Laplacian VIII Escuela-Taller 10 / 40

  7. Orthogonal group Recall that the orthogonal group is O ( n ) = { T ∈ M n ( R ) : T t T = TT t = I } Group 3 Fractional Laplacian VIII Escuela-Taller 11 / 40

  8. Orthogonal group Recall that the orthogonal group is O ( n ) = { T ∈ M n ( R ) : T t T = TT t = I } The usual Laplacian satisfies ∆ ( u ◦ T ) = ∆ u ◦ T for every T ∈ O ( n ) . What about the fractional Laplacian? Group 3 Fractional Laplacian VIII Escuela-Taller 11 / 40

  9. Orthogonal group Recall that the orthogonal group is O ( n ) = { T ∈ M n ( R ) : T t T = TT t = I } The usual Laplacian satisfies ∆ ( u ◦ T ) = ∆ u ◦ T for every T ∈ O ( n ) . What about the fractional Laplacian? We say that a function f : R n → R has spherical symmetry if f ( x ) = f ∗ ( | x | ) for some f ∗ : R n → R or, equivalently, if f ( Tx ) = f ( x ) for every T ∈ O ( n ) and every x ∈ R Group 3 Fractional Laplacian VIII Escuela-Taller 11 / 40

  10. Orthogonal group Recall that the orthogonal group is O ( n ) = { T ∈ M n ( R ) : T t T = TT t = I } The usual Laplacian satisfies ∆ ( u ◦ T ) = ∆ u ◦ T for every T ∈ O ( n ) . What about the fractional Laplacian? We say that a function f : R n → R has spherical symmetry if f ( x ) = f ∗ ( | x | ) for some f ∗ : R n → R or, equivalently, if f ( Tx ) = f ( x ) for every T ∈ O ( n ) and every x ∈ R Proposition 2 Let u ∈ S ( R n ) (actually, it is enough that u ∈ C 2 ( R n ) ∩ L ∞ ( R n ) ) be a function with spherical symmetry. Then, ( − ∆ ) s has spherical symmetry. Group 3 Fractional Laplacian VIII Escuela-Taller 11 / 40

  11. Orthogonal group (Cont.) PROOF.- Group 3 Fractional Laplacian VIII Escuela-Taller 12 / 40

  12. Orthogonal group (Cont.) PROOF.- Let’s see that ( − ∆ ) s u ( Tx ) = ( − ∆ ) s u ( x ) for each T ∈ O ( n ) and each x ∈ R n Group 3 Fractional Laplacian VIII Escuela-Taller 12 / 40

  13. Orthogonal group (Cont.) PROOF.- Let’s see that ( − ∆ ) s u ( Tx ) = ( − ∆ ) s u ( x ) for each T ∈ O ( n ) and each x ∈ R n 2 u ∗ ( | Tx | ) − u ∗ ( | Tx + y | ) − u ∗ ( | Tx − y | ) ( − ∆ ) s u ( Tx ) = γ ( n , s ) ˆ dy | y | 2 n + s 2 R n 2 u ∗ ( | x | ) − u ∗ ( | x + T t y | ) − u ∗ ( | x − T t y | ) = γ ( n , s ) ˆ dy | y | 2 n + s 2 R n Group 3 Fractional Laplacian VIII Escuela-Taller 12 / 40

  14. Orthogonal group (Cont.) PROOF.- Let’s see that ( − ∆ ) s u ( Tx ) = ( − ∆ ) s u ( x ) for each T ∈ O ( n ) and each x ∈ R n 2 u ∗ ( | Tx | ) − u ∗ ( | Tx + y | ) − u ∗ ( | Tx − y | ) ( − ∆ ) s u ( Tx ) = γ ( n , s ) ˆ dy | y | 2 n + s 2 R n 2 u ∗ ( | x | ) − u ∗ ( | x + T t y | ) − u ∗ ( | x − T t y | ) = γ ( n , s ) ˆ dy | y | 2 n + s 2 R n Change of variable: z = T t y Group 3 Fractional Laplacian VIII Escuela-Taller 12 / 40

  15. Orthogonal group (Cont.) PROOF.- Let’s see that ( − ∆ ) s u ( Tx ) = ( − ∆ ) s u ( x ) for each T ∈ O ( n ) and each x ∈ R n 2 u ∗ ( | Tx | ) − u ∗ ( | Tx + y | ) − u ∗ ( | Tx − y | ) ( − ∆ ) s u ( Tx ) = γ ( n , s ) ˆ dy | y | 2 n + s 2 R n 2 u ∗ ( | x | ) − u ∗ ( | x + T t y | ) − u ∗ ( | x − T t y | ) = γ ( n , s ) ˆ dy | y | 2 n + s 2 R n Change of variable: z = T t y 2 u ∗ ( | x | ) − u ∗ ( | x + z | ) − u ∗ ( | x − z | ) = γ ( n , s ) ˆ dz | Tz | 2 n + s 2 R n Group 3 Fractional Laplacian VIII Escuela-Taller 12 / 40

  16. Orthogonal group (Cont.) PROOF.- Let’s see that ( − ∆ ) s u ( Tx ) = ( − ∆ ) s u ( x ) for each T ∈ O ( n ) and each x ∈ R n 2 u ∗ ( | Tx | ) − u ∗ ( | Tx + y | ) − u ∗ ( | Tx − y | ) ( − ∆ ) s u ( Tx ) = γ ( n , s ) ˆ dy | y | 2 n + s 2 R n 2 u ∗ ( | x | ) − u ∗ ( | x + T t y | ) − u ∗ ( | x − T t y | ) = γ ( n , s ) ˆ dy | y | 2 n + s 2 R n Change of variable: z = T t y 2 u ∗ ( | x | ) − u ∗ ( | x + z | ) − u ∗ ( | x − z | ) = γ ( n , s ) ˆ dz | Tz | 2 n + s 2 R n 2 u ∗ ( | x | ) − u ∗ ( | x + z | ) − u ∗ ( | x − z | ) = γ ( n , s ) ˆ dz = ( − ∆ ) s u ( x ) | z | 2 n + s 2 R n Group 3 Fractional Laplacian VIII Escuela-Taller 12 / 40

  17. Alternative expression for the fractional Laplacian Now we find a new pointwise expression for the fractional Laplacian which will be useful when we prove the equivalence of the different definitions. Group 3 Fractional Laplacian VIII Escuela-Taller 13 / 40

  18. Alternative expression for the fractional Laplacian Now we find a new pointwise expression for the fractional Laplacian which will be useful when we prove the equivalence of the different definitions. Theorem Let u ∈ S ( R n ) , then u ( x ) − u ( y ) ˆ ( − ∆ ) s u ( x ) = γ ( n , s ) P . V . | x − y | n + 2 s dy R n where P.V. means the Cauchy’s principal value, i.e. u ( x ) − u ( y ) u ( x ) − u ( y ) ˆ ˆ P . V . | x − y | n + 2 s dy = lim | x − y | n + 2 s dy ε → 0 + R n | x − y | > ε Group 3 Fractional Laplacian VIII Escuela-Taller 13 / 40

  19. Alternative expression for the fractional Laplacian (Cont.) PROOF.- Group 3 Fractional Laplacian VIII Escuela-Taller 14 / 40

  20. Alternative expression for the fractional Laplacian (Cont.) PROOF.- 2 u ( x ) − u ( x + y ) − u ( x − y ) ( − ∆ ) s u ( x ) = 1 ˆ 2 lim dy | y | n + 2 s ε → 0 + | y | > ε Group 3 Fractional Laplacian VIII Escuela-Taller 14 / 40

  21. Alternative expression for the fractional Laplacian (Cont.) PROOF.- 2 u ( x ) − u ( x + y ) − u ( x − y ) ( − ∆ ) s u ( x ) = 1 ˆ 2 lim dy | y | n + 2 s ε → 0 + | y | > ε = 1 u ( x ) − u ( x + y ) dy + 1 u ( x ) − u ( x − y ) ˆ ˆ 2 lim 2 lim dy | y | n + 2 s | y | n + 2 s ε → 0 + ε → 0 + | y | > ε | y | > ε Group 3 Fractional Laplacian VIII Escuela-Taller 14 / 40

  22. Alternative expression for the fractional Laplacian (Cont.) PROOF.- 2 u ( x ) − u ( x + y ) − u ( x − y ) ( − ∆ ) s u ( x ) = 1 ˆ 2 lim dy | y | n + 2 s ε → 0 + | y | > ε = 1 u ( x ) − u ( x + y ) dy + 1 u ( x ) − u ( x − y ) ˆ ˆ 2 lim 2 lim dy | y | n + 2 s | y | n + 2 s ε → 0 + ε → 0 + | y | > ε | y | > ε Changes of variables: x + y = z in the first integral x − y = z in the second integral Group 3 Fractional Laplacian VIII Escuela-Taller 14 / 40

  23. Alternative expression for the fractional Laplacian (Cont.) PROOF.- 2 u ( x ) − u ( x + y ) − u ( x − y ) ( − ∆ ) s u ( x ) = 1 ˆ 2 lim dy | y | n + 2 s ε → 0 + | y | > ε = 1 u ( x ) − u ( x + y ) dy + 1 u ( x ) − u ( x − y ) ˆ ˆ 2 lim 2 lim dy | y | n + 2 s | y | n + 2 s ε → 0 + ε → 0 + | y | > ε | y | > ε Changes of variables: x + y = z in the first integral x − y = z in the second integral = 1 u ( x ) − u ( z ) | z − x | n + 2 s dz + 1 u ( x ) − u ( z ) ˆ ˆ 2 lim 2 lim | x − z | n + 2 s dz ε → 0 + ε → 0 + | z − x | > ε | x − z | > ε Group 3 Fractional Laplacian VIII Escuela-Taller 14 / 40

  24. Alternative expression for the fractional Laplacian (Cont.) PROOF.- 2 u ( x ) − u ( x + y ) − u ( x − y ) ( − ∆ ) s u ( x ) = 1 ˆ 2 lim dy | y | n + 2 s ε → 0 + | y | > ε = 1 u ( x ) − u ( x + y ) dy + 1 u ( x ) − u ( x − y ) ˆ ˆ 2 lim 2 lim dy | y | n + 2 s | y | n + 2 s ε → 0 + ε → 0 + | y | > ε | y | > ε Changes of variables: x + y = z in the first integral x − y = z in the second integral = 1 u ( x ) − u ( z ) | z − x | n + 2 s dz + 1 u ( x ) − u ( z ) ˆ ˆ 2 lim 2 lim | x − z | n + 2 s dz ε → 0 + ε → 0 + | z − x | > ε | x − z | > ε u ( x ) − u ( z ) ˆ = lim | x − z | n + 2 s dz ε → 0 + | x − z | > ε Group 3 Fractional Laplacian VIII Escuela-Taller 14 / 40

  25. Another two definitions of the fractional Laplacian Group 3 Fractional Laplacian VIII Escuela-Taller 15 / 40

  26. Fourier transform We recall the definition of the Fourier transform, F , of a function f ∈ S ( R n ) : ˆ f ( ξ ) = ( 2 π ) − n / 2 R n f ( x ) e − ix · ξ dx , F ( f )( ξ ) = ˆ ξ ∈ R n , Group 3 Fractional Laplacian VIII Escuela-Taller 16 / 40

  27. Fourier transform We recall the definition of the Fourier transform, F , of a function f ∈ S ( R n ) : ˆ F ( f )( ξ ) = ˆ f ( ξ ) = ( 2 π ) − n / 2 R n f ( x ) e − ix · ξ dx , ξ ∈ R n , whose inverse function is given by ˆ F − 1 ( f )( x ) = ( 2 π ) − n / 2 R n f ( ξ ) e ix · ξ d ξ , x ∈ R n , Group 3 Fractional Laplacian VIII Escuela-Taller 16 / 40

  28. Fourier transform We recall the definition of the Fourier transform, F , of a function f ∈ S ( R n ) : ˆ R n f ( x ) e − 2 π ix · ξ dx , F ( f )( ξ ) = ˆ ξ ∈ R n , f ( ξ ) = whose inverse function is given by ˆ F − 1 ( f )( x ) = R n f ( ξ ) e 2 π ix · ξ d ξ , x ∈ R n , Group 3 Fractional Laplacian VIII Escuela-Taller 16 / 40

  29. Fourier transform We recall the definition of the Fourier transform, F , of a function f ∈ S ( R n ) : ˆ R n f ( x ) e − 2 π ix · ξ dx , F ( f )( ξ ) = ˆ ξ ∈ R n , f ( ξ ) = whose inverse function is given by ˆ F − 1 ( f )( x ) = R n f ( ξ ) e 2 π ix · ξ d ξ , x ∈ R n , so that ˆ f ( x ) = F − 1 ◦ F ( f )( x ) = ( 2 π ) − n / 2 f ( ξ ) e ix · ξ d ξ , ˆ x ∈ R n . R n Group 3 Fractional Laplacian VIII Escuela-Taller 16 / 40

  30. Definition of ( − ∆ ) s via the heat semigroup e t ∆ We will define ( − ∆ ) s f in terms of the heat semigroup e t ∆ , which is nothing but an operator such that maps every function f ∈ S ( R n ) to the solution of the heat equation with initial data given by f : ( x , t ) ∈ R n × ( 0, ∞ ) � v t = ∆ v , x ∈ R n . v ( x , 0 ) = f ( x ) , Using Fourier transform and its inverse and with a bit of magic, we can write ˆ R n e − t | ξ | 2 ˆ ˆ e t ∆ f ( x ) : = v ( x , t ) = ( 2 π ) − n / 2 f ( ξ ) e ix · ξ d ξ = R n W t ( x − z ) f ( z ) dz , where W t ( x ) = ( 4 π t ) − n / 2 e − | x | 2 x ∈ R n , 4 t , is the Gauss-Weierstrass kernel. Group 3 Fractional Laplacian VIII Escuela-Taller 17 / 40

  31. Definition of ( − ∆ ) s via the heat semigroup e t ∆ Inspired by the following numerical identity: for λ > 0, ˆ ∞ 1 ( e − t λ − 1 ) dt λ s = t 1 + s , 0 < s < 1, Γ ( − s ) 0 where ˆ ∞ ( e − r − 1 ) dr Γ ( − s ) = r 1 + s < 0; 0 we can think of ( − ∆ ) s as the following operator ˆ ∞ 1 ( e t ∆ f ( x ) − f ( x )) dt ( − ∆ ) s f ( x ) ∼ t 1 + s , 0 < s < 1. Γ ( − s ) 0 Group 3 Fractional Laplacian VIII Escuela-Taller 18 / 40

  32. Definition of ( − ∆ ) s via the Fourier Transform By the well-known properties of F with respect to derivatives, we have that, for f ∈ S ( R n ) , F [ − ∆ f ]( ξ ) = | ξ | 2 F ( f )( ξ ) , ξ ∈ R n , so it is reasonable to write something like ( − ∆ ) s f ( x ) ∼ F − 1 [ | · | 2 s F ( f )]( x ) , x ∈ R n , 0 < s < 1. Group 3 Fractional Laplacian VIII Escuela-Taller 19 / 40

  33. Definition of ( − ∆ ) s via the Fourier Transform By the well-known properties of F with respect to derivatives, we have that, for f ∈ S ( R n ) , F [ − ∆ f ]( ξ ) = | ξ | 2 F ( f )( ξ ) , ξ ∈ R n , so it is reasonable to write something like ( − ∆ ) s f ( x ) ∼ F − 1 [ | · | 2 s F ( f )]( x ) , x ∈ R n , 0 < s < 1. Group 3 Fractional Laplacian VIII Escuela-Taller 19 / 40

  34. Definition of ( − ∆ ) s via the Fourier Transform By the well-known properties of F with respect to derivatives, we have that, for f ∈ S ( R n ) , F [ − ∆ f ]( ξ ) = | 2 πξ | 2 F ( f )( ξ ) , ξ ∈ R n , so it is reasonable to write something like ( − ∆ ) s f ( x ) ∼ F − 1 [ | 2 π · | 2 s F ( f )]( x ) , x ∈ R n , 0 < s < 1. Group 3 Fractional Laplacian VIII Escuela-Taller 19 / 40

  35. ∼ is = Theorem (Lemma 2.1. P. Stinga’s PhD thesis) Given f ∈ S ( R n ) and 0 < s < 1, ˆ ∞ 1 ( e t ∆ f ( x ) − f ( x )) dt F − 1 [ | · | 2 s F ( f )]( x ) = x ∈ R n t 1 + s , Γ ( − s ) 0 and this two functions coincide in a pointwise way with ( − ∆ ) s f ( x ) when the constant γ ( n , s ) in its definition is given by γ ( n , s ) = 4 s Γ ( n / 2 + s ) − π n / 2 Γ ( − s ) > 0. Group 3 Fractional Laplacian VIII Escuela-Taller 20 / 40

  36. Everybody wants to be the fractional Laplacian Let x ∈ R n . By Fubini’s theorem and inverse Fourier formula, ˆ ∞ ˆ ∞ 1 1 ( e − t | ξ | 2 − 1 ) dt ( e t ∆ f ( x ) − f ( x )) dt ˆ f ( ξ ) e ix · ξ d ξ t 1 + s ˆ t 1 + s = Γ ( − s ) Γ ( − s ) 0 R n 0 ˆ ∞ 1 ˆ ( e − r − 1 ) dr r 1 + s | ξ | 2 s ˆ f ( y ) e ix · ξ dy = Γ ( − s ) R n 0 ˆ R n | ξ | 2 s ˆ f ( ξ ) e ix · ξ d ξ = F − 1 [ | · | 2 s F ( f )]( x ) . = Since f ∈ S ( R n ) , we have that ˆ ∞ | e t ∆ f ( x ) − f ( x ) | dt t 1 + s < ∞ , 0 and so Tonelli authorises us to apply Fubini’s theorem. Group 3 Fractional Laplacian VIII Escuela-Taller 21 / 40

  37. Everybody wants to be the fractional Laplacian Let x ∈ R n . By Fubini’s theorem and inverse Fourier formula, ˆ ∞ ˆ ∞ 1 1 ( e − t | ξ | 2 − 1 ) dt ( e t ∆ f ( x ) − f ( x )) dt ˆ f ( ξ ) e ix · ξ d ξ t 1 + s ˆ t 1 + s = Γ ( − s ) Γ ( − s ) 0 R n 0 ˆ ∞ 1 ˆ ( e − r − 1 ) dr r 1 + s | ξ | 2 s ˆ f ( y ) e ix · ξ dy = Γ ( − s ) R n 0 ˆ R n | ξ | 2 s ˆ f ( ξ ) e ix · ξ d ξ = F − 1 [ | · | 2 s F ( f )]( x ) . = Since f ∈ S ( R n ) , we have that ˆ ∞ | e t ∆ f ( x ) − f ( x ) | dt t 1 + s < ∞ , 0 and so Tonelli authorises us to apply Fubini’s theorem. Group 3 Fractional Laplacian VIII Escuela-Taller 21 / 40

  38. Everybody wants to be the fractional Laplacian Let x ∈ R n . By Fubini’s theorem and inverse Fourier formula, ˆ ∞ ˆ ∞ 1 1 ( e − t | ξ | 2 − 1 ) dt ( e t ∆ f ( x ) − f ( x )) dt ˆ f ( ξ ) e ix · ξ d ξ t 1 + s ˆ t 1 + s = Γ ( − s ) Γ ( − s ) 0 R n 0 ˆ ∞ 1 ˆ ( e − r − 1 ) dr r 1 + s | ξ | 2 s ˆ f ( y ) e ix · ξ dy = Γ ( − s ) R n 0 ˆ R n | ξ | 2 s ˆ f ( ξ ) e ix · ξ d ξ = F − 1 [ | · | 2 s F ( f )]( x ) . = Since f ∈ S ( R n ) , we have that ˆ ∞ | e t ∆ f ( x ) − f ( x ) | dt t 1 + s < ∞ , 0 and so Tonelli authorises us to apply Fubini’s theorem. Group 3 Fractional Laplacian VIII Escuela-Taller 21 / 40

  39. Everybody wants to be the fractional Laplacian Let x ∈ R n . By Fubini’s theorem and inverse Fourier formula, ˆ ∞ ˆ ∞ 1 1 ( e − t | ξ | 2 − 1 ) dt ( e t ∆ f ( x ) − f ( x )) dt ˆ f ( ξ ) e ix · ξ d ξ t 1 + s ˆ t 1 + s = Γ ( − s ) Γ ( − s ) 0 R n 0 ˆ ∞ 1 ˆ ( e − r − 1 ) dr r 1 + s | ξ | 2 s ˆ f ( y ) e ix · ξ dy = Γ ( − s ) R n 0 ˆ R n | ξ | 2 s ˆ f ( ξ ) e ix · ξ d ξ = F − 1 [ | · | 2 s F ( f )]( x ) . = Since f ∈ S ( R n ) , we have that ˆ ∞ | e t ∆ f ( x ) − f ( x ) | dt t 1 + s < ∞ , 0 and so Tonelli authorises us to apply Fubini’s theorem. Group 3 Fractional Laplacian VIII Escuela-Taller 21 / 40

  40. Everybody wants to be the fractional Laplacian Let x ∈ R n . By Fubini’s theorem and inverse Fourier formula, ˆ ∞ ˆ ∞ 1 1 ( e − t | ξ | 2 − 1 ) dt ( e t ∆ f ( x ) − f ( x )) dt ˆ f ( ξ ) e ix · ξ d ξ t 1 + s ˆ t 1 + s = Γ ( − s ) Γ ( − s ) 0 R n 0 ˆ ∞ 1 ˆ ( e − r − 1 ) dr r 1 + s | ξ | 2 s ˆ f ( y ) e ix · ξ dy = Γ ( − s ) R n 0 ˆ R n | ξ | 2 s ˆ f ( ξ ) e ix · ξ d ξ = F − 1 [ | · | 2 s F ( f )]( x ) . = Since f ∈ S ( R n ) , we have that ˆ ∞ | e t ∆ f ( x ) − f ( x ) | dt t 1 + s < ∞ , 0 and so Tonelli authorises us to apply Fubini’s theorem. Group 3 Fractional Laplacian VIII Escuela-Taller 21 / 40

  41. Everybody wants to be the fractional Laplacian Next, we will see that ˆ ∞ t 1 + s = 4 s Γ ( n / 2 + s ) f ( x ) − f ( z ) 1 ( e t ∆ f ( x ) − f ( x )) dt ˆ x ∈ R n . − π n / 2 Γ ( − s ) P.V. | x − z | n + 2 s dz , Γ ( − s ) 0 R n Let ε > 0. Using that � W t ( x − · ) � L 1 ( R n ) = 1 for any x ∈ R n , ˆ ∞ ˆ ∞ ( e t ∆ f ( x ) − f ( x )) dt ˆ R n W t ( x − z )( f ( z ) − f ( x )) dz dt t 1 + s = t 1 + s 0 0 = I ε + II ε . Group 3 Fractional Laplacian VIII Escuela-Taller 22 / 40

  42. Everybody wants to be the fractional Laplacian Next, we will see that ˆ ∞ t 1 + s = 4 s Γ ( n / 2 + s ) f ( x ) − f ( z ) 1 ( e t ∆ f ( x ) − f ( x )) dt ˆ x ∈ R n . − π n / 2 Γ ( − s ) P.V. | x − z | n + 2 s dz , Γ ( − s ) 0 R n Let ε > 0. Using that � W t ( x − · ) � L 1 ( R n ) = 1 for any x ∈ R n , ˆ ∞ ˆ ∞ ( e t ∆ f ( x ) − f ( x )) dt ˆ R n W t ( x − z )( f ( z ) − f ( x )) dz dt t 1 + s = t 1 + s 0 0 = I ε + II ε . Group 3 Fractional Laplacian VIII Escuela-Taller 22 / 40

  43. Everybody wants to be the fractional Laplacian Next, we will see that ˆ ∞ t 1 + s = 4 s Γ ( n / 2 + s ) f ( x ) − f ( z ) 1 ( e t ∆ f ( x ) − f ( x )) dt ˆ x ∈ R n . − π n / 2 Γ ( − s ) P.V. | x − z | n + 2 s dz , Γ ( − s ) 0 R n Let ε > 0. Using that � W t ( x − · ) � L 1 ( R n ) = 1 for any x ∈ R n , ˆ ∞ ˆ ∞ ( e t ∆ f ( x ) − f ( x )) dt ˆ R n W t ( x − z )( f ( z ) − f ( x )) dz dt t 1 + s = t 1 + s 0 0 = I ε + II ε . Group 3 Fractional Laplacian VIII Escuela-Taller 22 / 40

  44. Everybody wants to be the fractional Laplacian Next, we will see that ˆ ∞ t 1 + s = 4 s Γ ( n / 2 + s ) f ( x ) − f ( z ) 1 ( e t ∆ f ( x ) − f ( x )) dt ˆ x ∈ R n . − π n / 2 Γ ( − s ) P.V. | x − z | n + 2 s dz , Γ ( − s ) 0 R n Let ε > 0. Using that � W t ( x − · ) � L 1 ( R n ) = 1 for any x ∈ R n , ˆ ∞ ˆ ∞ ( e t ∆ f ( x ) − f ( x )) dt ˆ R n W t ( x − z )( f ( z ) − f ( x )) dz dt t 1 + s = t 1 + s 0 0 = I ε + II ε . Group 3 Fractional Laplacian VIII Escuela-Taller 22 / 40

  45. Everybody wants to be the fractional Laplacian Next, we will see that ˆ ∞ t 1 + s = 4 s Γ ( n / 2 + s ) f ( x ) − f ( z ) 1 ( e t ∆ f ( x ) − f ( x )) dt ˆ x ∈ R n . − π n / 2 Γ ( − s ) P.V. | x − z | n + 2 s dz , Γ ( − s ) 0 R n Let ε > 0. Using that � W t ( x − · ) � L 1 ( R n ) = 1 for any x ∈ R n , ˆ ∞ ˆ ∞ ( e t ∆ f ( x ) − f ( x )) dt ˆ R n W t ( x − z )( f ( z ) − f ( x )) dz dt t 1 + s = t 1 + s 0 0 = I ε + II ε . Group 3 Fractional Laplacian VIII Escuela-Taller 22 / 40

  46. Everybody wants to be the fractional Laplacian Using Fubini’s theorem, ˆ ∞ ˆ ( 4 π t ) − n / 2 e − | x − z | 2 ( f ( z ) − f ( x )) dt I ε = t 1 + s dz 4 t | x − z | > ε 0 ˆ ∞ ˆ ( 4 π t ) − n / 2 e − | x − z | 2 dt = ( f ( z ) − f ( x )) t 1 + s dz 4 t | x − z | > ε 0 ( f ( x ) − f ( z )) 4 s Γ ( n / 2 + s ) 1 ˆ = | x − z | n + 2 s dz − π n / 2 | x − z | > ε where we used the change of variables r = | x − z | 2 . 4 t Observe that I ε converges absolutely for any ε > 0 since f is bounded, so the use of Fubini’s theorem is licit. Group 3 Fractional Laplacian VIII Escuela-Taller 23 / 40

  47. Everybody wants to be the fractional Laplacian Using Fubini’s theorem, ˆ ∞ ˆ ( 4 π t ) − n / 2 e − | x − z | 2 ( f ( z ) − f ( x )) dt I ε = t 1 + s dz 4 t | x − z | > ε 0 ˆ ∞ ˆ ( 4 π t ) − n / 2 e − | x − z | 2 dt = ( f ( z ) − f ( x )) t 1 + s dz 4 t | x − z | > ε 0 ( f ( x ) − f ( z )) 4 s Γ ( n / 2 + s ) 1 ˆ = | x − z | n + 2 s dz − π n / 2 | x − z | > ε where we used the change of variables r = | x − z | 2 . 4 t Observe that I ε converges absolutely for any ε > 0 since f is bounded, so the use of Fubini’s theorem is licit. Group 3 Fractional Laplacian VIII Escuela-Taller 23 / 40

  48. Everybody wants to be the fractional Laplacian Using Fubini’s theorem, ˆ ∞ ˆ ( 4 π t ) − n / 2 e − | x − z | 2 ( f ( z ) − f ( x )) dt I ε = t 1 + s dz 4 t | x − z | > ε 0 ˆ ∞ ˆ ( 4 π t ) − n / 2 e − | x − z | 2 dt = ( f ( z ) − f ( x )) t 1 + s dz 4 t | x − z | > ε 0 ( f ( x ) − f ( z )) 4 s Γ ( n / 2 + s ) 1 ˆ = | x − z | n + 2 s dz − π n / 2 | x − z | > ε where we used the change of variables r = | x − z | 2 . 4 t Observe that I ε converges absolutely for any ε > 0 since f is bounded, so the use of Fubini’s theorem is licit. Group 3 Fractional Laplacian VIII Escuela-Taller 23 / 40

  49. Everybody wants to be the fractional Laplacian Using Fubini’s theorem, ˆ ∞ ˆ ( 4 π t ) − n / 2 e − | x − z | 2 ( f ( z ) − f ( x )) dt I ε = t 1 + s dz 4 t | x − z | > ε 0 ˆ ∞ ˆ ( 4 π t ) − n / 2 e − | x − z | 2 dt = ( f ( z ) − f ( x )) t 1 + s dz 4 t | x − z | > ε 0 ( f ( x ) − f ( z )) 4 s Γ ( n / 2 + s ) 1 ˆ = | x − z | n + 2 s dz − π n / 2 | x − z | > ε where we used the change of variables r = | x − z | 2 . 4 t Observe that I ε converges absolutely for any ε > 0 since f is bounded, so the use of Fubini’s theorem is licit. Group 3 Fractional Laplacian VIII Escuela-Taller 23 / 40

  50. Everybody wants to be the fractional Laplacian Using polar coordinates, ˆ ∞ ˆ W t ( x − z )( f ( z ) − f ( x )) dz dt II ε = t 1 + s 0 | x − z | < ε ˆ ∞ ˆ ε ( f ( x + rz ′ ) − f ( z )) dS ( z ′ ) dr dt e − r 2 ˆ ( 4 π t ) − n / 2 4 t r n − 1 = t 1 + s . | z ′ | = 1 0 0 By Taylor’s theorem, using the symmetry of the sphere, ˆ ( f ( x + rz ′ ) − f ( z )) dS ( z ′ ) = C n r 2 ∆ f ( x ) + O ( r 3 ) , | z ′ | = 1 thus ˆ ∞ e − r 2 ˆ ε dt 4 t r n + 1 | II ε | ≤ C n , ∆ f ( x ) t n / 2 + s t 0 0 ˆ ε r n + 1 C n , s r − n − 2 s dr = C n , ∆ f ( x ) , s ε 2 ( 1 − s ) . = C n , ∆ f ( x ) 0 Group 3 Fractional Laplacian VIII Escuela-Taller 24 / 40

  51. Everybody wants to be the fractional Laplacian Using polar coordinates, ˆ ∞ ˆ W t ( x − z )( f ( z ) − f ( x )) dz dt II ε = t 1 + s 0 | x − z | < ε ˆ ∞ ˆ ε ( f ( x + rz ′ ) − f ( z )) dS ( z ′ ) dr dt e − r 2 ˆ ( 4 π t ) − n / 2 4 t r n − 1 = t 1 + s . | z ′ | = 1 0 0 By Taylor’s theorem, using the symmetry of the sphere, ˆ ( f ( x + rz ′ ) − f ( z )) dS ( z ′ ) = C n r 2 ∆ f ( x ) + O ( r 3 ) , | z ′ | = 1 thus ˆ ∞ e − r 2 ˆ ε dt 4 t r n + 1 | II ε | ≤ C n , ∆ f ( x ) t n / 2 + s t 0 0 ˆ ε r n + 1 C n , s r − n − 2 s dr = C n , ∆ f ( x ) , s ε 2 ( 1 − s ) . = C n , ∆ f ( x ) 0 Group 3 Fractional Laplacian VIII Escuela-Taller 24 / 40

  52. Everybody wants to be the fractional Laplacian Using polar coordinates, ˆ ∞ ˆ W t ( x − z )( f ( z ) − f ( x )) dz dt II ε = t 1 + s 0 | x − z | < ε ˆ ∞ ˆ ε ( f ( x + rz ′ ) − f ( z )) dS ( z ′ ) dr dt e − r 2 ˆ ( 4 π t ) − n / 2 4 t r n − 1 = t 1 + s . | z ′ | = 1 0 0 By Taylor’s theorem, using the symmetry of the sphere, ˆ ( f ( x + rz ′ ) − f ( z )) dS ( z ′ ) = C n r 2 ∆ f ( x ) + O ( r 3 ) , | z ′ | = 1 thus ˆ ∞ e − r 2 ˆ ε dt 4 t r n + 1 | II ε | ≤ C n , ∆ f ( x ) t n / 2 + s t 0 0 ˆ ε r n + 1 C n , s r − n − 2 s dr = C n , ∆ f ( x ) , s ε 2 ( 1 − s ) . = C n , ∆ f ( x ) 0 Group 3 Fractional Laplacian VIII Escuela-Taller 24 / 40

  53. Everybody wants to be the fractional Laplacian Using polar coordinates, ˆ ∞ ˆ W t ( x − z )( f ( z ) − f ( x )) dz dt II ε = t 1 + s 0 | x − z | < ε ˆ ∞ ˆ ε ( f ( x + rz ′ ) − f ( z )) dS ( z ′ ) dr dt e − r 2 ˆ ( 4 π t ) − n / 2 4 t r n − 1 = t 1 + s . | z ′ | = 1 0 0 By Taylor’s theorem, using the symmetry of the sphere, ˆ ( f ( x + rz ′ ) − f ( z )) dS ( z ′ ) = C n r 2 ∆ f ( x ) + O ( r 3 ) , | z ′ | = 1 thus ˆ ∞ e − r 2 ˆ ε dt 4 t r n + 1 | II ε | ≤ C n , ∆ f ( x ) t n / 2 + s t 0 0 ˆ ε r n + 1 C n , s r − n − 2 s dr = C n , ∆ f ( x ) , s ε 2 ( 1 − s ) . = C n , ∆ f ( x ) 0 Group 3 Fractional Laplacian VIII Escuela-Taller 24 / 40

  54. Everybody wants to be the fractional Laplacian Using polar coordinates, ˆ ∞ ˆ W t ( x − z )( f ( z ) − f ( x )) dz dt II ε = t 1 + s 0 | x − z | < ε ˆ ∞ ˆ ε ( f ( x + rz ′ ) − f ( z )) dS ( z ′ ) dr dt e − r 2 ˆ ( 4 π t ) − n / 2 4 t r n − 1 = t 1 + s . | z ′ | = 1 0 0 By Taylor’s theorem, using the symmetry of the sphere, ˆ ( f ( x + rz ′ ) − f ( z )) dS ( z ′ ) = C n r 2 ∆ f ( x ) + O ( r 3 ) , | z ′ | = 1 thus ˆ ∞ e − r 2 ˆ ε dt 4 t r n + 1 | II ε | ≤ C n , ∆ f ( x ) t n / 2 + s t 0 0 ˆ ε r n + 1 C n , s r − n − 2 s dr = C n , ∆ f ( x ) , s ε 2 ( 1 − s ) . = C n , ∆ f ( x ) 0 Group 3 Fractional Laplacian VIII Escuela-Taller 24 / 40

  55. Everybody IS the fractional Laplacian This proves that II ε → 0 as ε → 0, so ˆ ∞ R n W t ( x − z )( f ( z ) − f ( x )) dz dt ˆ t 1 + s = l´ ε → 0 I ε + II ε ım 0 = 4 s Γ ( n / 2 + s ) f ( x ) − f ( z ) ˆ P.V. | x − z | n + 2 s dz − π n / 2 R n . . ⌣ This kind of computations (bearing in mind the exact expression of the constant γ ( n , s ) ) also prove the following pointwise convergence x ∈ R n as s → 0 + , ( − ∆ ) s f ( x ) → − ∆ f ( x ) , when f ∈ C 2 ( R n ) ∩ L ∞ ( R n ) (observe that, in S ( R n ) this is obvious by the definition via Fourier transform). Group 3 Fractional Laplacian VIII Escuela-Taller 25 / 40

  56. Everybody IS the fractional Laplacian This proves that II ε → 0 as ε → 0, so ˆ ∞ R n W t ( x − z )( f ( z ) − f ( x )) dz dt ˆ t 1 + s = l´ ε → 0 I ε + II ε ım 0 = 4 s Γ ( n / 2 + s ) f ( x ) − f ( z ) ˆ P.V. | x − z | n + 2 s dz − π n / 2 R n . . ⌣ This kind of computations (bearing in mind the exact expression of the constant γ ( n , s ) ) also prove the following pointwise convergence x ∈ R n as s → 0 + , ( − ∆ ) s f ( x ) → − ∆ f ( x ) , when f ∈ C 2 ( R n ) ∩ L ∞ ( R n ) (observe that, in S ( R n ) this is obvious by the definition via Fourier transform). Group 3 Fractional Laplacian VIII Escuela-Taller 25 / 40

  57. Everybody IS the fractional Laplacian This proves that II ε → 0 as ε → 0, so ˆ ∞ R n W t ( x − z )( f ( z ) − f ( x )) dz dt ˆ t 1 + s = l´ ε → 0 I ε + II ε ım 0 = 4 s Γ ( n / 2 + s ) f ( x ) − f ( z ) ˆ P.V. | x − z | n + 2 s dz − π n / 2 R n . . ⌣ This kind of computations (bearing in mind the exact expression of the constant γ ( n , s ) ) also prove the following pointwise convergence x ∈ R n as s → 0 + , ( − ∆ ) s f ( x ) → − ∆ f ( x ) , when f ∈ C 2 ( R n ) ∩ L ∞ ( R n ) (observe that, in S ( R n ) this is obvious by the definition via Fourier transform). Group 3 Fractional Laplacian VIII Escuela-Taller 25 / 40

  58. Everybody IS the fractional Laplacian This proves that II ε → 0 as ε → 0, so ˆ ∞ R n W t ( x − z )( f ( z ) − f ( x )) dz dt ˆ t 1 + s = l´ ε → 0 I ε + II ε ım 0 = 4 s Γ ( n / 2 + s ) f ( x ) − f ( z ) ˆ P.V. | x − z | n + 2 s dz − π n / 2 R n . . ⌣ This kind of computations (bearing in mind the exact expression of the constant γ ( n , s ) ) also prove the following pointwise convergence x ∈ R n as s → 0 + , ( − ∆ ) s f ( x ) → − ∆ f ( x ) , when f ∈ C 2 ( R n ) ∩ L ∞ ( R n ) (observe that, in S ( R n ) this is obvious by the definition via Fourier transform). Group 3 Fractional Laplacian VIII Escuela-Taller 25 / 40

  59. And last but not least Group 3 Fractional Laplacian VIII Escuela-Taller 26 / 40

  60. Extension Problem Let s ∈ ( 0, 1 ) and consider a = 1 − 2 s . We want to solve the extension problem  L a U ( x , y ) = div x , y ( y a ∇ x , y U ) = 0, x ∈ R n + , y > 0,  U ( x , 0 ) = u ( x ) , U ( x , y ) → 0 as y → ∞ .  The previous system can be written as  � � ∂ yy + a x ∈ R n − ∆ x U ( x , y ) = U ( x , y ) , + , y > 0, y ∂ y   (1) U ( x , 0 ) = u ( x ) ,  U ( x , y ) → 0 as y → ∞ .  Group 3 Fractional Laplacian VIII Escuela-Taller 27 / 40

  61. Extension Problem Let s ∈ ( 0, 1 ) and consider a = 1 − 2 s . We want to solve the extension problem  L a U ( x , y ) = div x , y ( y a ∇ x , y U ) = 0, x ∈ R n + , y > 0,  U ( x , 0 ) = u ( x ) , U ( x , y ) → 0 as y → ∞ .  The previous system can be written as  � � ∂ yy + a x ∈ R n − ∆ x U ( x , y ) = U ( x , y ) , + , y > 0, y ∂ y   (1) U ( x , 0 ) = u ( x ) ,  U ( x , y ) → 0 as y → ∞ .  Group 3 Fractional Laplacian VIII Escuela-Taller 27 / 40

  62. Extension Problem Let s ∈ ( 0, 1 ) and consider a = 1 − 2 s . We want to solve the extension problem  L a U ( x , y ) = div x , y ( y a ∇ x , y U ) = 0, x ∈ R n + , y > 0,  U ( x , 0 ) = u ( x ) , U ( x , y ) → 0 as y → ∞ .  The previous system can be written as  � � ∂ yy + a x ∈ R n − ∆ x U ( x , y ) = U ( x , y ) , + , y > 0, y ∂ y   (1) U ( x , 0 ) = u ( x ) ,  U ( x , y ) → 0 as y → ∞ .  Group 3 Fractional Laplacian VIII Escuela-Taller 27 / 40

  63. Extension Problem Theorem 1 (Extension Theorem) Let u ∈ S ( R n ) . Then, the solution U to the extension problem (1) is given by ˆ U ( x , y ) = ( P s ( · , y ) ⋆ u )( x ) = R n P s ( x − z , y ) u ( z ) dz , (2) where y 2 s P s ( x , y ) = Γ ( n / 2 + s ) (3) ( y 2 + | x | 2 ) ( n + 2 s ) / 2 π n / 2 Γ ( s ) is the Poisson Kernel for the extension problem in the half-space R n + 1 . For U as + in (2) one has ( − ∆ ) s u ( x ) = − 2 2 s − 1 Γ ( s ) y → 0 + y 1 − 2 s ∂ y U ( x , y ) . l´ ım (4) Γ ( 1 − s ) Group 3 Fractional Laplacian VIII Escuela-Taller 28 / 40

  64. Extension Problem Theorem 1 (Extension Theorem) Let u ∈ S ( R n ) . Then, the solution U to the extension problem (1) is given by ˆ U ( x , y ) = ( P s ( · , y ) ⋆ u )( x ) = R n P s ( x − z , y ) u ( z ) dz , (2) where y 2 s P s ( x , y ) = Γ ( n / 2 + s ) (3) ( y 2 + | x | 2 ) ( n + 2 s ) / 2 π n / 2 Γ ( s ) is the Poisson Kernel for the extension problem in the half-space R n + 1 . For U as + in (2) one has ( − ∆ ) s u ( x ) = − 2 2 s − 1 Γ ( s ) y → 0 + y 1 − 2 s ∂ y U ( x , y ) . l´ ım (4) Γ ( 1 − s ) Group 3 Fractional Laplacian VIII Escuela-Taller 28 / 40

  65. Extension Problem PROOF If we take a partial Fourier transform of (1) � U ( ξ , y ) + 1 − 2 s U ( ξ , y ) − 4 π 2 | ξ | 2 ˆ in R n + 1 ∂ yy ˆ ∂ y ˆ U ( ξ , y ) = 0 , + y ˆ ˆ x ∈ R n . U ( ξ , 0 ) = ˆ u ( ξ ) , U ( ξ , y ) → 0 as y → ∞ , If we fix ξ ∈ R n \ { 0 } and Y ( y ) = Y ξ ( y ) = ˆ U ( ξ , y ) , ′′ ( y ) + ( 1 − 2 s ) yY ′ ( y ) − 4 π 2 | ξ | 2 y 2 Y ( y ) = 0 y 2 Y � y in R + , Y ( 0 ) = ˆ u ( ξ ) , y ( y ) → 0 as y → ∞ , then it can be compared with the generalized modified Bessel equation: ′′ + ( 1 − 2 α ) yY ′ ( y ) + [ β 2 γ 2 y 2 γ + ( α − ν 2 γ 2 )] Y ( y ) = 0 y 2 Y (5) α = s , γ = 1, ν = s , β = 2 π | ξ | . Group 3 Fractional Laplacian VIII Escuela-Taller 29 / 40

  66. Extension Problem PROOF If we take a partial Fourier transform of (1) � U ( ξ , y ) + 1 − 2 s U ( ξ , y ) − 4 π 2 | ξ | 2 ˆ in R n + 1 ∂ yy ˆ ∂ y ˆ U ( ξ , y ) = 0 , + y ˆ ˆ x ∈ R n . U ( ξ , 0 ) = ˆ u ( ξ ) , U ( ξ , y ) → 0 as y → ∞ , If we fix ξ ∈ R n \ { 0 } and Y ( y ) = Y ξ ( y ) = ˆ U ( ξ , y ) , ′′ ( y ) + ( 1 − 2 s ) yY ′ ( y ) − 4 π 2 | ξ | 2 y 2 Y ( y ) = 0 y 2 Y � y in R + , Y ( 0 ) = ˆ u ( ξ ) , y ( y ) → 0 as y → ∞ , then it can be compared with the generalized modified Bessel equation: ′′ + ( 1 − 2 α ) yY ′ ( y ) + [ β 2 γ 2 y 2 γ + ( α − ν 2 γ 2 )] Y ( y ) = 0 y 2 Y (5) α = s , γ = 1, ν = s , β = 2 π | ξ | . Group 3 Fractional Laplacian VIII Escuela-Taller 29 / 40

  67. Extension Problem PROOF If we take a partial Fourier transform of (1) � U ( ξ , y ) + 1 − 2 s U ( ξ , y ) − 4 π 2 | ξ | 2 ˆ in R n + 1 ∂ yy ˆ ∂ y ˆ U ( ξ , y ) = 0 , + y ˆ ˆ x ∈ R n . U ( ξ , 0 ) = ˆ u ( ξ ) , U ( ξ , y ) → 0 as y → ∞ , If we fix ξ ∈ R n \ { 0 } and Y ( y ) = Y ξ ( y ) = ˆ U ( ξ , y ) , ′′ ( y ) + ( 1 − 2 s ) yY ′ ( y ) − 4 π 2 | ξ | 2 y 2 Y ( y ) = 0 y 2 Y � y in R + , Y ( 0 ) = ˆ u ( ξ ) , y ( y ) → 0 as y → ∞ , then it can be compared with the generalized modified Bessel equation: ′′ + ( 1 − 2 α ) yY ′ ( y ) + [ β 2 γ 2 y 2 γ + ( α − ν 2 γ 2 )] Y ( y ) = 0 y 2 Y (5) α = s , γ = 1, ν = s , β = 2 π | ξ | . Group 3 Fractional Laplacian VIII Escuela-Taller 29 / 40

  68. The general solutions of (5) are given by U ( ξ , y ) = Ay s I s ( 2 π | ξ | y ) + By s K s ( 2 π | ξ | y ) ˆ where I s and K s are the Bessel functions of second and third kind,both independent solutions of the modified Bessel equation of order s ′′ + z φ ′ − ( z 2 + s 2 ) φ = 0 z 2 φ (6) where ⇒ Y ( y ) = y α φ ( β y γ ) solution of (5). φ solution of (6) = Group 3 Fractional Laplacian VIII Escuela-Taller 30 / 40

  69. The general solutions of (5) are given by U ( ξ , y ) = Ay s I s ( 2 π | ξ | y ) + By s K s ( 2 π | ξ | y ) ˆ where I s and K s are the Bessel functions of second and third kind,both independent solutions of the modified Bessel equation of order s ′′ + z φ ′ − ( z 2 + s 2 ) φ = 0 z 2 φ (6) where ⇒ Y ( y ) = y α φ ( β y γ ) solution of (5). φ solution of (6) = Group 3 Fractional Laplacian VIII Escuela-Taller 30 / 40

  70. ( z / 2 ) s + 2 k ∑ ∞ k = 0 ( − 1 ) k J s ( z ) = Γ ( k + 1 ) Γ ( k + s + 1 ) , | z | < ∞ , | arg ( z ) | < π , ( z / 2 ) s + 2 k ∑ ∞ I s ( z ) = Γ ( k + 1 ) Γ ( k + s + 1 ) , | z | < ∞ , | arg ( z ) | < π , k = 0 I − s ( z ) − I s ( z ) π K s ( z ) = | arg ( z ) | < π . , 2 sin π s Group 3 Fractional Laplacian VIII Escuela-Taller 31 / 40

  71. Extension Problem The condition ˆ U ( ξ , y ) → 0 as y → ∞ forces A = 0. Using I s asymptotic behavior, y s I − s ( 2 π | ξ | y ) − y s I s ( 2 π | ξ | y ) By s K s ( 2 π | ξ | y ) = B π 2 sin π s B π 2 s − 1 π Γ ( 1 − s ) sin π s ( 2 π | ξ | ) − s = � � → Γ ( s ) Γ ( s − 1 ) = sin π s = B 2 s − 1 Γ ( s )( 2 π | ξ | ) − s . In order to fulfill the condition ˆ U ( ξ , 0 ) = ˆ u ( ξ ) , we impose U ( ξ , y ) = ( 2 π | ξ | ) s ˆ u ( ξ ) ˆ y s K s ( 2 π | ξ | y ) . (7) 2 s − 1 Γ ( s ) Group 3 Fractional Laplacian VIII Escuela-Taller 32 / 40

  72. Extension Problem The condition ˆ U ( ξ , y ) → 0 as y → ∞ forces A = 0. Using I s asymptotic behavior, y s I − s ( 2 π | ξ | y ) − y s I s ( 2 π | ξ | y ) By s K s ( 2 π | ξ | y ) = B π 2 sin π s B π 2 s − 1 π Γ ( 1 − s ) sin π s ( 2 π | ξ | ) − s = � � → Γ ( s ) Γ ( s − 1 ) = sin π s = B 2 s − 1 Γ ( s )( 2 π | ξ | ) − s . In order to fulfill the condition ˆ U ( ξ , 0 ) = ˆ u ( ξ ) , we impose U ( ξ , y ) = ( 2 π | ξ | ) s ˆ u ( ξ ) ˆ y s K s ( 2 π | ξ | y ) . (7) 2 s − 1 Γ ( s ) Group 3 Fractional Laplacian VIII Escuela-Taller 32 / 40

  73. Extension Problem The condition ˆ U ( ξ , y ) → 0 as y → ∞ forces A = 0. Using I s asymptotic behavior, y s I − s ( 2 π | ξ | y ) − y s I s ( 2 π | ξ | y ) By s K s ( 2 π | ξ | y ) = B π 2 sin π s B π 2 s − 1 π Γ ( 1 − s ) sin π s ( 2 π | ξ | ) − s = � � → Γ ( s ) Γ ( s − 1 ) = sin π s = B 2 s − 1 Γ ( s )( 2 π | ξ | ) − s . In order to fulfill the condition ˆ U ( ξ , 0 ) = ˆ u ( ξ ) , we impose U ( ξ , y ) = ( 2 π | ξ | ) s ˆ u ( ξ ) ˆ y s K s ( 2 π | ξ | y ) . (7) 2 s − 1 Γ ( s ) Group 3 Fractional Laplacian VIII Escuela-Taller 32 / 40

  74. We want to prove U ( x , y ) = ( P s ( · , y ) ⋆ u )( x ) . Taking inverse Fourier transform and using (7), we have to show that � ( 2 π | ξ | ) s y 2 s � = Γ ( n / 2 + s ) F − 1 2 s − 1 Γ ( s ) y s K s ( 2 π | ξ | y ) ( y 2 + | x | 2 ) ( n + 2 s ) / 2 . ξ → x π n / 2 Γ ( s ) Since the function in the left hand-side of (33) is spherically symmetric , proving (33) is equivalent to establishing the follow identity y 2 s F ξ → x ( 2 π s | ξ | s y s K s ( 2 π | ξ | y )) = Γ ( n / 2 + s ) ( y 2 + | x | 2 ) ( n + 2 s ) / 2 . π n / 2 (Hankel transform : H ≡ H − 1 for radial functions.) Group 3 Fractional Laplacian VIII Escuela-Taller 33 / 40

  75. We want to prove U ( x , y ) = ( P s ( · , y ) ⋆ u )( x ) . Taking inverse Fourier transform and using (7), we have to show that � ( 2 π | ξ | ) s y 2 s � = Γ ( n / 2 + s ) F − 1 2 s − 1 Γ ( s ) y s K s ( 2 π | ξ | y ) ( y 2 + | x | 2 ) ( n + 2 s ) / 2 . ξ → x π n / 2 Γ ( s ) Since the function in the left hand-side of (33) is spherically symmetric , proving (33) is equivalent to establishing the follow identity y 2 s F ξ → x ( 2 π s | ξ | s y s K s ( 2 π | ξ | y )) = Γ ( n / 2 + s ) ( y 2 + | x | 2 ) ( n + 2 s ) / 2 . π n / 2 (Hankel transform : H ≡ H − 1 for radial functions.) Group 3 Fractional Laplacian VIII Escuela-Taller 33 / 40

  76. We want to prove U ( x , y ) = ( P s ( · , y ) ⋆ u )( x ) . Taking inverse Fourier transform and using (7), we have to show that � ( 2 π | ξ | ) s y 2 s � = Γ ( n / 2 + s ) F − 1 2 s − 1 Γ ( s ) y s K s ( 2 π | ξ | y ) ( y 2 + | x | 2 ) ( n + 2 s ) / 2 . ξ → x π n / 2 Γ ( s ) Since the function in the left hand-side of (33) is spherically symmetric , proving (33) is equivalent to establishing the follow identity y 2 s F ξ → x ( 2 π s | ξ | s y s K s ( 2 π | ξ | y )) = Γ ( n / 2 + s ) ( y 2 + | x | 2 ) ( n + 2 s ) / 2 . π n / 2 (Hankel transform : H ≡ H − 1 for radial functions.) Group 3 Fractional Laplacian VIII Escuela-Taller 33 / 40

  77. We want to prove U ( x , y ) = ( P s ( · , y ) ⋆ u )( x ) . Taking inverse Fourier transform and using (7), we have to show that � ( 2 π | ξ | ) s y 2 s � = Γ ( n / 2 + s ) F − 1 2 s − 1 Γ ( s ) y s K s ( 2 π | ξ | y ) ( y 2 + | x | 2 ) ( n + 2 s ) / 2 . ξ → x π n / 2 Γ ( s ) Since the function in the left hand-side of (33) is spherically symmetric , proving (33) is equivalent to establishing the follow identity y 2 s F ξ → x ( 2 π s | ξ | s y s K s ( 2 π | ξ | y )) = Γ ( n / 2 + s ) ( y 2 + | x | 2 ) ( n + 2 s ) / 2 . π n / 2 (Hankel transform : H ≡ H − 1 for radial functions.) Group 3 Fractional Laplacian VIII Escuela-Taller 33 / 40

  78. Theorem 2 (Fourier-Bessel Representation) Let u ( x ) = f ( | x | ) , and suppose that t �→ t n / 2 f ( t ) J n / 2 − 1 ( 2 π | ξ | t ) ∈ L 1 ( R n ) . Then, ˆ ∞ u ( ξ ) = 2 π | ξ | − n / 2 + 1 t n / 2 f ( t ) J n / 2 − 1 ( 2 π | ξ | t ) dt . ˆ 0 Then, the latter identity (33) is equivalent to ˆ ∞ 2 2 π s + 1 y s t n / 2 + s K s ( 2 π yt ) J n / 2 − 1 ( 2 π | ξ | t ) dt | x | n / 2 − 1 0 y 2 s = Γ ( n / 2 + s ) ( y 2 + | x | 2 ) ( n + 2 s ) / 2 . π n / 2 Γ ( s ) Group 3 Fractional Laplacian VIII Escuela-Taller 34 / 40

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