Lie series An old concept going back to Sophus Lie, but already used by Newton and made rigorous by Cauchy. Widely exploited, e.g.

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1 ÄÁ Ë ÊÁ Ë Æ ÄÁ ÌÊ ÆË ÇÊÅË Lie erie An old concept going back to Sophu Lie, but already ued by Newton and made rigorou by Cauchy Widely exploited, eg, in differential geometry Ued a a method for numerical integration after Gröbner (957) Lie tranform Introduced by Hori (966) and Deprit (969) a a formal tool in perturbation theory Nowaday widely ued in Celetial Mechanic Formal theory: diregard the convergence of the erie produced, in the pirit of claical Celetial Mechanic Analytic tool: Cauchy etimate and generalization; convergence of the erie

2 3 Formal definition and propertie Developing perturbation tool diregardig convergence of erie 3 Formal expanion Ue a example the equation L V ϕ = λ j ϕ with V a vector field and ϕ a function, a in the problem of Poincaré and Siegel Conider the truncated polynomial ϕ(x) = ϕ 0 (x)++ϕ r (x), non homogeneou of degree r+ with arbitrary r We ay that the equation i atified in formal ene if by replacing the truncated polynomial we get ie, the difference i at leat of degre r +2 Write the full olution (if any) a L V (ϕ 0 ++ϕ r ) λ j (ϕ 0 ++ϕ r ) = O(x r+2 ), ϕ(x) = ϕ 0 (x)+(x)+ϕ r (x)+f (r), F (r) (x) = O( x r+2 ) The erie ϕ(x) = 0 ϕ (x) i aid to be aymptotic in cae F (r) (x) 0 for x 0 A convergent erie i alo aymptotic, but an aymptotic erie need not be convergent (ee next page) Expanion in a parameter ε ϕ(x) = ϕ 0 (x)+εϕ (x)+ε 2 ϕ 2 (x)+ (the coefficient need not be polynomial): the definition of aymptotic erie i traightforward: jut refer to power of ε Lie erie and Lie tranform 46

3 Example 3: A claical example The function Integrate by part: x Hence, formally, t ex t dt = x f(x) = x x t ex t dt, x > 0, t R t 2ex t dt,, f(x) = ( ) k k! x x, k k 0 x t kex t dt = x k k x a divergent erie for all x > 0, but, after r integration, we get the aymptotic expreion for x f (r) (x) = ( ) r! ++( )r, x x x r f(x) f (r) (x) = (r+)! dt < (r+)! x t k+ex t If, eg, x > r, then (by Stirling formula) f(x) f (r) (x) < r! r r e r which make the truncated expreion very ueful for an actual calculation x dt r! < tr+ x r t k+ex t dt Lie erie and Lie tranform 47

4 32 Lie derivative (refrehing) Let X a holomorphic vector field over an open domain D C n Lie derivative of a differentiable function f and of a differentiable vector field V: the linear operator L X f = d dt ( φ t X f ) t=0, L X V = d dt ( φ t X V ) t=0 Propertie: let f and g be differentiable function, V and W be differentiable vector field Leibniz rule: ( ) (L L X (fg) = fl X g +gl X f, L j X(fg) = X j f)( L j X g), (imilar for vector field) Commutator between vector field and between Lie derivative: Jacobi identity: alo written in either form j=0 L X V = {X,V}, [L X,L Y ] = L X L Y L Y L X {X,{V,W}}+{V,{W,X}}+{W,{X,V}} = 0, L X {V,W} = { L X V,W } + { V,L X W }, [L X,L Y ] = L {X,Y} In coordinate: L X f = n j= X j x j f, ( LX V ) j = n l= ( ) V j X j X l V l x l x l Lie erie and Lie tranform 48

5 33 Lie erie Let the vector field X be holomorphic Write the flow generated by X a: (3) (φ t x 0 ) j = x 0,j +tψ,j +t 2 ψ 2,j +, j =,,n, Actually a Taylor expanion, for ψ,j =! d dt (φt x 0 ) j t=0 But, ince ẋ = X(x), d dt (φt x 0 ) j d = X j,, t=0 x0 dt (φt x 0 ) j = L t=0 X X j, x0 Since L X x = X, write x0 (32) (φ t x0 x 0 ) j = x 0,j +tl X x j + t2 2! L2 X x j + t3 x0 3! L3 X x j + x0 namely an exponential erie Lie erie and Lie tranform 49

6 Example 32: Contant vector field Let F = C and X(x) = c C a contant vector field Then L X = c d dx, and o we get the tranlation φ t x 0 = x 0 +ct Example 33: Euler exponential function Let G = C and X(x) = x Hence L X = x d dx and ( ) φ t x 0 = x 0 +t+ t2 2! + t3 3! + = x 0 e t, namely the exponential function Example 34: Linear ytem Let G = C n and X(x) = Ax Then φ t x 0 = e x t x0 Exercie 35: Solve by erie the equation with initial data: (a) x(0) =, ẋ(0) = 0 (b) x(0) = 0, ẋ(0) = ẍ+x = 0 Lie erie and Lie tranform 50

7 Definition 36: The Lie erie operator i defined a the flow generated by a vector field exp(εl X ) = 0 ε! L X A linear operator acting on holomorphic function or vector field Further propertie: let f and g be holomorphic function, V and W be holomorphic vector field Preerve product: exp(εl X )(fg) = ( exp(εl X )f )( exp(εl X )g ) Local one parameter group: for arbitrary t, τ (formally, eay to prove) Preerve the commutator: exp(tl X ) exp(τl X ) = exp ( (t+τ)l X ) exp(εl X ){V,W} = {exp(εl X )V,exp(εL X )W} Invere: jut revere the vector field ( exp(εlx ) ) = exp( εl X ) The Lie erie i a particular cae of the Lie tranform Proof are included later for Lie tranform Lie erie and Lie tranform 5

8 34 Compoition of Lie erie Let X = {X,X 2,} be a generating equence of holomorphic vector field Define the equence of operator {S (0) X, S() X, S(2) X,} a (33) S (0) X =, S(r) X = exp( L Xr ) S (r ) X The compoition of Lie erie i defined (in formal ene) a the limit (34) S X = exp ( L Xr ) exp ( LX2 ) exp ( LX ) Inherit the propertie of Lie erie: linearity Preerve product: Preerve the commutator: S X (αf +βg) = αs X f +βs X g, S X (αv +βw) = αs X V +βs X V, S X (fg) = S X f ) S X g ) S X {V,W} = {S X V,S X W} Invere: the formal limit for r of the equence S (0) X =, S (r) X (r ) ( ) = S X exp LXr Lie erie and Lie tranform 52

9 35 Lie tranform Let X = {X,X 2,} be a generating equence of holomorphic vector field Definition 37: The Lie tranform operator i defined a (35) T X = 0E X with the equence E X of linear operator recurrently defined a (36) E0 X =, E X j = L X j E j X If X = {0,0,X k,0,0,} then (eay to check) j= T X = exp(l Xk ) Propoition 38: Let let f and g be holomorphic function, V and W be holomorphic vector field The Lie tranform i a linear operator with the following propertie: (i) Preerve product: T X (fg) = ( T X f )( T X g ) (ii) Preerve the commutator: (iii) Invere (ee later for a more practical method): ( ) TX = G X j, (37) T X {V,W} = {T X V,T X W} 0 G X 0 =, GX = j= j GX j L X j Lie erie and Lie tranform 53

10 Lemma 39: Let X = {X, X 2,}; f, g be function; V, W be vector field Then (i) E (fg) = j=0( Ej f )( E j g ) ; (ii) E L V = j=0 L E j VE j ; (iii) E {V,W} = j=0{ Ej V,E j W } ; Proof In all cae the claim i true for = For > ue induction (i) Calculate (hint at the end of ome line) m E (fg) = L X m E m (fg) = = = = = = m= m= m= j=0 m m L X m (E j f)(e m j g) * induction m m m= j=0 j=0 j=0 j= j m j=0 ( (LXm E j f ) (E m j g)+(e j f) ( L Xm E m j g )) * Leibniz rule m ( (LXm E m j f ) (E j g)+(e j f) ( L Xm E m j g )) * exchange j m j j m= m ( (LXm E j m f ) (E j g)+(e j f) ( L Xm E j m g )) * exchange um j j( (E j f)(e j g)+(e j f)(e j g) ) * definition of E j (E jf)(e j g)+ j=0 j (E j f)(e j g) = (E j f)(e j g) j= Lie erie and Lie tranform 54

11 (ii) Ue Jacobi identity a L V L W = L {V,W} +L W L V Calculate m E L V = L X m E m L V = = = = = = m= m= m= j=0 m m L X m L Ej VE m j m m m= j=0 j=0 j=0 j= j m j=0 (L LXm E jv +L Ej VL Xm ) E m j m ) (L LXm E j mve j +L Ej VL Xm E j m j m= m ) (L LXm j E j mve j +L Ej VL Xm E j m j ) (L E j VE j +L Ej VE j j L E j VE j + j=0 j L Ej VE j = L Ej VE j j=0 * induction * Jacobi identity * exchange j m j * exchange um * definition of E (iii) Ue (ii) on both ide: E {V,W} = E L V W = L Ej VE j W = {E j V,E j W} j=0 j=0 QED Lie erie and Lie tranform 55

12 Proof of propoition 38 The linearity i a trivial conequence of the linearity of E for 0 (i) Ue (i) of lemma 39 Calculate T X (fg) = E (fg) = (E j f)(e j g) 0 0 j=0 ( )( ) = E j f E k g = ( T X f )( T X g ) (ii) Ue (iii) of lemma 39 Calculate T X {V,W} = E {V,W} = {E j V,E j W} 0 0 j=0 { } = j V j 0E, E k W = {T X V,T X W} k 0 (iii) Uing the definition, mut prove that T X T X = j 0 k 0 ( ( ) G l ) E = 0 0 Split order by order, and prove that G 0 E 0 = (obviou) and G l E l = 0, Trivial for =, ince l 0 l=0 G l E l l=0 G 0 E +G E 0 = L X L X = 0 Lie erie and Lie tranform 56

13 For > ue induction Split the um a (38) G l E l = Show that the econd um cancel the firt one Elaborate l=0 l=0 l G l le l = l=0 m = = = m= m= l=0 m l=0 l l=0 m= = = l=0 l= which cancel the firt um in (38) l m= l= l G l le l + G le l, l=0 m l G ll Xm E l m m G ll Xm E l m m G l ml Xm E l m G l ml Xm E l l G le l l G le l, * definition of E l * exchange um * exchange l l m * exchange um * definition of G l * exchange l l QED Lie erie and Lie tranform 57

14 36 Near the identity tranformation of coordinate A family of holomorphic coordinate tranformation depending on the parameter ε: (39) y j = x j +εϕ,j (x)+ε 2 ϕ 2,j (x)+, j =,,n May be generated by a Lie erie: given an autonomou vector field X write the time ε flow a or, more explicitly y = exp ( εl X ) x y j = exp(εl X )x j = x j +εx j (x)+ ε2 2 L XX j (x)+, j =,,n May be generated by a Lie tranform: given the generating equence X = {X,X 2,} write y = T X x, ie, in coordinate, [ ] y j = x j +εx,j (x)+ε 2 2 L X X,j (x)+x 2,j (x) +, j =,,n For polynomial vector field the ε expanion i eaily replaced by a power erie expanion: jut tranform the coordinate, eg, a x = εξ But we may alo forget the parameter if X i mall of order in ome ene to be made precie (eg, with ome norm) Lie erie and Lie tranform 58

15 Lemma 30: Any near the identity tranformation of the form (39) may be written a a Lie tranform y = T X x with the generating equence (30) X,j = ϕ,j, X r,j = ϕ r,j r k= k r L X k E r k x j, j =,,n, r > 2 Lemma 3: Any near the identity tranformation of the form (39) may be repreented via a compoition of Lie erie of the form y = S X x with a generating equence {X, X 2,} that can be explicitly determined The contruction of the generating equence i traightforward, although there i no recurrent formula: jut proceed tep by tep The proof will be given later Cave canem: the generating equence for the compoition of Lie erie and for the Lie tranform are not the ame! Lemma 32: If the vector field X,X 2, are ymplectic (generated by a canonical vector field) then the tranformation generated by either the compoition of Lie erie or by the Lie tranform i ymplectic (canonical) Proof The tranformation preerve the commutator between vector field, ie, the Poion bracket between any two function QED Lie erie and Lie tranform 59

16 37 The exchange theorem (Gröbner, 960) Propoition 33: Let f be a function and V be a vector field Conider the near the identity tranformation y = T X x and denote by D the differential of the tranformation (namely, in coordinate, the jacobian matrix with element D j,k = y j x k ) Then f(y) = ( y=tx T x X f ) (x), D V(y) = ( T X V ) (x) y=tx x Similar tatement for the (compoition of) Lie erie Very ueful for practical calculation On the left member: ubtitute the tranformation y = x+εϕ (x)+ in the function f (vector field V multiplied by D, repectively); expand the reult in power of ε On the right member: Apply the Lie tranform operator to the function f (vector field V, repectively no multiplication by D ); jut rename the variable a x no further expanion The invere tranformation: on the left member: invert the tranformation (eg, by erie for holomorphic function) make the ubtitution, a above On the right member: Apply the algorithm for the invere to the function (vector field, repectively) ee next ection for a traightforward implementation Lie erie and Lie tranform 60

17 Hint for the formal proof Lie erie: it i quite natural Conider a function f Recall: a vector field generate a flow and a pull back ( y = φ ε x, φ ε f ) (x) = f ( φ ε (x) ) Expand φ ε f in power of ε and get f ( φ ε (x) ) = f(x)+ε df ε2 d 2 f (x)+ dε 2! dε +, d f 2 dε (x) = L X f x, Thi i the formal claim of the exchange theorem Cave canem: we have performed different operation, hence we hould produce a rigorou proof that the reult i the ame Lie tranform, defined in algebraic form Eay for polynomial A polynomial i contructed via product and um of function (the coordinate) The Lie tranform preerv both product and um A repeated ue of the latter propertie take the operator T X out of the ign of function, a claimed For power erie: the claim i true in formal ene, a for polynomial We hould produce proof of convergence For vector field: mut ue alo the property of preerving commutator (Poion bracket) The formal proof i not o hort See the note available on the web page, appendix A Lie erie and Lie tranform 6

18 38 Expanion in a mall parameter: the Lie triangle Problem: Let f = f 0 + εf + ε 2 f 2 + with holomorphic coefficient Want to determine the expanion g = g 0 +εg +ε 2 g 2 + of the tranformed function The triangular diagram for Lie erie g 0 f 0 g L X f 0 f g 2 2 L2 X f 0 L X f f 2 g 3 3! L3 Xf 0 2 L2 Xf L X f 2 f 3 g 4 4! L4 X f 0 3! L3 X f 2 L2 X f 2 L X f 3 f 4 Starting with the known function f 0,f, fill the triangle by column (a indicated by the arrow); align on the ame line all term with the ame power of ε a coefficient Determine g 0,g, by adding up all function on the ame line Lie erie and Lie tranform 62

19 The triangular diagram for a vector field of order ε 2 ha many empty cell: g 0 f 0 g f g 2 L X2 f 0 f 2 g 3 L X2 f f 3 g 4 2 L2 X 2 f 0 L X2 f 2 f 4 For a vector field of order ε a column ha one non zero element out of (pare triangle) General recurrent formula: g 0 = f 0, g r = r/k =0! L X f r k, where r/k denote the maximum integer not greater than (the floor of) r/k Lie erie and Lie tranform 63

20 The triangular diagram for the Lie tranform: imilar to the one for Lie erie g 0 f 0 g E f 0 f g 2 E 2 f 0 E f f 2 g 3 E 3 f 0 E 2 f E f 2 f 3 Recall E 0 =, j E = L χ j E j j= g 4 E 4 f 0 E 3 f E 2 f 2 E f 3 f 4 An effective algorithm for the invere Aume g = g 0 + εg + be known; want to determine the expanion f = f 0 +εf + of f = T X g From the firt line get f 0 = g 0, and fill the column for f 0 From the econd line get f = g E f 0, and fill the column for f From the third line get f 2 = g 2 E f E 2 f 0, and fill the column for f 2 And o on Lie erie and Lie tranform 64

21 32 Baic analytic tool A reviitation of the method of majorant developed by Cauchy 32 Cauchy inequalitie Let (0) be an open dik centered at the origin of C Let the function f be holomorphic in (0) and bounded on the cloed dik Uniform norm (alo called upremum norm, Chebihev norm, &c): f = up f(z) z (0) Cauchy etimate: f (0) f, f () (0)! f Follow from Cauchy integral formula f () (z) =! 2πi f(ζ) (ζ z) +dζ Extenion to a polydik (0) C n, with the ame uniform norm f (0) z j f, j n and (3) ++ f f z (0) z n n! n! f ++ n Lie erie and Lie tranform 65

22 322 Complex extenion of domain Let D be a ubet C n The complexification D i defined a D = (z) z D where (z) i the complex polydik of radiu > 0 with center z Uniform norm: f = up f(z) z D Lemma 34: For any poitive δ f p j δ δ f, j n D Proof Let z D δ We have δ (z) D, and o up z δ (z) f(z ) f By Cauchy inequality in a dik find f (z) p j δ f, j n for every z D δ Conclude f p j δ δ f, j n QED Lie erie and Lie tranform 66

23 323 Generalized Cauchy inequalitie for Lie derivative Define the uniform norm for a vector field V on an extended domain D C n a (32) V n = V j j= Lemma 35: Let X be an analytic bounded vector field on the complex domain D, and f a holomorphic bounded function on D ( d ) with 0 d < Then for 0 d < d < we have (33) Proof Pick x D ( d) and et L X f ( d) (d d ) X f ( d ) F x (τ) = f(x+τx(x)) Let τ X < (d d ) ; then x+τx(x) D ( d ) The dik τ < (d d ) / X i a ubet of D ( d ) F x (τ) i holomorphic in the dik τ < (d d ) / X and bounded by Fx (τ) f ( d ) By Cauchy inequality for derivative of a function of a ingle variable get df x dτ (0) (d d ) X f ( d ) By the definition of F x (τ) get Conclude ( LX f ) (x) for every x D ( d) d dτ F x = ( L X f ) (x), τ=0 (d d ) X f ( d ) QED Lie erie and Lie tranform 67

24 Lemma 36: Let X and V be holomorphic bounded vector field on D and on D ( d ),repectively, with 0 d < Then for 0 d < d < we have LX V ( d) Proof Recall that in coordinate we have ( LX V ) j = n l= 2 (d d ) X V ( d ) ( ) V j X j X l V l x l x l Apply lemma 35 to every component X j, V j n ( ) V j X j n ( X l V l x l x l (d d ) X l V j ( d ) + ) d V l ( d ) X j l= ( d) l= Ue the definition of the norm of a vector field Get ( L X V ) ( ) j X V (d d j ( d ) ) + V ( d ) X j, Ue again the definition of the norm of a vector field and get n ( L X V ) 2 j (d d ) X V ( d ) j= QED Lie erie and Lie tranform 68

25 Lemma 37: Let X,,X be holomorphic bounded vector field on D Let f and V be a function and a vector field, repectively, bounded on D ( d ) with 0 d < Then for 0 d < d < we have LX L X f ( d)! ( ) e X e (d d X f ( d ) ) LX L X V ( d)! ( ) 2e e (d d X X V ( d ) ) Proof True for = by lemma 35 and 36 Ue induction for > Set δ = (d d )/ Start with L X f ( d δ) δ X f ( d ) For j = 2,, conider the domain D ( d jδ) Apply j time lemma 35 with δ in place of d d, and get the recurrent formula LXj L X f δ X ( d jδ) j LXj L X f ( d (j )δ) Recall that δ = (d d )/, and get ( ) L X L X f ( d) X (d X f ( d d) ) Ue the elementary inequality e!, which prove the claim for the function f Same argument for the vector field V, with an extra factor 2 in the etimate of Lie derivative QED Lie erie and Lie tranform 69

26 324 Convergence of Lie erie Propoition 38: Let the function f and the analytic vector field X and V be bounded in D Then for every poitive < the following tatement hold true (i) The Lie erie exp(tl X )f i abolutely and uniformly convergent in the retricted domain D for all t T < e X (ii) The Lie erie exp(tl X )V i abolutely and uniformly convergent in the retricted domain D for all t T < 2e X ; ; Proof Ue the generalized Cauchy etimate of 37 For 0 < d < get the majorant 0 t! L X f ( d) f + f e >0 ( ) e X t d Thi i a geometric erie which i abolutely and uniformly convergent for t T < d e X Setting = d the claim follow Similar argument for the vector field QED Lie erie and Lie tranform 70

27 Corollary 39: The Lie erie operator atifie the following propertie: (i) It i bounded for t mall enough; eg, we have exp ( ) ( tl X f + 2 X t ) f for t T = 2e X exp ( ) ( tl X V + 2 X t ) V for t T 2 = 4e X (ii) With the ame limitation on t the remainder of the Lie erie i etimated a exp( ) r t j ( ) tl X f j! Lj X f r+ e X t 2 f, j=0 exp( ) r t j ( ) tl X V j! Lj X V r+ 2e X t 2 V, j=0, For the proof ue the majorant a in the proof of propoition 38 Lie erie and Lie tranform 7

28 325 Local flow and coordinate tranformation Propoition 320: Let the vector field X be analytic and bounded on a domain D with Then for every poitive δ < /2 there exit ε > 0 uch that the following hold true: for every ε < ε the Lie erie define a map φ ε x = exp(εl X )x atifying φ ε d d D 2δ φ ε (D δ ) D, D 2δ φ ε (D δ ) D An explicit etimate for ε i d ε = e e2 e δ X > δ 2e X and etting ε = δ we have 2e X φ ε x x < δ, φ ε x x < δ, x D δ φ ε d d The flow φ t of X i analytic in t at leat up to t = ε (the theorem of Cauchy on the olution of holomorphic differential equation) The flow at time ε i ued a a map, ie, a near the identity tranformation of coordinate The map induce an ε dependent deformation of coordinate For varying ε the relation ε = δ 2e X may be een a a good etimate of the deformation δ The proof i a variazione of the proof of propoition 38 and corollary 39 Lie erie and Lie tranform 72

29 326 Convergence of the compoition of Lie erie Propoition 32: Let the vector field X (), X (2), X (3), be analytic and bounded on a domain D If the convergence condition (34) X (j) < 4e j i atified, then the following tatement hold true: for every poitive δ atifying 2e j X(j) < δ < 2 the map Φ : D δ D defined a i analytic in D δ, and atifie there Φx = exp(l X (3)) exp(l X (2)) exp(l X ())x D 2δ Φ(D δ ) D, D 2δ Φ (D δ ) With the map Φ we may tranform any holomorphic function or vector field A direct proof of convergence of a compoition of Lie erie applied to either a function or a vector field i jut a variazione of the proof of thi propoition Lie erie and Lie tranform 73

30 Proof Define the equence of poitive number For every we have δ = δ X() j X(j) δ X () = δ j X(j) and propoition 320 applie with ε = Define by recurrence, o that δ = δ > 2e, ie, δ 2e X () >, 0 = +0 = δ, = δ, + = + +δ, o that + 2δ and + + Define the equence of map Φ, Φ 2, Φ 3, a Φ = exp(l X ()), Φ = exp(l X ()) Φ By recurrence, uing propoition 320, Φ i analytic on the domain D δ, and D Φ (D δ ) D + Start with D δ D δ and ditort it inward and outward by δ,δ 2,δ 3, a tated by the propoition Every Φ i holomorphic (compoition of holomorphic map) Foreveryx D δ wehave Φ x Φ x = exp(lx ())Φ x Φ x < δ Hencetheequence Φ x i abolutely convergent in D δ, becaue the um j δ j i abolutely convergent By Weiertra theorem, conclude that Φ i a holomorphic map QED Lie erie and Lie tranform 74

31 327 Convergence of the Lie tranform Propoition 322: Let the generating equence X, X 2, X 3, and the function f and the vector field V be analytic and bounded on a domain D Aume that the generating equence atifie X b Θ for ome b 0 and Θ > 0 Then for every poitive d < the following tatement hold true: (i) If the convergence condition eθ d +b < i atified then the Lie tranform T X f and T X f are abolutely and uniformly convergent in the retricted domain D ( d) (ii) If the convergence condition 2eΘ d +b < i atified then the Lie tranform T X V and T X V are abolutely and uniformly convergent in the retricted domain D ( d) We give the proof only for T X f For a vector field there i jut an extra factor 2 in the etimate of the Lie derivative For the invere exactly the ame etimate apply Lie erie and Lie tranform 75

32 Lemma 323: With the hypothee of propoition 322, for 0 < d < the operator E in the definition of the Lie tranform and of it invere atify ( ) ( ) eθ E f ( d) d +b Θ eθ, G f ( d) d f d +b Θ d f Proof Firt prove that with a equence {B } defined by recurrence a (35) B = Θ d, B = eθ d E V ( d) B f j= j + b j B j + b Θ d For = it i the etimate of lemma 35 For > look for a recurrent etimate May write E = α J c α F α where J i a et of indexe, {c α } α J a et of real coefficient, and {F α } α J a et of linear operator Every F α i a compoition of at mot operator L χj By lemma 37, F α f ( d)(,σ) A α f, with ome et of poitive contant {A α } α J Thu, E f ( d)(,σ) α J c α A α f Aume α J r c α A α B r (true for = 2) For > 2 write L Xj E j = α J j c α L Xj F α By lemma 37 we get (only the number of Lie derivative in F α count, no matter of the order; it i the ame for the operator G ) LXj E j f ( d) L Xj F α f ( d) e( j +) X j d Recalling X b Θ thi give (35) e( j +) X j A α f, hence d c α A α f α J j e( j +) X j B j f d σ Lie erie and Lie tranform 76

33 Prove that the equence (35) atifie B = Θ d, B = eθ d B < j= j + ( ) eθ d +b B b j B j + b Θ d Evident for =,2 For > 2 iolate the term j = in the um in (35), and change the ummation index j to j : B = eθ + ( )b eθ 2 ( ) j + b d B j ΘB j + b 2 Θ d d eθ + ( )b B ; d B (the term in quare bracket i B ) Thi i the wanted recurrent formula j= QED Proof of propoition 322 By lemma 323 we have ) d +b Θ, d f TX f ( d) E f ( d) ( eθ 0 0 ( ) eθ which i a convergent erie in view of hypothei d +b < By Weiertra theorem, the um of the erie T X f i analytic in the domain D ( d) QED Lie erie and Lie tranform 77

34 Lie erie and Lie tranform 78