Hadamard and Perron JWR. October 18, On page 23 of his famous monograph [2], D. V. Anosov writes

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1 Hadamard and Perron JWR October 18, 1999 On page 23 of his famous monograph [2], D. V. Anosov writes Every five years or so, if not more often, someone discovers the theorem of Hadamard and Perron proving it either by Hadamard s method or Perron s. I myself have been guilty of this. 1 Notation If X, d X and Y, d Y are metric spaces and T : X Y is a map then the Lipschitz constant of T is the quantity { } dy T x 1, T x 2 lipt = sup : x 1, x 2 X x 1 x 2. d X x 1, x 2 If X and Y are Banach spaces and q : X Y satisfies q0 = 0 then the Lipschitz constant at 0 is the quantity { } qx Y q = sup : x X \ 0 x X 2 Lipschitz Inverse Function Theorem Theorem 2.1. Suppose that X is a Banach space and has form Then f is a lipeomorphism and f : X X fx = x + rx lipr < 1. lipf 1 1 lipr 1. 1

2 Proof. Choose u X and define ξ u : X X by ξ u x = u rx. Thus fx = u if and only if ξ u x = x. As lipξ u = lipr < 1 there is a unique fixed point x for each choice of u by the Contraction Mapping Principle; thus f is bijective. To estimate the Lipschitz constant of f 1 we compute f 1 u f 1 v = x y = ξ u x ξ v y = u rx v ry u v + lipr x y x = f 1 u and y = f 1 v. This proves as required. = u v + lipr f 1 u f 1 v f 1 u f 1 v 1 lipr 1 u v 3 The Resolvent Inequality Lemma 3.1. Now suppose f i : X X for i = 1, 2 both have form f i x = x + r i x lipr i < 1 and now assume also that r i 0 = 0. Then f 1 1 f 1 2 c 1 c 2 r 1 r 2 c i = 1 lipr i 1. Proof. For u X we have f 1 i u = u r i f 1 i u so that and f u f2 u τ + σ! τ = r 1 f 1 1 u r 1f 1 2 u σ = r 1 r 2 f 1 2 u. 2

3 for τ we have so that! becomes τ lipr 1 f1 1 1 u f2 u For σ we have 1 lipr 1 f u f2 u σ. σ r 1 r 2 f 1 2 u. But lipf 1 2 c 2 by the Lipschitz Inverse Function Theorem 2.1 so f 1 2 u c 2 u as f2 1 0 = 0. Combining these inequalities gives the estimate as required. f1 1 1 u f2 u c 1c 2 r 1 r 2 u For linear transformations the Lipschitz constant and the Lipschitz constant from 0 are both equal to the operator norm and the inequality just proved takes the form T1 1 T2 1 T1 1 T2 1 T 1 T 2. This in turn follows immediately from the identity When T 1 and T 2 commute this is T 1 1 T 1 2 = T 1 1 T 2 T 1 T 1 2. T1 1 T2 1 = T 2 T 1 T1 1 T2 1 and a special case of this is the Resolvent Identity R λ R µ = µ λr λ R µ T 1 = λ T, T 2 = µ T, R λ = T 1 1, and R µ = T Graph Transform Lemma The arguments in this section come from [4]. Let X, Y, U, V be Banach spaces ultimately we take X = U and Y = V and have the form F : X Y U V F x, y = Ax + Rx, y, Hx, y A : X U, R : X Y U, H : X Y V. 3

4 Lemma 4.1. Assume that the map A : X U is bijective and Then for each α : X Y satisfying lipa 1 lipr < 1 lipα 1 there exists a necessarily unique because the graph of a function determines the function uniquely β : U V satisfying graphβ = F graphα The map β is called the graph transform of α by F. Proof. We can rewrite this as follows. Define depending on α by Then we must solve φ : X U φ = A + R 1, α. β φ = H 1, α. This suggests inverting φ. Now φ has the form φ = 1 + R 1, α A 1 A. A is bijective. The term R 1, α A 1 has a Lipschitz constant satisfying lipr 1, α A 1 liprlipa 1 < 1 since lip1, α = 1. The norm on U V is u, v = max u, v. According to the Lipschitz Inverse Function Theorem 2.1, the map φ is a bijection so β is given by β = H 1, α φ 1. 5 Hadamard The arguments in this section come from [4]. Let X and Y be Banach spaces and F : X Y X Y have the form F x, y = Ax + Rx, y, Hx, y A : X X, R : X Y X, H : X Y Y. 4

5 Theorem 5.1. Assume that the map A : X X is bijective, that and that Assume also lipa 1 liph < 1 1 A0 = 0, R0, 0 = 0, H0, 0 = 0. 2 lipr < ɛ 3 ɛ > 0 is a small constant depending only on liph and lipa 1 defined explicitly in the proof. Then there exists a unique α : X Y satisfying and F graphα = graphα α0 = 0, lipα 1. Proof. According to the Graph Transform Lemma 4.1 each α : X Y with lipα 1 determines a unique β : X Y satisfying F graphα = graphβ. This graph transform is defined by β := H 1, α φ 1, φ := A + R 1, α. 4 To satisfy the hypothesis of Lemma 4.1 we require only ɛ < lipa 1 1. Denote by G the space of all α : X Y such that α0 = 0 and lipα 1. We first show that the graph transform maps G to itself. By hypothesis 2 we have F 0, 0 = 0, 0 so β0 = 0 follows from α0 = 0. Equation 4 gives the estimate lipβ liphlipφ 1. But φ = 1 + R 1, α A 1 A and hence by the Lipschitz Inverse Function Theorem 2.1 we have the estimate Thus lipφ 1 lipa 1 C C = 1 liprlipa lipβ liphlipa 1 C. Now C is near 1 by 3 so that lipβ 1 follows from the hypothesis 1. Next we show that the map which assigns to α G its graph transform β G is a strict contraction in the norm, i.e. there is a constant λ < 1 such that β 1 β 2 λ α 1 α 2 6 whenever α i 0 = 0 and lipα 1 1 for i = 1, 2 and β i is the graph transform of α i. Abbreviate φ i := A + R 1, α i, r i = R 1, α i A 1, f i = 1 + r i, 5

6 so φ i = f i A and hence φ 1 i and = A 1 f 1 i. For u X we have β 1 u β 2 u σ + τ σ := H 1, α 1 φ 1 1 u H 1, α 1 φ 1 2 u τ := H 1, α 1 φ 1 2 u H 1, α 2 φ 1 2 u. Recall that lip1, α = 1 and C = 1 liprlipa 1 1. We estimate σ: σ liph φ 1 1 u φ 1 2 u liphlipa 1 f1 1 1 u f2 liphlipa 1 C 2 r 1 u r 2 u We estimate τ: liphlipa 1 C 2 lipr α 1 A 1 u α 2 A 1 u liphlipa 1 C 2 lipr α 1 α 2 A 1 u liphlipa 1 2 C 2 lipr α 1 α 2 u τ liph α 1 α 2 φ 1 2 u liph α 1 α 2 φ 1 2 u liph α 1 α 2 lipa 1 C u Combining the estimates for σ and τ gives β 1 u β 2 u λ α 1 α 2 u λ < 1 for lipr small enough. This proves 6. The space G is a complete metric space in the metric dα 1, α 2 = α 1 α 2. This is because convergence in the norm implies pointwise convergence and uniform Lipschitz estimates are preserved under pointwise limits. Hence the theorem follows from the Contraction Mapping Principle. 6 Perron The arguments in this section come from [8]. Let X and Y be Banach spaces and F : X Y X Y have the form F x, y = Kx, y, By + Sx, y K : X Y X, B : Y Y, S : X Y Y. 6

7 Theorem 6.1. Assume that the map B : Y Y is linear and invertible, that B 1 lipk + lips < 1, 7 and that K0, 0 = 0, S0, 0 = 0. Choose δ 0, 1 so small that λ := B δlipk + lips < 1 8 and assume that lips is so small that Then there exists a unique α : X Y satisfying and B 1 δ lipk + lips < δ. 9 F graphα = graphα α0 = 0, lipα δ. Proof. The condition F graphα = graphα may be written in the form which is in turn equivalent to Bα + S 1, α = α K 1, α α = Γα Γα : X Y is defined by Γα = B α 1 K 1, α S 1, α. 10 Lemma 6.2. Suppose α : X Y satisfies α0 = 0 and lipα δ Then β = Γα satisfies these same conditions: β0 = 0 and lipβ δ. Proof. Since K0, 0 = 0, S0, 0 = 0, B 1 is linear, and α0 = 0 it follows that β0 = 0 from the definition of β. For the Lipschitz estimate we compute lipβ B 1 lipα K 1, α S 1, α B 1 lipα K 1, α + lips 1, α B 1 lipαlipklip1, α + lipslip1, α = B 1 lipαlipk + lips B 1 δ lipk + lips δ. 7

8 Here we have use the fact that lip1, α = 1 from lipα δ 1. completes the proof of lemma 6.2. This Lemma 6.3. The map Γ is a contraction in the norm. More precisely, if α i : X Y for i = 1, 2 satisfies α i 0 = 0 and lipα i δ and if β i = Γα i, then β 1 β 2 λ α 1 α 2. Proof. Choose x X. Then For σ we have For τ we have β 1 x β 2 x B 1 σ + τ + ρ σ = α 1 Kx, α 1 x α 1 Kx, α 2 x τ = α 1 Kx, α 2 x α 2 Kx, α 2 x ρ = Sx, α 1 x Sx, α 2 x σ lipα 1 lipk α 1 x α 2 x δlipk α 1 x α 2 x τ α 1 α 2 lipk max x, α 2 x = α 1 α 2 lipk x since α 2 0 = 0,K0, 0 = 0, and lipα 2 δ 1. For ρ we have ρ lips α 1 x α 2 x Now divide by x and take the supremum over x. This completes the proof of lemma 6.3. Now the space G δ of all α : X Y with α0 = 0 and lipα δ form a complete metric space in the metric dα 1, α 2 = α 1 α 2. This is because convergence in the norm implies pointwise convergence and uniform Lipscitz estimates are preserved under pointwise limits. According to lemmas 6.2 and 6.3 the map Γ is a strict contraction on this space. This completes the proof of Theorem Smoothness Lemma 7.1 Fiber Contraction Principle. Let G be a topological space, H be a complete metric space, and Φ : G H G H a map of form Assume Φg, h = Γg, g h. 8

9 i Γ : G G has an attractive fixed point. ii For each h H the map G H : g g h is continuous. iii There exists ρ [0, 1 such that d g h 1, g h 2 ρdh 1, h 2 for all g G and h 1, h 1 H. Then Φ : H H has an attractive fixed point. Proof. Let ḡ be the fixed point of Γ and h be the fixed point of the contraction map ḡ. Choose g, h H H and let g n, h n = Φ n g, h, and σ n = dh n, h. Then g n ḡ as n by i; we must show σ n 0 as n. Now by iii dh n+1, h = d gn h n, h d gn h n, gn h + d gn h, h ρdh n, h + d gn h, h i.e. σ n+1 ρσ n + δ n δ n = d gn h, h. Hence by induction For 0 < m < n we have σ n ρ n σ 0 + C n 1 σ n ρ n σ 0 + ρ n 1 k δ k. m ρ n 1 k + c k=0 n 1 k=0 k=m+1 ρ n 1 k ρ n σ 0 + Cρn 1 m + c 1 ρ C = max 0 k m δ k max k δ k and c = max m<k n δ k max m<k δ k. But ḡ h = h so δ n = d gn h, ḡ h so δ n 0 as n by i and ii. Hence σ n is small if n > 2m and m is large. Theorem 7.2. Assume that F : X Y X Y is a C k diffeomorphism on a product of Banach spaces of form F x, y = Ax + Rx, y, By + Sx, y R0, 0 = 0, S0, 0 = 0, dr0, 0 = 0, ds0, 0 = 0 and A LX, X and B LY, Y are invertible linear maps satisfying the spectral gap condition A 1 i B < 1, i = 1, 2,..., k. 11 Then for each k = 1, 2,... there exists δ k > 0 such that if R k + S k < δ k, then the hypotheses of Theorem 5.1 hold and the map α : X Y defined in Theorem 5.1 is C k. 9

10 Proof. First we show that with Hx, y = By + Sx, y the hypotheses of Theorem 5.1 hold for R 1 + S 1 < δ 1 and δ = δ 1 sufficiently small. This follows easily from lipr R 1, lips S 1, lipa 1 = A 1, lipb = B. and the fact that 11 with i = 1 implies that 1 and 3 hold for R = 0 and S = 0. Let G be the metric space of all maps α : X Y with α0 = 0 and lipα 1 with metric dα 1, α 2 = α 1 α 2 as in the proof of Theorem 5.1. Define Γ : G G by Γα := Bα φ 1 + S 1, α φ 1, φ := A + R 1, α, i.e. Γα is the graph transform of α as in 4 in the proof of Theorem 5.1. Assume that α is smooth. Then so is β = Γα. Differentiating βx with respect to x gives dβx = Bdαx + dsx, y 1, dαx dφx 1 x = φ 1 x and y = αφ 1 x. From dφ 1 x = dφφ 1 x 1 we get dφ 1 x = 1 A + drx, y 1, dαx In both formulas the derivative of α is evaluated at x so combining gives a formula of form dβx = J 1 x, αx, dαx. Successive differentiation gives a formula of form j k indicates the k-jet, i.e. j k βx = J k x, αx, j k αx j k αx = dαx, d 2 αx,, d k αx. The Banach space of bounded linear maps from X to Y is denoted by LX, Y, the space L k X, Y bounded k-multilinear maps from X k to Y is defined inductively by the equations L 0 X, Y = Y, L k+1 X, Y = LX, L k X, Y and the subspace of symmetric maps is denoted by L k symx, Y L k X, Y. Define P k X, Y = LX, Y L 2 symx, Y L k symx, Y so that the k-jet of a C k map α : X Y is a map j k α : X P k X, Y. Let H be the Banach space of all bounded continuous maps h : X P k X, Y with the sup norm. For α G define α : H H by α hx = J k x, αx, hx. 10

11 If R and S vanish identically, we have that Γαx = BαA 1 x and hence α hx = BA ha 1 x A : P k X, Y P k X, Y is defined by A p 1, p 2,..., p k ˆx = p 1 A 1ˆx, p 2 A 1ˆx 2,..., p k A 1ˆx k. The spectral gap condition 11 guarantees that α is a contraction uniformly in α when R and S vanish identically and hence if R k znd S k are sufficiently small. By the Fiber Contraction Principle the map Φα, h = Γα, α h has an attractive fixed point. But if α n, h n = Φ n α, h, α is C k. and h = j k α, then h n = j k α n, so the first k derivatives of α n converge uniformly and the limit is also C k. Remark 7.3. A similar proof is given in [8] using the map Γ from the proof of Theorem 6.1 in place of the graph transform. Example 7.4. We show that Theorem 7.2 fails for k =. Choose a smooth function S : R R with S0 = ds0 = 0. Take X = R 2, Y = R, µ = 1/2, and let F : X Y X Y be a diffeomorphism such that F l, x, y = l, lx, µy + Sx in a neighborhood U of l, x, y = 1, 0, 0. It follows that this point is a fixed point of F. We show that F has no invariant manifold graphα α : X Y is smooth and α1, 0 = 0. The condition F graphα = graphα implies that µαl, x + hx = αl, lx for l, x, αl, x U. If α C this implies µ n α x n l, x + dn Sx = l n n α l, lx xn Taking l, x = µ 1/n, 0 and n is so large that l, x, αl, x U shows that d n S0 = 0. Thus if ds n 0 0 for infinitely many n no such smooth α exists. By multiplying the nonlinear terms by a smooth cuttoff function and rescaling we can construct, for any positive integer k, a smooth diffeomorphism F l, x, y = l, x, µy + Ql, x, y Q vanishes to first order at the fixed point, the C k norm of Q is as small as desired, and the invariant manifold y = αl, x of Theorem 7.2 is not smooth. This example comes from Lanford s lecture notes [7] it was adapted it from [9]. Now we sketch the proof of smoothness from [5]. It is a discrete version of the proofs of [3] and [6], but unlike those, it explicitly uses the infinite dimensional implicit function theorem. Let F be a C r map r 1 from a Banach space to itself with a hyperbolic fixed point at the origin. Thus after a suitable linear change of co-ordinates and 11

12 choice of norms, the Banach space is a product X Y of closed subspaces and F : X Y X Y has the form F x, y = Ax + Rx, y, By + Sx, y A is a linear contraction on X, B 1 is a linear contraction on Y, and R, S, DR, DS vanish at the origin. Give X Y the product norm and let B X Y = B X B Y be a small ball about the origin. The stable manifold theorem asserts the existence of a necessarily unique C r map α: B X B Y such that the graph of α is precisely the set of all z B X Y such that F n z B X Y for all n 0. It follows that F n z 0 as n for z graphα. Choose z = x, αx graphα and define x n, y n = z n = F n z so that It follows easily that x n+1 = Ax n + Rz n, y n+1 = By n + Sz n. n 1 x n = A n x + A n 1 ν Rz ν, ν=0 y n = B n 1 ν Sz ν. ν=n The former is by induction and the first formula in. The latter follows by letting m go to infinity in the formula n+m 1 y n = B m y n+m B n 1 ν Sz ν which in turn follows by induction from the formula y n = B 1 y n+1 B 1 Sz n, a consequence of the second formula in. The equations may be written in the form Γx, γx = 0, γx denotes the sequence {z n } and, for x B X and any sequence σ of elements of B X Y such that σ n 0 as n, Γx, σ is the sequence defined by n 1 Γx, σ n = σ n A n x + A n 1 ν Rσ ν, B n 1 ν Sσ ν. ν=0 For a suitable Banach space S of sequences, Γ: B X S S is C r and Γ0, 0 = 0 and D 2 Γ0, 0 = identity. Thus by the implicit function theorem can be solved and γ is C r. It it easily shown that α is given by γx 0 = x, αx, so that α is C r by the smoothness of the evaluation map. References [1] R. Abraham & J. Robbin: Transversal mappings and flows, W. A, Benjamin, ν=n ν=n 12

13 [2] D. V. Anosov: Geodesic flows on closed Riemann manifolds with negative curvature, Proceedings of the Steklov Institute AMS [3] E. Coddington & N. Levinson, Theory of ordinary differential equations, McGraw-Hill, [4] M. Hirsch, J. Palis, C. Pugh, M. Shub: Neighborhoods of Hyperbolic Sets, Inv. Math. 9, [5] M. C. Irwin: On the stable manifold theorem. Bull. London Math. Soc [6] A. Kelley: The stable, center-stable, center, center-unstable manifolds, Appendix C in [1]. [7] O. E. Lanford, Lecture notes on dynamical systems, Part 2: Invariant manifolds, [8] J. Robbin, Stable Manifolds of Semi-Hyperbolic Fixed Points, Ill. J. of Math. 15, [9] S. J. van Strien, Center manifolds are not C, Math. Z