CHAPTER 1 A Special case of the Grobman-Hartman Theorem

This is a homework assignment where you will prove a special case of the Grobman-Hartman theorem. In class (and the handwritten notes I handed out) we have done the case where the linear map is a contraction. Here you will do the case where it is an expansion.

1.1. The Setup and Notation.

Let C^{0}(R^{n}, R^{n}) be the set of all bounded continuous functions f : R^{n} → R^{n} with the norm

∥f∥_{C^{0}} := sup_{x∈R^{n}} ∥f(x)∥.

Let C^{1}(R^{n}, R^{n}) be the set of all bounded continuously differentiable functions f : R^{n} → R^{n} so that the norm

∥f∥_{C^{1}} := ∥f∥_{C^{0}} + sup_{x∈R^{n}} ∥f′(x)∥_{op}

is finite. Here ∥ ∥_{op} is the operator norm on linear maps. That is if L : R^{n} → R^{n} is linear then

∥L∥_{op} := sup_{x≠0} \frac{∥Ax∥}{∥x∥}.

In particular this implies

∥Lx∥ ≤ ∥L∥_{op}∥x∥.

Exercise 1.1.1. Let f ∈ C^{1}(R^{n}, R^{n}). Then for any x_{0}, x_{2} ∈ R^{n}

∥f(x_{1}) − f(x_{2})∥ ≤ ∥f∥_{C^{1}}∥x_{1} − x_{0}∥.

Let A : R^{n} → R^{n} be linear and assume that there is a constant λ > 1 so that

∥Ax∥ ≥ λ∥x∥  (1.1)

for all x ∈ R^{n}. (This is dual to the case we did in class where ∥Ax∥ ≤ ρ∥x∥ for some ρ < 1.) Let F : R^{n} → R^{n} be of the form

F (x) = Ax + f(x)  with  f ∈ C^{1}(R^{n}, R^{n}).  (1.2)

and let H : R^{n} → R^{n} be of the form

H(x) := x + h(x)  with  h ∈ C^{0}(R^{n}, R^{n}).  (1.3)

We wish to show that given F of this form with ∥F − A∥_{C^{1}} = ∥f∥_{C^{1}} small enough that it is possible to find H of the given form so that

F ∘ H = H ∘ A.  (1.4)

Exercise 1.1.2. Show that the equation (1.4) is equivalent to

f(Ax) − Ah(x) = f(x + h(x)).  (1.5)

We will also be interested in the linearized version of this equation:

h(Ax) − Ah(x) − f(x).  (1.6)

1.2. Solution of the Linearized Problem

Define a linear map S : C^{0}(R^{n}, R^{n}) → C^{0}(R^{n}, R^{n}) by
(Sg)(x) := g(Ax).

Exercise 1.2.1. The map S : C^{0}(R^{n}, R^{n}) → C^{0}(R^{n}, R^{n}) is invertible with

(S^{−1}g)(x) = g(A^{−1}x).

Also show

∥S∥_{op} = ∥S^{−1}∥_{op} = 1.

We can extend the linear map A : R^{n} → R^{n} as a map A : C^{0}(R^{n}, R^{n}) → C^{0}(R^{n}, R^{n}) in the obvious way. That is (Ag)(x) := Ag(x).

Exercise 1.2.2. Use the inequality (1.1) to show that A^{−1} has operator norm satisfying

∥A^{−1}∥_{op} ≤ \frac{1}{λ}.

Note that the linearized equation (1.6) can be written as

(S − A)h = f.

Thus to show the linearized equation is solvable we want to show the linear operator S − A is invertible.

Exercise 1.2.3. Show that S − A is invertible and

∥(S − A)^{−1}∥_{op} ≤ \frac{1}{λ − 1}.

Hint: Write S − A = (−A)(I − A^{−1}S). Then ∥A^{−1}S∥_{op} ≤ 1/λ < 1 and thus (I − A^{−1}S)^{−1} can be computed by using the power series (1 − x)^{−1} = 1 + x + x^{2} + x^{3} + · · ·. .

Exercise 1.2.4. Show that for all f ∈ C^{0}(R^{n}, R^{n}) equation (1.6) has a unique solution h. Moreover this solution satisfies

∥h∥_{C^{0}} ≤ \frac{1}{λ − 1}∥f∥_{C^{0}}.

1.4. PROOF THAT H IS A HOMEOMORPHISM 3

1.3. Reduction to a Fixed Point Problem

For f ∈ C^{1}(R^{n}, R^{n}) define a mapping Φ_{f} : C^{0}(R^{n}, R^{n}) → C^{0}(R^{n}, R^{n}) by

Φ_{f}[h](x) = (S − A)^{−1}f(x + h(x)).

Exercise 1.3.1. Given f ∈ C^{1}(R^{n}, R^{n}) show h ∈ C^{0}(R^{n}, R^{n}) is a fixed point of Φ_{f} (that is Φ_{f}[h] = h) if and only if h is a solution to the equation (1.5)

Exercise 1.3.2. For f ∈ C^{1}(R^{n}, R^{n}) the inequality

∥Φ_{f}[h_{1}] − Φ_{f}[h_{0}]∥_{C^{0}} ≤ \frac{∥f∥_{C^{1}}}{λ − 1} ∥h_{1} − h_{0}∥_{C_{0}}

holds for all h_{0}, h_{1} ∈ C^{0}(R^{n}, R^{n}). Thus if ∥f∥_{C^{0}} ≤ (λ − 1)/2 then Φ_{f} is a contraction with contraction factor 1/2.

Exercise 1.3.3. Show that if F : R^{n} → R^{n} is a C^{1} diffeomorphism so that ∥F − A∥_{C_{1}} ≤ (λ − 1)/2 then there is a unique function H of the form (1.3) with F ∘ H = H ∘ A.

Exercise 1.3.4 (Optional). With notation as in the last exercise show and letting I be the identity map on R^{n}

∥H − I∥_{C^{0}} ≤ \frac{2∥F − A∥_{C^{0}}}{λ − 1} .

Hint: Let h_{0} = 0, h_{1} := Φ_{f}[h_{0}] and recursively h_{k+1} = Φ_{f}[h_{k}]. Then from the proof of the Banach fixed point theorem we have that h = lim_{k→∞} h_{k} and that (using that the contraction factor is 1/2) ∥h∥_{C_{0}} = ∥h − h_{0}∥_{C^{0}} ≤ ∥h_{1} − h_{0}∥_{C^{0}}/(1 − 1/2) = 2∥h∥_{C^{0}} and h_{1}(x) = Φ_{f} [0](x) = (S − A)f(x).

1.4. Proof that H is a Homeomorphism

In this section H will be as in Exercise 1.3.3.

Exercise 1.4.1. Show H : R^{n} → R^{n} is injective. Hint: Use a modification of the argument we used in class in the case where ∥Ax∥ ≤ ρ∥x∥.

Proposition 1.4.2. The map H : R^{n} → R^{n} is an open map. That is if U ⊂ R^{n} is open then H[U] is open in R^{n}. Thus the image of H is open in R^{n}.

Proof. This is a special case of the theorem of invariance of domain which says (in one form) that if V ⊆ R^{n} is open and φ : V → R is injective, then the image φ[V] is open in R^{n}.

Exercise 1.4.3. The image of H is closed in R^{n} and thus H is surjective.

Exercise 1.4.4. The inverse of H^{−1} : R^{n} → R^{n} is continuous and thus H is a homeomorphism. Hint: The function H^{−1} continuous iff for all open U ⊆ R^{n} the preimage (H^{−1})^{−1}[U] = H[U] is open.

1.5. The Theorem

Putting all the above together we get:

Theorem 1.5.1 (Grobman-Hartman). Let A : R^{n} → R^{n} be a linear map so that (1.1) holds. Then for any C^{1} diffeomorphism F : R^{n} → R^{n} with

∥A − F ∥_{C^{1}} ≤ \frac{λ − 1}{2}

there is a unique homeomorphism H : R^{n} → R^{n} of the form (1.3) so that F ∘ H = H ∘ A holds. That is F = H ∘ A ∘ H^{−1}. Finally if I is the identity map on R^{n} then

∥H − I∥_{C^{0}} ≤ \frac{2∥A − F ∥_{C^{0}}}{λ − 1}