1 Problem 8 from slides

The Jacobian matrix of f at a point (x,y), call it J, is

Then det(J) = -e^x(e^-y - e^y), implying that J is not invertible whenever e^-y - e^y = 0, i.e. whenever y = 0. Since f is continuously differentiable, then the inverse function theorem tells us that f is locally invertible whenever y≠0 since J will be invertible at those points.

Compute the Jacobian matrix of the inverse map

and

then

is the Jacobian of the inverse of f at the point f(x,y).

2 Problem 9 from slides

Now because

then the Jacobian of g, denote it by J_g, is

Furthermore, because

then the Jacobian of f, denote it by J_f, is

Finally, to check that indeed J_f and J_g are inverses of each other, we perform the following two multiplications for these matrices evaluated at the points (u,v) = f(x,y) and (x,y), respectively and then ensure the result of each is the identity matrix.

3 Problem 10 from slides

which is

at the point (0,0,1). The matrix above has determinant of 1 (it’s the identity matrix). Hence f′(0,0,1) is invertible. Furthermore, the fact that each of y₁, y₂, and y₃ are polynomials makes f continuously differentiable. These two facts therefore tell us that f is locally invertable at (0,0,1), by the inverse function theorem.

4 Problem 11 from slides

(4.1)

for all t and some real value c. Define F : R² → R² by

Prove F is injective

which can be rewritten as

The property in 4.1 then informs us that

Therefore |x - x′| = |x - x′| (whether or not c = 0), which implies x = x′. Furthermore we can similarly conclude y = y′ using the same argument after subsituting |x-x′| from equation 4.2 into equation 4.3. Hence we have (x,y) = (x′,y′), and since our only assumption was that F(x,y) = F(x′,y′), then F is injective.

Prove F is surjective

5 Outline and describe the steps and ideas in Rudin’s proof of the Inverse Function Theorem, without giving all the details, to the extent that satisfies you personally.

I’m not really satisfied, personally, with a proof that is missing details, so I will provide a full proof.

Let f : R → Rⁿ be a continuously differentiable mapping of an open set E, and f′(a) be invertible for some a ∈ E.

Define an open set U ⊂ E and prove f is injective on it

Denote f′(a) by just A, and choose λ such that

(5.4)

Note that this is not ill-defined as A is invertible by the hypothesis. Furthermore f′ is continuous at a, being that f is continuously differentiable, which implies that there is an open ball U ⊂ E centered at a such that

(5.5)

for all x ∈ U.

We now create a helpful function that makes our proof easier; followed by a helpful lemma. For each y ∈ Rⁿ define φ_y : Rⁿ → Rⁿ by φ_y(x) = x - A^-1(y - f(x)).

It also turns out that this function is a contraction mapping.

Lemma 5.1. For any y ∈ Rⁿ, the function φ_y has that

for all x₁,x₂ ∈ U.

Proof. We see that the derivative of φ_y is φ_y′(x) = I -A^-1f′(x) = A^-1A-A^-1f′(x) = A^-1(A-f′(x)). Equations 5.4 and 5.5 thus imply |φ_y′(x)| = |A^-1||A - f′(x)| < |A^-1|λ = . Therefore Rudin’s theorem 9.19 tells us that

for all x₁,x₂ ∈ U, as desired. __

Becuase lemma 5.1 indicates that φ_y is a contraction mapping, then it can have at most one fixed point. In light of remark 5.1, we then know that f must be injective on U. We next move on to proving that f(U) is open.

Prove V = f(U) is open

Denote f(U) by V and choose a y₀ ∈ V . Since f is certainly onto its image, then we of course have the existence of some x₀ ∈ U with f(x₀) = y₀. Define B to be an open ball centered at x₀ with a radius r > 0 small enough so that the closure of B, denote it by B, is contained in U. To show the openess of V , we show that the open ball of radius λr centered at y₀ is contained in V . So let y ∈ V such that |y - y₀| < λr. Then we have

by the definition of λ in equation 5.4. Thus if x ∈B, Lemma 5.1 paired with the above equation imply

Hence φ_y(x) ∈ B, i.e. φ_y is a contraction of B into itself. Since B is closed, it is complete, and the contraction mapping principle (Rudin Theorem 9.23) therefore implies the existence of one and only one fixed point of φ_y, which Remark 5.1 tells us is an x ∈B where f(x) = y. In other words, y ∈ f(B) ⊂ f(U) = V . Hence V is open.