Inverse and implicit function theorems

Theorem 8.5.1. Inverse function theorem.

Let

U \subset \R^n

be an open set and let

f \colon U \to \R^n

be a continuously differentiable function. Suppose

p \in U

and

f'(p)

is invertible (that is,

J_f(p) \neq 0

). Then there exist open sets

V, W \subset \R^n

such that

p \in V \subset U\text{,}

f(V) = W\text{,}

and

f|_V

is one-to-one. Hence a function

g \colon W \to V

exists such that

g(y) \coloneqq (f|_V)^{-1}(y)\text{.}

Furthermore,

g

is continuously differentiable and

\begin{equation*} g'(y) = {\bigl(f'(x)\bigr)}^{-1}, \qquad \text{for all } x \in V, y = f(x). \end{equation*}

Proof.

Write

A = f'(p)\text{.}

f'

is continuous, there is an open ball

V

centered at

p

such that

\begin{equation*} \bnorm{A-f'(x)} < \frac{1}{2\snorm{A^{-1}}} \qquad \text{for all } x \in V. \end{equation*}

Consequently, the derivative

f'(x)

is invertible for all

x \in V

by Proposition 8.2.6.

Given

y \in \R^n\text{,}

define

\varphi_y \colon V \to \R^n

\begin{equation*} \varphi_y (x) \coloneqq x + A^{-1}\bigl(y-f(x)\bigr) . \end{equation*}

A^{-1}

is one-to-one,

\varphi_y(x) = x

(

x

is a fixed point) if and only if

y-f(x) = 0\text{,}

or in other words

f(x)=y\text{.}

Using the chain rule, we obtain

\begin{equation*} \varphi_y'(x) = I - A^{-1} f'(x) = A^{-1} \bigl( A-f'(x) \bigr) . \end{equation*}

So for

x \in V\text{,}

we have

\begin{equation*} \bnorm{\varphi_y'(x)} \leq \snorm{A^{-1}} \, \bnorm{A-f'(x)} < \nicefrac{1}{2} . \end{equation*}

V

is a ball, it is convex. Hence,

\begin{equation*} \bnorm{\varphi_y(x_1)-\varphi_y(x_2)} \leq \frac{1}{2} \snorm{x_1-x_2} \qquad \text{for all } x_1,x_2 \in V. \end{equation*}

In other words,

\varphi_y

is a contraction defined on

V\text{,}

though we so far do not know what is the range of

\varphi_y\text{.}

We cannot yet apply the fixed point theorem, but we can say that

\varphi_y

has at most one fixed point in

V\text{:}

\varphi_y(x_1) = x_1

and

\varphi_y(x_2) = x_2\text{,}

then

\snorm{x_1-x_2} = \bnorm{\varphi_y(x_1)-\varphi_y(x_2)} \leq \frac{1}{2} \snorm{x_1-x_2}\text{,}

x_1 = x_2\text{.}

That is, there exists at most one

x \in V

such that

f(x) = y\text{,}

and so

f|_V

is one-to-one.

Let

W \coloneqq f(V)

and let

g \colon W \to V

be the inverse of

f|_V\text{.}

We need to show that

W

is open. Take a

y_0 \in W\text{.}

There is a unique

x_0 \in V

such that

f(x_0) = y_0\text{.}

Let

r > 0

be small enough such that the closed ball

C(x_0,r) \subset V

(such

r > 0

exists as

V

is open).

Suppose

y

is such that

\begin{equation*} \snorm{y-y_0} < \frac{r}{2\snorm{A^{-1}}} . \end{equation*}

If we show that

y \in W\text{,}

then we have shown that

W

is open. If

x_1 \in C(x_0,r)\text{,}

then

\begin{equation*} \begin{split} \bnorm{\varphi_y(x_1)-x_0} & \leq \bnorm{\varphi_y(x_1)-\varphi_y(x_0)} + \bnorm{\varphi_y(x_0)-x_0} \\ & \leq \frac{1}{2}\snorm{x_1-x_0} + \bnorm{A^{-1}(y-y_0)} \\ & \leq \frac{1}{2}r + \snorm{A^{-1}} \, \snorm{y-y_0} \\ & < \frac{1}{2}r + \snorm{A^{-1}} \frac{r}{2\snorm{A^{-1}}} = r . \end{split} \end{equation*}

\varphi_y

takes

C(x_0,r)

into

B(x_0,r) \subset C(x_0,r)\text{.}

It is a contraction on

C(x_0,r)

and

C(x_0,r)

is complete (closed subset of

\R^n

is complete). Apply the contraction mapping principle to obtain a fixed point

x\text{,}

i.e.

\varphi_y(x) = x\text{.}

That is,

f(x) = y\text{,}

and

y \in f\bigl(C(x_0,r)\bigr) \subset f(V) = W\text{.}

Therefore,

W

is open.

Next we need to show that

g

is continuously differentiable and compute its derivative. First, let us show that it is differentiable. Let

y \in W

and

k \in \R^n\text{,}

k \neq 0\text{,}

such that

y+k \in W\text{.}

Because

f|_V

is a one-to-one and onto mapping of

V

onto

W\text{,}

there are unique

x \in V

and

h \in \R^n\text{,}

h \neq 0

and

x+h \in V\text{,}

such that

f(x) = y

and

f(x+h) = y+k\text{.}

In other words,

g(y) = x

and

g(y+k) = x+h\text{.}

See Figure 8.12.

A diagram of two sets and mappings between them. On the left is a shaded set V with dotted boundary and the points x plus h and x marked. On the right is a shaded set W with the points y plus k and y marked. Two arrows labeled f go from left to right one going from the point x plus h to the point y plus k and one going from the point x to the point y. Two arrows labeled g go the opposite way between the same points. — Figure 8.12. Proving that $g$ is differentiable.

We can still squeeze some information from the fact that

\varphi_y

is a contraction.

\begin{equation*} \varphi_y(x+h)-\varphi_y(x) = h + A^{-1} \bigl( f(x)-f(x+h) \bigr) = h - A^{-1} k . \end{equation*}

\begin{equation*} \snorm{h-A^{-1}k} = \bnorm{\varphi_y(x+h)-\varphi_y(x)} \leq \frac{1}{2}\snorm{x+h-x} = \frac{\snorm{h}}{2}. \end{equation*}

By the inverse triangle inequality,

\snorm{h} - \snorm{A^{-1}k} \leq \frac{1}{2}\snorm{h}\text{.}

\begin{equation*} \snorm{h} \leq 2 \snorm{A^{-1}k} \leq 2 \snorm{A^{-1}} \, \snorm{k}. \end{equation*}

In particular, as

k

goes to 0, so does

h\text{.}

x \in V\text{,}

we find that

f'(x)

is invertible. Let

B \coloneqq \bigl(f'(x)\bigr)^{-1}\text{,}

which is what we think the derivative of

g

y

is. Then

\begin{equation*} \begin{split} \frac{\bnorm{g(y+k)-g(y)-Bk}}{\snorm{k}} & = \frac{\snorm{h-Bk}}{\snorm{k}} \\ & = \frac{\bnorm{h-B\bigl(f(x+h)-f(x)\bigr)}}{\snorm{k}} \\ & = \frac{\bnorm{B\bigl(f(x+h)-f(x)-f'(x)h\bigr)}}{\snorm{k}} \\ & \leq \snorm{B} \frac{\snorm{h}}{\snorm{k}}\, \frac{\bnorm{f(x+h)-f(x)-f'(x)h}}{\snorm{h}} \\ & \leq 2\snorm{B} \, \snorm{A^{-1}} \frac{\bnorm{f(x+h)-f(x)-f'(x)h}}{\snorm{h}} . \end{split} \end{equation*}

k

goes to 0, so does

h\text{.}

So the right-hand side goes to 0 as

f

is differentiable, and hence the left-hand side also goes to 0. And

B

is precisely what we wanted

g'(y)

to be.

We have

g

is differentiable, let us show it is

C^1(W)\text{.}

The function

g \colon W \to V

is continuous (it is differentiable),

f'

is a continuous function from

V

L(\R^n)\text{,}

and

X \mapsto X^{-1}

is a continuous function on the set of invertible operators. As

g'(y) = {\bigl( f'\bigl(g(y)\bigr)\bigr)}^{-1}

is the composition of these three continuous functions, it is continuous.

Proof.

Define

F \colon U \to \R^{n+m}

F(x,y) \coloneqq \bigl(x,f(x,y)\bigr)\text{.}

It is clear that

F

C^1\text{,}

and we want to show that its derivative at

(p,q)

is invertible. Let us compute the derivative. The quotient

\begin{equation*} \frac{\bnorm{f(p+h,q+k) - f(p,q) - A_x h - A_y k}}{\bnorm{(h,k)}} \end{equation*}

goes to zero as

\bnorm{(h,k)} = \sqrt{\snorm{h}^2+\snorm{k}^2}

goes to zero. But then so does

\begin{gathered} \frac{\bnorm{F(p+h,q+k)-F(p,q) - (h,A_x h+A_y k)}}{\bnorm{(h,k)}} \\ \begin{aligned} & = \frac{\bnorm{\bigl(h,f(p+h,q+k)-f(p,q)\bigr) - (h,A_x h+A_y k)}}{\bnorm{(h,k)}} \\ & = \frac{\bnorm{f(p+h,q+k) - f(p,q) - A_x h - A_y k}}{\bnorm{(h,k)}} . \end{aligned} \end{gathered}

So the derivative of

F

(p,q)

takes

(h,k)

(h,A_x h+A_y k)\text{.}

In block matrix form, it is

\left[\begin{smallmatrix}I & 0\\A_x & A_y\end{smallmatrix}\right]\text{.}

(h,A_x h+A_y k) = (0,0)\text{,}

then

h=0\text{,}

and so

A_y k = 0\text{.}

A_y

is one-to-one,

k=0\text{.}

Thus

F'(p,q)

is one-to-one, and hence invertible. We apply the inverse function theorem.

That is, there exists an open set

V \subset \R^{n+m}

with

F(p,q) = (p,0) \in V\text{,}

and a

C^1

mapping

G \colon V \to \R^{n+m}\text{,}

such that

F\bigl(G(x,s)\bigr) = (x,s)

for all

(x,s) \in V\text{,}

G

is one-to-one, and

G(V)

is open. Write

G = (G_1,G_2)

(the first

n

and the next

m

components of

G

). Then

\begin{equation*} F\bigl(G_1(x,s),G_2(x,s)\bigr) = \Bigl(G_1(x,s),f\bigl(G_1(x,s),G_2(x,s) \bigr)\Bigr) = (x,s) . \end{equation*}

x = G_1(x,s)

and

f\bigl(G_1(x,s),G_2(x,s)\bigr) = f\bigl(x,G_2(x,s)\bigr) = s\text{.}

Plugging in

s=0\text{,}

we obtain

\begin{equation*} f\bigl(x,G_2(x,0)\bigr) = 0 . \end{equation*}

As the set

G(V)

is open and

(p,q) \in G(V)\text{,}

there exist some open sets

\widetilde{W}

and

W'

such that

\widetilde{W} \times W' \subset G(V)

with

p \in \widetilde{W}

and

q \in W'\text{.}

Take

W \coloneqq \bigl\{ x \in \widetilde{W} : G_2(x,0) \in W' \bigr\}\text{.}

The function that takes

x

G_2(x,0)

is continuous and therefore

W

is open. Define

g \colon W \to \R^m

g(x) \coloneqq G_2(x,0)\text{,}

which is the

g

in the theorem. The fact that

g(x)

is the unique point in

W'

follows because

W \times W' \subset G(V)

and

G

is one-to-one.

Next, differentiate

\begin{equation*} x\mapsto f\bigl(x,g(x)\bigr) \end{equation*}

p\text{,}

which is the zero map, so its derivative is zero. Using the chain rule,

\begin{equation*} 0 = A\bigl(h,g'(p)h\bigr) = A_xh + A_yg'(p)h \end{equation*}

for all

h \in \R^{n}\text{,}

and we obtain the desired derivative for

g\text{.}

Theorem 8.5.1. Inverse function theorem.

Proof.

Theorem 8.5.6. Implicit function theorem.

Proof.