CAL1_L21

Functions with non-zero Jacobian determinant

Theorem 1

Theorem 1(Theorem).

Let $B=B_r(\mathbf{a})\subseteq {\mathbb{R}}^n$, and let $\mathbf{f}$ be a mapping of $\overline{B}$ into ${\mathbb{R}}^n$. Assume:

1. $\mathbf{f}$ is continuous on $\overline{B}$;
2. all partial derivatives $D_j{f}_i(\mathbf{x})$ in $B$;
3. $\mathbf{f}(\mathbf{x})\ne \mathbf{f}(\mathbf{a})$ for all $\mathbf{x}\in \overline{B}\setminus B$;
4. ${\mathbf{f}}^{\prime }(\mathbf{x})$ is invertible for all $\mathbf{x}\in B$.

Then, $\mathbf{f}(B)$ contains a neighborhood of $\mathbf{f}(\mathbf{a})$.

Proof.

Let $\partial B$ denote the boundary $\overline{B}\setminus B$ of the ball $B$.

Define $g$ on $\partial B$ by $g(\mathbf{x})=\Vert \mathbf{f}(\mathbf{x})-\mathbf{f}(\mathbf{a})\Vert$. Observe that $g$ is continuous. Since $\partial B$ is compact, the follows from the extreme value theorem that $g$ attains a minimum value $m$ on $\partial B$. $m>0$ by hypothesis.

Let $T\equiv B_{m/2}(\mathbf{f}(\mathbf{a}))$. We will show that $T\subseteq f(B)$. Let $\mathbf{y}\in T$. Define $h$ on $\overline{B}$ by

$$h(\mathbf{x})=\Vert \mathbf{f}(\mathbf{x})-\mathbf{y}\Vert .$$

Again, $h$ is continuous on a compact set, and hence attains a minimum value. Note that $h(\mathbf{a})<m/2$. Thus, the minimum value of $h$ must also be less than $m/2$. For any $\mathbf{x}\in \partial B$,

$$\begin{array}{rl}\Vert \mathbf{f}(\mathbf{x})-\mathbf{y}\Vert & \ge \Vert \mathbf{f}(\mathbf{x})-\mathbf{f}(\mathbf{a})\Vert -\Vert \mathbf{f}(\mathbf{a})-\mathbf{y}\Vert \ge m-\frac{m}{2}=\frac{m}{2}.\end{array}$$

Thus, $h$ must attain its minimum at a point $\mathbf{c}\in B$.

At this point $h^2$ also has a minimum. Since

$$\begin{array}{r}{h}^2(\mathbf{x})=\Vert \mathbf{f}(\mathbf{x})-\mathbf{y}{\Vert }^2=\sum _{i=1}^n(f_i(\mathbf{x})-y_i{)}^2,\end{array}$$

and since each partial derivative $D_k(h^2)$ must be zero at $\mathbf{c}$, we have

$$\begin{array}{r}\sum _{i=1}^n(f_i(\mathbf{c})-y_i)D_j{f}_i(\mathbf{c})=0\text{ for }j=1,\dots ,n.\end{array}$$

This is equivalent to the matrix equation

$$\mathbf{Df}(\mathbf{c})(\mathbf{f}(\mathbf{c})-\mathbf{y})=0.$$

Since $\det \mathbf{Df}(\mathbf{c})\ne 0$, it follows that $\mathbf{f}(\mathbf{c})=\mathbf{y}$. Therefore, $\mathbf{y}\in \mathbf{f}(B)$.□

Theorem 2

Theorem 2(Theorem).

Let $\mathbf{f}$ be a mapping of an open set $A\subseteq {\mathbb{R}}^n$ into ${\mathbb{R}}^n$. Assume:

1. $\mathbf{f}$ is continuous on $A$ and all partial derivatives $D_j{f}_i$ exist on $A$;
2. $\mathbf{f}$ is injective on $A$;
3. ${\mathbf{f}}^{\prime }(\mathbf{x})$ is invertible for every $\mathbf{x}\in A$,

Then $\mathbf{f}$ is an open mapping on $A$.

Proof.

Let $E\subseteq A$ be open. If $\mathbf{b}\in \mathbf{f}(E)$, then $\mathbf{b}=\mathbf{f}(\mathbf{a})$ for some $\mathbf{a}\in E$. There is a ball $B_r(\mathbf{a})\subseteq E$ on which $\mathbf{f}$ satisfies the hypotheses of Theorem 1. Thus, $\mathbf{f}(E)$ contains a neighborhood of $\mathbf{b}$.□

Theorem 3

Theorem 3(Theorem).

Let $\mathbf{f}$ be a mapping of an open set $S\subseteq {\mathbb{R}}^n$ into ${\mathbb{R}}^n$. Assume:

1. $\mathbf{f}$ is a $C^1$ mapping on $S$;
2. ${\mathbf{f}}^{\prime }(\mathbf{a})$ is invertible for some $\mathbf{a}\in S$.

Then there is an $n$-ball $B(\mathbf{a})$ on which $\mathbf{f}$ is injective.

Proof.

Let ${\mathbf{Z}}_1,\dots ,{\mathbf{Z}}_n$ be $n$ points in $S$ and let $\mathbf{Z}\equiv ({\mathbf{Z}}_1;\dots ;{\mathbf{Z}}_n)$ denote the points in ${\mathbb{R}}^{n^2}$ whose first $n$ components are the components of ${\mathbf{Z}}_1$, whose next $n$ components are the components of ${\mathbf{Z}}_2$, and so on. Define a real valued function $h$ as follows:

$$h(\mathbf{Z})=\det [D_j{f}_i({\mathbf{Z}}_i)].$$

Note that $h$ is defined on $S^n$ and continuous on $S^n$, because each $D_j{f}_i$ is continuous on $S$ and a determinant is a polynomial in the entries of the matrix. Let $\mathbf{Z}$ be the special point in ${\mathbb{R}}^{n^2}$ obtained by putting

$${\mathbf{Z}}_1={\mathbf{Z}}_2=\cdots ={\mathbf{Z}}_n=\mathbf{a}.$$

Then $h(\mathbf{Z})=\det \mathbf{Df}(\mathbf{a})\ne 0$, and hence, by continuity, there is some $B_{\delta }(\mathbf{Z})\subseteq {\mathbb{R}}^{n^2}$ such that $\mathbf{W}\in B(\mathbf{Z})$ implies $h(\mathbf{W})\ne 0$. Now, If ${\mathbf{W}}_1...,{\mathbf{W}}_n\in B(\mathbf{a})\equiv B_{\delta /\sqrt{n}}(\mathbf{a})$,

$$\begin{array}{rl}\Vert \mathbf{W}-\mathbf{Z}\Vert & =\sqrt{\Vert {\mathbf{W}}_1-{\mathbf{Z}}_1{\Vert }^2+\cdots +\Vert {\mathbf{W}}_n-{\mathbf{Z}}_n{\Vert }^2}\\ & =\sqrt{\Vert {\mathbf{W}}_1-\mathbf{a}{\Vert }^2+\cdots +\Vert {\mathbf{W}}_n-\mathbf{a}{\Vert }^2}\\ & \le \delta ,\end{array}$$

so $\mathbf{W}\in B(\mathbf{Z})$. So, ${\mathbf{W}}_1...,{\mathbf{W}}_n\in B(\mathbf{a})$ implies $\det [D_j{f}_i({\mathbf{W}}_i)]\ne 0$.

We will prove that $\mathbf{f}$ is injective on $B(\mathbf{a})$. Assume the contrary, that is, assume $\mathbf{f}(\mathbf{x})=\mathbf{f}(\mathbf{y})$ for some $\mathbf{x}\ne \mathbf{y}\in B(\mathbf{a})$. Since $B(\mathbf{a})$ is convex, $L(\mathbf{x},\mathbf{y})\subseteq B(\mathbf{a})$, and we can apply the mean value theorem (recall that $\mathbf{f}$ is differentiable on $S$) to write for every $\mathbf{\alpha }\in {\mathbb{R}}^n$,

$$\begin{array}{rl}0=\mathbf{\alpha }\cdot (\mathbf{f}(\mathbf{y})-\mathbf{f}(\mathbf{x}))& =\mathbf{\alpha }\cdot ({\mathbf{f}}^{\prime }(\mathbf{z})(\mathbf{y}-\mathbf{x}))\end{array}$$

for some $\mathbf{z}\in L(\mathbf{x},\mathbf{y})$. Taking $\mathbf{\alpha }={\mathbf{e}}_1,\dots ,{\mathbf{e}}_n$, we get the equations

$$0=\nabla f_i({\mathbf{Z}}_i)\cdot (\mathbf{y}-\mathbf{x})\text{ for }i=1,\dots ,n,$$

where each ${\mathbf{Z}}_i\in L(\mathbf{x},\mathbf{y})\subseteq B(\mathbf{a})$. This is equivalent to writing

$$\left[\begin{array}{c}\nabla f_1({\mathbf{Z}}_1)\\ ⋮\\ \nabla f_n({\mathbf{Z}}_n)\end{array}\right](\mathbf{y}-\mathbf{x})=0$$

or

$$[D_j{f}_i({\mathbf{Z}}_i)](\mathbf{y}-\mathbf{x})=0.$$

However, $\det [D_j{f}_i({\mathbf{Z}}_i)]\ne 0$, which is a contradiction.

.$A\equiv {\mathbf{f}}^{\prime }(\mathbf{a})$, and choose $\lambda$ so that $2\lambda \Vert A^{-1}\Vert =1$. Since ${\mathbf{f}}^{\prime }$ is continuous at $\mathbf{a}$, there is an open ball $U\subseteq E$, with center $\mathbf{a}$, such that

$$\Vert {\mathbf{f}}^{\prime }(\mathbf{x})-A\Vert <\lambda (\mathbf{x}\in U).$$

We associate to each $\mathbf{y}\in {\mathbb{R}}^n$ a function $\phi$, defined by

$$\phi (\mathbf{x})=\mathbf{x}+A^{-1}(\mathbf{y}-\mathbf{f}(\mathbf{x}))(\mathbf{x}\in E).$$

Note that $\mathbf{f}(\mathbf{x})=\mathbf{y}$ iff $\mathbf{x}$ is a fixed point of $\phi$.

Since ${\phi }^{\prime }(\mathbf{x})=I-A^{-1}({\mathbf{f}}^{\prime }(\mathbf{x}))=A^{-1}(A-{\mathbf{f}}^{\prime }(\mathbf{x}))$, we have for $\mathbf{x}\in U$:

$$\begin{array}{rl}\Vert {\phi }^{\prime }(\mathbf{x})\Vert & =\Vert A^{-1}(A-{\mathbf{f}}^{\prime }(\mathbf{x}))\Vert \\ & \le \Vert A^{-1}\Vert \Vert A-{\mathbf{f}}^{\prime }(\mathbf{x})\Vert \\ & <\frac{1}{2}(\mathbf{x}\in U).\end{array}$$

Using the first result from here,

$$\Vert \phi ({\mathbf{x}}_1)-\phi ({\mathbf{x}}_2)\Vert \le \frac{1}{2}\Vert {\mathbf{x}}_1-{\mathbf{x}}_2\Vert ({\mathbf{x}}_1,{\mathbf{x}}_2\in U).$$

It follows that $\phi$ cannot have more than one fixed point (Note that this does not require the Banach contraction principle; assuming the existence of two fixed points easily yields a contradiction). Thus, $\mathbf{f}(\mathbf{x})=\mathbf{y}$ for at most one $\mathbf{x}\in U$. Thus, $\mathbf{f}$ is injective in $U$.

Theorem 4

Theorem 4(Theorem).

Let $\mathbf{f}$ be a mapping of an open set $A\subseteq {\mathbb{R}}^n$ into ${\mathbb{R}}^n$. Assume:

1. $\mathbf{f}$ is a $C^1$ mapping on $A$;
2. ${\mathbf{f}}^{\prime }(\mathbf{x})$ is invertible for all $\mathbf{x}\in A$.

Then $\mathbf{f}$ is an open mapping on $A$.

Proof.

Let$S$ be any open subset of $A$. If $\mathbf{x}\in S$ there is a ball $B(\mathbf{x})\subseteq S$ in which $\mathbf{f}$ is injective by Theorem 3. By Theorem 2, $\mathbf{f}(B(\mathbf{x}))$ is open in ${\mathbb{R}}^n$. But we can write $S={\bigcup }_{\mathbf{x}\in S}B(\mathbf{x})$. Applying $\mathbf{f}$ on both sides, we find $\mathbf{f}(S)={\bigcup }_{\mathbf{x}\in S}\mathbf{f}(B(\mathbf{x}))$, so $\mathbf{f}(S)$ is open.□

The hypotheses made in this corollary ensure that each point $\mathbf{x}\in E$ has a neighborhood in which $\mathbf{f}$ is injective. This may be expressed by saying that $\mathbf{f}$ is locally injective in $E$. But this does not imply that $\mathbf{f}$ is injective on $E$!

Inverse function theorem

The inverse function theorem roughly states that a continuously differentiable mapping $\mathbf{f}$ is invertible in a neighborhood of any point $\mathbf{x}$ at which the linear transformation ${\mathbf{f}}^{\prime }(\mathbf{x})$ is invertible.

Theorem 5(Theorem).

Suppose $\mathbf{f}$ is a $C^1$ mapping of an open set $E\subseteq {\mathbb{R}}^n$ into ${\mathbb{R}}^n$, and ${\mathbf{f}}^{\prime }(\mathbf{a})$ is invertible for some $\mathbf{a}\in E$. Then,

1. there exist open sets $U$ and $V$ in ${\mathbb{R}}^n$ such that $\mathbf{a}\in U$, $\mathbf{f}(\mathbf{a})\in V$, $\mathbf{f}$ is injective on $U$, and $\mathbf{f}(U)=V$;
2. the inverse $\mathbf{g}$ of $\mathbf{f}$ (which exists by $(1)$), defined in $V$ by $\mathbf{g}(\mathbf{f}(\mathbf{x}))=\mathbf{x}$ for $\mathbf{x}\in U$, is a $C^1$ mapping.

[!Proof]-
Let $U$ be the open ball centered at $\mathbf{a}$ on which $\mathbf{f}$ is injective, as given by Theorem 3.

Next, put $V\equiv f(U)$, and pick ${\mathbf{y}}_0\in V$. Then ${\mathbf{y}}_0=\mathbf{f}({\mathbf{x}}_0)$ for some ${\mathbf{x}}_0\in U$. Let $B$ be an open ball with center at ${\mathbf{x}}_0$ and radius $r>0$, so small that its closure $\overline{B}$ lies in $U$. We will show that $\mathbf{y}\in V$ whenever $|\mathbf{y}-{\mathbf{y}}_0|<\lambda r$. This will prove that $V$ is open.

[!Proof]-
The map $\mathbf{x}\mapsto \det {\mathbf{f}}^{\prime }(\mathbf{x})$ is continuous on $S$ since each $D_j{f}_i$ is continuous on $\mathbf{x}$. Since $\det {\mathbf{f}}^{\prime }(\mathbf{a})\ne 0$, there exists a ball $B_1(\mathbf{a})$ in which ${\mathbf{f}}^{\prime }(\mathbf{x})$ is invertible for each $\mathbf{x}$. Also, from Theorem 3, there exists a ball $B(\mathbf{a})\subseteq B_1(\mathbf{a})$ on which $\mathbf{f}$ is injective. Let $B$ be a ball with center $\mathbf{a}$ and radius less than $B(\mathbf{a})$. Then, by Theorem 1, $\mathbf{f}(B)$ contains a ball $V$ with center at $\mathbf{f}(\mathbf{a})$. Define $U\equiv {\mathbf{f}}^{-1}(V)\cap B$, which is open since both ${\mathbf{f}}^{-1}(V)$ and $B$ are open. Since $U\subseteq \overline{B}$ and $V\subseteq \mathbf{f}(\overline{B})$, $(1)$ is proved.

$\mathbf{f}$ is injective and continuous on $\overline{B}$, a compact set. It follows that $\mathbf{g}\equiv {\mathbf{f}}^{-1}:\mathbf{f}(\overline{B})\to \overline{B}$ is continuous on its domain.