LEC CAL1 21

Functions with non-zero Jacobian determinant

See Apostol (1985, p. 369)

Theorem 1

Theorem 190.1(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 $D_j{f}_i(\mathbf{x})$ exist 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 190.2(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 190.3(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.

Proof.

Define $A\equiv {\mathbf{f}}^{\prime }(\mathbf{a})$, 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 190.4(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 $E$!

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Inverse function theorem

The inverse function theorem roughly states that a continuously differentiable mapping $\mathbf{f}$ is a $C^1$ diffeomorphism in a neighborhood of any $\mathbf{a}$ at which ${\mathbf{f}}^{\prime }(\mathbf{a})$ is invertible.

Here’s the one variable version:

Example 190.5(Example).

Suppose $f$ is a $C^1$ mapping of an interval $(\alpha ,\beta )$ into $\mathbb{R}$, and $f^{\prime }(a)$ is nonzero for some $a\in (\alpha ,\beta )$. Then,

1. there exist open sets $U\subseteq (\alpha ,\beta )$ and $V\subseteq \mathbb{R}$ such that $a\in U$, $f(a)\in V$, $f$ is injective on $U$, and $f(U)=V$.

2. the inverse $g$ of $f$, defined in $V$ by $g(f(x))=x$ for $x\in U$, is a $C^1$ mapping.

Proof.

Since$f^{\prime }$ is continuous at $a$, we can pick a neighborhood $U$ of $a$ on which $f^{\prime }$ is non-zero. Let $V=f(U)$. is open since strictly monotone continuous maps are open. $f$ is monotone on $U$, so $f$ is injective on $U$. $f^{-1}$ is clearly continuous, so $f$ is a $C^1$ homeomorphism. By Pugh (2015) 4.15, we are done.□

Theorem 190.6(Rudin (1976) 9.24).

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 the previous point), defined in $V$ by $\mathbf{g}(\mathbf{f}(\mathbf{x}))=\mathbf{x}$ for $\mathbf{x}\in U$, is a $C^1$ mapping, with its derivative given by ${\mathbf{g}}^{\prime }(\mathbf{y})={\mathbf{f}}^{\prime }(\mathbf{g}(\mathbf{y}){)}^{-1}$ for $\mathbf{y}\in V$.

Proof.

Put ${\mathbf{f}}^{\prime }(\mathbf{a})=A$, and let

$$\lambda =\frac{1}{2\Vert A^{-1}\Vert }.$$E1

Since ${\mathbf{f}}^{\prime }$ is continuous at $\mathbf{a}$, there is an open ball $U\subset E$ with center $\mathbf{a}$ such that

$$\Vert {\mathbf{f}}^{\prime }(\mathbf{x})-A\Vert <\lambda (\mathbf{x}\in U).$$E2

$f{|}_U$ is an injection

Associate with each $\mathbf{y}\in {\mathbb{R}}^n$ a function ${\phi }_{\mathbf{y}}$, defined by

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

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

${\phi }_{\mathbf{y}}^{\prime }(\mathbf{x})=I-A^{-1}{\mathbf{f}}^{\prime }(\mathbf{x})=A^{-1}(A-{\mathbf{f}}^{\prime }(\mathbf{x}))$. By (E2),

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

Applying the mean value theorem yields

$$\begin{array}{r}|{\phi }_{\mathbf{y}}({\mathbf{x}}_1)-{\phi }_{\mathbf{y}}({\mathbf{x}}_2)|\le \frac{1}{2}|{\mathbf{x}}_1-{\mathbf{x}}_2|({\mathbf{x}}_1,{\mathbf{x}}_2\in U).\end{array}$$E3

It follows that ${\phi }_{\mathbf{y}}$ has at most one fixed point in $U$, so $\mathbf{f}(\mathbf{x})=\mathbf{y}$ for at most one $\mathbf{x}\in U$.

$f(U)$ is open in ${\mathbb{R}}^n$

Let $V=\mathbf{f}(U)$. Pick ${\mathbf{y}}_0\in V$. Then ${\mathbf{y}}_0=\mathbf{f}({\mathbf{x}}_0)$ for some ${\mathbf{x}}_0\in U$. Let $B=B_r({\mathbf{x}}_0)$ be such that $\overline{B_r({\mathbf{x}}_0)}\subset U$. We will show that $B_{\lambda r}({\mathbf{y}}_0)\subset V$.

Let $\mathbf{y}\in B_{\lambda r}({\mathbf{y}}_0)$.

$$\begin{array}{r}|{\phi }_{\mathbf{y}}({\mathbf{x}}_0)-{\mathbf{x}}_0|=|A^{-1}(\mathbf{y}-{\mathbf{y}}_0)|<\Vert A^{-1}\Vert \lambda r=\frac{r}{2}.\end{array}$$E4

If $\mathbf{x}\in \overline{B}$, by (E3) and (E4),

$$\begin{array}{rl}|{\phi }_{\mathbf{y}}(\mathbf{x})-{\mathbf{x}}_0|& \le |{\phi }_{\mathbf{y}}(\mathbf{x})-{\phi }_{\mathbf{y}}({\mathbf{x}}_0)|+|{\phi }_{\mathbf{y}}({\mathbf{x}}_0)-{\mathbf{x}}_0|\\ & <\frac{1}{2}|\mathbf{x}-{\mathbf{x}}_0|+\frac{r}{2}\le r;\end{array}$$

hence ${\phi }_{\mathbf{y}}(\mathbf{x})\in B$.

Thus ${\phi }_{\mathbf{y}}$ is a contraction of $\overline{B}$ into $\overline{B}$. Being a closed subset of ${\mathbb{R}}^n$, $\overline{B}$ is complete. By the Banach contraction principle, ${\phi }_{\mathbf{y}}$ has a unique fixed point $\mathbf{x}\in \overline{B}$, for which we must have $\mathbf{f}(\mathbf{x})=\mathbf{y}$. Thus $\mathbf{y}\in \mathbf{f}(\overline{B})\subset \mathbf{f}(U)=V$.

$\mathbf{g}$ is a $C^1$ mapping

Let $\mathbf{y},\mathbf{y}+\mathbf{k}\in V$. There exist $\mathbf{x},\mathbf{x}+\mathbf{h}\in U$ such that $\mathbf{y}=\mathbf{f}(\mathbf{x})$, $\mathbf{y}+\mathbf{k}=f(\mathbf{x}+\mathbf{h})$.

$$\begin{array}{r}{\phi }_{\mathbf{y}}(\mathbf{x}+\mathbf{h})-{\phi }_{\mathbf{y}}(\mathbf{x})=\mathbf{h}+A^{-1}[\mathbf{f}(\mathbf{x})-\mathbf{f}(\mathbf{x}+\mathbf{h})]=\mathbf{h}-A^{-1}\mathbf{k}.\end{array}$$

By (E3), $|\mathbf{h}-A^{-1}\mathbf{k}|\le |\mathbf{h}|/2$. Hence $|A^{-1}\mathbf{k}|\ge |\mathbf{h}|/2$ and

$$|\mathbf{h}|\le 2\Vert A^{-1}\Vert |\mathbf{k}|={\lambda }^{-1}|\mathbf{k}|.$$E5

Next, note that by (E1), (E2), and Proposition 172.4, ${\mathbf{f}}^{\prime }(\mathbf{x})$ has an inverse, say $T$. We will show that $T$ is ${\mathbf{g}}^{\prime }(\mathbf{y})$. Since

$$\begin{array}{rl}\mathbf{g}(\mathbf{y}+\mathbf{k})-\mathbf{g}(\mathbf{y})-T\mathbf{k}& =\mathbf{h}-T\mathbf{k}\\ & =-T[\mathbf{f}(\mathbf{x}+\mathbf{h})-\mathbf{f}(\mathbf{x})-{\mathbf{f}}^{\prime }(\mathbf{x})\mathbf{h}],\end{array}$$

(E5) implies

$$\begin{array}{r}\frac{|\mathbf{g}(\mathbf{y}+\mathbf{k})-\mathbf{g}(\mathbf{y})-T\mathbf{k}|}{|\mathbf{k}|}\le \frac{\Vert T\Vert }{\lambda }\frac{|\mathbf{f}(\mathbf{x}+\mathbf{h})-\mathbf{f}(\mathbf{x})-{\mathbf{f}}^{\prime }(\mathbf{x})\mathbf{h}|}{|\mathbf{h}|}.\end{array}$$

As $\mathbf{k}\to 0$, $\mathbf{h}\to 0$ by (E5). Thus, the right side of the inequality tends to $0$ as $\mathbf{k}\to 0$. Hence the same is true of the left. By the Definition of the derivative, we have proven that

$${\mathbf{g}}^{\prime }(\mathbf{y})=T={\mathbf{f}}^{\prime }(\mathbf{x}{)}^{-1}={\mathbf{f}}^{\prime }(\mathbf{g}(\mathbf{y}){)}^{-1}(\mathbf{y}\in V).$$

Since $\mathbf{g}$ is differentiable on $V$, it is continuous on $V$. If we denote the set of all invertible elements of $\mathcal{L}({\mathbb{R}}^n)$ by $\Omega$, ${\mathbf{f}}^{\prime }:U\to \Omega$ is continuous, and the inversion map $\Omega \to \Omega$ is also continuous by Proposition 172.4. Thus, ${\mathbf{g}}^{\prime }$ is continuous, and $g$ is $C^1$.□

Corollary 190.7(Corollary).

If $\mathbf{f}$ is a $C^1$ mapping of an open set $E\subset {\mathbb{R}}^n$ into ${\mathbb{R}}^n$ and if ${\mathbf{f}}^{\prime }(\mathbf{x})$ is invertible for every $\mathbf{x}\in E$, then $\mathbf{f}(W)$ is an open subset of ${\mathbb{R}}^n$ for every open set $W\subset E$. In other words, $\mathbf{f}$ is an open mapping of $E$ into ${\mathbb{R}}^n$.

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References

Apostol, T. M. (1985). Mathematical Analysis (2d ed). Narosa.

Pugh, C. C. (2015). Real Mathematical Analysis. Springer International Publishing. https://doi.org/10.1007/978-3-319-17771-7

Rudin, W. (1976). Principles of Mathematical Analysis (3d ed). McGraw-Hill.