# Change of Variables Formula — MTH 429H

Suppose ${f\in C_{c}(\mathbb{R}^{d})}$, so ${f}$ is continuous on ${\mathbb{R}^{d}}$ and it support,

$\displaystyle \mathrm{supp}f=\overline{\left\{ \mathbf{x}\ :\ f(\mathbf{x})>0\right\} },$

is compact. Recall that we defined

$\displaystyle \int_{\mathbb{R}^{d}}f\left(\mathbf{x}\right)d\mathbf{x}:=\int_{[-M,M]^{d}}f\left(\mathbf{x}\right)d\mathbf{x}$

with ${M}$ large enough that ${\mathrm{supp}f\subset[-M,M]^{d}}$. Our main goal in these notes is to prove the change of variables formula:

Theorem 1 Let ${T\in C^{1}\left(U,\mathbb{R}^{d}\right)}$ be a one-to-one map such that ${\det T'(\mathbf{x})\neq0}$ for all ${\mathbf{x}\in U}$. If ${f\in C(\mathbb{R}^{d})}$ has compact support with ${\mathrm{supp}f\subset T(U)}$ then

$\displaystyle \int_{\mathbb{R}^{d}}f(T(\mathbf{x}))\left|\det dT(\mathbf{x})\right|d\mathbf{x}=\int_{\mathbb{R}^{d}}f(\mathbf{x})d\mathbf{x}, \ \ \ \ \ (1)$

where we set ${f(T(\mathbf{x}))\left|\det dT(\mathbf{x})\right|=0}$ for ${\mathbf{x}\not\in U}$.

Open Mapping Theorem

In proving the change of variables formula it will be useful to know that ${T(U)}$ is in fact an open set. This is, in fact, a simple consequence of the inverse function theorem:

Thm 2 (Open Mapping Theorem) Let ${T\in C^{1}(U,\mathbb{R}^{d})}$ where ${U\subset\mathbb{R}^{d}}$ is open and suppose ${\det dT(\mathbf{x})\neq0}$ for all ${\mathbf{x}\in U}$. Then ${T(U)}$ is open.

Remark 3 In the applications below, ${T}$ will be one-to-one, however that is not required for the open mapping theorem.

Proof: Let ${\mathbf{y}_{0}\in T(U)}$. Suppose ${\mathbf{y}_{0}=T(\mathbf{x}_{0})}$. Since ${dT(\mathbf{x}_{0})}$ is invertible, by the inverse function theorem, there are open sets ${U',V}$ such that ${\mathbf{x}_{0}\in U'}$, ${\mathbf{y_{0}}\in V}$, and ${T}$ is a one-to-one map of ${U'}$ onto ${V}$. In particular we see that ${\mathbf{y}_{0}\in V\subset T(U)}$. Thus ${T(U)}$ is open. $\Box$

Special cases of the Change of Variables Formula

For some special types of maps the for which (1) is easily seen to hold. For example, if ${T}$ is a translation

$\displaystyle T(\mathbf{x})=\mathbf{x}+\mathbf{a}$

where ${\mathbf{a}}$ is a fixed vector in ${\mathbb{R}^{d}}$ then ${dT\left(\mathbf{x}\right)=\mathrm{I}}$ so ${\det dT\left(\mathbf{x}\right)=1}$. Since our definition of the integral implies

$\displaystyle \int_{\mathbb{R}^{d}}f\left(T\left(\mathbf{x}\right)\right)d\mathbf{x}=\int_{\mathbb{R}^{d}}f\left(\mathbf{x}+\mathbf{a}\right)d\mathbf{x}=\int_{\mathbb{R}^{d}}f\left(\mathbf{x}\right)d\mathbf{x} \ \ \ \ \ (2)$

we see that (1) holds in this case.

Another important special case of(1) follows from the fact that the integral does not depend on the order in which we integrate the variables. Recall that a permutation map is a linear map ${P_{\sigma}\in L\left(\mathbb{R}^{k}\right)}$ such that ${P_{\sigma}\left(\mathbf{e}_{i}\right)=\mathbf{e}_{\sigma(i)}}$ for some permutation ${\sigma}$ of ${\left\{ 1,\ldots,k\right\} }$.

Lemma 4 Let ${P_{\sigma}\in L\left(\mathbb{R}^{d}\right)}$ be a permuation map then (1) holds.

Proof: Let ${f\in C_{c}\left(\mathbb{R}^{d}\right)}$. In a previous lecture we proved that

$\displaystyle \int_{P_{\sigma}^{-1}\left(I_{d}\right)}f\left(P_{\sigma}\left(\mathbf{x}\right)\right)d\mathbf{x}=\int_{P_{\sigma}^{-1}\left(I_{d}\right)}f\left(\mathbf{x}\right)d\mathbf{x}$

for any ${d}$-cell. If ${I_{d}=[-M,M]^{d}}$ then ${P_{\sigma}^{-1}\left([-M,M]^{d}\right)=[-M,M]^{d}}$ so with ${M}$ large enough we see that

$\displaystyle \int_{\mathbb{R}^{d}}f\left(P_{\sigma}\left(\mathbf{x}\right)\right)d\mathbf{x}=\int_{\mathbb{R}^{d}}f\left(\mathbf{x}\right)d\mathbf{x}.$

Since ${P_{\sigma}}$ is linear ${dP_{\sigma}\left(\mathbf{x}\right)=P_{\sigma}}$ for each ${\mathbf{x}}$ and so the proof will be complete once we show that ${\det P_{\sigma}=\pm1}$. Thus the proof is completed by the following

Exercise 5 Let ${P_{\sigma}\in L\left(\mathbb{R}^{d}\right)}$ be a permutation map. Prove that ${\det P_{\sigma}=\mathrm{sgn}\sigma}$.

$\Box$

Another special case for which we know 1 is in 1D:

Lemma 6 Suppose ${T\in C^{1}(U)}$ and ${T'(x)\neq0}$ for all ${x\in U}$ where ${U\subset\mathbb{R}}$ is open. If ${f\in C_{c}\left(\mathbb{R}\right)}$ and ${\mathrm{supp}f\subset T(U)}$, then ${T^{-1}\left(x\right)}$ is finite for each ${x\in\mathrm{supp}f}$ and

$\displaystyle \int_{\mathbb{R}}f(T(x))\left|T'(x)\right|dx=\int_{\mathbb{R}}f(x)n(x)dx, \ \ \ \ \ (3)$

where ${n(x)=\#T^{-1}\left(x\right)}$, the number of points mapped to ${x}$ by ${T}$. In particular, if ${T}$ is one-to-one then the change of variables formula (1) holds for any ${f\in C_{c}\left(\mathbb{R}\right)}$ with ${\mathrm{supp}f\subset T\left(U\right)}$.

Proof: Since ${U}$ is an open subset of ${\mathbb{R}}$ it is a countable or finite union of disjoint open intervals,

$\displaystyle U=\bigcup_{i}I_{i}$

Note that ${T(U)=\cup_{i}T(I_{i})}$. Furthermore, ${T}$ is one-to-one on ${I_{i}}$ and the image ${T(I_{i})}$ is an open interval for each ${i}$ (because ${T'(x)\neq0}$ for all ${x\in I_{i}}$). Since ${\mathrm{supp}f\subset T(U)}$ is compact, there are finitely many intervals ${I_{i}}$ such that ${T(I_{i})}$ cover ${ }$. Hence, without loss of generality we may assume that ${U=\bigcup_{i=1}^{n}I_{i}}$ with ${n<\infty}$ and ${I_{i}}$ pairwise disjoint. Note that ${n(x)\le n}$ for each ${x\in\mathrm{supp}f}$.

Because ${\mathrm{supp}f}$ is compact it is closed. Thus for each ${i=1,\ldots,n}$ there is a closed interval ${[a_{i},b_{i}]\subset I_{i}}$ such that ${\mathrm{supp}f\subset\bigcup_{i=1}^{n}T\left([a_{i},b_{i}]\right)}$. Thus

$\displaystyle \int_{\mathbb{R}}f(T(x))\left|T'(x)\right|dx=\sum_{i=1}^{n}\int_{a_{i}}^{b_{i}}f(T(x))\left|T'(x)\right|dx. \ \ \ \ \ (4)$

Fix ${i\in\left\{ 1,\ldots,n\right\} }$. Since ${T'(x)\neq0}$ on ${[a_{i},b_{i}]}$, there are two cases:

1. ${T'(x)>0}$ for all ${x\in[a_{i},b_{i}]}$. Then ${\left|T'(x)\right|=T'(x)}$ on ${[a_{i},b_{i}]}$ and ${T(a_{i}), so

$\displaystyle \int_{a_{i}}^{b_{i}}f(T(x))\left|T'(x)\right|dx=\int_{a_{i}}^{b_{i}}f(T(x))T'(x)dx=\int_{T(a_{i})}^{T(b_{i})}f(y)dy=\int_{T([a_{i},b_{i}])}f(y)dy.$

2. ${T'(x)<0}$ for all ${x\in[a_{i},b_{i}]}$. Then ${\left|T'(x)\right|=-T'(x)}$ on ${[a_{i},b_{i}]}$ and ${T(a_{i})>T(b_{i})}$ so

$\displaystyle \begin{array}{rcl} \int_{a_{i}}^{b_{i}}f(T(x))\left|T'(x)\right|dx & = & -\int_{a_{i}}^{b_{i}}f(T(x))T'(x)\\ & = & -\int_{T(a_{i})}^{T(b_{i})}f(y)dy=\int_{T(b_{i})}^{T(a_{i})}f(y)dy=\int_{T([a_{i},b_{i}])}f(y)dy. \end{array}$

In either case,

$\displaystyle \int_{a_{i}}^{b_{i}}f(T(x))\left|T'(x)\right|dx=\int_{T([a_{i},b_{i}])}f(y)dy.$

Since ${\mathrm{supp}f\subset\bigcup_{i=1}^{n}T([a_{i},b_{i}])}$, this identity and (4) imply (3).$\Box$

Remark 7 If ${I=[a,b]}$ is an interval with ${a, the integral ${\int_{I}f(x)dx}$ is equal to ${\int_{a}^{b}f(x)dx}$. This integral is “unoriented” — it does not depend on the “orientation” of the interval ${[a,b]}$ and is positive whenever ${f}$ is. By contrast the usual notation ${\int_{a}^{b}f(x)dx}$ refers to an “oriented” integral: if ${b it is defined to be ${-\int_{b}^{a}f(x)dx}$. In 1D this is rather trivial, but such distinctions will be important when we look at integration of differential forms.

The 1D change of variables formula rather directly shows a change of variables formula for a special class of maps on ${\mathbb{R}^{d}}$.

Definition 8 A function ${T\in C^{1}\left(U,\mathbb{R}^{d}\right)}$ with ${U}$ open in ${\mathbb{R}^{d}}$ is called a primitive map if there is ${g\in C(U,\mathbb{R})}$ and ${i\in\left\{ 1,\ldots,d\right\} }$ such that

$\displaystyle T\left(\mathbf{x}\right)=\mathbf{x}+\left[g(\mathbf{x})-x_{i}\right]\mathbf{e}_{i}.$

Here ${\mathbf{e}_{i}}$ denotes the ${i}$-th elementary basis vector and ${x_{i}=\mathbf{e}_{i}\cdot\mathbf{x}}$ is the ${i}$-th coordinate of ${x}$. Thus a primitive map alters only one coordinate of ${\mathbf{x}}$.

Lemma 9 Let ${T\in C^{1}\left(U,\mathbb{R}^{d}\right)}$ be a one-to-one primitive map with non-singular derivative and let ${f\in C_{c}(\mathbb{R}^{d})}$ have support in ${U}$. Then (1) holds.

Proof: First, suppose we have obtained (1) whenever ${T}$ is a primitive mapping that alters only the ${d}$-th coordinate. Given a primitive mapping that alters only the ${i}$-th coordinate with ${i, let ${P_{\tau}}$ be the permutation map with ${\tau}$ the permutation ${(i,d)}$ that interchanges ${i}$ and ${d}$ but leaves all other indices unchanged. Note that ${\widetilde{T}=P_{\tau}\circ T\circ P_{\tau}}$ is a primitive map that alters only the ${d}$-th coordinate. Since

$\displaystyle \det d\widetilde{T}\left(\mathbf{x}\right)=\left(\det P_{\tau}\right)^{2}\det dT(P_{\tau}\mathbf{x})=\det dT(P_{\tau}\mathbf{x}),$

we see that

$\displaystyle \begin{array}{rcl} \int_{\mathbb{R}^{d}}f(T(\mathbf{x}))\left|\det dT(\mathbf{x})\right|d\mathbf{x} & = & \int_{\mathbb{R}^{d}}f(P_{\tau}\widetilde{T}(P_{\tau}\mathbf{x}))\left|\det d\widetilde{T}(P_{\tau}\mathbf{x})\right|d\mathbf{x}\\ & = & \int_{\mathbb{R}^{d}}f(P_{\tau}\widetilde{T}\left(\mathbf{x}\right))\left|\det d\widetilde{T}(\mathbf{x})\right|d\mathbf{x}\\ & = & \int_{\mathbb{R}^{d}}f(P_{\tau}\mathbf{x})d\mathbf{x}=\int_{\mathbb{R}^{d}}f\left(\mathbf{x}\right)d\mathbf{x}, \end{array}$

where we have used the permutation invariance of the integral and (1) for ${\widetilde{T}}$.

Thus we may assume without loss of generality that

$\displaystyle T(\mathbf{x})=\mathbf{x}+\left[g(\mathbf{x})-x_{d}\right]\mathbf{e}_{d}.$

What are the partial derivatives of ${T}$? Clearly, for ${i\neq d}$ we have ${\frac{\partial T_{i}}{\partial x_{i}}=1}$ and ${\frac{\partial T_{i}}{\partial x_{j}}=0}$ for ${j\neq i}$. On the other hand

$\displaystyle \frac{\partial T_{d}}{\partial x_{j}}=\frac{\partial g}{\partial x_{j}}$

for each ${j}$. It follows that

$\displaystyle \det dT\left(\mathbf{x}\right)=\frac{\partial g}{\partial x_{d}}\left(\mathbf{x}\right).$

Since ${\det dT\left(\mathbf{x}\right)\neq0}$ we see that ${\frac{\partial g}{\partial x_{d}}(\mathbf{x})\neq0}$ for all ${\mathbf{x}\in U}$.

Now fix ${x_{1},\ldots,x_{d-1}\in\mathbb{R}}$ and consider the following continuous functions of a single variable

$\displaystyle \phi(t)=g(x_{1},\ldots,x_{d-1},t)$

and

$\displaystyle h(s)=f((x_{1},\ldots,x_{d-1},s)).$

Here ${h}$ is defined for any ${s\in\mathbb{R}}$ and has compact support (because ${f}$ does). However, ${\phi}$ is defined on the set

$\displaystyle U_{0}=\left\{ t\ :\ \left(x_{1},\ldots,x_{d-1},t\right)\in U\right\}$

which is an open set in ${\mathbb{R}}$ by virtue of the fact that ${U}$ is open in ${\mathbb{R}^{d}}$. Note that

$\displaystyle h\circ\phi(t)=f(T(x_{1},\ldots,x_{d-1},t))$

and

$\displaystyle \phi'(t)=\frac{\partial g}{\partial x_{d}}\left(x_{1},\ldots,x_{d-1},t\right)=\det dT(x_{1},\ldots,x_{d-1},t).$

Since ${T}$ and hence ${\phi}$ is one-to-one, we have

$\displaystyle \begin{array}{rcl} \int_{\mathbb{R}}f(T(x_{1},\ldots,x_{d-1},t))\left|\det T'(x_{1},\ldots,x_{d-1},t)\right|dt & = & \int_{\mathbb{R}}h(\phi(t))\left|\phi'(t)\right|dt\\ & = & \int_{\mathbb{R}}h(s)ds\\ & = & \int_{\mathbb{R}}f(x_{1},\ldots,x_{d-1},s)ds, \end{array}$

by the 1D change of variables formula (3). Integrating over the remaining ${d-1}$ variables yields (1). $\Box$

The primitive map factorization theorem, partitions of unity and the general change of variables formula

The idea of the proof of Theorem 1 is to use what we know so far — that the identity holds for translations, permutations and primitive maps — to obtain the result for general maps ${T}$. The key behind this proof is that the class of maps ${T}$ for which (1) holds is closed under composition:

Lemma 10 Let ${T_{1}\in C^{1}(U_{1},\mathbb{R}^{d})}$ and ${T_{2}\in C^{1}(U_{2},\mathbb{R}^{d})}$ be one-to-one maps such that, for ${j=1,2}$, ${\det dT_{j}(\mathbf{x})\neq0}$ for all ${\mathbf{x}\in U_{j}}$ and ${T_{1}(U_{1})\subset U_{2}}$. If, for ${j=1,2}$, (1) holds with ${T=T_{j}}$ for all ${f}$ with compact support in ${T_{j}(U_{j})}$ then (1) holds also with ${T=T_{2}\circ T_{1}}$ for all ${f}$ with compact support in ${T_{2}(T_{1}(U_{1}))}$.

Proof: Let ${T=T_{2}\circ T_{1}}$ and let ${f\in C(\mathbb{R}^{d})}$ have compact support in ${T(U_{1})}$. By the chain rule

$\displaystyle \det dT(\mathbf{x})=\det dT_{2}(T_{1}(\mathbf{x}))\det dT_{1}(\mathbf{x}),$

so we have

$\displaystyle \begin{array}{rcl} \int_{\mathbb{R}^{k}}f(T(\mathbf{x}))\det dT(\mathbf{x})d\mathbf{x} & = & \int_{\mathbb{R}^{k}}f(T_{2}(T_{1}(\mathbf{x})))\det dT_{2}\left(T_{1}\left(\mathbf{x}\right)\right)\det dT_{1}\left(\mathbf{x}\right)d\mathbf{x}\\ & = & \int_{\mathbb{R}^{k}}f\left(T_{2}\left(\mathbf{x}\right)\right)\det dT_{2}\left(\mathbf{x}\right)d\mathbf{x}=\int_{\mathbb{R}^{k}}f(\mathbf{x})d\mathbf{x}. \end{array}$

$\Box$

So the question is: Can every ${C^{1}}$ one-to-one map with non-vanishing Jacobian be written as the composition of flips and primitive mappings? The correct but limited answer is: “No.” A more useful observation is, “Not quite, but almost.” Namely, we can do this locally.

Thm 11 (Primitive map factorization theorem) Let ${T\in C^{1}(U,\mathbb{R}^{d})}$ with ${U}$ open in ${\mathbb{R}^{d}}$. If ${\det dT(\mathbf{x}_{0})\neq0}$ for some ${\mathbf{x}_{0}\in U}$ then there is an open set ${U_{0}\subset U}$ containing ${\mathbf{x}_{0}}$ such that f

$\displaystyle T(\mathbf{x})=T_{\mathbf{b}}\circ P\circ G_{l}\circ\cdots\circ G_{1}\circ T_{\mathbf{a}}(\mathbf{x}) \ \ \ \ \ (5)$

or ${\mathbf{x}\in U_{0}}$, where ${T_{\mathbf{a}}}$ and ${T_{\mathbf{b}}}$ are translations, ${T_{\mathbf{a}}(\mathbf{x})=\mathbf{x}+\mathbf{a}}$ and ${T_{\mathbf{b}}(\mathbf{x})=\mathbf{x}+\mathbf{b}}$, ${P}$ is a permutation map, and for each ${j=1,\ldots,l}$, ${G_{j}}$ is a one-to-one primitive map defined on an open set ${U_{j}}$. Furthermore, the number of primitive maps ${l}$ is less than or equal to ${d}$.

Remark 12 It is implied by (5) that the domains ${U_{j}}$ satisfy ${T_{a}(U_{0})\subset U_{1}}$ and ${G_{j}(U_{j})\subset U_{j+1}}$ for ${j=1,\ldots,l-1}$.

Let us defer the proof of Theorem 11. Suppose we fix ${T}$ with ${\det dT(\mathbf{x})\neq0}$ throughout it’s domain ${U}$. If we are lucky enough to have a continuous function ${f}$ with compact support in a neighborhood ${U_{0}}$ on which (5) holds then Lemma 10 immediately implies that (1) holds. However, what do we do if we are not so lucky? The answer is to use a partition of unity:

Thm 13 (Partition of Unity Theorem) Suppose ${K\subset\mathbb{R}^{d}}$ is compact and ${\left\{ U_{\alpha}\right\} }$ is an open cover of ${K}$. Then there exist functions ${\psi_{1},\ldots,\psi_{m}\in C(\mathbb{R}^{k})}$ such that

$\displaystyle \sum_{l=1}^{m}\psi_{l}(\mathbf{x})=1\quad\mbox{for each }\mathbf{x}\in K$

and, for each ${l=1,\ldots,m}$,

1. ${0\le\psi_{l}\le1}$, and
2. there is ${\alpha_{l}}$ such that ${\mathrm{supp}\psi_{l}\subset U_{\alpha_{l}}}$.

Remark 14 We call the functions ${\left\{ \psi_{l}\ :\ l=1,\ldots,m\right\} }$ a partition of unity on ${K}$ subordinate to the open cover ${\left\{ U_{\alpha}\right\} }$. We won’t need it here, but it is possible to take the functions ${\psi_{l}\in C^{r}}$ for any ${r}$ or even in ${C^{\infty}}$!

Partitions of unity are a very important tool throughout analysis for localizing proofs. An example of this is the following proof of the Change of Variables Theorem: Proof: .} Let ${f}$ and ${T}$ be as in the statement of the theorem and let ${K=\mathrm{supp}f}$. For each ${\mathbf{x}\in K}$ let ${U_{\mathbf{x}}\subset U}$ be an open set on which the factorization (5) holds and such that ${\mathbf{x}\in T(U_{\mathbf{x}})}$. Let ${\left\{ \psi_{l}\ :\ l=1,\ldots,m\right\} }$ be a partition of unity on ${K}$ subordinate to the open cover ${\left\{ T(U_{\mathbf{x}})\ :\ \mathbf{x}\in K\right\} }$ — note that this is an open cover by the open mapping theorem. Then

$\displaystyle \int_{\mathbb{R}^{d}}f(T(\mathbf{x}))\left|\det dT(\mathbf{x})\right|d\mathbf{x}=\sum_{l=1}^{m}\int_{\mathbb{R}^{d}}\psi_{l}(T(\mathbf{x}))f(T(\mathbf{x}))\left|\det dT(\mathbf{x})\right|d\mathbf{x}.$

However, since ${\mathrm{supp}\psi_{l}f\subset T(U_{\mathbf{x}_{l}})}$ and the factorization (5) holds on ${U_{\mathbf{x}_{l}}}$ we see by Lemma 10 that

$\displaystyle \int_{\mathbb{R}^{d}}\psi_{l}\left(T\left(\mathbf{x}\right)\right)f\left(T\left(\mathbf{x}\right)\right)\left|\det dT\left(\mathbf{x}\right)\right|d\mathbf{x}=\int_{\mathbb{R}^{k}}\psi_{l}\left(\mathbf{x}\right)f(\mathbf{x})d\mathbf{x}.$

Summing over ${l}$, we see that 1 holds! $\Box$

It remains to prove the existence of a partition of unity (Theorem 13) and the primitive map factorization theorem (Theorem 11). We start with the partition of unity: Proof: } For each ${\mathbf{x}\in K}$ there is ${\alpha(\mathbf{x})}$ such that ${\mathbf{x}\in U_{\alpha(\mathbf{x})}}$. Since ${U_{\alpha(\mathbf{x})}}$ are open, then we can find radii ${r(\mathbf{x}) such that

$\displaystyle \overline{B_{r(\mathbf{x})}(\mathbf{x})}\subset B_{s(\mathbf{x})}(\mathbf{x})\subset U_{\alpha(\mathbf{x})}.$

Since ${\left\{ B_{s(\mathbf{x})}(\mathbf{x})\ :\ \mathbf{x}\in K\right\} }$ is an open cover of ${K}$ there are finitely many points ${\mathbf{x}_{1},\ldots,\mathbf{x}_{m}}$ such that

$\displaystyle K\subset B_{s(\mathbf{x}_{1})}(\mathbf{x}_{1})\cup\cdots\cup B_{s(\mathbf{x}_{m})}(\mathbf{x}_{m}).$

For each ${l=1,\ldots,m}$, I claim that we can find ${\phi_{l}\in C(\mathbb{R}^{d})}$ such that

1. ${0\le\phi_{l}(\mathbf{x})\le1}$ for all ${\mathbf{x}\in\mathbb{R}^{d}}$,
2. ${\phi_{l}(\mathbf{x})=1}$ if ${|\mathbf{x}-\mathbf{x}_{l}|\le r(\mathbf{x}_{l})}$,
3. ${\phi_{l}(\mathbf{x})=0}$ if ${\left|\mathbf{x}-\mathbf{x}_{l}\right|>s(\mathbf{x}_{l})}$.

Indeed, simply set

$\displaystyle \phi_{l}(\mathbf{x})=\begin{cases} 1 & \text{ if }\left|\mathbf{x}-\mathbf{x}_{l}\right|\le r(\mathbf{x}_{l})\\ \frac{s(\mathbf{x}_{l})-\left|\mathbf{x}-\mathbf{x}_{l}\right|}{s(\mathbf{x}_{l})-r(\mathbf{x}_{l})} & \text{ if }r(\mathbf{x}_{l})<\left|\mathbf{x}-\mathbf{x}_{l}\right|

The functions ${\psi_{l}}$ can be defined recursively by taking ${\psi_{1}=\phi_{1}}$ and

$\displaystyle \psi_{l}=\left(1-\sum_{j=1}^{l-1}\psi_{j}\right)\phi_{l}, \ \ \ \ \ (7)$

for ${l=2,\cdots,m}$. Clearly ${ }$. Suppose we have proved that

$\displaystyle 0\le\sum_{j=1}^{l}\psi_{j}\le1. \ \ \ \ \ (8)$

Then

$\displaystyle \sum_{j=1}^{l+1}\psi_{j}=\phi_{l+1}+(1-\phi_{l+1})\sum_{j=1}^{l}\psi_{l} \ \ \ \ \ (9)$

from which it follows that ${0\le\sum_{j=1}^{l+1}\psi_{j}\le1}$. Since ${0\le\psi_{1}\le1}$ we see by induction that (8) holds for each ${l=1,\ldots,m}$ and thus, by (7), that ${0\le\psi_{l}\le1}$ for each ${l}$. Finally for ${\mathbf{x}\in K}$ we have ${\phi_{l}(\mathbf{x})=1}$ for some ${l=1,\ldots,m}$. Thus by (9)

$\displaystyle \sum_{j=1}^{l}\psi_{j}\left(\mathbf{x}\right)=1,$

from which it follows that ${\sum_{j=1}^{m}\psi_{j}(\mathbf{x})=1}$. $\Box$

Exercise 15 Modify the definition (6) so as to produce a partion of unity with ${\psi_{j}\in C^{\infty}.}$

Proof of the Primitive Map Factorization Theorem

It remains to prove Theorem 11: Proof: First suppose ${T}$ is a map as in the statement of the Theorem and let ${\mathbf{a}=-\mathbf{x}_{0}}$ and ${\mathbf{b}=T(\mathbf{x}_{0})}$. Then

$\displaystyle T_{\mathbf{b}}^{-1}\circ T\circ T_{\mathbf{a}}^{-1}(\mathbf{0})=\mathbf{0}.$

Thus it suffices to prove (5) with ${\mathbf{a}=\mathbf{b}=0}$ but under the additional assumption that ${\mathbf{0}\in U}$ and ${T(\mathbf{0})=\mathbf{0}}$. That is, it suffices to prove

Claim 16 For each ${T\in\mathcal{C}}$, where

$\displaystyle \mathcal{C}=\left\{ T\in C^{1}(U)\ :\ U\text{ is open, }\textbf{0}\in U,\ T\left(\textbf{0}\right)=\mathbf{0},\ \det dT(\textbf{0})\neq0\right\} ,$

there is an open neigheborhood ${U_{0}}$ of ${\mathbf{0}}$ such that for ${\mathbf{x}\in U_{0}}$ we have

$\displaystyle T(\mathbf{x})=PG_{l}\circ\cdots\circ G_{1}(\mathbf{x}) \ \ \ \ \ (10)$

where ${P}$ is a permutation map, for each ${j=1,\ldots,l}$, ${G_{j}}$ is a one-to-one primitive map defined on a neighborhood of ${\mathbf{0}}$ and satisfying ${G(\mathbf{0})=\mathbf{0}}$, and the number of primitive maps ${l\le d}$.

Let ${P_{0}\in L(\mathbb{R}^{d})}$ denote the zero map and for each ${m=1,\ldots,d}$ let ${P_{m}\in L(\mathbb{R}^{d})}$ be the projection onto the first ${m}$ coordinates in ${\mathbb{R}^{d}}$,

$\displaystyle P_{m}\mathbf{x}=x_{1}\mathbf{e}_{1}+\cdots+x_{m}\mathbf{e}_{m}.$

Note that ${P_{d}=\mathrm{I}_{d}}$. Now define for each ${m=0,\ldots,d}$ the following class of maps

$\displaystyle \mathcal{C}_{m}=\left\{ T\in\mathcal{C}:\ P_{m}T(\mathbf{x})=P_{m}\mathbf{x}\text{ for all \textbf{x} in the domain of }T\right\} .$

Clearly ${\mathcal{C}=\mathcal{C}_{0}\supset\mathcal{C}_{1}\supset\mathcal{C}_{2}\supset\cdots\supset\mathcal{C}_{d}=\left\{ \mathrm{I}_{d}\right\} }$. We prove the claim by using backwards induction on ${m}$ to show that any ${T\in\mathcal{C}_{m}}$ can be factored as in (10) with the number of primitive maps ${l\le d-m}$. The base case ${m=d}$ is trivial since the only map in d is the identity ${I_{d}}$, which is a permutation map. Likewise the case ${m=d-1}$ is trivial since any map ${T\in\mathcal{C}_{d-1}}$ is primitive.

Now suppose we have verified for some ${1\le m\le d}$ that (10) holds with ${l\le d-m}$ primitive maps for any ${T\in\mathcal{C}_{m}}$. Let ${T\in\mathcal{C}_{m-1}}$, so

$\displaystyle T\left(\mathbf{x}\right)=P_{m-1}\mathbf{x}+\sum_{j=m}^{d}\alpha_{j}\left(\mathbf{x}\right)\mathbf{e}_{j}$

for some functions ${\alpha_{m},\ldots,\alpha_{d}}$. Thus

$\displaystyle dT(\mathbf{0})\mathbf{e}_{m}=\sum_{j=m}^{d}\frac{\partial\alpha_{j}}{\partial x_{m}}\left(\mathbf{0}\right)\mathbf{e}_{j}.$

Since ${dT\left(\mathbf{0}\right)\mathbf{e}_{m}\neq0}$ we must have ${\frac{\partial\alpha_{j}}{\partial x_{m}}(\mathbf{0})\neq0}$ for some ${j\in\left\{ m,\ldots,d\right\} }$. Let ${F}$ be the permutation map that interchanges ${\mathbf{e}_{m}}$ and ${\mathbf{e}_{j}}$, taking ${F=\mathrm{I}}$ if ${m=j}$, and let

$\displaystyle G\left(\mathbf{x}\right)=\mathbf{x}+\left(\alpha_{j}\left(\mathbf{x}\right)-x_{m}\right)\mathbf{e}_{m}.$

So ${G}$ is primitive, ${G(\mathbf{0})=0}$ and, because ${\frac{\partial\alpha_{j}}{\partial x_{m}}(\mathbf{0})\neq0}$, ${dG(\mathbf{0})}$ is invertible. Thus by the inverse function theorem there is an open neighborhood ${U_{m}}$ of ${\mathbf{0}}$ such that ${G}$ is a one-to-one map of ${U_{m}}$ onto a neighborhood ${V_{m}}$ of ${\mathbf{0}}$ on which ${G^{-1}}$ is continuosly differentiable. Consider the function ${S=F\circ T\circ G^{-1}}$. Clearly ${S\in C^{1}(V_{m})}$, ${S(\mathbf{0})=\mathbf{0}}$ and ${\det dS(\mathbf{0})\neq0}$ (by the chain rule), so ${S\in\mathcal{C}}$. Furthermore

$\displaystyle \begin{array}{rcl} P_{m}S\left(G(\mathbf{x})\right) & = & P_{m}FT(\mathbf{x})\\ & = & P_{m}F\left(P_{m-1}\mathbf{x}+\sum_{l=m}^{d}\alpha_{l}(\mathbf{x})\mathbf{e}_{l}\right)\\ & = & P_{m-1}\mathbf{x}+\alpha_{j}(\mathbf{x})\mathbf{e}_{m}\\ & = & P_{m}G(\mathbf{x}). \end{array}$

Thus ${P_{m}S\left(\mathbf{x}\right)=P_{m}\mathbf{x}}$ for all ${x\in V_{m}}$ and we see that ${S\in\mathcal{C}_{m}}$. By the induction hypothesis, ${S}$ factors as

$\displaystyle S=P\circ G_{1}\circ\cdots\circ G_{j}$

with ${P}$ a permutation map and ${G_{1},\ldots,G_{j}}$ primitive maps with ${j\le d-m}$. Thus for ${\mbox{\textbf{x}}\in U_{m}}$

$\displaystyle T\left(\mathbf{x}\right)=F\circ S\circ G\left(\mathbf{x}\right)=\widetilde{P}\circ G_{1}\circ\cdots\circ G_{j}\circ G$

where ${\widetilde{P}=FP}$ is a permutation map and the number of primitive maps satisfies

$\displaystyle j+1\le d-m+1=d-(m-1).$

$\Box$