MAT257Y1/Unit 10
Unit 10 · Analysis II

Partitions of Unity & Change of Variables

Two tools that look like technical apparatus and are not. A partition of unity chops a global problem into local ones and glues the answers back together with no seams; it is the device that will make integration on manifolds possible at all. Change of variables explains where the Jacobian factor in every polar and spherical integral comes from: it is the local volume distortion of the substitution.

By the end you can
Why it matters & where it comes from
Why it exists

Local statements are the ones you can prove; global ones are the ones you want. A partition of unity is the standard bridge, and it is available in smooth geometry precisely because smooth bump functions exist — something that fails in the analytic or holomorphic categories.

In the real world

Finite element methods build global basis functions exactly this way. Blending weights in graphics, mixture models, and domain decomposition in numerical PDE all use the same partition-of-unity idea. Change of variables is behind every density transformation in probability.

Story & open questions

Euler and Lagrange used substitution freely; a correct general proof took until the nineteenth century, and it is still one of the harder theorems in an undergraduate course. Partitions of unity became standard with Whitney and the rise of differential topology in the 1930s.

Bump functions

Theorem A smooth function that dies

Let $\varphi(t)=e^{-1/t}$ for $t>0$ and $\varphi(t)=0$ for $t\le0$. Then $\varphi$ is $C^\infty$ on $\mathbb{R}$.

Every derivative on $t>0$ has the form $p(1/t)e^{-1/t}$ for a polynomial $p$, and $e^{-1/t}$ beats every polynomial as $t\to0^{+}$. So all derivatives extend continuously by $0$. Note $\varphi$ is not analytic at $0$: its Taylor series there is identically zero and does not represent it.

Theorem Bump on a ball

From $\varphi$ build $\psi(x)=\varphi(1-\|x\|^2)$: smooth, strictly positive on the open unit ball, and identically zero outside it. Rescaling and translating gives a smooth bump supported in any ball you like, and normalising makes it $1$ on a smaller concentric ball.

This is why smooth is easier than analytic. A nonzero analytic function cannot vanish on an open set. Everything in this unit — and everything about manifolds later — depends on smooth functions being able to.

Partitions of unity

Definition

Let $A\subseteq\mathbb{R}^n$ and let $\mathcal{O}$ be an open cover of $A$. A $C^\infty$ partition of unity on $A$ subordinate to $\mathcal{O}$ is a collection $\Phi$ of smooth functions $\rho:\mathbb{R}^n\to[0,1]$ such that

(1) each $x\in A$ has a neighbourhood on which all but finitely many $\rho$ vanish;
(2) $\sum_{\rho\in\Phi}\rho(x)=1$ for every $x\in A$ — a finite sum at each point, by (1);
(3) each $\rho$ has support inside some member of $\mathcal{O}$.
Theorem Existence

For any $A\subseteq\mathbb{R}^n$ and any open cover, such a partition exists.

Sketch: exhaust $A$ by compact pieces, cover each piece by finitely many balls each inside a member of the cover, take a bump function for each ball, and normalise by dividing by the (locally finite) sum. Local finiteness is what makes the sum a genuine function.

Example What it buys

To define $\int_Af$ when $A$ is not a box: write $\int_Af=\sum_\rho\int\rho f$. Each $\rho f$ is supported in one small set where you already know what to do, the sum is locally finite, and the answer is independent of the partition chosen. This is exactly the argument Unit 13 will use to integrate over a manifold covered by charts.

Change of variables

Theorem Change of variables

Let $U\subseteq\mathbb{R}^n$ be open and $g:U\to\mathbb{R}^n$ be an injective $C^1$ map with $\det Jg(x)\ne0$ on $U$. Then for every integrable $f$ on $g(U)$,

$\displaystyle\int_{g(U)}f=\int_U(f\circ g)\,|\det Jg|.$

The absolute value matters: volume is unsigned, and orientation is not tracked here. Unit 13 will drop the absolute value once forms give it a place to record the sign.

Why the Jacobian The linear case first
For a linear $T$, the image of the unit cube is a parallelepiped of volume $|\det T|$ — that is what the determinant measures, and it can be proved by reducing $T$ to elementary matrices, each of which scales volume in an obvious way.
A $C^1$ map is locally close to its derivative, so a small box near $x$ maps to something of volume approximately $|\det Jg(x)|$ times the box's volume.
Summing over a fine partition and passing to the limit gives the theorem. The genuine proof does this carefully with a partition of unity to reduce to small neighbourhoods, plus the inverse function theorem to make $g$ a diffeomorphism there.
Example Polar coordinates

$g(r,\theta)=(r\cos\theta,r\sin\theta)$ has $\det Jg=r$, so

$\displaystyle\int_{D}f\,dx\,dy =\int_0^{2\pi}\!\!\int_0^{R}f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta$

for the disc of radius $R$. The factor $r$ is not a convention: it is the rate at which polar boxes stretch as you move outwards.

Example The Gaussian integral

Let $I=\int_{\mathbb{R}}e^{-x^2}dx$. Then $I^2$ is a double integral over the plane, and in polar coordinates it becomes $\int_0^{2\pi}\int_0^\infty e^{-r^2}r\,dr\,d\theta=2\pi\cdot\tfrac12=\pi$. So $I=\sqrt\pi$. The whole trick is the $r$ from the Jacobian, which makes the radial integral elementary.

Example Spherical coordinates

With $g(\rho,\varphi,\theta)=(\rho\sin\varphi\cos\theta,\rho\sin\varphi\sin\theta,\rho\cos\varphi)$ the determinant is $\rho^2\sin\varphi$, so the volume of the ball of radius $R$ is $\int_0^{2\pi}\int_0^{\pi}\int_0^{R}\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta =\tfrac43\pi R^3$.

Hypotheses can fail on a measure-zero set. Polar coordinates are not injective at $r=0$ and spherical ones degenerate on the polar axis. Both are measure zero, so the formula survives — a standard extension of the theorem, and the reason nobody worries about it in practice.
Common pitfalls

• Dropping the absolute value on $\det Jg$. Without it, an orientation-reversing substitution flips the sign of a volume.
• Substituting into the integrand but forgetting the Jacobian factor.
• Using a non-injective substitution over a set of positive measure. The formula then counts part of the image twice.
• Forgetting to transform the region as well as the integrand. The new limits are $g^{-1}$ of the old region.
• Treating a partition of unity as an infinite sum problem. Local finiteness makes every evaluation a finite sum.

Practice

Original problems written for this guide in the style of Spivak and Munkres exercises. Write a full solution before opening the answer.

Practice 1 · Area of an ellipse

Use $x=au$, $y=bv$ to find the area of $x^2/a^2+y^2/b^2\le1$.

Solution
$\det Jg=ab$, and the region pulls back to the unit disc of area $\pi$. So the area is $\pi ab$.
Practice 2 · A linear substitution

Evaluate $\int_S(x+y)\,dA$ where $S$ is the parallelogram with vertices $(0,0),(1,1),(2,1),(1,0)$.

Solution
Put $u=x-y$, $v=y$, so $x=u+v$, $y=v$ and $\det=1$. The region becomes $0\le u\le1$, $0\le v\le1$, and the integrand is $u+2v$. The integral is $\int_0^1\int_0^1(u+2v)\,du\,dv=\tfrac12+1=\tfrac32$.
Practice 3 · Bump function

Construct a smooth function that is $1$ on $\|x\|\le1$ and $0$ for $\|x\|\ge2$.

Solution
With $\varphi$ as above, set $h(t)=\dfrac{\varphi(2-t)}{\varphi(2-t)+\varphi(t-1)}$. The denominator is never zero, $h=1$ for $t\le1$ and $h=0$ for $t\ge2$. Take $x\mapsto h(\|x\|)$; smoothness at the origin holds because $h$ is constant near $t=0$.
Practice 4 · Volume of an $n$-ball

Set up the integral for the volume of the unit ball in $\mathbb{R}^n$ and find it for $n=4$.

Solution
Slice: $V_n=\int_{-1}^{1}V_{n-1}(1-t^2)^{(n-1)/2}dt$. With $V_2=\pi$ this gives $V_3=\tfrac43\pi$ and $V_4=\tfrac{\pi^2}{2}$.
Practice 5 · Where injectivity fails

Why is the polar map still usable despite failing injectivity?

Solution
It fails only at $r=0$ and along the identified edge $\theta=0$ versus $2\pi$, both of measure zero. Integrals are blind to measure-zero sets, so the formula holds on the complement and extends.
Practice 6 · A hyperbolic region

Evaluate $\int_S xy\,dA$ over the region bounded by $xy=1$, $xy=4$, $y=x$, $y=4x$ in the first quadrant.

Solution
Put $u=xy$, $v=y/x$. Then $\partial(u,v)/\partial(x,y)=2y/x=2v$, so the Jacobian of the inverse is $1/(2v)$. The integral becomes $\int_1^4\int_1^4 u\cdot\tfrac{1}{2v}\,dv\,du =\tfrac{15}{2}\cdot\tfrac12\ln4=\tfrac{15}{2}\ln2$.
Practice 7 · Locally finite

Why must a partition of unity be locally finite?

Solution
Otherwise $\sum\rho(x)$ is an infinite series whose convergence and smoothness are not automatic. Local finiteness makes it a finite sum near each point, so it is smooth term by term.
Practice 8 · Density of a transformed variable

If $X$ has density $p$ on $\mathbb{R}^n$ and $Y=g(X)$ for a diffeomorphism $g$, what is the density of $Y$?

Solution
$q(y)=p(g^{-1}(y))\,|\det Jg^{-1}(y)|$. It is the change of variables formula with $f$ the density, and it is the reason normalising flows in machine learning track log-determinants.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT257Y1 topic map. It should cover the following items before you move on:

Textbook/source alignment

Spivak proves change of variables in full and it is one of the longest arguments in the book; Munkres splits it into digestible lemmas. Both build partitions of unity first. Check the syllabus for which proof is examinable.

Core theorem map

Bump functions give partitions of unity, which localise the change of variables proof and later make integration on manifolds well defined. The inverse function theorem supplies the local diffeomorphisms that both arguments run on.

Study routine. Recompute the polar, cylindrical and spherical Jacobians from scratch until they are automatic. Those three determinants account for most of the computational marks in the integration half of the course.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · Role of the determinant

What does $|\det Jg(x)|$ measure?

Guide
The factor by which $g$ scales volume near $x$, to first order.
Extension 2 · Smooth but not analytic

Why can no analytic function be a bump function?

Guide
An analytic function vanishing on an open set vanishes identically on a connected domain, by the identity theorem.
Extension 3 · Subordinate

What does "subordinate to a cover" mean?

Guide
Each function of the partition has support contained in one member of the cover.
Extension 4 · Absolute value

Why the absolute value on the determinant?

Guide
Because volume is unsigned. Forms in Unit 11 keep the sign instead and integrate over oriented regions.
← Unit 09The Riemann integral in $\mathbb{R}^n$ Unit 11 →Tensors & differential forms