第 06 章

柯西收敛定理：不用猜极限也能证明存在

先证柯西数列必有界，再借 Bolzano–Weierstrass 定理证明柯西收敛定理；随后用它证明压缩映射定理（不动点存在唯一），并以牛顿法为例；最后给出连续性的 ε-δ 定义与最值定理。

The Cauchy Convergence Theorem: Proving Existence Without Guessing the Limit

We prove that Cauchy sequences are bounded, deduce the Cauchy convergence theorem from Bolzano–Weierstrass, use it to prove the contraction mapping theorem (existence and uniqueness of fixed points), illustrate with Newton's method, and close with ε-δ continuity and the extreme value theorem.

首发于 2025.09.02中 / EN

系统课程real-analysiscauchy-sequencecontraction-mappingcontinuityextreme-value-theorem

篇内目录

为什么需要柯西判据
时间轴
柯西数列
柯西收敛定理
压缩映射
迭代序列为什么是柯西列
压缩映射定理
牛顿法：一个应用
Bolzano–Weierstrass 定理
连续性与极限交换
最值定理
本节结论
理解检验
与前后课的关系

Why a Cauchy criterion
Timeline
Cauchy sequences
The Cauchy convergence theorem
Contractions
Why the iteration is Cauchy
The contraction mapping theorem
Newton's method: an application
The Bolzano–Weierstrass theorem
Continuity and exchanging limits
The extreme value theorem
What this lecture establishes
Comprehension check
Relation to neighbouring lectures

为什么需要柯西判据

到目前为止，证明一个数列收敛的标准做法是：先猜出极限，再用 $\varepsilon$ - $N$ 定义验证。但很多极限根本写不出闭式——微分方程的解、压缩映射的不动点都是这样。

柯西判据把「极限存在」改写成一条只涉及数列本身的条件：远处各项彼此靠拢。它的价值在于：不必知道极限是什么，也能断定极限存在。

时间轴

时间	内容
0:00	回顾单调收敛定理与柯西数列
3:53	定理：柯西数列必有界
9:27	柯西收敛定理；压缩映射定理
14:41	压缩映射至多一个不动点
21:36	迭代序列 $a_{n+1}=T(a_n)$ 的柯西估计
28:09	用柯西收敛定理证明压缩映射定理
38:06	牛顿法作为应用
50:17	Bolzano–Weierstrass 定理（下讲证明）
59:08	连续性、最值定理

柯西数列

定义。 数列 $(x_n)$ 是柯西数列，如果

\forall\varepsilon>0\ \exists N\ \forall n,m\ge N:\quad |x_n-x_m|<\varepsilon .

与收敛的定义相比，这里没有出现候选极限 $a$ 。直观上：收敛说「远处的项都聚在 $a$ 附近」，柯西说「远处的项彼此聚在一起」。

定理。 柯西数列必有界。

证明。 在柯西定义中取 $\varepsilon=1$ ，得到 $N$ ，使 $n,m\ge N$ 时 $|x_n-x_m|<1$ 。

固定 $m=N$ 。则对一切 $n\ge N$ ，

|x_n|=|x_n-x_N+x_N|\le|x_n-x_N|+|x_N|<1+|x_N| .

前面只剩 $x_1,\dots,x_{N-1}$ 这有限多项。令

M=\max\{|x_1|,\dots,|x_{N-1}|,\,1+|x_N|\},

则 $|x_n|\le M$ 对一切 $n$ 成立。 $\square$

手法与第 4 讲「收敛必有界」完全一样：固定一个参考点 $m=N$ ，用三角不等式把尾巴压住，头部取最大值。区别只在于参考点从极限 $a$ 换成了数列自身的第 $N$ 项。

柯西收敛定理

定理。 在 $\mathbb R$ 中，数列收敛当且仅当它是柯西数列。

「收敛 ⇒ 柯西」容易：若 $x_n\to x$ ，给定 $\varepsilon>0$ ，取 $N$ 使 $n\ge N$ 时 $|x_n-x|<\varepsilon/2$ 。则 $n,m\ge N$ 时

|x_n-x_m|\le|x_n-x|+|x-x_m|<\frac\varepsilon2+\frac\varepsilon2=\varepsilon .

反方向「柯西 ⇒ 收敛」需要完备性。课程用 Bolzano–Weierstrass 定理（下一讲证明）来推：

有界数列必有收敛子列。

证明（柯西 ⇒ 收敛）。 设 $(x_n)$ 是柯西数列。

由上一节， $(x_n)$ 有界。由 Bolzano–Weierstrass 定理，存在子列 $x_{n_k}\to x$ 。

现在证明整条数列也收敛到 $x$ 。给定 $\varepsilon>0$ 。

由柯西性，存在 $N_1$ ，使 $n,m\ge N_1$ 时 $|x_n-x_m|<\varepsilon/2$ 。
由子列收敛，存在 $N_2$ ，使 $k\ge N_2$ 时 $|x_{n_k}-x|<\varepsilon/2$ 。

取 $k\ge N_2$ 使 $n_k\ge N_1$ （这样的 $k$ 存在，因为 $n_k$ 严格递增趋于无穷）。令 $N=\max\{N_1,n_k\}$ 。当 $n\ge N$ 时，

|x_n-x|\le|x_n-x_{n_k}|+|x_{n_k}-x|<\frac\varepsilon2+\frac\varepsilon2=\varepsilon .

所以 $x_n\to x$ 。 $\square$

这里的逻辑是：子列给出一条「已知极限的锚」，柯西性保证整条数列与这条子列不能分道扬镳。子列收敛靠 Bolzano–Weierstrass（其证明依赖完备性），所以这条定理本质上是完备性的另一种表述。

压缩映射

定义。 映射 $T:\mathbb R\to\mathbb R$ 称为压缩映射，如果存在常数 $c$ ，满足

$c\ge0$ ；
$c<1$ ；
$|T(x)-T(y)|\le c|x-y|$ 对一切 $x,y$ 成立。

第三点强调「同一个 $c$ 」：不能对这一对点用一个常数、对另一对点换一个。

定义。 $x$ 是 $T$ 的不动点，如果 $T(x)=x$ 。

引理。 压缩映射至多有一个不动点。

证明。 设 $x,y$ 都是不动点。则

|x-y|=|T(x)-T(y)|\le c|x-y| .

移项得 $(1-c)|x-y|\le0$ 。因为 $c<1$ ，有 $1-c>0$ ，所以 $|x-y|\le0$ ，即 $x=y$ 。 $\square$

证明只有一行，关键是 $c<1$ ：若 $c=1$ ， $(1-c)|x-y|\le0$ 不提供任何信息，结论就不成立（平移 $T(x)=x+1$ 没有不动点）。

迭代序列为什么是柯西列

定理只说「至多一个」，还要证明「至少一个」。构造方法是迭代：任取 $x$ ，令

a_0=x, \qquad a_{n+1}=T(a_n)=T^{n+1}(x).

第一步：相邻两项的距离按几何速度缩小。 由压缩性，

|a_{n+1}-a_n|=|T(a_n)-T(a_{n-1})|\le c|a_n-a_{n-1}| .

反复使用：

|a_{n+1}-a_n|\le c^n|a_1-a_0| .

第二步：任意两项的距离可控。 设 $m>n$ ，用三角不等式把中间各项串起来：

|a_m-a_n|\le\sum_{k=n}^{m-1}|a_{k+1}-a_k| \le\sum_{k=n}^{m-1}c^k|a_1-a_0| .

几何级数求和：

\sum_{k=n}^{m-1}c^k\le\sum_{k=n}^{\infty}c^k=\frac{c^n}{1-c} .

所以

|a_m-a_n|\le\frac{c^n}{1-c}\,|a_1-a_0| .

因为 $0\le c<1$ ， $c^n\to0$ 。给定 $\varepsilon>0$ ，取 $n$ 足够大使右端小于 $\varepsilon$ ，则所有 $m>n$ 都满足 $|a_m-a_n|<\varepsilon$ 。故 $(a_n)$ 是柯西数列。

这里两个条件都用上了： $c<1$ 让几何级数收敛， $c\ge0$ 让 $c^n$ 单调趋于 $0$ 。若 $c=1$ ，估计变成 $(m-n)|a_1-a_0|$ ，随 $m$ 增大而无界，迭代点可能一路漂走。

压缩映射定理

定理。 设 $T:\mathbb R\to\mathbb R$ 是压缩映射，则 $T$ 有唯一不动点。

证明。 唯一性已由引理给出。存在性：

任取 $a_0\in\mathbb R$ ，令 $a_{n+1}=T(a_n)$ 。上一节证明了 $(a_n)$ 是柯西数列。由柯西收敛定理，存在 $a$ 使 $a_n\to a$ 。

证明 $T(a)=a$ 。由压缩性，

|T(a_n)-T(a)|\le c|a_n-a|\to0,

所以 $T(a_n)\to T(a)$ 。另一方面 $T(a_n)=a_{n+1}\to a$ 。数列 $(T(a_n))$ 只能有一个极限，故

T(a)=a .

所以 $a$ 是不动点。 $\square$

这条定理的构造性很强：它不仅断言不动点存在，还给出了求法——从任意点出发反复迭代。即使写不出不动点的闭式，也能任意逼近。课程说，正是这一点让它在微分方程（ODE 与 PDE）里极其有用：把方程改写成 $x=T(x)$ 的形式，验证 $T$ 压缩，就得到解的存在唯一性，第 24 讲会这样做。

牛顿法：一个应用

求方程 $f(x)=0$ 的根。若 $f$ 可导且 $f'(x)\ne0$ ，定义

T(x)=x-\frac{f(x)}{f'(x)} .

这是牛顿迭代：从当前点沿切线走到横轴。

先看不动点意味着什么。若 $T(x_0)=x_0$ ，则

x_0=x_0-\frac{f(x_0)}{f'(x_0)} \implies \frac{f(x_0)}{f'(x_0)}=0 \implies f(x_0)=0 .

所以 $T$ 的不动点恰是 $f$ 的零点。

再看 $T$ 何时压缩。求导（用商法则）：

T'(x)=1-\frac{(f')^2-ff''}{(f')^2} =\frac{ff''}{(f')^2} .

由中值定理，对 $x\ne y$ ，存在 $\xi$ 介于两者之间，使

|T(x)-T(y)|=|T'(\xi)|\,|x-y| .

因此只要在考虑的区间上

\left|\frac{f(x)f''(x)}{f'(x)^2}\right|\le c<1,

$T$ 就是压缩映射，牛顿迭代收敛到唯一的零点。

「好的初值」由此有了精确含义：起点附近 $|ff''|/(f')^2$ 要小。但课程也提醒，牛顿法本身只保证「在一个压缩区域内」收敛，还必须另外论证迭代不会跑出这个区域——这一步定理并不自动提供。

Bolzano–Weierstrass 定理

定理。 有界数列必有收敛子列。

下一讲会给出证明。本讲已经两次用到它：证明柯西收敛定理、证明最值定理。它与完备性等价，是实数系「没有空隙」的另一种说法。

连续性与极限交换

柯西收敛定理的另一个应用是最值定理，先要定义连续。

定义。 $f$ 在 $x_0$ 处连续，如果

\forall\varepsilon>0\ \exists\delta>0:\quad |x-x_0|<\delta\Rightarrow|f(x)-f(x_0)|<\varepsilon .

$f$ 连续指它在定义域内每一点连续。

定理（序列刻画）。 若 $f$ 在 $x$ 处连续且 $x_n\to x$ ，则 $f(x_n)\to f(x)$ 。

证明。 给定 $\varepsilon>0$ ，由连续性取对应的 $\delta>0$ 。由 $x_n\to x$ ，存在 $N$ ，使 $n\ge N$ 时 $|x_n-x|<\delta$ 。于是对这些 $n$ ， $|f(x_n)-f(x)|<\varepsilon$ 。 $\square$

这条定理把「连续」翻译成「可以与数列极限交换」： $\lim f(x_n)=f(\lim x_n)$ 。它是后面所有最值、介值论证的桥梁。

最值定理

定理。 设 $f$ 在闭区间 $[a,b]$ 上连续，则 $f$ 在 $[a,b]$ 上取得最大值与最小值。即存在 $x_M,x_m\in[a,b]$ ，使

f(x_M)=\sup_{[a,b]}f, \qquad f(x_m)=\inf_{[a,b]}f .

证明（最大值部分）。 令 $S=\sup_{[a,b]}f$ （暂允许 $S=+\infty$ ）。由上确界的定义，存在数列 $x_n\in[a,b]$ 使

f(x_n)\to S .

（取 $\varepsilon=1/n$ ：因 $S-1/n$ 不是上界，存在 $x_n$ 使 $f(x_n)>S-1/n$ 。）

因为 $a\le x_n\le b$ ， $(x_n)$ 有界。由 Bolzano–Weierstrass 定理，存在子列 $x_{n_k}\to X$ 。由于 $[a,b]$ 是闭区间， $X\in[a,b]$ （极限保持非严格不等式： $a\le x_{n_k}\le b$ 推出 $a\le X\le b$ ）。

由序列刻画， $f(x_{n_k})\to f(X)$ 。另一方面 $(f(x_{n_k}))$ 是收敛到 $S$ 的数列的子列，故也收敛到 $S$ 。极限唯一，所以

S=f(X).

于是 $S$ 是有限实数，且被 $f$ 在某点取到，即上确界就是最大值。最小值对 $-f$ 用同样论证。 $\square$

闭区间这个条件不能省。 若区间开，例如 $f(x)=x$ 在 $(0,1)$ 上，上确界 $1$ 取不到。证明中「 $X\in[a,b]$ 」正是靠端点被包含。

本节结论

柯西数列必有界；证明手法与「收敛必有界」相同，参考点换成 $x_N$ 。
在 $\mathbb R$ 中柯西 ⇔ 收敛；证明借助 Bolzano–Weierstrass 定理。
压缩映射：同一个 $c<1$ 控制所有点对；至多一个不动点，因为 $(1-c)|x-y|\le0$ 。
迭代序列满足 $|a_m-a_n|\le\dfrac{c^n}{1-c}|a_1-a_0|$ ，故为柯西列；由此得到不动点存在。
牛顿法把 $f(x)=0$ 化为 $T(x)=x-f/f'$ 的不动点问题， $T'=ff''/(f')^2<1$ 时收敛。
连续 ⇔ 可与数列极限交换；由此推出闭区间上连续函数取得最值。

理解检验

1. 证明柯西数列有界时，为什么要在 $\varepsilon=1$ 的定义里固定 $m=N$ ？

2. 柯西收敛定理的证明为什么需要 Bolzano–Weierstrass 定理？只用柯西性本身能推出收敛吗？

参考答案：柯西性只说「远处各项彼此靠近」，它并不能凭空产生一个极限点。Bolzano–Weierstrass 定理利用有界性给出一个收敛子列，从而提供一个候选极限 $x$ ；再用柯西性把整条数列与这个子列绑在一起，证明整条数列也收敛到 $x$ 。只用柯西性在 $\mathbb Q$ 中就会失败（ $\sqrt2$ 的有理截断是柯西列但不收敛），所以必须借助完备性；而 Bolzano–Weierstrass 正是完备性的等价形式。

3. 压缩映射为什么要求对所有 $x,y$ 用同一个 $c$ ，并且 $c<1$ ？如果 $c=1$ 会怎样？

参考答案：同一个 $c$ 保证误差每一步按固定比例缩小，从而把相邻距离压成几何级数。若 $c=1$ ，相邻距离估计变为 $|a_{n+1}-a_n|\le|a_n-a_{n-1}|$ ，只能说明距离不增，无法趋于零；例如 $T(x)=x+1$ 满足 $|T(x)-T(y)|=|x-y|$ ，却没有任何不动点。唯一性证明也依赖 $c<1$ ： $(1-c)|x-y|\le0$ 只有在 $1-c>0$ 时才能推出 $|x-y|=0$ 。

4. 迭代序列的柯西估计中，几何级数是怎么出现的？哪一步用到 $c\ge0$ ？

参考答案：由 $|a_{k+1}-a_k|\le c^k|a_1-a_0|$ ，再用三角不等式把 $|a_m-a_n|$ 拆成从 $k=n$ 到 $m-1$ 的相邻差之和，得到 $\sum_{k=n}^{m-1}c^k|a_1-a_0|$ ，这就是几何级数；它被 $\frac{c^n}{1-c}|a_1-a_0|$ 控制。 $c\ge0$ 保证 $c^n$ 单调递减趋于 $0$ ，并且幂次估计中不会出现符号翻转；若 $c<0$ ，压缩性不等式本身仍可能成立，但通常把 $c$ 换成 $|c|$ 处理。

5. 最值定理为什么必须要求闭区间？把证明中哪一步去掉就会失效？

参考答案：证明中先取 $x_n\in[a,b]$ 使 $f(x_n)\to\sup f$ ，再用 Bolzano–Weierstrass 得到子列 $x_{n_k}\to X$ ，然后断言 $X\in[a,b]$ ，最后用连续性得 $\sup f=f(X)$ 。若区间是开的，例如 $(a,b)$ ， $X$ 可能等于端点 $a$ 或 $b$ ，而函数在端点没有定义， $f(X)$ 无从谈起，结论也可能不成立（ $f(x)=x$ 在 $(0,1)$ 上取不到上确界 $1$ ）。

6. 用序列刻画证明：若 $f$ 在 $x$ 处连续， $x_n\to x$ 且所有 $x_n$ 在定义域内，则 $f(x_n)\to f(x)$ 。指出 $\delta$ 与 $N$ 的依赖关系。

与前后课的关系

本讲把完备性包装成两个可直接使用的工具：柯西收敛定理与压缩映射定理。下一讲证明 Bolzano–Weierstrass 定理，并把它与「闭区间套」「列紧性」联系起来。第 8 讲起处理级数，柯西判据会变成级数收敛的判据；第 21–24 讲的一致收敛与 Picard–Lindelöf 定理，则把本讲的压缩映射迭代直接用于微分方程。

一条线索贯穿：完备性 → 柯西判据 → 不动点与最值 → 后续所有存在性定理。

Why a Cauchy criterion

Until now, proving convergence meant guessing the limit and verifying it with the $\varepsilon$ - $N$ definition. But many limits have no closed form — solutions of differential equations and fixed points of contractions, for example.

The Cauchy criterion rewrites "a limit exists" as a condition about the sequence alone: distant terms crowd together. Its value is that one can prove existence without knowing the limit.

Timeline

Time	Topic
0:00	Review: monotone convergence and Cauchy sequences
3:53	Theorem: Cauchy sequences are bounded
9:27	The Cauchy convergence theorem; the contraction mapping theorem
14:41	A contraction has at most one fixed point
21:36	The Cauchy estimate for the iteration $a_{n+1}=T(a_n)$
28:09	Proving the contraction mapping theorem via Cauchy convergence
38:06	Newton's method as an application
50:17	The Bolzano–Weierstrass theorem (proved next lecture)
59:08	Continuity and the extreme value theorem

Cauchy sequences

Definition. A sequence $(x_n)$ is Cauchy if

\forall\varepsilon>0\ \exists N\ \forall n,m\ge N:\quad |x_n-x_m|<\varepsilon .

Unlike convergence, no candidate limit $a$ appears. Intuitively: convergence says distant terms cluster around $a$ ; Cauchy says distant terms cluster around each other.

Theorem. Every Cauchy sequence is bounded.

Proof. Apply the Cauchy condition with $\varepsilon=1$ , obtaining $N$ such that $|x_n-x_m|<1$ whenever $n,m\ge N$ .

Fix $m=N$ . Then for every $n\ge N$ ,

|x_n|=|x_n-x_N+x_N|\le|x_n-x_N|+|x_N|<1+|x_N| .

Only $x_1,\dots,x_{N-1}$ remain, finitely many. Let

M=\max\{|x_1|,\dots,|x_{N-1}|,\,1+|x_N|\},

so $|x_n|\le M$ for all $n$ . $\square$

The device is identical to "convergent implies bounded" from Lecture 4: fix a reference point $m=N$ , control the tail with the triangle inequality, bound the head by a maximum. Only the reference changes from the limit $a$ to the $N$ -th term itself.

The Cauchy convergence theorem

Theorem. In $\mathbb R$ , a sequence converges if and only if it is Cauchy.

"Convergent ⇒ Cauchy" is easy: if $x_n\to x$ , given $\varepsilon>0$ choose $N$ with $|x_n-x|<\varepsilon/2$ for $n\ge N$ . Then for $n,m\ge N$ ,

|x_n-x_m|\le|x_n-x|+|x-x_m|<\frac\varepsilon2+\frac\varepsilon2=\varepsilon .

The converse needs completeness. The lecture derives it from the Bolzano–Weierstrass theorem (proved next lecture):

Every bounded sequence has a convergent subsequence.

Proof (Cauchy ⇒ convergent). Let $(x_n)$ be Cauchy.

By the previous theorem it is bounded. By Bolzano–Weierstrass there is a subsequence $x_{n_k}\to x$ .

We show the whole sequence converges to $x$ . Given $\varepsilon>0$ :

Cauchy gives $N_1$ with $|x_n-x_m|<\varepsilon/2$ for $n,m\ge N_1$ .
Subsequence convergence gives $N_2$ with $|x_{n_k}-x|<\varepsilon/2$ for $k\ge N_2$ .

Choose $k\ge N_2$ with $n_k\ge N_1$ (possible since $n_k$ strictly increases to infinity). Let $N=\max\{N_1,n_k\}$ . For $n\ge N$ ,

|x_n-x|\le|x_n-x_{n_k}|+|x_{n_k}-x|<\frac\varepsilon2+\frac\varepsilon2=\varepsilon .

Hence $x_n\to x$ . $\square$

The logic: the subsequence supplies an anchor with a known limit, and the Cauchy property prevents the rest of the sequence from drifting away from it. The subsequence's convergence rests on Bolzano–Weierstrass, whose proof uses completeness — so the theorem is another face of completeness.

Contractions

Definition. A map $T:\mathbb R\to\mathbb R$ is a contraction if there is a constant $c$ with

$c\ge0$ ;
$c<1$ ;
$|T(x)-T(y)|\le c|x-y|$ for all $x,y$ .

The third point insists on a single $c$ : one cannot use one constant for one pair and another constant for another.

Definition. $x$ is a fixed point of $T$ if $T(x)=x$ .

Lemma. A contraction has at most one fixed point.

Proof. Let $x,y$ both be fixed points. Then

|x-y|=|T(x)-T(y)|\le c|x-y| .

Rearranging, $(1-c)|x-y|\le0$ . Since $c<1$ , $1-c>0$ , so $|x-y|\le0$ , i.e. $x=y$ . $\square$

One line, and the key is $c<1$ : with $c=1$ the inequality $(1-c)|x-y|\le0$ gives nothing, and the claim fails (the translation $T(x)=x+1$ has no fixed point).

Why the iteration is Cauchy

"At most one" still leaves existence open. The construction is iteration: pick any $x$ and set

a_0=x, \qquad a_{n+1}=T(a_n)=T^{n+1}(x).

Step one: consecutive distances shrink geometrically. By contractivity,

|a_{n+1}-a_n|=|T(a_n)-T(a_{n-1})|\le c|a_n-a_{n-1}| .

Iterating,

|a_{n+1}-a_n|\le c^n|a_1-a_0| .

Step two: any two terms are controlled. For $m>n$ , chain the intermediate differences:

|a_m-a_n|\le\sum_{k=n}^{m-1}|a_{k+1}-a_k| \le\sum_{k=n}^{m-1}c^k|a_1-a_0| .

Summing the geometric series,

\sum_{k=n}^{m-1}c^k\le\sum_{k=n}^{\infty}c^k=\frac{c^n}{1-c},

hence

|a_m-a_n|\le\frac{c^n}{1-c}\,|a_1-a_0| .

Since $0\le c<1$ , $c^n\to0$ . Given $\varepsilon>0$ , choose $n$ large enough that the right side is below $\varepsilon$ ; then all $m>n$ satisfy $|a_m-a_n|<\varepsilon$ . So $(a_n)$ is Cauchy.

Both hypotheses are used: $c<1$ makes the geometric series converge; $c\ge0$ makes $c^n$ decrease to $0$ . With $c=1$ the estimate becomes $(m-n)|a_1-a_0|$ , unbounded in $m$ , and the iterates may wander off.

The contraction mapping theorem

Theorem. A contraction $T:\mathbb R\to\mathbb R$ has a unique fixed point.

Proof. Uniqueness is the lemma. For existence:

Pick any $a_0\in\mathbb R$ and set $a_{n+1}=T(a_n)$ . By the previous section $(a_n)$ is Cauchy. By the Cauchy convergence theorem, $a_n\to a$ for some $a$ .

Show $T(a)=a$ . Contractivity gives

|T(a_n)-T(a)|\le c|a_n-a|\to0,

so $T(a_n)\to T(a)$ . But $T(a_n)=a_{n+1}\to a$ . A sequence has only one limit, so

T(a)=a .

Thus $a$ is a fixed point. $\square$

The theorem is constructive: it asserts existence and simultaneously gives a method — iterate from any starting point. Even without a closed form for the fixed point, one can approximate it arbitrarily well. That is exactly what makes it valuable for differential equations (ODE and PDE): rewrite the equation as $x=T(x)$ , verify contractivity, and obtain existence and uniqueness. Lecture 24 carries this out.

Newton's method: an application

To solve $f(x)=0$ with $f$ differentiable and $f'(x)\ne0$ , define

T(x)=x-\frac{f(x)}{f'(x)} .

This is Newton's iteration: follow the tangent line down to the axis.

What does a fixed point mean? If $T(x_0)=x_0$ , then

x_0=x_0-\frac{f(x_0)}{f'(x_0)} \implies \frac{f(x_0)}{f'(x_0)}=0 \implies f(x_0)=0 .

So fixed points of $T$ are exactly zeros of $f$ .

When is $T$ a contraction? Differentiate with the quotient rule:

T'(x)=1-\frac{(f')^2-ff''}{(f')^2} =\frac{ff''}{(f')^2} .

By the mean value theorem, for $x\ne y$ there is $\xi$ between them with

|T(x)-T(y)|=|T'(\xi)|\,|x-y| .

Hence if on the relevant interval

\left|\frac{f(x)f''(x)}{f'(x)^2}\right|\le c<1,

then $T$ is a contraction and Newton's iteration converges to the unique zero.

A "good initial guess" now has a precise meaning: near the starting point, $|ff''|/(f')^2$ must be small. But the lecture warns that Newton's method only guarantees convergence inside a contracting region; one must also argue that the iterates never leave it, which the theorem alone does not supply.

The Bolzano–Weierstrass theorem

Theorem. Every bounded sequence has a convergent subsequence.

The proof comes in the next lecture. It has already been used twice here: to derive the Cauchy convergence theorem and to prove the extreme value theorem. It is equivalent to completeness — another way of saying that the real line has no gaps.

Continuity and exchanging limits

The other application is the extreme value theorem, which needs continuity first.

Definition. $f$ is continuous at $x_0$ if

\forall\varepsilon>0\ \exists\delta>0:\quad |x-x_0|<\delta\Rightarrow|f(x)-f(x_0)|<\varepsilon .

$f$ is continuous if it is continuous at every point of its domain.

Theorem (sequential characterisation). If $f$ is continuous at $x$ and $x_n\to x$ , then $f(x_n)\to f(x)$ .

Proof. Given $\varepsilon>0$ , continuity supplies $\delta>0$ . Since $x_n\to x$ , there is $N$ with $|x_n-x|<\delta$ for $n\ge N$ . For those $n$ , $|f(x_n)-f(x)|<\varepsilon$ . $\square$

This translates continuity into "limits may pass through $f$ ": $\lim f(x_n)=f(\lim x_n)$ . It is the bridge behind every extreme-value and intermediate-value argument.

The extreme value theorem

Theorem. If $f$ is continuous on the closed interval $[a,b]$ , then $f$ attains its maximum and minimum there: there are $x_M,x_m\in[a,b]$ with

f(x_M)=\sup_{[a,b]}f, \qquad f(x_m)=\inf_{[a,b]}f .

Proof (maximum). Let $S=\sup_{[a,b]}f$ , temporarily allowing $S=+\infty$ . By the definition of supremum there is a sequence $x_n\in[a,b]$ with

f(x_n)\to S .

(Take $\varepsilon=1/n$ : since $S-1/n$ is not an upper bound, some $x_n$ has $f(x_n)>S-1/n$ .)

Because $a\le x_n\le b$ , the sequence $(x_n)$ is bounded. Bolzano–Weierstrass gives a subsequence $x_{n_k}\to X$ . Since $[a,b]$ is closed, $X\in[a,b]$ (limits preserve non-strict inequalities: $a\le x_{n_k}\le b$ gives $a\le X\le b$ ).

By the sequential characterisation, $f(x_{n_k})\to f(X)$ . On the other hand $(f(x_{n_k}))$ is a subsequence of a sequence converging to $S$ , hence also converges to $S$ . Limits are unique, so

S=f(X).

Thus $S$ is a finite real number attained by $f$ : the supremum is a maximum. The minimum follows by applying the result to $-f$ . $\square$

The closed interval cannot be dropped. On $(0,1)$ , $f(x)=x$ has supremum $1$ , which is never attained. In the proof, " $X\in[a,b]$ " is exactly where endpoints being included matters.

What this lecture establishes

Cauchy sequences are bounded; the proof mirrors "convergent implies bounded" with $x_N$ as the reference point.
In $\mathbb R$ , Cauchy ⇔ convergent, via Bolzano–Weierstrass.
A contraction uses one $c<1$ for all pairs; it has at most one fixed point since $(1-c)|x-y|\le0$ .
The iteration satisfies $|a_m-a_n|\le\dfrac{c^n}{1-c}|a_1-a_0|$ , hence is Cauchy, yielding existence of a fixed point.
Newton's method turns $f(x)=0$ into the fixed-point equation $T(x)=x-f/f'$ , contracting when $T'=ff''/(f')^2<1$ .
Continuity ⇔ limits may pass through $f$ ; hence a continuous function on a closed interval attains its extrema.

Comprehension check

1. In the boundedness proof for Cauchy sequences, why fix $m=N$ after taking $\varepsilon=1$ ?

Reference answer: The Cauchy condition gives $|x_n-x_m|<1$ for all $n,m\ge N$ . To obtain an estimate for a single $x_n$ , a fixed reference point must be introduced; taking $m=N$ gives $|x_n-x_N|<1$ for every $n\ge N$ , and the triangle inequality then yields $|x_n|<1+|x_N|$ . Without fixing $m$ , the estimate would involve two quantities both varying with $n$ and no uniform bound would follow.

2. Why does the proof of the Cauchy convergence theorem need Bolzano–Weierstrass? Can Cauchy-ness alone produce convergence?

Reference answer: Cauchy-ness only says distant terms are close to each other; it cannot manufacture a limit point. Bolzano–Weierstrass uses boundedness to produce a convergent subsequence, supplying a candidate limit $x$ ; the Cauchy property then ties the whole sequence to that subsequence. Cauchy-ness alone fails in $\mathbb Q$ — the rational truncations of $\sqrt2$ form a Cauchy sequence with no limit — so completeness must enter, and Bolzano–Weierstrass is one of its equivalent forms.

3. Why must a contraction use one $c$ for all $x,y$ , with $c<1$ ? What happens when $c=1$ ?

Reference answer: A single $c$ makes each step shrink the error by a fixed factor, turning successive differences into a geometric series. With $c=1$ the estimate becomes $|a_{n+1}-a_n|\le|a_n-a_{n-1}|$ , which only says distances do not grow and cannot force them to zero; for example $T(x)=x+1$ satisfies $|T(x)-T(y)|=|x-y|$ yet has no fixed point. Uniqueness also needs $c<1$ : $(1-c)|x-y|\le0$ yields $|x-y|=0$ only when $1-c>0$ .

4. Where does the geometric series come from in the Cauchy estimate, and where is $c\ge0$ used?

Reference answer: From $|a_{k+1}-a_k|\le c^k|a_1-a_0|$ , the triangle inequality expands $|a_m-a_n|$ into a sum of consecutive differences over $k=n,\dots,m-1$ , giving $\sum c^k|a_1-a_0|$ — a geometric series bounded by $\frac{c^n}{1-c}|a_1-a_0|$ . The condition $c\ge0$ makes $c^n$ decrease monotonically to $0$ and keeps the power estimates free of sign changes; for negative $c$ one usually replaces $c$ by $|c|$ .

5. Why must the extreme value theorem require a closed interval? Which step of the proof fails otherwise?

Reference answer: The proof chooses $x_n\in[a,b]$ with $f(x_n)\to\sup f$ , extracts $x_{n_k}\to X$ by Bolzano–Weierstrass, asserts $X\in[a,b]$ , then uses continuity to get $\sup f=f(X)$ . On an open interval $(a,b)$ , $X$ might equal an endpoint where $f$ is undefined, so $f(X)$ makes no sense and the conclusion can fail: $f(x)=x$ on $(0,1)$ never attains its supremum $1$ .

6. Using the sequential characterisation, prove that continuity at $x$ and $x_n\to x$ imply $f(x_n)\to f(x)$ . State how $\delta$ and $N$ depend on each other.

Reference answer: Given $\varepsilon>0$ , continuity gives $\delta>0$ depending only on $\varepsilon$ (and on $x$ ), such that $|y-x|<\delta$ implies $|f(y)-f(x)|<\varepsilon$ . Since $x_n\to x$ , for this fixed $\delta$ there is $N$ with $|x_n-x|<\delta$ for $n\ge N$ . Then $|f(x_n)-f(x)|<\varepsilon$ for $n\ge N$ . The dependency chain is $\varepsilon\mapsto\delta\mapsto N$ : $N$ depends on $\varepsilon$ through $\delta$ and never on $n$ .

Relation to neighbouring lectures

This lecture packages completeness into two immediately usable tools: the Cauchy convergence theorem and the contraction mapping theorem. The next lecture proves Bolzano–Weierstrass and connects it with nested intervals and sequential compactness. From Lecture 8 onward series are treated, with the Cauchy criterion becoming a convergence test; Lectures 21–24 on uniform convergence and Picard–Lindelöf apply the contraction iteration directly to differential equations.

One thread runs through: completeness → Cauchy criterion → fixed points and extrema → every later existence theorem.

来源

视频Lecture 6: Cauchy Convergence TheoremTobias Holck Colding, MIT OpenCourseWare · 2025-09-02