上节课留下的两个收敛

设区间 $I$ 上有一列函数 $f_n$ 与一个函数 $f$。

• 逐点收敛：固定每个 $x$，数列 $f_n(x)$ 收敛到 $f(x)$。
• 一致收敛：对每个 $\varepsilon>0$，存在 $N$，使得 $n\ge N$ 时

$$|f_n(x)-f(x)|<\varepsilon \quad\text{对一切 }x\in I\text{ 成立}.$$

区别只在最后那句「对一切 $x$」：逐点收敛允许 $N$ 依赖 $x$，一致收敛不允许。

课程反复使用的反例是 $[0,1]$ 上的 $f_n(x)=x^n$。它逐点收敛到

$$f(x)= \begin{cases} 0,&0\le x<1,\\ 1,&x=1. \end{cases}$$

每个 $f_n$ 都连续，极限 $f$ 却在 $1$ 处跳跃。为什么不一致？由介值定理，对每个 $n$ 都存在 $x_n$ 使 $x_n^n=\tfrac12$，于是

$$|f_n(x_n)-f(x_n)|=\tfrac12 .$$

这个误差不随 $n$ 变小，所以 $\sup_x|f_n(x)-f(x)|\ge\tfrac12$，不可能趋于 $0$。

逐点收敛不足以保住连续性；一致收敛可以。

时间轴

定理：一致极限保持连续

定理。 设 $f_n$ 在区间 $I$ 上连续，且 $f_n\to f$ 一致。则 $f$ 在 $I$ 上连续。

完整证明

任取 $x_0\in I$，证明 $f$ 在 $x_0$ 连续。给定 $\varepsilon>0$。

第一步：用一致收敛固定一个下标。 因为 $f_n\to f$ 一致，存在 $N$，使 $n\ge N$ 时对所有 $x$ 都有

$$|f_n(x)-f(x)|<\frac{\varepsilon}{3}.$$

只需取 $n=N$：

$$|f_N(x)-f(x)|<\frac{\varepsilon}{3} \quad\text{对一切 }x\in I\text{ 成立}.$$

这一步允许我们把 $f$ 换成固定的 $f_N$，而且换的代价在所有点上一致地小。

第二步：用 $f_N$ 的连续性找 $\delta$。 因为 $f_N$ 在 $x_0$ 连续，存在 $\delta>0$，使

$$|x-x_0|<\delta \Rightarrow |f_N(x)-f_N(x_0)|<\frac{\varepsilon}{3}.$$

第三步：三角不等式。 对满足 $|x-x_0|<\delta$ 的 $x$，插入 $f_N(x)$ 与 $f_N(x_0)$：

$$\begin{aligned} |f(x)-f(x_0)| &\le |f(x)-f_N(x)| \\ &\quad + |f_N(x)-f_N(x_0)| \\ &\quad + |f_N(x_0)-f(x_0)| \\ &< \frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3} =\varepsilon . \end{aligned}$$

所以 $f$ 在 $x_0$ 连续。由于 $x_0$ 任意，$f$ 在 $I$ 上连续。$\square$

三段的来源要分清：第一段和第三段来自一致收敛（同一个点、同一个下标 $N$，误差对所有 $x$ 统一），中间一段来自 $f_N$ 的连续性（同一个函数、两个邻近的点）。逐点收敛做不到第一段和第三段统一，因为那里的 $N$ 会随 $x$ 变大。

立刻可用的推论

设

$$E_n(x)=\sum_{k=0}^{n}\frac{x^k}{k!}, \qquad E(x)=\sum_{k=0}^{\infty}\frac{x^k}{k!}.$$

$E_n$ 是多项式，连续。由 Weierstrass $M$ 判别法（上一讲证明），在任意固定区间 $[-L,L]$ 上，级数被 $\sum L^k/k!<\infty$ 控制，因此 $E_n\to E$ 在 $[-L,L]$ 上一致。由上面的定理，$E$ 在每个 $[-L,L]$ 上连续，从而在 $\mathbb R$ 上连续。

注意「一致」的范围：不是在整个 $\mathbb R$ 上一致，而是在每个有界区间上一致。这类收敛通常叫「内闭一致收敛」。

一致收敛就是 sup 度量下的收敛

$C([a,b])$ 表示闭区间上全体连续函数。它带有天然的距离

$$d(f,g)=\max_{x\in[a,b]}|f(x)-g(x)| .$$

这个 $\max$ 确实取得到：$f-g$ 连续，取绝对值后仍连续，由最值定理在有界闭区间上达到最大值。

命题。 $f_n\to f$ 一致，当且仅当 $d(f_n,f)\to 0$。

证明。 两个方向都是重写定义。

若 $f_n\to f$ 一致：给定 $\varepsilon>0$，存在 $N$，使 $n\ge N$ 时 $|f_n(x)-f(x)|<\varepsilon$ 对一切 $x$ 成立。对固定 $n$ 取上确界（此时一致收敛已保证误差对所有 $x$ 一致地小），得到 $d(f_n,f)\le\varepsilon$。故 $d(f_n,f)\to0$。

反之，若 $d(f_n,f)\to0$：给定 $\varepsilon>0$，取 $N$ 使 $n\ge N$ 时 $d(f_n,f)<\varepsilon$。由于每个点上的值不超过最大值，

$$|f_n(x)-f(x)|\le d(f_n,f)<\varepsilon \quad\text{对一切 }x$$

同时成立。这正是 $f_n\to f$ 一致。$\square$

这个改写把「函数列的一致收敛」变成了「度量空间中点列的收敛」，于是度量空间的全部工具（Cauchy 列、完备性、压缩映射）都可以直接使用。这不是记号上的方便，而是后面解常微分方程的关键入口。

$C([a,b])$ 是完备的

定理。 度量空间 $\bigl(C([a,b]),d\bigr)$ 是 Cauchy 完备的：每个 Cauchy 列都收敛到该空间中的某个函数。

设 $(f_n)$ 是 Cauchy 列。要证明两件事：先构造候选极限 $f$，再证明它在空间里并且收敛。

第一步：逐点构造极限

固定 $x\in[a,b]$。由

$$|f_n(x)-f_m(x)| \le \max_{z\in[a,b]}|f_n(z)-f_m(z)| =d(f_n,f_m),$$

可知实数列 $(f_n(x))$ 是 Cauchy 列。实数系完备，所以它收敛。定义

$$f(x)=\lim_{n\to\infty}f_n(x).$$

每个 $x$ 给出唯一一个实数，因此 $f$ 是一个函数。此时只知道逐点收敛，还不知道 $f$ 连续。

第二步：把逐点提升为一致

给定 $\varepsilon>0$。因为 $(f_n)$ 在度量意义下是 Cauchy 列，存在 $N$，使 $n,m\ge N$ 时

$$d(f_n,f_m)\le\frac{\varepsilon}{2}.$$

于是对每个 $x$，

$$|f_n(x)-f_m(x)|\le\frac{\varepsilon}{2}.$$

固定 $n\ge N$，让 $m\to\infty$。左边趋于 $|f_n(x)-f(x)|$，右边是常数，于是

$$|f_n(x)-f(x)|\le\frac{\varepsilon}{2} \quad\text{对一切 }x\text{ 成立}.$$

这一步只能得到「不超过 $\varepsilon/2$」，因为不等式在取极限时会保留 $\le$。现在对所有 $x$ 取上确界：

$$\sup_{x}|f_n(x)-f(x)|\le\frac{\varepsilon}{2}<\varepsilon .$$

这里必须用 $\sup$ 而不是 $\max$：此刻还不知道 $f$ 连续，最值定理还不能用，最大值可能取不到。等会儿证明 $f$ 连续之后，$\sup$ 就等于 $\max$。

结论是 $f_n\to f$ 一致，即 $d(f_n,f)\to0$。

第三步：极限仍在空间里

$f_n$ 都连续，且 $f_n\to f$ 一致，由本节第一个定理，$f$ 连续。所以 $f\in C([a,b])$。结合第二步，$f_n\to f$ 在度量 $d$ 下成立。$\square$

整个证明的分工值得记住：完备性只在第一步用到（实数的完备性），一致收敛性在第二步建立，闭包性（连续性）在第三步补上。三步各管一件事，缺一不可。

定理：积分与一致极限可交换

设 $f_n$ 在 $[a,b]$ 上可积，且 $f_n\to f$ 一致。定理断言：

1. $f$ 可积；
2. $\displaystyle\lim_{n\to\infty}\int_a^b f_n=\int_a^b f$，特别地，右边的极限存在。

注意这里只假设 $f_n$ 可积，没有假设连续。极限 $f$ 也可能不连续（如 $x^n$ 的例子），因此不能靠「连续函数可积」蒙混过去。

证明思路

Riemann 可积的判据是：对每个 $\varepsilon>0$，存在分割 $P$，使得

$$U(f,P)-L(f,P)<\varepsilon ,$$

其中 $U,L$ 分别是上、下和。

思路是：先用一致收敛把 $f$ 的上下和夹在 $f_N$ 的上下和之间，误差是 $\pm\varepsilon/3$；再对 $f_N$ 选一个使上下和接近的分割；两者合成，得到 $f$ 的上下和之差小于 $\varepsilon$。

上、下和的逐项估计

设分割 $P$ 的分点为 $x_0<\cdots<x_m$，小区间 $I_i=[x_{i-1},x_i]$，长度 $\Delta x_i$。记

$$M_i(f)=\sup_{I_i}f, \qquad m_i(f)=\inf_{I_i}f .$$

因为 $f_n\to f$ 一致，取 $n$ 足够大，使

$$|f_n(x)-f(x)|<\frac{\varepsilon}{3(b-a)} \quad\text{对一切 }x .$$

固定这样一个 $n$。对任意 $x$，

$$f(x)\le f_n(x)+|f(x)-f_n(x)|<f_n(x)+\frac{\varepsilon}{3(b-a)}.$$

在小区间 $I_i$ 上取上确界：

$$M_i(f)\le M_i(f_n)+\frac{\varepsilon}{3(b-a)}.$$

同理，从 $f(x)\ge f_n(x)-|f(x)-f_n(x)|$ 取下确界得

$$m_i(f)\ge m_i(f_n)-\frac{\varepsilon}{3(b-a)}.$$

把两条不等式分别乘 $\Delta x_i$ 并对 $i$ 求和。注意 $\sum_i\Delta x_i=b-a$，常数项正好消掉分母：

$$L(f_n,P)-\frac{\varepsilon}{3} \le L(f,P) \le U(f,P) \le U(f_n,P)+\frac{\varepsilon}{3}.$$

这一步对任意分割 $P$ 都成立，只用到一致收敛。

选分割，得到可积性

上面那个 $n$ 固定后，$f_n$ 是可积的。于是存在分割 $P$，使

$$U(f_n,P)-L(f_n,P)<\frac{\varepsilon}{3}.$$

对这个 $P$，

$$\begin{aligned} U(f,P)-L(f,P) &\le \Bigl(U(f_n,P)+\frac{\varepsilon}{3}\Bigr) -\Bigl(L(f_n,P)-\frac{\varepsilon}{3}\Bigr)\\ &=\bigl(U(f_n,P)-L(f_n,P)\bigr)+\frac{2\varepsilon}{3}\\ &<\frac{\varepsilon}{3}+\frac{2\varepsilon}{3}=\varepsilon . \end{aligned}$$

由可积判据，$f$ 可积。这就完成了第一件事。

积分的收敛

对任何分割 $P$，积分被上下和夹住：

$$L(f,P)\le\int_a^b f\le U(f,P).$$

取上面那个使 $U(f_N,P)-L(f_N,P)<\varepsilon/3$ 的 $P$（把 $n$ 取成 $N$）。由上、下和的夹逼，

$$L(f_N,P)-\frac{\varepsilon}{3}\le\int_a^b f\le U(f_N,P)+\frac{\varepsilon}{3},$$

同时

$$L(f_N,P)\le\int_a^b f_N\le U(f_N,P).$$

所以 $\int_a^b f$ 与 $\int_a^b f_N$ 都落在长度为 $\varepsilon$ 的区间

$$\Bigl[L(f_N,P)-\frac{\varepsilon}{3},\;U(f_N,P)+\frac{\varepsilon}{3}\Bigr]$$

内（其长度为 $\bigl(U(f_N,P)-L(f_N,P)\bigr)+\tfrac{2\varepsilon}{3}<\varepsilon$）。于是

$$\left|\int_a^b f-\int_a^b f_N\right|<\varepsilon .$$

对任意 $\varepsilon>0$ 找到这样的 $N$，就证明了 $\int f_n\to\int f$。$\square$

课程末尾回应了一个提问：证明中有些地方写 $\le$、有些写 $<$，这无关紧要，因为 $\varepsilon$ 是任取的，把常数取成 $\varepsilon/4$ 之类的更小量即可得到严格不等号。

用幂级数算积分

回到 $E_n(x)=\sum_{k=0}^n x^k/k!$，它在 $[a,b]$ 上一致收敛到 $E$。$E_n$ 是多项式，可以逐项积分：

$$\int_a^b E_n(x)\,dx =\sum_{k=0}^{n}\int_a^b\frac{x^k}{k!}\,dx .$$

由积分交换定理，左边趋于 $\int_a^b E$，于是

$$\int_a^b E(x)\,dx =\sum_{k=0}^{\infty}\int_a^b\frac{x^k}{k!}\,dx .$$

每一项可以直接算。找 $x^{k+1}/(k+1)!$ 求导：

$$\frac{d}{dx}\frac{x^{k+1}}{(k+1)!} =\frac{(k+1)x^{k}}{(k+1)!} =\frac{x^k}{k!}.$$

所以

$$\int_a^b \frac{x^k}{k!}\,dx =\frac{b^{k+1}-a^{k+1}}{(k+1)!}.$$

结论：一致收敛把「对无穷级数积分」合法地变成「对每一项积分再求和」。有限和当然可以随便交换，无限和不行——这里的一致性正是许可证。

本节结论

• 连续函数的一致极限连续；逐点极限不一定连续（$x^n$ 是反例）。
• 连续性的证明固定一个 $f_N$，用 $\varepsilon/3$ 三段估计。
• 一致收敛等价于 sup 度量下的收敛；$\sup$ 与 $\max$ 的区别在于极限函数是否已知连续。
• $C([a,b])$ 在 sup 度量下 Cauchy 完备。
• 可积函数的一致极限可积，且积分与极限可交换。
• 幂级数在其收敛区间内可以逐项积分；逐项求导留到下一讲。

理解检验

1. 用 $f_n(x)=x^n$ 说明：为什么逐点收敛不能保证极限连续？请写出使一致收敛失败的那组点。

参考答案：逐点极限是 $f(0\le x<1)=0$、$f(1)=1$，在 $1$ 处不连续，而每个 $f_n$ 连续，所以逐点收敛不保连续性。取 $x_n=2^{-1/n}\in(0,1)$，则 $f_n(x_n)=1/2$，而 $f(x_n)=0$，故 $|f_n(x_n)-f(x_n)|=1/2$。于是 $\sup_x|f_n(x)-f(x)|\ge1/2$ 对所有 $n$ 成立，不趋于 $0$，一致收敛失败。

2. 证明「$f_n\to f$ 一致」等价于「$d(f_n,f)\to0$」，并指出证明中哪一步用了「对所有 $x$ 同时成立」。

参考答案：两个方向都只是把定义换个写法。若一致收敛，给定 $\varepsilon$，存在 $N$ 使 $n\ge N$ 时 $|f_n(x)-f(x)|<\varepsilon$ 对所有 $x$ 同时成立；这个同一个 $\varepsilon$ 上界对所有 $x$ 有效，所以取上确界得 $d(f_n,f)\le\varepsilon$。反过来，若 $d(f_n,f)<\varepsilon$，因为每个点的值不超过最大值，$|f_n(x)-f(x)|\le d(f_n,f)<\varepsilon$ 对所有 $x$ 同时成立。关键就是「同一个 $\varepsilon$ 界对所有 $x$ 有效」；逐点收敛时 $\varepsilon$ 界依赖于 $x$，无法取上确界。

3. 在完备性证明中，已经得到 $f_n(x)\to f(x)$（逐点），为什么还必须再证一致收敛？

参考答案：逐点收敛只能说明每个点上的值收敛，还不足以保证 $f$ 连续，也不足以说明 $(f_n)$ 在度量 $d$ 下收敛。完备性要求的是：Cauchy 列在该度量下收敛到空间内的点，而 $d$ 度量的是全体点上误差的上确界。只有把逐点收敛提升为 $\sup_x|f_n-f|\to0$，才能得到 $d(f_n,f)\to0$；也只有再用一次「一致极限保连续」，才能确认 $f\in C([a,b])$，即极限没有跑出空间。

4. 积分定理的证明里，$\varepsilon/(3(b-a))$ 中的 $3$ 和 $(b-a)$ 各起什么作用？

参考答案：$(b-a)$ 用来抵消求和后出现的区间总长：常数误差在长度为 $b-a$ 的区间上积分，总量是 $\frac{\varepsilon}{3(b-a)}(b-a)=\frac{\varepsilon}{3}$，于是上下和各自偏移 $\varepsilon/3$。$3$ 用来分配总误差：$f$ 与 $f_n$ 的上下和之间共偏移 $2\varepsilon/3$，再留 $\varepsilon/3$ 给 $f_n$ 自身的上下和之差（可积性），合计小于 $\varepsilon$。换成 $\varepsilon/4$ 或别的常数同样可行，只要三部分之和小于 $\varepsilon$。

5. 定理只假设 $f_n$ 可积，没有假设连续。为什么不能直接说「$f$ 是连续函数的极限所以可积」？

参考答案：因为一致极限虽然保持连续，但这里的 $f_n$ 未必连续，而且即便 $f_n$ 连续，逐点极限也可能不连续（$x^n$），所以「极限连续」这条路线本身就不总可用。定理的结论是 $f$ 可积，这必须用 Riemann 判据直接证明：对给定 $\varepsilon$ 构造分割 $P$ 使 $U(f,P)-L(f,P)<\varepsilon$。证明中把 $f$ 的上下和夹在 $f_N$ 的上下和之间，正是绕开连续性、只依赖可积性与一致收敛的做法。

与前后课的关系

上一讲建立了 Weierstrass $M$ 判别法与逐点/一致收敛的区别；本讲把它们组织成一套工具：连续性的保持、度量观点、完备性、积分交换。下一讲处理剩下的那块——求导能否与极限交换。答案需要更强的假设（导数一致收敛），得到之后就能证明幂级数在收敛区间内可以逐项求导任意多次，并导出指数函数满足 $E'=E$。

再往后，$C([a,b])$ 的完备性与 sup 度量会被直接用来证明 Picard–Lindelöf 定理：把微分方程的解写成积分算子的不动点。本讲看似抽象的三步证明，正是那条主线的地基。

Two notions of convergence, recalled

On an interval $I$ we have a sequence of functions $f_n$ and a function $f$.

• Pointwise convergence: for each fixed $x$, the numerical sequence $f_n(x)$ converges to $f(x)$.
• Uniform convergence: for every $\varepsilon>0$ there is $N$ such that for $n\ge N$,

$$|f_n(x)-f(x)|<\varepsilon \quad\text{for all }x\in I .$$

The only difference is that last clause: pointwise convergence lets $N$ depend on $x$, uniform convergence does not.

The example used repeatedly in the course is $f_n(x)=x^n$ on $[0,1]$. It converges pointwise to

$$f(x)= \begin{cases} 0,&0\le x<1,\\ 1,&x=1. \end{cases}$$

Every $f_n$ is continuous, yet $f$ jumps at $1$. Why is the convergence not uniform? By the intermediate value theorem, for each $n$ there is $x_n$ with $x_n^n=\tfrac12$, so

$$|f_n(x_n)-f(x_n)|=\tfrac12 .$$

This error does not shrink with $n$; hence $\sup_x|f_n(x)-f(x)|\ge\tfrac12$ and cannot tend to $0$.

Pointwise convergence does not preserve continuity; uniform convergence does.

Timeline

Theorem: uniform limits preserve continuity

Theorem. Let $f_n$ be continuous on an interval $I$ and suppose $f_n\to f$ uniformly. Then $f$ is continuous on $I$.

The full proof

Fix $x_0\in I$; we show $f$ is continuous at $x_0$. Let $\varepsilon>0$.

Step 1: freeze an index using uniform convergence. Since $f_n\to f$ uniformly, there is $N$ such that for every $n\ge N$ and every $x$,

$$|f_n(x)-f(x)|<\frac{\varepsilon}{3}.$$

Take $n=N$:

$$|f_N(x)-f(x)|<\frac{\varepsilon}{3} \quad\text{for all }x\in I .$$

This lets us replace $f$ by the single fixed function $f_N$, at a cost that is uniformly small over all points.

Step 2: use continuity of $f_N$ to find $\delta$. Since $f_N$ is continuous at $x_0$, there is $\delta>0$ such that

$$|x-x_0|<\delta \Rightarrow |f_N(x)-f_N(x_0)|<\frac{\varepsilon}{3}.$$

Step 3: triangle inequality. For $|x-x_0|<\delta$, insert $f_N(x)$ and $f_N(x_0)$:

$$\begin{aligned} |f(x)-f(x_0)| &\le |f(x)-f_N(x)| \\ &\quad + |f_N(x)-f_N(x_0)| \\ &\quad + |f_N(x_0)-f(x_0)| \\ &< \frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3} =\varepsilon . \end{aligned}$$

So $f$ is continuous at $x_0$. Since $x_0$ was arbitrary, $f$ is continuous on $I$. $\square$

Note where each piece comes from. The first and third terms come from uniform convergence: same point, same index $N$, error controlled for all $x$ at once. The middle term comes from continuity of $f_N$: same function, two nearby points. Pointwise convergence cannot supply the first and third terms uniformly, because there the index $N$ would grow with $x$.

An immediate corollary

Let

$$E_n(x)=\sum_{k=0}^{n}\frac{x^k}{k!}, \qquad E(x)=\sum_{k=0}^{\infty}\frac{x^k}{k!}.$$

Each $E_n$ is a polynomial, hence continuous. By the Weierstrass $M$-test (proved in the previous lecture), on any fixed interval $[-L,L]$ the series is dominated by $\sum L^k/k!<\infty$, so $E_n\to E$ uniformly on $[-L,L]$. The theorem then gives continuity of $E$ on every $[-L,L]$, hence on all of $\mathbb R$.

Watch the scope: the convergence is not uniform on $\mathbb R$, only on each bounded interval. This is usually called locally uniform convergence.

Uniform convergence is convergence in the sup metric

Let $C([a,b])$ be the set of continuous functions on a closed interval, with

$$d(f,g)=\max_{x\in[a,b]}|f(x)-g(x)| .$$

The maximum exists: $f-g$ is continuous, so is $|f-g|$, and the extreme value theorem gives a maximum on the compact interval.

Proposition. $f_n\to f$ uniformly if and only if $d(f_n,f)\to0$.

Proof. Both directions just rewrite the definitions.

If $f_n\to f$ uniformly: given $\varepsilon>0$, there is $N$ such that for $n\ge N$, $|f_n(x)-f(x)|<\varepsilon$ for all $x$. This single bound is valid for every $x$, so taking the supremum gives $d(f_n,f)\le\varepsilon$, hence $d(f_n,f)\to0$.

Conversely, if $d(f_n,f)\to0$: given $\varepsilon>0$, choose $N$ with $d(f_n,f)<\varepsilon$ for $n\ge N$. Since a value at one point never exceeds the maximum,

$$|f_n(x)-f(x)|\le d(f_n,f)<\varepsilon \quad\text{for all }x,$$

which is exactly uniform convergence. $\square$

This rewriting turns a statement about sequences of functions into a statement about sequences of points in a metric space. All the metric machinery — Cauchy sequences, completeness, contractions — becomes available. That is not a notational convenience; it is the doorway used later to solve differential equations.

$C([a,b])$ is complete

Theorem. The metric space $\bigl(C([a,b]),d\bigr)$ is Cauchy complete: every Cauchy sequence converges to a function in the space.

Let $(f_n)$ be Cauchy. Two things must be done: build a candidate limit $f$, then show it lies in the space and that convergence holds.

Step 1: build the limit pointwise

Fix $x\in[a,b]$. Since

$$|f_n(x)-f_m(x)| \le \max_{z\in[a,b]}|f_n(z)-f_m(z)| =d(f_n,f_m),$$

the sequence of real numbers $(f_n(x))$ is Cauchy. The reals are complete, so it converges. Define

$$f(x)=\lim_{n\to\infty}f_n(x).$$

Each $x$ gives exactly one real number, so $f$ is a function. At this stage only pointwise convergence is known; continuity of $f$ is not yet established.

Step 2: upgrade pointwise to uniform

Let $\varepsilon>0$. Since $(f_n)$ is Cauchy in the metric, there is $N$ such that for $n,m\ge N$,

$$d(f_n,f_m)\le\frac{\varepsilon}{2}.$$

Hence for every $x$,

$$|f_n(x)-f_m(x)|\le\frac{\varepsilon}{2}.$$

Fix $n\ge N$ and let $m\to\infty$. The left side tends to $|f_n(x)-f(x)|$ and the right side is constant, so

$$|f_n(x)-f(x)|\le\frac{\varepsilon}{2} \quad\text{for all }x .$$

Only "at most $\varepsilon/2$" survives, because limits preserve non-strict inequalities. Taking the supremum over $x$,

$$\sup_{x}|f_n(x)-f(x)|\le\frac{\varepsilon}{2}<\varepsilon .$$

Here one must use $\sup$, not $\max$: continuity of $f$ is not yet known, so the extreme value theorem does not apply and the maximum might not be attained. Once continuity is proved, $\sup=\max$.

Thus $f_n\to f$ uniformly, i.e. $d(f_n,f)\to0$.

Step 3: the limit stays in the space

Each $f_n$ is continuous and $f_n\to f$ uniformly, so by the first theorem of this lecture, $f$ is continuous. Hence $f\in C([a,b])$, and Step 2 gives $f_n\to f$ in the metric $d$. $\square$

The division of labour is worth remembering: completeness is used only in Step 1 (completeness of the reals), uniform convergence is established in Step 2, and closure (continuity) is supplied in Step 3. Each step handles one thing.

Theorem: integration commutes with uniform limits

Let $f_n$ be integrable on $[a,b]$ and $f_n\to f$ uniformly. The theorem asserts:

1. $f$ is integrable;
2. $\displaystyle\lim_{n\to\infty}\int_a^b f_n=\int_a^b f$; in particular the limit on the left exists.

Only integrability of $f_n$ is assumed, not continuity. The limit $f$ may be discontinuous (as with $x^n$), so the conclusion cannot be smuggled in through "continuous implies integrable".

Idea of the proof

The Riemann criterion says: for every $\varepsilon>0$ there is a partition $P$ with

$$U(f,P)-L(f,P)<\varepsilon ,$$

where $U,L$ are the upper and lower sums. The plan is to squeeze the upper and lower sums of $f$ between those of $f_N$, with an error of $\pm\varepsilon/3$, and then choose for $f_N$ a partition whose upper and lower sums are close.

Estimating the upper and lower sums

Let $P$ have division points $x_0<\cdots<x_m$, with subintervals $I_i=[x_{i-1},x_i]$ of length $\Delta x_i$. Write

$$M_i(f)=\sup_{I_i}f, \qquad m_i(f)=\inf_{I_i}f .$$

Because $f_n\to f$ uniformly, choose $n$ large enough that

$$|f_n(x)-f(x)|<\frac{\varepsilon}{3(b-a)} \quad\text{for all }x .$$

Fix this $n$. For every $x$,

$$f(x)\le f_n(x)+|f(x)-f_n(x)|<f_n(x)+\frac{\varepsilon}{3(b-a)}.$$

Taking the supremum over $I_i$,

$$M_i(f)\le M_i(f_n)+\frac{\varepsilon}{3(b-a)}.$$

Likewise, from $f(x)\ge f_n(x)-|f(x)-f_n(x)|$, taking the infimum gives

$$m_i(f)\ge m_i(f_n)-\frac{\varepsilon}{3(b-a)}.$$

Multiply by $\Delta x_i$ and sum over $i$. Since $\sum_i\Delta x_i=b-a$, the constant terms cancel the denominator:

$$L(f_n,P)-\frac{\varepsilon}{3} \le L(f,P) \le U(f,P) \le U(f_n,P)+\frac{\varepsilon}{3}.$$

This holds for every partition $P$ and uses only uniform convergence.

Choose a partition; get integrability

With that $n$ fixed, $f_n$ is integrable, so there is a partition $P$ with

$$U(f_n,P)-L(f_n,P)<\frac{\varepsilon}{3}.$$

For this $P$,

$$\begin{aligned} U(f,P)-L(f,P) &\le \Bigl(U(f_n,P)+\frac{\varepsilon}{3}\Bigr) -\Bigl(L(f_n,P)-\frac{\varepsilon}{3}\Bigr)\\ &=\bigl(U(f_n,P)-L(f_n,P)\bigr)+\frac{2\varepsilon}{3}\\ &<\frac{\varepsilon}{3}+\frac{2\varepsilon}{3}=\varepsilon . \end{aligned}$$

By the integrability criterion, $f$ is integrable.

Convergence of the integrals

For any partition $P$, the integral is trapped between the sums:

$$L(f,P)\le\int_a^b f\le U(f,P).$$

Take the partition $P$ for which $U(f_N,P)-L(f_N,P)<\varepsilon/3$. The squeezing estimates give

$$L(f_N,P)-\frac{\varepsilon}{3}\le\int_a^b f\le U(f_N,P)+\frac{\varepsilon}{3},$$

while also

$$L(f_N,P)\le\int_a^b f_N\le U(f_N,P).$$

So both $\int f$ and $\int f_N$ lie in

$$\Bigl[L(f_N,P)-\frac{\varepsilon}{3},\;U(f_N,P)+\frac{\varepsilon}{3}\Bigr],$$

an interval of length $\bigl(U(f_N,P)-L(f_N,P)\bigr)+\tfrac{2\varepsilon}{3}<\varepsilon$. Therefore

$$\left|\int_a^b f-\int_a^b f_N\right|<\varepsilon .$$

Since $\varepsilon>0$ was arbitrary, $\int f_n\to\int f$. $\square$

At the end of the lecture a student asked about strict versus non-strict inequalities. The lecturer's answer: it does not matter, because $\varepsilon$ is arbitrary; choosing constants like $\varepsilon/4$ produces strict inequalities whenever desired.

Integrating a power series term by term

Return to $E_n(x)=\sum_{k=0}^n x^k/k!$, which converges uniformly to $E$ on $[a,b]$. Each $E_n$ is a polynomial, so

$$\int_a^b E_n(x)\,dx =\sum_{k=0}^{n}\int_a^b\frac{x^k}{k!}\,dx .$$

By the interchange theorem the left side tends to $\int_a^b E$, hence

$$\int_a^b E(x)\,dx =\sum_{k=0}^{\infty}\int_a^b\frac{x^k}{k!}\,dx .$$

Each term is computed by finding an antiderivative. Since

$$\frac{d}{dx}\frac{x^{k+1}}{(k+1)!} =\frac{(k+1)x^{k}}{(k+1)!} =\frac{x^k}{k!},$$

we get

$$\int_a^b \frac{x^k}{k!}\,dx =\frac{b^{k+1}-a^{k+1}}{(k+1)!}.$$

So uniform convergence licenses swapping an infinite sum with an integral. Finite sums may be swapped freely; infinite sums may not — uniform convergence is the licence.

What this lecture establishes

• Uniform limits of continuous functions are continuous; pointwise limits need not be ($x^n$ is the counterexample).
• The proof freezes one $f_N$ and uses a three-part $\varepsilon/3$ estimate.
• Uniform convergence is convergence in the sup metric; the $\sup$ versus $\max$ distinction tracks whether the limit is known to be continuous.
• $C([a,b])$ is Cauchy complete in the sup metric.
• Uniform limits of integrable functions are integrable, and integration commutes with the limit.
• Power series may be integrated term by term inside the radius of convergence; term-by-term differentiation is postponed to the next lecture.

Comprehension check

1. Use $f_n(x)=x^n$ to explain why pointwise convergence does not preserve continuity. Write down the points that make uniform convergence fail.

Reference answer: The pointwise limit is $f(0\le x<1)=0$, $f(1)=1$, which is discontinuous at $1$, although every $f_n$ is continuous. Take $x_n=2^{-1/n}\in(0,1)$; then $f_n(x_n)=1/2$ while $f(x_n)=0$, so $|f_n(x_n)-f(x_n)|=1/2$. Hence $\sup_x|f_n(x)-f(x)|\ge1/2$ for every $n$, which does not tend to $0$; uniform convergence fails.

2. Prove that $f_n\to f$ uniformly if and only if $d(f_n,f)\to0$, and identify where the proof uses "for all $x$ simultaneously".

Reference answer: Both directions merely rewrite the definition. If convergence is uniform, then for a given $\varepsilon$ there is $N$ such that for $n\ge N$, $|f_n(x)-f(x)|<\varepsilon$ holds for all $x$ at once; because one bound works for every $x$, taking the supremum gives $d(f_n,f)\le\varepsilon$. Conversely, if $d(f_n,f)<\varepsilon$, then since a pointwise value never exceeds the maximum, $|f_n(x)-f(x)|\le d(f_n,f)<\varepsilon$ for all $x$ simultaneously. The key is exactly that one $\varepsilon$-bound holds for all $x$; under pointwise convergence the bound depends on $x$, and no supremum can be taken.

3. In the completeness proof we already know $f_n(x)\to f(x)$ pointwise. Why must we still prove uniform convergence?

Reference answer: Pointwise convergence only controls each point separately. It does not ensure that $f$ is continuous, nor that $(f_n)$ converges in the metric $d$. Completeness requires convergence in that metric to a point of the space, and $d$ measures the supremum of the error over all points. Only upgrading to $\sup_x|f_n-f|\to0$ gives $d(f_n,f)\to0$; and only applying "uniform limits preserve continuity" once more confirms $f\in C([a,b])$, i.e. that the limit has not left the space.

4. In the integration theorem, what roles do the $3$ and the $(b-a)$ in $\varepsilon/(3(b-a))$ play?

Reference answer: $(b-a)$ cancels the total length produced by summing: a constant error of $\varepsilon/(3(b-a))$ integrated over an interval of length $b-a$ totals $\varepsilon/3$, so each of the upper and lower sums shifts by $\varepsilon/3$. The $3$ splits the total error budget: the two shifts between $f$ and $f_n$ cost $2\varepsilon/3$, leaving $\varepsilon/3$ for the gap $U(f_n,P)-L(f_n,P)$ supplied by integrability, for a total below $\varepsilon$. Any constants with the same total work, e.g. quarters.

5. The theorem assumes only that $f_n$ is integrable, not continuous. Why can we not simply say "the limit of continuous functions is integrable"?

Reference answer: Because a uniform limit of continuous functions is continuous, but here the $f_n$ need not be continuous; and even if they were, the pointwise limit can be discontinuous ($x^n$), so the route through continuity is not always available. The conclusion that $f$ is integrable must be proved directly from the Riemann criterion: construct, for a given $\varepsilon$, a partition $P$ with $U(f,P)-L(f,P)<\varepsilon$. The proof squeezes the upper and lower sums of $f$ between those of $f_N$, which is exactly how it avoids continuity and uses only integrability plus uniform convergence.

Relation to neighbouring lectures

The previous lecture established the Weierstrass $M$-test and the pointwise/uniform distinction; this lecture turns them into a toolkit: preservation of continuity, the metric viewpoint, completeness, and interchange with integrals. The next lecture handles the remaining piece — whether differentiation commutes with limits — which needs a stronger hypothesis (uniform convergence of the derivatives). With it, power series can be differentiated term by term arbitrarily often, and the exponential satisfies $E'=E$.

Further on, the completeness of $C([a,b])$ in the sup metric is used directly to prove the Picard–Lindelöf theorem, by writing the solution of a differential equation as a fixed point of an integral operator. The abstract three-step proof in this lecture is the foundation of that line.