上一讲留下的两笔债

上一讲用幂级数定义

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

并在两处借用了尚未证明的加法公式 $E(x+y)=E(x)E(y)$：一次用来证明负 $x$ 时 $E(x)>0$，一次用来验证 $e^{m/n}=E(m/n)$。这一讲把债还上。

同时还有两件事被布置成习题（pset 5），课上不展开：

• $E$ 在 $\mathbb R$ 上连续；
• 两个在 $\mathbb R$ 上连续的函数，若在 $\mathbb Q$ 上处处相等，则在整个 $\mathbb R$ 上相等。

这两条合起来给出：$E$ 是定义在 $\mathbb Q$ 上的那个 $e^x$ 的唯一连续延拓。也就是说，「指数函数」这个概念本身是被这两条性质钉死的——它必须是幂级数给出的那一个。

时间轴

非负项级数的乘积定理

定理。 设 $\sum a_n$、$\sum b_n$ 都是非负项级数且收敛。定义

$$c_n=\sum_{i=0}^{n}a_i b_{n-i},$$

（下标之和恰为 $n$）。则 $\sum c_n$ 收敛，且

$$\sum_{n=0}^{\infty}c_n=\left(\sum_{n=0}^{\infty}a_n\right)\left(\sum_{n=0}^{\infty}b_n\right).$$

记部分和

$$S_n^a=\sum_{i=0}^{n}a_i,\qquad S_n^b=\sum_{i=0}^{n}b_i,\qquad S_n^c=\sum_{k=0}^{n}c_k .$$

证明

非负项级数收敛等价于部分和有界，所以下面两个不等式一夹就够。

上界：$S_n^c\le S_n^a S_n^b$。

把乘积 $S_n^aS_n^b$ 展开，得到所有 $a_ib_j$（$0\le i,j\le n$）。按 $i+j=k$ 分组：

• 当 $k\le n$ 时，$k$ 的所有分拆 $(i,k-i)$ 的两个下标都不超过 $n$，因此 $c_k=\sum_{i=0}^{k}a_ib_{k-i}$ 的每一项都在乘积里；
• 当 $k>n$ 时，有些项缺席（例如 $a_{n+1}b_0$ 根本不在乘积中）。

所以 $S_n^a S_n^b$ 至少包含 $c_0,c_1,\dots,c_n$ 的全部项，还可能多出一些非负项。于是

$$S_n^c\le S_n^a S_n^b .$$

下界：$S_n^aS_n^b\le S_{2n}^c$。

乘积展开中的每一项 $a_ib_j$ 都满足 $i+j\le 2n$，所以它必定出现在 $c_{i+j}$ 中，而 $i+j\le 2n$ 意味着 $c_{i+j}$ 被计入 $S_{2n}^c$。因此

$$S_n^aS_n^b\le S_{2n}^c .$$

夹逼。 合并两个不等式：

$$S_n^c\le S_n^aS_n^b\le S_{2n}^c .$$

令 $n\to\infty$。左边与右边都趋于 $\sum c_n$（$(S_n^c)$ 单调递增收敛），中间趋于 $(\sum a_n)(\sum b_n)$。由夹逼，

$$\sum_{n=0}^{\infty}c_n=\left(\sum a_n\right)\left(\sum b_n\right). \qquad\square$$

这个证明的关键在于：两个不等式都只用到非负性，不需要任何估计常数，也不需要知道和是多少。

推论：绝对收敛的情形

若 $\sum a_n$、$\sum b_n$ 绝对收敛，则 $\sum c_n$ 也绝对收敛，且和仍等于两个和之积。

证明：对 $|c_n|$ 用三角不等式，

$$|c_n|=\left|\sum_{i=0}^{n}a_ib_{n-i}\right|\le\sum_{i=0}^{n}|a_i|\,|b_{n-i}| .$$

右边正是把非负项级数 $\sum|a_n|$、$\sum|b_n|$ 按乘积定理组合得到的级数，它收敛。由比较判别法，$\sum|c_n|$ 收敛。

这个推论正是把加法公式从「$x,y\ge0$」推广到任意符号的关键。

为什么必须要求非负：顺序会改变和

对非负项级数，求和顺序无关紧要：可以先把第 5 项、第 7 项、第 119 项加起来，再回头补第 1 项，和不变。

一旦出现负项且不绝对收敛，顺序就变得致命。以交错调和级数

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$$

为例：它收敛（上一讲由交错级数判别法给出），但不绝对收敛。课程指出，若重新排列它的项，可以把它变成发散，也可以让它收敛到任意指定的实数——先把足够多的负项加起来，和会掉到很低，再补几个正项拉回来，如此反复，极限完全由排列方式决定。

补充：这个结论通常称为 Riemann 重排定理，课上只作口头说明，未给证明。

这解释了乘积定理为何必须限制在非负项：证明中要比较「同一批项的不同分组」，而只有在非负情形下这种比较才是安全的。

指数函数的加法公式

定理。 对任意实数 $x,y$，

$$E(x+y)=E(x)E(y).$$

课程只证 $x,y\ge0$ 的情形，其余由上述推论补足。

证明（$x,y\ge0$）。 取

$$a_n=\frac{x^n}{n!},\qquad b_n=\frac{y^n}{n!},\qquad c_n=\frac{(x+y)^n}{n!}.$$

前两个级数对所有实数收敛（比值判别法），并求和得 $E(x)$、$E(y)$。$c_n$ 也是非负的。若能证明

$$c_n=\sum_{i=0}^{n}a_ib_{n-i},$$

就可以直接套用乘积定理，得到 $\sum c_n=E(x)E(y)$，而左边就是 $E(x+y)$。

验证这一步只需二项式定理：

$$(x+y)^n=\sum_{i=0}^{n}\binom{n}{i}x^i y^{n-i}.$$

把组合数写成阶乘：

$$\binom{n}{i}=\frac{n!}{i!\,(n-i)!}.$$

于是

$$c_n=\frac{(x+y)^n}{n!} =\frac{1}{n!}\sum_{i=0}^{n}\frac{n!}{i!\,(n-i)!}x^iy^{n-i} =\sum_{i=0}^{n}\frac{x^i}{i!}\cdot\frac{y^{n-i}}{(n-i)!} =\sum_{i=0}^{n}a_ib_{n-i}.$$

正是乘积定理要求的形状。$\square$

一般的符号则由推论给出：$E(x+y)$、$E(x)E(y)$ 都可以写成绝对收敛级数的乘积形式，逐项对应后两项相等。课程未展开这一推广，只说「不难」。

补充：得到加法公式后，上一讲用到的两条性质就都成立了——负 $x$ 时 $E(x)=1/E(-x)>0$，以及 $e^{m/n}=E(m/n)$。

连续性的回顾

$f$ 在定义域中的点 $x_0$ 连续，意思是：对每个 $\varepsilon>0$，存在 $\delta>0$，使得只要 $x$ 在定义域内且 $|x-x_0|<\delta$，就有

$$|f(x)-f(x_0)|<\varepsilon .$$

定义对定义域完全没有要求：可以是区间，可以是一堆离散点，可以是 $\mathbb Q$。这一点在构造反例时很方便。

一个处处不连续的函数

Dirichlet 函数。

$$f(x)= \begin{cases} 0, & x\in\mathbb Q,\\[2pt] 1, & x\notin\mathbb Q. \end{cases}$$

定理。 $f$ 在每一点都不连续。

有理点处

设 $x_0\in\mathbb Q$，则 $f(x_0)=0$。取任意 $\delta>0$。因为 $\sqrt2\notin\mathbb Q$，对任意正整数 $n$，数

$$x=x_0+\frac{\sqrt2}{n}$$

都是无理数（有理数加无理数仍是无理数），且当 $n$ 足够大时 $|x-x_0|=\dfrac{\sqrt2}{n}<\delta$。但这个点上的函数值是 $1$，于是

$$|f(x)-f(x_0)|=|1-0|=1 .$$

取 $\varepsilon=\tfrac12<1$，无论 $\delta$ 多小都找不到满足要求的邻域。所以 $f$ 在 $x_0$ 不连续。

无理点处

设 $x_0\notin\mathbb Q$，则 $f(x_0)=1$。任何开区间内都有有理数（有理数的稠密性），所以对任意 $\delta>0$，存在有理数 $x$ 满足 $|x-x_0|<\delta$。此时

$$|f(x)-f(x_0)|=|0-1|=1,$$

同样取 $\varepsilon=\tfrac12$ 即可否定连续性。

直观

图像由两条「水平线」组成：有理点上取 $0$，无理点上取 $1$。两组点稠密地交错在一起，画不出来，但它清楚地说明了：连续性不是「在同类点之间取值接近」就够的——要求的是邻域内所有点，而不只是某一类点。这也正是学生在课上提出的疑问的答案：在有理点之间看，函数当然「像常数」；但连续性不让你挑邻居。

两条代数规则的证明

上一讲列出了五条规则，其中两条被单独拿出来证明。

复合（规则 5）

设 $f:A\to B$、$g:B\to\mathbb R$ 连续，$h=g\circ f$。要证 $h$ 在 $x_0$ 连续。

记 $y_0=f(x_0)$。

给定 $\varepsilon>0$。因为 $g$ 在 $y_0$ 连续，存在 $\delta_g>0$，使得

$$|y-y_0|<\delta_g\ \Rightarrow\ |g(y)-g(y_0)|<\varepsilon .$$

现在把 $\delta_g$ 当作「$\varepsilon$」用到 $f$ 上：因为 $f$ 在 $x_0$ 连续，存在 $\delta_f>0$，使得

$$|x-x_0|<\delta_f\ \Rightarrow\ |f(x)-f(x_0)|=|f(x)-y_0|<\delta_g .$$

于是当 $|x-x_0|<\delta_f$ 时，把 $y=f(x)$ 代回第一段：

$$|h(x)-h(x_0)|=|g(f(x))-g(f(x_0))|<\varepsilon .$$

所以 $h$ 在 $x_0$ 连续。$\square$

证明的方向是从外向内：先由 $g$ 的连续性拿到 $\delta_g$，再把它当作目标精度去找 $f$ 的 $\delta_f$。这与数列复合的论证同型，只是把「$N$」换成了「$\delta$」。

倒数（规则 4）

设 $f$ 在 $x_0$ 连续且 $f(x_0)\ne0$。要证 $1/f$ 在 $x_0$ 连续。

先算

$$\left|\frac{1}{f(x)}-\frac{1}{f(x_0)}\right| =\frac{|f(x_0)-f(x)|}{|f(x)|\,|f(x_0)|}.$$

分子可控，麻烦在分母的 $|f(x)|$ 可能很小。办法是先保证 $f(x)$ 远离零。

因为 $f(x_0)\ne0$ 且 $f$ 连续，取「$\varepsilon$」为 $\dfrac{|f(x_0)|}{2}$，得 $\delta_1>0$，使得当 $|x-x_0|<\delta_1$ 时

$$|f(x)-f(x_0)|<\frac{|f(x_0)|}{2}.$$

由三角不等式（反向形式），

$$|f(x_0)|=|f(x_0)-f(x)+f(x)|\le|f(x_0)-f(x)|+|f(x)|,$$

移项得

$$|f(x)|\ge|f(x_0)|-|f(x_0)-f(x)|>\frac{|f(x_0)|}{2}.$$

于是在 $|x-x_0|<\delta_1$ 时，

$$\left|\frac{1}{f(x)}-\frac{1}{f(x_0)}\right| <\frac{2}{|f(x_0)|^2}\,|f(x)-f(x_0)| .$$

再取 $\delta_2>0$，使 $|x-x_0|<\delta_2$ 时 $|f(x)-f(x_0)|<\dfrac{\varepsilon|f(x_0)|^2}{2}$。令 $\delta=\min\{\delta_1,\delta_2\}$，则

$$\left|\frac{1}{f(x)}-\frac{1}{f(x_0)}\right|<\varepsilon .$$

所以 $1/f$ 在 $x_0$ 连续。$\square$

这里的套路与乘积规则中的「先证有界」完全一致：先把分母从零推开，再分配误差。区别只是这次用的是下界而不是上界。

本节结论

• 非负项级数的乘积定理：$c_n=\sum_{i=0}^{n}a_ib_{n-i}$，则 $\sum c_n=(\sum a_n)(\sum b_n)$；证明靠 $S_n^c\le S_n^aS_n^b\le S_{2n}^c$ 夹逼。
• 绝对收敛级数的乘积级数绝对收敛。
• 非负性不可省：重排会改变非绝对收敛级数的和，甚至可以让它收敛到任意值。
• 二项式定理把 $((x+y)^n/n!)$ 写成乘积级数，于是 $E(x+y)=E(x)E(y)$。
• Dirichlet 函数在有理点与无理点上都不连续，因为两类点互相稠密。
• 复合的证明从外向内传递 $\delta$；倒数的证明先用 $|f(x_0)|/2$ 把分母推开。

理解检验

1. 乘积定理的证明中，两个不等式各自用到了什么？为什么缺一不可？

参考答案：$S_n^c\le S_n^aS_n^b$ 说明乘积「至少」包含前 $n$ 组，用的是逐项非负（多出来的项不会让和变小）；$S_n^aS_n^b\le S_{2n}^c$ 说明乘积展开的每一项都落在某个 $c_k$（$k\le2n$）里，同样依赖非负性。只有上界，无法排除 $\sum c_n$ 比乘积小；只有下界，无法排除它比乘积大。两者夹逼才能断言相等。若去掉非负性，两边都可能被符号抵消，夹逼失效。

2. 为什么证明 $E(x+y)=E(x)E(y)$ 时要先做二项式展开？它在乘积定理里对应哪个对象？

参考答案：乘积定理处理的是 $c_n=\sum_{i=0}^{n}a_ib_{n-i}$，而 $E(x+y)$ 的通项是 $(x+y)^n/n!$。二项式展开给出

$$\frac{(x+y)^n}{n!}=\sum_{i=0}^{n}\frac{x^i}{i!}\cdot\frac{y^{n-i}}{(n-i)!},$$

正好把 $c_n$ 写成 $a_i b_{n-i}$ 的形式，其中 $a_i=x^i/i!$、$b_j=y^j/j!$。对应上之后，乘积定理直接给出 $\sum c_n=(\sum a_n)(\sum b_n)=E(x)E(y)$。

3. 证明 Dirichlet 函数在有理点不连续时，为什么构造 $x_0+\sqrt2/n$？

参考答案：需要找一列无理数趋近有理点 $x_0$。$x_0$ 有理、$\sqrt2$ 无理，所以 $x_0+\sqrt2/n$ 无理（有理数与无理数之和为无理数）；当 $n\to\infty$ 时它趋于 $x_0$。于是无论 $\delta$ 多小，邻域内都存在函数值为 $1$ 的点，而 $f(x_0)=0$，差为 $1$，取 $\varepsilon<1$ 即否定连续性。在无理点处则相反，用有理数的稠密性找有理点。

4. 证明 $1/f$ 连续时，为什么要先取 $\delta_1$ 使 $|f(x)-f(x_0)|<|f(x_0)|/2$？

参考答案：估计式

$$\left|\frac1{f(x)}-\frac1{f(x_0)}\right|=\frac{|f(x_0)-f(x)|}{|f(x)||f(x_0)|}$$

的分母含 $|f(x)|$，它在 $x_0$ 附近可能任意小，直接控制分子不够。取 $\delta_1$ 保证 $|f(x)-f(x_0)|<|f(x_0)|/2$ 后，由反向三角不等式得 $|f(x)|>|f(x_0)|/2$，分母就有了与 $x$ 无关的正下界。之后才能把整个表达式压到 $|f(x)-f(x_0)|$ 的常数倍，再分配 $\varepsilon$。

5. 连续性的定义为什么允许定义域是 $\mathbb Q$ 这样的集合？这对本节有什么用处？

参考答案：$\varepsilon$-$\delta$ 定义只要求「定义域内」满足 $|x-x_0|<\delta$，不要求定义域是区间或连通集。于是「$E$ 限制在 $\mathbb Q$ 上」也可以谈连续性。用处在于：先把 $e^x$ 在 $\mathbb Q$ 上定义好，再证明 $E$ 在 $\mathbb R$ 上连续且与它在 $\mathbb Q$ 上一致；由「两个连续函数若在 $\mathbb Q$ 上相等则处处相等」，$E$ 就是那个唯一延拓。

与前后课的关系

加法公式证明之后，指数函数作为一个「由幂级数定义、满足 $E(x+y)=E(x)E(y)$」的对象就完整了。它同时说明了一件事：幂级数不只是形式上的无穷多项式，它可以承担函数的定义。

连续性这边，本讲补上了两条较难的代数规则，并给出一个反例说明连续性有多强。下一讲转向连续函数的两条基本定理——极值与介值定理，并把讨论搬到度量空间，为后面的紧性做准备。第一讲提出的介值定理，届时将用完备性真正证明出来。

Two debts from the last lecture

The previous lecture defined

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

and borrowed the addition formula $E(x+y)=E(x)E(y)$ twice without proof: once to show $E(x)>0$ for negative $x$, once to verify $e^{m/n}=E(m/n)$. This lecture repays the debt.

Two further facts are assigned as exercises (problem set 5) rather than proved in class:

• $E$ is continuous on $\mathbb R$;
• two continuous functions on $\mathbb R$ that agree on $\mathbb Q$ agree everywhere.

Together they say that $E$ is the unique continuous extension of the function $e^x$ first defined on $\mathbb Q$. In other words, the exponential function is pinned down by these two properties — it must be the one given by the power series.

Timeline

Product theorem for non-negative series

Theorem. Let $\sum a_n$ and $\sum b_n$ be convergent series of non-negative terms. Define

$$c_n=\sum_{i=0}^{n}a_i b_{n-i},$$

so that the two indices always sum to $n$. Then $\sum c_n$ converges and

$$\sum_{n=0}^{\infty}c_n=\left(\sum_{n=0}^{\infty}a_n\right)\left(\sum_{n=0}^{\infty}b_n\right).$$

Write the partial sums as

$$S_n^a=\sum_{i=0}^{n}a_i,\qquad S_n^b=\sum_{i=0}^{n}b_i,\qquad S_n^c=\sum_{k=0}^{n}c_k .$$

Proof

For non-negative series, convergence is equivalent to boundedness of partial sums, so two inequalities suffice.

Upper bound: $S_n^c\le S_n^a S_n^b$.

Expand $S_n^aS_n^b$, obtaining every $a_ib_j$ with $0\le i,j\le n$. Group by $i+j=k$:

• for $k\le n$, every split $(i,k-i)$ has both indices at most $n$, so each term of $c_k=\sum_{i=0}^{k}a_ib_{k-i}$ appears in the product;
• for $k>n$, some terms are absent (for instance $a_{n+1}b_0$ never appears).

So $S_n^aS_n^b$ contains all of $c_0,\dots,c_n$ and possibly extra non-negative terms:

$$S_n^c\le S_n^aS_n^b .$$

Lower bound: $S_n^aS_n^b\le S_{2n}^c$.

Every product term $a_ib_j$ has $i+j\le2n$, so it appears inside $c_{i+j}$, which is counted in $S_{2n}^c$:

$$S_n^aS_n^b\le S_{2n}^c .$$

Squeeze. Combining,

$$S_n^c\le S_n^aS_n^b\le S_{2n}^c .$$

As $n\to\infty$, the left and right sides both tend to $\sum c_n$ (monotone convergence), while the middle tends to $(\sum a_n)(\sum b_n)$. Hence

$$\sum_{n=0}^{\infty}c_n=\left(\sum a_n\right)\left(\sum b_n\right). \qquad\square$$

The key point: both inequalities use only non-negativity. No estimate of constants is needed, and neither sum must be known.

Corollary: the absolutely convergent case

If $\sum a_n$ and $\sum b_n$ converge absolutely, then $\sum c_n$ converges absolutely, with the same product formula.

Proof: by the triangle inequality,

$$|c_n|=\left|\sum_{i=0}^{n}a_ib_{n-i}\right|\le\sum_{i=0}^{n}|a_i|\,|b_{n-i}| .$$

The right side is exactly the product series formed from the non-negative series $\sum|a_n|$ and $\sum|b_n|$, which converges by the theorem. Comparison gives convergence of $\sum|c_n|$.

This corollary is precisely what extends the addition formula beyond $x,y\ge0$.

Why non-negativity matters: order changes the sum

For a non-negative series the order of summation is irrelevant: one may add the fifth, seventh and hundred-and-nineteenth terms first and fill in the first term later, and the sum is unchanged.

Once negative terms appear and absolute convergence fails, order becomes decisive. Take the alternating harmonic series

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n},$$

which converges (by the alternating series test) but not absolutely. The lecture notes that rearranging its terms can make it diverge, or make it converge to any prescribed real number: add enough negative terms to push the sum far down, then a few positive ones to pull it back, and repeat — the limit is determined entirely by the ordering.

Note: this statement is usually called the Riemann rearrangement theorem; the lecture mentions it only in words.

This explains why the product theorem restricts to non-negative terms: the proof compares different groupings of the same terms, and only in the non-negative case is such a comparison safe.

The exponential addition formula

Theorem. For all real $x,y$,

$$E(x+y)=E(x)E(y).$$

The lecture proves the case $x,y\ge0$; the general case follows from the corollary.

Proof ($x,y\ge0$). Set

$$a_n=\frac{x^n}{n!},\qquad b_n=\frac{y^n}{n!},\qquad c_n=\frac{(x+y)^n}{n!}.$$

The first two series converge for every real argument (ratio test), with sums $E(x)$ and $E(y)$; the $c_n$ are non-negative. If we show

$$c_n=\sum_{i=0}^{n}a_ib_{n-i},$$

the product theorem yields $\sum c_n=E(x)E(y)$, and the left side is $E(x+y)$ by definition.

That identity is just the binomial theorem:

$$(x+y)^n=\sum_{i=0}^{n}\binom{n}{i}x^i y^{n-i}, \qquad \binom{n}{i}=\frac{n!}{i!\,(n-i)!}.$$

Therefore

$$c_n=\frac{(x+y)^n}{n!} =\frac{1}{n!}\sum_{i=0}^{n}\frac{n!}{i!\,(n-i)!}x^iy^{n-i} =\sum_{i=0}^{n}\frac{x^i}{i!}\cdot\frac{y^{n-i}}{(n-i)!} =\sum_{i=0}^{n}a_ib_{n-i},$$

exactly the required shape. $\square$

For general signs the corollary applies: both sides can be written as products of absolutely convergent series, and matching terms gives equality. The lecture does not spell this out, saying only that it is not hard.

Note: with the addition formula in hand, both borrowed facts from the previous lecture follow — $E(x)=1/E(-x)>0$ for negative $x$, and $e^{m/n}=E(m/n)$.

Review of continuity

$f$ is continuous at a point $x_0$ of its domain if for every $\varepsilon>0$ there is $\delta>0$ such that whenever $x$ lies in the domain and $|x-x_0|<\delta$,

$$|f(x)-f(x_0)|<\varepsilon .$$

Nothing is assumed about the domain: it may be an interval, a scattered set, or $\mathbb Q$. That freedom is what makes the next counterexample possible.

A function continuous nowhere

Dirichlet function.

$$f(x)= \begin{cases} 0, & x\in\mathbb Q,\\[2pt] 1, & x\notin\mathbb Q. \end{cases}$$

Theorem. $f$ is continuous at no point.

At rational points

Let $x_0\in\mathbb Q$, so $f(x_0)=0$, and let $\delta>0$ be arbitrary. Since $\sqrt2\notin\mathbb Q$, for every positive integer $n$ the number

$$x=x_0+\frac{\sqrt2}{n}$$

is irrational (rational plus irrational is irrational), and $|x-x_0|=\dfrac{\sqrt2}{n}<\delta$ for $n$ large enough. But $f(x)=1$, so

$$|f(x)-f(x_0)|=|1-0|=1 .$$

Taking $\varepsilon=\tfrac12<1$, no neighbourhood of any size works. So $f$ is discontinuous at $x_0$.

At irrational points

Let $x_0\notin\mathbb Q$, so $f(x_0)=1$. Every open interval contains a rational number (density of $\mathbb Q$), so for any $\delta>0$ there is $x\in\mathbb Q$ with $|x-x_0|<\delta$. Then

$$|f(x)-f(x_0)|=|0-1|=1,$$

and again $\varepsilon=\tfrac12$ disproves continuity.

Intuition

The graph consists of two "horizontal lines": one at $0$ over the rationals, one at $1$ over the irrationals. The two sets are densely interleaved, so the graph cannot be drawn — but the example makes a sharp point: continuity is not about behaving well among points of the same kind, but about all points in a neighbourhood. This answers the student's question: among rationals the function does look constant, but continuity does not let you choose your neighbours.

Two algebraic rules proved

The previous lecture listed five rules; two are proved here.

Composition (rule 5)

Let $f:A\to B$ and $g:B\to\mathbb R$ be continuous, and let $h=g\circ f$. Show $h$ is continuous at $x_0$; write $y_0=f(x_0)$.

Given $\varepsilon>0$. Since $g$ is continuous at $y_0$, there is $\delta_g>0$ with

$$|y-y_0|<\delta_g\ \Rightarrow\ |g(y)-g(y_0)|<\varepsilon .$$

Now use $\delta_g$ as the "$\varepsilon$" for $f$: since $f$ is continuous at $x_0$, there is $\delta_f>0$ with

$$|x-x_0|<\delta_f\ \Rightarrow\ |f(x)-f(x_0)|=|f(x)-y_0|<\delta_g .$$

Then for $|x-x_0|<\delta_f$, putting $y=f(x)$ into the first statement,

$$|h(x)-h(x_0)|=|g(f(x))-g(f(x_0))|<\varepsilon .$$

So $h$ is continuous at $x_0$. $\square$

The proof runs from the outside in: the continuity of $g$ supplies $\delta_g$, which then becomes the target accuracy for $f$. This is the same structure as for compositions of sequences, with $N$ replaced by $\delta$.

Reciprocal (rule 4)

Let $f$ be continuous at $x_0$ with $f(x_0)\ne0$. Show $1/f$ is continuous at $x_0$.

Compute

$$\left|\frac{1}{f(x)}-\frac{1}{f(x_0)}\right| =\frac{|f(x_0)-f(x)|}{|f(x)|\,|f(x_0)|}.$$

The numerator is controllable; the difficulty is that $|f(x)|$ may be small. So first keep $f(x)$ away from zero.

Since $f(x_0)\ne0$ and $f$ is continuous, applying continuity with "$\varepsilon$" equal to $\dfrac{|f(x_0)|}{2}$ gives $\delta_1>0$ such that $|x-x_0|<\delta_1$ implies

$$|f(x)-f(x_0)|<\frac{|f(x_0)|}{2}.$$

The reverse triangle inequality,

$$|f(x_0)|=|f(x_0)-f(x)+f(x)|\le|f(x_0)-f(x)|+|f(x)|,$$

gives

$$|f(x)|\ge|f(x_0)|-|f(x_0)-f(x)|>\frac{|f(x_0)|}{2}.$$

Hence for $|x-x_0|<\delta_1$,

$$\left|\frac{1}{f(x)}-\frac{1}{f(x_0)}\right| <\frac{2}{|f(x_0)|^2}\,|f(x)-f(x_0)| .$$

Choose $\delta_2>0$ so that $|f(x)-f(x_0)|<\dfrac{\varepsilon|f(x_0)|^2}{2}$ whenever $|x-x_0|<\delta_2$, and set $\delta=\min\{\delta_1,\delta_2\}$. Then

$$\left|\frac{1}{f(x)}-\frac{1}{f(x_0)}\right|<\varepsilon .$$

So $1/f$ is continuous at $x_0$. $\square$

The pattern matches the product rule's "first prove boundedness": push the denominator away from zero, then allocate the error. Only this time a lower bound, not an upper bound, is needed.

What this lecture establishes

• Product theorem: for non-negative series, $c_n=\sum_{i=0}^{n}a_ib_{n-i}$ gives $\sum c_n=(\sum a_n)(\sum b_n)$, proved by $S_n^c\le S_n^aS_n^b\le S_{2n}^c$.
• Products of absolutely convergent series converge absolutely.
• Non-negativity cannot be dropped: rearrangements change the sum of a conditionally convergent series, even arbitrarily.
• The binomial theorem writes $((x+y)^n/n!)$ as a product series, giving $E(x+y)=E(x)E(y)$.
• The Dirichlet function is discontinuous at rational and irrational points alike, because each set is dense in the other.
• Compositions pass $\delta$ from outside in; reciprocals are handled by first pushing the denominator below $|f(x_0)|/2$ in absolute value.

Comprehension check

1. Which inequality does each half of the product-theorem proof use, and why is neither alone enough?

Reference answer: $S_n^c\le S_n^aS_n^b$ shows the product contains at least the first $n$ blocks, using termwise non-negativity (extra terms cannot shrink the sum). $S_n^aS_n^b\le S_{2n}^c$ shows every term of the expanded product sits in some $c_k$ with $k\le2n$, again using non-negativity. With only the upper bound, $\sum c_n$ could be smaller than the product; with only the lower bound, larger. Squeezing gives equality. Without non-negativity, cancellation can break both comparisons.

2. Why expand $(x+y)^n$ binomially before applying the product theorem? Which objects correspond?

Reference answer: The product theorem handles $c_n=\sum_{i=0}^{n}a_ib_{n-i}$, while the term of $E(x+y)$ is $(x+y)^n/n!$. The binomial theorem gives

$$\frac{(x+y)^n}{n!}=\sum_{i=0}^{n}\frac{x^i}{i!}\cdot\frac{y^{n-i}}{(n-i)!},$$

so $c_n$ takes the required shape with $a_i=x^i/i!$ and $b_j=y^j/j!$. Matching done, the theorem yields $\sum c_n=(\sum a_n)(\sum b_n)=E(x)E(y)$.

3. In the Dirichlet-function proof at rational points, why construct $x_0+\sqrt2/n$?

Reference answer: One needs a sequence of irrational numbers approaching the rational point $x_0$. Since $x_0$ is rational and $\sqrt2$ irrational, $x_0+\sqrt2/n$ is irrational (a rational plus an irrational is irrational), and it tends to $x_0$ as $n\to\infty$. Hence every neighbourhood, however small, contains a point with value $1$ while $f(x_0)=0$; the gap is $1$, so any $\varepsilon<1$ disproves continuity. At irrational points one instead uses density of the rationals.

4. Why take $\delta_1$ making $|f(x)-f(x_0)|<|f(x_0)|/2$ in the reciprocal proof?

Reference answer: In

$$\left|\frac1{f(x)}-\frac1{f(x_0)}\right|=\frac{|f(x_0)-f(x)|}{|f(x)||f(x_0)|}$$

the denominator contains $|f(x)|$, which can be arbitrarily small near $x_0$; controlling the numerator alone is not enough. With $|f(x)-f(x_0)|<|f(x_0)|/2$, the reverse triangle inequality gives $|f(x)|>|f(x_0)|/2$, a positive lower bound independent of $x$. Only then can the expression be bounded by a constant multiple of $|f(x)-f(x_0)|$, and $\varepsilon$ allocated.

5. Why does the definition of continuity allow a domain such as $\mathbb Q$, and what is that good for here?

Reference answer: The $\varepsilon$-$\delta$ condition only quantifies over points of the domain within $\delta$ of $x_0$; the domain need not be an interval or connected. So continuity of $E$ restricted to $\mathbb Q$ makes sense. This allows one to define $e^x$ first on $\mathbb Q$, then prove $E$ continuous on $\mathbb R$ and agreeing with it on $\mathbb Q$; since two continuous functions agreeing on $\mathbb Q$ agree everywhere, $E$ is the unique extension.

Relation to neighbouring lectures

With the addition formula proved, the exponential is complete as an object: defined by a power series and satisfying $E(x+y)=E(x)E(y)$. It also makes a broader point: a power series is not merely a formal infinite polynomial — it can carry a function's definition.

On continuity, the lecture supplies two of the harder algebraic rules and a counterexample showing how strong the notion is. The next lecture turns to the two fundamental theorems about continuous functions — extreme values and intermediate values — and moves the discussion into metric spaces, preparing for compactness. The intermediate value theorem announced in the very first lecture will then be genuinely proved.