数学分析原理答案——第七章 习题12
【第七章 习题12】
假设ggg与fn(n=1,2,3,…)f_{n}(n = 1,2,3,\ldots)fn(n=1,2,3,…)都定义在(0,∞)(0,\infty)(0,∞)上，只要0<t<T<∞0 < t < T < \infty0<t<T<∞，便都在[t,T]\lbrack t,T\rbrack[t,T]上Riemann可积分。∣fn∣≤g\left| f_{n} \right| \leq g∣fn∣≤g，在(0,∞)(0,\infty)(0,∞)的每个紧子集上fn→ff_{n} \rightarrow ffn→f是一致的，而且
∫0∞g(x)dx<∞\int_{0}^{\infty}{g(x)dx} < \infty∫0∞g(x)dx<∞
试证
limn→∞∫0∞fn(x)dx=∫0∞f(x)dx\lim_{n \rightarrow \infty}{\int_{0}^{\infty}{f_{n}(x)dx}} = \int_{0}^{\infty}{f(x)dx}n→∞lim∫0∞fn(x)dx=∫0∞f(x)dx
【解】
(a) 首先由于在(0,∞)(0,\infty)(0,∞)的每个紧子集上fn→ff_{n} \rightarrow ffn→f是一致，所以在[t,T]\lbrack t,T\rbrack[t,T]上f(x)f(x)f(x)，Riemann可积分。
(b) 然后根据极限的定义可知
∫0∞g(x)dx=limt→∞∫0tg(x)dx\int_{0}^{\infty}{g(x)dx} = \lim_{t \rightarrow \infty}{\int_{0}^{t}{g(x)dx}}∫0∞g(x)dx=t→∞lim∫0tg(x)dx
翻译成ε−δ\varepsilon - \deltaε−δ语言就是
∃A∈R,∀ε>0, ∃t>0, ∀T≥t ∣∫0Tg(x)dx−A∣<ε\exists A\mathbb{\in R,\forall}\varepsilon > 0,\ \exists t > 0,\ \forall T \geq t\ \ \left| \int_{0}^{T}{g(x)dx} - A \right| < \varepsilon∃A∈R,∀ε>0, ∃t>0, ∀T≥t ∫0Tg(x)dx−A<ε
从数理逻辑来看，等价于（注意此处不是柯西序列）
∀ε>0, ∃t>0, ∀T≥t ∣∫0Tg(x)dx−∫0tg(x)dx∣<ε⇔∣∫tTg(x)dx∣<ε\forall\varepsilon > 0,\ \exists t > 0,\ \forall T \geq t\ \ \left| \int_{0}^{T}{g(x)dx} - \int_{0}^{t}{g(x)dx} \right| < \varepsilon \Leftrightarrow \left| \int_{t}^{T}{g(x)dx} \right| < \varepsilon∀ε>0, ∃t>0, ∀T≥t ∫0Tg(x)dx−∫0tg(x)dx<ε⇔∫tTg(x)dx<ε
这是因为：前者的否命题
∀A∈R,∃ε>0, ∀t>0, ∃T≥t ∣∫0Tg(x)dx−A∣≥ε\forall A\mathbb{\in R,\exists}\varepsilon > 0,\ \forall t > 0,\ \exists T \geq t\ \ \left| \int_{0}^{T}{g(x)dx} - A \right| \geq \varepsilon∀A∈R,∃ε>0, ∀t>0, ∃T≥t ∫0Tg(x)dx−A≥ε
选取
A=∫0tg(x)dxA = \int_{0}^{t}{g(x)dx}A=∫0tg(x)dx
此时与后者命题有矛盾，所以后者能推出前者；
另外容易看出，前者可直接推出后者。
(c) 由于
∣fn∣≤g\left| f_{n} \right| \leq g∣fn∣≤g
所以
∀ε>0, ∃t>0, ∀T≥t,∀n, ∣∫tTfn(x)dx∣≤∫tTg(x)dx<ε\forall\varepsilon > 0,\ \exists t > 0,\ \forall T \geq t,\forall n,\ \ \left| \int_{t}^{T}{f_{n}(x)dx} \right| \leq \int_{t}^{T}{g(x)dx} < \varepsilon∀ε>0, ∃t>0, ∀T≥t,∀n, ∫tTfn(x)dx≤∫tTg(x)dx<ε
根据（b）的结论，也就是说极限
∫0∞fn(x)dx=limt→∞∫0tfn(x)dx\int_{0}^{\infty}{f_{n}(x)dx} = \lim_{t \rightarrow \infty}{\int_{0}^{t}{f_{n}(x)dx}}∫0∞fn(x)dx=t→∞lim∫0tfn(x)dx
是存在的。
(d) 由于
∣∫tTf(x)dx∣≤∣∫tT[f(x)−fn(x)]dx∣+∣∫tTfn(x)dx∣\left| \int_{t}^{T}{f(x)dx} \right| \leq \left| \int_{t}^{T}{\left\lbrack f(x) - f_{n}(x) \right\rbrack dx} \right| + \left| \int_{t}^{T}{f_{n}(x)dx} \right|∫tTf(x)dx≤∫tT[f(x)−fn(x)]dx+∫tTfn(x)dx
而在(0,∞)(0,\infty)(0,∞)的每个紧子集上fn→ff_{n} \rightarrow ffn→f是一致的，所以无论t、Tt、Tt、T如何取值，都有
∀t∈R,∀T≥t,∀ε>0,∃N>0,∀n>N ∣∫tT[f(x)−fn(x)]dx∣<ε(T−t)\forall t\mathbb{\in R,\forall}T \geq t,\forall\varepsilon > 0,\exists N > 0,\forall n > N\ \ \left| \int_{t}^{T}{\left\lbrack f(x) - f_{n}(x) \right\rbrack dx} \right| < \varepsilon(T - t)∀t∈R,∀T≥t,∀ε>0,∃N>0,∀n>N ∫tT[f(x)−fn(x)]dx<ε(T−t)
根据（c）的结论
∀ε>0, ∃t>0, ∀T≥t ∣∫tTfn(x)dx∣<ε\forall\varepsilon > 0,\ \exists t > 0,\ \forall T \geq t\ \ \left| \int_{t}^{T}{f_{n}(x)dx} \right| < \varepsilon∀ε>0, ∃t>0, ∀T≥t ∫tTfn(x)dx<ε
即
∀ε>0, ∃t>0, ∀T≥t ∣∫tTf(x)dx∣≤∣∫tT[f(x)−fn(x)]dx∣+∣∫tTfn(x)dx∣<ε(T−t)+ε{\forall\varepsilon > 0,\ \exists t > 0,\ \forall T \geq t\ \ }{\left| \int_{t}^{T}{f(x)dx} \right| \leq \left| \int_{t}^{T}{\left\lbrack f(x) - f_{n}(x) \right\rbrack dx} \right| + \left| \int_{t}^{T}{f_{n}(x)dx} \right| < \varepsilon(T - t) + \varepsilon}∀ε>0, ∃t>0, ∀T≥t ∫tTf(x)dx≤∫tT[f(x)−fn(x)]dx+∫tTfn(x)dx<ε(T−t)+ε
根据（b）的结论，也就是说极限
∫0∞f(x)dx=limt→∞∫0tf(x)dx\int_{0}^{\infty}{f(x)dx} = \lim_{t \rightarrow \infty}{\int_{0}^{t}{f(x)dx}}∫0∞f(x)dx=t→∞lim∫0tf(x)dx
是存在的
又因为
∣∫0∞f(x)dx−∫0∞fn(x)dx∣≤∣∫0∞f(x)dx−∫0tf(x)dx∣+∣∫0tf(x)dx−∫0tfn(x)dx∣+∣∫0tfn(x)dx−∫0∞fn(x)dx∣{\left| \int_{0}^{\infty}{f(x)dx} - \int_{0}^{\infty}{f_{n}(x)dx} \right| }{\leq \left| \int_{0}^{\infty}{f(x)dx} - \int_{0}^{t}{f(x)dx} \right| + \left| \int_{0}^{t}{f(x)dx} - \int_{0}^{t}{f_{n}(x)dx} \right| + \left| \int_{0}^{t}{f_{n}(x)dx} - \int_{0}^{\infty}{f_{n}(x)dx} \right|}∫0∞f(x)dx−∫0∞fn(x)dx≤∫0∞f(x)dx−∫0tf(x)dx+∫0tf(x)dx−∫0tfn(x)dx+∫0tfn(x)dx−∫0∞fn(x)dx
所以
∀ε>0,∃t>0, ∀T≥t,∃N>0,∀n>N ∣∫0∞f(x)dx−∫0∞fn(x)dx∣≤ε3+ε3+ε3=ε{\forall\varepsilon > 0,\exists t > 0,\ \forall T \geq t,\exists N > 0,\forall n > N\ }{\ \left| \int_{0}^{\infty}{f(x)dx} - \int_{0}^{\infty}{f_{n}(x)dx} \right| \leq \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon}∀ε>0,∃t>0, ∀T≥t,∃N>0,∀n>N ∫0∞f(x)dx−∫0∞fn(x)dx≤3ε+3ε+3ε=ε
(先选ttt，再选nnn，顺序很重要!!!)也就是
limn→∞∫0∞fn(x)dx=∫0∞f(x)dx\lim_{n \rightarrow \infty}{\int_{0}^{\infty}{f_{n}(x)dx}} = \int_{0}^{\infty}{f(x)dx}n→∞lim∫0∞fn(x)dx=∫0∞f(x)dx
备注：
翻译ε−δ\varepsilon - \deltaε−δ语言成极限操作就是
limn→∞∫0∞fn(x)dx−∫0∞f(x)dx=limn→∞limt→∞∫0tfn(x)dx−limt→∞∫0tf(x)dx=limn→∞limt→∞∫0tfn(x)dx−limt→∞limn→∞∫0tfn(x)dx{\lim_{n \rightarrow \infty}{\int_{0}^{\infty}{f_{n}(x)dx}} - \int_{0}^{\infty}{f(x)dx} }{= \lim_{n \rightarrow \infty}{\lim_{t \rightarrow \infty}{\int_{0}^{t}{f_{n}(x)dx}}} - \lim_{t \rightarrow \infty}{\int_{0}^{t}{f(x)dx}} }{= \lim_{n \rightarrow \infty}{\lim_{t \rightarrow \infty}{\int_{0}^{t}{f_{n}(x)dx}}} - \lim_{t \rightarrow \infty}{\lim_{n \rightarrow \infty}{\int_{0}^{t}{f_{n}(x)dx}}} }n→∞lim∫0∞fn(x)dx−∫0∞f(x)dx=n→∞limt→∞lim∫0tfn(x)dx−t→∞lim∫0tf(x)dx=n→∞limt→∞lim∫0tfn(x)dx−t→∞limn→∞lim∫0tfn(x)dx其实在(0,∞)(0,\infty)(0,∞)的每个紧子集上fn→ff_{n} \rightarrow ffn→f是一致的，可以得到在(0,∞)(0,\infty)(0,∞)的每个紧子集上
∫0tfn(x)dx→∫0tf(x)dx\int_{0}^{t}f_{n}(x)dx \rightarrow \int_{0}^{t}f(x)dx∫0tfn(x)dx→∫0tf(x)dx
也是一致的。
沃尔特·鲁丁（Walter Rudin）所著《数学分析原理》中第七章
习题12的个人答案，仅供参考