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## Solution to Principles of Mathematical Analysis Chapter 7 Part A

### Chapter 7 Sequences and Series of Functions

#### Exercise 1

(By analambanomenos) Let $\{f_n\}$ be a uniformly convergent sequence of bounded functions on a set $E$. For each $n$, there is a number $M_n$ such that $\big|f_n(x)\big|<M_n$ for all $x\in E$. By Theorem 7.8, there is an integer $N$ such that $\big|f_n(x)-f_N(x)\big|<1$ if $n\ge N$ for all $x\in E$. Let $$M=\max\{M_1,\ldots,M_N\}.$$ Then for $n\le N$ and $x\in E$ we have $\big|f_n(x)\big|<M+1$, and for $n\ge N$ and $x\in E$ we have $$\big|f_n(x)\big|\le\big|f_n(x)-f_N(x)\big|+\big|f_N(x)\big|<M+1.$$ That is, the $f_n$ are uniformly bounded by $M+1$ in $E$.

#### Exercise 2

(By analambanomenos) Let $\varepsilon>0$. By Theorem 7.8 there is an integer $N$ such that for all $n,m\ge N$ and $x\in E$, $$\big|f_n(x)-f_m(x)\big|<\varepsilon/2\quad\quad\big|g_n(x)-g_m(x)\big|<\varepsilon/2.$$ Hence for all $n,m\ge N$ and $x\in E$, $$\big|(f_n+g_n)(x)-(f_m+g_m)(x)\big|\le\big|f_n(x)-f_m(x)\big|+\big|g_n(x)-g_m(x)\big|<\varepsilon.$$ Hence, also by Theorem 7.8, $\{f_n+g_n\}$ converges uniformly on $E$.

If $\{f_n\}$ and $\{g_n\}$ are uniformly convergent sequences of bounded functions, then by Exercise 1 they are uniformly bounded. That is, there is a number $M$ such that $\big|f_n(x)\big|<M$ and $\big|g_n(x)\big|<M$ for all $n$ and for all $x\in E$. By Theorem 7.8 there is an integer $N$ such that for all $n\ge N$ and for all $x\in E$ we have $$\big|f_n(x)-f_m(x)\big|<\varepsilon/M \quad\quad\big|g_n(x)-g_m(x)\big|<\varepsilon/M.$$ Hence for all $n,m\ge N$ and $x\in E$,
\begin{align*}
\big|f_n(x)g_n(x)-f_m(x)g_m(x)\big| &\le \big|f_n(x)g_n(x)-f_m(x)g_n(x)\big|+\big|f_m(x)g_n(x)-f_m(x)g_m(x)\big| \\
&\le M\big|f_n(x)-f_m(x)\big|+M\big|g_n(x)-g_m(x)\big| \\
&\le \varepsilon
\end{align*}Hence, also by Theorem 7.8, $\{f_ng_n\}$ converges uniformly on $E$.

#### Exercise 3

(By analambanomenos) Let $f_n(x)=x^{-1}+n^{-1}$ on $(0,1)$. Then $f_n$ converges uniformly to $x^{-1}$ on $(0,1)$, and $$f_n^2(x)=\frac{1}{x^2}+\frac{2}{nx}+\frac{1}{n^2}$$ converges to $x^{-2}$ on $(0,1)$, but not uniformly, since the difference $f_n^2(x)-x^{-2}=2/(nx)+1/n^2$ is arbitrarily large as $x\rightarrow 0$.

#### Exercise 4

(By analambanomenos) Let $f_n(x)$ be the $n$-th term of the series. Since $f_n(0)=1$ for all $n$, the series diverges for $x=0$. If $x>0$, then $$f(x)=\sum_{n=1}^\infty\frac{1}{1+n^2x}< \frac{1}{x}\sum_{n=1}^\infty\frac{1}{n^2}<\infty$$ by Theorem 3.28, so the series converges if $x>0$. The convergence is absolute since $f_n(x)>0$ for $x>0$, and since $f_n(x)\rightarrow \infty$ as $x\rightarrow 0+$, $f(x)$ is not bounded on $(0,\infty)$. If $a>0$, then since $f_n(x)\le 1/(an^2)$ for $a\le x$ and all $n$, the series converges uniformly to $f(x)$ on $[a,\infty)$ by Theorem 7.10. Since the partial sums are continuous, $f(x)$ is continuous on $[a,\infty)$ by Theorem 7.12, so it is continuous on all of $(0,\infty)$. The series does not converge uniformly on $(0,\infty)$ since the difference between $f(x)$ and a partial sum is a series $\sum_N^\infty f_n(x)>1/(1+N^2x)$, which is arbitrarily large as $x\rightarrow 0+$.

The case $x<0$ is more complicated. First note that $f_n(x)$ is not even defined for $x=-n^{-2}$, so we are limited to considering the intervals $(-\infty,-1)$ and $\bigl(-n^{-2},-(n+1)^{-2}\bigr)$ for $n=1,2,\ldots$ For $-n^{-2}<x<-(n+1)^{-2}$ and $m>n+1$ we have $$\frac{1}{1-\bigl(m/(n+1)\bigr)^2}<f_m(x)<\frac{1}{1-(m/n)^2}<0$$ so that
$$\big|f_m(x)\big|<\frac{1}{\bigl(m/(n+1)\bigr)^2-1}.$$ Similarly, for $x<-1$ and $m>1$, we get $$\big|f_m(x)\big|<\frac{1}{m^2-1},$$ which is the same inequality, with $n=0$. The series
$$\sum_{m=n+2}^\infty\frac{1}{\bigl(m/(n+1)\bigr)^2-1}$$ converges, which can be shown either by the integral test or the “limit comparison test,” which was not covered in Rudin’s text. Hence the series converges uniformly and absolutely for $-n^{-2}<x<(n+1)^{-2}$, or $x<-1$ by Theorem 7.10. Since the partial sums are continuous, $f(x)$ is continuous on this interval by Theorem 7.12. It is not bounded, since $f_n(x)\rightarrow -\infty$ as $x\rightarrow -n^{-2}$, while the rest of the series (everything except the $n$-th term) sums to a finite value.

#### Exercise 5

(By analambanomenos) For $x\le 0$, we have $f_n(x)=0$ for all $n$, and for $x>0$ and $n$ large enough so that $n^{-1}<x$, we have $f_n(x)=0$. Hence $f_n(x)$ converges to the constant function $f(x)=0$. Since for all $n$ there is an $x$ such that $f_n(x)=1$ (viz. $x=2/(2n+1)$), for any $0<\varepsilon<1$ there is no $N$ such that $\big|f_n(x)-f(x)\big|=\big|f_n(x)\big|<\varepsilon$ for all $x$ and all integers $n>N$, so $\{f_n(x)\}$ does not converge uniformly to $f(x)$.

Let $$F_N(x)=\sum_{n=1}^N f_n(x)= \begin{cases} 0 & \displaystyle \biggl(x<\frac{1}{N+1}\biggr), \\ \displaystyle \sin^2\frac{\pi}{x} & \displaystyle\biggl(\frac{1}{n+1}\le x\biggr). \end{cases}$$ Then $F_N(x)$ converges to $$F(x)= \begin{cases} 0 & \displaystyle (x\le0), \\ \displaystyle \sin^2\frac{\pi}{x} & \displaystyle(0<x). \end{cases}$$ Since for all $N$ there is a positive number $x<(N+1)^{-1}$ such that $\big|F(x)-F_N(x)\big|=1$, the convergence is not uniform.

#### Exercise 6

(By analambanomenos) Since $$\lim_{n\rightarrow\infty}\frac{x^2+n}{n^2}=0,$$ the alternating series converges for all $x$ by Theorem 3.43. It doesn’t converge absolutely for any $x$ since $$\sum_{n=1}^\infty \frac{x^2+n}{n^2}\ge\sum_{n=1}^\infty\frac{1}{n}$$ which diverges. The partial sums are $$f_m(x)=\sum_{n=1}^m(-1)^n\frac{x^2+n}{n^2}=x^2\sum_{n=1}^m\frac{(-1)^n}{n^2}+ \sum_{n=1}^m\frac{(-1)^n}{n}=A_mx^2+B_m\rightarrow Ax^2+B.$$ Let $\varepsilon>0$ and let $N$ be large enough so that $|A-A_n|<\varepsilon$ and $|B-B_n|<\varepsilon$ for all $n>N$. Let $[a,b]$ be a bounded interval, and let $c=\max\bigl(|a|,|b|\bigr)$. Then for $n>N$ and $x\in[a,b]$ we have $$\big|f(x)-f_n(x)\big|<c^2\varepsilon+\varepsilon,$$ so that $\{f_n\}$ converges uniformly to $f(x)$ on $[a,b]$.

#### Exercise 7

(By analambanomenos) Calculating the minimum and maximum values of $f_n$ using elementary calculus, we get that $$\big|f_n(x)\big|\le\frac{1}{2\sqrt{n}},$$ so that $f_n(x)$ converges uniformly to the constant function $f(x)=0$ on $\mathbf R$. For $x\ne0$, $$f_n’(x)=\frac{1-nx^2}{(1+nx^2)^2}$$ converges to $f’(x)=0$, but $f_n’(0)=1$ does not.

#### Exercise 8

(By analambanomenos) Let $f_n(x)$ be the partial sum $\sum_{k=1}^nc_kI(x-x_k)$. Then $$\big|f_n(x)-f_m(x)\big|\le\sum_{k=m+1}^n\big|c_kI(x-x_k)\big|\le\sum_{k=m+1}^n|c_k|.$$ Since $\sum|c_k|$ converges, by Theorem 3.22 for any $\varepsilon>0$ there is an integer $N$ such that for $m\ge N$ and $n\ge N$ we have $$\big|f_n(x)-f_m(x)\big|\le\sum_{k=m+1}^n|c_k|<\varepsilon.$$ Hence by Theorem 7.8 the series converges uniformly.

Let $x\in(a,b)$ such that $x\ne x_n$ for any $n$. Then the partial sum $f_n$ is constant in any neighborhood of $x$ not containing any of $x_1,\ldots,x_n$. Hence by Theorem 7.11,
$$\lim_{t\rightarrow x}f(x)=\lim_{n\rightarrow\infty}f_n(x)=f(x),$$ that is, $f$ is continuous at $x$.

#### Exercise 9

(By analambanomenos) The problem didn’t state it explicitly, but in order to apply the Chapter’s Theorems let’s assume that $E$ is a set in a metric space.

Let $\varepsilon>0$. Since $\{f_n\}$ converges uniformly to $f$, there is an integer $N_1$ such that if $n\ge N_1$ then for any $y\in E$ we have $$\big|f_n(y)-f(y)\big|<\frac{\varepsilon}{2}.$$
Since $f$ is continuous by Theorem 7.12, there is an integer $N_2$ such that if $n\ge N_2$ then $$\big|f(x_n)-f(x)\big|<\frac{\varepsilon}{2}.$$ Hence, if $n\ge \max(N_1,N_2)$, we have
$$\big|f_n(x_n)-f(x)\big|\le\big|f_n(x_n)-f(x_n)\big|+\big|f(x_n)-f(x)\big|<\varepsilon.$$

The converse is not true. For example, let $E=[0,1)$ and let $f_n(x)=x^n$. Then this sequence converges to the constant function $f(x)=0$, but not uniformly. If $\{x_n\}$ is a sequence of points of $E$ converging to a point $x$ in $E$, then it is contained in a closed subinterval $[0,a]$ where $a<1$. Since $f_n$ converges uniformly on this subinterval, $f_n(x_n)$ converges to $f(x)$ by the first part of this problem. Hence this property is insufficient for determining that the convergence is uniform.

#### Exercise 10

(By analambanomenos) The discontinuities of the $n$-th term, $f_n(x)=(nx)/n^2$, occur at the rational numbers $m/n$ for any integer $m$. Let $y=p/q$ be a rational number where $p$ and $q$ have no common divisors. Then $f_n$ is discontinuous at $y$ if and only if $n$ is a multiple of $q$. Let $g_q=\sum_m(mqx)/(mq)^2$, which is discontinuous at $y$, and let $h_q=f-g_q$. Then the partial sums of $h_q$ are all continuous at $y$, hence by Theorem 7.11, $h_q$ is also continuous at $y$, and so $f=g_q+h_q$ is discontinuous at $y$. If the real number $z$ is irrational, then the partial sums of $f$ are all continuous at $z$, hence by again applying Theorem 7.11 we have $f$ is continuous at $z$. Therefore the discontinuities of $f$ are the rational numbers, a countable dense subset of the real numbers.

The partial sums of $f$ converge uniformly to $f$ and are all Riemann-integrable on bounded intervals. Hence by Theorem 7.16, $f$ is also Riemann-integrable on bounded intervals.

#### Exercise 11

(By analambanomenos) Theorem 3.42 gives us pointwise convergence of the series. Uniform convergence follows by repeating the proof of that Theorem in this context, as follows.

Let $F_n(x)$ be the partial sums of $\sum f_n(x)$. Choose $M$ such that $\big|F_n(x)\big|<M$ for all $n$ and all $x\in E$. Given $\varepsilon>0$ there is an integer $N$ such that $g_N(x)\le \varepsilon/(2M)$ for all $x\in E$. For $N\le p\le q$ we have by Theorem 3.41
\begin{align*}
\Bigg|\sum_{n=p}^qf_n(x)g(x)\Bigg| &= \Bigg|\sum_{n=p}^{q-1}F_n(x)\bigl(g_n(x)-g_{n+1}(x)\bigr)+F_q(x)g_q(x)-F_{p-1}(x)g_p(x)\Bigg| \\
&\le M\Bigg|\sum_{n=p}^{q-1}\bigl(g_n(x)-g_{n+1}(x)\bigr)+g_q(x)+g_{p}\Bigg| \\
&= 2Mg_p(x)\le 2Mg_N(x)\le\varepsilon.
\end{align*}Uniform convergence follows from Theorem 7.8 (the Cauchy condition for uniform convergence).

#### Exercise 12

(By analambanomenos) Fix $t>0$ and let $k(T)=\int_t^T\big|f(x)\big|\,dx$. Then $k(T)$ is a monotonically increasing function which is bounded by $\int_t^\infty g(x)\,dx$. Hence by Theorem 3.14 (that theorem refers to sequences, but can be easily extended to this case), $k(T)$ converges as $T\rightarrow\infty$, and since $\big|\int_t^Tf(x)\,dx\big|\le k(T)$, $\int_t^\infty f(x)\,dx$ converges. Similarly, $\int_t^\infty f_n(x)\,dx$ also converges.

For $t>0$ define the functions $$h_n(t)=\int_t^\infty f_n(x)\,dx\quad\quad h(t)=\int_t^\infty f(x)\,dx.$$ The problem is to show that $$\lim_{n\rightarrow\infty}\biggl(\lim_{t\rightarrow0} h_n(t)\biggr)=\lim_{t\rightarrow0}h(t).$$ By Theorem 7.11, this follows if we can show that $h_n(t)$ converges uniformly to $h(t)$ on $(0,\infty)$.

Let $\varepsilon>0$. Since $\int_0^\infty g(x)\,dx$ converges, there are numbers $0<T_1<T_2<\infty$ such that $$\int_0^{T_1}g(x)\,dx<\frac{\varepsilon}{2}\quad\quad\int_{T_2}^\infty g(x)\,dx <\frac{\varepsilon}{2}.$$ Since $f_n$ converges to $f$ uniformly on $[T_1,T_2]$, there is a positive integer $N$ such that for $n>N$ and $T_1\le x\le T_2$ we have $$\big|f_n(x)-f(x)\big|< \frac{\varepsilon}{T_2-T_1}.$$ Hence, for $0<t<T_1$ and $n>N$, we have
\begin{align*}
\big|h_n(t)-h(t)\big| &\le \int_t^{T_1}\big|f_n(x)-f(x)\big|\,dx + \int_{T_1}^{T_2}\big|f_n(x)-f(x)\big|\,dx + \int_{T_2}^\infty\big|f_n(x)-f(x)\big|\,dx \\
&\le 2\int_0^{T_1}g(x)\,dx + \frac{\varepsilon}{T_2-T_1}(T_2-T_1) + 2\int_{T_2}^\infty g(x)\,dx \\
&< 3\varepsilon.
\end{align*}Similar arguments yielding similar results can be made for $t\ge T_1$. This shows that $h_n(t)$ converges uniformly to $h(t)$ on $(0,\infty)$.

Baby Rudin 数学分析原理完整第七章习题解答