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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

Linearity

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