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Solution to Principles of Mathematical Analysis Chapter 7 Part B


Chapter 7 Sequences and Series of Functions

Exercise 13

(By analambanomenos)

(a) Following the steps of the hint: (i) Show that some subsequence $\{f_{n_i}\}$ converges at all rational points $r$, say, to $f(r)$.

Let $\{r_m\}$ be an enumeration of the rational points of $\mathbf R$. Since $\{f_n(r_1)\}\subset[0,1]$, a compact set, there is a subsequence $\{f_{1,n_j}(r_1)\}$ which converges to $f(r_1)$, by Theorem 2.41. Similarly, the $\{f_{1,n_j}(r_2)\}$ has a subsequence $\{f_{2,n_j}(r_2)\}$ which converges to $f(r_2)$. Continuing in this fashion, we get a sequence of subsequences $\{f_{m,n_j}\}$ such that each $\{f_{m,n_j}\}$ is a subsequence of $\{f_{m-1,n_j}\}$ and $f_{m,n_j}(r_m)$ converges to $f(r_m)$. Letting $f_{n_i}=f_{i,n_i}$, we get a subsequence of $\{f_n\}$ which converges on all rational points $r$ of $\mathbf R$ to $f(r)$.

(ii) Define $f(x)$, for any $x\in\mathbf R$, to be $\sup f(r)$, the sup being taken over all rational points $r\le x$.

First we need to show that this new definition of $f(r)$ agrees with the previous one; let $f^*(r)$ be the new definition of $f(r)$. It is clear from the definition that $f^*(r)\ge f(r)$. Suppose $f^*(r)>f(r)$, that is, there is a rational point $q<r$ and a real number $a$ such that $f(q)>a>f(r)$. Since $f(q)=\lim f_{n_i}(q)$, and $f(r)=\lim f_{n_i}(r)$, there is an integer $N$ such that for all $i\ge N$ we have $f_{n_i}(q)>a>f_{n_i}(r)$, but this contradicts the fact that these functions are monotonically increasing.

It is clear from the definition that $f$ is monotonically increasing.

(iii) Show that $f_{n_i}(x)\rightarrow f(x)$ at every $x$ at which $f$ is continuous.

Suppose $f$ is continuous at $x$ and let $\varepsilon>0$. Then there are rational numbers $r_1<x<r_2$ such that $$f(x)-\frac{\varepsilon}{2}<f(r_1)\le f(x)\le f(r_2)<f(x)+\frac{\varepsilon}{2}.$$ Since $f_{n_i}(r_1)\rightarrow f(r_1)$ and $f_{n_i}(r_2)\rightarrow f(r_2)$, there is an integer $N$ such that for all $i>N$ we have $$\big|f_{n_i}(r_1)-f(r_1)\big|<\frac{\varepsilon}{2}\quad\hbox{and}\quad\big|f_{n_i}(r_2)-f(r_2)\big|<\frac{\varepsilon}{2}.$$ And since $f_{n_i}$ is monotonically increasing, we have for all $i>N$ $$f(x)-\varepsilon<f(r_1)-\frac{\varepsilon}{2}<f_{n_i}(r_1)\le f_{n_i}(x)\le f_{n_i}(r_2)<f(r_2)+\frac{\varepsilon}{2}<f(x)+\varepsilon.$$ That is, $\big|f(x)-f_{n_i}(x)\big|<\varepsilon$, so $f_{n_i}(x)\rightarrow f(x)$ as $i\rightarrow
\infty$.

(iv) Show that a subsequence of $\{f_{n_i}\}$ converges at every point of discontinuity of $f$ since there are at most countably many such points.

Since $f$ is monotonically increasing, $f$ has at most countably many simple points of discontinuity by Theorems 4.29 and 4.30. Let $\{x_m\}$ be an enumeration of the points of discontinuity of $f$. Repeating the argument of step (i), there is a subsequence $\{f_{1,n_j}(x_1)\}$ of $\{f_{n_i}(x_1)$ which converges to a value $a$ between 0 and 1. We can continue this way for all the $x_m$, getting a sequence of subsequences $\{f_{m,n_j}\}$ of $\{f_{n_i}\}$ such that each $\{f_{m,n_j}\}$ is a subsequence of $\{f_{m-1,n_j}\}$ for which $f_{m,n_j}(x_m)$ converges. Letting $f_{n_k}=f_{k,n_k}$, we get a subsequence of $\{f_n\}$ which converges at all the points $\{x_m\}$. We can redefine $f(x_i)$ at these points to be those values.

Since we have already shown that $\{f_{n_k}\}$ converges at the points where $f$ is continuous, we have $f_{n_k}(x)$ converging to $f(x)$ for all real values $x$.

(b) Let $K$ be a compact subset of $\mathbf R$ and let $\varepsilon>0$. It was shown in step (iii) of part (a) that, since $f$ is everywhere continuous, $f(x)=\lim f_{n_k}(x)$ for all points $x$. Also, we are using the initial definition of $f$, so that is it monotonically increasing. For each $x\in K$ there is a $\delta_x>0$ such that for all $y\in[x-\delta_x,x+\delta_x]$ we have
$$\big|f(y)-f(x)\big|<\frac{\varepsilon}{4}.$$Also, there is an integer $N_x$ such that for all $k>N_x$ $$\big|f_{n_k}(x-\delta_x)\big|<\frac{\varepsilon}{2}\quad\hbox{and}\quad\big|f_{n_k}(x+\delta_x)\big|<\frac{\varepsilon}{4}.$$ Since $f_{n_k}$ is monotonically increasing, we must have, for all $y\in(x-\delta_x,x+\delta_x)$ and all $k>N_x$, $$f(x)-\frac{\varepsilon}{2}<f(x-\delta_x)-\frac{\varepsilon}{4}<f_{n_k}(x-\delta_x)\le f_{n_k}(y)\le f_{n_k}(x+\delta_x)+\frac{\varepsilon}{4}<f(x)+\frac{\varepsilon}{2},$$ that is, $\big|f_{n_k}(y)-f(x)\big|<\varepsilon/2$. Hence, for $y\in(x-\delta_x,x+\delta_x)$ and $k>N_x$, $$\big|f_{n_k}(y)-f(y)\big|\le\big|f_{n_k}(y)-f(x)\big|+\big|f(x)-f(y)\big|<\varepsilon.$$ Since the open neighborhoods $(x-\delta_x,
x+\delta_x)$ cover the compact set $K$, a finite subset of these neighborhoods, centered at $x_1,\ldots,x_M$, also covers $K$. Letting $N=\max(N_{x_1},\ldots,N_{x_M})$, we have for all $y\in K$ and all $k>N$ that $\big|f_{n_k}(y)-f(y)\big|\le\varepsilon$, that is, $\{f_{n_k}\}$ converges uniformly to $f$ on $K$.


Exercise 14

(By analambanomenos) Since for each $n$ we have $\big|2^{-n}f(3^{2n-1}t)\big|\le2^{-n}$ and $\big|2^{-n}f(3^{2n}t)\big|\le2^{-n}$, the series defining $x(t)$ and $y(t)$ converge uniformly since $\sum 2^{-n}$ converges, by Theorem 7.10. And since the partial sums are continuous functions, $x(t)$ and $y(t)$ are continuous functions, by Theorem 7.12. Hence $\Phi$ is continuous.

Following the hint, let $(x_0,y_0)\in I^2$, and let $$x_0=\sum_{n=1}^\infty 2^{-n}a_{2n-1},\quad\quad y_0=\sum_{n=1}^\infty 2^{-n}a_{2n}$$ be the binary expansions of $x_0$ and $y_0$, where each $a_i$ is 0 or 1. Let $$t_0=\sum_{i=1}^\infty 3^{-i-1}(2a_i).$$ By Exercise 3.19, the set of all such $t_0$ is precisely the Cantor set.

We have for $k=1,2,3,\ldots,$
\begin{align*}
3^kt_0 &= \sum_{i=1}^\infty 3^{k-i-1}(2a_i) \\
&= 2\sum_{n=0}^{k-2}3^na_{k-1-n}+\frac{2}{3}\,a_k+\frac{2}{3}\sum_{n=1}^\infty 3^{-n}a_{k+n}
\end{align*}Note that the last term lies between 0 and $$\frac{2}{3}\sum_{n=1}^\infty 3^{-n} = \frac{2}{3}\biggl(\frac{1/3}{1-(1/3)}\biggr)=\frac{1}{3}.$$ Also, the first term is an even integer, so that, since $f(t+2)=f(t)$, we have $$f(3^kt_0)=f\Biggl(\frac{2}{3}\,a_k+\frac{2}{3}\sum_{n=1}^\infty 3^{-n}a_{k+n}\Biggr).$$ If $a_k=0$, then the expression in the argument of $f$ lies between 0 and $\frac{1}{3}$, so that $f(3^kt_0)=0$. And if $a_k=1$, then the expression in the argument of $f$ lies between $\frac{2}{3}$ and 1, so that $f(3^kt_0)=1$. In either case, we have $f(3^kt_0)=a_k$. Hence $\Phi(t_0)=\bigl(x(t_0),y(t_0)\bigr)=(x_0,y_0)$.


Exercise 15

(By analambanomenos) The function $f$ must be constant on $[0,\infty]$. Let $0\le x<y$ and let $\varepsilon>0$. Then there is a $\delta>0$ such that for all $n=1,2,3,\ldots,$ we have $\big|f_n(z)-f_n(w)\big|<
\varepsilon$ if $0\le w,z\le 1$ and $|w-z|<\delta$. For large enough $N$, we have $$\bigg|\frac{y}{N}-\frac{x}{N}\bigg|=\frac{|y-x|}{N}<\delta,\quad\quad 0\le\frac{x}{N}<\frac{y}{N}\le1.$$Hence $$\big|f(y)-f(x)\big|=\Bigg|f_N\biggl(\frac{y}{N}\biggr)-f_N\biggl(\frac{x}{N}\biggr)\Bigg|<\varepsilon.$$ Since $\varepsilon>0$ was arbitrary, this shows that $f(x)=f(y)$.

By the way, it is easy to extend this argument to show that if $\{f_n\}$ were equicontinuous on $[-1,1]$, then $f$ would be constant on all of $\mathbf R$.


Exercise 16

(By analambanomenos) Let $\varepsilon>0$ and let $\delta>0$ such that if $d(x,y)<\delta$ then $$\big|f_n(x)-f_n(y)\big|<\varepsilon/3$$ for all $n$. Let $x\in K$ and let $N_x$ be an integer large enough so that if $m>N_x$ and $n>N_x$, then $\big|f_m(x)-f_n(x)\big|<\varepsilon$. Then for all $m>N_x$ and $n>N_x$ and all $y\in K$ such that $d(x,y)<\delta$, we have
$$\big|f_n(y)-f_m(y)\big|\le\big|f_n(y)-f_n(x)\big|+\big|f_n(x)-f_m(x)\big|+\big|f_m(x)-f_m(y)\big|<\varepsilon.$$ Since $K$ is compact, there are a finite number of points $x_1,\ldots,x_M$ in $K$ such that the neighborhoods of radius $\delta$ centered at the $x_i$ cover $K$. So if we let $N=\max(N_{x_1},\ldots,N_{x_M})$, then for $m>N$ and $n>N$ and for all $y\in K$, we have $\big|f_m(y)-f_n(y)\big|<\varepsilon$. Hence by Theorem 7.8, $\{f_n\}$ converges uniformly on $K$.


Exercise 17

(By analambanomenos) The extensions of the definitions, and the statements and proofs of Theorems 7.8 through 7.11, are trivial, so I will simply copy them from the text, italicizing the changes.

Definition 7.7 We say that a sequence of mappings $\{f_n\}$, $n=1,2,3,\ldots,$ into a metric space $F$ converges uniformly on $E$ to a function $f$ if for every $\varepsilon>0$ there is an integer $N$ such that $n\ge N$ implies \begin{equation}\label{7.17.1}d\bigl(f_n(x),f(x)\bigr)\le\varepsilon \end{equation} for all $x\in E$.

Definition 7.22 A family $\mathscr F$ of mappings $f$ defined on a set $E$ in a metric space $X$ with values in a metric space $Y$ is said to be equicontinuous on $E$ if for every $\varepsilon>0$ there exists a $\delta>0$ such that $$d_Y\bigl(f(x),f(y)\bigr)<\varepsilon$$ whenever $d_X(x,y)\le\delta$, $x\in E$, $y\in E$, and $f\in\mathscr F$.

Theorem 7.8 The sequence of mappings $\{f_n\}$ into a complete metric space $F$, defined on $E$, converges uniformly on $E$ if and only if for every $\varepsilon>0$ there exists an integer $N$ such that $m\ge N$, $n\ge N$, $x\in E$ implies $d\bigl(f_n(x),f_m(x)\bigr)\le\varepsilon$.

Proof: Suppose $\{f_n\}$ converges uniformly on $E$, and let $f$ be the limit function. There there is an integer $N$ such that $n\ge E$ implies $d\bigl(f_n(x),f(x)\bigr)\le
\varepsilon/2$ so that $$d\bigl(f_n(x),f_m(x)\bigr)\le d\bigl(f_n(x),f(x)\bigr)+d\bigl(f(x),f_m(x)\bigr)\le\varepsilon$$ if $n\ge N$, $m\ge M$, $x\in E$.

Conversely, suppose the Cauchy condition holds. Since $F$ is a complete metric space, the sequence $\{f_n(x)\}$ converges, for every $x$, to a limit which we may call $f(x)$. Thus the sequence $\{f_n\}$ converges on $E$, to $f$. We have to prove that the convergence is uniform.

Let $\varepsilon>0$ be given, and choose $N$ such that \eqref{7.17.1} holds. Fix $n$, and let $m\rightarrow\infty$ in \eqref{7.17.1}. Since $f_m(x)\rightarrow f(x)$ as $m\rightarrow\infty$, this gives $d\bigl(f_n(x),f(x)\bigr)\le\varepsilon$ for every $n\ge N$ and every $x\in E$, which completes the proof.

Theorem 7.9 Suppose $\{f_n\}$ {\it is a sequence of mappings from $E$ into a metric space $F$ such that} $\lim f_n(x)=f(x)$ for $x\in E$. Put $$M_n=\sup_{x\in E}d\bigl(f_n(x),f(x)\bigr).$$ Then $f_n\rightarrow f$ uniformly on $E$ if and only if $M_n\rightarrow0$ as $n\rightarrow\infty$.

Proof: Since this is an immediate consequence of Definition 7.7, we omit the details of the proof.

Theorem 7.10 Suppose $\{f_n\}$ is a sequence of vector-valued functions with values in $\mathbf R^k$ defined on $E$, and suppose $$\big|\big|f_n(x)\big|\big|\le M_n\quad\quad
(x\in E,\, n=1,2,3,\ldots).$$ Then $\sum f_n$ converges uniformly on $E$ if $\sum M_n$ converges.

Proof: If $\sum M_n$ converges, then, for arbitrary $\varepsilon>0$, $$\Bigg|\Bigg|\sum_{i=n}^mf_i(x)\Bigg|\Bigg|\le\sum_{i=n}^m M_i\le\varepsilon\quad\quad(x\in E),$$ provided $m$ and $n$ are large enough. Uniform convergence now follows from Theorem 7.8.

Theorem 7.11 Suppose $\{f_n\}$, a sequence of mappings with values in a complete metric space $F$, converges uniformly on a set $E$ in a metric space. Let $x$ be a limit point
of $E$, and suppose that $$\lim_{t\rightarrow x}f_n(t)=A_n\quad\quad(n=1,2,3,\ldots).$$ Then $\{A_n\}$ converges and \begin{equation}\label{7.17.2}\lim_{t\rightarrow x}f(t)=\lim_{n\rightarrow\infty}A_n.\end{equation}

Proof: Let $\varepsilon>0$ be given. By the uniform convergence of $\{f_n\}$, there exists $N$ such that $n\ge N$, $m\ge N$, $t\in E$ imply \begin{equation}\label{7.17.3} d\bigl(f_n(t),f_m(t)\bigr)\le\varepsilon.\end{equation} Letting $t\rightarrow x$ in \eqref{7.17.3}, we obtain $$d(A_n,A_m)\le\varepsilon$$ for $n\ge N$, $m\ge M$, so that $\{A_n\}$ is a Cauchy sequence and therefore converges, say to $A$.

Next, \begin{equation}\label{7.17.4}d\bigl(f(t),A\bigr)\le d\bigl(f(t),f_n(t)\bigr)+d\bigl(f_n(t),A_n\bigr)+d\bigl(A_n,A\bigr)\end{equation} We first choose $n$ such that \begin{equation}\label{7.17.5} d\bigl(f(t),f_n(t)\bigr)\le\frac{\varepsilon}{3}\end{equation} for all $t\in E$ (this is possible by the uniform convergence), and such that \begin{equation}\label{7.17.6}d\bigl(A_n,A\bigr)\le\frac{\varepsilon}{3}.\end{equation} Then, for this $n$ we choose a neighborhood $V$ of $x$ such that \begin{equation}\label{7.17.7}d\bigl(f_n(t)-A_n\bigr)\le\frac{\varepsilon}{3},\end{equation} if $t\in V\cap E$, $t\ne x$.

Substituting the inequalities \eqref{7.17.5}, \eqref{7.17.6}, \eqref{7.17.7} into \eqref{7.17.4}, we see that $d\bigl(f(t),A\bigr)\le\varepsilon$, provided $t\in V\cap E$, $t\ne x$. This is equivalent to \eqref{7.17.2}.

Theorem 7.12 If $\{f_n\}$ is a sequence of continuous {\it mappings on $E$ into a metric space} $F$, and if $f_n\rightarrow f$ uniformly on $E$, then $f$ is continuous on $E$.

Proof: (In the text, Theorem 7.12 is an immediately corollary of Theorem 7.11. Here, that would require that $F$ be complete, which is not necessarily the case. So the following proof is entirely new.)

By Theorem 4.8, to show that $f$is continuous, it suffices to show that $f^{-1}(V)$ is open for any open set $V\subset F$. If $x\in f^{-1}(V)$, we need to find a neighborhood $N_x$ of $x$ which is contained in $f^{-1}(V)$, for then $f^{-1}(V)$ would be the union of the open neighborhoods $N_x$, $x\in f^{-1}(V)$, and so be open.

Let $\varepsilon>0$ be small enough so that the neighborhood $M_y$ of $y=f(x)$ of radius $\varepsilon$ is contained in $V$. Let $n$ be large enough so that $d\bigl(f_n(z),f(z)\bigr)
\le\varepsilon/3$ for all $z\in E$, which is possible by the assumption of uniform convergence. Let $\delta>0$ such that $d\bigl(f_n(z),f_n(x)\bigr)<\varepsilon/3$ for all $z$ in the neighborhood $N_x$ of $x$ of radius $\delta$, which is possible since $f_n$ is continuous. Then, for all $z\in N_x$, $$d\bigl(f(z),f(x)\bigr)\le d\bigl(f(z),f_n(z)\bigr)+d\bigl(f_n(z),f_n(x)\bigr)+
d\bigl(f_n(x),f(x)\bigr)\le\varepsilon.$$ That is, $f(z)\in M_y\subset V$, or $z\in f^{-1}(V)$, so that $N_x\subset f^{-1}(V)$.

The remaining Theorems, for mappings into $\mathbf R^k$, are largely corollaries of their scalar counterparts in the text.

Theorem 7.16 Let $\alpha$ be monotonically increasing on $[a,b]$. Suppose $\mathbf f_n=(f_{n1},\ldots,f_{nk})\in\mathscr R(\alpha)$ on $[a,b]$, for $n=1,2,3,\ldots,$ and suppose $\mathbf f_n \rightarrow\mathbf f=(f_1,\ldots,f_k)$ uniformly on $[a,b]$. Then $\mathbf f\in\mathscr R(\alpha)$ on $[a,b]$, and $$\int_a^b\mathbf f\,d\alpha=\lim_{n\rightarrow\infty}\int_a^b
\mathbf f_n\,d\alpha.$$

Theorem 7.17 Suppose $\big\{\mathbf f_n=(f_{n1},\ldots,f_{nk})\big\}$ is a sequence of vector-valued functions, differentiable on $[a,b]$ and such that $\big\{\mathbf f(x_0)\big\}$ converges for some point $x_0$ on $[a,b]$. If $\{\mathbf f_n’\}$ converges uniformly on $[a,b]$, to a function $\mathbf f=(f_1,\ldots,f_k)$, and $$\mathbf f’(x)=\lim_{n\rightarrow\infty}f_n’(x)\quad\quad(a\le x\le b).$$

Proofs: If $\mathbf x=(x_1,\ldots,x_k)\in\mathbf R^k$, then $|x_i|\le||\mathbf x||\le\sum_i^k|x_i|$. Hence $\mathbf f_n$ convergences uniformly if and only if the component functions $f_{n1},\ldots,f_{nk}$ converge uniformly. And since the integers or derivatives, and integrability or differentiability, of vector-valued functions are defined by component, the vector-valued versions of the Theorems 7.16 and 7.17 are immediate corollaries of the scalar versions in the text.

Theorem 7.24 If $K$ is a compact metric space, if $\mathbf f_n=(f_{n1},\ldots,f_{nk})\in\mathscr C(K)$ for $n=1,2,3,\ldots,$ and if $\{\mathbf f_n\}$ converges uniformly on $K$, then $\{\mathbf f_n\}$ is equicontinuous on $K$.

Proof: If $\mathbf x=(x_1,\ldots,x_k)\in\mathbf R^k$, then $|x_i|\le||\mathbf x||\le\sum_i|x_i|$. Hence $\{\mathbf f_n\}$ is equicontinuous if and only if each of the $\{f_{ni}\}$, $i=1,\ldots,k$ are equicontinuous. Hence the vector-valued version of Theorem 7.24 is an immediate corollary of the scalar version in the text.

Theorem 7.25 If $K$ is compact, if $\mathbf f_n=(f_{n1},\ldots,f_{nk})$ is pointwise bounded and equicontinuous on $K$, then

(a) $\{\mathbf f_n\}$ is uniformly bounded on $K$,

(b)$\{\mathbf f_n\}$ contains a uniformly convergent subsequence.

Proof: Since each of the sequences $\{f_{ni}\}$, $i=1,\ldots,k$, is pointwise bounded and equicontinuous on $K$, by the scalar version of Theorem 7.25 in the text each of them is uniformly bounded on $K$, so $\{\mathbf f_n\}$ is uniformly bounded on $K$. Also, $\{f_{n1}\}$ contains a uniformly convergent subsequence, $\{f_{n_j1}\}$. Since the subsequence $\{f_{n_j2}\}$ is also pointwise bounded and equicontinuous on $K$, it also has a uniformly convergent subsequence such that the corresponding subsequence of $\{f_{n_j1}\}$ also converges uniformly. Continuing in this manner, after $k$ steps we have a subsequence $\{\mathbf f_{n_j}\}$ whose component functions all converge uniformly, so $\{\mathbf f_{n_j}\}$ also converges uniformly.

Baby Rudin 数学分析原理完整第七章习题解答

Linearity

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