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Solution to Principles of Mathematical Analysis Chapter 4 Part C

Chapter 4 Continuity

Exercise 19

(By analambanomenos) Suppose $f$ is not continuous at $x_0$. Then there is a sequence of real numbers $y_n$ converging to $x_0$ such that $f(y_n)$ doesn’t converge to $f(x_0)$, that is, there is a $\delta>0$ such that $(f(x_0)-\delta,f(x_0)+\delta)$ contains only a finite number of the $f(y_n)$. Hence there is an infinite subsequence $x_n$ of the $y_n$ such that either all the $f(x_n)$ are greater than $f(x_0)+\delta$ or less than $f(x_0)-\delta$. Suppose the first case is true (the second case can be handled by considering the function $-f$, which also satisfies the hypotheses). Following the hint, let $r$ be a rational number such that $f(x_n)>r>f(x_0)$ for all $n$. Then by the intermediate value property satisfied by $f$, for each $n$ there is a number $t_n$ such that $t_n$ lies between $x_0$ and $x_n$ and $f(t_n)=r$. Since $t_n$ converges to $x_0$, $x_0$ is in the closure of the set of all $x$ such that $f(x)=r$. This set is closed by hypothesis, so we must have $f(x_0)=r$, contradicting the assumption that $f(x_0)<r$.

Exercise 20

(By analambanomenos)

(a): The condition $\rho_E(x)=0$ is equivalent to having every neighborhood of $x$ of radius $\delta>0$ having some element $z$ of $E$, that is, $x\in\bar E$.

(b): Following the hint, if $z\in E$, then $\rho_E(x)\le d(x,y)+d(y,z)$ by the triangle inequality, so that $\rho_E(x)\le d(x,y)+\rho_E(y)$. Symmetrically, $\rho_E(y)\le d(x,y)+\rho_E(x)$, so that $$\big|\rho_E(x)-\rho_E(y)\big|\le d(x,y)$$ for all $x\in X$, $y\in X$.

Exercise 21

(By analambanomenos) Since $\rho_F$ is a continuous function on the compact set $K$, by Theorem 4.16 it must attain its minimum value on $K$ at some $x\in K$. This minimum value $\delta$ must be nonzero, since if $\rho_F(x)=0$ then by Exercise 20(a) $x\in\bar F=F$ which contradicts the disjointness of $K$ and $F$.

An example where this fails for two planar closed sets is
F_1 &= \{(x,y)\in{\mathbf R}^2|x<0,y\ge 1/x^2\} \\
F_2 &= \{(x,y)\in{\mathbf R}^2|x>0,y\ge 1/x^2\}
\end{align*}A one-dimensional example is
F_1 &= \hbox{set of positive integers} \\
F_2 &= \cup_{n=0}^\infty \bigl[n+2^{-(n+2)},n+1-2^{-(n+2)}\bigr] \\
&= \textstyle [\frac{1}{4},\frac{3}{4}]\cup[1\frac{1}{8},1\frac{7}{8}]\cup[2\frac{1}{16},2\frac{15}{16}]\cup\cdots

Exercise 22

(By analambanomenos) If $\rho_A(p)+\rho_B(p)=0$, then $\rho_A(p)=\rho_B(p)=0$. By Exercise 20(a) this implies that $p\in\bar A\cap\bar B=A\cap B$. But $A$ and $B$ are disjoint, hence $\rho_A(p)+\rho_B(p)>0$. Hence, since $\rho_A$ and $\rho_B$ are continuous on $X$, $f(p)$ is also continuous on $X$ by Theorem 4.9. Also, $f(p)=0$ if and only if $\rho_A(p)=0$, that is, $p\in\bar A=A$, and $f(p)=1$ if and only if $\rho_B(p)=0$, that is, $p\in\bar B=B$.

Since $[0,1/2)$ and $(1/2,1]$ are disjoint open sets in $[0,1]$ and $f$ is continuous on $X$, $V$ and $W$ are disjoint open sets of $X$ by Theorem 4.8.

Exercise 23

(By analambanomenos) Letting $a<s<t<u<b$, then $t=\lambda u+(1+\lambda)s$ for $\lambda=(t-s)/(u-s)$. Hence
f(t) &\le \lambda f(u)+(1-\lambda)f(s) \\
f(t)-f(s) &\le \lambda\bigl(f(u)-f(s)\bigr) \\
\frac{f(t)-f(s)}{t-s} &\le \frac{f(u)-f(s)}{u-s}
\end{align*}Also, $t=\lambda’ s+(1-\lambda’)u$ for $\lambda’ =(u-t)/(u-s)$. Hence
f(t) &\le \lambda’ f(s)+(1-\lambda’)f(u) \\
\lambda’\bigl(f(u)-f(s)\bigr) &\le f(u)-f(t) \\
\frac{f(u)-f(s)}{u-s} &\le \frac{f(u)-f(t)}{u-t}
\end{align*}The first inequality can be rewritten $f(t)-f(s)=K(t-s)$ for $s<t<u$, which shows that $f$ is continuous from the right at $s$. Similarly, the second inequality can be rewritten $$f(u)-f(t)=K’(u-t)$$ for $s<t<u$, which shows that $f$ is continuous from the left at $u$. Since $s$ and $u$ were arbitrary elements of $(a,b)$, we see that $f$ is continuous on $(a,b)$.

Let $f$ be a convex function defined in $(a,b)$ and let $g$ be an increasing convex function defined on the range of $f$. Then for $a<x<b$, $a<y<b$, $0<\lambda<0$, we have
g\Bigl(f\bigl(\lambda x+(1-\lambda)y\bigr)\Bigr) &\le g\bigl(\lambda f(x)+(1-\lambda)f(y)\bigr) \\
&\le \lambda g\bigl(f(x)\bigr)+(1-\lambda)g\bigl(f(y)\bigr).
\end{align*}Hence $g\circ f$ is convex on $(a,b)$.

Exercise 24

(By analambanomenos) Fix $x,y\in(a,b)$. We can show by induction that $$f(\lambda x+(1-\lambda)y)\le\lambda f(x)+(1-\lambda)f(y)$$ for all rational $\lambda$ between 0 and 1 with denominator a power of 2. This is true for $\lambda=1/2$ by assumption. Assume that the inequality holds for all $\lambda=j/2^m$ for $m<n$ and $1<j<2^m$, and let $\lambda=j/2^n$.
If $j$ is even, then the inequality holds by the induction hypothesis, so assume $j=2k+1$ is odd. Let $$x_1=\biggl(\frac{k}{2^{n-1}}\biggr)x+\biggl(1-\frac{k}{2^{n-1}}\biggr)y\quad\hbox{and}\quad y_1=\biggl(\frac{k+1}{2^{n-1}}\biggr)x+\biggl(1-\frac{k+1}{2^{n-1}}\biggr)y.$$Then $\lambda x+(1-\lambda)y=(x_1+y_1)/2$, so we have
f(\lambda x+(1-\lambda)y) &\le \frac{f(x_1)+f(y_1)}{2} \\
&\le \frac{1}{2}\biggl(\biggl(\frac{k}{2^{n-1}}\biggr)f(x)+\biggl(1-\frac{k}{2^{n-1}}\biggr)f(y)\biggr) +
\frac{1}{2}\biggl(\biggl(\frac{k+1}{2^{n-1}}\biggr)f(x)+\biggl(1-\frac{k+1}{2^{n-1}}\biggr)f(y)\biggr) \\
&= \lambda f(x)+(1-\lambda)f(y)
\end{align*}Both sides of the inequality $$f(\lambda x+(1-\lambda)y)\le\lambda f(x)+(1-\lambda)f(y)$$ are continuous for $\lambda\in(0,1)$, and the inequality holds for a dense subset of $(0,1)$. Hence it holds for all $\lambda\in(0,1)$.

Exercise 25

(By analambanomenos)

(a): Following the hint, take $\mathbf z\notin K+C$ and put $F=\{\mathbf z\}-C$. $K$ and $F$ are disjoint since if there were a $\mathbf y\in F\cap K$ then there would be $\mathbf x\in C$ such that $\mathbf{y=z-x}$, but then $\mathbf{z=y+x}$ contradicts $\mathbf z\notin K+C$. $F$ is closed since it is the inverse image of $C$ by the continuous map $\mathbf{x\mapsto z-x}$. Hence by Exercise 21 there is a $\delta>0$ such that $|\mathbf{x-y}|>\delta$ if $\mathbf x\in K$ and $\mathbf y\in F$. If $\mathbf{x+w}\in K+C$, then $\big|\mathbf{(x+w)-z}\big|=\big|\mathbf{x-(z-w)}\big|>\delta$, so that the open neighborhood of $\mathbf z$ of radius $\delta$ is disjoint from $K+C$. Hence the complement of $K+C$ is open.

(b): Each element of $C_1+C_2$ can be associated with a unique pair of integers, since if $$n_1+n_2\alpha=m_1+m_2\alpha$$ for $n_2\neq m_2$, then this equation can be rewritten to express $\alpha$ as a rational number. Since the collection of such integer pairs is countable, $C_1+C_2$ is countable.

Consider the fractional parts of the integer multiples of $\alpha$, the set of $p\alpha-[p\alpha]$ for $p$ an integer, which are all in $C_1+C_2$. These are all distinct by the argument above. Let $n$ be any positive integer. By the pigeonhole principle, there are at least two such fractional parts in one of the subintervals $(0,1/n),(1/n,2/n),\ldots,((n-1)/n,1)$ of $(0,1)$. That is, there are integers $p,q$ such that $$0<(p-q)\alpha-([p\alpha]-[q\alpha])<1/n,$$ so that $C_1+C_2\cap(0,1/n)$ is nonempty.

Let $x$ be any real number and let $\epsilon>0$. Let $n$ be a large enough positive integer such that $1/n<\epsilon$, and let $y\in C_1+C_2\cap(0,1/n)$. Then some multiple of $y$ lies in $[x,x+1/n)$, so that some element of $C_1+C_2$ is within $\epsilon$ of $x$. Hence the closure of $C_1+C_2$ is $\mathbf R$, and since it is a proper subset of $\mathbf R$, it is not closed.

Exercise 26

(By analambanomenos) Since $g$ is one-to-one, it defines a function $g^{-1}$ from $g(Y)$ to $Y$, which is continuous by Theorem 4.17. If $h$ is continuous, then $f=g^{-1}\circ h$ is continuous by Theorem 4.7, and if h is uniformly continuous, then $f$ is uniformly continuous by Exercise 12.

Let $X=Z=[0,2]$ and $Y=[0,1)\cup[2,3]$. Let $f(x)=x$ for $0\le x<1$ and $f(x)=x+1$ for $1\le x\le 2$. Let $g(x)=x$ for $0\le x<1$ and $g(x)=x-1$ for $2\le x\le 3$. Then $X$ and $Z$ are compact, $Y$ is not compact, $g$ is continuous and one-to-one, $h(x)=g(f(x))=x$ for $0\le x\le 2$ is uniformly continuous, but $f$ is discontinuous at $x=1$.

Baby Rudin 数学分析原理完整第四章习题解答


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