Exercise 4.3.1
See Understanding Analysis Instructors’ Solution Manual Exercise 4.3.1.
Exercise 4.3.2
Solution:
(a) Let $f(x)\equiv 0$ (constant function).
(b) Let $f(x)=x$. Please check that this function is not onetinuous.
(c) Let $f(x)=2x$ and take $\delta=\epsilon/2$.
(d) A lesstinuous function is continuous by definition. A continuous function is also lesstinuous. Suppose $f$ is continuous, then for all $\epsilon>0$, there exists $\delta >0$ such that $|f(x)-f(c)|<\epsilon$ whenever $|x-c|<\delta$.
- If $\delta <\epsilon$, then $f$ is lesstinuous at $c$.
- If $\delta > \epsilon$, then we can take $\tilde\delta=\epsilon/2$. Thus we have $|f(x)-f(c)|<\epsilon$ whenever $|x-c|<\tilde\delta$ and $0< \tilde\delta < \epsilon$. This implies that $f$ is lesstinuous at $c$
Exercise 4.3.3
See Understanding Analysis Instructors’ Solution Manual Exercise 4.3.2.
Exercise 4.3.4
(a) Let $f(x)=1$ be a constant function. Define $$g(x)=\begin{cases}1,& \text{ if }x\ne 1;\\ 0,& \text{ if }x= 1.\end{cases}$$Then we have\[\lim_{x\to 0}f(x)=1,\quad \lim_{x\to 1}g(x)=1.\]But $$\lim_{x\to 0}g(f(x))=\lim_{x\to 0}g(1)=0\ne 1.$$(b) If $f$ and $g$ are continuous, then so is $g\circ f$. Moreover, $q=f(p)$ and $r=g(q)$. Hence $$\lim_{x\to p}g\circ f(x)=g\circ f(p)=g(f(p))=g(q)=r.$$(c) If $f$ is continuous, then (a) still may be false, see the previous example. But if we assume $g$ is continuous, then (a) holds. If $g$ is continuous, then $$g(q)=\lim_{x\to q}g(x)=r.$$For any $\epsilon$, there exists $\delta>0$ such that $|g(x)-r|<0$ provided $|x-q|<\delta$ (not $0<|x-q|<\delta$). Since $$\lim_{x\to p}f(x)=q,$$ there exists $\tilde \delta>0$ such that $|f(x)-q|<\delta$ whenever $0<|x-p|<\tilde\delta$. Hence if $0<|x-p|<\tilde\delta$, we have $|f(x)-q|<\delta$ and hence \[|g(f(x))-r|<\epsilon.\]Therefore (a) holds in this case. Please think it carefully why this holds.
Exercise 4.3.5
Solution: Recall from Definitions 3.2.4 and 3.2.6, if $c$ is an isolated point of $A\subset \mathbf R$, then there exist a $\delta$-neighborhood of $c$ that intersects $A$ only at $c$. Hence for all $\epsilon >0$, taking this $\delta>0$, there is only one point in $A$ such that $|x-c|<\delta$, namely $c$. Therefore, we have $|f(x)-f(c)|=0<\epsilon$ for $x\in A$ such that $|x-c|<\delta$ ($x$ can only be $c$). Hence $f:A\to\mathbf R$ is continuous at $c$.
Exercise 4.3.6
Solution:
(a) Let $f(x)=-g(x)$ defined by $f(x)=1$ if $x$ is rational and $f(x)=-1$ if $x$ is irrational. Then neither $f(x)$ nor $g(x)$ is continuous at 0 but $f(x)g(x)=-1$ and $f(x)+g(x)=0$ are constant functions and hence are continuous at 0.
(b) Impossible. If $f(x)+g(x)$ and $f(x)$ are continuous at 0, then $g(x)=(f(x)+g(x))-f(x)$ is also continuous at 0 by Theorem 4.3.4 (Algebraic Continuity Theorem).
(c) Let $f(x)=0$ be a constant function. Let $g(x)$ be any function which is not continuous at 0, then $f(x)g(x)=0$ is continuous at 0.
(d) Let $f(x)=2+\sqrt{3}$ if $x\ne 0$ and $f(0)=2-\sqrt{3}$. Then $f(x)$ is not continuous at $0$. But $f(x)+\dfrac{1}{f(x)}=4$ is a constant function and hence continuous at 0.
(e) Impossible. Suppose $[f(x)]^3$ is continuous at 0.
If $f(0)=0$, then for all $\epsilon >0$ there exist $\delta>0$ such that $$|f(x)|^3=|f^3(x)-f^3(0)|<\epsilon^3$$ provided $|x-0|<\delta$. Hence $|f(x)|<\epsilon$, namely $$|f(x)-f(0)|=|f(x)|<\epsilon$$ for all $|x-0|<\delta$. Therefore $f(x)$ is continuous at 0.
If $f(0)\ne 0$, let $c=|f(0)|>0$. Then for all $\epsilon >0$ there exist $\delta>0$ such that $$|f^3(x)-f^3(0)|<\frac{3c^2}{4}\epsilon$$ provided $|x-0|<\delta$. This implies that \begin{equation}\label{4.3.6.1}|f(x)-f(0)||f^2(x)-f(x)f(0)+f^2(0)|<\frac{3c^2}{4}\epsilon.\end{equation}Note that \begin{equation}\label{4.3.6.2}f^2(x)-f(x)f(0)+f^2(0)=
(f(x)-f(0)/2)^2+\frac{3}{4}f^2(0)\geqslant \frac{3c^2}{4}.\end{equation}Combining \eqref{4.3.6.1} and \eqref{4.3.6.2}, we conclude that $|f(x)-f(0)|<\epsilon$ for all $|x-0|<\delta$. Therefore $f(x)$ is continuous at 0 as well.
Or use the composition functions. Let $g(x)=\sqrt[3]{x}$, then $g(x)$ is continuous, see Exercise 4.3.1. If $[f(x)]^3$ is continuous, then $f(x)=g([f(x)]^3)$ is also continuous. Hence no such function exists.
Exercise 4.3.7
See Understanding Analysis Instructors’ Solution Manual Exercise 4.3.6.
Exercise 4.3.8
Solution:
(a) True. Note that $g(x)$ is continuous at $1$, by Theorem 4.3.2, we have $g(1)=\lim_{n\to \infty}g(x_n)$ for sequence $x_n=1-\dfrac{1}{n}$. Since $g(x_n)\geqslant 0$ for all $n\in\mathbf N$, by Theorem 2.3.4 (Order Limit Theorem), we have $g(1)\geqslant 0$ as well.
(b) True. Let $x\in\mathbf R$ be fixed. By Theorem 3.2.10 (Density of $\mathbf Q$ in $\mathbf R$), there exists a sequence of rational numbers $r_n$ such that $(r_n)\to x$. Since $g(x)$ is continuous at $x$, it follows from Theorem 4.3.2 that $$g(x)=\lim_{n\to\infty }g(r_n)=0.$$Hence the statement is true.
(c) True. Let $c=g(x_0)>0$. Since $g(x)$ is continuous at $x_0$, there exists $\delta>0$ such that $|f(x)-f(x_0)|<c/2$ for all $|x-x_0|<\delta$. Hence for all $|x-x_0|<\delta$ we have\[-\frac{c}{2}<f(x)-c<\frac{c}{2}.\]Therefore $$\frac{c}{2}< f(x) <\frac{3c}{2}$$ for all $|x-x_0|<\delta$. In particular, $f(x)>0$ for all $|x-x_0|<\delta$. Hence the statement is true.
Exercise 4.3.9
See Understanding Analysis Instructors’ Solution Manual Exercise 4.3.7.
Exercise 4.3.10
Solution: We first show that if $f(x)$ is continuous then so is $|f(x)|$. Suppose $f(x)$ is continuous at $c$, then for all $\epsilon$ there exists $\delta>0$ such that $|f(x)-f(c)|<\epsilon$ for all $|x-c|<\epsilon$. Note that we also have $$||f(x)|-|f(c)||\leqslant |f(x)-f(c)|<\epsilon$$ for all $|x-c|<\epsilon$, see Exercise 1.2.6. Therefore, $|f(x)|$ is continuous at $c$.
(a) We proceed by induction on $n$. For $n=2$, since $f_1(x)$ and $f_2(x)$ are continuous, so are $f_1(x)-f_2(x)$ and $f_1(x)+f_2(x)$. Thus $|f_1(x)-f_2(x)|$ is continuous as well. Then$$\max\{f_1(x),f_2(x)\}=\frac{1}{2}\big((f_1(x)+f_2(x))+|f_1(x)-f_2(x)|\big)$$ is also continuous. Suppose the statement is true for $n$, wwe consider the case $n+1$. Note that \begin{equation}\label{4.3.10.1}\max\{f_1(x),f_2(x),\dots,f_{n}(x),f_{n+1}(x)\}=\max\{\max\{f_1(x),f_2(x),\dots,f_{n}(x)\},f_{n+1}(x)\}.\end{equation}Since $f_1(x),f_2(x),\dots,f_{n}(x)$ are continuous, by induction hypothesis, we have that $\max\{f_1(x),f_2(x),\dots,f_{n}(x)\}$ is continuous. Considering the statement for $n=2$ which is proved above and using \eqref{4.3.10.1}, we conclude that $\max\{f_1(x),f_2(x),\dots,f_{n}(x),f_{n+1}(x)\}$ is continuous. Hence the statement is true for $n+1$. We are done.
(b) It is clear that $h(x)=1$ if $x\ne 0$ and $h(0)=0$. Hence $h(x)$ is not continuous. This implies this statement is false for infinite case.
Exercise 4.3.11
See Understanding Analysis Instructors’ Solution Manual Exercise 4.3.9.
Exercise 4.3.13
See Understanding Analysis Instructors’ Solution Manual Exercise 4.3.10.
