#### 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.