If you find any mistakes, please make a comment! Thank you.

Solution to Understanding Analysis Exercise 4.4


Exercise 4.4.1

See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.1.


Exercise 4.4.2

(a) No, let $x_n=\dfrac{1}{n}$ and $y_n=\dfrac{1}{2n}$, then\[|x_n-y_n|=\frac{1}{2n}\to 0,\quad |f(x_n)-f(y_n)|=n\geqslant 1.\]Hence by Theorem 4.4.5, $f(x)$ is not uniformly continuous on $(0,1)$.

(b) Yes. Note that $g(x)$ is continuous on $[0,1]$. Since $[0,1]$ is bounded and closed, $[0,1]$ is compact. By Theorem 4.4.7, $g(x)$ is uniformly continuous on $[0,1]$. Therefore, $g(x)$ is also uniformly continuous on $(0,1)$.

(c) Yes. We extend the function $h(x)$ by defining $h(0)=0$, then it is clear that $h(x)$ is continuous on $[0,1]$. Since $[0,1]$ is bounded and closed, $[0,1]$ is compact. By Theorem 4.4.7, $h(x)$ is uniformly continuous on $[0,1]$. Therefore, $h(x)$ is also uniformly continuous on $(0,1)$.


Exercise 4.4.3

See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.2.


Exercise 4.4.4

(a) See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.4.

(b) True. We argue it by contradiction. Suppose $f(A)$ is not bounded. Then there exists a sequence $(x_n)$ such that $x_n\in A$ and $|f(x_{n+1})|-|f(x_n)|\geqslant 1$ for all $n\geqslant 1$ (this sequence can be obtained inductively). Since $A$ is bounded, by Theorem 2.5.5, then we have a convergent subsequence  $(x_{n_k})$. Let $y_{n_k}=x_{n_{k+1}}$, then\[|x_{n_k}-y_{n_k}|\to 0,\] but\[|f(x_{n_k})-f(y_{n_{k}})|\geqslant |f(y_{n_{k}})|-|f(x_{n_k})|\geqslant 1.\]Hence by Theorem 4.4.5, $f$ is not uniformly continuous, which is a contradiction. Therefore, we are done.

(c) False. Define a function $f(x)$ as follows,\[f(x)=\begin{cases}0,&\text{ if }x\in \mathbf Q;\\ 1, &\text{ otherwise. }\end{cases}\]Then $f(K)$ is always a finite set and hence compact. But $f$ is not continuous on $\mathbf R$.


Exercise 4.4.5

See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.7.


Exercise 4.4.6

(a) Let $f(x)=1/x$ for $x\in (0,1)$. Then $f$ is continuous on $(0,1)$. Let $x_n=1/n$, then $(x_n)$ is Cauchy. But $f(x_n)=n$, so the sequence $(f(x_n))$ is not Cauchy.

(b) Impossible. Since $f$ is uniformly continuous on $(0,1)$, for any $\epsilon>0$, there exists $\delta >0$ such that $|f(x)-f(y)|<\epsilon$ provided $|x-y|<\delta$. Since $(x_n)$ is Cauchy, there exists $N>0$ such that $|x_n-x_m|<\delta$ whenever $n,m>N$. Hence for $n,m>N$, we have\[|f(x_n)-f(x_m)|<\epsilon.\]This implies that $(f(x_n))$ is Cauchy.

(c) Impossible. Since $(x_n)$ is Cauchy, $(x_n)$ is bounded by Lemma 2.6.3. Let $|x_n|\leqslant M$ where $M>0$ for all $n\geqslant 1$. Then we can consider the restriction of $f$ on the interval $[0,M]$. Because $[0,M]$ is compact, by Theorem 4.4.7, $f$ is uniformly continuous on $[0,M]$. Using the same argument as in (b), you can show that $(f(x_n))$ is Cauchy.


Exercise 4.4.7

See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.8 (b).


Exercise 4.4.8

(a) Impossible. Because $f$ is continuous and $[0,1]$ is compact, by Theorem 4.4.1, the range is compact as well. But $(0,1)$ is not compact.

(b) Let $f(x)=\left|\sin\dfrac{1}{x}\right|$, see Example 4.2.6.

(c) Let\[f(x)=\frac{|\sin\frac{1}{x}|+x}{1+2x}.\]


Exercise 4.4.9

See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.9.


Exercise 4.4.10

The functions $f(x)+g(x)$ and $f(g(x))$ are necessarily uniformly continuous.

$f(x)+g(x)$ is uniformly continuous. This follows from the triangle inequality. Let $\epsilon>0$. Because $f(x)$ and $g(x)$ are uniformly continuous on $A$, there exist $\delta_1>0$ and $\delta_2>0$ such that $$|f(x_1)-f(y_1)|<\epsilon/2,\quad |g(x_2)-g(y_2)|<\epsilon/2$$for all $|x_1-y_1|<\delta_1$ and $|x_2-y_2|<\delta_2$ on $A$. Let $\delta=\min\{\delta_1,\delta_2\}$. Then for all $x,y\in A$ and $|x-y|<\delta$, we have\begin{align*}|(f(x)+g(x))-(f(y)+g(y))|&\leqslant |f(x)-f(y)|+|g(x)-g(y)|\\ &< \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.\end{align*}Thus $f(x)+g(x)$ is uniformly continuous.

$f(x)g(x)$ may bot be uniformly continuous. Let $f(x)=g(x)=x$, then $f(x)g(x)=x^2$ is not uniformly continuous. Because we can choose $x_n=n$ and $y_n=n+1/n$, then\[|x_n-y_n|\to 0,\quad |x_n^2-y_n^2|=|2+\frac{1}{n^2}|\geqslant 2.\]By Theorem 4.4.5, $x^2$ is not uniformly continuous.

$f(x)/g(x)$ may bot be uniformly continuous. Let $f(x)=1$ and $g(x)=x$ defined on $(0,1)$, then $f(x)/g(x)=1/x$ is not uniformly continuous, see Exercise 4.4.2.

$f(g(x))$ is uniformly continuous. Let $\epsilon>0$. Since $f(x)$ is uniformly continuous, there exists $\delta_1>0$ such that for all $x,y\in A$ and $|x-y|<\delta_1$ we have $$|f(x)-f(y)|<\epsilon.$$ Since $g(x)$ is uniformly continuous, there exists $\delta>0$ such that for all $x,y\in A$ and $|x-y|<\delta$ we have $$|g(x)-g(y)|<\delta.$$ Therefore, if $x,y\in A$ and $|x-y|<\delta$, we have $|g(x)-g(y)|<\delta_1$ and therefore $$|f(g(x))-f(g(y))|<\epsilon.$$Thus $f(g(x))$ is uniformly continuous.


Exercise 4.4.11

See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.11.


Exercise 4.4.12

(a) (b) (c) may be false. Consider the example $f(x)=1$ for all $x\in\mathbf R$. Let $B=K=A$, then $f^{-1}(B)=f^{-1}(K)=f^{-1}(A)=\mathbf R$ which is clear not finite, not compact, not bounded.

(d) is true. To show $f^{-1}(F)$ is closed, it suffices to show that $f^{-1}(F)^c$ is open. We show that $f^{-1}(F)^c=f^{-1}(F^c)$. Then because $F^c$ is open, by definition of Exercise 4.4.11, $f^{-1}(F)^c=f^{-1}(F^c)$ is also open.

Now we show $f^{-1}(F)^c=f^{-1}(F^c)$. First we show $f^{-1}(F)^c\subset f^{-1}(F^c)$. Let $a\in f^{-1}(F)^c$, then $a\not\in f^{-1}(F)$, i.e. $f(a)\not\in F$. Therefore $f(a)\in F^c$, hence $a\in f^{-1}(F^c)$.

Then we show $f^{-1}(F^c)\subset f^{-1}(F)^c$. Let $a\in f^{-1}(F^c)$, then $f(a)\in F^c$. Thus $f(a)\not\in F$ and $a\not\in f^{-1}(F)$. Therefore $a\in f^{-1}(F)^c$, this implies $f^{-1}(F^c)\subset f^{-1}(F)^c$.

Now we have $f^{-1}(F)^c=f^{-1}(F^c)$.


Exercise 4.4.13

See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.13.

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

Leave a Reply

Close Menu