- Let $f:[a, b] \rightarrow \mathbb{R}$ be continuous. Suppose that $f(a)<f(b)$ and that $f$ is a 1-1 mapping. Use Intermediate Value Theorem to show that $f(a)<f(x)<f(b)$ for all $x \in(a, b)$. Hence or otherwise prove that $f$ is strictly increasing on $[a,b]$.

*Solution.*

Since $f$ is 1-1, $f(x)≠f(a)$ and $f(x)≠f(b)$. If $f(x)<f(a)$, by intermediate value theorem, $∃x_0∈(x,b),f(x_0)=f(a)$, a contradiction; if $f(x)>f(b)$, by intermediate value theorem, $∃x_0∈(a,x):f(x_0)=f(b)$, a contradiction. Therefore $f(a)<f(x)<f(b)$.

For any $x_1,x_2∈[a,b]$, $x_1<x_2$, since $f$ is continuous on $(a,x_2)$, by the same argument as above, $f(x_1)<f(x_2)$, so $f$ is strictly increasing on $[a,b]$. - The function $g$ is defined by $g(x)=\frac{x}{1-|x|}$ for $-1<x<1$. Show that $g$ is 1-1, find $g^{-1}$ and determine its domain. Are $g$ and $g^{-1}$ continuous?

*Solution.*

$g(x)>0$ for $0<x<1$ and $g(x)<0$ for $-1<x<0$. $g$ is strictly increasing on (-1,0] and [0,1), therefore $g$ is strictly increasing on (-1,1), so $g$ is 1-1. The domain of $g^{-1}$ is $\mathbb R$.

Let $y=\frac{x}{1-|x|}$, if $y>0$, $y=\frac x{1-x}$⇒$x=\frac y{1+y}$; if $y<0$, $y=\frac x{1+x}$⇒$x=\frac y{1-y}$, therefore $g^{-1}(x)=\frac x{1+|x|}$.

$g$ and $g^{-1}$ are continuous.

Error! Click to view log. - (a) Which of the following real-valued functions $f$, defined on [-1,1] by (i) and (ii) below, have inverses $f^{-1}:[f(-1),f(1)] \rightarrow[-1,1]$? Which have continuous inverses? Given brief reasons.

(i) $f(x)=(x+1)^2$;

(ii) $f(x)=x$ for $x \in[-1,0]$ and $f(x)=x+1$ for $x \in(0,1]$.

(b) Let $f:(a, b] \rightarrow(c, d]$ be strictly increasing and onto, where $a<b, c<d$, and $b,d$ are two real numbers. Show that $f$ has a continuous inverse mapping from $(c, d]$ to $(a, b]$.

*Solution.*

(a)(i) $f(x)$ has continuous inverse $f^{-1}(x)=\sqrt x-1\;(0≤x≤4)$.

(ii)$f(x)$ doesn't have continuous inverse, because $f([-1,1])=[-1,0]∪[1,2]≠[f(-1),f(1)]$.

(b)Since $f$ is strictly increasing and onto, $f$ is a bijection, $f^{-1}$ is strictly increasing, we need to prove $f^{-1}$ is continuous. If there were $x_0∈(c,d]$ such that $f^{-1}$ were not continuous at $y_0$, let $x_1=f^{-1}(y_0-),x_2=f^{-1}(y_0+)$, then $x_1<x_2$, so $y_0=f(x_1)<f(x_2)=y_0$, a contradiction. - (a) Let $a>0$. Show that $f(x)=\frac1x$ is uniformly continuous on $[a, \infty)$.

(b) Show that $f(x)=x^2$ is not uniformly continuous on $[0, \infty)$.

*Proof.*

(a) For $x,y∈[a,∞)$ and $ε>0$, if $|x-y|<a^2ε$, then $\left|\frac1x-\frac1y\right|=\frac{|x-y|}{xy}<\frac{a^2ε}{a^2}=ε$, therefore $f$ is uniformly continuous on $[a,∞)$.

(b) For every $n = 1,2,⋯$, let $x_n = n+\frac1n$ and $y_n = n$ then $x_n -y_n =\frac1n$ tends to zero as $n → ∞$, but $\left|f\left(x_{n}\right)-f\left(y_{n}\right)\right|=2+\frac{1}{n^{2}}>2$. so $f$ is not uniformly continuous. - (a) Suppose that $h$ is continuous on $[0, \infty)$ and suppose that $h$ is uniformly continuous on $[a, \infty)$ for some positive number $a$. Show that $h$ is uniformly continuous on $[0, \infty)$.

(b) Show that $f(x)=x^{1 / 3}$ is uniformly continuous on $\mathbb{R}$. Is it Lipschitz continuous?

*Proof.*

(a)Because $h$ is continuous on $[0,a]$, it is uniformly continuous on $[0,a]$. $∀ε>0$, for $x,y∈[0,a]$, we can find $δ_1$ such that $|x-y|<δ_1$ implies $|h(x)-h(y)|<ε/2$; for $x,y∈[a,∞)$, we can find $δ_2$ such that $|x-y|<δ_2$ implies $|h(x)-h(y)|<ε/2$; for $x<a<y$, $|x-y|<\min\{δ_1,δ_2\}$ implies $|x-a|<δ_1$ and $|y-a|<δ_2$, so $|h(x)-h(y)|≤|h(x)-h(a)|+|h(y)-h(a)|<ε/2+ε/2=ε$, therefore $h$ is uniformly continuous on $[a,∞)$.

(b)First we show that $\sqrt[3]x-\sqrt[3]y<\sqrt[3]{x-y}$ for $x>y$. Cubing both sides, $x-y-3\sqrt[3]{xy}(\sqrt[3]x-\sqrt[3]y)<x-y$ is true.

∀ε>0, let $δ=\sqrt[3]{ε}$,∀$x_1>x_2$ such that $x_1-x_2<δ$: using above inequality, $\sqrt[3]{x_1}-\sqrt[3]{x_2}<\sqrt[3]{x_1-x_2}<𝜀$, so $f$ is uniformly continuous on $\mathbb R$.

Suppose $f$ is Lipschitz continuous, then there is a constant $M$ such that $|f(x)-f(y)|≤M|x-y|$ for any $x,y∈\mathbb R$, but $x^{1/3}-y^{1/3}= \frac{x-y}{x^{2/3}+(xy)^{1/3}+y^{2/3}}>M(x-y)$ for $0<y<x<\left(3M\right)^{-3/2}$, a contradiction, so $f$ is not Lipschitz continuous. - (a) Let $a<b$ be two real numbers. Suppose that $f:(a, b] \rightarrow \mathbb{R}$ is continuous and suppose that the limit of $f$ as $x \rightarrow a$ exists. Show that $f$ is uniformly continuous on $(a, b]$.

(b) Suppose now $g:(a, b] \rightarrow \mathbb{R}$ is uniformly continuous.

(i) Show that if $\left(x_{n}\right) \subset(a, b]$ is a Cauchy sequence, then $\left(g\left(x_{n}\right)\right)$ is also a Cauchy sequence.

(ii) Suppose $x_{n} \in(a, b]$ and $y_{n} \in(a, b]$ (where $n=1,2, \cdots$ ) are two sequences and $x_{n} \rightarrow a, y_{n} \rightarrow a$ as $n \rightarrow \infty$. Show that $\left(g\left(x_{n}\right)\right)$ and $\left(g\left(y_{n}\right)\right)$ converge to the same limit. Deduce that $g(x)$ has a limit as $x \rightarrow a$.

*Proof.*

(a)Let $f^*(x)=\begin{cases}f(x),&a<x≤b\\\lim_{x\to a}f(x),&x=a\end{cases}$, then $f^*$ is continuous on $[a,b]$, so $f^*$ is uniformly continuous on $[a,b]$, so $f$ is uniformly continuous on $(a,b]$.

(b)(i)For any given ε>0, since $g$ is uniformly continuous on $(a,b]$, there is δ>0, whenever $x,y∈(a,b]$ such that $|x-y|<δ$ we have $|g(x)-g(y)|<ε$. Since $(x_n)$ is Cauchy, there is $N>0$ such that ∀$n,m≥N,|x_n-x_m|<δ$. Hence $|g(x_n)-g(x_m)|<ε∀n,m≥N$. Therefore $(g(x_n))$ is a Cauchy sequence.

(ii)Since $g$ is uniformly continuous, $∀ε>0∃δ>0:$ for any $x,y∈(a,b]$ such that $|x-y|<δ$, $|g(x)-g(y)|<ε$. Since $x_n\rightarrow a,y_n\rightarrow a$ as $n \rightarrow \infty$, $∃N_1∈\mathbb N∀n>N_1:x_n-a<δ$ and $∃N_2∈\mathbb N∀n>N_2:y_n-a<δ$, therefore $|x_n-y_n|<δ∀n>\max\{N_1,N_2\}$⇒$|g(x_n)-g(y_n)|<ε$, therefore $\lim g(x_n)-g(y_n)=0$. By Theorem 1.1.17, $g(x)$ has a limit as $x→a$.

Another approach: let $z_m=\begin{cases}y_n&m=2n\\x_n&m=2n-1\end{cases}$. Even and odd subsequences of $z_n$ both converge to $a$, therefore $z_n$ converges to $a$. By part (a), $g(z_n)$ is a Cauchy sequence, therefore even and odd subsequences of $g(z_n)$ converges to same limit, therefore $g(x_n)$ and $g(y_n)$ converge to same limit.