Elementary Classical Analysis, Marsden, 2nd ed, Chapter 4, Problem 21

Problem

Which of the following functions on $\mathbb{R}$ are uniformly continuous?

(a) $f(x)=\frac{1}{(x^2 + 1)}$,

(b) $f(x)=\cos^3x$,

(c) $f(x)=\frac{x^2}{(x^2+2)}$,

(d) $f(x)=x \sin x$.

Answer

(a) For $\epsilon > 0$, choose $\delta = \epsilon$. Then for $|x-a|<\delta$, $|f(x)-f(a)| = |\frac{1}{x^2+1}-\frac{1}{a^2+1}| = |\frac{a^2 - x^2}{(x^2+1)(a^2+1)}| = |x-a||\frac{a+x}{(x^2+1)(a^2+1)}| \leq |x-a||\frac{a}{a^2+1} + \frac{x}{x^2+1}| < \delta = \epsilon$.

(b) For $\epsilon > 0$, choose $\delta = \frac{1}{3}\epsilon$. Then for $|x-a|<\delta$, $|f(x)-f(a)| = |\cos^3x - \cos^3a| = |\cos x - \cos a||\cos^2x + \cos x \cos a + \cos^2 a| \leq |-2\sin(\frac{x+a}{2})\sin(\frac{x-a}{2})|\cdot 3 \leq 6 |\sin\frac{x-a}{2}| \leq 3|x-a| \leq 3\delta = \epsilon$.

(c) For $\epsilon > 0$, choose $\delta = \frac{\epsilon}{\sqrt{2}}$. Then for $|x-a|<\delta$, $|f(x) - g(x)| = |(1-\frac{2}{x^2+2}) - (1 - \frac{2}{a^2+2})| = |\frac{2}{x^2+2} - \frac{2}{a^2 + 2}| = |\frac{2(a^2 - x^2)}{(x^2+2)(a^2 + 2)}| = |x-a||\frac{2(a+x)}{(x^2+2)(a^2+2)}| \leq 2|x-a||\frac{x}{x^2+2} + \frac{a}{a^2 + 2}|\leq \sqrt{2} \delta = \epsilon$.

(d) We will prove that $f(x)$ is not uniformly continuous by contradiction. Let $\epsilon = 1$. For any $\delta >0$, choose $n > \frac{1}{\pi}\sin\delta/2,x=n\pi + \delta/2,a=n\pi$. Then $|x-a|=\delta/2<\delta$, but $|f(x)-f(a)|=|x\sin x - a\sin a|=|(n \pi + \delta/2)\sin\delta/2| \geq n\pi\sin\delta/2 > 1=\epsilon$.

Comments

Post a Comment

Popular posts from this blog

Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 5, Problem 15

Problem Suppose $a \in \mathbb{R}^1$, $f$ is a twice-differentiable real function on $(a,\infty)$, and $M_0,M_1,M_2$ are the least upper bounds of $|f(x)|,|f'(x)|,|f''(x)|$, respectively, on $(a,\infty)$. Prove that $$ M_1^2 \leq 4M_0M_2.$$ Does $M_1^2 \leq 4M_0M_2$ hold for vector-valued functions too? Answer For $x \in (a,\infty)$ and $h>0$, $f(x+2h)=f(x)+2f'(x)h + \frac{1}{2}f''(\xi)(2h)^2$ for some $\xi$ between $x$ and $x+2h$ by Theorem 5.15. Then $|f'(x)| \leq \frac{|f(x+2h) - f(x)|}{2h} + |f''(\xi)h| \leq \frac{M_0}{h} + M_2h$. This holds for all $x$ and $h$ and $$ \inf_h(\frac{M_0}{h} + M_2h) = 2\sqrt{M_0M_2}$$ Hence, we have $M_1 \leq 2\sqrt{M_0M_2}$  $\Rightarrow$  $M_1^2 \leq 4M_0M_2.$ This inequality also holds for vector-valued functions $f=(f_1,\cdots,f_k)$. We may assume that $M_1 \neq 0$. For a fixed point $y \in (a,\infty)$, define a function $$ g(x):=f_1'(y)f_1(x) + \cdots + f_k'(y)f_k(x). $$ Then we have $$|g'(x)|^2 ...
Read more

Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 2, Problem 14

Problem Give an example of an open cover of the segment $(0,1)$ which has no finite subcover. Answer Consider a family of sets $\{(1/n,1)\}_{n=2}^\infty$. For any $x \in (0,1)$, there exists $N \in \mathbb{N}$ such that $1/N < x$, which means $x \in (1/N,1)$. This means that $\{(1/n,1)\}_{n=2}^\infty$ covers $(0,1)$. Consider an arbitrary finite subcover $\{(1/n_1,1),\cdots,(1/n_K)\}$ of $\{(1/n,1)\}_{n=2}^\infty$. Then $\bigcup_{k=1}^K(1/n_k,1)=(1/n_\ast,1)$ where $n_\ast=\max(n_1,\cdots,n_K)$, which cannot cover $(0,1)$.
Read more

Elementary Classical Analysis, Marsden, 2nd ed, Chapter 4, Problem 28

Problem Let $f:(0,1) \rightarrow \mathbb{R}$ be uniformly continuous. Must $f$ be bounded? Answer Since $f$ is a uniformly continuous, there exists a $\delta >0$ such that $|x-y|<\delta$  $\Rightarrow$  $|f(x)-f(y)|<1$. Suppose $f$ is not bounded. Then, we can choose a sequence $x_n \in (0,1)$ such that $f(x_{n+1}) > f(x_n) + 1$. Since $x_n$ is a sequence in the compact set $[0,1]$, there exists a convergent subsequence $x_{n_k}$. Since $x_{n_k}$ is a Cauchy sequence, there exists $K \in \mathbb{N}$ such that $|x_{n_k} - x_{n_l}|<\delta$ for all $k,l \geq K$. However, $|f(x_{n_K}) - f(x_{n_{K+1}})|>1$, which is contradiction.
Read more