Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 3, Problem 8

Get link
Facebook
X
Pinterest
Email
Other Apps

June 05, 2022

Problem

If $\sum a_n$ converges, and if $\{b_n\}$ is monotonic and bounded, prove that $\sum a_nb_n$ converges.

Answer

Since $b_n$ is monotonic and bounded, $b_n \rightarrow b$ . If $b_n$ is increasing, then $\sum a_n (b-b_n)$ converges by Theorem 3.42. Then $\sum a_nb_n = \sum a_n b - \sum a_n (b - b_n)$ also converges.

If $b_n$ is decreasing, then $\sum a_n (b_n-b)$ converges by Theorem 3.42. Then $\sum a_nb_n = \sum a_n b + \sum a_n (b_n - b)$ also converges.

Get link
Facebook
X
Pinterest
Email
Other Apps

Comments

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

April 18, 2022

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

$(0,1)$ .

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

June 02, 2022

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 ...

Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 4, Problem 4

June 02, 2022

Problem Let

$f$ and

$g$ be continuous mappings of a metric space

$X$ into a metric space

$Y$ , and let

$E$ be a dense subset of

$X$ . Prove that

$f(E)$ is dense in

$f(X)$ . If

$g(p)=f(p)$ for all

$p \in E$ , prove that

$g(p)=f(p)$ for all

$p \in X$ . (In other words, a continuous mapping is determined by its values on a dense subset of its domain.) Answer For

$y \in f(X)$ , there exists

$x \in X$ such that

$y=f(x)$ . For

$\epsilon > 0$ , there exists

$\delta > 0$ such that

$d_X(x,x') < \delta$

$\Rightarrow$

$d_Y(f(x),f(x'))<\epsilon$ . Since

$E$ is dense in

$X$ , there exists

$z \in E$ such that

$z \in N_\delta(x)$ , which implies

$f(x) \in N_\epsilon(f(z))$ . This means

$f(E)$ is dense in

$f(X)$ . For

$p \in X$ and

$\epsilon >0$ , there exists

$\delta > 0$ such that

$d_X(x,x') < \delta$

$\Rightarrow$

$d_Y(f(x),f(x'))<\epsilon/2, d_Y(g(x),g(x'))<\epsilon/2$ . For

$q \in E$ satisfying

$q \in N_\delta(p)$ , $d_Y(g(p),f(p)) \leq d_Y(g(p),g(q)) + d_Y(g(q)...

Search This Blog

Calculus TA