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

Problem

For any two real sequences $\{a_n\}$, $\{b_n\}$, prove that

$$ \limsup_{n \rightarrow \infty} (a_n + b_n) \leq \limsup_{n \rightarrow \infty} a_n + \limsup_{n \rightarrow \infty} b_n $$

Answer

Note that there exists $n_k$ such that $a_{n_k} + b_{n_k} \rightarrow \limsup_{n \rightarrow \infty} (a_n + b_n)$.

$\limsup_{n \rightarrow \infty} (a_n + b_n) = \lim_{k \rightarrow \infty} (a_{n_k} + b_{n_k}) \leq \lim_{k \rightarrow \infty} (\sup_{m \geq k}a_{n_m} + \sup_{m \geq k}b_{n_m}) = \lim_{k \rightarrow \infty}\sup_{m \geq k}a_{n_m} + \lim_{k \rightarrow \infty}\sup_{m \geq k}b_{n_m} = \limsup_{n \rightarrow \infty} a_n + \limsup_{n \rightarrow \infty} b_n$.

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

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

Problem Suppose $g$ is a real function on $\mathbb{R}^1$, with bounded derivative (say $|g'| \leq M$). Fix $\epsilon >0$, and define $f(x) = x + \epsilon g(x)$. Prove that $f$ is one-to-one if $\epsilon$ is small enough. (A set of admissible values of $\epsilon$ can be determined which depends only on $M$.) Answer It is enough to show that $f'>0$ for a small $\epsilon>0$. Note that $$f'(x) = 1+ \epsilon g'(x) \geq 1 - \epsilon M.$$ Hence, if we choose $\epsilon < 1/M$, $f$ is one-to-one.
Read more