Elementary Classical Analysis, Marsden, 2nd ed, Introduction, Problem 3

Problem

Let $f:A \rightarrow B$ be a function, $C_1,C_2 \subset B$, and $D_1,D_2 \subset A$. Prove

(a) $f^{-1}(C_1 \cup C_2) = f^{-1}(C_2) \cup f^{-1}(C_2)$

(b) $f(D_1 \cup D_2) = f(D_1) \cup f(D_2)$

(c) $f^{-1}(C_1 \cap C_2) = f^{-1}(C_1) \cap f^{-1}(C_2)$

(d) $f(D_1 \cap D_2) \subset f(D_1) \cap f(D_2)$


Answer

(a) $x \in f^{-1}(C_1 \cup C_2)$  $\Leftrightarrow$  there exists $y \in C_1 \cup C_2$ such that $y = f(x)$  $\Leftrightarrow$  there exists $y_1 \in C_1$ such that $y_1 = f(x)$ or there exists $y_2 \in C_2$ such that $y_2 = f(x)$  $\Leftrightarrow$  $x \in f^{-1}(C_1)$ or $x \in f^{-1}(C_2)$  $\Leftrightarrow$  $x \in f^{-1}(C_1) \cup f^{-1}(C_2)$

(b) $y \in f(D_1 \cup D_2)$  $\Leftrightarrow$  $y=f(x)$ for some $x \in D_1 \cup D_2$  $\Leftrightarrow$  $y=f(x_1)$ for some $x_1 \in D_1$ or $y=f(x_2)$ for some $x_2 \in D_2$  $\Leftrightarrow$  $y \in f(D_1)$ or $y \in f(D_2)$  $\Leftrightarrow$  $y \in f(D_1) \cup f(D_2)$

(c) $x \in f^{-1}(C_1 \cap C_2)$  $\Rightarrow$  there exists $y \in C_1 \cap C_2$ such that $y = f(x)$  $\Rightarrow$  $x \in f^{-1}(C_1)$ and $x \in f^{-1}(C_2)$  $\Rightarrow$  $x \in f^{-1}(C_1) \cap f^{-1}(C_2)$

$x \in f^{-1}(C_1) \cap f^{-1}(C_2)$  $\Rightarrow$  $x \in f^{-1}(C_1)$ and $x \in f^{-1}(C_2)$  $\Rightarrow$  there exists $y_1 \in C_1$ and $y_2 \in C_2$ such that $y_1 = f(x) = y_2$  $\Rightarrow$  there exists $y \in C_1 \cap C_2$ such that $y = f(x)$  $\Rightarrow$  $x \in f^{-1}(C_1 \cap C_2)$

(d) $y \in f(D_1 \cap D_2)$  $\Rightarrow$  $y=f(x)$ for $x \in D_1 \cap D_2$  $\Rightarrow$  $y \in f(D_1)$ and $y \in f(D_2)$  $\Rightarrow$  $y \in f(D_1) \cap f(D_2)$

Comments

Post a Comment

Popular posts from this blog

Elementary Classical Analysis, Marsden, 2nd ed, Chapter 2, Problem 1

Problem Discuss whether the following sets are open or closed: (a) (1,2) in $\mathbb{R}$ (b) [2,3] in $\mathbb{R}$ (c) $\bigcap_{n=1}^\infty [-1,1/n)$ in $\mathbb{R}$ (d) $\mathbb{R}^n$ in $\mathbb{R}^n$ (e) A hyperplane in $\mathbb{R}^n$ (f) $\{r \in (0,1) \;|\; \text{$r$ is rational} \}$ in $\mathbb{R}$ (g) $\{(x,y) \in \mathbb{R}^2 \;|\; 0<x \leq 1\}$ in $\mathbb{R}^2$ (h) $\{x \in \mathbb{R}^n \;|\; \|x\|=1\}$ in $\mathbb{R}^n$. Answer (a) Open: For $x \in (1,2)$, we can choose $\epsilon = \min(x-1,2-x)$ so that $D(x,\epsilon) \subset (1,2)$. (b) Closed: For $x \in [2,3]^c$, we can choose $\epsilon = \min(2-x,x-3)$ so that $D(x,\epsilon) \subset [2,3]^c$. (c) Closed: We first prove that $[-1,0] = A:=\bigcap_{n=1}^\infty [-1,1/n)$. Since $[-1,0] \subset [-1,1/n)$ for all $n \in \mathbb{N}$, $[-1,0] \subset A$. To prove $A \subset [-1,0]$, we assume that $x \in A$. Then obviously, $x \geq -1$. If $x > 0$, then there exists $n \in \mathbb{N}$ such that $x > 1/n$, which means
Read more

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