Elementary Classical Analysis, Marsden, 2nd ed, Chapter 6, Problem 15

Problem

If $f:\mathbb{R} \rightarrow \mathbb{R}$ be differentiable. Assume there is no $x \in \mathbb{R}$ such that $f(x)=0=f'(x)$. Show that $S=\{x\;|\;0\leq x \leq 1,f(x)=0\}$ is finite.

Answer

Note that $S=f^{-1}(0)$ is closed since $f$ is continuous and $\{0\}$ is closed. Also, $S$ is bounded since $S \subset [0,1]$. These imply $S$ is compact. If $S$ is infinite, there is a sequence $\{x_n\} \subset S$ and its subsequence $x_{n_k} \rightarrow x \in S$. Then

$$ f'(x) = \lim_{k \rightarrow \infty} \frac{f(x) - f(x_{n_k})}{x-x_{n_k}} =0. $$

Hence, $S$ has to be finite.

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