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

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