%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%% exercise set 6 - solutions %%% October 16, 2009 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \documentclass{amsart} \usepackage{graphicx} \usepackage{amssymb} \usepackage{epstopdf} \usepackage{nicefrac} \DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png} % \input hw.sty \textwidth = 6 in \textheight = 9 in \oddsidemargin = 0.0 in \evensidemargin = 0.0 in \topmargin = 0.0 in \headheight = 0.0 in \headsep = 0.0 in \parskip = 6pt \parindent = 0.0in % \hoffset=-.7in % \voffset=-.7in \newcommand{\ds}{\displaystyle} \newcommand{\mA}{{\mathbb A}} \newcommand{\mB}{{\mathbb B}} \newcommand{\mD}{{\mathbb D}} \newcommand{\mN}{{\mathbb N}} \newcommand{\mQ}{{\mathbb Q}} \newcommand{\mR}{{\mathbb R}} \newcommand{\mS}{{\mathbb S}} \newcommand{\mZ}{{\mathbb Z}} \newcommand{\cA}{{\mathcal A}} \newcommand{\cB}{{\mathcal B}} \newcommand{\cC}{{\mathcal C}} \newcommand{\cE}{{\mathcal E}} \newcommand{\cI}{{\mathcal I}} \newcommand{\cP}{{\mathcal P}} \newcommand{\cT}{{\mathcal T}} \newcommand{\cU}{{\mathcal U}} \begin{document} \begin{center} Math 445, Fall 2009 \hfill Exercise Set \#6 \hfill Solutions \end{center} \bigskip {\bf 1.} [\#1, page 93] ~ Prove that the space $X = \cB(\mN, \mR)$ of bounded sequences with the ``sup norm'' is complete. [This is Example 7, page 119 of Appendix 1 in the text.] The metric is defined by, for $x = \{a_n\} \in \cB(\mN, \mR)$ and $y = \{b_n\} \in \cB(\mN, \mR)$ $$D(x,y) ~ = ~ \sup_{n \in \mN} ~ |a_n - b_n| $$ \proof A sequence $x = \{a_n \mid n = 1,2, \ldots \}$ can be written in function notation as $x \colon \mN \to \mR$ where $x(n) = a_n$. A sequence of sequences $\{x_i \mid i = 1,2, \ldots\}$ assigns to each index $i$ the sequence $x_i = \{a_n^{(i)} \mid n = 1,2, \ldots\}$. Let $\{x_i \mid i = 1,2, \ldots\} \subset \cB(\mN, \mR)$ be a Cauchy sequence in $ \cB(\mN, \mR)$. We show that it converges to a sequence $x_*$ and that $x_*$ is bounded, so that $x_* \in \cB(\mN, \mR)$. To define the sequence $x_*$ we need to define $x_*(m)$ for each $m \geq 1$. The claim is that for $m$ fixed, the sequence $\{x_i(m) \mid i =1,2, \ldots \}$ is Cauchy. Given $\epsilon > 0$, there exists $N_{\epsilon}$ such that $i, j \geq N_{\epsilon}$ implies $D(x_i, x_j) < \epsilon$. By the definitions, this implies the Cauchy estimate $$ | x_i(m) - x_j(m) | ~ \leq ~ \sup_{n \in \mN} ~ |x_i(n) - x_j(n)| ~ = ~ D(x_i , x_j) ~ < ~ \epsilon$$ Set $\ds x_*(m) = \lim_{i \to \infty} ~ x_i(m)$. This defines the limit sequence $x_* = \{x_*(n) \mid n =1,2, \ldots \}$. We next must show that $x_*$ is a bounded sequence. Fix $\epsilon > 0$ say $\epsilon =1$. Then by the Cauchy condition on $\{x_i\}$ there exists $N_1$ such that $i, j \geq N_1$ implies $D(x_i, x_j) < 1$. Pick any $i_1 \geq N_1$. Let $C = \| x_{i_1} \| = D(0, x_{i_1}) < \infty$. Then for all $j \geq N_1$ we have $D(x_{i_1} , x_j) < 1$ which implies by the Triangle Inequality that $$C - 1 = \| x_{i_1} \| -1 ~ < ~ \| x_{j} \| ~ < ~ \| x_{i_1} \| +1 = C + 1$$ Thus, for each $m$ we have $C -1 < |x_j(m) | < C+1$ for all $j \geq N_1$. Thus, the limit $x_*(m)$ satisfies $C-1 \leq x_*(m) \leq C+1$. This implies $\|x_* \| \leq C+1$ so $x_* \in \cB(\mN, \mR)$. \endproof \bigskip {\bf 2.} [\#5, page 93] ~ If every countable closed subset of a metric space $M$ is complete, prove that $M$ is complete. \proof First, a closed subset $F \subset M$ of a complete metric space is complete. [Note - we proved this in class, but no problem to show it again.] Let $\{x_n\} \subset F$ be a Cauchy sequence, then $M$ complete implies there is a limit $x_n \to x_* \in M$. Since $F$ is closed, it contains all of its limit points, so $x_* \in F$. Thus, every Cauchy sequence in $F$ has a limit in $F$, showing that $F$ is complete. We show the converse by contradiction: assume that $M$ is not complete, then we construct a countable closed subset $F \subset M$ which is not complete. Let $\{x_n\} \subset M$ be a Cauchy sequence for which there is no limit point. This means that every $y \in M$ is not a limit point of $\{x_n\}$, so for some $\epsilon_y > 0$ the intersection $B(y,\epsilon_y) \cap \{x_n\}$ is a finite set. Claim: the set $F = \{x_n\}$ is closed. Let $y \in M - F$. Let $\epsilon_y > 0$ be such that the intersection $B(y,\epsilon_y) \cap \{x_n\}$ is a finite set. Since $y \not\in F$ the lower bound $\delta_y = \min \{D(y, x_n) \mid n =1,2, \ldots\} > 0$. Thus, $B(y, \delta_y) \subset M - F$. This shows that $M-F$ is open. Now $F$ is closed and contains a Cauchy sequence $\{x_n\}$ which has no limit point, so $F$ is not complete, a contradiction. \endproof \bigskip {\bf 3.} [\#6, page 93] ~ Let $\{M, d\}$ be a metric space. If for every $u \in M$ and $\epsilon > 0$, the closed ball $D(u,\epsilon) = \{ y \in M \mid d(u,y) \leq \epsilon\}$ is complete, prove that $\{M, d\}$ is complete. \proof Let $\{x_n\} \subset M$ be a Cauchy sequence. Set $\epsilon =1$, the by the Cauchy condition there exists $N_1$ such that $m,n \geq N_1$ implies $D(x_m , x_m) < 1$. In particular, for all $n \geq N_1$ we have $D(x_{N_1} , x_n) < 1$ That is, $$\{ x_n \mid n \geq N_1\} \subset B(x_{N_1} , 1) \subset D(x_{N_1},1)$$ It is given that $D(x_{N_1},1)$ is complete, so there exists $x_* \in D(x_{N_1},1)$ such that $x_n \to x_*$. But then $x_n \to x_* \in M$ also, so $M$ is complete. \endproof \bigskip {\bf 4.} [\#11, page 93] ~ In the space of real numbers $\mR$, give an example of a descending sequence of non-empty closed sets with empty intersection. That is, find $F_1 \supset F_2 \supset \cdots $ where each $F_n \subset \mR$ is closed, and $\ds \bigcap_{n=1}^{\infty} ~ F_n ~ = ~ \emptyset$. \proof There are lots of examples, but no examples that are \emph{bounded}. For example, let $F_n = [n, \infty)$. The complement $\mR - F_n = (- \infty, n)$ is open, so $F_n$ is closed. \endproof {\bf 5.} [\#12, page 93] ~ The following functions are continuous from the real numbers to the real numbers. Which are \emph{uniformly} continuous? a) $ f(x) = x^2$ \quad b) $g(x) = |x|$ \quad c) $\ds h(x) = \frac{1}{1 + x^2}$ \proof a) $ f(x) = x^2$ is not uniformly continuous. Given any $\delta > 0$ and $x > 0$ consider $$|f(x + \delta) - f(x)| = | (x + \delta)^2 - x^2| = 2x \delta \longrightarrow \infty$$ That is, for $\epsilon > 0$ there is no $\delta > 0$ so that $|f(x + \delta) - f(x)| < \epsilon$ for all $x \in \mR$. b) $g(x) = |x|$ is uniformly continuous. Given $\epsilon > 0$ set $\delta = \epsilon$. Then $|x-y| < \delta$ implies (by the Triangle Inequality) that $$|g(x) - g(y) | = | |x| - |y| | \leq |x -y | < \delta = \epsilon$$. c) $\ds h(x) = \frac{1}{1 + x^2}$ is is uniformly continuous. Given $x,y \in \mR$, consider $$ | h(x) - h(y) | = \left| \frac{1}{1 + x^2} - \frac{1}{1 + y^2} \right| = \frac{| y^2 - x^2 |}{(1 + x^2)(1 + y^2)} = \frac{| y - x | \cdot |x+y|}{(1 + x^2)(1 + y^2)} $$ The claim is that there is a finite upper bound $$C = \sup \left\{ \frac {|x+y|}{(1 + x^2)(1 + y^2)} \mid x,y \in \mR \right\} < \infty$$ Then for $\epsilon > 0$ set $\delta = \epsilon/C$. Th show that $C$ is finite is just standard calculus. The quotient is bounded by $|x+y|$ for $|x| \leq 1$ or $|y| \leq 1$. And if $|x| \geq 1$ and $|y| \geq 1$ then the quotient is bounded by $|x+y|/(x^2 + y^2) \leq 1$. \endproof {\bf 6.} [\#13, page 93] ~ Let $\mR$ be the real line with the standard metric, $d(x,y) = |x-y|$. Let $f \colon \mR \to \mR$ be a function which has derivative $f'(x)$ for every $x \in \mR$. Assume there exists $K \geq 0$ such that $| f'(x)| \leq K$ for all $x \in \mR$. Prove that $f$ is uniformly continuous. [Hint: Mean Value Theorem] \proof Let $x < y \in \mR$ then the Mean Value Theorem states there exists $x < \xi < y$ for which $f(y) - f(x) = f'(\xi) \cdot (y-x)$. Then $$| f(y) - f(x)| = | f'(\xi) \cdot (y-x)| \leq K \cdot |y-x|$$ Given $\epsilon > 0$ set $\delta = \epsilon/K$. \endproof \bigskip \vfill \end{document}