%amslatex file
%\documentstyle[12pt,amssymb]{amsart}
\documentclass[12pt]{amsart} \usepackage{amssymb} \usepackage{mathabx} \usepackage{bm} \usepackage{wasysym} \usepackage{graphicx}
\pagenumbering{gobble}
\newenvironment{pf}{\proof[\proofname]}{\endproof}

\renewcommand\baselinestretch{1.1}
\renewcommand\theequation{\arabic{section}.\arabic{equation}}

\newcommand{\wrlab}[1]{\label{#1}}
\renewcommand{\arraystretch}{1}
\newtheorem{thm}{THEOREM}[section]
\newtheorem{thmmain}{\tref{thm-main}} \renewcommand{\thethmmain}{}
\newtheorem{cormain}{\cref{cor-main}} \renewcommand{\thecormain}{}
\newtheorem{lem}[thm]{LEMMA} \newtheorem{rmk}[thm]{REMARK}
\newtheorem{cor}[thm]{COROLLARY} \newtheorem{prop}[thm]{PROPOSITION}
\newtheorem{conj}[thm]{CONJECTURE} \newtheorem{quest}[thm]{QUESTION}
\newtheorem{hyp}[thm]{HYPOTHESES} \newtheorem{thm*}{THEOREM}[]
\newtheorem{defn}[thm]{DEFINITION} \newtheorem{ex}[thm]{EXAMPLE}

\newcommand{\secref}[1]{\S\ref{#1}}

\newcommand{\tref}[1]{Theorem~\ref{#1}}
\newcommand{\cref}[1]{Corollary~\ref{#1}}
\newcommand{\pref}[1]{Proposition~\ref{#1}}
\newcommand{\dref}[1]{Definition~\ref{#1}}
\newcommand{\lref}[1]{Lemma~\ref{#1}}
\newcommand{\rref}[1]{Remark~\ref{#1}}
\newcommand{\exref}[1]{Example~\ref{#1}}

\newfam\msbfam 
\font\tenmsb=msbm10 \textfont\msbfam=\tenmsb  
\font\sevenmsb=msbm7 \scriptfont\msbfam=\sevenmsb

\def\N{{\mathbb N}} \def\Z{{\mathbb Z}} \def\Q{{\mathbb Q}}
\def\R{{\mathbb R}} \def\C{{\mathbb C}} \def\T{{\mathbb T}}
\def\P{{\mathbb P}}

\def\bfzero{{\bf0}}

% CAL 
\def\Cal{\mathcal} \def\scra{{\mathcal A}} \def\scrb{{\mathcal B}}
\def\scrd{{\mathcal D}} \def\scrf{{\mathcal F}} \def\scrg{{\mathcal G}}
\def\scri{{\mathcal I}} \def\scrk{{\mathcal K}} \def\scrl{{\mathcal L}}
\def\scrm{{\mathcal M}} \def\scrp{{\mathcal P}} \def\scrq{{\mathcal Q}}
\def\scrr{{\mathcal R}} \def\scrs{{\mathcal S}} \def\scrt{{\mathcal T}}
\def\scru{{\mathcal U}} \def\scrv{{\mathcal V}} \def\scrw{{\mathcal W}}
\def\scre{{\mathcal E}} \def\scrx{{\mathcal X}} \def\scry{{\mathcal Y}}
\def\scrz{{\mathcal Z}} \def\scrh{{\mathcal H}} \def\scrc{{\mathcal C}}
\def\scrn{{\mathcal N}} \def\scro{{\mathcal O}}

\newcommand\scrlito{
  \mathchoice
    {{\scriptstyle\mathcal{O}}}% \displaystyle
    {{\scriptstyle\mathcal{O}}}% \textstyle
    {{\scriptscriptstyle\mathcal{O}}}% \scriptstyle
    {\scalebox{.7}{$\scriptscriptstyle\mathcal{O}$}}%\scriptscriptstyle
  }

\def\ds{\displaystyle}
\def\drta{\dashrightarrow}

%operators
\def\OB{\hbox{OB}}
\def\opstar{\mathop{*}}
\def\Ell{{E\ell\ell}}
\def\Cl{{\hbox{C}\ell}}
\def\fincap{{\hbox{{\rm fin}}\,\cap}}
\def\fincup{{\hbox{{\rm fin}}\,\cup}}
\def\id{{\rm id}}
\def\LP{{\rm LP}}
\def\LIMS{{\rm LIMS}}
\def\LINS{{\rm LINS}}
\def\opLIMS{\mathop{\LIMS}}
\def\diam{{\rm diam}}
\def\linf{{\rm liminf\,}}
\def\lsup{{\rm limsup\,}}
\def\acc{{\rm acc\,}}
\def\dom{{\rm dom\,}}
\def\im{{\rm im\,}}
\def\barR{{\overline{\R}}}
\def\add{{\hbox{add}}}
\def\mult{{\hbox{mult}}}
\def\dd{{\hbox{dd}}}
\def\sym{{\rm sym}}
\def\Sym{{\rm Sym}}
\def\as{{\rm as}}
\def\SBF{\hbox{SBF}}
\def\STF{\hbox{STF}}
\def\SM{\hbox{SM}}
\def\dct{\hbox{dct}}
\def\dlin{\hbox{dlin}}
\def\Int{{\rm Int}}
\def\IntD{{\rm IntD}}
\def\adj{{\rm adj}}
\def\LPD{{\rm LPD}}
\def\ELT{{\rm ELT}}
\def\slp{{\rm sl}}
\def\Det{\hbox{Det\,}}
\def\UB{{\rm UB}}
\def\LB{{\rm LB}}

% HAT
\def\hatT{{\widehat{T}}}
\def\bigo{{\widehat{\scro}}}

% CHECK
\def\ckT{{\widecheck{T}}}
\def\lito{{\widecheck{\scrlito}}}

\begin{document}

\centerline{\bf Solutions for MATH 4604 (Advanced Calculus II)}
\centerline{\bf Spring 2018}
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 14: Due on Tuesday 1 May}
\vskip.1in\hrule\vskip.1in\noindent
14-1. Let $W$ be a normed vector space, let $f:\R\drta W$ and let $p\in\R$.
Show: $(f_p^T)'(0)=f'(p)$.
\vskip.1in\noindent
{\it Proof:}
Let $g:=f_p^T$.
Want: $g'(0)=f'(p)$.
As $\ds{g'(0)=\lim_0\,SS_g^0}$ and $\ds{f'(p)=\lim_0\,SS_f^p}$,
it suffices to show: $SS_g^0=SS_f^p$.
Want:
$\forall h\in\R$, $(SS_g^0)(h)=(SS_f^p)(h)$.
Given $h\in\R$.
Want: $(SS_g^0)(h)=(SS_f^p)(h)$.

We have
\begin{eqnarray*}
[\,\,g(0+h)\,\,]\,\,-\,\,[\,\,g(0)\,\,]&=&[\,\,g(h)\,\,]\,\,-\,\,[\,\,g(0)\,\,]\\
&=&[\,\,(f_p^T)(h)\,\,]\,\,-\,\,[\,\,(f_p^T)(0)\,\,]\\
&=&[\,\,[f(p+h)]\,-\,[f(p)]\,\,]\\
&&\qquad-\,\,[\,\,[f(p+0)]\,\,-\,\,[f(p)]\,\,]\\
&=&[\,\,f(p+h)\,\,]\,\,-\,\,[\,\,f(p)\,\,].
\end{eqnarray*}
Then
\begin{eqnarray*}
(SS_g^0)(h)&=&\frac{[\,\,g(0+h)\,\,]\,\,-\,\,[\,\,g(0)\,\,]}{h}\\
&=&\frac{[\,\,f(p+h)\,\,]\,\,-\,\,[\,\,f(p)\,\,]}{h}\,\,\,=\,\,\,(SS_f^p)(h),
\end{eqnarray*}
as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
14-2. Let $V$, $W$ be finite dimensional vector spaces,
$\alpha,\beta:V\drta W$, $p\ge0$.
Assume: $\alpha=\beta$ near $0_V$ and $\alpha\in\lito_p(V,W)$.
Show: $\beta\in\lito_p(V,W)$.
\vskip.1in\noindent
{\it Proof:}
By assumption, choose $U\in\scrn_V(0_V)$
s.t.~$\alpha=\beta$ on $U$.
Also,
\begin{itemize}
\item[(A)]$\dom[\alpha]\in\scrn_V(0_V)$,
\item[(B)]$\alpha(0_V)=0_W$ \qquad and
\item[(C)]$\alpha$ is continuous at $0_V$.
\end{itemize}
Let $D:=\dom[\alpha]$.
By (A), we have $D\in\scrn_V(0_V)$.
As $D,U\in\scrn_V(0_V)$, we get $D\cap U\in\scrn_V(0_V)$.
We wish to show:
\begin{itemize}
\item[(1)]$\dom[\beta]\in\scrn_V(0_V)$,
\item[(2)]$\beta(0_V)=0_W$ \qquad and
\item[(3)]$\beta$ is continuous at $0_V$.
\end{itemize}

{\it Proof of (1):}
Since we have $D\cap U\in\scrn_V(0_V)$,
it suffices to show: $D\cap U\subseteq\dom[\beta]$.
We want:
$\forall x\in D\cap U$, $x\in\dom[\beta]$.
Let $x\in D\cap U$ be given.
We want to show: $x\in\dom[\beta]$.

We have: $x\in D$ and $x\in U$.
Since $x\in D$, we get $\alpha(x)\ne\frownie$.
Since $x\in U$, we get $\alpha(x)=\beta(x)$.
Since $\beta(x)=\alpha(x)\ne\frownie$,
we get $x\in\dom[\beta]$, as desired.
{\it End of proof of (1).}

{\it Proof of (2):}
Since $U\in\scrn_V(0_V)$, $0_V\in U$.
So, since $\alpha=\beta$ on~$U$,
we get $\alpha(0_V)=\beta(0_V)$.
By (B), we have $\alpha(0_V)=0_W$.

Then $\beta(0_V)=\alpha(0_V)=0_W$, as desired.
{\it End of proof of (2).}

{\it Proof of (3):}
Choose $|\bullet|_V\in\scrn(V)$ and $|\bullet|_W\in\scrn(W)$.
We wish to~show:
$\forall\varepsilon>0$, $\exists\delta>0$ such that, $\forall x\in\dom[\beta]$,
$$[\,\,0<|x-0_V|_V<\delta\,\,]\quad\Rightarrow\quad[\,\,|[\beta(x)]-[\beta(0_V)]|_W<\varepsilon\,\,].$$
Let $\varepsilon>0$ be given.
We wish to show: $\exists\delta>0$ such that, $\forall x\in\dom[\beta]$,
$$[\,\,0<|x-0_V|_V<\delta\,\,]\quad\Rightarrow\quad[\,\,|[\beta(x)]-[\beta(0_V)]|_W<\varepsilon\,\,].$$

Since $D\cap U\in\scrn_V(0_V)$ and since $\scrb_V(0_V)$ is a neighborhood base at~$0_V$ in~$V$,
choose $\rho>0$ such that $B_V(0_V,\rho)\subseteq D\cap U$.
By (C), choose $\tau>0$ such that, $\forall x\in D$,
$$[\,\,0<|x-0_V|_V<\tau\,\,]\quad\Rightarrow\quad[\,\,|[\alpha(x)]-[\alpha(0_V)]|_W<\varepsilon\,\,].$$
Let $\delta:=\min\{\rho,\tau\}$.
Then $\delta>0$, $\delta\le\rho$ and $\delta\le\tau$.
Want: $\forall x\in\dom[\beta]$,
$$[\,\,0<|x-0_V|_V<\delta\,\,]\quad\Rightarrow\quad[\,\,|[\beta(x)]-[\beta(0_V)]|_W<\varepsilon\,\,].$$
Let $x\in\dom[\beta]$ be given.
We wish to show:
$$[\,\,0<|x-0_V|_V<\delta\,\,]\quad\Rightarrow\quad[\,\,|[\beta(x)]-[\beta(0_V)]|_W<\varepsilon\,\,].$$
Assume $0<|x-0_V|_V<\delta$.
Want: $|[\beta(x)]-[\beta(0_V)]|_W$.

We have $0<|x-0_V|<\rho$ and $0<|x-0_V|<\tau$.
Since $|x-0_V|_V<\rho$, we get $x\in B_V(0_V,\rho)$.
So, since $B_V(0_V,\rho)\subseteq D\cap U$, we see that $x\in D\cap U$.
Then $x\in D$ and $x\in U$.
Since $x\in U$, by choice of $U$, we have $\alpha(x)=\beta(x)$.
By (2), we have $\beta(0_V)=0_W$.
So, by (B), we have $\beta(0_V)=\alpha(0_V)$.
Since $x\in D$ and since $0<|x-0_V|_V<\tau$, by choice of~$\tau$,
we conclude that $|[\alpha(x)]-[\alpha(0_V)]|_W<\varepsilon$.
Then
$$|\,[\beta(x)]\,-\,[\beta(0_V)]\,|_W\quad=\quad|\,[\alpha(x)]\,-\,[\alpha(0_V)]\,|_W\quad<\quad\varepsilon,$$
as desired.
{\it End of proof of (3).}
QED
\vskip.1in\hrule\vskip.1in\noindent
14-3. Let $\delta>0$, let $I:=(-\delta,\delta)$ and let $\alpha:\R\drta\R$.
Assume: $\forall x\in I$, $\alpha(x)\in[0|x]$.
Show that $\alpha\in\bigo_1(\R,\R)$.
\vskip.1in\noindent
{\it Proof:}
Let $|\bullet|\in\scrn(\R)$ denote absolute value.
We wish to show: $\alpha\in[|\bullet|]\cdot[\bigo(\R,\R)]$.
Let $\beta:=\adj_0^0(\alpha/|\bullet|)$.
It suffices to show:
\begin{itemize}
\item[(1)]$\beta\in\bigo(\R,\R)$ \qquad and
\item[(2)]$\alpha=[|\bullet|]\cdot\beta$.
\end{itemize}

{\it Proof of (1):}
We wish to show: $\exists U\in\scrn_\R(0)$ such that
$$[\,\,U\subseteq\dom[\beta]\,\,]\qquad\hbox{and}\qquad[\,\,\sup|\beta_*(U)|<\infty\,\,].$$

Since $I=(-\delta,\delta)$, we see that $I$ is open in $\R$.
So since $0\in I$, we get $I\in\scrn_\R(0)$.
Let $U:=I$.
We wish to show:
\begin{itemize}
\item[(A)]$U\subseteq\dom[\beta]$\qquad and
\item[(B)]$\sup|\beta_*(U)|<\infty$.
\end{itemize}

{\it Proof of (A):}
Since $U=I$, we wish to show $I\subseteq\dom[\beta]$.
Since $\beta(0)=0\ne\frownie$, we get $0\in\dom[\beta]$.
It remains to show: $I_0^\times\subseteq\dom[\beta]$.

We wish to show: $\forall x\in I_0^\times$, $x\in\dom[\beta]$.
Let $x\in I_0^\times$ be given.
We wish to show: $x\in\dom[\beta]$.

Since $x\in I_0^\times$, we get $x\in I$ and $x\ne0$.
Since $x\ne0$, we get $|x|\ne0$ and $\beta(x)=[|\alpha(x)|]/[|x|]$.
Since $x\in I$, we get $\alpha(x)\in[0|x]\subseteq\R$.
So, since $|x|\ne0$, we get $[\alpha(x)]/[|x|]\in\R$.
Then $\beta(x)=[|\alpha(x)|]/[|x|]\in\R$, so $\beta(x)\ne\frownie$,
and so $x\in\dom[\beta]$, as desired.
{\it End of proof of (A).}

{\it Proof of (B):}
It suffices to show $\sup|\beta_*(U)|\le1$.
We wish to show: $|\beta_*(U)|\le1$.
As $U=I$, we wish to show: $|\beta_*(I)|\le1$.
Since
$$|\beta(0)|\quad=\quad|0|\quad=\quad0\quad\le\quad1,$$
it remains to show: $|\beta_*(I_0^\times)|\le1$.

We wish to show: $\forall x\in\dom[\beta]$,
$$[\,\,x\in I_0^\times\,\,]\quad\Rightarrow\quad[\,\,|\beta(x)|\le1\,\,].$$
Let $x\in\dom[\beta]$ be given.
We wish to show:
$$[\,\,x\in I_0^\times\,\,]\quad\Rightarrow\quad[\,\,|\beta(x)|\le1\,\,].$$
Assume: $x\in I_0^\times$.
Want: $|\beta(x)|\le1$.

Since $x\in I_0^\times$, we get $x\in I$ and $x\ne0$.
Let $y:=|x|$.
As $x\ne0$, $y>0$.
Since $x\ne 0$, $\beta(x)=[\alpha(x)]/[|x|]$.
That is, $\beta(x)=[\alpha(x)]/y$.
Since $y>0$, we get $|y|=y$.
Then $|\beta(x)|=[|\alpha(x)|]/[|y|]=[|\alpha(x)|]/y$.
Since $x\in I$, we get $\alpha(x)=[0|x]$.
We have $-|x|\le x\le|x|$, {\it i.e.}, $-y\le x\le y$.
Since $-y\le 0$ and $-y\le x$ we get $-y\le\min\{0,x\}$.
Since $0\le y$ and $x\le y$, we get $\max\{0,x\}\le y$.
Then
$$[0|x]\,\,\,=\,\,\,[\,\min\{0,x\}\,,\,\max\{0,x\}\,]\,\,\,\subseteq\,\,\,[-y,y].$$
Then $\alpha(x)\in[0|x]=[-y,y]$,
so $-y\le\alpha(x)\le y$, so $|\alpha(x)|\le y$.
Then $|\alpha(x)|/y\le1$.
Then $|\beta(x)|=|\alpha(x)|/y\le1$, as desired.
{\it End of proof of~(B).}
{\it End of proof of (1).}

{\it Proof of (2):}
We have $0\in(-\delta,\delta)=I$, so $\alpha(0)\in[0|0]=\{0\}$,
so $\alpha(0)=0$.
Also $\beta(0)=0$.
Then $$\alpha(0)\,\,=\,\,0\,\,=\,\,0\cdot0\,\,=\,\,[|0|]\cdot[\beta(0)|].$$
It remains to show: $\alpha=[|\bullet|]\cdot\beta$ on $\R_0^\times$.
We wish to show: $\forall x\in\R_0^\times$, $\alpha(x)=[|x|]\cdot[\beta(x)]$.
Let $x\in\R_0^\times$ be given.
Want: $\alpha(x)=[|x|]\cdot[\beta(x)]$.

Since $x\ne0$, it follows that $\beta(x)=[\alpha(x)]/[|x|]$.
We conclude that $\alpha(x)=[|x|]\cdot[\beta(x)]$, as desired.
{\it End of proof of (2).}
QED
\vskip.1in\hrule\vskip.1in\noindent
14-4. Let $k\in\N_0$ and let $f\in\lito_k(\R,\R)$.
Let $g:\R\drta\R$.
Assume that $g'=f$ near $0$.
Assume that $g(0)=0$.
Show that $g\in\lito_{k+1}(\R,\R)$.
\vskip.1in\noindent
{\it Proof:}
Let $|\bullet|\in\scrn(\R)$ denote absolute value.
Let $\psi:=\adj_0^0(g/[|\bullet|^{k+1}])$.
So, since $\psi=g/[|\bullet|^{k+1}]$ on $\R_0^\times$,
we get $g=|\bullet|^{k+1}\cdot\psi$ on $\R_0^\times$.
So, since $g(0)=0=[|0|^{k+1}]\cdot[\psi(0)]$,
we see that $g=|\bullet|^{k+1}\cdot\psi$ on $\R$.
Then $g=|\bullet|^{k+1}\cdot\psi$.
We wish to show: $g\in[|\bullet|^{k+1}]\cdot[\lito(\R,\R)]$.
It therefore suffices to show: $\psi\in\lito(\R,\R)$.
Since $\psi(0)=0$,
we need to show:
$$[\,\,\dom[\psi]\in\scrn_\R(0)\,\,]\qquad\hbox{and}\qquad[\,\,\psi\hbox{ is continuous at }0\,\,].$$

We have $\lito_k(\R,\R)=[|\bullet|]\cdot[\lito(\R,\R)]$,
so, since $f\in\lito_k(\R,\R)$,
choose $\phi\in\lito(\R,\R)$ s.t.~$f=|\bullet|\cdot\phi$.
Then $\dom[f]=(\dom[|\bullet|])\cap(\dom[\phi])$,
so, since $\dom[\phi]\subseteq\R=\dom[|\bullet|]$,
we get $\dom[f]=\dom[\phi]$.
Since $g=|\bullet|^{k+1}\cdot\psi$,
we get $\dom[g]=(\dom[|\bullet|])\cap(\dom[\psi])$,
so, since $\dom[\psi]\subseteq\R=\dom[|\bullet|]$,
we get $\dom[g]=\dom[\psi]$.

Since $g'=f$ near $0$,
choose $U\in\scrn_\R(0)$ s.t.~$f=g'$ on $U$.
Let $D:=\dom[\phi]$.
Since $\phi\in\lito(\R,\R)$,
we have $D\in\scrn_\R(0)$.
So, since $U\in\scrn_\R(0)$,
we get $D\cap U\in\scrn_\R(0)$.
So, since $\scrb_\R(0)$ is a neighborhood base at $0$ in $\R$,
choose $\rho>0$ s.t.~$B_\R(0,\rho)\subseteq D\cap U$.
Let $I:=(-\rho,\rho)$.
Then $I\in\scrn_\R(0)$.
Also, $I=B_\R(0,\rho)\subseteq D\cap U$,
so $I\subseteq D$ and $I\subseteq U$.

We have $I\subseteq D=\dom[\phi]=\dom[f]$.
Since $f=g$ on $U$ and since $I\subseteq U$,
it follows that $f=g'$ on $I$.
So, since $I\subseteq\dom[f]$, we get $I\subseteq\dom[g']$.
Then $I\subseteq\dom[g']\subseteq\dom[g]=\dom[\psi]$.
So, since $I\in\scrn_\R(0)$,
it follows that $\dom[\psi]\in\scrn_\R(0)$.
It only remains to show that $\psi$ is continuous at $0$.
We want: $\forall\varepsilon>0$,
$\exists\delta>0$ s.t., $\forall x\in\dom[\psi]$,
$$[\,\,|x-0|<\delta\,\,]\quad\Rightarrow\quad[\,\,|[\psi(x)]-[\psi(0)]|<\varepsilon\,\,].$$
Let $\varepsilon>0$ be given.
We wish to show: $\exists\delta>0$ s.t., $\forall x\in\dom[\psi]$,
$$[\,\,0<|x-0|<\delta\,\,]\quad\Rightarrow\quad[\,\,|[\psi(x)]-[\psi(0)]|<\varepsilon\,\,].$$

As $\phi\in\lito(\R,\R)$,
$\phi$ is continuous at $0$.
Choose $\tau>0$ s.t., $\forall\alpha\in D$,
$$[\,\,0<|\alpha-0|<\tau\,\,]\quad\Rightarrow\quad[\,\,|[\phi(\alpha)]-[\phi(0)]|<\varepsilon\,\,].$$
Let $\delta:=\min\{\rho,\tau\}$.
Then $\delta\le\rho$ and $\delta\le\tau$.
We want: $\forall x\in\dom[\psi]$,
$$[\,\,0<|x-0|<\delta\,\,]\quad\Rightarrow\quad[\,\,|[\psi(x)]-[\psi(0)]|<\varepsilon\,\,].$$
Let $x\in\dom[\psi]$ be given.
We wish to show:
$$[\,\,0<|x-0|<\delta\,\,]\quad\Rightarrow\quad[\,\,|[\psi(x)]-[\psi(0)]|<\varepsilon\,\,].$$
Assume: $0<|x-0|<\delta$.
We wish to show: $|[\psi(x)]-[\psi(0)]|<\varepsilon$.
Since $\psi(0)=0$, we wish to prove: $|\psi(x)|<\varepsilon$.

Since $I$ is an interval and $0\in(-\delta,\delta)=I$,
we know: $[0|x]\subseteq I$.
Then $[0|x]\subseteq I\subseteq\dom[g']$, so $g$ is c/d on $[0|x]$.
So, by the Mean Value Theorem,
choose $\alpha\in(0|x)$ s.t.~$g'(\alpha)=DQ_g(0,x)$.
Since $g(0)=0$, we get $DQ_g(0,x)=[g(x)]/x$.
We have $\alpha\in(0|x)\subseteq[0|x]\subseteq I$.
So, since $g'=f$ on $I$,
we get $g'(\alpha)=f(\alpha)$.
Recall that $f=|\bullet|^k\cdot\phi$ and that $g=|\bullet|^{k+1}\cdot\psi$.
Then
\begin{eqnarray*}
|\alpha|^k\cdot[\phi(\alpha)]&=&f(\alpha)\,\,\,=\,\,\,g'(\alpha)\,\,\,=\,\,\,DQ_g(0,x)\\
&=&[g(x)]\,/\,x\,\,\,=\,\,\,|x|^{k+1}\cdot[\psi(x)]\,/\,x.
\end{eqnarray*}
Let $y:=|x|$.
Taking absolute values gives
$|\alpha|^k\cdot|\phi(\alpha)|=y^{k+1}\cdot|\psi(x)|/y$.
Then $|\psi(x)|=[|\alpha|/y]^k\cdot|\phi(\alpha)|$.
Want: $[|\alpha|/y]^k\cdot|\phi(\alpha)|<\varepsilon$.

We have $-|x|\le x\le|x|$, {\it i.e.}, $-y\le x\le y$.
Since $-y\le 0$ and $-y\le x$, we see that $-y\le\min\{0,x\}$.
Since $0\le y$ and $x\le y$, see that $\max\{0,x\}\le y$.
Then
$[0|x]=[\min\{0,x\},\max\{0,x\}]\subseteq[-y,y]$.
Then $\alpha\in(0|x)\subseteq[0|x]=[-y,y]$,
so $-y\le\alpha\le y$, so $|\alpha|\le y$.
Then $[|\alpha|/y]^k\le1$.
It therefore suffices to show: $|\phi(\alpha)|<\varepsilon$.

Since $\alpha\in I=(-\delta,\delta)$,
we see that $|\alpha|<\delta$.
Since $\alpha\in(0|x)$, we see that $\alpha\ne0$, so $|\alpha|>0$.
Then $0<|\alpha|<\delta\le\tau$.
Then $0<|\alpha-0|<\tau$.
So, since $\alpha\in I\subseteq D$, by choice of $\tau$,
we get $|[\phi(\alpha)]-[\phi(0)]|<\varepsilon$.
Since $\phi\in\lito(\R,\R)$,
we have $\phi(0)=0$.
Then $|\phi(\alpha)|<\varepsilon$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
14-5. Let $V$,  $W$ be finite dimensional vector spaces.
Let $f,g:V\drta W$.
Let $p\in V$.
Assume: $f=g$ near $p$.
Show: $\LINS_pf\subseteq\LINS_pg$.
\vskip.1in\noindent
{\it Proof:}
We wish to show: $\forall L\in\LINS_pf$, $L\in\LINS_pg$.
Let $L\in\LINS_pf$ be given.
We wish to show:  $L\in\LINS_pg$.

As $L\in\LINS_pf$, we get $f_p^T-L\in\lito_1(V,W)$.
Want: $g_p^T-L\in\lito_1(V,W)$.
By HW\#14-2, it suffices to show: $f_p^T-L=g_p^T-L$ near $0_V$.

Since $f=g$ near $p$,
choose $U\in\scrn_V(p)$ s.t.~$f=g$ on $U$.
Choose $|\bullet|\in\scrn(V)$.
Since $\scrb_V(p)$ is a neighborhood base at $p$ in $V$,
choose $\delta>0$ s.t.~$B_V(p,\delta)\subseteq U$.
Let $A:=B_V(0_V,\delta)$.
Then $A\in\scrn_V(0_V)$,
so it suffices to show:
$f_p^T-L=g_p^T-L$ on $A$.
We wish to show: $\forall h\in A$,
$(f_p^T-L)(h)=(g_p^T-L)(h)$.
Let $h\in A$ be given.
We wish to show: $(f_p^T-L)(h)=(g_p^T-L)(h)$.
Want: $[f_p^T(h)]-[L(h)]=[g_p^T(h)]-[L(h)]$.
It suffices to prove:
$f_p^T(h)=g_p^T(h)$.

Since $h\in A=B_V(0_V,\delta)$,
we get $|h-0_V|<\delta$.
Let $x:=p+h$.
Then $|x-p|=|h|=|h-0_V|<\delta$,
so $x\in B_V(p,\delta)$.
So, since $B_V(p,\delta)\subseteq U$,
we get $x\in U$.
So, since $f=g$ on $U$,
we get $f(x)=g(x)$.
As $U\in\scrn_V(p)$, we get $p\in U$.
So, since $f=g$ on $U$,
we get $f(p)=g(p)$.
Then
\begin{eqnarray*}
(f_p^T)(h)&=&[f(p+h)]-[f(p)]\\
&=&[f(x)]-[f(p)]\\
&=&[g(x)]-[g(p)]\\
&=&[g(p+h)]-[g(p)]\,\,\,=\,\,\,(g_p^T)(h),
\end{eqnarray*}
as desired.
QED
\vskip.1in\noindent\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 13: Due on Tuesday 24 April}
\vskip.1in\hrule\vskip.1in\noindent
13-1. Let $V$, $W$ be finite dimensional VSs.
Show: $L(V,W)\subseteq\bigo_1(V,W)$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall T\in L(V,W)$, $T\in\bigo_1(V,W)$.
Let $T\in L(V,W)$ be given.
Want: $T\in\bigo_1(V,W)$.

Since $T\in L(V,W)$, it follows that $T(0_V)=0_W$.
Let $|\bullet|_V\in\scrn(V)$, $|\bullet|_W\in\scrn(W)$.
Let $\alpha:=\adj_{0_V}^{0_W}(T/|\bullet|_V)$.
Then $T=|\bullet|_V\cdot\alpha$ on~$V_{0_V}^\times$.
So, since $T(0_V)=0_W=0\cdot 0_W=|0_V|_V\cdot[\alpha(0_V)]$,
we conclude that $T=|\bullet|_V\cdot\alpha$.
It suffices to show: $\alpha\in\bigo(V,W)$.
We wish to show:
$\exists U\in\scrn_V(0_V)$ such that:
$[\,\,U\subseteq\dom[\alpha]\,\,]\,\,\hbox{and}\,\,[\,\,\sup|\alpha_*(U)|_W<\infty\,\,]$.

We have $V\in\scrn_V(0_V)$.
Let $U:=V$.
We wish to show:
$$[\,\,U\subseteq\dom[\alpha]\,\,]\quad\hbox{and}\quad[\,\,\sup|\alpha_*(U)|_W<\infty\,\,].$$
Since $U=V\subseteq V=\dom[\alpha]$,
it remains to show: $\sup|\alpha_*(U)|_W<\infty$.

By (1) of Theorem 46.2 p.~331,
$T:V\to W$ is bounded.
We define $M:=\hatT_{VW}$.
Then $0\le M<\infty$.
Also, $T:V\to W$ is $M$-bounded.
It suffices to show: $\sup|\alpha_*(U)|_W\le M$.
We wish to show: $|\alpha_*(U)|_W\le M$.
We wish to show:
$\forall x\in\dom[\alpha]$,
$[\,\,x\in U\,\,]\,\,\Rightarrow\,\,[\,\,|\alpha(x)|_W\le M\,\,]$.
Let $x\in\dom[\alpha]$ be given.
We want:
$[\,\,x\in U\,\,]\,\,\Rightarrow\,\,[\,\,|\alpha(x)|_W\le M\,\,]$.
Assume: $x\in U$.
We wish to show: $|\alpha(x)|_W\le M$.

One of the following must be true:
\begin{itemize}
\item[(1)]$x=0_V$\qquad or
\item[(2)]$x\ne0_V$.
\end{itemize}

{\it Case (1):}
As $\alpha(x)=\alpha(0_V)=0_W$,
$|\alpha(x)|_W=|0_W|_W=0\le M$, as desired.
{\it End of case (1).}

{\it Case (2):}
We have $\alpha(x)=[T(x)]/[|x|_V]$.
Since $T$ is $M$-bounded, we have $|T(x)|_W\le M\cdot|x|_V$.
Then
$$|\alpha(x)|_W\quad=\quad\left|\frac{T(x)}{|x|_V}\right|_W\quad=\quad\frac{|T(x)|_W}{|x|_V}\quad\le\quad M,$$
as desired.
{\it End of case (2).}
QED
\vskip.1in\hrule\vskip.1in\noindent
13-2. Let $m,n\in\N$.
Let $V:=(\R^m,|\bullet|_{m,1})$ and let $W:=(\R^n,|\bullet|_{n,1})$.
Let $Z$ be a normed vector space.
Let $*\in B(V,W,Z)$.
Show: $\exists K\ge0$ such that, $\forall v\in V$, $\forall w\in W$,
\quad $|v*w|_Z\,\le\,K\cdot|v|_V\cdot|w|_W$.
\vskip.1in\noindent
{\it Proof:}
Let $K:=\max\{|e_i^m*e_j^n|_Z\,|\,i\in[1..m],j\in[1..n]\}$.
We wish to show:
$\forall v\in V$, $\forall w\in W$,
$|v*w|_Z\,\le\,K\cdot|v|_V\cdot|w|_W$.
Let $v\in V$, $w\in W$ be given.
We wish to show: $|v*w|_Z\,\le\,K\cdot|v|_V\cdot|w|_W$.

Since $|\bullet|_V=|\bullet|_{m,1}$,
we see that $\ds{|v|_V=\sum_{i=1}^m|v_i|}$.
Since $|\bullet|_W=|\bullet|_{n,1}$,
we see that $\ds{|w|_W=\sum_{j=1}^n|w_j|}$.
Since $\ds{v=\sum_{i=1}^m\,v_ie_i^m}$ and $\ds{w=\sum_{j=1}^n\,w_je_j^n}$,
we get $\ds{v*w=\sum_{i=1}^m\sum_{j=1}^n\,v_iw_j(e_i^m*e_j^n)}$.
Then, by subadditivity of $|\bullet|_Z$,
we have $\ds{|v*w|_Z\le\sum_{i=1}^m\sum_{j=1}^n\,|v_iw_j(e_i^m*e_j^n)|_Z}$.
We have: $\forall i\in[1..m]$, $\forall j\in[1..n]$,
$$|v_iw_j(e_i^m*e_j^n)|_Z\,\,\,=\,\,\,|v_i|\cdot|w_j|\cdot|e_i^m*e_j^n|_Z\,\,\,\le\,\,\,|v_i|\cdot|w_j|\cdot K.$$
Then
\begin{eqnarray*}
|v*w|_Z&\le&\sum_{i=1}^m\sum_{j=1}^n\,(\,|v_i|\cdot|w_j|\cdot K\,)\\
&=&K\cdot\sum_{i=1}^m\sum_{j=1}^n\,(\,|v_i|\cdot|w_j|\,)\\
&=&K\cdot\left[\sum_{i=1}^m\,|v_i|\right]\cdot\left[\sum_{j=1}^n\,|w_j|\right]\\
&=&K\cdot|v|_V\cdot|w|_W,
\end{eqnarray*}
as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
13-3. Let $\ell,m,n\in\N$.
Let $U:=(\R^\ell,|\bullet|_{\ell,1})$, let $V:=(\R^m,|\bullet|_{m,1})$ and let $W:=(\R^n,|\bullet|_{n,1})$.
Let $Z$ be a normed vector space.
Let $F\in T(U,V,W,Z)$.
Show: $\exists K\ge0$ such that, $\forall u\in U$, $\forall v\in V$, $\forall w\in W$,
$$|F(u,v,w)|_Z\quad\le\quad K\cdot|u|_U\cdot|v|_V\cdot|w|_W.$$
\vskip.1in\noindent
{\it Proof:}
Let $K:=\max\{|F(e_h^\ell,e_i^m,e_j^n)|_Z\,|\,h\in[1..\ell],i\in[1..m],j\in[1..n]\}$.
We wish to show:
$\forall u\in U$, $\forall v\in V$, $\forall w\in W$,
$$|F(u,v,w)|_Z\quad\le\quad K\cdot|u|_U\cdot|v|_V\cdot|w|_W.$$
Given $u\in U$, $v\in V$, $w\in W$.
Want: $|F(u,v,w)|_Z\,\le\,K\cdot|u|_U\cdot|v|_V\cdot|w|_W$.

Since $|\bullet|_U=|\bullet|_{\ell,1}$,
we see that $\ds{|u|_U=\sum_{h=1}^\ell|u_h|}$.
Since $|\bullet|_V=|\bullet|_{m,1}$,
we see that $\ds{|v|_V=\sum_{i=1}^m|v_i|}$.
Since $|\bullet|_W=|\bullet|_{n,1}$,
we see that $\ds{|w|_W=\sum_{j=1}^n|w_j|}$.
Since $\ds{u=\sum_{h=1}^m\,u_he_h^\ell}$ and $\ds{v=\sum_{i=1}^m\,v_ie_i^m}$ and $\ds{w=\sum_{j=1}^n\,w_je_j^n}$,
we get
$$F(u,v,w)\quad=\quad\sum_{h=1}^\ell\sum_{i=1}^m\sum_{j=1}^n\,u_hv_iw_j(F(e_h^\ell,e_i^m,e_j^n)).$$
Then, by subadditivity of $|\bullet|_Z$,
we have
$$|F(u,v,w)|_Z\quad\le\quad\sum_{h=1}^\ell\sum_{i=1}^m\sum_{j=1}^n\,|u_hv_iw_j(F(e_h^\ell,e_i^m,e_j^n))|_Z.$$
We have: $\forall h\in[1..\ell]$, $\forall i\in[1..m]$, $\forall j\in[1..n]$,
\begin{eqnarray*}
|u_hv_iw_j(F(e_h^\ell,e_i^m,e_j^n))|_Z&=&|u_h|\cdot|v_i|\cdot|w_j|\cdot|F(e_h^\ell,e_i^m,e_j^n)|_Z\\
&\le&|u_h|\cdot|v_i|\cdot|w_j|\cdot K.
\end{eqnarray*}
Then
\begin{eqnarray*}
|F(u,v,w)|_Z&\le&\sum_{h=1}^\ell\sum_{i=1}^m\sum_{j=1}^n\,(\,|u_h|\cdot|v_i|\cdot|w_j|\cdot K\,)\\
&=&K\cdot\sum_{h=1}^\ell\sum_{i=1}^m\sum_{j=1}^n\,(\,|u_h|\cdot|v_i|\cdot|w_j|\,)\\
&=&K\cdot\left[\sum_{h=1}^\ell\,|u_h|\right]\cdot\left[\sum_{i=1}^m\,|v_i|\right]\cdot\left[\sum_{j=1}^n\,|w_j|\right]\\
&=&K\cdot|u|_U\cdot|v|_V\cdot|w|_W,
\end{eqnarray*}
as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
13-4. Let $U$, $V$ and $W$ be finite dimensional normed vector spaces.
Let $Z$ be a normed vector space.
Let $F\in T(U,V,W,Z)$.
Show: $\exists K\ge0$ such that, $\forall u\in U$, $\forall v\in V$, $\forall w\in W$,
$$|F(u,v,w)|_Z\quad\le\quad K\cdot|u|_U\cdot|v|_V\cdot|w|_W.$$
\vskip.1in\noindent
{\it Proof:}
Let $\ell:=\dim U$, $m:=\dim V$ and $n:=\dim W$.
Choose
$$A\in\OB(U),\qquad B\in\OB(V)\qquad\hbox{and}\qquad C\in\OB(W).$$
Let $U':=(\R^\ell,|\bullet|_{\ell,1})$,
let $V':=(\R^m,|\bullet|_{m,1})$
and let $W':=(\R^n,|\bullet|_{n,1})$.
Define $F'\in T(U',V',W',Z)$
by
$$F'\,(\,u\,,\,v\,,\,w\,)\quad=\quad F\,(\,L_A(u)\,,\,L_B(v)\,,\,L_C(w)\,).$$
By HW\#13-3, choose $K'\ge0$ s.t., $\forall u'\in U'$, $\forall v'\in V'$, $\forall w'\in W'$,
$$|F'(u',v',w')|_Z\quad\le\quad K'\cdot|u'|_{U'}\cdot|v'|_{V'}\cdot|w'|_{W'}.$$
By (1) of Theorem 46.2 p.~331,
the maps
$$L_A^{-1}:U\to U',\qquad L_B^{-1}:V\to V'\qquad\hbox{and}\qquad L_C^{-1}:W\to W'$$
are all bounded.
Let
$$Q:=(L_A^{-1})_{U,U'}\!\!\!\!\!\!\!\!\!\widehat{\phantom{x}}\quad,\quad
R:=(L_B^{-1})_{V,V'}\!\!\!\!\!\!\!\!\!\widehat{\phantom{x}}\quad\quad\hbox{and}\quad
S:=(L_C^{-1})_{W,W'}\!\!\!\!\!\!\!\!\!\!\!\widehat{\phantom{x}}\,\,\quad.$$
Then $L_A^{-1}:U\to U'$ is $Q$-bounded,
$L_B^{-1}:V\to V'$ is $R$-bounded
and $L_C^{-1}:W\to W'$ is $S$-bounded.
Let $K:=K'\cdot Q\cdot R\cdot S$.
We wish to~show: $\forall u\in U$, $\forall v\in V$, $\forall w\in W$,
$$|F(u,v,w)|_Z\quad\le\quad K\cdot|u|_U\cdot|v|_V\cdot|w|_W.$$
Let $u\in U$, $v\in V$ and $w\in W$ be given.
We wish to show:
$$|F(u,v,w)|_Z\quad\le\quad K\cdot|u|_U\cdot|v|_V\cdot|w|_W.$$

Since $L_A^{-1}:U\to U'$ is $Q$-bounded,
since $L_B^{-1}:V\to V'$ is $R$-bounded
and since $L_C^{-1}:W\to W'$ is $S$-bounded,
we get
\begin{eqnarray*}
|L_A^{-1}(u)|_{U'}&\le&Q\cdot|u|_U,\\
|L_B^{-1}(v)|_{V'}&\le&R\cdot|v|_V\qquad\hbox{and}\\
|L_C^{-1}(w)|_{W'}&\le&S\cdot|w|_W.
\end{eqnarray*}
Let $u':=(L_A^{-1})(u)$, $v':=(L_B^{-1})(v)$ and $w':=(L_C^{-1})(w)$.
Then
$$|u'|_{U'}\,\le\,Q\cdot|u|_U,\quad
|v'|_{V'}\,\le\,R\cdot|v|_V,\quad\hbox{and}\quad
|w'|_{W'}\,\le\,S\cdot|w|_W.$$
By choice of $K'$, we have
$$|F'(u',v',w')|_Z\quad\le\quad K'\cdot|u'|_{U'}\cdot|v'|_{V'}\cdot|w'|_{W'}.$$
Since $L_A(u')=u$, $L_B(v')=v$ and $L_C(w')=w$, by definition of $F'$,
we get $F'(u',v',w')=F(u,v,w)$.
Then
\begin{eqnarray*}
|F(u,v,w)|_Z&\le&|F'(u',v',w')|_Z\\
&\le&K'\cdot|u'|_{U'}\cdot|v'|_{V'}\cdot|w'|_{W'}\\
&\le&K'\cdot Q\cdot|u|_U\cdot R\cdot|v|_V\cdot S\cdot|w|_W\\
&=&K'\cdot Q\cdot R\cdot S\cdot|u|_U\cdot|v|_V\cdot|w|_W\\
&=&K\cdot|u|_U\cdot|v|_V\cdot|w|_W,
\end{eqnarray*}
as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
13-5. Let $S$ $V$, $W$ and $Z$ all be finite dimensional vector spaces,
and let $*\in B(V,W,Z)$.
Show: $\ds{[\bigo(S,V)]\,\opstar_S\,[\bigo(S,W)]\,\,\subseteq\,\,\bigo(S,Z)}$.
\vskip.1in\noindent
{\it Proof:}
We wish to show: $\forall\alpha\in\bigo(S,V)$,
$\forall\beta\in\bigo(S,W)$,
we have: $\ds{\alpha\opstar_S\beta\in\bigo(S,Z)}$.
Let $\alpha\in\bigo(S,V)$ and $\beta\in\bigo(S,W)$ be given.
We wish to show: $\ds{\alpha\opstar_S\beta\in\bigo(S,Z)}$.

Choose $|\bullet|_S\in\scrn(S)$ and $|\bullet|_V\in\scrn(V)$
and $|\bullet|_W\in\scrn(W)$ and $|\bullet|_Z\in\scrn(Z)$.
By Theorem 49.1 p.~359,
choose $C\ge0$ such that,
$\forall v\in V$, $\forall w\in W$,
$|v*w|_Z\le C\cdot|v|_V\cdot|w|_W$.

Since $\alpha\in\bigo(S,V)$,
choose $P\in\scrn_S(0_S)$ such that
\begin{itemize}
\item[]$P\subseteq\dom[\alpha]$ \qquad and \qquad $\sup|\alpha_*(P)|_V<\infty$.
\end{itemize}
Let $K:=\sup|\alpha_*(P)|_V$.
Then $0\le K<\infty$ and $|\alpha_*(P)|_V\le K$.

Since $\beta\in\bigo(S,W)$,
choose $Q\in\scrn_S(0_S)$ such that
\begin{itemize}
\item[]$Q\subseteq\dom[\beta]$ \qquad and \qquad $\sup|\beta_*(Q)|_W<\infty$.
\end{itemize}
Let $L:=\sup|\beta_*(Q)|_W$.
Then $0\le L<\infty$ and $|\beta_*(Q)|_W\le L$.

Let $\ds{\gamma:=\alpha\,\opstar_S\,\beta}$.
We wish to show: $\gamma\in\bigo(S,Z)$.
We wish to show: $\exists R\in\scrn_S(0_S)$ such that
\begin{itemize}
\item[]$R\subseteq\dom[\gamma]$ \qquad and \qquad $\sup|\gamma_*(R)|_Z<\infty$.
\end{itemize}

Since $P\in\scrn_S(0_S)$ and $Q\in\scrn_S(0_S)$,
it follows that $P\cap Q\in\scrn_S(0_S)$.
Let $R:=P\cap Q$.
We wish to show:
\begin{itemize}
\item[]$R\subseteq\dom[\gamma]$ \qquad and \qquad $\sup|\gamma_*(R)|_Z<\infty$.
\end{itemize}

Since $\ds{\gamma=\alpha\,\opstar_S\,\beta}$,
it follows that $\dom[\gamma]=(\dom[\alpha])\cap(\dom[\beta])$.
Since $P\subseteq\dom[\alpha]$ and $Q\subseteq\dom[\beta]$,
it follows that
$$P\cap Q\qquad\subseteq\qquad(\dom[\alpha])\,\,\cap\,\,(\dom[\beta]).$$
Then
$R=P\cap Q\subseteq(\dom[\alpha])\cap(\dom[\beta])=\dom[\gamma]$.
It remains to~show that
$\sup|\gamma_*(R)|_Z<\infty$.

Let $M:=CKL$.
Then $0\le M<\infty$.
It suffices to show that
$\sup|\gamma_*(R)|_Z\le M$.
We wish to show:
$\forall x\in\dom[\gamma]$,
$$[\,\,x\in R\,\,]\quad\Rightarrow\quad[\,\,|\gamma(x)|_Z\le M\,\,].$$
Let $x\in\dom[\gamma]$ be given.
We wish to show:
$$[\,\,x\in R\,\,]\quad\Rightarrow\quad[\,\,|\gamma(x)|_Z\le M\,\,].$$
Assume: $x\in R$.
Want: $|\gamma(x)|_Z\le M$.

We have
$x\in\dom[\gamma]\subseteq\dom[\alpha]$ and $x\in R\subseteq P$,
so $\alpha(x)\in\alpha_*(P)$.
Then $|\alpha(x)|_V\in|\alpha_*(P)|_V\le K$.
Let $v:=\alpha(x)$.
Then $|v|_V\le K$.
We have
$x\in\dom[\gamma]\subseteq\dom[\beta]$ and $x\in R\subseteq Q$,
so $\beta(x)\in\beta_*(Q)$.
Then $|\beta(x)|_W\in|\beta_*(Q)|_W\le L$.
Let $w:=\beta(x)$.
Then $|w|_W\le L$.

We have $\ds{\gamma(x)=(\alpha\,\opstar_S\,\beta)(x)=[\alpha(x)]*[\beta(x)]=v*w}$.
By the choice of $C$, we have $|v*w|_Z\le C\cdot|v|_V\cdot|w|_W$.

Then $|\gamma(x)|_Z=|v*w|_Z\le C\cdot|v|_V\cdot|w|_W\le CKL=M$.
QED
\vskip.1in\noindent\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 12: Due on Tuesday 17 April}
\vskip.1in\hrule\vskip.1in\noindent
12-1. Let $V$ and $W$ both be finite dimensional vector spaces,
and let $|\bullet|,\|\bullet\|\in\scrn(W)$.
Show: $\bigo(V,W,|\bullet|)\subseteq\bigo(V,W,\|\bullet\|)$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall\alpha\in\bigo(V,W,|\bullet|)$, $\alpha\in\bigo(V,W,\|\bullet\|)$.
Let $\alpha\in\bigo(V,W,|\bullet|)$ be given.
Want: $\alpha\in\bigo(V,W,\|\bullet\|)$.
Want: $\exists U\in\scrn_V(0_V)$ such that
$$[\,\,U\subseteq\dom[\alpha]\,\,]\qquad\hbox{and}\qquad[\,\,\sup\|\alpha_*(U)\|<\infty\,\,].$$

Since $\alpha\in\bigo(V,W,|\bullet|)$,
choose $U\in\scrn_V(0_V)$ such that
$$[\,\,U\subseteq\dom[\alpha]\,\,]\qquad\hbox{and}\qquad[\,\,\sup|\alpha_*(U)|<\infty\,\,].$$
Want: $\sup\|\alpha_*(U)\|<\infty$.

Let $M:=\sup|\alpha_*(U)|$.
Then $|\alpha_*(U)|\le M$.
By Theorem 46.3 p.~332, $|\bullet|\approx\|\bullet\|$, so $\|\bullet\|<<|\bullet|$,
so choose $K>0$ s.t.~$\|\bullet\|\le K\cdot|\bullet|$.
It suffices to~show: $\|\alpha_*(U)\|\le KM$.
We want: $\forall x\in\dom[\alpha]$,
$$[\,\,x\in U\,\,]\quad\Rightarrow\quad[\,\,\|\alpha(x)\|\le KM\,\,].$$
Let $x\in\dom[\alpha]$ be given.
We wish to prove:
$$[\,\,x\in U\,\,]\quad\Rightarrow\quad[\,\,\|\alpha(x)\|\le KM\,\,].$$
Assume that $x\in U$.
We wish to show: $\|\alpha(x)\|\le KM$.

Since $x\in\dom[\alpha]$ and $x\in U$,
we get $|\alpha(x)|\in|\alpha_*(U)|$.
So, since $|\alpha_*(U)|\le M$,
we get $|\alpha(x)|\le M$.
Since $\|\bullet\|\le K\cdot|\bullet|$,
we get $\|\alpha(x)\|\le K\cdot|\alpha(x)|$.
Then $\|\alpha(x)\|\le K\cdot|\alpha(x)|\le KM$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
12-2. Let $V$ and $W$ both be finite dimensional vector spaces,
and let $\alpha:V\dashrightarrow W$.
Assume that $\dom[\alpha]\in\scrn_V(0_V)$.
Assume that $\alpha$ is continuous at $0_V$.
Show: $\alpha\in\bigo(V,W)$.
\vskip.1in\noindent
{\it Proof:}
Choose $|\bullet|_V\in\scrn(V)$ and $|\bullet|_W\in\scrn(W)$.
We wish to show: $\exists U\in\scrn_V(0_V)$ such that
$$[\,\,U\subseteq\dom[\alpha]\,\,]\quad\hbox{and}\quad[\,\,\sup|\alpha_*(U)|_W<\infty\,\,].$$

Let $z:=\alpha(0_V)$.
Since $\alpha$~is continuous at $0_V$,
choose $\delta>0$ such that, $\forall x\in\dom[\alpha]$,
we have: $[\,|x|_V<\delta\,]\,\Rightarrow\,[\,|[\alpha(x)]-z|_W<1\,]$.
We define $U:=(\dom[\alpha])\cap(B_V(0_V,\delta))$.
Since $\dom[\alpha]\in\scrn_V(0_V)$
and since $B_V(0_V,\delta)\in\scrn_V(0_V)$,
it follows that
$U\in\scrn_V(0_V)$.
Moreover, we have both $U\subseteq\dom[\alpha]$ and $U\subseteq B_V(0_V,\delta)$.
It remains to~show: $\sup|\alpha_*(U)|_W<\infty$.
Let $M\,:=\,1\,+\,[\,|z|_W\,]$.
It suffices to~show: $|\alpha_*(U)|_W\le M$.
We wish to prove: $\forall x\in\dom[\alpha]$,
$$[\,\,x\in U\,\,]\quad\Rightarrow\quad[\,\,|\alpha(x)|_W\le M\,\,].$$
Let $x\in\dom[\alpha]$ be given.
We wish to prove:
$$[\,\,x\in U\,\,]\quad\Rightarrow\quad[\,\,|\alpha(x)|_W\le M\,\,].$$
Assume that $x\in U$.
We wish to prove: $|\alpha(x)|_W\le M$.

Since $x\in U\subseteq B_V(0_V,\delta)$,
we conclude that $|x|_V<\delta$.
So, since $x\in\dom[\alpha]$,
by choice of $\delta$,
we conclude that: $|[\alpha(x)]-z|_W<1$.
Let $y:=\alpha(x)$.
Then $|y-z|_W<1$,
and we wish to prove: $|y|_W\le M$.

Since $y=(y-z)+z$,
we get $|y|_W\,\le\,[\,|y-z|_W\,]\,+\,[\,|z|_W\,]$.
Then $|y|_W\,<\,1\,+\,[\,|z|_W\,]=M$,
as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
12-3. Let $V$ and $W$ both be finite dimensional vector spaces,
and let $|\bullet|,\|\bullet\|\in\scrn(V)$, and let $p>0$.
Show:
\begin{itemize}
\item[(1)]$\lito_p(V,W,|\bullet|)\subseteq\lito_p(V,W,\|\bullet\|)$ \qquad and
\item[(2)]$\bigo_p(V,W,|\bullet|)\subseteq\bigo_p(V,W,\|\bullet\|)$.
\end{itemize}
\vskip.1in\noindent
{\it Proof:}
Let $\alpha\,:=\,\adj_{0_V}^1\,(\,[\,\|\bullet\|^p\,]\,/\,[\,|\bullet|^p\,]\,)$,
$\beta\,:=\,\adj_{0_V}^1\,(\,[\,|\bullet|^p\,]\,/\,[\,\|\bullet\|^p\,]\,)$.
Then $\alpha,\beta:V\to\R$.
Also, for all $x\in V\backslash\{0_V\}$, we have:
$$(\alpha\beta)(x)\quad=\quad\left[\,\frac{\|x\|^p}{|x|^p}\,\right]\cdot\left[\,\frac{|x|^p}{\|x\|^p}\,\right]\quad=\quad1\quad=\quad C_V^1(x).$$
So, since $(\alpha\beta)(0_V)=1\cdot1=1=C_V^1(0_V)$,
we conclude that $\alpha\beta=C_V^1$.

By Theorem 46.3 p.~332, $|\bullet|\approx\|\bullet\|$, so $|\bullet|<<\|\bullet\|$,
so choose $C>0$ such that $|\bullet|\le C\cdot\|\bullet\|$.
Then, for all $x\in V\backslash\{0_V\}$,
$$\beta(x)\quad=\quad\frac{|x|^p}{\|x\|^p}\quad=\quad\left[\frac{|x|}{\|x\|}\right]^p\quad\le\quad\left[\frac{C\cdot\|x\|}{\|x\|}\right]^p\quad=\quad C^p.$$
Let $K:=\max\{C^p,1\}$.
Then $K\ge1$, so $K>0$.
Also, for all $x\in V$,
we have $0\le\beta(x)\le K$,
and so $|\beta(x)|=\beta(x)\le K$.
Choose $|||\bullet|||\in\scrn(W)$.

{\it Claim A:}
$\lito(V,W)\subseteq\alpha\cdot[\,\lito(V,W)\,]$.
{\it Proof of Claim A:}
We wish to~show:
$\forall\gamma\in\lito(V,W)$,
$\gamma\in\alpha\cdot[\,\lito(V,W)\,]$.
Let $\gamma\in\lito(V,W)$ be given.
Want: $\gamma\in\alpha\cdot[\,\lito(V,W)\,]$.
Since $\gamma=\gamma\cdot C_V^1=\gamma\cdot[\alpha\beta]=\alpha\cdot[\beta\gamma]$,
it suffices to show that $\beta\gamma\in\lito(V,W)$.
We wish to show:
\begin{itemize}
\item$\dom[\beta\gamma]\in\scrn_V(0_V)$,
\item$(\beta\gamma)(0_V)=0_W$ \qquad and
\item$\beta\gamma:V\drta W$ is continuous at $0_V$.
\end{itemize}

Since $\gamma\in\lito(V,W)$,
it follows that:
\begin{itemize}
\item$\dom[\gamma]\in\scrn_V(0_V)$,
\item$\gamma(0_V)=0_W$ \qquad and
\item$\gamma:V\drta W$ is continuous at $0_V$.
\end{itemize}
Since $\dom[\beta]=V$ and since $\dom[\gamma]\subseteq V$,
we get $\dom[\beta\gamma]=\dom[\gamma]$.
Then $\dom[\beta\gamma]=\dom[\gamma]\in\scrn_V(0_V)$.
Also, $(\beta\gamma)(0_V)=1\cdot0_W=0_W$.
It remains to show:
$\beta\gamma:V\drta W$ is continuous at $0_V$.
We wish to show: $\forall\varepsilon>0$,
$\exists\delta>0$ such that, for all $x\in\dom[\beta\gamma]$,
$$[\,\,|x|<\delta\,\,]\quad\Rightarrow\quad[\,\,|||(\beta\gamma)(x)|||<\varepsilon\,\,].$$
Let $\varepsilon>0$ be given.
Want:
$\exists\delta>0$ such that, for all $x\in\dom[\beta\gamma]$,
$$[\,\,|x|<\delta\,\,]\quad\Rightarrow\quad[\,\,|||(\beta\gamma)(x)|||<\varepsilon\,\,].$$

As $\gamma:V\drta W$ is continuous at $0_V$,
choose $\delta>0$ s.t., $\forall x\in\dom[\gamma]$,
$$[\,\,|x|<\delta\,\,]\quad\Rightarrow\quad[\,\,|||\gamma(x)|||<\varepsilon/K\,\,].$$
We wish to show: $\forall x\in\dom[\beta\gamma]$,
$$[\,\,|x|<\delta\,\,]\quad\Rightarrow\quad[\,\,|||(\beta\gamma)(x)|||<\varepsilon\,\,].$$
Let $x\in\dom[\beta\gamma]$ be given.
We wish to show:
$$[\,\,|x|<\delta\,\,]\quad\Rightarrow\quad[\,\,|||(\beta\gamma)(x)|||<\varepsilon\,\,].$$
Assume that $|x|<\delta$.
We wish to show: $|||(\beta\gamma)(x)|||<\varepsilon$.

We have $x\in\dom[\beta\gamma]\subseteq\dom[\gamma]$.
So, since $|x|<\delta$, by choice of $\delta$,
we get: $|||\gamma(x)|||<\varepsilon/K$.

We have $x\in\dom[\beta\gamma]\subseteq\dom[\gamma]$,
so, as $x\in U$, $\gamma(x)\in\gamma_*(U)$.
Then
$$|||\gamma(x)|||\quad\in\quad|||\gamma_*(U)|||\quad\le\quad\sup|||\gamma_*(U)|||\quad=\quad M.$$
Recall: $|\beta(x)|\le K$.
Then
\begin{eqnarray*}
|||(\beta\gamma)(x)|||&=&|||[\beta(x)]\cdot[\gamma(x)]|||\\
&=&|\beta(x)|\cdot|||\gamma(x)|||\\
&<&K\cdot(\varepsilon/K)\,\,=\,\,\varepsilon,
\end{eqnarray*}
as desired.
{\it End of proof of Claim A.}

{\it Claim B:}
$\bigo(V,W)\subseteq\alpha\cdot[\,\bigo(V,W)\,]$.
{\it Proof of Claim B:}
We wish to~show:
$\forall\gamma\in\bigo(V,W)$,
$\gamma\in\alpha\cdot[\,\bigo(V,W)\,]$.
Let $\gamma\in\bigo(V,W)$ be given.
Want: $\gamma\in\alpha\cdot[\,\bigo(V,W)\,]$.
Since $\gamma=\gamma\cdot C_V^1=\gamma\cdot[\alpha\beta]=\alpha\cdot[\beta\gamma]$,
it suffices to show that $\beta\gamma\in\bigo(V,W)$.
Want:
$\exists U\in\scrn_V(0_V)$ such that
\begin{itemize}
\item[]$U\subseteq\dom[\beta\gamma]$ \qquad and \qquad $\sup|||(\beta\gamma)_*(U)|||<\infty$.
\end{itemize}

Since $\gamma\in\bigo(V,W)=\bigo(V,W,|\bullet|)$,
choose $U\in\scrn_V(0_V)$ such that
\begin{itemize}
\item[]$U\subseteq\dom[\gamma]$ \qquad and \qquad $\sup|||\gamma_*(U)|||<\infty$.
\end{itemize}
We wish to show:
\begin{itemize}
\item[]$U\subseteq\dom[\beta\gamma]$ \qquad and \qquad $\sup|||(\beta\gamma)_*(U)|||<\infty$.
\end{itemize}

As $\dom[\beta]=V$ and $\dom[\gamma]\subseteq V$,
we get $\dom[\beta\gamma]=\dom[\gamma]$.
Then $U\subseteq\dom[\gamma]=\dom[\beta\gamma]$.
It remains to show: $\sup|||(\beta\gamma)_*(U)|||<\infty$.

Let $M:=\sup|||\gamma_*(U)|||$.
It suffices to show: $|||(\beta\gamma)_*(U)|||\le KM$.
We wish to show: $\forall x\in\dom[\beta\gamma]$,
$$[\,\,x\in U\,\,]\quad\Rightarrow\quad[\,\,|||(\beta\gamma)(x)|||\le KM\,\,].$$
Let $x\in\dom[\beta\gamma]$ be given.
We wish to show:
$$[\,\,x\in U\,\,]\quad\Rightarrow\quad[\,\,|||(\beta\gamma)(x)||\le KM\,\,].$$
Assume that $x\in U$.
Want: $|||(\beta\gamma)(x)|||\le KM$

We have $x\in\dom[\beta\gamma]\subseteq\dom[\gamma]$,
so, as $x\in U$, $\gamma(x)\in\gamma_*(U)$.
Then
$$|||\gamma(x)|||\quad\in\quad|||\gamma_*(U)|||\quad\le\quad\sup|||\gamma_*(U)|||\quad=\quad M.$$
Recall: $|\beta(x)|\le K$.
Then
$$|||(\beta\gamma)(x)|||\,\,=\,\,|||[\beta(x)]\cdot[\gamma(x)]|||\,\,=\,\,|\beta(x)|\cdot|||\gamma(x)|||\,\,\le\,\,KM,$$
as desired.
{\it End of proof of Claim B.}

{\it Claim C:}
$[\,|\bullet|^p\,]\cdot\alpha=\|\bullet\|^p$.
{\it Proof of Claim C:}
On $V\backslash\{0_V\}$, we have
$[\,|\bullet|^p\,]\cdot\alpha=[\,|\bullet|^p\,]\cdot[\,\|\bullet\|^p\,/\,|\bullet|^p\,]=\|\bullet\|^p$,
so it suffices to show that
$(\,[\,|\bullet|^p\,]\cdot\alpha\,)(0_V)=(\,\|\bullet\|^p\,)(0_V)$.

We have
$(\,[\,|\bullet|^p\,]\cdot\alpha\,)(0_V)=0^p\cdot1=0^p=(\,\|\bullet\|^p\,)(0_V)$,
as desired.
{\it End of proof of Claim C.}

{\it Proof of (1):}
By Claim A, 
$[\,|\bullet|^p\,]\cdot[\,\lito(V,W)\,]\subseteq[\,|\bullet|^p\,]\cdot\alpha\cdot[\,\lito(V,W)\,]$.
By Claim C,
$[\,|\bullet|^p\,]\cdot\alpha\cdot[\,\lito(V,W)\,]=\|\bullet\|^p\cdot[\,\lito(V,W)\,]$.
Then
\begin{eqnarray*}
\lito_p(V,W,|\bullet|)&=&[\,|\bullet|^p\,]\cdot[\,\lito(V,W)\,]\\
&\subseteq&[\,|\bullet|^p\,]\cdot\alpha\cdot[\,\lito(V,W)\,]\\
&=&[\,\|\bullet\|^p\,]\cdot[\,\lito(V,W)\,]\\
&=&\lito_p(V,W,\|\bullet\|),
\end{eqnarray*}
as desired.
{\it End of proof of (1).}

{\it Proof of (2):}
By Claim B, 
$[\,|\bullet|^p\,]\cdot[\,\bigo(V,W)\,]\subseteq[\,|\bullet|^p\,]\cdot\alpha\cdot[\,\bigo(V,W)\,]$.
By Claim C,
$[\,|\bullet|^p\,]\cdot\alpha\cdot[\,\bigo(V,W)\,]=\|\bullet\|^p\cdot[\,\bigo(V,W)\,]$.
Then
\begin{eqnarray*}
\bigo_p(V,W,|\bullet|)&=&[\,|\bullet|^p\,]\cdot[\,\bigo(V,W)\,]\\
&\subseteq&[\,|\bullet|^p\,]\cdot\alpha\cdot[\,\bigo(V,W)\,]\\
&=&[\,\|\bullet\|^p\,]\cdot[\,\bigo(V,W)\,]\\
&=&\bigo_p(V,W,\|\bullet\|),
\end{eqnarray*}
as desired.
{\it End of proof of (2).}
QED
\vskip.1in\hrule\vskip.1in\noindent
12-4. Let $V$ and $W$ both be finite dimensional vector spaces,
and let $p,q\ge0$.
Assume $p<q$.
Show: $\lito_p(V,W)\supseteq\bigo_q(V,W)$.
\vskip.1in\noindent
{\it Proof:}
We want:
$\forall\gamma\in\bigo_q(V,W)$, $\gamma\in\lito_p(V,W)$.
Let $\gamma\in\bigo_q(V,W)$ be given.
We wish to show: $\gamma\in\lito_p(V,W)$.

Chose $|\bullet|\in\scrn(V)$.
Then $\bigo_q(V,W)=[|\bullet|^q]\cdot[\bigo(V,W)]$
and $\lito_p(V,W)=[|\bullet|^p]\cdot[\lito(V,W)]$.
We have $\gamma\in[|\bullet|^q]\cdot[\bigo(V,W)]$.
We wish to show: $\gamma\in[|\bullet|^p]\cdot\lito(V,W)$.

Since $\gamma\in\bigo_q(V,W)=[|\bullet|^q]\cdot[\bigo(V,W)]$,
choose $\alpha\in\bigo(V,W)$ such that $\gamma=[|\bullet|^q]\cdot\alpha$.
Choose $\|\bullet\|\in\scrn(W)$.
Since $\alpha\in\bigo(V,W)$, choose $U\in\scrn_V(0_V)$ such that
\begin{itemize}
\item[]$U\subseteq\dom[\alpha]$ \qquad and \qquad $\sup\|\alpha_*(U)\|<\infty$.
\end{itemize}
Since $\dom[\alpha]\supseteq U\in\scrn_V(0_V)$,
we see that $\dom[\alpha]\in\scrn_V(0_V)$.

Let $\beta:=[|\bullet|^{q-p}]\cdot\alpha$.
Then $[|\bullet|^p]\cdot\beta=[|\bullet|^q]\cdot\alpha$.
Then
$$\gamma\quad=\quad[\,|\bullet|^q\,]\,\cdot\,\alpha\quad=\quad[\,|\bullet|^p\,]\,\cdot\,\beta,$$
and so it suffices to show: $\beta\in\lito(V,W)$.
We wish to show:
\begin{itemize}
\item$\dom[\beta]\in\scrn_V(0_V)$,
\item$\beta(0_V)=0_W$ \qquad and
\item$\beta:V\drta W$ is continuous at $0_V$.
\end{itemize}

Because $\dom[|\bullet|^{q-p}]=V$
and because $\dom[\alpha]\subseteq V$,
we conclude that $\dom[\,[|\bullet|^{q-p}]\cdot\alpha\,]=\dom[\alpha]$.
Then
$$\dom[\beta]\quad=\quad\dom\,[\,[|\bullet|^{q-p}]\,\cdot\,\alpha\,]\quad=\quad\dom[\alpha].$$
Then $\dom[\beta]=\dom[\alpha]\in\scrn_v(0_V)$.
Also,
$$\beta(0_V)\,\,\,=\,\,\,([|\bullet|^{q-p}]\cdot\alpha)(0_V)\,\,\,=\,\,\,0^{q-p}\cdot[\alpha(0_V)]\,\,\,=\,\,\,0_W.$$
It remains to~show: $\beta:V\drta W$ is continuous at $0_V$.
We wish to~show: $\forall\varepsilon>0$, $\exists\delta>0$ such that, $\forall x\in\dom[\beta]$,
$$[\,\,|x|<\delta\,\,]\quad\Rightarrow\quad[\,\,\|\beta(x)\|<\varepsilon\,\,].$$
Let $\varepsilon>0$ be given.
We wish to show: $\exists\delta>0$ such that, $\forall x\in\dom[\beta]$,
$$[\,\,|x|<\delta\,\,]\quad\Rightarrow\quad[\,\,\|\beta(x)\|<\varepsilon\,\,].$$

Since $U\in\scrn_V(0_V)$ and since $\scrb_V(0_V)$ is a neighborhood base at $0_V$ in~$V$,
choose $\rho>0$ such that $B(0_V,\rho)\subseteq U$.
Let $M:=(\sup\|\alpha_*(U)\|)+1$.
Then $M>0$ and $\|\alpha_*(U)\|<M$.
Let $\tau:=(\varepsilon/M)^{1/(q-p)}$.
We define $\delta:=\min\{\rho,\tau\}$.
We wish to show: $\forall x\in\dom[\beta]$,
$$[\,\,|x|<\delta\,\,]\quad\Rightarrow\quad[\,\,\|\beta(x)\|<\varepsilon\,\,].$$
Let $x\in\dom[\beta]$ be given.
We wish to show:
$$[\,\,|x|<\delta\,\,]\quad\Rightarrow\quad[\,\,\|\beta(x)\|<\varepsilon\,\,].$$
Assume that $|x|<\delta$.
We wish to show: $\|\beta(x)\|<\varepsilon$.

Since $|x|<\delta\le\rho$,
we get $x\in B_V(0_V,\rho)$.
So, since $B_V(0_V,\rho)\subseteq U$,
$x\in U$.
So, since $x\in\dom[\beta]=\dom[\alpha]$,
we have $\alpha(x)\in\alpha_*(U)$.
Then $\|\alpha(x)\|\in\|\alpha_*(U)\|<M$.
Since $\beta=[|\bullet|^{q-p}]\cdot\alpha$, we conclude that
$\beta(x)=[|x|^{q-p}]\cdot[\alpha(x)]$.
Since $|x|<\delta\le\tau$, we get $|x|^{q-p}<\tau^{q-p}$.
Then $\|\beta(x)\|=[|x|^{q-p}]\cdot\|\alpha(x)\|<[\tau^{q-p}]\cdot M=[\varepsilon/M]\cdot M=\varepsilon$.
QED
\vskip.1in\hrule\vskip.1in\noindent
12-5. Let $p,q\ge0$.
Show:
\begin{itemize}
\item[](\,\,[\,\,$(\lito_p(\R,\R))\cdot(\lito_q(\R,\R))\subseteq\lito_{p+q}(\R,\R)$\,\,]\qquad and\\
\phantom{x}\,[\,\,$(\lito_q(\R,\R))\circ(\lito_p(\R,\R))\,\subseteq\,\lito_{qp}(\R,\R)$\,\,]\,\,).
\end{itemize}
\vskip.1in\noindent
{\it Proof:}
Let $\lito:=\lito(\R,\R)$.
By (1) of Fact 47.10 p.~342, we have $\lito\cdot\lito\subseteq\lito$.
Then
\begin{eqnarray*}
(\lito_p(\R,\R))\cdot(\lito_q(\R,\R))
&=&|\bullet|^p\cdot\lito\cdot|\bullet|^q\cdot\lito\\
&=&|\bullet|^p\cdot|\bullet|^q\cdot\lito\cdot\lito\\
&\subseteq&|\bullet|^{p+q}\cdot\lito\\
&=&\lito_{p+q}(\R,\R).
\end{eqnarray*}
It remains to show: $(\lito_q(\R,\R))\circ(\lito_p(\R,\R))\,\subseteq\,\lito_{qp}(\R,\R)$.
We want:
$\forall f\in\lito_p(\R,\R)$,
$\forall g\in\lito_q(\R,\R)$,
$g\circ f\in\lito_{qp}(\R,\R)$.
Let $f\in\lito_p(\R,\R)$ and $g\in\lito_q(\R,\R)$ be given.
We wish to prove: $g\circ f\in\lito_{qp}(\R,\R)$.

Since $f\in\lito_p(\R,\R)=|\bullet|^p\cdot\lito$,
choose $\delta\in\lito$ s.t.~$f=|\bullet|^p\cdot\delta$.
Since $g\in\lito_p(\R,\R)=|\bullet|^q\cdot\lito$,
choose $\varepsilon\in\lito$ s.t.~$g=|\bullet|^q\cdot\varepsilon$.
For all $x\in\R$,
\begin{eqnarray*}
g(f(x))&=&(|\bullet|^q\cdot\varepsilon)(f(x))
\,\,\,=\,\,\,\big|f(x)\big|^q\cdot[\varepsilon(f(x))]\\
&=&\big|\,(|\bullet|^p\cdot\delta)(x)\,\big|^q\cdot[\varepsilon(f(x))]\\
&=&\big|\,|x|^p\cdot[\delta(x)]\,\big|^q\cdot[\varepsilon(f(x))]\\
&=&|x|^{qp}\cdot|\delta(x)|^q\cdot[\varepsilon(f(x))].
\end{eqnarray*}
Then $g\circ f=|\bullet|^{qp}\cdot|\delta|^q\cdot[\varepsilon\circ f]$.

By Fact 47.12 p.~342, (with $V$ and $W$ both replaced by $\R$, $\|\bullet\|$ by $|\bullet|$
and $p$ by $q$),
$|\lito|^q\subseteq\lito$.
Then $|\delta|^q\in|\lito|^q\subseteq\lito$.

We have $f\in\lito_p(\R,\R)\subseteq\lito_0(\R,\R)=\lito$.
By (1) of Fact 47.11 p.~342 (with $V$, $W$ and $X$ all replaced by $\R$),
$\lito\circ\lito\subseteq\lito$.
Then $\varepsilon\circ f\in\lito\circ\lito\subseteq\lito$.

Recall that $\lito\cdot\lito\subseteq\lito$.
Then $|\delta|^q\cdot[\varepsilon\circ f]\in\lito\cdot\lito\subseteq\lito$.
It follows that $g\circ f=|\bullet|^{qp}\cdot|\delta|^q\cdot[\varepsilon\circ f]\in|\bullet|^{qp}\cdot\lito=\lito_{qp}(\R,\R)$,
as desired.
QED
\vskip.1in\noindent\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 11: Due on Tuesday 10 April}
\vskip.1in\hrule\vskip.1in\noindent
11-1. Let $S\subseteq\R$.
Assume that $S$ has a minimum.
(That is, assume: $\exists a\in S$ s.t.~$a\le S$.)
Show that $\inf S=\min S\in S$.
\vskip.1in\noindent
{\it Proof:}
Choose $a\in S$ s.t.~$a\le S$.
Since $a\le S$, we get $a\in\LB(S)$.
Since $a\in S$ and $a\in\LB(S)$,
we get $a\in S\cap[\LB(S)]$.

{\it Claim:} $S\cap[\LB(S)]=\{a\}$.
{\it Proof of Claim:}
Since $a\in S\cap[\LB(S)]$,
we have $\{a\}\subseteq S\cap[\LB(S)]$.
We wish to show: $S\cap[\LB(S)]\subseteq\{a\}$.
We wish to show: $\forall z\in S\cap[\LB(S)]$, $z\in\{a\}$.
Let $z\in S\cap[\LB(S)]$ be given.
We wish to show: $z\in\{a\}$.
We wish to show: $z=a$.

We have both $z\in S$ and $z\in\LB(S)$.
Also, since $a\le S$, it follows that $S\ge a$.
Since $z\in S\ge a$,
we get $z\ge a$.
It remains to show: $z\le a$.

Since $z\in\LB(S)$, it follows that $z\le S$.
So, since $a\in S$, we conclude that $z\le a$, as desired.
{\it End of proof of Claim.}

By the Claim, $\ELT(S\cap[\LB(S)])=\ELT\{a\}$.
Then
$$\min S\quad=\quad\ELT\,(\,S\,\cap\,[\LB(S)]\,)\quad=\quad\ELT\{a\}\quad=\quad a.$$
Since $a\in S$, we get $a\ne\frownie$.
By Fact 8.13 p.~69, $\inf S\,=^*\,\min S$.
Then $\inf S\,=^*\,\min S=a\ne\frownie$,
so $\inf S=a$.
Then $\inf S=a=\min S$.
It remains to show: $\min S\in S$.

We have $\min S=a\in S$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
11-2. Let $U$ be a vector space,
let $|\bullet|,\|\bullet\|\in\scrn(V)$
and let $S\subseteq U$.
Let $V:=(U,|\bullet|)$ and let $W:=(U,\|\bullet\|)$.
Assume both that $|\bullet|<<\|\bullet\|$,
and that $S$ is bounded in $W$.
Show that $S$ is bounded in $V$.
\vskip.1in\noindent
{\it Proof:}
Since $S$ is bounded in $W$,
choose $A\in\scrb_W$ s.t.~$S\subseteq A$.
Since $A\in\scrb_W$, choose $p\in W$ and $r>0$
s.t.~$A=B_W(p,r)$.
Since $|\bullet|<<\|\bullet\|$,
choose $K>0$ s.t.~$|\bullet|\le K\cdot\|\bullet\|$.
It suffices to show: $S\subseteq B_V(p,Kr)$.
We wish to show:
$\forall q\in S$, $q\in B_V(p,Kr)$.
Let $q\in S$ be given.
We wish to show: $q\in B_V(p,Kr)$.
That is, we wish to show: $|q-p|_V<Kr$.

Since $q\in S\subseteq A=B_W(p,r)$,
it follows that $|q-p|_W<r$.
So, since $K>0$,
we get $K\cdot|q-p|_W<Kr$.
We have $|\bullet|_V=|\bullet|$ and $|\bullet|_W=\|\bullet\|$.
Then $|q-p|_V=|q-p|$ and $|q-p|_W=\|q-p\|$.

Since $|\bullet|\le K\cdot\|\bullet\|$,
we get $|q-p|\le K\cdot\|q-p\|$.
Then
$$|q-p|_V\,\,\,=\,\,\,|q-p|\,\,\,\le\,\,\,K\cdot\|q-p\|\,\,\,=\,\,\,K\cdot|q-p|_W\,\,\,<\,\,\,Kr,$$
as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
11-3. Let $V$ and $W$ be normed vector spaces,
and let $T\in L(V,W)$.
Assume that $T:V\to W$ is bounded below.
Show that $T$ is 1-1.
\vskip.1in\noindent
{\it Proof:}
By (3) of Fact 44.3 p.~319,
it suffices to show: $\ker[T]=\{0_V\}$.
By (1) of Fact 44.3 p.~319,
$\ker[T]$ is a vector subspace of $V$,
and so $\{0_V\}\subseteq\ker[T]$.
We wish to show: $\ker[T]\subseteq\{0_V\}$.
We wish to show: $\forall x\in\ker[T]$, $x\in\{0_V\}$.
Given $x\in\ker[T]$.
Want: $x\in\{0_V\}$.

Since $x\in\ker[T]=T^*(\{0_W\})$,
we get $Tx\in\{0_W\}$.
Then $Tx=0_W$.

Since $T$ is bounded below,
we see that $\ckT>0$.
Let $\varepsilon:=\ckT$.
Then $\varepsilon>0$
and $T$ is $\varepsilon$-bounded below.
Then, by HW\#10-1,
$|Tx|_W\ge\varepsilon|x|_V$.
Then $\varepsilon|x|_V\le|Tx|_W=|0_W|_W=0$.
Since $\varepsilon>0$ and $\varepsilon|x|_V\le0$,
we see that $|x|_V\le0$.
So, since $|x|_V\ge0$,
we get $|x|_V=0$.
It follows that $x=0_V$, so $x\in\{0_V\}$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
11-4. Let $T\in L(\R^4,\R^4)$ be defined by $Tx=(6x_1,5x_2,8x_3,7x_4)$,
and define $V:=(\R^4,|\bullet|_{4,2})$.
Show that $\ckT_{VV}=5$
and that $\hatT_{VV}=8$.
\vskip.1in\noindent
{\it Proof:}
We wish to show:
\begin{itemize}
\item[(1)]$\ckT_{VV}\le5$,
\item[(2)]$\ckT_{VV}\ge5$,
\item[(3)]$\hatT_{VV}\le8$ \qquad and
\item[(4)]$\hatT_{VV}\ge8$.
\end{itemize}

Let $e_\bullet$ be the standard basis of $\R^4$.
Then
$$e_1=(1,0,0,0),\,\,\,
e_2=(0,1,0,0),\,\,\,
e_3=(0,0,1,0),\,\,\,
e_4=(0,0,0,1).$$
So, since $|\bullet|_V=|\bullet|_{4,2}$,
we get
$$|e_1|_V\quad=\quad|e_2|_V\quad=\quad|e_3|_V\quad=\quad|e_4|_V\quad=\quad1,$$
so $e_1,e_2,e_3,e_4\in S_V$.
So, since $e_1,e_2,e_3,e_4\in V=\dom[T]$,
we conclude that $T(e_1),T(e_2),T(e_3),T(e_4)\in T_*(S_V)$.
Let $E:=\Ell_T$.
Then
$$E=T_*(S_V),\qquad
\ckT_{VV}=\inf|E|_V,\qquad
\hatT_{VV}=\sup|E|_V.$$
We have $T(e_1),T(e_2),T(e_3),T(e_4)\in T_*(S_V)=E$.

{\it Proof of (1):}
We have $T(e_2)=5e_2$.
Also, $|5e_2|_V=5\cdot|e_2|_V=5\cdot1=5$.
Then $5e_2=T(e_2)\in E$.
Then $5=|5e_2|_V\in|E|_V\ge\inf|E_V|=\ckT_{VV}$.
Then $\ckT_{VV}\le 5$, as desired.
{\it End of proof of (1).}

{\it Proof of (2):}
We want: $T:V\to V$ is $5$-bounded below.
We want: $\forall x\in V$, $|Tx|_V\ge5\cdot|x|_V$.
Given $x\in V$.
Want: $|Tx|_V\ge5\cdot|x|_V$.

We have
\begin{eqnarray*}
|Tx|_V^2&=&|(6x_1,5x_2,8x_3,7x_4)|_V^2\,\,=\,\,|(6x_1,5x_2,8x_3,7x_4)|_{4,2}^2\\
&=&(6x_1)^2+(5x_2)^2+(8x_3)^2+(7x_4)^2\\
&=&36x_1^2+25x_2^2+64x_3^2+49x_4^2\\
&\ge&25x_1^2+25x_2^2+25x_3^2+25x_4^2\\
&=&25(x_1^2+x_2^2+x_3^2+x_4^2)\\
&=&25\cdot|x|_{4,2}^2\,\,=\,\,25\cdot|x|_V^2.
\end{eqnarray*}
So, since $25\cdot|x|_V^2\ge0$, we get $\sqrt{|Tx|_V^2}\ge\sqrt{25\cdot|x|_V^2}$.
Then
$$|Tx|_V\,\,=\,\,\sqrt{|Tx|_V^2}\,\,\ge\,\,\sqrt{25\cdot|x|_V^2}\,\,=\,\,\sqrt{25}\cdot\sqrt{|x|_V^2}\,\,=\,\,5\cdot|x|_V,$$
as desired.
{\it End of proof of (2).}

{\it Proof of (3):}
We wish to show: $T:V\to V$ is $8$-bounded.
We want: $\forall x\in V$, $|Tx|_V\le8\cdot|x|_V$.
Given $x\in V$.
Want: $|Tx|_V\le8\cdot|x|_V$.

We have
\begin{eqnarray*}
|Tx|_V^2&=&|(6x_1,5x_2,8x_3,7x_4)|_V^2\,\,=\,\,|(6x_1,5x_2,8x_3,7x_4)|_{4,2}^2\\
&=&(6x_1)^2+(5x_2)^2+(8x_3)^2+(7x_4)^2\\
&=&36x_1^2+25x_2^2+64x_3^2+49x_4^2\\
&\le&64x_1^2+64x_2^2+64x_3^2+64x_4^2\\
&=&64(x_1^2+x_2^2+x_3^2+x_4^2)\\
&=&64\cdot|x|_{4,2}^2\,\,=\,\,64\cdot|x|_V^2.
\end{eqnarray*}
So, since $0\le|Tx|_V^2$, we get $\sqrt{|Tx|_V^2}\le\sqrt{64\cdot|x|_V^2}$.
Then
$$|Tx|_V\,\,=\,\,\sqrt{|Tx|_V^2}\,\,\le\,\,\sqrt{64\cdot|x|_V^2}\,\,=\,\,\sqrt{64}\cdot\sqrt{|x|_V^2}\,\,=\,\,8\cdot|x|_V,$$
as desired.
{\it End of proof of (3).}

{\it Proof of (4):}
We have $T(e_3)=8e_3$.
Also, $|8e_3|_V=8\cdot|e_3|_V=8\cdot1=8$.
Then $8e_3=T(e_3)\in E$.
Then $8=|8e_3|_V\in|E|_V\le\sup|E_V|=\hatT_{VV}$.
Then $\ckT_{VV}\ge 8$, as desired.
{\it End of proof of (4).}
QED
\vskip.1in\hrule\vskip.1in\noindent
11-5. Let $V$ and $W$ be normed vector spaces
and let $T:V\hookrightarrow>W$ be a vector space isomorphism.
Assume $T:V\to W$ is bounded below.
Show that $T^{-1}:W\to V$ is bounded.
\vskip.1in\noindent
{\it Proof:}
As $T:V\to W$ is bounded below, $\ckT_{VW}>0$.
Let $\varepsilon:=\ckT_{VW}$.
Then $\varepsilon>0$ and $T:V\to W$ is $\varepsilon$-bounded below.
Let $K:=1/\varepsilon$.
Then $K>0$, $K\varepsilon=1$, and it suffices to show: $T^{-1}:W\to V$ is $K$-bounded.
We wish to show: $\forall y\in W$, $|T^{-1}(y)|_V\le K\cdot|y|_W$.
Let~$y\in W$ be given.
We wish to show: $|T^{-1}(y)|_V\le K\cdot|y|_W$.

Let $x:=T^{-1}(y)$.
Then $Tx=y$.
Since $T:V\to W$ is $\varepsilon$-bounded below,
we get $|Tx|_W\ge\varepsilon\cdot|x|_V$.
So, since $K>0$, $K\cdot|Tx|_W\ge(K\varepsilon)\cdot|x|_V$.
So, since $K\varepsilon=1$,
we get $K\cdot|Tx|_W\ge|x|_V$.
That is, $|x|_V\le K\cdot|Tx|_W$.
Then $|T^{-1}(y)|_V=|x|_V\le K\cdot|Tx|_W=K\cdot|y|_W$,
as desired.
QED
\vskip.1in\noindent\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 10: Due on Tuesday 3 April}
\vskip.1in\hrule\vskip.1in\noindent
10-1. Let $V$ and $W$ be normed vector spaces,
$T\in L(V,W)$ and $\varepsilon>0$.
Show:
$[\,T\hbox{ is }\varepsilon\hbox{-bounded below}\,]\,\Leftrightarrow\,
[\,\forall x\in V,\,|Tx|_W\ge\varepsilon|x|_V\,]$.
\vskip.1in\noindent
{\it Proof:}
{\it Proof of $\Rightarrow$:}
Assume that $T$ is $\varepsilon$-bounded below.
We want: $\forall x\in V,\,|Tx|_W\ge\varepsilon|x|_V$.
Let $x\in V$ be given.
Want: $|Tx|_W\ge\varepsilon|x|_V$.

By Fact 42.5 p.~306, choose $a\ge0$ and $u\in S_V$ s.t.~$x=au$.
Since $a\ge0$, we get $|a|=a$.
Since $u\in S_V$, it follows that $|u|_V=1$.
Then
$|x|_V=|au|_V=|a|\cdot|u|_V=a\cdot1=a$.
Want: $|Tx|_W\ge\varepsilon a$.

We have $u\in S_V\subseteq V=\dom[T]$.
So, as $u\in S_V$, we get $Tu\in T_*(S_V)$.
So, since $T_*(S_V)=\Ell_T$,
we get $Tu\in\Ell_T$.
Since $T$ is $\varepsilon$-bounded below,
we have $\ckT\ge\varepsilon$.
Then
$$|Tu|_W\quad\in\quad|\Ell_T|_W\quad\ge\quad\inf\,|\Ell_T|_W\quad=\quad\ckT\quad\ge\quad\varepsilon,$$
so, since $|a|=a$, we get $|a|\cdot|Tu|_W\ge\varepsilon a$.
Then
$$|Tx|_W\,\,=\,\,|T(au)|_W\,\,=\,\,|a\cdot[Tu]|_W\,\,=\,\,|a|\cdot|Tu|_W\,\,\ge\,\,\varepsilon a,$$
as desired.
{\it End of proof of $\Rightarrow$.}

{\it Proof of $\Leftarrow$:}
Assume that $\forall x\in V,\,|Tx|_W\ge\varepsilon|x|_V$.
We wish to show: $T$ is $\varepsilon$-bounded below.
Want: $\ckT\ge\varepsilon$.
So, since $\ckT=\inf|\Ell_T|_W$,
it suffices to show: $|\Ell_T|_W\ge\varepsilon$.
We wish to show: $\forall s\in|\Ell_T|_W$, $s\ge\varepsilon$.
Let $s\in|\Ell_T|_W$ be given.
We wish to show: $s\ge\varepsilon$.

Since $s\in|\Ell_T|_W$,
choose $y\in\Ell_T$ such that $s=|y|_W$.
Since $y\in\Ell_T=T_*(S_V)$,
choose $u\in S_V$ such that $y=Tu$.
By assumption, $|Tu|_W\ge\varepsilon|u|_V$.
Since $u\in S_V$, we conclude that $|u|_V=1$.
Then
$s=|y|_W=|Tu|_W\ge\varepsilon|u|_V=\varepsilon\cdot1=\varepsilon$.
{\it End of proof of $\Leftarrow$.}
QED
\vskip.1in\hrule\vskip.1in\noindent
10-2. Let $m\in\N$.
Show: $|\bullet|_{m,\infty}\le|\bullet|_{m,2}\le|\bullet|_{m,1}\le m\cdot|\bullet|_{m,\infty}$.
\vskip.1in\noindent
{\it Proof:}
We wish to show:
\begin{itemize}
\item[(1)]$|\bullet|_{m,\infty}\le|\bullet|_{m,2}$,
\item[(2)]$|\bullet|_{m,2}\le|\bullet|_{m,1}$, \qquad and
\item[(3)]$|\bullet|_{m,1}\le m\cdot|\bullet|_{m,\infty}$.
\end{itemize}

{\it Proof of (1):}
Want: $\forall x\in\R^m$,
$|x|_{m,\infty}\le|x|_{m,2}$.
Let $x\in\R^m$ be given.
We wish to prove: $|x|_{m,\infty}\le|x|_{m,2}$.

Let $a:=|x|_{m,2}$.
Want: $|x|_{m,\infty}\le a$.
As $|x|_{m,\infty}=\max\{|x_1|,\ldots,|x_m|\}$,
it suffices to show $\{|x_1|,\ldots,|x_m|\}\le a$.
Want: $\forall t\in\{|x_1|,\ldots,|x_m|\}$, $t\le a$.
Let $t\in\{|x_1|,\ldots,|x_m|\}$ be given.
Want: $t\le a$.

Since $t\in\{|x_1|,\ldots,|x_m|\}$,
choose $j\in[1..m]$ such that $t=|x_j|$.
We have $0\le x_j^2\le x_1^2+\cdot+x_m^2$.
Then $\sqrt{x_j^2}\le\sqrt{x_1^2+\cdot+x_m^2}$.
Then
$$t\,\,\,\,=\,\,\,\,|x_j|\,\,\,\,=\,\,\,\,\sqrt{x_j^2}\,\,\,\,\le\,\,\,\,\sqrt{x_1^2+\cdot+x_m^2}\,\,\,\,=\,\,\,\,|x|_2\,\,\,\,=\,\,\,\,a,$$
as desired.
{\it End of proof of (1).}

{\it Proof of (2):}
Want: $\forall x\in\R^m$,
$|x|_{m,2}\le|x|_{m,1}$.
Let $x\in\R^m$ be given.
We wish to prove: $|x|_{m,2}\le|x|_{m,1}$.

Let $I:=[1..m]$.
For all $j\in I$, let $a_j:=|x_j|$.
For all $j\in I$, we have $a_j^2=|x_j|^2=x_j^2$.
Then $|x|_{m,2}=\sqrt{x_1^2+\cdots+x_m^2}=\sqrt{a_1^2+\cdots+a_m^2}$.
Also, we have $|x|_{m,1}=|x_1|+\cdots+|x_m|=a_1+\cdots+a_m$.

Let $K:=\{r\in I^2\,|\,r_1\ne r_2\}$.
For all $j\in I$, we have $a_j=|x_j|\ge0$.
Then, for all $r\in K$, we have $a_{r_1}a_{r_2}\ge0$.
It follows that $\ds{\sum_{r\in K}\,a_{r_1}a_{r_2}\ge0}$.
Let $\ds{S:=\sum_{r\in K}\,a_{r_1}a_{r_2}}$.
Then $S\ge0$.
We have
$$(a_1\,+\,\cdots\,+\,a_m)^2\quad=\quad(a_1^2\,+\,\cdots\,+\,a_m^2)\,\,+\,\,S.$$
So, since $S\ge0$, we get
$(a_1+\cdots+a_m)^2\ge a_1^2+\cdots+a_m^2$.
Since $0\le a_1^2+\cdots+a_m^2\le(a_1+\cdots+a_m)^2$,
we see that
$$\sqrt{a_1^2+\cdots+a_m^2}\quad\le\quad\sqrt{(a_1+\cdots+a_m)^2}.$$
Recall: $\forall j\in I$, $a_j\ge0$.
It follows that $a_1+\cdots+a_m\ge0$.
Then we have $|a_1+\cdots+a_m|=a_1+\cdots+a_m$.
Then
\begin{eqnarray*}
|x|_{m,2}&=&\sqrt{a_1^2+\cdots+a_m^2}\\
&\le&\sqrt{(a_1+\cdots+a_m)^2}\\
&\le&|a_1+\cdots+a_m|\\
&=&a_1+\cdots+a_m\,\,=\,\,|x|_{m,1},
\end{eqnarray*}
as desired.
{\it End of proof of (2).}

{\it Proof of (3):}
Want: $\forall x\in\R^m$,
$|x|_{m,1}\le m\cdot|x|_{m,\infty}$.
Let $x\in\R^m$ be given.
Want: $|x|_{m,1}\le m\cdot|x|_{m,\infty}$.
Let $a:=|x|_{m,\infty}$.
Want: $|x|_{m,1}\le ma$.

For all $j\in[1..m]$,
we have $|x_j|\le\max\{|x_1|,\ldots,|x_m|\}=|x|_{m,\infty}=a$.
Then $|x|_{m,1}=|x_1|+\cdots+|x_m|\le ma$.
{\it End of proof of (3).}
QED
\vskip.1in\hrule\vskip.1in\noindent
10-3. Let $V$ be a normed vector space, let $Z$ be a topological space, let $\mu:V\drta Z$ and let $p\in V$.
Let $\lambda:=\mu(p+\bullet)$.
Show: $\ds{\lim_{0_V}\,\lambda=\lim_p\,\mu}$.
\vskip.1in\noindent
{\it Proof:}
It suffices to show:
$\ds{\opLIMS_{0_V}\,\lambda=\opLIMS_p\,\mu}$.

{\it Proof of $\subseteq$:}
Want:
$\ds{\forall y\in\opLIMS_{0_V}\,\lambda}$, $\ds{y\in\opLIMS_p\,\mu}$.
Given $\ds{y\in\opLIMS_{0_V}\,\lambda}$.
Want: $\ds{y\in\opLIMS_p\,\mu}$.
Know: $\lambda\to y$ near $0_V$.
Want: $\mu\to y$ near $p$.
We wish to show: $\forall U\in\scrn_Z(y)$, $\exists\delta>0$ s.t., $\forall x\in\dom[\mu]$,
$$[\,\,0<|x-p|_V<\delta\,\,]\quad\Rightarrow\quad[\,\,\mu(x)\in U\,\,].$$
Let $U\in\scrn_Z(y)$ be given.
We wish to show: $\exists\delta>0$ s.t., $\forall x\in\dom[\mu]$,
$$[\,\,0<|x-p|_V<\delta\,\,]\quad\Rightarrow\quad[\,\,\mu(x)\in U\,\,].$$

Since $\lambda\to y$ near $0_V$, choose $\delta>0$ s.t., $\forall h\in\dom[\lambda]$,
$$[\,\,0<|h|_V<\delta\,\,]\quad\Rightarrow\quad[\,\,\lambda(h)\in U\,\,].$$
We wish to show: $\forall x\in\dom[\mu]$,
$$[\,\,0<|x-p|_V<\delta\,\,]\quad\Rightarrow\quad[\,\,\mu(x)\in U\,\,].$$
Let $x\in\dom[\mu]$ be given.
We wish to show:
$$[\,\,0<|x-p|_V<\delta\,\,]\quad\Rightarrow\quad[\,\,\mu(x)\in U\,\,].$$
Assume that $0<|x-p|_V<\delta$.
We wish to show: $\mu(x)\in V$.

Let $h:=x-p$.
Then $h\in(\dom[\mu])-p=\dom[\lambda]$.
Also, $0<|h|_V<\delta$, so, by choice of $\delta$,
we have $\lambda(h)\in U$.
By definition of $\lambda$, we have $\lambda(h)=\mu(p+h)$.
So, since $p+h=x$, we get $\lambda(h)=\mu(x)$.
Then $\mu(x)=\lambda(h)\in U$.
{\it End of proof of $\subseteq$.}

{\it Proof of $\supseteq$:}
Want:
$\ds{\forall y\in\opLIMS_p\,\mu}$, $\ds{y\in\opLIMS_{0_V}\,\lambda}$.
Given $\ds{y\in\opLIMS_p\,\mu}$.
Want: $\ds{y\in\opLIMS_{0_V}\,\lambda}$.
Know: $\mu\to y$ near $p$.
Want: $\lambda\to y$ near $0_V$.
We wish to show: $\forall U\in\scrn_Z(y)$, $\exists\delta>0$ s.t., $\forall h\in\dom[\lambda]$,
$$[\,\,0<|h|_V<\delta\,\,]\quad\Rightarrow\quad[\,\,\lambda(h)\in U\,\,].$$
Let $U\in\scrn_Z(y)$ be given.
We wish to show: $\exists\delta>0$ s.t., $\forall h\in\dom[\lambda]$,
$$[\,\,0<|h|_V<\delta\,\,]\quad\Rightarrow\quad[\,\,\lambda(h)\in U\,\,].$$

Since $\mu\to y$ near $p$, choose $\delta>0$ s.t., $\forall x\in\dom[\mu]$,
$$[\,\,0<|x-p|_V<\delta\,\,]\quad\Rightarrow\quad[\,\,\mu(x)\in U\,\,].$$
We wish to show: $\forall h\in\dom[\lambda]$,
$$[\,\,0<|h|_V<\delta\,\,]\quad\Rightarrow\quad[\,\,\lambda(h)\in U\,\,].$$
Let $h\in\dom[\lambda]$ be given.
We wish to show:
$$[\,\,0<|h|_V<\delta\,\,]\quad\Rightarrow\quad[\,\,\lambda(h)\in U\,\,].$$
Assume that $0<|h|_V<\delta$.
We wish to show: $\lambda(h)\in U$.

Let $x:=p+h$.
Then $x\in(\dom[\lambda])+p=\dom[\mu]$.
Also, $h=x-p$
Then $0<|x-p|_V<\delta$, so, by choice of $\delta$,
we have $\mu(x)\in U$.
By definition of $\lambda$, we have $\lambda(h)=\mu(p+h)$.
So, since $p+h=x$, we get $\lambda(h)=\mu(x)$.
Then $\lambda(h)=\mu(x)\in U$.
{\it End of proof of $\supseteq$.}
QED
\vskip.1in\hrule\vskip.1in\noindent
10-4. Let $X$ be a metric space, let $D\subseteq X$ and let $p\in\Int_XD$.
Show that there exists $B\in\scrb_X(p)$ such that $B\subseteq D$.
\vskip.1in\noindent
{\it Proof:}
Since $p\in\Int_XD$, choose an open subset $U$ of $X$ such that $p\in U\subseteq D$.
Since $U$ is open in $X$, we have $U\in\scrt_X$.
We have $\scrt_X=\langle\scrb_X\rangle_\cup$.
Then $p\in U\in\langle\scrb_X\rangle_\cup$.
Choose $C\in\scrb_X$ such that $p\in C\subseteq U$.
By the Recentering Down Lemma (Lemma 14.2 p.~104),
choose $B\in\scrb_X(p)$ such that $B\subseteq C$.
We wish to show: $B\subseteq D$.

We have $B\subseteq C\subseteq U\subseteq D$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
10-5. Let $g:\R\drta\R$.
Let $p,q\in\R$.
Assume $g$ has a local unique min at $p$ in $\R$.
Show: $g+C_\R^q$ has a local unique min at $p$ in~$\R$.
\vskip.1in\noindent
{\it Proof:}
Let $h:=g+C_\R^q$.
We wish to show:
$\exists V\in\scrn_X(p)$~s.t.
$$(\,\,V\subseteq\dom[h]\,\,)\qquad\hbox{and}\qquad(\,\,h(p)<h_*(V_p^\times)\,\,).$$

Since $g$ has a local unique min at $p$ in $\R$,
choose $V\in\scrn_X(p)$~s.t.
$$(\,\,V\subseteq\dom[g]\,\,)\qquad\hbox{and}\qquad(\,\,g(p)<g_*(V_p^\times)\,\,).$$
We wish to show:
$$(\,\,V\subseteq\dom[h]\,\,)\qquad\hbox{and}\qquad(\,\,h(p)<h_*(V_p^\times)\,\,).$$
We have $\dom[h]=\dom[g+C_\R^q]=(\dom[g])\cap(\dom[C_\R^q])$.
So, since $\dom[g]\subseteq\R=\dom[C_\R^q]$, we get $\dom[h]=\dom[g]$.
Then we have $V\subseteq\dom[g]=\dom[h]$.
It remains to show: $h(p)<h_*(V_p^\times)$.
We wish to show: $\forall y\in h_*(V_p^\times)$, $h(p)<y$.
Let $y\in h_*(V_p^\times)$ be given.
We wish to~show: $h(p)<y$.
Since $y\in h_*(V_p^\times)$, choose $x\in\dom[h]$ such that: both $x\in V_p^\times$ and $y=h(x)$.
We wish to show $h(p)<h(x)$.

Since $x\in\dom[h]=\dom[g]$ and since $x\in V_p^\times$,
it follows that $g(x)\in g_*(V_p^\times)$.
So, since $g(p)<g_*(V_p^\times)$, we get $g(p)<g(x)$.
Then $[g(p)]+q<[g(x)]+q$.
So since
\begin{eqnarray*}
h(p)&=&(g+C_\R^q)(p)\,\,\,=\,\,\,[g(p)]\,+\,(C_\R^q)(p)]\,\,\,=\,\,\,[g(p)]\,+\,q\\
\hbox{and }\,\,\,h(x)&=&(g+C_\R^q)(x)\,\,\,=\,\,\,[g(x)]\,+\,(C_\R^q)(x)]\,\,\,=\,\,\,[g(x)]\,+\,x,
\end{eqnarray*}
we conclude that $h(p)<h(x)$, as desired.
QED
\vskip.1in\noindent\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 9: Due on Tuesday 27 March}
\vskip.1in\hrule\vskip.1in\noindent
9-1. Let $V$ and $W$ be normed vector spaces, $T\in L(V,W)$ and $K\ge0$.
Show: \qquad
$[\,\,T\hbox{ is }K\hbox{-bounded}\,\,]\,\,\,\,\Leftrightarrow\,\,\,\,[\,\,\forall x\in V,\,\,\,|Tx|_W\,\le\,K\cdot|x|_V\,\,]$.
\vskip.1in\noindent
{\it Proof:}
{\it Proof of $\Rightarrow$:}
Assume that $T$ is $K$-bounded.
We wish to show: $\forall x\in V,\,|T(x)|_W\le K\cdot|x|_V$.
Given $x\in V$.
Want: $|T(x)|_W\le K\cdot|x|_V$.

Since $T$ is $K$-bounded, we have $\hatT\le K$.
By definition of $\hatT$, we have $|\Ell_T|_W\le\hatT$.
By the polar decomposition (Fact 42.5),
choose $a\ge0$ and $u\in S_V$ s.t.~$x=au$.
Since $u\in S_V$, we get $|u|_V=1$.
Then $K\cdot|x|_V=K\cdot|au|_V=K\cdot|a|\cdot|u|_V=K\cdot|a|\cdot1=|a|\cdot K$.

We have $|T(u)|_W\in|T_*(S_V)|_W=|\Ell_T|_W\le\hatT\le K$.
Then
\begin{eqnarray*}
|T(x)|_W&=&|T(au)|_W\,\,\,=\,\,\,|a\cdot[T(u)]|_W\\
&=&|a|\cdot|T(u)|_W\,\,\,\le\,\,\,|a|\cdot K\,\,\,=\,\,\,K\cdot|x|_V,
\end{eqnarray*}
as desired.
{\it End of proof of $\Rightarrow$.}

{\it Proof of $\Leftarrow$:}
Assume: $\forall x\in V,\,|T(x)|_W\le K\cdot|x|_V$.
We wish to show that $T$ is $K$-bounded.
Want: $\hatT\le K$.
Want: $\sup|\Ell_T|_W\le K$.
We wish to show: $|\Ell_T|_W\le K$.
Want: $\forall r\in|\Ell_T|_W$, $\,$ $r\le K$.
Let $r\in|\Ell_T|_W$ be given.
We want: $r\le K$.

Choose $y\in\Ell_T$ s.t.~$r=|y|_W$.
Since $y\in\Ell_T=T_*(S_V)$,
choose $u\in S_V$ s.t.~$y=T(u)$.
We have $|T(u)|_W\le K\cdot|u|_V$.
Since $u\in S_V$, we have $|u|_V=1$.
Then $r=|y|_W=|T(u)|_W\le K\cdot|u|_V=K\cdot1=K$,
as desired.
{\it End of proof of $\Leftarrow$.}
QED
\vskip.1in\hrule\vskip.1in\noindent
9-2. Let $V$ and $W$ be normed vector spaces and let $K\ge0$.
Let $T\in L(V,W)$ be $K$-bounded.
Show: $T$ is $K$-Lipschitz.
That is, show:
$$\forall x,y\in V,\qquad\qquad|\,\,[T(x)]\,\,-\,\,[T(y)]\,\,|_W\,\,\le\,\,K\cdot|\,x\,-\,y\,|_V.$$
\vskip.1in\noindent
{\it Proof:}
Let $x,y\in V$ be given.
Want: $|\,[T(x)]\,-\,[T(y)]\,|_W\,\le\,K\cdot|x-y|_V$.

By HW\#9-1, we know: $\forall z\in V$, $|T(z)|_W\le K\cdot|z|_V$.
It follows that $|T(x-y)|_W\le K\cdot|x-y|_V$.
By linearity, $T(x-y)=[T(x)]-[T(y)]$.
Then
$|\,[T(x)]\,-\,[T(y)]\,|_W\,=\,|T(x-y)|_W\,\le\,K\cdot|x-y|_V$.
QED
\vskip.1in\hrule\vskip.1in\noindent
9-3. Let $Y$ and $Z$ be metric spaces and let $K>0$.
Let $f:Y\dashrightarrow Z$ be a $K$-Lipschitz function.
Let $x\in\dom[f]$ and let $r>0$.
Show that: $f_*(B_Y(x,r))\subseteq B_Z(f(x),Kr)$.
\vskip.1in\noindent
{\it Proof:}
We want to show: $\forall p\in\dom[f]$,
$$[\,\,p\in B_Y(x,r)\,\,]\quad\Rightarrow\quad[\,\,f(p)\in B_Z(f(x),Kr)\,\,].$$
Let $p\in\dom[f]$ be given.
We want to show:
$$[\,\,p\in B_Y(x,r)\,\,]\quad\Rightarrow\quad[\,\,f(p)\in B_Z(f(x),Kr)\,\,].$$
Assume $p\in B_Y(x,r)$.
We want to show: $f(p)\in B_Z(f(x),Kr)$.

Since $p\in B_Y(x,r)$, we have $d_Y(p,x)<r$.
So, since $f$ is $K$-Lipschitz,
$d_Z(f(p),f(x))<Kr$.
Then $f(p)\in B_Z(f(x),Kr)$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
9-4. Let $V$ and $W$ be normed vector spaces and let $T\in L(V,W)$.
Assume that $T$ is continuous at $0_V$.
Show that $T$ is bounded.
\vskip.1in\noindent
{\it Proof:}
Since $T$ is linear, $T(0_V)=0_W$.
So, since $T$ is continuous at $0_V$,
choose $\delta>0$ s.t., for all $x\in V$,
$$[\,\,|x|_V<\delta\,\,]\quad\Rightarrow\quad[\,\,|Tx|_W<1\,\,].$$
Let $K:=2/\delta$.
We want: $T$ is $K$-bounded.
By $\Leftarrow$ of Remark 43.1 p.~311,
it suffices to show: $\forall u\in S_V$, $|Tu|_W\le K$.
Let $u\in V$ be given:
We want to prove: $|Tu|_W\le K$.
It suffices to show: $|Tu|_W<K$.

Since $K=2/\delta>0$,
we get $|K|=K$.
Also, $|1/K|=|\delta/2|=\delta/2$.
Since $u\in S_V$, we have $|u|_V=1$.
Let $x:=(1/K)u$.
Then we have $|x|_V=|1/K|\cdot|u|_V=(\delta/2)\cdot1=\delta/2<\delta$.
So, by choice of $\delta$,
we see that $|Tx|_W<1$.
Then $K\cdot|Tx|_W<K\cdot1=K$.
Since $x=(1/K)u$, we get $u=Kx$,
and so, by linearity of $T$, we have $Tu=K\cdot(Tx)$.

Then $|Tu|_W=|K\cdot(Tx)|_W=|K|\cdot|Tx|_W=K\cdot|Tx|_W<K$.
QED
\vskip.1in\hrule\vskip.1in\noindent
9-5. Let $f,g:\R\dashrightarrow\R$.
Assume: $\exists M\in\R$ s.t.~$(M,\infty)\subseteq\dom[g'/f'\,]$.
Assume: $\ds{\lim_\infty\,f=0=\lim_\infty\,g}$.
Show: $\ds{\lim_\infty\,(g/f)\,\,=^*\,\,\lim_\infty\,(g'/f')}$.
\vskip.1in\noindent
{\it Proof:}
Define $\phi,\psi:(0,\infty)\drta\R$
by $\phi(t)=f(1/t)$ and $\psi(t)=g(1/t)$.
Then, for all $t\in(0,\infty)$,
we have $(\psi/\phi)(t)=(g/f)(1/t)$.
It follows,
by~(1) of~Fact 42.19,
that $\ds{\lim_\infty\,(g/f)=\lim_{0^+}\,(\psi/\phi)}$.

{\it Claim 1:} $\forall t\in(0,\infty)$,
$(\psi'/\phi')(t)=(g'/f')(1/t)$.
{\it Proof of Claim 1:}
Let $t\in(0,\infty)$ be given.
We wish to show: $(\psi'/\phi')(t)=(g'/f')(1/t)$.

By (2) of Fact 42.19, we have both
$$\phi'(t)\,\,=\,\,-\,\frac{f'(1/t)}{t^2}\qquad\hbox{and}\qquad\psi'(t)\,\,=\,\,-\,\frac{g'(1/t)}{t^2}.$$
Then $[\psi'(t)]/[\phi'(t)]=[g'(t)]/[f'(t)]$,
{\it i.e.}, $(\psi'/\phi')(t)=(g'/f')(1/t)$,
as desired.
{\it End of proof of Claim 1.}

By Claim 1, for all $t\in(0,\infty)$,
we have $(\psi'/\phi')(t)=(g'/f')(1/t)$.
It follows,
by~(1) of~Fact 42.19,
that $\ds{\lim_\infty\,(g'/f')=\lim_{0^+}\,(\psi'/\phi')}$.

By assumption, choose $M\in\R$ such that $(M,\infty)\subseteq\dom[g'/f'\,]$.

{\it Claim 2:} $(0,1/M)\in\dom[\psi'/\phi']$.
{\it Proof of Claim 2:}
We wish to~show:
$\forall t\in(0,1/M),\,t\in\dom[\psi'/\phi']$.
Let $t\in(0,1/M)$ be given.
We wish to~show: $t\in\dom[\psi'/\phi']$.

Since $t\in(0,1/M)$, $1/t\in(M,\infty)$.
Then $1/t\in(M,\infty)\subseteq\dom[g'/f']$,
so $(g'/f')(1/t)\ne\frownie$.
Since $t\in(0,1/M)\subseteq(0,\infty)$,
by Claim 1,
we get $(\psi'/\phi')(t)=(g'/f')(1/t)$.
Then $(\psi'/\phi')(t)=(g'/f')(1/t)\ne\frownie$,
and so $t\in\dom[\psi'/\phi']$, as desired.
{\it End of proof of Claim 2.}

Since $\ds{\lim_\infty\,f=0=\lim_\infty\,g}$,
it follows, from (1) of Fact 42.19,
that $\ds{\lim_{0^+}\,\phi=0=\lim_{0^+}\,\psi}$.
By Claim 2, we have $(0,1/M)\in\dom[\psi'/\phi']$.
Therefore,
by Theorem 42.18 (with $p$ replaced by~$0$, $f$ by $\phi$ and $g$ by $\psi$),
we conclude that $\ds{\lim_{0^+}\,(\phi/\psi)\,=^*\,\lim_{0^+}\,(\phi'/\psi')}$.

Then $\ds{\lim_\infty\,(g/f)=\lim_{0^+}\,(\phi/\psi)\,=^*\,\lim_{0^+}\,(\phi'/\psi')=\lim_\infty\,(g'/f')}$.
QED
\vskip.1in\noindent\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 8: Due on Tuesday 20 March}
\vskip.1in\hrule\vskip.1in\noindent
8-1. Let $W$ be a vector space and let $u,v,v'\in W$.
Assume that $\R v=\R v'$ and that $u\|v$.
Show that $u\|v'$.
\vskip.1in\noindent
{\it Proof:}
We wish to show: $u\in\R v'$ or $v'\in\R u$.
Since $u\|v$, at least one of the following must be true:
\begin{itemize}
\item[(1)]$u\in\R v$ \qquad\qquad or
\item[(2)]$v\in\R u$.
\end{itemize}

{\it Case (1):}
We wish to show: $u\in\R v'$.
Since $u\in\R v$, $\R u\subseteq\R v$.
So, as $\R v=\R v'$, $\R u\subseteq\R v'$.
Then $u\in\R u\subseteq\R v'$.
{\it End of Case (1).}

{\it Case (2):}
We wish to show: $v'\in\R u$.
Since $v\in\R u$, $\R v\subseteq\R u$.
So, as $\R v=\R v'$, $\R v'\subseteq\R u$.
Then $v'\in\R v'\subseteq\R u$.
{\it End of Case (2).}
QED
\vskip.1in\hrule\vskip.1in\noindent
8-2. Let $u,v\in\R^2$.
Show:
$$[\,\,u\|v\,\,]\quad\Leftrightarrow\quad[\,\,(u=0_2)\hbox{ or }(v=0_2)\hbox{ or }(\slp\,u=\slp\,v)\,\,].$$
\vskip.1in\noindent
{\it Proof:}
{\it Proof of $\Rightarrow$:}
Assume that $u\|v$.
We wish to prove:
$$(u=0_2)\quad\hbox{or}\quad(v=0_2)\quad\hbox{or}\quad(\slp\,u=\slp\,v).$$
Equivalently, we want: \quad
$[\,\,(u\ne0_2)\hbox{ and }(v\ne0_2)\,\,]\,\,\Rightarrow\,\,[\,\,\slp\,u=\slp\,v\,\,]$.

Assume $(u\ne0_2)\hbox{ and }(v\ne0_2)$.
We wish to prove: $\slp\,u=\slp\,v$.

Since $u\|v$, at least one of the following is true:
\begin{itemize}
\item[(1)]$u\in\R v$ \qquad or
\item[(2)]$v\in\R u$.
\end{itemize}

{\it Case (1):}
Choose $a\in\R$ such that $u=av$.
Since $av=u\ne0_2=0\cdot v$,
we conclude that $a\ne0$.
Since $(u_1,u_2)=u=av=(av_1,av_2)$,
we see that $u_1=av_1$ and that $u_2=av_2$.
Then
$$\slp\,u\quad=\quad\frac{u_2}{u_1}\quad=\quad\frac{av_2}{av_1}\quad=\quad\frac{v_2}{v_1}\quad=\quad\slp\,v,$$
as desired.
{\it End of Case (1).}

{\it Case (2):}
Choose $a\in\R$ such that $v=au$.
Since $au=v\ne0_2=0\cdot u$,
we conclude that $a\ne0$.
Since $(v_1,v_2)=v=au=(au_1,au_2)$,
we see that $v_1=au_1$ and that $v_2=au_2$.
Then
$$\slp\,u\quad=\quad\frac{u_2}{u_1}\quad=\quad\frac{au_2}{au_1}\quad=\quad\frac{v_2}{v_1}\quad=\quad\slp\,v,$$
as desired.
{\it End of~Case (2).}
{\it End of proof of $\Rightarrow$.}
\vskip.1in\noindent
{\it Proof of $\Leftarrow$:}
Assume
\begin{itemize}
\item[$(*)$] \qquad\qquad $(u=0_2)\quad\hbox{or}\quad(v=0_2)\quad\hbox{or}\quad(\slp\,u=\slp\,v)$.
\end{itemize}
We wish to prove: $u\|v$.
We want: $u\in\R v$ or $v\in\R u$.

At least one of the following must be true:
\begin{itemize}
\item[(1)]$u_1=0=v_1$\qquad or
\item[(2)]$u_1\ne0=v_1$\qquad or
\item[(3)]$u_1=0\ne v_1$\qquad or
\item[(4)]$u_1\ne0\ne v_1$.
\end{itemize}

{\it Case (1):}
We wish to prove that: $\,\,$ $[\,\,u\notin\R v\,\,]\,\,\Rightarrow\,\,[\,\,v\in\R u\,\,]$.
Assume that: $u\notin\R v$.
We wish to prove that: $v\in\R u$.

Since $u\notin\R v$ and $0_2=0\cdot v\in\R v$, we get: $u\ne0_2$.
Then we have: $(0,u_2)=(u_1,u_2)=u\ne0_2=(0,0)$,
so $u_2\ne0$.
Let $a:=v_2/u_2$.
Then $au_2=v_2$.
Then $au=(au_1,au_2)=(a\cdot0,v_2)=(0,v_2)=(v_1,v_2)=v$.
Then $v=au\in\R u$, as desired.
{\it End of Case (1).}

{\it Case (2):}
We wish to show: $v\in\R u$.

Since $u_1\ne0$, we have $(u_1,u_2)\ne(0,0)$, {\it i.e.}, $u\ne0_2$.
Since $u_1\ne0$, we get $u_2/u_1\ne\frownie$.
Since $v_1=0$, we get $v_2/v_1=\frownie$.
Then
$$\slp\,u\quad=\quad u_2/u_1\quad\ne\quad\frownie\quad=\quad v_2/v_1\quad=\quad\slp\,v.$$
Since $u\ne0_2$ and $\slp\,u\ne\slp\,v$,
we see, by $(*)$, that $v=0_2$.
Then $v=0_2=0\cdot u\in\R u$, as desired.
{\it End of Case (2).}

{\it Case (3):}
We wish to show: $u\in\R v$.

Since $v_1\ne0$, we have $(v_1,v_2)\ne(0,0)$, {\it i.e.}, $v\ne0_2$.
Since $v_1\ne0$, we get $v_2/v_1\ne\frownie$.
Since $u_1=0$, we get $u_2/u_1=\frownie$.
Then
$$\slp\,u\quad=\quad u_2/u_1\quad=\quad\frownie\quad\ne\quad v_2/v_1\quad=\quad\slp\,v.$$
Since $v\ne0_2$ and $\slp\,u\ne\slp\,v$,
we see, by $(*)$, that $u=0_2$.
Then $u=0_2=0\cdot v\in\R v$, as desired.
{\it End of~Case (3).}

{\it Case (4):}
We wish to show: $v\in\R u$.

Since $u_1\ne0$, we have $(u_1,u_2)\ne(0,0)$, {\it i.e.}, $u\ne0_2$.
Since $v_1\ne0$, we have $(v_1,v_2)\ne(0,0)$, {\it i.e.}, $v\ne0_2$.
Since $u\ne0_2$ and $v\ne0_2$,
we see, by $(*)$, that $\slp\,u=\slp\,v$.
That is, $u_2/u_1=v_2/v_1$.
Let $a:=v_1/u_1$.
Then $au_1=v_1$
and $au_2=(v_1/u_1)u_2=(u_2/u_1)v_1=(v_2/v_1)v_1=v_2$.
Then $au=(au_1,au_2)=(v_1,v_2)=v$.
Then $v=au\in\R u$, as desired.
{\it End of~Case (4).}
{\it End of proof of $\Leftarrow$.}
QED
\vskip.1in\hrule\vskip.1in\noindent
8-3. Let $u,v\in\R^2$.
Show:
$$(\,\,u\|v\,\,)\quad\Leftrightarrow\quad
\left(\,\,\Det\left[\begin{array}{c}u\\v\end{array}\right]=0\,\,\right).$$
\vskip.1in\noindent
{\it Proof:}
{\it Proof of $\Rightarrow$:}
Assume: $u\|v$.
Want: $\ds{\Det\left[\begin{array}{c}u\\v\end{array}\right]=0}$.

Since $u\|v$, at least one of the following is true:
\begin{itemize}
\item[(1)]$u\in\R v$ \qquad or
\item[(2)]$v\in\R u$.
\end{itemize}

{\it Case (1):}
Choose $a\in\R$ such that $u=av$.
Then
\begin{eqnarray*}
\Det\left[\begin{array}{c}u\\v\end{array}\right]&=&\Det\left[\begin{array}{c}av\\v\end{array}\right]\\
&=&\Det\left[\begin{array}{c}(av_1,av_2)\\(v_1,v_2)\end{array}\right]\\
&=&(av_1)\cdot v_2-(av_2)\cdot v_1\,\,\,=\,\,\,0.
\end{eqnarray*}
{\it End of Case (1).}

{\it Case (2):}
Choose $a\in\R$ such that $v=au$.
Then
\begin{eqnarray*}
\Det\left[\begin{array}{c}u\\v\end{array}\right]&=&\Det\left[\begin{array}{c}u\\au\end{array}\right]\\
&=&\Det\left[\begin{array}{c}(u_1,u_2)\\(au_1,au_2)\end{array}\right]\\
&=&u_1\cdot(au_2)-u_2\cdot(au_1)\,\,\,=\,\,\,0.
\end{eqnarray*}
{\it End of Case (2).}
{\it End of proof of $\Rightarrow$.}
\vskip.1in\noindent
{\it Proof of $\Leftarrow$:}
Assume: $\ds{\Det\left[\begin{array}{c}u\\v\end{array}\right]=0}$.
We wish to prove: $u\|v$.
We want to show: $u\in\R v$ or $v\in\R u$.

Since
$\ds{0=\Det\left[\begin{array}{c}u\\v\end{array}\right]=u_1v_2-u_2v_1}$,
we see that $u_1v_2=u_2v_1$.
At least one of the following is true:
\begin{itemize}
\item[(1)]$u_1\ne0$\qquad or
\item[(2)]$u_2\ne0$\qquad or
\item[(3)]$u_1=0=u_2$.
\end{itemize}

{\it Case (1):}
Want: $v\in\R u$.
Let $a:=v_1/u_1$.
Want: $v=au$.

We have both $au_1=v$ and $au_2=u_2v_1/u_1=u_1v_2/u_1=v_2$.
Then $v=(v_1,v_2)=(au_1,au_2)=au$, as desired.
{\it End of Case (1).}

{\it Case (2):}
Want: $v\in\R u$.
Let $a:=v_2/u_2$.
Want: $v=au$.

We have both $au_1=u_1v_2/u_2=u_2v_1/u_2=v_1$ and $au_2=v_2$.
Then $v=(v_1,v_2)=(au_1,au_2)=au$, as desired.
{\it End of Case (2).}

{\it Case (3):}
Want: $u\in\R v$.
Let $a:=0$.
Want $u=av$.

Since $u_1=0=u_2$, we get $u=0_2$.
Since $a=0$, we get $av=0_2$.
Then $u=0_2=av$, as desired.
{\it End of Case (3).}
{\it End of proof of $\Leftarrow$.}
QED
\vskip.1in\hrule\vskip.1in\noindent
8-4. Let $V$ and $W$ be normed vector spaces.
Let $x:\R\dashrightarrow V$ and $y:\R\dashrightarrow W$.
Let $p\in\R$.
Assume that $x$ and $y$ are both continuous at~$p$.
Show that $(x,y)$ is continuous at $p$.
\vskip.1in\noindent
{\it Proof:}
Let $z:=(x,y)$ and let $U:=V\times W$.
Then $z:\R\dashrightarrow U$
and $\dom[z]\subseteq\dom[x]$
and $\dom[z]\subseteq\dom[y]$.

We want: $z$ is continuous at $p$.
We wish to show: $\forall U_0\in\scrn_U(z(p))$,
$\exists A\in\scrn_\R(p)$ s.t.~$z_*(A)\subseteq U_0$.
Let $U_0\in\scrn_U(z(p))$ be given.
We wish to~prove: $\exists A\in\scrn_\R(p)$ s.t.~$z_*(A)\subseteq U_0$.

Choose $V_0\in\scrn_V(x)$ and $W_0\in\scrn_W(y)$ s.t.~$V_0\times W_0\subseteq U_0$.
Since $x$ is continuous at $p$,
choose $B\in\scrn_\R(p)$ s.t.~$x_*(B)\subseteq V_0$.
Since $y$ is continuous at $p$,
choose $C\in\scrn_\R(p)$ s.t.~$y_*(C)\subseteq W_0$.
As $B,C\in\scrn_\R(p)$,
we get $B\cap C\in\scrn_\R(p)$.
Let $A:=B\cap C$.
We wish to show: $z_*(A)\subseteq U_0$.
We wish to show: $\forall q\in\dom[z]$,
$$[\,\,q\in A\,\,]\quad\Rightarrow\quad[\,\,z(q)\in U_0\,\,].$$
Let $q\in\dom[z]$ be given.
We wish to show:
$$[\,\,q\in A\,\,]\quad\Rightarrow\quad[\,\,z(q)\in U_0\,\,].$$
Assume: $q\in A$.
We wish to show: $z(q)\in U_0$.

Since $q\in\dom[z]\subseteq\dom[x]$
and since $q\in A=B\cap C\subseteq B$,
we have $x(q)\in x_*(B)$.
So, by choice of $B$,
we have $x(q)\in V_0$.
Since $q\in\dom[z]\subseteq\dom[y]$
and since $q\in A=B\cap C\subseteq C$,
we have $y(q)\in y_*(C)$.
So, by choice of $C$,
we have $y(q)\in W_0$.

Then $z(q)=(x(q),y(q))\in V_0\times W_0$.
Then, by choice of $V_0$ and $W_0$,
we conclude that $z(q)\in U_0$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
8-5. Let $V$ and $W$ be normed vector spaces.
Let $x:\R\dashrightarrow V$ and $y:\R\dashrightarrow W$.
Let $p\in\LPD_\R(x,y)$.
Show: $(x,y)'(p)\,\,=^*\,\,(\,x'(p)\,,\,y'(p)\,)$.
\vskip.1in\noindent
{\it Proof:}
Let $\phi:=SS_x^p$, $\psi:=SS_y^p$.
Then $\ds{x'(p)=\lim_0\phi}$ and $\ds{y'(p)=\lim_0\psi}$.
Also, $\ds{(x,y)'(p)=\lim_0\,SS_{(x,y)}^p}$.
For all $h\in\R$, we have
\begin{eqnarray*}
SS_{(x,y)}^p(h)&=&\frac{[(x,y)(p+h)]\,\,-\,\,[(x,y)(p)]}{h}\\
&=&\frac{(\,x(p+h)\,,\,y(p+h)\,)\,\,-\,\,(\,x(p)\,,\,y(p)\,)}{h}\\
&=&\frac{(\,\,\,[x(p+h)]\,\,-\,\,[x(p)]\,\,\,,\,\,\,[y(p+h)]\,\,-\,\,[y(p)]\,\,\,)}{h}\\
&=&\left(\,\,\frac{[x(p+h)]-[x(p)]}{h}\,\,,\,\,\frac{[y(p+h)]-[y(p)]}{h}\,\,\right)\\
&=&(\,\,SS_x^p(h)\,\,,\,\,SS_y^p(h)\,\,)\\
&=&(\,\,SS_x^p\,\,,\,\,SS_y^p\,\,)\,(h)\,\,\,=\,\,\,(\phi,\psi)(h).
\end{eqnarray*}
Then $SS_{(x,y)}^p=(\phi,\psi)$.
Since $p\in\LPD_\R(x,y)$,
by (3) of Remark 26.7 p.~197,
we get $0\in\LPD_\R(SS_{(x,y)}^p)$.
So, since $SS_{(x,y)}^p=(\phi,\psi)$,
we conclude that $0\in\LPD_\R(\phi,\psi)$.
Then, by Remark 40.4 p.~295, we have
$\ds{\lim_0\,(\phi,\psi)\,\,=^*\,\,(\,\lim_0\phi\,,\,\lim_0\psi\,)}$.
Then
\begin{eqnarray*}
(x,y)'(p)&=&\lim_0\,SS_{(x,y)}^p\,\,\,=\,\,\,\lim_0\,(\phi,\psi)\\
&=^*&(\,\lim_0\phi\,,\,\lim_0\psi\,)\,\,\,=\,\,\,(\,x'(p)\,,\,y'(p)\,),\\
\end{eqnarray*}
as desired.
QED
\vskip.1in\noindent\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 7: Due on Tuesday 6 March}
\vskip.1in\hrule\vskip.1in\noindent
7-1. Let $f:\R\dashrightarrow\R$, $p\in\R$ and $\delta>0$.
Assume
\begin{itemize}
\item[(1)]$(p-\delta,p+\delta)\,\,\subseteq\,\,\dom[f]$,
\item[(2)]$f$ is strictly decreasing on $(p-\delta,p]$ \qquad\qquad and
\item[(3)]$f$ is strictly increasing on $[p,p+\delta)$.
\end{itemize}
Show that $f$ has a local unique minimum at $p$ in $\R$.
\vskip.1in\noindent
{\it Proof:}
We want:
$\exists V\in\scrn_\R(p)$ s.t.~$V\subseteq\dom[f]$ and $f_*(V_p^\times)>f(p)$.

Since $(p-\delta,p+\delta)$ is open in $\R$
and since $p\in(p-\delta,p+\delta)$,
it follows that $(p-\delta,p+\delta)\in\scrn_\R(p)$.
Let $V:=(p-\delta,p+\delta)$.
We wish to show: $V\subseteq\dom[f]$ and $f_*(V_p^\times)>f(p)$.

Since $V=(p-\delta,p+\delta)$, by (1),
we have $V\subseteq\dom[f]$.
It remains to show: $f_*(V_p^\times)>f(p)$.
We wish to show: $\forall y\in f_*(V_p^\times)$, $y>f(p)$.
Let $y\in f_*(V_p^\times)$ be given.
We wish to show: $y>f(p)$.

Since $y\in f_*(V_p^\times)$,
choose $t\in V_p^\times$ such that $y=f(t)$.
We have $t\in V_p^\times=(p-\delta,p)\cup(p,p+\delta)$,
so one of the following must be true:
\begin{itemize}
\item[(A)]$t\in(p-\delta,p)$\qquad\qquad or
\item[(B)]$t\in(p,p+\delta)$.
\end{itemize}

{\it Case (A):}
We have $t,p\in(p-\delta,p]$ and $t<p$.
So, by (2), $f(t)>f(p)$.
So, as $y=f(t)$, we get $y>f(p)$.
{\it End of Case (A).}

{\it Case (B):}
We have $t,p\in[p,p+\delta)$ and $t>p$.
So, by (3), $f(t)>f(p)$.
So, as $y=f(t)$, we get $y>f(p)$.
{\it End of Case (B).}
QED
\vskip.1in\hrule\vskip.1in\noindent
7-2. Let $g:\R\dashrightarrow\R$ and $p\in\IntD_\R g$.
Assume both that $g(p)=0$ and that $g'(p)>0$.
Show: $\exists\delta>0$ such that
all three of the following hold:
\begin{itemize}
\item[(A)]$(p-\delta,p+\delta)\,\,\subseteq\,\,\dom[g]$,
\item[(B)]$g<0$ on $(p-\delta,p)$ \qquad\qquad and
\item[(C)]$g>0$ on $(p,p+\delta)$.
\end{itemize}
\vskip.1in\noindent
{\it Proof:}
By Lemma 32.1 p.~240,
choose $U\in\scrn_\R(0)$ s.t.~$(SS_g^p)_*(U)>0$.
Then $U+p\in\scrn_\R(p)$.
Let $D:=\dom[g]$.
Since $p\in\IntD_\R g=\Int_\R D$,
$D\in\scrn_\R(p)$.
As $U+p,D\in\scrn_\R(p)$,
we get $(U+p)\cap D\in\scrn_\R(p)$.
So, since $\scrb_\R(p)$ is a neighborhood base at $p$ in $\R$,
choose $\delta>0$ such that $B_\R(p,\delta)\subseteq(U+p)\cap D$.
We wish to show: (A) and (B) and (C).

{\it Proof of (A):}
We have
$$(p-\delta,p+\delta)\,\,=\,\,B_\R(p,\delta)\,\,\subseteq\,\,(U+p)\cap D\,\,\subseteq\,\,D\,\,=\,\,\dom[g],$$
as desired.
{\it End of proof of (A).}

{\it Proof of (B):}
We wish to show: $\forall t\in(p-\delta,p)$, $g(t)<0$.
Let $t\in(p-\delta,p)$ be given.
We wish to show: $g(t)<0$.

We have
$t\in(p-\delta,p)\subseteq B_\R(p,\delta)\subseteq(U+p)\cap D$,
so $t\in U+p$ and $t\in D$.
Since $t\in U+p$, we get $t-p\in U$.
Let $h:=t-p$.
Then $h\in U$ and $p+h=t$.
Also, since $t<p$, we get $h<0$.
In particular, $h\ne0$.
We have $\dom[SS_g^p]=[(\dom[g])-p]_0^\times=[D-p]_0^\times$.
So, since $h=t-p\in D-p$ and since $h\ne0$,
we get $h\in\dom[SS_g^p]$.
So, since $h\in U$,
we get $(SS_g^p)(h)\in(SS_g^p)_*(U)$.
So, since $(SS_g^p)_*(U)>0$,
we get $(SS_g^p)(h)>0$.
So, since $h<0$,
we get $h\cdot[(SS_g^p)(h)]<0$.
So, since $h\cdot[(SS_g^p)(h)]=[g(p+h)]-[g(p)]$,
we get $g(p+h)<g(p)$.
So, since $p+h=t$,
we get $g(t)<g(p)$.
By assumption, $g(p)=0$.
Then $g(t)<0$, as desired.
{\it End of proof of (B).}

{\it Proof of (C):}
We wish to show: $\forall t\in(p,p+\delta)$, $g(t)>0$.
Let $t\in(p,p+\delta)$ be given.
We wish to show: $g(t)>0$.

We have
$t\in(p,p+\delta)\subseteq B_\R(p,\delta)\subseteq(U+p)\cap D$,
so $t\in U+p$ and $t\in D$.
Since $t\in U+p$, we get $t-p\in U$.
Let $h:=t-p$.
Then $h\in U$ and $p+h=t$.
Also, since $t>p$, we get $h>0$.
In particular, $h\ne0$.
We have $\dom[SS_g^p]=[(\dom[g])-p]_0^\times=[D-p]_0^\times$.
So, since $h=t-p\in D-p$ and since $h\ne0$,
we get $h\in\dom[SS_g^p]$.
So, since $h\in U$,
we get $(SS_g^p)(h)\in(SS_g^p)_*(U)$.
So, since $(SS_g^p)_*(U)>0$,
we get $(SS_g^p)(h)>0$.
So, since $h>0$,
we get $h\cdot[(SS_g^p)(h)]>0$.
So, since $h\cdot[(SS_g^p)(h)]=[g(p+h)]-[g(p)]$,
we get $g(p+h)>g(p)$.
So, since $p+h=t$,
we get $g(t)>g(p)$.
By assumption, $g(p)=0$.
Then $g(t)>0$, as desired.
{\it End of proof of (C).}
QED
\vskip.1in\hrule\vskip.1in\noindent
7-3. Let $f:\R\dashrightarrow\R$ and let $p\in\IntD_\R(f')$.
Assume both that $f'(p)=0$ and that $f''(p)>0$.
Show: $f$ has a local unique minimum at $p$ in~$\R$.
\vskip.1in\noindent
{\it Proof:}
Let $g:=f'$.
Then $p\in\IntD_\R g$ and $g(p)=0$ and $g'(p)>0$.
So, by HW\#7-2,
choose $\delta>0$ such that
all three of the following hold:
\begin{itemize}
\item[(A)]$(p-\delta,p+\delta)\,\,\subseteq\,\,\dom[g]$,
\item[(B)]$g<0$ on $(p-\delta,p)$ \qquad\qquad and
\item[(C)]$g>0$ on $(p,p+\delta)$.
\end{itemize}
By HW\#7-1, it suffices to show:
\begin{itemize}
\item[(1)]$(p-\delta,p+\delta)\,\,\subseteq\,\,\dom[f]$,
\item[(2)]$f$ is strictly decreasing on $(p-\delta,p]$ \qquad\qquad and
\item[(3)]$f$ is strictly increasing on $[p,p+\delta)$.
\end{itemize}

{\it Proof of (1):}
Since $\dom[g]=\dom[f']\subseteq\dom[f]$,
by (A), we see that $(p-\delta,p+\delta)\subseteq\dom[f]$, as desired.
{\it End of proof of (1).}

{\it Proof of (2):}
Let $I:=(p-\delta,p]$.
We want: $f|I$ is strictly decreasing.

Since $g=f'$, by (A), we see that $f$ is differentiable on $(p-\delta,p+\delta)$.
So, as $I\subseteq(p-\delta,p+\delta)$, $f$ is differentiable on $I$.
Then $f$ is c/d on~$I$.
Since $g=f'$ and $\Int_\R I=(p-\delta,p)$, by (B), we get: $f'<0$ on~$\Int_\R I$.
Then, by (5) of Corollary 34.9 p.~259,
we see that $f|I$ is strictly decreasing, as desired.
{\it End of proof of (2).}

{\it Proof of (3):}
Let $I:=[p,p+\delta)$.
We want: $f|I$ is strictly increasing.

Since $g=f'$, by (A), we see that $f$ is differentiable on $(p-\delta,p+\delta)$.
So, as $I\!\subseteq\!(p-\delta,p+\delta)$, $f$ is differentiable on $I$.
Then $f$ is c/d on~$I$.
Since $g=f'$ and $\Int_\R I=(p,p+\delta)$, by (B), we get: $f'>0$ on~$\Int_\R I$.
Then, by (3) of Corollary 34.9 p.~259,
we see that $f|I$ is strictly increasing, as desired.
{\it End of proof of (3).}
QED
\vskip.1in\hrule\vskip.1in\noindent
7-4. Let $f:\R\dashrightarrow\R$ and let $p\in\IntD_\R(f')$.
Assume both that $f'(p)=0$ and that $f''(p)<0$.
Show: $f$ has a local unique maximum at $p$ in~$\R$.
\vskip.1in\noindent
{\it Proof:}
We wish to show $\exists V\in\scrn_\R(p)$ such that
$V\subseteq\dom[f]$ and $f_*(V_p^\times)<f(p)$.

Let $f_0:=-f$.
Then $\dom[f_0]=\dom[f]$.
Also, $f'_0=-f'$.
Then $\dom[f'_0]=\dom[f']$,
so $\IntD_\R(f'_0)=\IntD_\R(f')$.
Also, $f_0''=-f''$.
Then
\begin{itemize}
\item$f_0:\R\dashrightarrow\R$,
\item$p\in\IntD_\R(f')=\IntD_\R(f'_0)$,
\item$f'_0(p)=(-f')(p)=-0=0$ \qquad\qquad and
\item$f''_0(p)=(-f'')(p)=-(f''(p))>0$.
\end{itemize}
Then, by HW\#7-3,
$f_0$ has a local unique minimum at $p$ in $\R$.
So choose $V\in\scrn_\R(p)$ such that
$V\subseteq\dom[f_0]$ and such that $(f_0)_*(V_p^\times)>f_0(p)$.
As $V\subseteq\dom[f_0]=\dom[f]$,
it remains only to show: $f_*(V_p^\times)<f(p)$.
We wish to show: $\forall t\in\dom[f]$,
$$[\,\,t\in V_p^\times\,\,]\quad\Rightarrow\quad[\,\,f(t)<f(p)\,\,].$$
Let $t\in\dom[f]$ be given.
We wish to show:
$$[\,\,t\in V_p^\times\,\,]\quad\Rightarrow\quad[\,\,f(t)<f(p)\,\,].$$
Assume: $t\in V_p^\times$.
We wish to prove: $f(t)<f(p)$.

Since $t\in\dom[f]=\dom[f_0]$,
and $t\in V_p^\times$,
we get $f_0(t)\in(f_0)_*(V_p^\times)$.
So, since $(f_0)_*(V_p^\times)>f_0(p)$,
we get $f_0(t)>f_0(p)$.
Then
$$-\,(f(t))\,\,=\,\,(-f)(t)\,\,=\,\,f_0(t)\,\,>\,\,f_0(p)\,\,=\,\,(-f)(p)\,\,=\,\,-\,(f(p)).$$
Multiplying by $-1$, we get $f(t)<f(p)$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
7-5. Let $f:\R\dashrightarrow\R$.
Asume both that $0\in\IntD_\R f$ and that $f(0)=0$.
Define $g:\R\to\R$ by $g(x)=x^4$.
Assume that $f/g\to1$ near $0$.
Show that $f$ has a local unique minimum at $0$ in $\R$.
\vskip.1in\noindent
{\it Proof:}
Want: $\exists V\in\scrn_\R(0)$ s.t.~$V\subseteq\dom[f]$ and $f(0)<f_*(V_0^\times)$.

Let $D:=\dom[f]$.
Then $0\in\IntD_\R f=\Int_\R D$.
Then $D\in\scrn_\R(0)$.
So, since $\scrb_\R(0)$ is a neighborhood base at $0$ in $\R$,
choose $\alpha>0$ s.t. $B_\R(0,\alpha)\subseteq D$.
Since $f/g\to1$ near $0$,
choose $\beta>0$ s.t., $\forall t\in\dom[f/g]$,
$$[\,\,0<|t|<\beta\,\,]\quad\Rightarrow\quad[\,\,|\,[(f/g)(t)]\,-\,1\,|\,<\,1/2\,\,].$$
Let $\delta:=\min\{\alpha,\beta\}$.
Then $\delta\le\alpha$, so $B(0,\delta)\subseteq B(0,\alpha)$.
Also, $\delta\le\beta$, so $B(0,\delta)\subseteq B(0,\beta)$.
Since $B_\R(0,\delta)$ is open in $\R$
and since $0\in B_\R(0,\delta)$,
it follows that $B_\R(0,\delta)\in\scrn_\R(0)$.
Let $V:=B_\R(0,\delta)$.
We wish to show:
$V\subseteq\dom[f]$ and $f(0)<f_*(V_0^\times)$.

We have $V=B_\R(0,\delta)\subseteq B_\R(0,\alpha)\subseteq D=\dom[f]$.
It remains to~show: $f(0)<f_*(V_0^\times)$.
We wish to show: $\forall y\in f_*(V_0^\times)$, $f(0)<y$.
Let $y\in f_*(V_0^\times)$ be given.
We wish to show: $f(0)<y$.
By assumption, $f(0)=0$.
We wish to show: $0<y$.
Since $y\in f_*(V_0^\times)$, choose $t\in V_0^\times$ such that $y=f(t)$.
We wish to show: $0<f(t)$.

Since $t\in V_0^\times=V\backslash\{0\}$,
we see that $t\ne0$.
Then $t^4>0$ and $0<|t|$.
We have $t\in V_0^\times\subseteq V\subseteq\dom[f]$.
Also, $t\in\R=\dom[g]$.
Also, $g(t)=t^4>0$, so, in particular, $g(t)\ne0$.
Then $t\in\dom[f/g]$.
Also, we have
$t\in V_0^\times\subseteq V= B_\R(0,\delta)\subseteq B_\R(0,\beta)$,
so $|t|<\beta$.
Since $t\in\dom[f/g]$ and since $0<|t|<\beta$,
by the choice of~$\beta$,
$|\,[(f/g)(t)]\,-\,1\,|\,<\,1/2$.
Then $1-(1/2)<(f/g)(t)<1+(1/2)$.
Then $1/2<(f/g)(t)$.
Multiplying this inequality by $g(t)$, since $g(t)>0$,
we get $[1/2]\cdot[g(t)]<[(f/g)(t)]\cdot[g(t)]$.
Since $g(t)>0$,
we conclude that $0<[1/2]\cdot[g(t)]$.
By definition of $f/g$,
since $t\in\dom[f/g]$,
we see that $[(f/g)(t)]\cdot[g(t)]=f(t)$.
Then we have
$0<[1/2]\cdot[g(t)]<[(f/g)(t)]\cdot[g(t)]= f(t)$,
as desired.
QED
\vskip.1in\noindent\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 6: Due on Tuesday 27 February}
\vskip.1in\hrule\vskip.1in\noindent
6-1. Let $f:\R\dashrightarrow\R$ be continuous, and let $I\subseteq\dom[f]$.
Assume that $I$ is an interval.
Show: $f_*(I)$ is an interval.
\vskip.1in\noindent
{\it Proof:}
Since $I$ is an interval, $I\ne\emptyset$.
So, since $I\subseteq\dom[f]$,
we get $f_*(I)\ne\emptyset$.
So, by $\Leftarrow$ of Fact 37.1 p.~276, we need only show:
$\forall a,b\in f_*(I)$, $[a|b]\subseteq f_*(I)$.
Given $a,b\in f_*(I)$.
Want: $[a|b]\subseteq f_*(I)$.

Since $a,b\in f_*(I)$,
choose $s,t\in I$ such that $a=f(s)$ and $b=f(t)$.
Since $s,t\in I$ and $I$ is an interval,
we conclude, by $\Rightarrow$ of Fact 37.1 p.~276, that $[s|t]\subseteq I$.
Then $f_*([s|t])\subseteq f_*(I)$.
Also, $[s|t]\subseteq I\subseteq\dom[f]$,
so, as $f$ is continuous,
$f$ is continuous on $[s|t]$.
Then, by the Intermediate Value Theorem (Theorem 29.7 p.~218),
$[f(s)|f(t)]\subseteq f_*([s|t])$.
Then $[a|b]=[f(s)|f(t)]\subseteq f_*([s|t])\subseteq f_*(I)$,
as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
6-2. Let $X,Y\subseteq\R$.
Let $f:X\hookrightarrow>Y$ be continuous.
Let $X_0:=\Int_\R X$.
Show that $f^{-1}$ is continuous on $f_*(X_0)$.
\vskip.1in\noindent
{\it Proof:}
We wish to show:
$\forall q\in f_*(X_0)$,
$f^{-1}$ is continuous at $q$.
Let $q\in f_*(X_0)$ be given.
We wish to show:
$f^{-1}$ is continuous at $q$.
Let $g:=f^{-1}$.
We wish to show: $g$ is continuous at $q$.
We wish to show: $\forall U\in\scrn_\R(g(q))$,
$\exists V\in\scrn_\R(q)$
s.t.~$g_*(V)\subseteq U$.
Let $U\in\scrn_\R(g(q))$ be given.
We wish to show:
$\exists V\in\scrn_\R(q)$
s.t.~$g_*(V)\subseteq U$.

Since $q\in f_*(X_0)$,
choose $p\in X_0$ s.t.~$q=f(p)$.
Then $g(q)=p$.
Then $U\in\scrn_\R(g(q))=\scrn_\R(p)$.
That is, $U$ is a neighborhood of $p$ in $X$.
Choose an open $U_0$ in $\R$ s.t.~$p\in U_0\subseteq U$.
Let $V:=f_*(U_0\cap X_0)$.
We wish to show: both $V\in\scrn_\R(q)$
and $g_*(V)\subseteq U$.

Since $g=f^{-1}$ and since $U_0\cap X_0\subseteq X_0\subseteq X=\dom[f]$,
it follows that $g_*(f_*(U_0\cap X_0))=U_0\cap X_0$.
Then
$$g_*(V)\,\,\,=\,\,\,g_*(f_*(U_0\cap X_0))\,\,\,=\,\,\,U_0\cap X_0\,\,\,\subseteq\,\,\,U_0\,\,\,\subseteq\,\,\,U.$$
It remains to show: $V\in\scrn_\R(q)$.

Since $X_0=\Int_\R X$,
we get: $X_0$ is open in $\R$.
So, since $U_0$ is open in $\R$,
we see that $U_0\cap X_0$ is open in $\R$.
Also, $U_0\cap X_0\subseteq X_0\subseteq X=\dom[f]$.
Also, $f:\R\dashrightarrow\R$ is 1-1 and continuous.
Therefore, by Invariance of~Domain (Theorem 30.3 p.~227),
$f_*(U_0\cap X_0)$ is open in $\R$.
That is, $V$ is open in $\R$.
Since $p\in U_0$ and $p\in X_0$,
we get $p\in U_0\cap X_0$.
So, since $p\in X_0\subseteq X=\dom[f]$,
we get $f(p)\in f_*(U_0\cap X_0)$.
Then $q=f(p)\in f_*(U_0\cap X_0)=V$.
By Remark 16.4 p.~117,
any open set is a~neighborhood of each of its points.
So, since $q\in V$ and since $V$ is open in $\R$,
it follows that $V\in\scrn_\R(q)$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
6-3. Let $f:\R\dashrightarrow\R$ be str.~increasing.
Show: $f^{-1}$ is str.~increasing.
\vskip.1in\noindent
{\it Proof:}
Let $g:=f^{-1}$.
We wish to show: $g$ is str.~increasing.
We wish to show:
$\forall s,t\in\dom[g]$,
$(\,[s<t]\,\Rightarrow\,[g(s)<g(t)]\,)$.
Let $s,t\in\dom[g]$ be given.
We wish to show:
$(\,[s<t]\,\Rightarrow\,[g(s)<g(t)]\,)$.
Assume that $s<t$.
We want: $g(s)<g(t)$.
Let $a:=g(s)$, $b:=g(t)$.
We want: $a<b$.
Assume that $a\ge b$.
We aim for a contradiction.

Since $f$ is strictly increasing,
it follows that $f$ is semiincreasing.
So, since $a\ge b$,
we get $f(a)\ge f(b)$.
Since $a=g(s)$, we get $f(a)=s$.
Since $b=g(t)$, we get $f(b)=t$.
Then $s=f(a)\ge f(b)=t$, so $t\le s$.
Then $t\le s<t$, so $t<t$.
Contradiction.
QED
\vskip.1in\hrule\vskip.1in\noindent
6-4. Let $X,Y\subseteq\R$.
Let $f:X\hookrightarrow>Y$ be strictly increasing.
Let $a\in\R$.
Assume: $X=[a,\infty)$.
Show: $f^{-1}$ is continuous at $f(a)$.
\vskip.1in\noindent
{\it Proof:}
Let $g:=f^{-1}$ and let $q=f(a)$.
We wish to show: $g$ is continuous at $q$.
We wish to show:
$\forall\varepsilon>0$, $\exists\delta>0$ s.t., $\forall t\in\dom[g]$,
$$[\,\,|t-q|<\delta\,\,]\quad\Rightarrow\quad[\,\,|[g(t)]-[g(q)]|<\varepsilon\,\,].$$
Let $\varepsilon>0$ be given.
We wish to show: $\exists\delta>0$ s.t., $\forall t\in\dom[g]$,
$$[\,\,|t-q|<\delta\,\,]\quad\Rightarrow\quad[\,\,|[g(t)]-[g(q)]|<\varepsilon\,\,].$$

We have $a,a+\varepsilon\in[a,\infty)=X=\dom[f]$.
So, since $f$ is strictly increasing
and $a<a+\varepsilon$,
we get $f(a)<f(a+\varepsilon)$.
We conclude that $0<[f(a+\varepsilon)]-[f(a)]$.
Let $\delta:=[f(a+\varepsilon)]-[f(a)]$.
Want: $\forall t\in\dom[g]$,
$$[\,\,|t-q|<\delta\,\,]\quad\Rightarrow\quad[\,\,|[g(t)]-[g(q)]|<\varepsilon\,\,].$$
Let $t\in\dom[g]$ be given.
We wish to show:
$$[\,\,|t-q|<\delta\,\,]\quad\Rightarrow\quad[\,\,|[g(t)]-[g(q)]|<\varepsilon\,\,].$$
Assume that $|t-q|<\delta$.
We wish to show: $|[g(t)]-[g(q)]|<\varepsilon$.

Let $s:=g(t)$.
Since $q=f(a)$, we get $g(q)=a$.
We want: $|s-a|<\varepsilon$.
Equivalently, we wish to show:
$a-\varepsilon<s<a+\varepsilon$.

Since $t\in\dom[g]$, we get $g(t)\in\im[g]$.
Since $g=f^{-1}$, we get $\im[g]=\dom[f]$.
Then $s=g(t)\in\im[g]=\dom[f]=X=[a,\infty)$,
so $s\ge a$.
Then $a-\varepsilon<a\le s$.
It remains to show: $s<a+\varepsilon$.

Since $|t-q|<\delta$, we see that $q-\delta<t<q+\delta$.
Then
$$t\,\,\,<\,\,\,q+\delta\,\,\,=\,\,\,[f(a)]+[f(a+\varepsilon)]-[f(a)]\,\,\,=\,\,\,f(a+\varepsilon).$$
Since $a+\varepsilon\in\dom[f]$,
we get $f(a+\varepsilon)\in\im[f]$.
Since $g=f^{-1}$, we get $\dom[g]=\im[f]$.
Then $f(a+\varepsilon)\in\im[f]=\dom[g]$.
By HW\#6-3, we see that $f^{-1}$ is strictly increasing.
That is, $g$ is strictly increasing.
So, since $t<f(a+\varepsilon)$, since $t\in\dom[g]$ and since $f(a+\varepsilon)\in\dom[g]$,
we get $g(t)<g(f(a+\varepsilon))$.
Since $a+\varepsilon\in\dom[f]$
and $g=f^{-1}$,
we get $g(f(a+\varepsilon))=a+\varepsilon$.
Then $s=g(t)<g(f(a+\varepsilon))=a+\varepsilon$.
QED
\vskip.1in\hrule\vskip.1in\noindent
6-5. Let $X,Y\subseteq\R$.
Let $f:X\hookrightarrow>Y$ be strictly increasing.
Let $a\in\R$.
Assume: $\exists b\in(a,\infty)$ s.t.~$X=[a,b)$.
Show: $f^{-1}$ is continuous at $f(a)$.
\vskip.1in\noindent
{\it Proof:}
Let $g:=f^{-1}$ and let $q=f(a)$.
We wish to show: $g$ is continuous at $q$.
We wish to show:
$\forall\varepsilon>0$, $\exists\delta>0$ s.t., $\forall t\in\dom[g]$,
$$[\,\,|t-q|<\delta\,\,]\quad\Rightarrow\quad[\,\,|[g(t)]-[g(q)]|<\varepsilon\,\,].$$
Let $\varepsilon>0$ be given.
We wish to show: $\exists\delta>0$ s.t., $\forall t\in\dom[g]$,
$$[\,\,|t-q|<\delta\,\,]\quad\Rightarrow\quad[\,\,|[g(t)]-[g(q)]|<\varepsilon\,\,].$$

Choose $b\in(a,\infty)$ s.t.~$X=[a,b)$.
Let $c:=(a+b)/2$.
Since $b\in(a,\infty)$,
it follows that $a<c<b$.
Then $c-a>0$.
Let $\varepsilon_0:=\min\{\varepsilon,c-a\}$.
Then $\varepsilon_0>0$, so $a<a+\varepsilon_0$.
Also, $\varepsilon_0\le c-a$.
Then $a+\varepsilon_0\le a+c-a=c<b$.
Then $a<a+\varepsilon_0<b$.
Then $a+\varepsilon_0\in(a,b)\subseteq[a,b)$.

We have $a,a+\varepsilon_0\in[a,b)=X=\dom[f]$.
So, since $f$ is strictly increasing
and since $a<a+\varepsilon_0$,
it follows that $f(a)<f(a+\varepsilon_0)$.
Therefore, we have $0<[f(a+\varepsilon_0)]-[f(a)]$.
Let $\delta:=[f(a+\varepsilon_0)]-[f(a)]$.
We wish to show: $\forall t\in\dom[g]$,
$$[\,\,|t-q|<\delta\,\,]\quad\Rightarrow\quad[\,\,|[g(t)]-[g(q)]|<\varepsilon\,\,].$$
Let $t\in\dom[g]$ be given.
We wish to show:
$$[\,\,|t-q|<\delta\,\,]\quad\Rightarrow\quad[\,\,|[g(t)]-[g(q)]|<\varepsilon\,\,].$$
Assume that $|t-q|<\delta$.
We wish to show: $|[g(t)]-[g(q)]|<\varepsilon$.

Let $s:=g(t)$.
Since $q=f(a)$, we get $g(q)=a$.
We want: $|s-a|<\varepsilon$.
Equivalently, we wish to show:
$a-\varepsilon<s<a+\varepsilon$.

Since $t\in\dom[g]$, we get $g(t)\in\im[g]$.
Since $g=f^{-1}$, we get $\im[g]=\dom[f]$.
Then $s=g(t)\in\im[g]=\dom[f]=X=[a,b)$,
so $s\ge a$.
Then $a-\varepsilon<a\le s$.
It remains to show: $s<a+\varepsilon$.

Since $|t-q|<\delta$, we see that $q-\delta<t<q+\delta$.
Then
$$t\,\,\,<\,\,\,q+\delta\,\,\,=\,\,\,[f(a)]+[f(a+\varepsilon_0)]-[f(a)]\,\,\,=\,\,\,f(a+\varepsilon_0).$$
Since $a+\varepsilon_0\in\dom[f]$,
we get $f(a+\varepsilon_0)\in\im[f]$.
Since $g=f^{-1}$, we get $\dom[g]=\im[f]$.
Then $f(a+\varepsilon_0)\in\im[f]=\dom[g]$.
By HW\#6-3, we see that $f^{-1}$ is strictly increasing.
That is, $g$ is strictly increasing.
So, since $t<f(a+\varepsilon_0)$, since $t\in\dom[g]$ and since $f(a+\varepsilon_0)\in\dom[g]$,
we get $g(t)<g(f(a+\varepsilon_0))$.
Since $a+\varepsilon_0\in\dom[f]$
and $g=f^{-1}$,
we get $g(f(a+\varepsilon_0))=a+\varepsilon_0$.
Then $s=g(t)<g(f(a+\varepsilon_0))=a+\varepsilon_0$.
Since $a+\varepsilon_0\in\dom[f]$
and $g=f^{-1}$,
we get $g(f(a+\varepsilon_0))=a+\varepsilon_0$.
Then $s=g(t)<g(f(a+\varepsilon_0))=a+\varepsilon_0$.
We have $\varepsilon_0=\min\{\varepsilon,c-a\}\le\varepsilon$.
Then $a+\varepsilon_0\le a+\varepsilon$.
Then $s<a+\varepsilon_0\le a+\varepsilon$, as desired.
QED
\vskip.1in\noindent\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 5: Due on Tuesday 20 February}
\vskip.1in\hrule\vskip.1in\noindent
5-1. Show: $\forall w,x\in\R$,\quad
$\cos(w+x)\,\,=\,\,(\cos w)\cdot(\cos x)\,-\,(\sin w)\cdot(\sin x)$.
\vskip.1in\noindent
{\it Proof:}
Let $w\in\R$ be given.
We wish to show: $\forall x\in\R$,
$$\cos(w+x)\quad=\quad(\cos w)\cdot(\cos x)\,-\,(\sin w)\cdot(\sin x).$$
Define $\tau:\R\to\R$ by $\tau(x)=w+x$.

{\it Claim 1:} $\forall x\in\R$, $\tau'(x)=1$.
{\it Proof of Claim 1:}
Let $x\in\R$ be given.
We wish to show: $\tau'(x)=1$.
We wish to show: $\ds{\lim_0\,SS_\tau^x=1}$.

For all $h\in\R_0^\times$, we have
\begin{eqnarray*}
(SS_\tau^x)(h)&=&\frac{[\tau(x+h)]-[\tau(x)]}{h}\\
&=&\frac{[w+x+h]-[w+x]}{h}\\
&=&\frac{h}{h}\,\,=\,\,1\,\,=\,\,C_\R^1(h).\\
\end{eqnarray*}
Then $SS_\tau^x=C_\R^1$ on $\R_0^\times$,
so $\ds{\lim_0\,SS_\tau^x=\lim_0\,C_\R^1}$.
So, since $\ds{\lim_0\,C_\R^1=1}$,
we get $\ds{\lim_0\,SS_\tau^x=1}$,
as desired.
{\it End of proof of Claim 1.}

Let $f:=\cos\circ\,\tau$.
Then, for all $x\in\R$,
we have
$$f(x)\,\,=\,\,(\cos\circ\,\tau)(x)\,\,=\,\,\cos(\tau(x))\,\,=\,\,\cos(w+x).$$
We therefore wish to show: $\forall x\in\R$,
$$f(x)\quad=\quad(\cos w)\cdot(\cos x)\,-\,(\sin w)\cdot(\sin x).$$

{\it Claim 2:} $\forall x\in\R$, $f'(x)=(-\sin)(w+x)$.
{\it Proof of Claim 2:}
Let $x\in\R$ be given.
We wish to show: $f'(x)=-[\sin(w+x)]$.

We have $\dom[\cos\circ\,\tau]=\R$.
Then
$$x\,\,\in\,\,\R\,\,=\,\,\LP_\R\R\,\,=\,\,\LP_\R(\dom[\exp\circ\,\tau])\,\,=\,\,\LPD_\R(\cos\circ\,\tau).$$
So, since $f=\cos\circ\,\tau$, by the Chain Rule,
$f'(x)\,=^*\,[\cos'(\tau(x))]\,[\tau'(x)]$.
By Claim 1, $\tau'(x)=1$.
Then $f'(x)\,=^*\,\cos'(\tau(x))$.

We have $\cos'=-\sin$ and $\tau(x)=w+x$.
Then
$$f'(x)\,\,=^*\,\,(-\sin)(w+x)\,\,\ne\,\,\frownie.$$
Then $f'(x)\,=\,(-\sin)(w+x)$, as desired.
{\it End of proof of Claim 2.}

{\it Claim 3:} $\forall x\in\R$, $f''(x)=(-\cos)(w+x)$.
{\it Proof of Claim 3:}
By Claim 2, $f'=(-\sin)\circ\,\tau$.
Let $x\in\R$ be given.
We wish to show: $f''(x)=-[\cos(w+x)]$.

We have $\dom[(-\sin)\circ\,\tau]=\R$.
Then
$$x\,\,\in\,\,\R\,\,=\,\,\LP_\R\R\,\,=\,\,\LPD_\R((-\sin)\circ\,\tau).$$
So, since $f'=(-\sin)\circ\,\tau$, by the Chain Rule,
we conclude that
$f''(x)\,=^*\,[(-\sin)'(\tau(x))]\,[\tau'(x)]$.
By Claim 1, $\tau'(x)=1$.
Then $f''(x)\,=^*\,(-\sin)'(\tau(x))$.

We have $(-\sin)'=-\cos$ and $\tau(x)=w+x$.
Then
$$f'(x)\,\,=^*\,\,(-\cos)(w+x)\,\,\ne\,\,\frownie.$$
Then $f'(x)\,=\,(-\cos)(w+x)$, as desired.
{\it End of proof of Claim 3.}

{\it Claim 4:} $f''=-f$.
{\it Proof of Claim 4:}
We wish to show: $\forall x\in\R$, $f''(x)=(-f)(x)$.
Let $x\in\R$ be given.
Want: $f''(x)=(-f)(x)$.

By Claim 3, $f''(x)=(-\cos)(w+x)$.
Then $f''(x)=-[\cos(w+x)]$.
By definition of $f$, we have $f(x)=(\cos\circ\,\tau)(x)$.
Then
$$f(x)\,\,=\,\,\cos(\tau(x))\,\,=\,\,\cos(w+x).$$
We conclude that
$f''(x)=-[\cos(w+x)]=-[f(x)]=(-f)(x)$,
as desired.
{\it End of proof of Claim 4.}

Let $a:=f(0)$ and let $b:=f'(0)$.
By Claim 4 and Theorem 35.2,
$f=a\cdot\cos+\,b\cdot\sin$.
By definition of $f$, $f(0)=\cos(w+0)$.
Then $a=f(0)=\cos(w+0)=\cos w$.
By Claim 2, $f'(0)=(-\sin)(w+0)$.
Then $b=f'(0)=(-\sin)(w+0)=-(\sin w)$.
Then: $\forall x\in\R$,
\begin{eqnarray*}
f(x)&=&(a\cdot\cos+\,b\cdot\sin)(x)\\
&=&a\cdot(\cos x)+b\cdot(\sin x)\\
&=&(\cos w)\cdot(\cos x)\,-\,(\sin w)\cdot(\sin x),
\end{eqnarray*}
as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
5-2. Show: $\forall x\in\R$,
\begin{eqnarray*}
\sin(2x)&=&2\cdot(\sin x)\cdot(\cos x)\qquad\qquad\hbox{and}\\
\cos(2x)&=&(\cos^2 x)\,-\,(\sin^2x).
\end{eqnarray*}
\vskip.1in\noindent
{\it Proof:}
Let $x\in\R$ be given.
We wish to show:
\begin{eqnarray*}
\sin(2x)&=&2\cdot(\sin x)\cdot(\cos x)\qquad\qquad\hbox{and}\\
\cos(2x)&=&(\cos^2 x)\,-\,(\sin^2x).
\end{eqnarray*}

Let $w:=x$.
Then $\sin w=\sin x$ and $\cos w=\cos x$.
By Theorem 35.3,
we have
$\sin(w+x)\,=\,(\sin w)\cdot(\cos x)\,+\,(\cos w)\cdot(\sin x)$.
Then
\begin{eqnarray*}
\sin(2x)&=&\sin(x+x)\,\,=\,\,\sin(w+x)\\
&=&(\sin w)\cdot(\cos x)\,+\,(\cos w)\cdot(\sin x)\\
&=&(\sin x)\cdot(\cos x)\,+\,(\cos x)\cdot(\sin x)\\
&=&2\cdot(\sin x)\cdot(\cos x).
\end{eqnarray*}
It remains to show:
$\cos(2x)\,=\,(\cos^2 x)\,-\,(\sin^2x)$.

By HW\#5-1,
$\cos(w+x)\,=\,(\cos w)\cdot(\cos x)\,-\,(\sin w)\cdot(\sin x)$.
Then
\begin{eqnarray*}
\cos(2x)&=&\cos(x+x)\,\,=\,\,\cos(w+x)\\
&=&(\cos w)\cdot(\cos x)\,-\,(\sin w)\cdot(\sin x)\\
&=&(\cos x)\cdot(\cos x)\,-\,(\sin x)\cdot(\sin x)\\
&=&(\cos^2x)\,-\,(\sin^2x),
\end{eqnarray*}
as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
5-3. Let $f:\R\to\R$, $a:=f(0)$.
Assume $f'=f$.
Show: $f=a\cdot\exp$.
\vskip.1in\noindent
{\it Proof:}
Let $\phi:=f-(a\cdot\exp)$.
We wish to show: $\phi=C_\R^0$.
We have
$$\phi(0)\,\,=\,\,[f(0)]\,-\,a\cdot[\exp(0)]\,\,=\,\,[a]\,-\,a\cdot[1]\,\,=\,\,0.$$
So, by Theorem 35.14, it suffices to show:
$\phi'=\phi$.
We wish to show:
$\forall x\in\R$, $\phi'(x)=\phi(x)$.
Let $x\in\R$ be given.
Want: $\phi'(x)=\phi(x)$.

We have $\dom[f]=\R$
and $\dom[\exp]=\R$,
so $\dom[\phi]=\R$.
Then $x\in\R=\dom[\phi]=\dom[f-(a\cdot\exp)$.
So, by linearity of differentiation
$$(f-(a\cdot\exp))'(x)\,\,=^*\,\,[f'(x)]\,-\,a\cdot[\exp'(x)].$$
So, since $f-(a\cdot\exp)=\phi$, we get  $\phi'(x)\,=^*\,[f'(x)]\,-\,a\cdot[\exp'(x)]$.
By~assumption, we have $f'=f$.
Then, since $\exp'=\exp$, we see that
$\phi'(x)\,=^*\,[f(x)]\,-\,a\cdot[\exp(x)]$.
So, since $\phi(x)=[f(x)]\,-\,a\cdot[\exp(x)]$,
we get $\phi'(x)\,=^*\,\phi(x)$.
Since $x\in\R=\dom[\phi]$, we get $\phi(x)\ne\frownie$.
Then $\phi'(x)\,=^*\,\phi(x)\,\ne\,\frownie$,
and so $\phi'(x)=\phi(x)$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
5-4. Show:
$\forall w,x\in\R$, \quad
$\exp(w+x)\,\,=\,\,(\exp w)\cdot(\exp x)$.
\vskip.1in\noindent
{\it Proof:}
Let $w\in\R$ be given.
We wish to show: $\forall x\in\R$,
$$\exp(w+x)\quad=\quad(\exp w)\cdot(\exp x).$$
Define $\tau:\R\to\R$ by $\tau(x)=w+x$.

{\it Claim 1:} $\forall x\in\R$, $\tau'(x)=1$.
{\it Proof of Claim 1:}
Let $x\in\R$ be given.
We wish to show: $\tau'(x)=1$.
We wish to show: $\ds{\lim_0\,SS_\tau^x=1}$.

For all $h\in\R_0^\times$, we have
\begin{eqnarray*}
(SS_\tau^x)(h)&=&\frac{[\tau(x+h)]-[\tau(x)]}{h}\\
&=&\frac{[w+x+h]-[w+x]}{h}\\
&=&\frac{h}{h}\,\,=\,\,1\,\,=\,\,C_\R^1(h).\\
\end{eqnarray*}
Then $SS_\tau^x=C_\R^1$ on $\R_0^\times$,
so $\ds{\lim_0\,SS_\tau^x=\lim_0\,C_\R^1}$.
So, since $\ds{\lim_0\,C_\R^1=1}$,
we get $\ds{\lim_0\,SS_\tau^x=1}$,
as desired.
{\it End of proof of Claim 1.}

Let $f:=\exp\circ\,\tau$.
Then, for all $x\in\R$,
we have
$$f(x)\,\,=\,\,(\exp\circ\,\tau)(x)\,\,=\,\,\exp(\tau(x))\,\,=\,\,\exp(w+x).$$
We therefore wish to show: $\forall x\in\R$,
$$f(x)\quad=\quad(\exp w)\cdot(\exp x).$$

{\it Claim 2:} $\forall x\in\R$, $f'(x)=\exp(w+x)$.
{\it Proof of Claim 2:}
Let $x\in\R$ be given.
We wish to show: $f'(x)=\exp(w+x)$.

We have $\dom[\exp\circ\,\tau]=\R$.
Then
$$x\,\,\in\,\,\R\,\,=\,\,\LP_\R\R\,\,=\,\,\LP_\R(\dom[\exp\circ\,\tau])\,\,=\,\,\LPD_\R(\exp\circ\,\tau).$$
So, since $f=\exp\circ\,\tau$, by the Chain Rule,
$f'(x)\,=^*\,[\exp'(\tau(x))]\,[\tau'(x)]$.
By Claim 1, $\tau'(x)=1$.
Then $f'(x)\,=^*\,\exp'(\tau(x))$.

We have $\exp'=\exp$ and $\tau(x)=w+x$.
Then
$$f'(x)\,\,=^*\,\,\exp(w+x)\,\,\ne\,\,\frownie.$$
Then $f'(x)\,=\,\exp(w+x)$, as desired.
{\it End of proof of Claim 2.}

{\it Claim 3:} $f'=f$.
{\it Proof of Claim 3:}
We wish to show: $\forall x\in\R$, $f'(x)=f(x)$.
Let $x\in\R$ be given.
Want: $f'(x)=f(x)$.

By Claim 2, we have $f'(x)=\exp(w+x)$.
By definition of $f$, we have $f(x)=(\exp\circ\,\tau)(x)$.
Then $f(x)\,=\,\exp(\tau(x))\,=\,\exp(w+x)$.
We conclude: $f'(x)=\exp(w+x)=f(x)$.
{\it End of proof of Claim 3.}

Let $a:=f(0)$.
By Claim 3 and HW\#5-3, $f=a\cdot\exp$.
By definition of $f$, $f(0)=\exp(w+0)$.
Then $a=f(0)=\exp(w+0)=\exp w$.
Then: $\forall x\in\R$,
$f(x)=(a\cdot\exp)(x)=a\cdot(\exp x)=(\exp w)\cdot(\exp x)$.
QED
\vskip.1in\hrule\vskip.1in\noindent
5-5. Let $f:\R\to\R$.
Assume that $f'=f$.
Let $g:=f^2$.
Show: $g'=2g$.
\vskip.1in\noindent
{\it Proof:}
Define $P:\R\to\R$ by $P(x)=x^2$.
By Theorem 16.3, for all $x\in\R$, we have $P'(x)=2x$.
Also, for all $x\in\R$, we have
$$(P\circ f)(x)\,\,=\,\,P(f(x))\,\,=\,\,[f(x)]^2\,\,=\,\,f^2(x)\,\,=\,\,g(x).$$
Then $g=P\circ f$.
We wish to show: $\forall x\in\R$, $g'(x)=(2g)(x)$.
Let $x\in\R$ be given.
We wish to show: $g'(x)=(2g)(x)$.

We have $\dom[P\circ f]=\R$.
so $\LPD_\R(P\circ f)=\LP_\R\R$.
Then we have $x\in\R=\LP_\R\R=\LPD_\R(P\circ f)$.
So, by the Chain Rule, we see that
$(P\circ f)'(x)\,=^*\,[P'(f(x))]\cdot[f'(x)]$.
Then $g'(x)\,=^*\,[P'(f(x))]\cdot[f'(x)]$.
We have $P'(f(x))=2\cdot[f(x)]$.
By assumption, $f'=f$, so $f'(x)=f(x)$.
Then $g'(x)\,=^*\,2\cdot[f(x)]^2\,\ne\,\frownie$,
so $g'(x)\,=\,2\cdot[f(x)]^2$.
We have $g(x)=f^2(x)=[f(x)]^2$.
Then $g'(x)=2\cdot[g(x)]=(2g)(x)$.
QED
\vskip.1in\noindent\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 4: Due on Tuesday 13 February}
\vskip.1in\hrule\vskip.1in\noindent
4-1. Let $m\in\R$.
Define $\lambda:\R\to\R$ by $\lambda(x)=mx$.
Show: $\lambda'=C_\R^m$.
\vskip.1in\noindent
{\it Proof:}
We want:
$\forall x\in\R$, $\lambda'(x)=C_\R^m(x)$.
Let $x\in\R$ be given.
We want: $\lambda'(x)=C_\R^m(x)$.
We have $\C_\R^m(x)=m$.
We want: $\lambda'(x)=m$.

Let $\phi:=C_{\R_0^\times}^m$.
Since $\phi=C_\R^m$ on $\R_0^\times$,
it follows that $\ds{\lim_0\phi=\lim_0C_\R^m}$.
So, since $\ds{\lim_0C_\R^m=m}$, we get $\ds{\lim_0\,\phi=m}$.
We want: $\ds{\lim_0\,SS_\lambda^x=m}$.
It therefore suffices to show: $\ds{SS_\lambda^x=\phi}$.

As $\R_0^\times$ is a common superdomain of $SS_\lambda^x$ and $\phi$,
it suffices to show:
$\forall h\in\R_0^\times$, $SS_\lambda^x(h)=\phi(h)$.
Let $h\in\R_0^\times$ be given.
Want: $SS_\lambda^x(h)=\phi(h)$.

We have $\phi(h)=C_{\R_0^\times}^m(h)=m$.
Then
\begin{eqnarray*}
SS_\lambda^x(h)&=&\frac{[\lambda(x+h)]-[\lambda(x)]}{h}\,\,=\,\,\frac{[m(x+h)]-[mx]}{h}\\
&=&\frac{mx+mh-mx}{h}\,\,=\,\,\frac{mh}{h}\,\,=\,\,m\,\,=\phi(h),\\
\end{eqnarray*}
as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
4-2. Let $S\subseteq\R$.
Show: $-(\LP_\R S)=\LP_\R(-S)$.
\vskip.1in\noindent
{\it Proof:}
Define $\lambda:\R\to\R$ by $\lambda(x)=-x$.
Then $\lambda:\R\to\R$ is continuous and one-to-one.
Also, $\lambda^{-1}=\lambda$.
Then $\lambda^{-1}:\R\to\R$ is continuous.
Then $\lambda:\R\to\R$ is a homeomorphism.
Then $\lambda_*(\LP_\R S)=\LP_\R(\lambda_*(S))$.
That is, $-(\LP_\R S)=\LP_\R(-S)$, as desried.
QED
\vskip.1in\hrule\vskip.1in\noindent
4-3. Let $f:\R\dashrightarrow\R$, $p\in\R$.
Define $f_0:\R\dashrightarrow\R$ by $f_0(x)=f(-x)$.
Let $p_0:=-p$.
Assume that $f$ has a local maximum at $p$ in $\R$.
Show that $f_0$ has a local maximum at $p_0$ in $\R$.
\vskip.1in\noindent
{\it Proof:}
Choose $V\in\scrn_\R(p)$ s.t.~$V\subseteq\dom[f]$ and $f(p)\ge f_*(V)$.
We wish to show:
$\exists U_0\in\scrn_\R(p_0)$ s.t.~$U_0\subseteq\dom[f_0]$ and $f_0(p_0)\ge(f_0)_*(U_0)$.

Since $\scrb_\R(p)$ is a neighborhood base at $p$ in $\R$,
choose $\delta>0$ such that $B_\R(p,\delta)\subseteq V$.
Let $U:=B_\R(p,\delta)$.
Then $U\subseteq V$.
Let $U_0:=B_\R(p_0,\delta)$.
Since $\scrb_\R(p_0)$ is a neighborhood base at $p_0$ in $\R$,
we get $U_0\in\scrn_\R(p_0)$.
Want: $U_0\subseteq\dom[f_0]$ and $f_0(p_0)\ge(f_0)_*(U_0)$.
Since $U\subseteq V\subseteq\dom[f]$, it follows that $-U\subseteq-(\dom[f])$.
We have
$$U_0\,\,=\,\,B_\R(p_0,\delta)\,\,=\,\,B_\R(-p,\delta)\,\,=\,\,-[B_\R(p,\delta)]\,\,=\,\,-U.$$
Then $U_0=-U\subseteq-(\dom[f])=\dom[f_0]$.
Want: $f_0(p_0)\ge(f_0)_*(U_0)$.
Want: $\forall y_0\in(f_0)_*(U_0)$, $f_0(p_0)\ge y_0$.
Let $y_0\in(f_0)_*(U_0)$ be given.
We wish to show: $f_0(p_0)\ge y_0$.

Since $y_0\in(f_0)_*(U_0)$,
choose $x_0\in U_0$ s.t.~$y_0=f_0(x_0)$.
Let $x:=-x_0$.
We have $f_0(p_0)=f(-p_0)=f(p)$
and $y_0=f_0(x_0)=f(-x_0)=f(x)$.
We wish to show that $f(p)\ge f(x)$,
or, equivalently, that $f(x)\le f(p)$.

We have $x=-x_0\in-U_0=-(-U)=U\subseteq V\subseteq\dom[f]$.
Then $x\in U$ and $x\in\dom[f]$,
and so $f(x)\in f_*(U)$.
Since $U\subseteq V$,
it follows that $f_*(U)\subseteq f_*(V)$.
By choice of $V$, we have $f_*(V)\le f(p)$.
We conclude that
$f(x)\in f_*(U)\subseteq f_*(V)\le f(p)$,
as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
4-4. Let $f:\R\dashrightarrow\R$, $p\in\R$.
Define $f_1:\R\dashrightarrow\R$ by $f_1(x)=-[f(x)]$.
Assume that $f$ has a local minimum at $p$ in $\R$.
Show that $f_1$ has a local maximum at $p$ in $\R$.
\vskip.1in\noindent
{\it Proof:}
Want: $\exists V\in\scrn_\R(p)$ s.t.~$V\subseteq\dom[f_1]$ and $f(p)\le(f_1)_*(V)$.
Choose $V\in\scrn_\R(p)$ s.t.~$V\subseteq\dom[f]$ and $f(p)\le f_*(V)$.
We want:
$V\subseteq\dom[f_1]$ and $f_1(p)\ge(f_1)_*(V)$.
As $V\subseteq\dom[f]=\dom[f_1]$,
it remains to show: $f_1(p)\ge(f_1)_*(V)$.
We wish to show:
$\forall y_1\in(f_1)_*(V)$, $f_1(p)\ge y_1$.
Let $y_1\in(f_1)_*(V)$ be given.
We wish to show: $f_1(p)\ge y_1$.

Since $y_1\in(f_1)_*(V)$,
choose $x\in V$ such that $y_1=f_1(x)$.
Since $x\in V\subseteq\dom[f]$,
we get $f(x)\in f_*(V)$.
Then, by the choice of $V$,
$f(p)\le f(x)$.
Then $f_1(p)=-[f(p)]\ge-[f(x)]=f_1(x)=y$.
QED
\vskip.1in\hrule\vskip.1in\noindent
4-5. Let $f:\R\dashrightarrow\R$, $p\in\dom[f']$.
Assume that $f$ has a local extremum at $p$.
Show that $f'(p)=0$.
\vskip.1in\noindent
{\it Proof:}
At least one of the following must be true:
\begin{itemize}
\item[(1)]$f$ has a local maximum at $p$ in $\R$ \qquad or
\item[(2)]$f$ has a local minimum at $p$ in $\R$.
\end{itemize}

{\it Case (1):}
By Lemma 33.16,
$f'(p)=0$, as desired.
{\it End of Case (1).}

{\it Case (2):}
Define $f_1:\R\dashrightarrow\R$ by $f_1(x)=-[f(x)]$.
By HW\#4-4, we see that $f_1$~has a local maximum at $p$ in $\R$.
Define $\lambda:\R\to\R$ by~$\lambda(x)=-x$.
Then $f_1=\lambda\circ f$.
Since $f_1$ has a local maximum at $p$ in~$\R$,
by (3) of Remark 33.9, we get $p\in\LPD_\R f_1$.
Then $p\in\LPD_\R(\lambda\circ f)$, so, by the Chain Rule, we get
$(\lambda\circ f)'(p)\,=^*\,[\lambda'(f(p))][f'(p)]$.
Since $p\in\dom[f']\subseteq\dom[f]$, we get $f(p)\in\im[f]$.
So, since $\im[f]\subseteq\R$, we get $f(p)\in\R$.
By HW\#4-2, $\lambda'(f(p))=-1$.
Then
$$f_1'(p)\,\,=\,\,(\lambda\circ f)'(p)\,\,=^*\,\,[\lambda'(f(p))][f'(p)]\,\,=\,\,-[f'(p)],$$
so $f_1'(p)\,=^*\,-[f'(p)]$.
Since $p\in\dom[f']$, we get $f'(p)\in\im[f]$.
So, since $\im[f]\subseteq\R$, we get $f'(p)\in\R$.
Then $-[f'(p)]\ne\frownie$.

Since $f_1'(p)\,=^*\,-[f'(p)]\ne\frownie$,
we conclude that $f_1'(p)=-[f'(p)]$.
Then $f_1'(p)\ne\frownie$, so $p\in\dom[f_1']$.
So, since $f_1$ has a local maximum at~$p$ in $\R$,
it follows, from Lemma 33.16, that $f_1'(p)=0$.

Then $0=f_1'(p)=-[f'(p)]$.
Then $f'(p)=0$.
{\it End of Case (2).}
QED
\vskip.1in\noindent\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 3: Due on Tuesday 6 February}
\vskip.1in\hrule\vskip.1in\noindent
3-1. Let $X$ be a topological space, let $S\subseteq X$ and let $p\in X$.
Show:
$$[\,\,p\in\Int_XS\,\,]\quad\Leftrightarrow\quad[\,\,S\in\scrn_X(p)\,\,].$$
\vskip.1in\noindent
{\it Proof:}
{\it Proof of $\Rightarrow$:}
Assume: $p\in\Int_XS$.
We wish to show: $S\in\scrn_X(p)$.
We wish to show: $\exists$open $U$ in $X$ s.t.~$p\in U\subseteq S$.

Let $U:=\Int_XS$.
Then $U$ is open in $X$, and we wish to show: $p\in U\subseteq S$.
We have $p\in\Int_XS=U$, and it remains to show: $U\subseteq S$.

We have $U=\Int_XS\subseteq S$, as desired.
{\it End of proof of $\Rightarrow$.}

{\it Proof of $\Leftarrow$:}
Assume: $S\in\scrn_X(p)$.
We wish to show: $p\in\Int_XS$.

Since $S\in\scrn_X(p)$, choose an open $U$ in $X$ s.t.~$p\in U\subseteq S$.
Since $U$~is open in $X$, we get $\Int_XU=U$.
Since $U\subseteq S$, $\Int_XU\subseteq\Int_XS$.
Then $p\in U=\Int_XU\subseteq\Int_XS$, as desired.
{\it End of proof of $\Leftarrow$.}
QED
\vskip.1in\hrule\vskip.1in\noindent
3-2. Let $\phi:\R\dashrightarrow\R$, let $p\in\R$ and let $q>0$.
Assume: $\phi\to q$ near $p$.
Show: $\exists U\in\scrn_\R^\times(p)$ s.t.~$\phi_*(U)>0$.
\vskip.1in\noindent
{\it Proof:}
Let $V:=(0,\infty)$.
Then $V$ is open in $\R$ and $q\in V$.
Therefore, by~Remark 16.4, we conclude that $V\in\scrn_\R(q)$.
So, since $\phi\to q$ near $p$,
choose $U\in\scrn_\R^\times(p)$ s.t.~$\phi_*(U)\subseteq V$.
We wish to show: $\phi_*(U)>0$.

We have $\phi_*(U)\subseteq V=(0,\infty)>0$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
3-3. Let $\alpha,\beta,\gamma:\R\dashrightarrow\R$.
Let $S\subseteq\dom[(\alpha,\beta,\gamma)]$.
Let $p\in\Int_\R S$ and let $q\in\R$.
Assume that $\alpha\le\beta\le\gamma$ on $S$.
Assume that $\alpha(p)=\gamma(p)$.
Assume that $\alpha'(p)=q=\gamma'(p)$.
Show: $\beta'(p)=q$.
\vskip.1in\noindent
{\it Proof:}
We wish to show: $\ds{\lim_0SS_\beta^p=q}$.

{\it Claim:} $\forall h\in (S-p)_0^\times$, $(SS_\beta^p)(h)\in\,[\,(SS_\alpha^p)(h)\,|\,(SS_\gamma^p)(h)\,]$.
{\it Proof of~Claim:}
Given $h\in(S-p)_0^\times$.
Want: $(SS_\beta^p)(h)\in\,[\,(SS_\alpha^p)(h)\,|\,(SS_\gamma^p)(h)\,]$.

We have $h\in(S-p)_0^\times\subseteq S-p$.
Then $p+h\in S$.
Then
$$\alpha(p+h)\quad\le\quad\beta(p+h)\quad\le\quad\gamma(p+h).$$
We have $p\in\Int_RS\subseteq S$.
It follows that $\alpha(p)\le\beta(p)\le\gamma(p)$.
So, since $\alpha(p)=\gamma(p)$,
we conclude that $\alpha(p)=\beta(p)=\gamma(p)$.
So, since $\alpha(p+h)\le\beta(p+h)\le\gamma(p+h)$,
we get:
$$[\alpha(p+h)]-[\alpha(p)]\le[\beta(p+h)]-[\beta(p)]\le[\gamma(p+h)]-[\gamma(p)].$$
Dividing this by $h$, we get:
\begin{eqnarray*}
(\,\,h>0\,\,)&\Rightarrow&(\,\,(SS_\alpha^p)(h)\le(SS_\beta^p)(h)\le SS_\gamma^p(h)\,\,)\quad\hbox{and}\\
(\,\,h<0\,\,)&\Rightarrow&(\,\,(SS_\alpha^p)(h)\ge(SS_\beta^p)(h)\ge SS_\gamma^p(h)\,\,).
\end{eqnarray*}
Then
\begin{eqnarray*}
(\,\,h>0\,\,)&\Rightarrow&(\,\,(SS_\beta^p)(h)\,\in\,[\,(SS_\alpha^p)(h)\,|\,(SS_\gamma^p)(h)\,]\,\,)\quad\hbox{and}\\
(\,\,h<0\,\,)&\Rightarrow&(\,\,(SS_\beta^p)(h)\,\in\,[\,(SS_\alpha^p)(h)\,|\,(SS_\gamma^p)(h)\,]\,\,).
\end{eqnarray*}
Since $h\in(S-p)_0^\times$, we get $h\ne0$, so: either $h>0$ or $h<0$.
Then $(SS_\beta^p)(h)\in\,[\,(SS_\alpha^p)(h)\,|\,(SS_\gamma^p)(h)\,]$, as desired.
{\it End of proof of Claim.}

Let $D:=\dom[\beta]$.
Then $\Int_\R D\subseteq\LP_\R D=\LPD_\R\beta$.
We have $S\subseteq\dom[(\alpha,\beta,\gamma)]\subseteq D$.
Then $p\in\Int_\R S\subseteq\Int_\R D\subseteq\LPD_\R\beta$.
It therefore suffices to show: $SS_\beta^p\to q$ near $0$.
We wish to show: $\forall\varepsilon>0$, $\exists\delta>0$
s.t., $\forall h\in\dom[SS_\beta^p]$,
$$[\,\,0<|h|<\delta\,\,]\quad\Rightarrow\quad[\,\,|[(SS_\beta^p)(h)]-q|<\varepsilon\,\,].$$
Let $\varepsilon>0$ be given.
We wish to show: $\exists\delta>0$
s.t., $\forall h\in\dom[SS_\beta^p]$,
$$[\,\,0<|h|<\delta\,\,]\quad\Rightarrow\quad[\,\,|[(SS_\beta^p)(h)]-q|<\varepsilon\,\,].$$

Since $p\in\Int_\R S$, by HW\#3-1, we get $S\in\scrn_\R(p)$.
So, since $\scrb_\R(p)$ is a neighborhood base at $p$ in $\R$,
choose $\eta>0$ s.t.~$B_\R(p,\eta)\subseteq S$.
Since $\alpha'(p)=q$,
it follows that $SS_\alpha^p\to q$ near $p$.
Choose $\lambda>0$ s.t., $\forall h\in\dom[SS_\alpha^p]$,
$$[\,\,0<|h|<\lambda\,\,]\quad\Rightarrow\quad[\,\,|[(SS_\alpha^p)(h)]-q|<\varepsilon\,\,].$$
Since $\gamma'(p)=q$,
it follows that $SS_\gamma^p\to q$ near $p$.
Choose $\mu>0$ s.t., $\forall h\in\dom[SS_\gamma^p]$,
$$[\,\,0<|h|<\mu\,\,]\quad\Rightarrow\quad[\,\,|[(SS_\gamma^p)(h)]-q|<\varepsilon\,\,].$$
Let $\delta:=\min\{\eta,\lambda,\mu\}$.
We wish to show: $\forall h\in\dom[SS_\beta^p]$,
$$[\,\,0<|h|<\delta\,\,]\quad\Rightarrow\quad[\,\,|[(SS_\beta^p)(h)]-q|<\varepsilon\,\,].$$
Let $h\in\dom[SS_\beta^p]$ be given.
We wish to show:
$$[\,\,0<|h|<\delta\,\,]\quad\Rightarrow\quad[\,\,|[(SS_\beta^p)(h)]-q|<\varepsilon\,\,].$$
Assume: $0<|h|<\delta$.
We wish to prove: $|[(SS_\beta^p)(h)]-q|<\varepsilon$.
Let $I:=(q-\varepsilon,q+\varepsilon)$.
We wish to prove $(SS_\beta^q)(h)\in I$

Since $0<|h|$, it follows that $h\ne0$.
Since $|(p+h)-p|=|h|<\delta\le\eta$,
we see that $p+h\in B_\R(p,\eta)$.
By the choice of $\eta$, we have $B_\R(p,\eta)\subseteq S$.
Since $p+h\in B_\R(p,\eta)\subseteq S$,
we get $h\in S-p$.
So, since $h\ne0$, we have $h\in(S-p)_0^\times$.
Then, by the Claim,
$(SS_\beta^p)(h)\in\,[\,(SS_\alpha^p)(h)\,|\,(SS_\gamma^p)(h)\,]$.

Since $0<|h|<\delta\le\lambda$, by choice of $\lambda$,
we get: $|[(SS_\alpha^p)(h)]-q|<\varepsilon$.
Then $(SS_\alpha^p)(h)\in I$.
Since $0<|h|<\delta\le\mu$, by choice of $\mu$,
we get: $|[(SS_\gamma^p)(h)]-q|<\varepsilon$.
Then $(SS_\gamma^p)(h)\in I$.
Since $I$ is an interval
and $(SS_\alpha^p)(h)\in I$
and $(SS_\gamma^p)(h)\in I$,
we get
$[\,(SS_\alpha^p)(h)\,|\,(SS_\gamma^p)(h)\,]\subseteq I$.

Then
$(SS_\beta^p)(h)\in\,[\,(SS_\alpha^p)(h)\,|\,(SS_\gamma^p)(h)\,]\subseteq I$,
as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
3-4. Let $W$ be a normed vector space.
Let $f,g:\R\dashrightarrow W$.
Let $U$ be an open subset of $\R$.
Assume: $U\subseteq\dom[(f,g)]$.
Assume: $f=g$ on $U$.
Show: $f'=g'$ on $U$.
\vskip.1in\noindent
{\it Proof:}
We wish to show: $\forall p\in U$, $f'(p)=g'(p)$.
Let $p\in U$ be given.
We wish to show: $f'(p)=g'(p)$.
We wish to show: $\ds{\lim_0SS_f^p=\lim_0SS_g^p}$.

Since $U$ is open in $\R$,
it follows that $U-p$ is open in $\R$.
So, since $0\in U-p$, we get $U-p\in\scrn_\R(0)$.
Let $V:=(U-p)_0^\times$.
Then $V\in\scrn_\R^\times(0)$.

{\it Claim:} $SS_f^p=SS_g^p$ on $V$.
{\it Proof of Claim:}
We want: $\forall h\in V$, $(SS_f^p)(h)=(SS_g^p)(h)$.
Given $h\in V$.
We want: $(SS_f^p)(h)=(SS_g^p)(h)$.

Since $h\in(U-p)_0^\times\subseteq U-p$,
we get $p+h\in U$.
Since $p,p+h\in U$ and since $f=g$ on $U$,
we get $f(p)=g(p)$ and $f(p+h)=g(p+h)$.
Then
$$(SS_f^p)(h)=\frac{[f(p+h)]-[f(p)]}{h}
=\frac{[g(p+h)]-[g(p)]}{h}=(SS_g^p)(h),$$
as desired.
{\it End of proof of Claim:}

Since $V\in\scrn_\R^\times(0)$,
by the Claim and by Theorem 20.8,
we have both $\ds{\opLIMS_0\,(SS_f^p)\subseteq\opLIMS_0\,(SS_g^p)}$
and $\ds{\opLIMS_0\,(SS_g^p)\subseteq\opLIMS_0\,(SS_f^p)}$.
Then $\ds{\opLIMS_0\,(SS_g^p)=\opLIMS_0\,(SS_f^p)}$.
Then
$$\lim_0\,SS_f^p=\ELT(\opLIMS_0(SS_f^p))
=\ELT(\opLIMS_0(SS_g^p))=\lim_0\,SS_g^p,$$
as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
3-5. Define $f:\R\dashrightarrow\R$ by
$f(x)=[x^2][\sin(x^{-3})]$.
Let $\beta:=\adj_0^0f$.
Show: $\beta'(0)=0$.
\vskip.1in\noindent
{\it Proof:}
Define $\alpha,\gamma:\R\to\R$ by $\alpha(x)=-x^2$ and $\gamma(x)=x^2$.
For all $x\in\R_0^\times$, we have
both $-1\le\sin(x^{-3})\le1$ and $x^2\ge0$, and so
$[x^2][-1]\le[x^2][\sin(x^{-3})]\le[x^2][1]$,
and so $\alpha(x)\le f(x)\le\gamma(x)$.
Then $\alpha\le f\le\gamma$ on $\R_0^\times$.
So, since $f=\beta$ on $\R_0^\times$,
we get: $\alpha\le\beta\le\gamma$ on $\R_0^\times$.
So, since $\alpha(0)=\beta(0)=\gamma(0)=0$,
we get $\alpha\le\beta\le\gamma$ on $\R$.

We have $\alpha(0)=0=\gamma(0)$.
By the Power Rule, for all $x\in\R$, we have $\gamma'(x)=2x$.
Then $\gamma'(0)=0$.
So, since $\alpha=-\gamma$, by linearity of~differentiation,
we have $\alpha'(0)=^*-[\gamma'(0)]$.
Then
$$\alpha'(0)\,\,=^*\,\,-\,[\gamma'(0)]\,\,=\,\,-0\,\,=\,\,0\,\,\ne\,\,\frownie.$$
Then $\alpha'(0)=0$.
Then $\alpha'(0)=0=\gamma'(0)$.
Let $p:=0$ and $q:=0$.
Then $\alpha(p)=\gamma(p)$ and $\alpha'(p)=q=\gamma'(p)$.
Then, by HW\#3-3,
we get $\beta'(p)=q$.
That is, $\beta'(0)=0$.
QED
\vskip.1in\noindent\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 2: Due on Tuesday 30 January}
\vskip.1in\hrule\vskip.1in\noindent
2-1. Let $X$ and $Y$ be topological spaces.
Let $X_0\subseteq X$ and let $Y_0\subseteq Y$.
Let $f:X_0\dashrightarrow Y_0$.
Show:
$$[\,\,f\hbox{ is }(X,Y)\hbox{-continuous}\,\,]\quad\Rightarrow\quad[\,\,f\hbox{ is }(X_0,Y_0)\hbox{-continuous}\,\,].$$
\vskip.1in\noindent
{\it Proof:}
Assume: $f$ is $(X,Y)$-continuous.
Want: $f$ is $(X_0,Y_0)$-continuous.
Want: $\forall p\in\dom[f]$, $f$ is $(X_0,Y_0)$-continuous at $p$.
Given $p\in\dom[f]$.
Want: $f$ is $(X_0,Y_0)$-continuous at~$p$.
Since $f$ is $(X,Y)$-continuous and $p\in\dom[f]$,
it follows that $f$ is $(X,Y)$-continuous at $p$.
Want:
$\forall V_0\in\scrn_{Y_0}(f(p))$, $\exists U_0\in\scrn_{X_0}(p)$ s.t.~$f_*(U_0)\subseteq V_0$.
Let $V_0\in\scrn_{Y_0}(f(p))$ be given.
We wish to show:
$\exists U_0\in\scrn_{X_0}(p)$ s.t.~$f_*(U_0)\subseteq V_0$.

By extension, choose $V\in\scrn_Y(f(p))$ s.t.~$V\cap Y_0=V_0$.
Since $f$ is $(X,Y)$-continuous at $p$,
choose $U\in\scrn_X(p)$ s.t.~$f_*(U)\subseteq V$.
We define $U_0:=U\cap X_0$.
By restriction, $U_0\in\scrn_{X_0}(p)$.
We wish to show: $f_*(U_0)\subseteq V_0$.
We wish to show:
$\forall x\in\dom[f]$,
$$[\,\,x\in U_0\,\,]\quad\Rightarrow\quad[\,\,f(x)\in V_0\,\,].$$
Let $x\in\dom[f]$ be given.
We wish to show:
$$[\,\,x\in U_0\,\,]\quad\Rightarrow\quad[\,\,f(x)\in V_0\,\,].$$
Assume: $x\in U_0$.
We wish to show: $f(x)\in V_0$.

We have $x\in U_0=U\cap X_0\subseteq U$.
So, since $x\in\dom[f]$,
we get $f(x)\in f_*(U)$.
Then $f(x)\in f_*(U)\subseteq V$.
Since $f:X_0\dashrightarrow Y_0$,
it follows that $\im[f]\subseteq Y_0$.
Then $f(x)\in\im[f]\subseteq Y_0$.
So, since $f(x)\in V$,
we get $f(x)\in V\cap Y_0$.
Then $f(x)\in V\cap Y_0=V_0$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
2-2. Let $X$, $Y$ and $Z$ be topological spaces.
Let $f:X\dashrightarrow Y$ and let $g:X\dashrightarrow Z$.
Let $a\in X$, $b\in Y$ and $c\in Z$.
Assume:
\begin{itemize}
\item[(1)]$f\to b$ in $Y$ near $a$ in $X$ \qquad and
\item[(2)]$g\to c$ in $Z$ near $a$ in $X$.
\end{itemize}
Show: $(f,g)\to(b,c)$ in $Y\times Z$ near $a$ in $X$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall W\in\scrn_{Y\times Z}((b,c))$,
$\exists R\in\scrn_X^\times(a)$
s.t.~$(f,g)_*(R)\subseteq W$.
Let $W\in\scrn_{Y\times Z}((b,c))$ be given.
Want: $\exists R\in\scrn_X^\times(a)$
s.t.~$(f,g)_*(R)\subseteq W$.

Choose $U\in\scrn_Y(b)$ and $V\in\scrn_Z(c)$ s.t.~$U\times V\subseteq W$.
Since $f\to b$ in~$Y$ near $a$ in $X$ and since $U\in\scrn_Y(b)$,
choose $P\in\scrn_X^\times(a)$ s.t.~$f_*(P)\subseteq U$.
Since $g\to c$ in $Z$ near $a$ in $X$  and since $V\in\scrn_Z(c)$,
choose $Q\in\scrn_X^\times(a)$ s.t.~$f_*(Q)\subseteq V$.
As $P,Q\in\scrn_X^\times(a)$,
$P\cap Q\in\scrn_X^\times(a)$.
Let $R:=P\cap Q$.
Then $R\in\scrn_x^\times(a)$.
Want: $(f,g)_*(R)\subseteq W$.
Want: $\forall x\in\dom[(f,g)]$,
$$[\,\,x\in R\,\,]\quad\Rightarrow\quad[\,\,(f,g)(x)\in W\,\,].$$
Let $x\in\dom[(f,g)]$ be given.
We wish to show:
$$[\,\,x\in R\,\,]\quad\Rightarrow\quad[\,\,(f,g)(x)\in W\,\,].$$
Assume: $x\in R$.
We wish to show: $(f,g)(x)\in W$.

We have $x\in\dom[(f,g)]=(\dom[f])\cap(\dom[g])$,
so $x\in\dom[f]$ and $x\in\dom[g]$.
Since $x\in\dom[f]$ and since $x\in R=P\cap Q\subseteq P$,
we get $f(x)\in f_*(P)$.
Since $x\in\dom[f]$ and since $x\in R=P\cap Q\subseteq Q$,
we get $g(x)\in g_*(P)$.
Then $f(x)\in f_*(P)\subseteq U$
and $g(x)\in g_*(P)\subseteq V$.
Then $(f,g)(x)=(f(x),g(x))\in U\times V\subseteq W$,
as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
2-3. Show: $\forall z\in\R$, $\exists!x\in\R$ s.t.~$x^5+x^3=z$.
\vskip.1in\noindent
{\it Proof:}
Let $z\in\R$ be given.
We wish to show: $\exists!x\in\R$ s.t.~$x^5+x^3=z$.

Define $f:\R\to\R$ by $f(x)=x^5+x^3$.
Then $f$ is continuous.
We wish to show: $\exists!x\in\R$ s.t.~$f(x)=z$.
We wish to show:
\begin{itemize}
\item[(1)]$\exists x\in\R$ s.t.~$f(x)=z$.
\item[(2)]$\forall w,x\in\R$, ( $[\,f(w)=z=f(x)\,]\,\,\Rightarrow\,\,[\,w=x\,]$ ).
\end{itemize}

{\it Proof of (1):}
As $z\in\R$, $-|z|\le z\le|z|$.
Let $b:=\max\{|z|/2,1\}$.
Since $b\ge1$, we get $b^5\ge b$ and $b^3\ge b$.
Then $b^5\ge b\ge|z|/2$ and $b^3\ge b\ge|z|/2$.
Then
$f(b)=b^5+b^3\ge(|z|/2)+(|z|/2)=|z|$.
Then $-[f(b)]\le-|z|$.
We have
$f(-b)=(-b)^5+(-b)^3=-(b^5+b^3)=-[f(b)]\le-|z|$.
Then $f(-b)\le-|z|\le z\le|z|\le f(b)$.
Then $z\in[f(-b),f(b)]$.
Also, since $b\ge 1$,
we see that $-b\le -1$.
Then $-b\le -1<1\le b$.
It suffices to~show: $\exists x\in[-b,b]$ s.t.~$f(x)=z$.

By the Intermediate Value Theorem,
$[f(-b),f(b)]\subseteq f_*([-b,b])$.
Then $z\in[f(-b),f(b)]\subseteq f_*([-b,b])$,
so $\exists x\in[-b,b]$ s.t.~$f(x)=z$, as desired.
{\it End of proof of (1).}

{\it Proof of (2):}
Given $w,x\in\R$.
Want: $[f(w)=z=f(x)]\,\Rightarrow\,[w=x]$.
Assume: $f(w)=z=f(x)$.
We wish to show: $w=x$.
Assume: $w\ne x$.
We aim for a contradiction.

Let $p:=\min\{w,x\}$ and let $q:=\max\{w,x\}$.
Since $w\ne x$, it follows that $p<q$.
Also, $f(p)\in\{f(w),f(x)\}=\{z\}$, so $f(p)=z$.
Also, $f(q)\in\{f(w),f(x)\}=\{z\}$, so $f(q)=z$.
Since $p<q$, we get $p^5<q^5$ and $p^3<q^3$.
Then $p^5+p^3<q^5+q^3$.
Then
$$f(p)\,\,\,=\,\,\,p^5\,+\,p^3\,\,\,<\,\,\,q^5\,+\,q^3\,\,\,=\,\,\,f(q)\,\,\,=\,\,\,z\,\,\,=\,\,\,f(p),$$
so $f(p)<f(p)$.
Contradiction.
{\it End of proof of (2).}
QED
\vskip.1in\hrule\vskip.1in\noindent
2-4. Let $X$ and $Y$ be topological spaces.
Let $f:X\dashrightarrow Y$.
Let $X_0\subseteq\dom[f]$.
Assume that $f$ is $(X,Y)$-continuous on $X_0$.
Show that $f|X_0$ is $(X_0,Y)$-continuous.
\vskip.1in\noindent
{\it Proof:}
Let $g:=f|X_0$.
Then $\dom[g]=X_0$.
We wish to show: $g$ is $(X_0,Y)$-continuous.
By HW\#2-1 (with $Y_0$ replaced by $Y$),
want: $g$ is $(X,Y)$-continuous.
Want: $\forall p\in X_0$, $g$ is $(X,Y)$-continuous at~$p$.
Let $p\in X_0$ be given.
We wish to show: $g$ is $(X,Y)$-continuous at $p$.
Want: $\forall V\in\scrn_Y(g(p))$, $\exists U\in\scrn_X(p)$ s.t.~$g_*(U)\subseteq V$.
Let $V\in\scrn_Y(g(p))$ be given.
Want: $\exists U\in\scrn_X(p)$ s.t.~$g_*(U)\subseteq V$.

Since $p\in X_0\subseteq\dom[f]$,
and since $f$ is $(X,Y)$-continuous,
it follows that $f$ is $(X,Y)$-continuous at $p$.
Since $p\in X_0\subseteq\dom[f]$, we have $(f|X_0)(p)=f(p)$.
Then $g(p)=(f|X_0)(p)=f(p)$.
Since $V\in\scrn_Y(g(p))=\scrn_Y(f(p))$ and since $f$ is $(X,Y)$-continuous at $p$,
choose $U\in\scrn_X(p)$ s.t.~$f_*(U)\subseteq V$.
We wish to show: $g_*(U)\subseteq V$.

Recall that $\dom[g]=W_0$.
We wish to show: $\forall w\in X_0$,
$$[\,\,w\in U\,\,]\quad\Rightarrow\quad[\,\,g(w)\in V\,\,].$$
Let $w\in X_0$ be given.
We wish to show:
$$[\,\,w\in U\,\,]\quad\Rightarrow\quad[\,\,g(w)\in V\,\,].$$
Assume: $w\in U$.
We wish to show: $g(w)\in V$.

As $w\in X_0\subseteq\dom[f]$ and as $w\in U$,
we get $f(w)\in f_*(U)$.
By~choice of $U$, we have $f_*(U)\subseteq V$.
Since $w\in X_0\subseteq\dom[f]$, $(f|X_0)(w)=f(w)$.
Then $g(w)=(f|X_0)(w)=f(w)\in f_*(U)\subseteq V$.
QED
\vskip.1in\hrule\vskip.1in\noindent
2-5. Let $f:\R\dashrightarrow\R$.
Let $\alpha,\beta\in\dom[f]$.
Assume that $f$ is semimonotone on $[\alpha|\beta]$.
Show: $f_*([\alpha|\beta])\subseteq[f(\alpha)|f(\beta)]$.
\vskip.1in\noindent
{\it Proof:}
Since $\alpha,\beta\in\dom[f]$,
we conclude that $\{\alpha,\beta\}\subseteq\dom[f]$.
Let $a:=\min\{\alpha,\beta\}$ and $b:=\max\{\alpha,\beta\}$.
Then $a\le b$.
Also, $[\alpha|\beta]=[a,b]$.
Then $f$ is semimonotone on $[a,b]$.
Also, $a,b\in\{\alpha,\beta\}\subseteq\dom[f]$.

Either ( ($a=\alpha$) and ($b=\beta$) )
or ( ($a=\beta$) and ($b=\alpha$) ).
Then either $[f(\alpha)|f(\beta)]=[f(a)|f(b)]$
or $[f(\alpha)|f(\beta)]=[f(b)|f(a)]$.
So, since $[f(b)|f(a)]=[f(a)|f(b)]$,
we see that $[f(\alpha)|f(\beta)]=[f(a)|f(b)]$.
We wish to show: $f_*([a,b])\subseteq[f(a)|f(b)]$.
We wish to show: $\forall x\in\dom[f]$,
$$(\,\,x\in[a,b]\,\,)\quad\Rightarrow\quad(\,\,f(x)\in[f(a)|f(b)]\,\,).$$
Let $x\in\dom[f]$ be given.
We wish to show:
$$(\,\,x\in[a,b]\,\,)\quad\Rightarrow\quad(\,\,f(x)\in[f(a)|f(b)]\,\,).$$
Assume: $x\in[a,b]$.
We wish to show: $f(x)\in[f(a)|f(b)]$.

Because $x\in[a,b]$, it follows that $a\le x\le b$.
Since $f$ is semimonotone on $[a,b]$,
at least one of the following must be true:
\begin{itemize}
\item[(1)]$f$ is semiincreasing on $[a,b]$\qquad or
\item[(2)]$f$ is semidecreasing on $[a,b]$.
\end{itemize}

{\it Case (1):}
Since $a,x,b\in\dom[f]$, since $a\le x\le b$
and since $f$~is semiincreasing on $[a,b]$,
it follows that $f(a)\le f(x)\le f(b)$.
Then $f(x)\in[f(a),f(b)]=[f(a)|f(b)]$, as desired.
{\it End of Case (1).}

{\it Case (2):}
Since $a,x,b\in\dom[f]$, since $a\le x\le b$
and since $f$~is semidecreasing on $[a,b]$,
it follows that $f(a)\ge f(x)\ge f(b)$.
Then $f(x)\in[f(b),f(a)]=[f(a)|f(b)]$, as desired.
{\it End of Case (2).}
QED
\vskip.1in\noindent\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 1: Due on Tuesday 23 January}
\vskip.1in\hrule\vskip.1in\noindent
1-1. Let $X$ and $Y$ be metric spaces, and let $f:X\to Y$.
Assume that $f$ is not uniformly continuous.
Show: $\exists\varepsilon>0$, $\exists p,q\in X^\N$ s.t., $\forall j\in\N$,
$$[\,\,d_X\,(\,p_j\,,\,q_j\,)\,\,<\,\,1/j\,\,]\quad\hbox{and}\quad[\,\,d_Y\,(\,f(p_j)\,,\,f(q_j)\,)\,\,\ge\,\,\varepsilon\,\,].$$
\vskip.1in\noindent
{\it Proof:}
As $f$ is not u.c.,
choose $\varepsilon>0$ s.t.~$\forall\delta>0$, $\exists p,q\in X$ s.t.
$$[\,\,d_X(p,q)<\delta\,\,]\quad\hbox{and}\quad[\,\,d_Y(f(p),f(q))\ge\varepsilon\,\,].$$
We wish to show: $\exists p,q\in X^\N$ s.t., $\forall j\in\N$,
$$[\,\,d_X\,(\,p_j\,,\,q_j\,)\,\,<\,\,1/j\,\,]\quad\hbox{and}\quad[\,\,d_Y\,(\,f(p_j)\,,\,f(q_j)\,)\,\,\ge\,\,\varepsilon\,\,].$$

{\it Claim:} $\forall j\in\N$, $\exists p,q\in X$ s.t.
$$[\,\,d_X(p,q)<1/j\,\,]\quad\hbox{and}\quad[\,\,d_Y(f(p),f(q))\ge\varepsilon\,\,].$$
{\it Proof of claim:}
Let $j\in\N$ be given.
By the choice of $\varepsilon$, $\exists p,q\in X$ s.t.
$$[\,\,d_X(p,q)<1/j\,\,]\quad\hbox{and}\quad[\,\,d_Y(f(p),f(q))\ge\varepsilon\,\,],$$
as desired.
{\it End of proof of claim.}

By the claim and the Axiom of Choice, $\exists p,q\in X^\N$ s.t.~$\forall j\in\N$,
$$[\,\,d_X(p_j,q_j)<1/j\,\,]\quad\hbox{and}\quad[\,\,d_Y(f(p_j),f(q_j))\ge\varepsilon\,\,],$$
as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
1-2. Let $K$ be a sequentially compact metric space,
let $Y$ be a metric space, and let $f:K\to Y$.
Assume that $f$ is not uniformly continuous.
Show: $\exists\varepsilon>0$, $\exists s,t\in K^\N$, $\exists u\in K$ s.t.
\begin{eqnarray*}
&&(\qquad[\,\,s_\bullet\to u\hbox{ in }K\,\,]\quad\hbox{and}\quad[\,\,t_\bullet\to u\hbox{ in }K\,\,]\quad\hbox{and}\\
&&\qquad\,\,\,[\,\,\forall j\in\N,\qquad\,d_Y\,(\,f(s_j)\,,\,f(t_j)\,)\,\,\ge\,\,\varepsilon\,\,]\qquad).
\end{eqnarray*}
\vskip.1in\noindent
{\it Proof:}
By HW\#1-1, choose $\varepsilon>0$ and $p,q\in K^\N$ s.t., $\forall j\in\N$,
$$[\,\,d_K\,(\,p_j\,,\,q_j\,)\,\,<\,\,1/j\,\,]\quad\hbox{and}\quad[\,\,d_Y\,(\,f(p_j)\,,\,f(q_j)\,)\,\,\ge\,\,\varepsilon\,\,].$$
Since $K$ is sequentially compact, $p_\bullet$ is subconvergent in $K$.
Choose a subsequence $s$ of $p_\bullet$ s.t.~$s_\bullet$ is convergent in $K$.
Choose $u\in K$ s.t.~$s_\bullet\to u$ in $K$.
Since $s_\bullet$ is a subsequence of $p_\bullet$,
choose a strictly increasing $\ell\in\N^\N$ s.t.~$s_\bullet=(p\circ\ell)_\bullet$.
Let $t:=(q\circ\ell)_\bullet$.
Then $s,t\in K^N$ and $u\in K$.
Want:
\begin{eqnarray*}
&&(\qquad[\,\,s_\bullet\to u\hbox{ in }K\,\,]\quad\hbox{and}\quad[\,\,t_\bullet\to u\hbox{ in }K\,\,]\quad\hbox{and}\\
&&\qquad\,\,\,[\,\,\forall j\in\N,\qquad\,d_Y\,(\,f(s_j)\,,\,f(t_j)\,)\,\,\ge\,\,\varepsilon\,\,]\qquad).
\end{eqnarray*}
By choice of $u$, we have $s_\bullet\to u$ in $K$, so it remains to show:
\begin{itemize}
\item[(1)]$t\to u$ in $K$ \qquad and
\item[(2)]$\forall j\in\N$, \qquad $d_Y\,(\,f(s_j)\,,\,f(t_j)\,)\,\ge\,\varepsilon$.
\end{itemize}

{\it Proof of (1):}
We wish to show: $\forall\eta>0$, $\exists I\in\N$ s.t., $\forall j\in\N$,
$$[\,\,j\ge I\,\,]\quad\Rightarrow\quad[\,\,d_K(t_j,u)<\eta\,\,].$$
Let $\eta>0$ be given.
We wish to show: $\exists I\in\N$ s.t., $\forall j\in\N$,
$$[\,\,j\ge I\,\,]\quad\Rightarrow\quad[\,\,d_K(t_j,u)<\eta\,\,].$$

Since $s_\bullet\to u$ in $K$, choose $R\in\N$ s.t., $\forall j\in\N$,
$$[\,\,j\ge R\,\,]\quad\Rightarrow\quad[\,\,d_K(s_j,u)<\eta/2\,\,].$$
By the Archimidean Principle, choose $S\in\N$ such that $S>2/\eta$.
Let $I:=\max\{R,S\}$.
We wish to show: $\forall j\in\N$,
$$[\,\,j\ge I\,\,]\quad\Rightarrow\quad[\,\,d_K(t_j,u)<\eta\,\,].$$
Let $j\in\N$ be given.
We wish to show:
$$[\,\,j\ge I\,\,]\quad\Rightarrow\quad[\,\,d_K(t_j,u)<\eta\,\,].$$
Assume: $j\ge I$.
We wish to show: $d_K(t_j,u)<\eta$.

Since $j\ge I\ge R$, by choice of $R$, we conclude that
$d_K(s_j,u)<\eta/2$.
Since $\ell\in\N^\N$ and since $\ell_\bullet$ is strictly increasing,
by Remark 23.12, we get $\ell_j\ge j$.
It follows that $1/\ell_j\le1/j$.
Since $j\ge I\ge S>2/\eta$, we get $1/j<\eta/2$.
By the choice of $p_\bullet$ and $q_\bullet$, we have
$d_K(p_{\ell_j},q_{\ell_j})<1/\ell_j$.
Then
$d_K(t_j,s_j)=d_K(s_j,t_j)=d_K(p_{\ell_j},q_{\ell_j})<1/\ell_j\le1/j<\eta/2$.
Then
$$d_K(t_j,u)\,\,\le\,\,[d_K(t_j,s_j)]\,+\,[d(s_j,u)]\,\,<\,\,(\eta/2)\,+\,(\eta/2)\,\,=\,\,\eta,$$
as desired.
{\it End of proof of (1).}

{\it Proof of (2):}
Let $j\in\N$ be given.
Want: $d_Y(f(s_j),f(t_j))\ge\varepsilon$.

By the choice of $p$ and $q$, we have
$d_Y(f(p_{\ell_j}),f(q_{\ell_j}))\ge\varepsilon$.
Then $d_Y(f(s_j),f(t_j))=d_Y(f(p_{\ell_j}),f(q_{\ell_j}))\ge\varepsilon$.
{\it End of proof of (2).}
QED
\vskip.1in\hrule\vskip.1in\noindent
1-3. Let $X$ be a metric space, and let $s\in X^\N$.
Assume that $s_\bullet$ is convergent in $X$.
Show that $s_\bullet$ is Cauchy in $X$.
\vskip.1in\noindent
{\it Proof:}
We wish to show: $\forall\varepsilon>0$, $\exists K\in\N$ s.t., $\forall i,j\in\N$,
$$[\,\,i,j\ge K\,\,]\quad\Rightarrow\quad[\,\,d_X(s_i,s_j)<\varepsilon\,\,].$$
Let $\varepsilon>0$ be given.
We wish to show: $\exists K\in\N$ s.t., $\forall i,j\in\N$,
$$[\,\,i,j\ge K\,\,]\quad\Rightarrow\quad[\,\,d_X(s_i,s_j)<\varepsilon\,\,].$$

Since $s_\bullet$ is convergent in $X$,
choose $u\in X$ s.t.~$s_\bullet\to u$ in $X$.
Choose $K\in\N$ s.t., $\forall j\in\N$,
$$[\,\,j\ge K\,\,]\quad\Rightarrow\quad[\,\,d_X(s_j,u)<\varepsilon/2\,\,].$$
We wish to show: $\forall i,j\in\N$,
$$[\,\,i,j\ge K\,\,]\quad\Rightarrow\quad[\,\,d_X(s_i,s_j)<\varepsilon\,\,].$$
Let $i,j\in\N$ be given.
We wish to show:
$$[\,\,i,j\ge K\,\,]\quad\Rightarrow\quad[\,\,d_X(s_i,s_j)<\varepsilon\,\,].$$
Assume $i,j\ge K$.
We wish to show: $d_X(s_i,s_j)<\varepsilon$.

Since $i\ge K$, by choice of $K$,
we have $d(s_i,u)<\varepsilon/2$.
Since $j\ge K$, by choice of $K$,
we have $d(s_j,u)<\varepsilon/2$.

Then $d_X(s_i,s_j)\le[d_X(s_i,u)]+[d_X(u,s_j)]<(\varepsilon/2)+(\varepsilon/2)=\varepsilon$.
QED
\vskip.1in\hrule\vskip.1in\noindent
1-4. Let $X$ be a metric space, and let $s\in X^\N$.
Assume that $s_\bullet$ is Cauchy and subconvergent in $X$.
Show that $s_\bullet$ is convergent in $X$.
\vskip.1in\noindent
{\it Proof:}
Since $s_\bullet$ is subconvergent in $X$, choose a subsequence $t$ of $s_\bullet$
such that $t_\bullet$ is convergent in $X$.
Choose $u\in X$ s.t.~$t_\bullet\to u$ in $X$.
We wish to show: $s_\bullet\to u$ in $X$.
Want: $\forall\varepsilon>0$, $\exists K\in\N$ s.t., $\forall j\in\N$,
$$[\,\,j\ge K\,\,]\quad\Rightarrow\quad[\,\,d_X(s_j,u)<\varepsilon\,\,].$$
Let $\varepsilon>0$ be given.
We wish to show: $\exists I\in\N$ s.t., $\forall j\in\N$,
$$[\,\,j\ge I\,\,]\quad\Rightarrow\quad[\,\,d_X(s_j,u)<\varepsilon\,\,].$$

Since $t_\bullet\to u$ in $X$, choose $L\in\N$ s.t., $\forall j\in\N$,
$$[\,\,j\ge L\,\,]\quad\Rightarrow\quad[\,\,d_X(t_j,u)<\varepsilon/2\,\,].$$
Since $s_\bullet$ is Cauchy, choose $M\in\N$ s.t., $\forall j,k\in\N$,
$$[\,\,j,k\ge M\,\,]\quad\Rightarrow\quad[\,\,d_X(s_j,s_k)<\varepsilon/2\,\,].$$
Let $I:=\max\{L,M\}$.
We wish to show: $\forall j\in\N$,
$$[\,\,j\ge I\,\,]\quad\Rightarrow\quad[\,\,d_X(s_j,u)<\varepsilon\,\,].$$
Let $j\in\N$ be given.
We wish to show:
$$[\,\,j\ge I\,\,]\quad\Rightarrow\quad[\,\,d_X(s_j,u)<\varepsilon\,\,].$$
Assume: $j\ge I$
We wish to show: $d_X(s_j,u)<\varepsilon$.

Since $j\ge I\ge L$, by choice of $L$,
we conclude that $d_X(t_j,u)<\varepsilon/2$.
By~Corollary 23.13,
choose $k\in[j..\infty]$ such that $t_j=s_k$.
Then we have $k\ge j\ge I\ge M$.
Then $j,k\ge M$, so, by choice of $M$,
we get: $d_X(s_j,s_k)<\varepsilon/2$.
Since $t_j=s_k$, we get $d(s_j,t_j)=d(s_j,s_k)$.
Then
\begin{eqnarray*}
d(s_j,u)&\le&[d(s_j,t_j)]\,+\,[d(t_j,u)]\\
&=&[d(s_j,s_k)]\,+\,[d(t_j,u)]\\
&<&[\varepsilon/2]\,+\,[\varepsilon/2]\,\,=\,\,\varepsilon,
\end{eqnarray*}
as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
1-5. Let $X$ be a metric space, and let $s\in X^\N$.
Assume that $s_\bullet$ is Cauchy in $X$.
Show that $s_\bullet$ is bounded in $X$.
\vskip.1in\noindent
{\it Proof:}
We wish to show: $\im[s_\bullet]$ is bounded in $X$.

Since $s_\bullet$ is Cauchy in $X$, choose $K\in\N$ s.t., $\forall i,j\in\N$,
$$[\,\,i,j\ge K\,\,]\quad\Rightarrow\quad[\,\,d_X(s_i,s_j)<1\,\,].$$
Let $p:=s_K$.
Let $a:=\max\{d_X(p,s_1),\ldots,d_X(p,s_{K-1})\}$.
Let $R:=1+a$.
We wish to show: $\im[s_\bullet]\subseteq B_X(p,R)$.
We wish to show:
$\forall q\in\im[s_\bullet]$, $q\in B_X(p,R)$.
Let $q\in\im[s_\bullet]$ be given.
We wish to show: $q\in B_X(p,R)$.
We wish to show $d_X(p,q)<R$.
Since $q\in\im[s_\bullet]$, choose $j\in\N$ such that $q=s_j$.
We wish to show: $d_X(p,s_j)<R$.

At least one of the follwing must be true:
\begin{itemize}
\item[(1)]$j<K$\qquad or
\item[(2)]$j\ge K$.
\end{itemize}

{\it Case (1):}
As $j\in[1..K)$,
$d_X(p,s_j)\in\{d_X(p,s_1),\ldots,d_X(p,s_{K-1})\}$.
Then $d_X(p,s_j)\le\max\{d_X(p,s_1),\ldots,d_X(p,s_{K-1})\}=a<1+a=R$,
as desired.
{\it End of Case (1).}

{\it Case (2):}
Since $p=s_K$, we conclude that $d_X(p,s_j)=d_X(s_K,s_j)$.
Also, since $K,j\ge K$, by~choice of $K$,
we see that $d_X(s_K,s_j)<1$.
By definition of $a$, we have $a\ge0$.
It follows that $1\le1+a$.
Then $d_X(p,s_j)=d_X(s_K,s_j)<1\le1+a=R$.
{\it End of Case (2).}
QED
\vskip.1in\hrule\vskip.1in\hrule
\end{document}
