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\newcommand{\tref}[1]{Theorem~\ref{#1}}
\newcommand{\aref}[1]{Axiom~\ref{#1}}
\newcommand{\dref}[1]{Definition~\ref{#1}}

% BLACK BOARD BOLD

\def\Bbb{\mathbb}  %needed in Nancy

\def\A{{\mathbb A}} \def\B{{\mathbb B}} \def\C{{\mathbb C}} \def\D{{\mathbb D}}
\def\N{{\mathbb N}} \def\I{{\mathbb I}} \def\Q{{\mathbb Q}} \def\R{{\mathbb R}}
\def\Z{{\mathbb Z}} 

\def\T{{\mathbb T}}

\def\P{{\mathbb P}} 

% BOLD
\def\bfzero{{\bf 0}}
\def\bfP{{\bf P}}
\def\bfQ{{\bf Q}}
\def\bfS{{\bf S}}
\def\bfT{{\bf T}}

% 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}}
\def\scrpf{{\scrp\scrf}}
\def\scral{{\scra\scrl}}
\def\scrsb{{\scrs\scrb}}

% HATTED ROMAN
\def\hatd{{\widehat{d}}}
\def\hatX{{\widehat{X}}}

% HATTEDCAL
\def\hatscrp{{\widehat{\scrp}}}
\def\hatscrp{{\widehat{\scrp}}}
\def\hatpi{{\widehat{\pi}}}

% HATTED BB BOLD
\def\hatbbR{{\widehat{\R}}}

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

\def\scrbigo{{\widehat{\scro}}}

% LIE ALGEBRAS

\font\tenolde=eufm10 at 10pt
\font\sevenolde=eufm7
\font\fiveolde=eufm5
\newfam\oldefam
\textfont\oldefam=\tenolde
\scriptfont\oldefam=\sevenolde
\scriptscriptfont\oldefam=\fiveolde
\def\fraktwo{\fam\oldefam\tenolde}

\newfam\cyrfam
\font\tencyr=wncyr10
\def\cyr{\fam\cyrfam\tencyr}

\newcommand\sbullet[1][.75]{\mathbin{\vcenter{\hbox{\scalebox{#1}{$\bullet$}}}}}

\def\aff{{\mathfrak{aff}(\R)}} \def\La{{\mathfrak a}}
\def\Lb{{\mathfrak b}} \def\Lc{{\mathfrak c}} \def\Lg{{\mathfrak g}}
\def\Lh{{\mathfrak h}} \def\Lk{{\mathfrak k}} \def\Ll{{\mathfrak l}}
\def\Lm{{\mathfrak m}} \def\Ln{{\mathfrak n}} \def\Lr{{\mathfrak r}}
\def\Ls{{\mathfrak s}} \def\Lt{{\mathfrak t}} \def\Lw{{\mathfrak w}}
\def\Lq{{\mathfrak q}} \def\Lu{{\mathfrak u}} \def\Lv{{\mathfrak v}}
\def\Lz{{\mathfrak z}} \def\Lsl{\mathfrak{sl}} \def\Lgl{\mathfrak{gl}}
\def\Lp{{\mathfrak p}} \def\Lf{{\mathfrak f}} \def\Lx{{\mathfrak x}}
\def\Lpsl{\mathfrak{psl}}
\def\Lso{{\mathfrak{so}}} 
\def\Lsu{{\mathfrak{su}}} 
\def\Lsp{{\mathfrak{sp}}} 
\def\orthog{{\mathfrak{o}}} 
\def\tilso{{\mathfrak{so}}}
\def\tilco{{\mathfrak{co}}} 
\def\co{\mathfrak{co}}

% OVERLINE
\def\barscru{{\overline{\scru}}}
\def\barPsi{{\overline{\Psi}}}
\def\barA{{\overline{A}}}
\def\barB{{\overline{B}}}
\def\barC{{\overline{C}}}
\def\barD{{\overline{D}}}
\def\barE{{\overline{E}}}
\def\barF{{\overline{F}}}
\def\barbarF{{\overline{\overline{F}}}}
\def\barG{{\overline{G}}}
\def\barH{{\overline{H}}}
\def\barI{{\overline{I}}}
\def\barJ{{\overline{J}}}
\def\barK{{\overline{K}}}
\def\barL{{\overline{L}}}
\def\barM{{\overline{M}}}
\def\barN{{\overline{N}}}
\def\barP{{\overline{P}}}
\def\barbarP{{\overline{\overline{P}}}}
\def\barQ{{\overline{Q}}}
\def\barR{{\overline{R}}}
\def\barS{{\overline{S}}}
\def\barT{{\overline{T}}}
\def\barU{{\overline{U}}}
\def\barV{{\overline{V}}}
\def\barW{{\overline{W}}}
\def\barbarW{{\overline{\overline{W}}}}
\def\barX{{\overline{X}}}
\def\barY{{\overline{Y}}}
\def\barZ{{\overline{Z}}}
\def\barLr{{\overline{\Lr}}}
\def\barLg{{\overline{\Lg}}}
\def\barLh{{\overline{\Lh}}}
\def\barLp{{\overline{\Lp}}}
\def\barLn{{\overline{\Ln}}}
\def\barLm{{\overline{\Lm}}}
\def\barLl{{\overline{\Ll}}}
\def\bara{{\overline{a}}}
\def\bare{{\overline{e}}}
\def\barf{{\overline{f}}}
\def\barg{{\overline{g}}}
\def\bark{{\overline{k}}}
\def\baru{{\overline{U}}}
\def\barv{{\overline{v}}}
\def\barw{{\overline{w}}}
\def\barz{{\overline{z}}}
\def\barbeta{{\overline{\beta}}}
\def\bargamma{{\overline{\gamma}}}
\def\barI{{\overline{I}}}

% TILDE
\def\tilvec{\widetilde{\vec}}

% HAT
\def\hatA{{\widehat{A}}}
\def\hatC{{\widehat{C}}}
\def\hatH{{\widehat{H}}}
\def\hatI{{\widehat{I}}}
\def\hatN{{\widehat{N}}}
\def\hatQ{{\widehat{Q}}}
\def\hatR{{\widehat{R}}}
\def\hatS{{\widehat{S}}}
\def\hatV{{\widehat{V}}}
\def\hatt{{\widehat{t}}}
\def\hatu{{\widehat{u}}}
\def\hatv{{\widehat{v}}}
\def\hatw{{\widehat{w}}}
\def\hatx{{\widehat{x}}}
\def\haty{{\widehat{y}}}
\def\hatz{{\widehat{z}}}
\def\hatalpha{{\widehat{\alpha}}}
\def\hatrho{{\widehat{\rho}}}
\def\hatkappa{{\widehat{\kappa}}}
\def\hatlambda{{\widehat{\lambda}}}
\def\hatsigma{{\widehat{\sigma}}}
\def\hatmu{{\widehat{\mu}}}
\def\hatnu{{\widehat{\nu}}}
\def\hattau{{\widehat{\tau}}}

% CHECK
\def\ckt{{\widecheck{t}}}
\def\ckw{{\widecheck{w}}}
\def\ckx{{\widecheck{x}}}
\def\cky{{\widecheck{y}}}
\def\ckz{{\widecheck{z}}}
\def\cksigma{{\widecheck{\sigma}}}
\def\cktau{{\widecheck{\tau}}}
\def\ckH{{\widecheck{H}}}
\def\ckN{{\widecheck{N}}}
\def\ckS{{\widecheck{S}}}

\def\ds{\displaystyle}
\def\bfindex[#1]{{\bf#1}\index[terms]{#1}}

\def\timestimes{{\times\!\!\times}}

%operators
\def\Qd{{\rm Qd}}
\def\IC{{\rm IC}}
\def\RC{{\rm RC}}
\def\TS{{\rm TS}}
\def\ES{{\rm ES}}
\def\PF{{\rm PF}}
\def\set{{\rm set}}
\def\bi{{\rm bi}}
\def\fn{{\rm fn}}
\def\ss{{\rm setset}}
\def\sf{{\rm setfn}}
\def\fs{{\rm fnset}}
\def\seq{{\rm seq}}
\def\EUCL{{\rm EUCL}}
\def\TNSR{{\rm TNSR}}
\def\starstar{{**}}
\def\fincap{{\hbox{{\rm fin}}\,\cap}}
\def\fincup{{\hbox{{\rm fin}}\,\cup}}
\def\ctblcap{{\hbox{{\rm ctbl}}\,\cap}}
\def\ctblcup{{\hbox{{\rm ctbl}}\,\cup}}
\def\finamalg{{\hbox{{\rm fin}}\,\amalg}}
\def\two{{\rm two}}
\def\three{{\rm three}}
\def\ten{{\rm ten}}
\def\Lin{{\rm Lin}}
\def\Bilin{{\rm Bilin}}
\def\Trilin{{\rm Trilin}}
\def\PS{{\rm PS}}
\def\UE{{\rm UE}}
\def\FV{{\rm FV}}
\def\CH{{\rm CH}}
\def\DQ{{\rm DQ}}
\def\SS{{\rm SS}}
\def\SUMS{{\rm SUMS}}
\def\res{{\rm res}}
\def\add{{\rm add\,}}
\def\FS{{\rm FS}}
\def\LIMS{{\rm LIMS}}
\def\NOT{{\rm NOT}}
\def\ORB{{\rm ORB}}
\def\dlin{{\rm dlin}}
\def\LINS{{\rm LINS}}
\def\opLIMS{\mathop{\LIMS}}
\def\IDS{{\rm IDS}}
\def\UL{{\rm UL}}
\def\UB{{\rm UB}}
\def\LB{{\rm LB}}
\def\LP{{\rm LP}}
\def\LPD{{\rm LPD}}
\def\Isol{{\rm Isol}}
\def\Cl{{\rm Cl}}
\def\sCl{{\rm sCl}}
\def\id{{\rm id}}
\def\dim{{\rm dim\,}}
\def\DIMS{{\rm DIMS\,}}
\def\Bas{{\rm Bas\,}}
\def\Int{{\rm Int}}
\def\IntD{{\rm IntD}}
\def\det{{\rm det\,}}
\def\exp{{\rm exp}}
\def\End{{\rm End}}
\def\SO{{\rm SO}}
\def\CO{{\rm CO}}
\def\O{{\rm O}}
\def\Stab{{\rm Stab}}
\def\Fix{{\rm Fix}}
\def\GL{{\rm GL}}
\def\Ad{{\rm Ad}}
\def\Aut{{\rm Aut}}
\def\der{{\rm der}}
\def\ad{{\rm ad}}
\def\diag{{\rm diag}}
\def\Mink{{\rm Mink}}
\def\Tay{{\rm Tay}}
\def\lin{{\rm lin}}
\def\switch{{\rm switch}}
\def\SOtwortwo{{\SO(2)\ltimes\R^2}}
\def\O{{\rm O}}
\def\OB{{\rm OB}}
\def\SO{{\rm SO}}
\def\SL{{\rm SL}}
\def\PSL{{\rm PSL}}
\def\SBF{{\rm SBF}}
\def\STF{{\rm STF}}
\def\SM{{\rm SM}}
\def\QF{{\rm QF}}
\def\Id{{\rm Id}}
\def\dom{{\rm dom}}
\def\im{{\rm im}}
\def\dd{{\rm dd}}
\def\dct{{\rm dct}}
\def\sym{{\rm sym}}
\def\Sym{{\rm Sym}}
\def\as{{\rm as}}
\def\mm{{\rm mm}}
\def\cof{{\rm cof}}
\def\adj{{\rm adj}}
\def\det{{\rm det}}
\def\DET{{\rm DET}}
\def\ADJ{{\rm ADJ}}
\def\DNZ{{\hbox{\rm DNZ}}}
\def\CVZ{{\hbox{\rm CVZ}}}
\def\BNZ{{\hbox{\rm BNZ}}}
\def\ev{{\rm ev}}
\def\sl{{\rm s\ell}}

%misc
\def\vep{{\varepsilon}}
\def\bksl{{\backslash}}
\def\sigmaalg{{\rm s.alg}}
\def\drta{\dashrightarrow}
\def\ltlt{{<\!\!<}}
\def\phx{\phantom{x}}

\def\Xint#1{\mathchoice
{\XXint\displaystyle\textstyle{#1}}%
{\XXint\textstyle\scriptstyle{#1}}%
{\XXint\scriptstyle\scriptscriptstyle{#1}}%
{\XXint\scriptscriptstyle\scriptscriptstyle{#1}}%
\!\int}
\def\XXint#1#2#3{{\setbox0=\hbox{$#1{#2#3}{\int}$ }
\vcenter{\hbox{$#2#3$ }}\kern-.6\wd0}}
\def\ddashint{\Xint=}
\def\dashint{\Xint-}

\begin{document}

\centerline{\bf Solutions for MATH 4604 (Advanced Calculus II)}
\centerline{\bf Spring 2020}
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 6: Due on Wednesday 18 March}
\vskip.1in\hrule\vskip.1in\noindent
6-1. Let $V,W\in\ES$, let $*\in\scrb^{VV}_W$ and let $\#:=\Sym^*$.\\
\phx\quad $\,$ Show: \qquad $\#\in\scrsb^V_W$ \qquad and \qquad $\Qd^\#=\Qd^*$.
\vskip.1in\noindent
{\it Proof:}
We have: $\forall u,v\in V$,\\
\phx\qquad $\ds{u\#v=\frac{u*v+v*u}{2}=\frac{u*v+v*u}{2}=\frac{v*u+u*v}{2}=v\#u}$.\\
Then $\#\in\scrsb^V_W$. \quad Want: $\Qd^\#=\Qd^*$. \quad Want: $\forall u\in V$, $\Qd^\#_u=\Qd^*_u$.\\
Given $u\in V$. \qquad Want: $\Qd^\#_u=\Qd^*_u$. \qquad We have:\\
\phx\quad $\ds{\Qd^\#_u=u\#u=\frac{u*u+u*u}{2}=u*u=\Qd^*_u}$, \quad as desired. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
6-2. Let $V,W\in\ES$, let $*\in\scrsb^V_W$ and let $Q:=\Qd^*$.\\
\phx\qquad\qquad Show: \quad $\forall u,v\in V$, \quad $\ds{u*v=\frac{Q_{u+v}-Q_u-Q_v}{2}}$.\\
\vskip.1in\noindent
{\it Proof:} Given $u,v\in V$. Want: $\ds{u*v=\frac{Q_{u+v}-Q_u-Q_v}{2}}$.\\
We have: $Q_{u+v}=\Qd^*_{u+v}=(u+v)*(u+v)=(u*u)+2\cdot(u*v)+(v*v)$\\
\phx\quad and \qquad\qquad\qquad\qquad\qquad\quad $Q_u=\Qd^*_u=u*u$, $Q_v=\Qd^*_v=v*v$,\\
\phx\quad so \qquad\qquad\qquad\qquad\qquad\qquad $Q_{u+v}-Q_u-Q_v=2\cdot(u*v)$.\\
Then \qquad\quad $\ds{\,u*v=\frac{2\cdot(u*v)}{2}=\frac{Q_{u+v}-Q_u-Q_v}{2}}$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
6-3. Let $V:=\R^2$ and $W:=\R$. \qquad Define $Q:V\to W$ by:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad  $\forall x,y\in\R$, \qquad $Q(x,y)=3x^2-4xy+5y^2$.\\
\phx\quad $\,$ Let $B\in\scrsb^V_W$. \quad Assume: $\Qd^B=Q$. \quad Show: $\ds{[B]=\left[\begin{array}{cc}3&-2\\-2&5\end{array}\right]}$.
\vskip.1in\noindent
{\it Proof:} Let $u:=\vep_1^V$ and $v:=\vep_2^V$.\\
Let $x:=B(u,u)$ and $y:=B(u,v)$ and $z:=B(v,v)$.\\
Then, as $B\in\scrsb^V_W$, we get: $y=B(v,u)$. \quad $\,$
Then $\ds{[B]=\left[\begin{array}{cc}x&y\\y&z\end{array}\right]}$.\\
We have $u=\vep_1^V=(1,0)$ and $v=\vep_2^V=(0,1)$. \quad Then $u+v=(1,1)$.\\
We have $\,\,\,\,\,Q(u)\,\,\,\,\,=Q(1,0)=3\cdot1^2-4\cdot1\cdot0+5\cdot0^2=3$.\\
We have $\,\,\,\,\,Q(v)\,\,\,\,\,=Q(0,1)=3\cdot0^2-4\cdot0\cdot1+5\cdot1^2=5$.\\
We have $Q(u+v)=Q(1,1)=3\cdot1^2-4\cdot1\cdot1+5\cdot1^2=3-4+5=4$.\\
By HW\#2, we have $\ds{B(u,v)=\frac{[Q(u+v)]-[Q(u)]-[Q(v)]}{2}}$.\\
\phx\qquad\quad Then \quad $\ds{\,y=B(u,v)=\frac{4-3-5}{2}=\frac{4-8}{2}=\frac{-4}{2}=-2}$.\\
Also, $x=B(u,u)=\Qd^B(u)=Q(u)=3$.\\
Also, $z=B(v,v)=\Qd^B(v)=Q(v)=5$.\\
Then $\ds{[B]=\left[\begin{array}{cc}x&y\\y&z\end{array}\right]=\left[\begin{array}{cc}3&-2\\-2&5\end{array}\right]}$, \quad as desired. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
6-4. Let $V:=\R^2$. \qquad Define $Q:V\to\R$ by:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\forall x,y\in\R$, \qquad $Q(x,y)=2x^2+6xy+5y^2$.\\
\phx\quad $\,$ Show: $Q>0$ on $V_{0_V}^\times$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall u\in V_{0_V}^\times$, $Q_u>0$.\\
Given $u\in V_{0_V}^\times$. \qquad Want: $Q_u>0$.\qquad Want: $2\cdot Q_u>0$.\\
Since $u\in V_{0_V}^\times=V\backslash\{0_V\}$, we conclude: \quad $u\ne0_V$.\\
Since $u\in V_{0_V}^\times\subseteq V=\R^2$, choose $x,y\in\R$ s.t.~$u=(x,y)$.\\
Then $Q_u=Q(u)=Q(x,y)=2x^2+6xy+5y^2$.\\
Then $2\cdot Q_v=4x^2+12xy+10y^2=(2x+3y)^2+y^2$.\\
Want: $(2x+3y)^2+y^2>0$. \quad Let $w:=2x+3y$. \quad Want: $w^2+y^2>0$.\\
Let $s:=w^2+y^2$. \qquad Want: $s>0$.\\
Assume $s\le0$. \qquad Want: Contradiction.\\
Since $s\le0$ and $y^2\ge0$, we get: $s-y^2\le0$.\\
Since $w^2=s-y^2\le0$ and $w^2\ge0$, we get: $w^2=0$. \quad Then $w=0$.\\
Then $s=w^2+y^2=0^2+y^2=y^2$, \qquad so \qquad $s=y^2$.\\
Since $s\le0$ and $s=y^2\ge0$, we get: $s=0$.\\
Then $y^2=s=0$. \quad Then $y=0$.\\
Then $0=w=2x+3y=2x+3\cdot0=2x$. \quad Then $x=0$.\\
Then $0_V\ne u=(x,y)=(0,0)=0_V$. \qquad Contradiction. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
6-5. Let $V:=\R^2$. \qquad Define $Q:V\to\R$ by:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\forall x,y\in\R$, \qquad $Q(x,y)=2x^2+6xy+4y^2$.\\
\phx\quad $\,$ Find $x,y\in\R$ s.t.~$Q(x,y)<0$.
\vskip.1in\noindent
{\it Solution:}
Let $x:=-3$ and $y:=2$. \quad Then $Q(x,y)=-2<0$.
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 5: Due on Tuesday 3 March}
\vskip.1in\hrule\vskip.1in\noindent
5-1. Let \quad $\alpha,\beta\in\N$, \quad $V:=\R^\alpha$, \quad $W:=\R^\beta$, \quad $L\in\scra\scrl^V_W$.\\
\phx\quad $\,$ Show: \quad $\Lin^{[L]}\,\,=\,\,L$.
\vskip.1in\noindent
{\it Proof:}
Let $S:=[L]$, $F:=\Lin^S$. \quad Then: $F=\Lin^{[L]}$. \quad Want: $F=L$.\\
Want: $\forall v\in V$, $F_v=L_v$. \qquad
Given $v\in V$. \quad Want: $F_v=L_v$.\\
We have: $\ds{F_v=\Lin^S(v)=\sum_{k\in\scri_W}\,\sum_{j\in\scri_V}\,v_j\cdot S_{kj}\cdot\vep_k^W=\sum_{j\in\scri_V}\,\sum_{k\in\scri_W}\,v_j\cdot S_{kj}\cdot\vep_k^W}$\\
\phx\qquad\qquad\qquad\qquad\quad $\ds{\,\,\,=\sum_{j\in\scri_V}\,v_j\cdot\left[\sum_{k\in\scri_W}\,S_{kj}\cdot\vep_k^W\right]}$.\\
\phx\qquad\qquad\qquad\quad $\,$ Want: $\ds{\sum_{j\in\scri_V}\,v_j\cdot\left[\sum_{k\in\scri_W}\,S_{kj}\cdot\vep_k^W\right]\,\,\,=\,\,\,L_v}$.\\
By the reproducing formula, \quad $\ds{\,\,\,\,\,v=\sum_{j\in\scri_V}\,v_j\cdot\vep_j^V}$.\\
So, as $L\in\scral^V_W$, we conclude: $\,$ $\ds{L_v=\sum_{j\in\scri_V}\,v_j\cdot[L(\vep_j^V)]}$.\\
\phx\quad Want: $\ds{\sum_{j\in\scri_V}\,v_j\cdot\left[\sum_{k\in\scri_W}\,S_{kj}\cdot\vep_k^W\right]\,=\,\sum_{j\in\scri_V}\,v_j\cdot[L(\vep_j^V)]}$.\\
\phx\qquad\quad Want: $\,$ $\forall j\in\scri_V$, $\ds{\,\,\sum_{k\in\scri_W}\,S_{kj}\cdot\vep_k^W=L(\vep_j^V)}$.\\
Given $j\in\scri_V$. \qquad Want: $\ds{\sum_{k\in\scri_W}\,S_{kj}\cdot\vep_k^W=L(\vep_j^V)}$.\\
Let $w:=L(\vep_j^V)$. \quad Want: $\ds{\sum_{k\in\scri_W}\,S_{kj}\cdot\vep_k^W=w}$.\\
By the reproducing formula, \quad $\ds{w=\sum_{k\in\scri_W}\,w_k\cdot\vep_k^W}$.\\
\phx\qquad\qquad\quad Want: $\ds{\,\sum_{k\in\scri_W}\,S_{kj}\cdot\vep_k^W=\sum_{k\in\scri_W}\,w_k\cdot\vep_k^W}$.\\
\phx\quad Want: \quad $\forall k\in\scri_W$, \quad $\ds{\,S_{kj}=w_k}$.\\
Given $k\in\scri_W$. $\,\,\,\,\,\,$ Want: $\ds{S_{kj}=w_k}$.\\
Recall that $S=[L]$ and that $w=L(\vep_j^V)$.\\
Then $S_{kj}=[L]_{kj}=(L(\vep_j^V))\sbullet\vep_k^W=w\sbullet\vep_k^W=w_k$, as desired. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
5-2. Let \quad $\alpha,\beta\in\N$, \quad $V:=\R^\alpha$, \quad $W:=\R^\beta$, \quad $S\in W\otimes V$.\\
\phx\quad $\,$ Show: \quad $[\,\Lin^S\,]\,\,=\,\,S$.
\vskip.1in\noindent
{\it Proof:} Since $S\in W\otimes V$, we get: $\Lin^S\in\scrl^V_W$.\\
Let $L:=\Lin^S$. \quad Want: $[L]=S$.\\
We have $L=\Lin^S\in\scrl^V_W$, so $[L]\in W\otimes V$. \quad Also, $S\in W\otimes V$.\\
Want: $\forall k\in\scri_W$, $\forall j\in\scri_V$, $[L]_{kj}=S_{kj}$.\\
Given $k\in\scri_W$, $j\in\scri_V$. \qquad Want: $[L]_{kj}=S_{kj}$.\\
We have $[L]_{kj}=(L(\vep_j^V))\sbullet\vep_k^W$. \qquad Want: $(L(\vep_j^V))\sbullet\vep_k^W=S_{kj}$.\\
Let $w:=L(\vep_j^V)$. \qquad Want: $w\sbullet\vep_k^W=S_{kj}$. \qquad Want: $w_k=S_{kj}$.\\
Let $v:=\vep_j^V$. \quad Then $\ds{w=L(v)=\Lin^S(v)=\sum_{K\in\scri_W}\,\sum_{J\in\scri_V}\,v_J\cdot S_{KJ}\cdot\vep_K^W}$.\\
Since $v=\vep_j^V$, we see that: \qquad $\forall J\in\scri_V$, \quad $v_J=\delta_J^j$.\\
Then $\ds{w=\sum_{K\in\scri_W}\,\sum_{J\in\scri_V}\,\delta_J^j\cdot S_{KJ}\cdot\vep_K^W=\sum_{K\in\scri_W}\,S_{Kj}\cdot\vep_K^W}$.\\
Then $\ds{w_k=\sum_{K\in\scri_W}\,S_{Kj}\cdot(\vep_K^W)_k=\sum_{K\in\scri_W}\,S_{Kj}\cdot\delta_K^k=S_{kj}}$, as desired. QED
\vskip.1in\hrule\vskip.1in\noindent
5-3. Let \quad $\alpha,\beta,\gamma\in\N$, \quad $V:=\R^\alpha$, \quad $W:=\R^\beta$, \quad $X:=\R^\gamma$.\\
\phx\quad $\,$ Let $L\in\scrl^V_W$ and let $M\in\scrl^W_X$. \qquad Show: \quad $[M\circ L]\,\,=\,\,[M]\cdot[L]$.
\vskip.1in\noindent
{\it Proof:}
We have $M\circ L\in\scrl^V_X$, so $[M\circ L]\in X\otimes V$. Then $[M\circ L]\in\R^{\gamma\times\alpha}$.\\
Since $M\in\scrl^W_X$ and $L\in\scrl^V_W$, we get: $[M]\in X\otimes W$ and $[L]\in W\otimes V$.\\
Then $[M]\in\R^{\gamma\times\beta}$ and $[L]\in\R^{\beta\times\alpha}$, and so $[M]\cdot[L]\in\R^{\gamma\times\alpha}$.\\
Want: \qquad $\forall k\in[1..\gamma]$, $\forall i\in[1..\alpha]$, \quad $[M\circ L]_{ki}=([M]\cdot[L])_{ki}$.\\
Given $k\in[1..\gamma]$, $i\in[1..\alpha]$. \qquad Want: $[M\circ L]_{ki}=([M]\cdot[L])_{ki}$.\\
By definition of matrix multiplication,\\
\phx\qquad\qquad we get: \quad $\ds{([M]\cdot[L])_{ki}\,\,=\,\,\sum_{j=1}^\beta\,[M]_{kj}\cdot[L]_{ji}}$.\\
Want: $\ds{[M\circ L]_{ki}=\sum_{j=1}^\beta\,[M]_{kj}\cdot[L]_{ji}}$.\\
Let \quad $w:=L(\vep_i^V)$ \quad and let \quad $x:=M(w)$.\\
Then: \quad $(M\circ L)(\vep_i^V)=M(L(\vep_i^V))=M(w)=x$.\\
Then: \quad $[M\circ L]_{ki}=((M\circ L)(\vep_i^V))\sbullet\vep_k^W=x\sbullet\vep_k^W=x_k$.\\
Want: $\ds{x_k=\sum_{j=1}^\beta\,[M]_{kj}\cdot[L]_{ji}}$.\\
By the reproducing formula, $\ds{w=\sum_{j=1}^\beta\,w_j\cdot\vep_j^W}$.\\
So, since $M\in\scrl^W_X\subseteq\scral^W_X$, we get $\ds{M(w)=\sum_{j=1}^\beta\,w_j\cdot(M(\vep_j^W))}$.\\
Then $\ds{x_k=(M(w))_k=\sum_{j=1}^\beta\,w_j\cdot[(M(\vep_j^W))_k]}$.\\
Want: $\ds{\sum_{j=1}^\beta\,w_j\cdot[(M(\vep_j^W))_k]=\sum_{j=1}^\beta\,[M]_{kj}\cdot[L]_{ji}}$.\\
Want: \qquad $\forall j\in[1..\beta]$, \quad $w_j\cdot[(M(\vep_j^W))_k]=[M]_{kj}\cdot[L]_{ji}$.\\
Given $j\in[1..\beta]$. \qquad Want: $w_j\cdot[(M(\vep_j^W))_k]=[M]_{kj}\cdot[L]_{ji}$.\\
We have $[M]_{kj}=(M(\vep_j^W))\sbullet\vep_k^X=(M(\vep_j^W))_k$.\\
Also, $[L]_{ji}=(L(\vep_i^V))\sbullet\vep_j^W=w\sbullet\vep_j^W=w_j$.\\
Then $w_j\cdot[(M(\vep_j^W))_k]=[L]_{ji}\cdot[M]_{kj}=[M]_{kj}\cdot[L]_{ji}$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
5-4. Let \quad $\alpha,\beta,\gamma\in\N$, \quad $V:=\R^\alpha$, \quad $W:=\R^\beta$, \quad $X:=\R^\gamma$.\\
\phx\quad $\,$ Let $B\in\scrb^{VW}_X$. \qquad Show: \quad $\Bilin^{[B]}=B$.
\vskip.1in\noindent
{\it Proof:}
Let $T:=[B]$, $F:=\Bilin^T$. \qquad Then: $F=\Bilin^{[B]}$.\\
Want: $F=B$. \qquad Want: $\forall v\in V$, $\forall w\in W$, $F_{(v,w)}=B_{(v,w)}$.\\
Given $v\in V$, $w\in W$. \qquad Want: $F_{(v,w)}=B_{(v,w)}$.\\
We have: $B_{(v,w)}=\Bilin^T_{(v,w)}$\\
\phx\qquad\qquad\qquad $\ds{\,\,\,\,=\sum_{k\in\scri_X}\,\sum_{i\in\scri_V}\,\sum_{j\in\scri_W}\,v_i\cdot w_j\cdot T_{kij}\cdot\vep_k^X}$\\
\phx\qquad\qquad\qquad $\ds{\,\,\,\,=\sum_{i\in\scri_V}\,\sum_{j\in\scri_W}\,\sum_{k\in\scri_X}\,v_i\cdot w_j\cdot T_{kij}\cdot\vep_k^X}$\\
\phx\qquad\qquad\qquad $\ds{\,\,\,\,=\sum_{i\in\scri_V}\,\sum_{j\in\scri_W}\,v_i\cdot w_j\cdot\left[\sum_{k\in\scri_X}\,T_{kij}\cdot\vep_k^X\right]}$.\\
\phx\qquad\qquad $\,$ Want: $\ds{\sum_{i\in\scri_V}\,\sum_{j\in\scri_W}\,v_i\cdot w_j\cdot\left[\sum_{k\in\scri_X}\,T_{kij}\cdot\vep_k^X\right]\,\,\,=\,\,\,B_{(v,w)}}$.\\
By the reproducing formula, \quad $\ds{v=\sum_{i\in\scri_V}\,v_i\cdot\vep_i^V}$ and $\ds{w=\sum_{j\in\scri_W}\,w_j\cdot\vep_j^W}$.\\
So, as $B\in\scrb^{VW}_X$, we conclude: \quad $\ds{B_{(v,w)}=\sum_{i\in\scri_V}\,\sum_{j\in\scri_W}\,v_i\cdot w_j\cdot[B(\vep_i^V,\vep_j^W)]}$.\\
Want: $\ds{\sum_{i\in\scri_V}\sum_{j\in\scri_W}v_i\cdot w_j\cdot\left[\sum_{k\in\scri_X}T_{kij}\cdot\vep_k^X\right]=\sum_{i\in\scri_V}\sum_{j\in\scri_W}v_i\cdot w_j\cdot[B(\vep_i^V,\vep_j^W)]}$.\\
\phx\qquad Want: $\,\,\,\,\forall i\in\scri_V$, $\forall j\in\scri_W$, $\ds{\,\,\,\sum_{k\in\scri_X}\,T_{kij}\cdot\vep_k^X=B(\vep_i^V,\vep_j^W)}$.\\
Given $i\in\scri_V$, $j\in\scri_W$. \qquad Want: $\ds{\sum_{k\in\scri_X}\,T_{kij}\cdot\vep_k^X=B(\vep_i^V,\vep_j^W)}$.\\
Let $w:=B(\vep_i^V,\vep_j^W)$. \qquad $\,$ Want: $\ds{\sum_{k\in\scri_X}\,T_{kij}\cdot\vep_k^X=w}$.\\
By the reproducing formula, $\ds{\,\,\,\,\,w=\sum_{k\in\scri_X}\,w_k\cdot\vep_k^X}$.\\
\phx\qquad\qquad\quad Want: $\ds{\,\sum_{k\in\scri_X}\,T_{kij}\cdot\vep_k^X=\sum_{k\in\scri_X}\,w_k\cdot\vep_k^X}$.\\
\phx\quad Want: \quad $\forall k\in\scri_X$, \quad $\ds{\,T_{kij}=w_k}$.\\
Given $k\in\scri_X$. $\,\,\,\,\,\,$ Want: $\ds{T_{kij}=w_k}$.\\
Recall that $T=[B]$ and that $w=B(\vep_i^V,\vep_j^W)$.\\
Then $T_{kij}=[B]_{kij}=(B(\vep_i^V,\vep_j^W))\sbullet\vep_k^X=w\sbullet\vep_k^X=w_k$. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
5-5. Let \quad $\alpha,\gamma\in\N$, \quad $V:=\R^\alpha$, \quad $X:=\R^\gamma$, \quad $*\in\scrb^{VV}_X$.\\
\phx\quad $\,$ Define $Q:V\to X$ by: $\forall v\in V$, $Q_v=v*v$. \quad Show: $Q\in\scrq^V_X$.
\vskip.1in\noindent
{\it Proof:} \qquad Let $T:=[*]$. \qquad Then $T\in X\otimes V\otimes V=\R^{\gamma\times\alpha\times\alpha}$.\\
For all $i,j\in\scri_V$, let $\ds{x_{ij}:=\sum_{k\in\scri_X}\,T_{kij}\cdot\vep_k^X}$ and let $M_{ij}:=C_V^{x_{ij}}\cdot\pi_i^V\cdot\pi_j^V$.\\
Then, $\forall i,j\in\scri_V$, we have: $x_{ij}\in X$, so $C_V^{x_{ij}}\in\scrc^V_X$, so $M_{ij}\in\scrm_2^{VX}$.\\
Then $\ds{\sum_{i\in\scri_V}\,\sum_{j\in\scri_V}\,M_{ij}\in\scrh_2^{VX}=\scrq^V_X}$. \quad Want: $\ds{Q=\sum_{i\in\scri_V}\,\sum_{j\in\scri_V}\,M_{ij}}$.\\
Want: \qquad $\forall v\in V$, \quad $\ds{\,\,Q(v)=\left(\sum_{i\in\scri_V}\,\sum_{j\in\scri_V}\,M_{ij}\right)(v)}$.\\
Given $v\in V$. \quad Want: $\ds{Q(v)=\left(\sum_{i\in\scri_V}\,\sum_{j\in\scri_V}\,M_{ij}\right)(v)}$.\\
\phx\qquad\qquad\qquad\quad Want: $\ds{\,Q_v\,\,=\,\,\,\sum_{i\in\scri_V}\,\sum_{j\in\scri_V}\,(M_{ij})_v}$.\\
We have: \quad $\forall i,j\in\scri_V$, $\,$ $(M_{ij})_v=(C_V^{x_{ij}}\cdot\pi_i^V\cdot\pi_j^V)_v$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\quad $\,=(C_V^{x_{ij}})_v\cdot(\pi_i^V)_v\cdot(\pi_j^V)_v$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\quad $\,=x_{ij}\cdot v_i\cdot v_j\,\,=\,\,v_i\cdot v_j\cdot x_{ij}$.\\
\phx\qquad\qquad\qquad\qquad\qquad Want: $\,\,\,$ $\ds{\,Q_v=\sum_{i\in\scri_V}\,\sum_{j\in\scri_V}\,v_i\cdot v_j\cdot x_{ij}}$.\\
By the reproducing formula, $\,$ $\ds{v=\sum_{i\in\scri_V}\,v_i\cdot\vep_i^V}$ $\,$ and $\,$ $\ds{v=\sum_{j\in\scri_V}\,v_j\cdot\vep_j^V}$.\\
Then: $\ds{\,\,Q_v=v*v=\left[\sum_{i\in\scri_V}\,v_i\cdot\vep_i^V\right]*\left[\sum_{j\in\scri_V}\,v_j\cdot\vep_j^V\right]}$\\
\phx\qquad\qquad $\ds{\,\,=\sum_{i\in\scri_V}\,\sum_{j\in\scri_V}\,v_i\cdot v_j\cdot[\vep_j^V*\vep_j^V]}$.\\
\phx\qquad Want: $\ds{\,\,\sum_{i\in\scri_V}\,\sum_{j\in\scri_V}\,v_i\cdot v_j\cdot[\vep_j^V*\vep_j^V]=\sum_{i\in\scri_V}\,\sum_{j\in\scri_V}\,v_i\cdot v_j\cdot x_{ij}}$.\\
\phx\qquad\quad Want: \quad $\forall i,j\in\scri_V$, $\,\,\,$ $\vep_i^V*\vep_j^V\,\,=\,\,x_{ij}$.\\
\phx\qquad Given $i,j\in\scri_V$. $\,$ Want: $\vep_i^V*\vep_j^V\,\,=\,\,x_{ij}$.\\
\phx\qquad\quad Want: \quad $\forall \ell\in\scri_X$, $\,\,\,\,$ $(\vep_i^V*\vep_j^V)_\ell=(x_{ij})_\ell$.\\
\phx\quad\quad Given $\ell\in\scri_X$. $\,$ Want: $(\vep_i^V*\vep_j^V)_\ell=(x_{ij})_\ell$.\\
Since $\ds{x_{ij}=\sum_{k\in\scri_X}\,T_{kij}\cdot\vep_k^X}$, we get
$\ds{(x_{ij})_\ell=\sum_{k\in\scri_X}\,T_{kij}\cdot(\vep_k^X)_\ell}$.\\
\phx\qquad\qquad\qquad\qquad\quad Then: \qquad $\ds{(x_{ij})_\ell=\sum_{k\in\scri_X}\,T_{kij}\cdot\delta^k_\ell\,\,=\,\,T_{\ell ij}}$.\\
Since $T=[*]$, we get $T_{\ell ij}=(\vep_i^V*\vep_j^V)\sbullet\vep_\ell^X$.\\
Then $(\vep_i^V*\vep_j^V)_\ell=(\vep_i^V*\vep_j^V)\sbullet\vep_\ell^X=T_{\ell ij}=(x_{ij})_\ell$, as desired.
QED
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 4: Due on Tuesday 25 February}
\vskip.1in\hrule\vskip.1in\noindent
4-1. Define $a\in\R^{\N_0}$ by: $\forall j\in\N_0$, $a_j=1/(j!)$. \qquad Show: $\RC_a=\infty$.
\vskip.1in\noindent
{\it Proof:}
Let $T:=\{s\ge0\,|\,TS_s^a\hbox{ is bounded in }\R\}$.\\
Then $\RC_a=\sup\,T$. \qquad Want: $\sup\,T=\infty$. \qquad Want: $T=[0;\infty)$.\\
By definition of $T$, we have $T\subseteq[0;\infty)$. \qquad Want: $[0;\infty)\subseteq T$.\\
Want: $\forall s\in[0;\infty)$, $s\in T$. \qquad Given $s\in[0;\infty)$. \quad Want: $s\in T$.\\
Since $s\ge0$, it suffices to show: $TS_s^a$ is bounded in $\R$.\\
Want: $\{a_0,a_1s,a_2s^2,a_3s^3,\ldots\}$ is bounded in $\R$.\\
Define $b\in\R^{\N_0}$ by: \quad $\forall j\in\N_0$, $\,$ $b_j=a_js^j$.\\
Then $\I_b=\{b_0,b_1,b_2,b_3,\ldots\}=\{a_0,a_1s,a_2s^2,a_3s^3,\ldots\}$.\\
Want: $\I_b$ is bounded in $\R$. \qquad Want: $b$ is convergent in $\R$.\\
Want: $b\to0$ in $\R$. \qquad Let $c:=1/2$. \qquad Then $c\in(0;1)$.\\
By Theorem 4.14.3, it suffices to show:\\
\phx\qquad $\exists k\in\N_0$ \quad s.t. \quad $\forall j\in[k..\infty)$, \quad $b_{j+1}/b_j\le c$.\\
By the Archimedean Principle, choose $k\in\N$ s.t. $k>2s-1$.\\
Then $k\in\N\subseteq\N_0$. \qquad
Want: $\forall j\in[k..\infty)$, $b_{j+1}/b_j\le c$.\\
Given $j\in[k..\infty)$. \qquad Want: $b_{j+1}/b_j\le c$.\\
We have \quad $b_{j+1}=a_{j+1}s^{j+1}=s^{j+1}/((j+1)!)=[s^j/(j!)]\cdot[s/(j+1)]$,\\
\phx\qquad\qquad\qquad $\,\,\,=b_j\cdot[s/(j+1)]$, \qquad\qquad so \quad $b_{j+1}/b_j=s/(j+1)$.\\
Since $j\in[k..\infty)\ge k>2s-1$, \quad we get $j>2s-1$, \quad so $j+1>2s$,\\
\phx\qquad so $(1/2)\cdot(j+1)>s$, \quad and so $1/2>s/(j+1)$.\\
Then $b_{j+1}/b_j=s/(j+1)<1/2=c$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
4-2. Let $\,$ $a\in\R^{\N_0}$, $\,$ $x,r\in\R$, $\,$ $i,j\in\N_0$, $\,$ $b:=P_\bullet SS_x^a$, $\,$ $c:=P_\bullet SS_r^{|a|}$.\\
\phx\quad $\,$ Assume: $|x|\le r$. \qquad Show: $|b_j-b_i|\le|c_j-c_i|$.
\vskip.1in\noindent
{\it Proof:}
Let \quad $k:=\min\{i,j\}$ \quad and \quad $\ell:=\max\{i,j\}$.\\
Then \quad $|b_j-b_i|=|b_\ell-b_k|$ \quad and \quad $|c_j-c_i|=|c_\ell-c_k|$.\\
Want: $|b_\ell-b_k|\le|c_\ell-c_k|$. \qquad
We have $k\le\ell$. \qquad Let $Q:=(k..\ell]$.\\
We have: \qquad $\forall n\in\N_0$, \quad $\ds{b_n=P_nSS_x^a=\sum_{m=1}^n\,a_m\cdot x^m}$.\\
Since \quad $\ds{b_k=\sum_{m=1}^k\,a_m\cdot x^m}$ \quad and \quad $\ds{b_\ell=\sum_{m=1}^\ell\,a_m\cdot x^m}$,\\
\phx\qquad it follows that: \quad $\ds{b_\ell-b_k=\sum_{m\in Q}\,a_m\cdot x^m}$.\\
We have: \quad $\forall n\in\N_0$, \quad $\ds{c_n=P_nSS_r^{|a|}=\sum_{m=1}^n\,|a|_m\cdot r^m=\sum_{m=1}^n\,|a_m|\cdot r^m}$.\\
Since \quad $\ds{c_k=\sum_{m=1}^k\,|a_m|\cdot r^m}$ \quad and \quad $\ds{c_\ell=\sum_{m=1}^\ell\,|a_m|\cdot r^m}$,\\
\phx\qquad it follows that: \quad $\ds{c_\ell-c_k=\sum_{m\in Q}\,|a_m|\cdot r^m}$.\\
Since $0\le|x|\le r$, we get $r\ge0$, and so $|r|=r$.\\
Then $\ds{\,|c_\ell-c_k|=\left|\sum_{m\in Q}\,|a_m|\cdot r^m\right|\le\sum_{m\in Q}\,|\,|a|_m\cdot r^m|}$\\
\phx\qquad\qquad\qquad $\ds{=\sum_{m\in Q}\,|a_m|\cdot|r^m|=\sum_{m\in Q}\,|a_m|\cdot|r|^m=\sum_{m\in Q}\,|a_m|\cdot r^m}$.\\
Since $|x|\le r$, we get: \qquad $\forall m\in\N_0$, \quad $|a_m|\cdot|x|^m\le|a_m|\cdot r^m$.\\
Then: \qquad $\ds{\sum_{m\in Q}\,|a_m|\cdot|x|^m\le\sum_{m\in Q}\,|a_m|\cdot r^m}$.\\
Then $\ds{\,|b_\ell-b_k|=\left|\sum_{m\in Q}\,a_m\cdot x^m\right|\le\sum_{m\in Q}\,|a_m\cdot x^m|=\sum_{m\in Q}\,|a_m|\cdot|x^m|}$\\
\phx\qquad\qquad\qquad $\ds{=\sum_{m\in Q}\,|a_m|\cdot|x|^m\le\sum_{m\in Q}\,|a_m|\cdot r^m=|c_\ell-c_k|}$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
4-3. Let $\,$ $\alpha\in[0;\infty)^{\N_0}$, $\,$ $r\in[0;\RC_\alpha)$, $\,$ $c:=P_\bullet SS_r^\alpha$.\\
\phx\quad $\,$ Show: $c$ is convergent in $\R$.
\vskip.1in\noindent
{\it Proof:}
We have: \quad $\forall n\in\N_0$, \quad $\ds{c_n=P_nSS_r^{\alpha}=\sum_{m=1}^n\,\alpha_m\cdot r^m}$.\\
\\
{\it Claim:} $c$ is semi-increasing.\\
{\it Proof of claim:}
\qquad Want: $\,\,\forall i\in\N$, $\,$ $c_i\ge c_{i-1}$.\\
\phx\qquad\qquad\qquad Given $i\in\N$. \quad Want: $c_i\ge c_{i-1}$.\\
Since \quad $\ds{c_{i-1}=\sum_{m=1}^{i-1}\,\alpha_m\cdot r^m}$ \quad and \quad $\ds{c_i=\sum_{m=1}^i\,\alpha_m\cdot r^m}$,\\
\phx\qquad it follows that: \quad $c_i-c_{i-1}=\alpha_i\cdot r^i$.\\
Since $\alpha\in[0;\infty)^{\N_0}$, \quad we get: $\alpha_i\in[0;\infty)$.\\
Since $\alpha_i\in[0;\infty)\ge0$ and $r\in[0;\infty)>0$, it follows that $\alpha_i\cdot r^i\ge0$.\\
Then $c_i-c_{i-1}=\alpha_i\cdot r^i\ge0$, and so $c_i\ge c_{i-1}$, as desired.\\
{\it End of proof of claim.}\\
\\
By Theorem 3.11.2, $\forall\tau\in\R^{\N_0}$,\\
\phx\qquad\qquad $[\,\,(\,\,\I_\tau\hbox{ is bounded in }\R\,\,)\,\,\&\,\,(\,\,\tau\hbox{ is semi-increasing}\,\,)\,\,]$\\
\phx\qquad\qquad\qquad\qquad $\Rightarrow\,\,[\,\,\tau\hbox{ is convergent in }\R\,\,]$.\\
So, by the claim, it suffices to show: \quad $\I_c$ is bounded in $\R$.\\
Since $r\in[0;\RC_\alpha),\RC_\alpha$, \quad by Theorem 4.16.16,\\
\phx\qquad\qquad\qquad choose $s>r$ s.t.~$\TS_s^\alpha$ is bounded in $\R$.\\
Choose $B\in\scrb_\R(0)$ s.t.~$\TS_s^\alpha\subseteq B$.\\
Choose $M>0$ s.t.~$B=B_\R(0,M)$.\\
We have $r\in[0;\RC_\alpha)\ge0$. \quad Then $0\le r<s$.\\
Let $u:=r/s$. \quad Then $0\le u<1$. \quad Then $u\in[0;1)$. \quad Also $r=su$.\\
We have $\TS_s^\alpha=\{\alpha_0,\alpha_1s^1,\alpha_2s^2,\alpha_3s^3,\ldots\}$.\\
Then: \quad $\forall j\in\N_0$, \quad $\alpha_js^j\in\TS_s^\alpha$.\\
Then: \quad $\forall j\in\N_0$, \quad $\alpha_js^j\in\TS_s^\alpha\subseteq B=B_\R(0,M)<M$.\\
Then: \quad $\forall j\in\N_0$, \quad $\alpha_js^j<M$.\\
Then, since $u\ge0$, we get: \quad $\forall j\in\N_0$, \quad $\alpha_js^ju^j\le Mu^j$.\\
Then: \quad $\,\,\forall j\in\N_0$, \quad $\alpha_jr^j=\alpha_j\cdot(su)^j=\alpha_js^ju^j\le Mu^j$.\\
Let $R:=M/(1-u)$. \quad As $M>0$ and $1-u>0$, we get: $R>0$.\\
Any ball in a metric space is bounded, so $B_\R(0,R)$ is bounded.\\
It therefore suffices to show: \qquad $\I_c\subseteq B_\R(0,R)$.\\
Want: \quad $\forall y\in\I_c$, $\,$ $y\in B_\R(0,R)$.\\
Given $y\in\I_c$. \qquad Want: $y\in B_\R(0,R)$. \qquad Want: $y\in(-R;R)$.\\
Want: $-R<y<R$. \qquad Want: $0\le y<R$.\\
Since $y\in\I_c$ and $\D_c=\N_0$, choose $k\in\N_0$ s.t.~$y=c_k$.\\
Then $y=\alpha_0+\alpha_1r+\alpha_2r^2+\alpha_3r^3+\cdots+\alpha_kr^k$.\\
So, since $\alpha_0,\ldots,\alpha_k\in\I_\alpha\subseteq[0;\infty)\ge0$ and since $r\in[0;\RC_\alpha)\ge0$,\\
\phx\qquad we conclude: $y\ge0$. \qquad It remains to show that: $y<R$.\\
Recall: $\,$ $\forall j\in\N_0$, $\,$ $\alpha_jr^j\le Mu^j$. \qquad\quad Then: $\,$ $\ds{\sum_{j=0}^k\,\alpha_jr^j\le\sum_{j=0}^k\,Mu^j}$.\\
By a theorem proved in class, $1+u+u^2+u^3+\cdots+u^k<1/(1-u)$.\\
So, as $M>0$, we get: $M\cdot(1+u+u^2+u^3+\cdots+u^k)<M/(1-u)$.\\
Then $y=\alpha_0+\alpha_1r+\alpha_2r^2+\alpha_3r^3+\cdots+\alpha_kr^k$\\
\phx\qquad\qquad\qquad $\ds{=\sum_{j=0}^k\,\alpha_jr^j\le\sum_{j=0}^k\,Mu^j=M\cdot\sum_{j=0}^k\,u^j}$\\
\phx\qquad\qquad\qquad $\ds{=M\cdot(1+u+u^2+u^3+\cdots+u^k)}$\\
\phx\qquad\qquad\qquad $<M/(1-u)=R$, \qquad as desired. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
4-4. Let $a\in\R^{\N_0}$ and let $\alpha:=|a|$. \qquad Show: $\RC_a=\RC_\alpha$.
\vskip.1in\noindent
{\it Proof:} We have: $\forall s\ge0$, \quad $|s|=s$.\\
Also: $\forall s\ge0$, \qquad $\TS_s^a=\{a_js^j~\hbox{s.t.}~j\in\N_0\}$.\\
Then: $\forall s\ge0$,\\
\phx\quad $\,\,\,\,|\TS_s^a|=\{|a_js^j|~\hbox{s.t.}~j\in\N_0\}$\\
\phx\qquad\qquad $\,\,=\{|a_j|\cdot|s^j|~\hbox{s.t.}~j\in\N_0\}$\\
\phx\qquad\qquad $\,\,=\{|a|_j\cdot|s|^j~\hbox{s.t.}~j\in\N_0\}$\\
\phx\qquad\qquad $\,\,=\{\alpha_j\cdot s^j~\hbox{s.t.}~j\in\N_0\}=\TS_s^\alpha$.\\
By a theorem from class, we know: $\forall s\ge0$,\\
\phx\qquad $(\,\,TS_s^a\hbox{ is bounded in }\R\,\,)\,\,\Leftrightarrow\,\,(\,\,|TS_s^a|\hbox{ is bounded in }\R\,\,)$.\\
Then $\{s\ge0~\hbox{s.t.}~\TS_s^a\hbox{ is bounded in }\R\}$\\
\phx\qquad\qquad $\,\,=\{s\ge0~\hbox{s.t.}~|\TS_s^a|\hbox{ is bounded in }\R\}$\\
\phx\qquad\qquad $\,\,=\{s\ge0~\hbox{s.t.}~\TS_s^\alpha\hbox{ is bounded in }\R\}$.\\
Then $\RC_a=\sup\,\{s\ge0~\hbox{s.t.}~\TS_s^a\hbox{ is bounded in }\R\}$\\
\phx\qquad\qquad $\,\,=\sup\,\{s\ge0~\hbox{s.t.}~\TS_s^\alpha\hbox{ is bounded in }\R\}=\RC_\alpha$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
4-5. Let $a\in\R^{\N_0}$. \qquad Show: $\RC_a=\RC_{a^*}$.
\vskip.1in\noindent
{\it Proof:}
Let $b:=a^*\,$. \quad Want: $\RC_a=\RC_b$.\\
Want: \quad (1) $\RC_a\ge\RC_b$ \quad and \quad (2) $\RC_b\ge\RC_a$.\\
\\
{\it Proof of (1):}
Assume: $\RC_a<\RC_b$. \qquad Want: Contradiction.\\
Since $\RC_a<\RC_b$, choose $s>\RC_a$ s.t.~$\TS_s^b$ is bounded in $\R$.\\
Since $s>\RC_a\ge0$, we get $s>0$. \qquad Then $|s|=s$.\\
Choose $C\in\scrb_\R(0)$ s.t.~$\TS_s^b\subseteq C$.\\
Choose $M>0$ s.t.~$C=B_\R(0,M)$.\\
We have $\TS_s^b=\{b_0,b_1s,b_2s^2,b_3s^3,\ldots\}=\{a_1,2a_2s,3a_3s^2,4a_4s^3,\ldots\}$.\\
Then $\{a_1,2a_2s,3a_3s^2,4a_4s^3,\ldots\}=TS_s^b\subseteq C=B_\R(0,M)$.\\
Then: \quad $\forall j\in\N$, \quad $|ja_js^{j-1}|<M$.\\
Then: \quad $\forall j\in\N$, \quad $|a_js^j|=|s|\cdot|ja_js^{j-1}|/|j|$\\
\phx\qquad\qquad\qquad\qquad\qquad $\,\,\,=s\cdot|ja_js^{j-1}|/j<s\cdot M/j\le s\cdot M$.\\
Then $\{a_1s,a_2s^2,a_3s^3,a_4s^4,\ldots\}\subseteq B_\R(0,sM)$.\\
Then $\{a_1s,a_2s^2,a_3s^3,a_4s^4,\ldots\}$ is bounded in $\R$.\\
So, since $\{a_0\}$ is bounded in $\R$,\\
\phx\qquad we get: $\{a_0\}\bigcup\{a_1s,a_2s^2,a_3s^3,a_4s^4,\ldots\}$ is bounded in $\R$.\\
Then $\{a_0,a_1s,a_2s^2,a_3s^3,a_4s^4,\ldots\}$ is bounded in $\R$.\\
We have $\TS_s^a=\{a_0,a_1s,a_2s^2,a_3s^3,\ldots\}$.\\
Then $\TS_s^a$ is bounded in $\R$. \qquad Recall: $s>0$.\\
Then $s\in\{r\ge0\,|\,\TS_r^a\hbox{ is bounded in }\R\}$.\\
Then $s\le\sup\{r\ge0\,|\,\TS_r^a\hbox{ is bounded in }\R\}$. \qquad
Recall that $s>\RC_a$.\\
Then $\RC_a<s\le\sup\{r\ge0\,|\,\TS_r^a\hbox{ is bounded in }\R\}=\RC_a$.\\
Then $\RC_a<\RC_a$. \qquad Contradiction.\\
{\it End of proof of (1).}\\
\\
{\it Proof of (2):}
Assume: $\RC_b<\RC_a$. \qquad Want: Contradiction.\\
Since $\RC_b<\RC_a$, choose $t>\RC_b$ s.t.~$\TS_t^a$ is bounded in $\R$.\\
Since $t>\RC_b\ge0$, we get $t>0$. \qquad Then $t\ge0$.\\
So, since $\TS_t^a$ is bounded in $\R$, we get $t\in\{r\ge0\,|\,TS_r^a\hbox{ is bounded in }\R\}$.\\
Then $t\le\sup\{r\ge0\,|\,TS_r^a\hbox{ is bounded in }\R\}=\RC_a$.\\
We have $\RC_b<t$. \quad Let $s:=(\RC_b+t)/2$. \quad Then $\RC_b<s<t$.\\
Then $0\le\RC_b<s<t\le\RC_a$.\\
Let $u:=s/t$. \quad Then $0<u<1$ and $s=tu$.\\
Since $TS_t^a$ is bounded in $\R$, choose $C\in\scrb_\R(0)$ s.t.~$\TS_t^a\subseteq C$.\\
Choose $M>0$ s.t.~$C=B_\R(0,M)$.\\
Since $s>\RC_b=\sup\{r\ge0\,|\,\TS_r^b\hbox{ is bounded in }\R\}$,\\
\phx\qquad we see that $s\notin\{r\ge0\,|\,\TS_r^b\hbox{ is bounded in }\R\}$,\\
\phx\qquad and so $\TS_s^b$ is not bounded in $\R$.\\
Define $\beta\in\R^{\N_0}$ by: \quad $\forall j\in\N_0$, $\,$ $\beta_j=b_js^j$.\\
Then $\TS_s^b=\{\beta_0,\beta_1,\beta_2,\ldots\}=\I_\beta$.\\
Then $\I_\beta$ is not bounded in $\R$,\\
\phx\qquad so $\beta$ is not convergent in $\R$,\\
\phx\qquad so $\neg(\beta\to0\hbox{ in }\R)$.\\
Choose $\vep>0$ s.t., $\forall\ell\in\N_0$, $\exists j\in[\ell..\infty)$ s.t. $d_\R(\beta_j,0)\ge\vep$.\\
Define $\gamma\in\R^{\N_0}$ by: \quad $\forall j\in\N_0$, $\,$ $\gamma_j=(j+1)\cdot u^j$.\\
By the Archimedean Principle, choose $k\in\N$ such that $k>2u/(1-u)$.\\
Recall: $u<1$. \qquad Let $c:=(u+1)/2$. \qquad Then $u<c<1$.\\
So, since $0\le u$, we get $0<c<1$, and so $c\in(0;1)$.\\
We have: $\forall j\in\N_0$, \qquad $\gamma_{j+1}/\gamma_j=[(j+1)/j]\cdot u=u+[u/j]$.\\
We have $u+[(1-u)/2]=(1-u+2u)/2=(u+1)/2=c$.\\
We have: $\forall j\in[k..\infty)$, \quad $\,\,\,j\ge k>2u/(1-u)$, \quad so $\,\,\,1-u>2u/j$,\\
\phx\qquad\qquad\qquad\qquad\quad so $\,\,\,(1-u)/2>u/j$, \quad so $\,\,\,u/j<(1-u)/2$,\\
\phx\qquad\qquad\qquad\qquad\quad so $\,\,\,u+[u/j]<u+[(1-u)/2]$, \quad so $\,\,\,\gamma_{j+1}/\gamma_j<c$.\\
Then, by Theorem 4.14.3, we get $\gamma\to0$ in $\R$.\\
Recall: \quad $t>0$ \quad and \quad $M>0$ \quad and \quad $\vep>0$. \quad Then $(t/M)\cdot\vep>0$.\\
Choose $\ell\in\N_0$ s.t., $\forall j\in[\ell..\infty)$, \qquad $d_\R(\gamma_j,0)<(t/M)\cdot\vep$.\\
By choice of $\vep$, choose $j\in[\ell..\infty)$ s.t. $d_\R(\beta_j,0)\ge\vep$.\\
Since $j\in[\ell..\infty)$, by choice of $\ell$, we have $d_\R(\gamma_j,0)<(t/M)\cdot\vep$.\\
Then $|\gamma_j|=|\gamma_j-0|=d_\R(\gamma_j,0)<(t/M)\cdot\vep$, \quad so $(M/t)\cdot|\gamma_j|<\vep$.\\
We have: $a_{j+1}t^{j+1}\in\TS_t^a\subseteq C\subseteq B_\R(0,M)$, \quad so $d_\R(a_{j+1}t^{j+1},0)<M$.\\
Then: \quad $|a_{j+1}|\cdot t^{j+1}=|a_{j+1}t^{j+1}|=|a_{j+1}t^{j+1}-0|=d_\R(a_{j+1}t^{j+1},0)<M$.\\
Then: \quad $|a_{j+1}|\cdot t^j=|a_{j+1}|\cdot t^{j+1}/t<M/t$.\\
Then: \quad $|a_{j+1}|\cdot t^j\cdot|\gamma_j|\le(M/t)\cdot|\gamma_j|$. \qquad Recall: $s=tu$.\\
We have: $\beta_j=b_js^j=b_j(tu)^j=a^*_jt^ju^j=(j+1)a_{j+1}t^ju^j=a_{j+1}t^j\cdot\gamma_j$,\\
\phx\quad so $|\beta_j|=|a_{j+1}|\cdot t^j\cdot|\gamma_j|\le(M/t)\cdot|\gamma_j|$.\\
Then $\vep\le d_\R(\beta_j,0)=|\beta_j-0|=|\beta_j|\le(M/t)\cdot|\gamma_j|<\vep$.\\
Then $\vep<\vep$. \qquad Contradiction.\\
{\it End of proof of (2).} \qquad
QED
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 3: Due on Tuesday 18 February}
\vskip.1in\hrule\vskip.1in\noindent
3-1. Let $X$ be a metric space, \quad $s\in X^\N$, \quad $q\in X$, \quad $m\in\N$.\\
\phx\quad $\,$ Define $t\in X^\N$ by: \quad $\forall j\in\N$, $\,$ $t_j=s_{j+m}$.\\
\phx\quad $\,$ Assume: $t\to q$ in $X$. \qquad Show: $s\to q$ in $X$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,j\ge K\,\,]\,\,\Rightarrow\,\,[\,\,d(s_j,q)<\vep\,\,]$.\\
Given $\vep>0$. \qquad Want: $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,j\ge K\,\,]\,\,\Rightarrow\,\,[\,\,d(s_j,q)<\vep\,\,]$.\\
Since $t\to q$, choose $L\in\N$ s.t., $\forall i\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,i\ge L\,\,]\,\,\Rightarrow\,\,[\,\,d(t_i,q)<\vep\,\,]$.\\
Since $L,m\in\N$, \quad we conclude: $L+m\in\N$.\\
Let $K:=L+m$. \qquad\qquad Then $K\in\N$.\\
\phx\quad Want: \qquad $\!\forall j\in\N$, \quad $[\,\,j\ge K\,\,]\,\,\Rightarrow\,\,[\,\,d(s_j,q)<\vep\,\,]$.\\
Given $j\in\N$. \qquad Want: $[\,\,j\ge K\,\,]\,\,\Rightarrow\,\,[\,\,d(s_j,q)<\vep\,\,]$.\\
\phx\qquad\qquad\quad $\,\,\,$ Assume: $j\ge K$. \quad Want: $d(s_j,q)<\vep$.\\
Let $i:=j-m$. \qquad Then $i+m=j$. \qquad Then $s_{i+m}=s_j$.\\
Since $i+m=j\ge K=L+m$, we get: $i\ge L$.\\
Since $j,m\in\N\subseteq\Z$, we get: $j-m\in\Z$.\\
Since $i=j-m\in\Z$ and $i\ge L\in\N>0$, we conclude: $i\in\N$.\\
Then, by definition of $t$, we have $t_i=s_{i+m}$.\\
Since $i\in\N$ and $i\ge L$, by choice of $L$, we get: $d(t_i,q)<\vep$.\\
So, since $t_i=s_{i+m}=s_j$, we get: $d(s_j,q)<\vep$, as desired. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
3-2. Let $X$ and $Y$ be metric spaces, \quad $s\in X^\N$, \quad $t\in Y^\N$.\\
\phx\quad $\,$ Show: \qquad\quad $\,[\,\,(s,t)\hbox{ is Cauchy in }X\times Y\,\,]$\\
\phx\qquad\qquad\qquad $\Leftrightarrow\,\,[\,\,(\,\,s\hbox{ is Cauchy in }X\,\,)\,\,\&\,\,(\,\,t\hbox{ is Cauchy in }Y\,\,)\,\,]$.
\vskip.1in\noindent
{\it Proof:}
{\it Proof of $\Rightarrow$:}\\
Assume: $(s,t)\hbox{ is Cauchy in }X\times Y$.\\
Want: \quad (1) $s\hbox{ is Cauchy in }X$ \quad and \quad (2) $t\hbox{ is Cauchy in }Y$.\\
\\
{\it Proof of (1):}\\
Want: $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\quad $\,\,\,\,[\,\,i,j\ge K\,\,]\,\,\Rightarrow\,\,[\,\,d(s_i,s_j)<\vep\,\,]$.\\
Given $\vep>0$. $\,$ Want: $\exists K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\quad $\,\,\,\,[\,\,i,j\ge K\,\,]\,\,\Rightarrow\,\,[\,\,d(s_i,s_j)<\vep\,\,]$.\\
Since $(s,t)\hbox{ is Cauchy in }X\times Y$, choose $K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\quad $\,\,\,\,[\,\,i,j\ge K\,\,]\,\,\Rightarrow\,\,[\,\,d(\,(s,t)_i\,,\,(s,t)_j\,)\,)\,<\,\vep\,\,]$.\\
Then $K\in\N$.\\
Want: $\forall i,j\in\N$, $\,\,\,\,\,[\,\,i,j\ge K\,\,]\,\,\Rightarrow\,\,[\,\,d(s_i,s_j)<\vep\,\,]$.\\
Given $i,j\in\N$.\\
\phx\qquad $\,\,\,\,\,\,\,$ Want: \quad $[\,\,i,j\ge K\,\,]\,\,\Rightarrow\,\,[\,\,d(s_i,s_j)<\vep\,\,]$.\\
\phx\qquad Assume: $\,i,j\ge K$. \quad $\,\,\,$ Want: $d(s_i,s_j)<\vep$.\\
Since $i,j\ge K$, by choice of $K$, we get: $d(\,(s,t)_i\,,\,(s,t)_j\,)\,)\,<\,\vep$.\\
Let $a:=d(s_i,s_j)$ and let $b:=d(t_i,t_j)$.\\
Then $d(\,(s_i,t_i)\,,\,(s_j,t_j)\,)\,=\,\sqrt{a^2+b^2}$.\\
Then $\sqrt{a^2+b^2}\,=\,d(\,(s_i,t_i)\,,\,(s_j,t_j)\,)\,=\,d(\,(s,t)_i\,,\,(s,t)_j\,)\,<\,\vep$.\\
Since $a=d(s_i,s_j)\ge0$, we get: $a=\sqrt{a^2}$.\\
Since $0\le a^2\le a^2+b^2$, we get $\sqrt{a^2}\le\sqrt{a^2+b^2}$.\\
Then $d(s_i,s_j)=a=\sqrt{a^2}\le\sqrt{a^2+b^2}<\vep$, as desired.\\
{\it End of proof of (1).}\\
\\
{\it Proof of (2):}\\
Want: $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\quad $\,\,\,\,[\,\,i,j\ge K\,\,]\,\,\Rightarrow\,\,[\,\,d(t_i,t_j)<\vep\,\,]$.\\
Given $\vep>0$. \qquad Want: $\exists K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\quad $\,\,\,\,[\,\,i,j\ge K\,\,]\,\,\Rightarrow\,\,[\,\,d(t_i,t_j)<\vep\,\,]$.\\
Since $(s,t)\hbox{ is Cauchy in }X\times Y$, choose $K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\quad $\,\,\,\,[\,\,i,j\ge K\,\,]\,\,\Rightarrow\,\,[\,\,d(\,(s,t)_i\,,\,(s,t)_j\,)\,)\,<\,\vep\,\,]$.\\
Then $K\in\N$.\\
Want: $\forall i,j\in\N$, $\,\,\,\,\,[\,\,i,j\ge K\,\,]\,\,\Rightarrow\,\,[\,\,d(t_i,t_j)<\vep\,\,]$.\\
Given $i,j\in\N$.\\
\phx\qquad $\,\,\,\,\,\,\,$ Want: \quad $[\,\,i,j\ge K\,\,]\,\,\Rightarrow\,\,[\,\,d(t_i,t_j)<\vep\,\,]$.\\
\phx\qquad Assume: $\,i,j\ge K$. \quad $\,\,\,$ Want: $d(t_i,t_j)<\vep$.\\
Since $i,j\ge K$, by choice of $K$, we get: $d(\,(s,t)_i\,,\,(s,t)_j\,)\,)\,<\,\vep$.\\
Let $a:=d(s_i,s_j)$ and let $b:=d(t_i,t_j)$.\\
Then $d(\,(s_i,t_i)\,,\,(s_j,t_j)\,)\,=\,\sqrt{a^2+b^2}$.\\
Then $\sqrt{a^2+b^2}\,=\,d(\,(s_i,t_i)\,,\,(s_j,t_j)\,)\,=\,d(\,(s,t)_i\,,\,(s,t)_j\,)\,<\,\vep$.\\
Since $b=d(t_i,t_j)\ge0$, we get: $b=\sqrt{b^2}$.\\
Since $0\le b^2\le a^2+b^2$, we get $\sqrt{b^2}\le\sqrt{a^2+b^2}$.\\
Then $d(t_i,t_j)=b=\sqrt{b^2}\le\sqrt{a^2+b^2}<\vep$, as desired.\\
{\it End of proof of (2).} \qquad {\it End of proof of $\Rightarrow$.}\\
\\
{\it Proof of $\Leftarrow$:}\\
Assume: $(\,\,s\hbox{ is Cauchy in }X\,\,)\,\,\&\,\,(\,\,t\hbox{ is Cauchy in }Y\,\,)$.\\
Want: $(s,t)\hbox{ is Cauchy in }X\times Y$.\\
Want: $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,i,j\ge K\,\,]\,\,\Rightarrow\,\,[\,\,d(\,(s,t)_i\,,\,(s,t)_j\,)\,<\,\vep\,\,]$.\\
Given $\vep>0$. \qquad Want: $\exists K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,i,j\ge K\,\,]\,\,\Rightarrow\,\,[\,\,d(\,(s,t)_i\,,\,(s,t)_j\,)\,<\,\vep\,\,]$.\\
Let $\sigma:=\vep/\sqrt{2}$. \qquad Then $\sigma>0$.\\
Since $s$ is Cauchy in $X$, choose $L\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,i,j\ge L\,\,]\,\,\Rightarrow\,\,[\,\,d(s_i,s_j)<\sigma\,\,]$.\\
Since $t$ is Cauchy in $X$, choose $M\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,i,j\ge M\,\,]\,\,\Rightarrow\,\,[\,\,d(t_i,t_j)<\sigma\,\,]$.\\
Let $K:=\max\{L,M\}$. \qquad Then $K\in\N$.\\
\phx\qquad $\,$ Want: \quad $\forall i,j\in\N$, \quad $[\,\,i,j\ge K\,\,]\,\,\Rightarrow\,\,[\,\,d(\,(s,t)_i\,,\,(s,t)_j\,)\,<\,\vep\,\,]$.\\
Given $i,j\in\N$. \qquad $\,$ Want: $\,[\,\,i,j\ge K\,\,]\,\,\Rightarrow\,\,[\,\,d(\,(s,t)_i\,,\,(s,t)_j\,)\,<\,\vep\,\,]$.\\
\phx\qquad\qquad Assume: $i,j\ge K$. \qquad\quad $\,\,\,\,\,\,$ Want: $d(\,(s,t)_i\,,\,(s,t)_j\,)\,<\,\vep$.\\
Since $i,j\ge K\ge L$, $\,$ by choice of $L$, $\,\,$ we get: $d(s_i,s_j)<\sigma$.\\
Since $i,j\ge K\ge M$, by choice of $M$, $\,$ we get: $d(t_i,t_j)<\sigma$.\\
Let $a:=d(s_i,s_j)$ and let $b:=d(t_i,t_j)$. \quad Then $a<\sigma$ and $b<\sigma$.\\
Also, \qquad $d(\,(s_i,t_i)\,,\,(s_j,t_j)\,)\,=\,\sqrt{a^2+b^2}$.\\
We have \quad $a=d(s_i,s_j)\ge0$ \quad and \quad $b=d(t_i,t_j)\ge0$.\\
Since $0\le a<\sigma$, we get $0\le a^2<\sigma^2$.\\
Since $0\le b<\sigma$, we get $0\le b^2<\sigma^2$.\\
Then $0\le a^2+b^2<2\cdot\sigma^2$. \qquad Then $\sqrt{a^2+b^2}<\sqrt{2\cdot\sigma^2}$.\\
Since $\sigma>0$, we get $\sqrt{\sigma^2}=\sigma$. \quad
Since $\sigma=\vep/\sqrt{2}$, we get $\sqrt{2}\cdot\sigma=\vep$.\\
Then $d(\,(s,t)_i\,,\,(s,t)_j\,)\,=\,d(\,(s_i,t_i)\,,\,(s_j,t_j)\,)\,=\,\sqrt{a^2+b^2}\,<\,\sqrt{2\cdot\sigma^2}$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $=\,\sqrt{2}\cdot\sqrt{\sigma^2}\,=\,\sqrt{2}\cdot\sigma\,=\,\vep$, \quad as desired.\\
{\it End of proof of $\Leftarrow$.} \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
3-3. Let $X$ and $Y$ be complete metric spaces.\\
\phx\quad $\,$ Show $X\times Y$ is complete.
\vskip.1in\noindent
{\it Proof:}
Let $Z:X\times Y$. \qquad Want $Z$ is complete.\\
Want: $\forall u\in Z^\N$, $\,$ $[\,\,u\hbox{ is Cauchy in }Z\,\,]\,\,\Rightarrow\,\,[\,\,u\hbox{ is convergent in }Z\,\,]$.\\
Given $u\in Z^\N$. \quad Want: $[\,\,u\hbox{ is Cauchy in }Z\,\,]\,\,\Rightarrow\,\,[\,\,u\hbox{ is convergent in }Z\,\,]$.\\
Assume: $u\hbox{ is Cauchy in }Z$. \qquad Want: $u\hbox{ is convergent in }Z$.\\
Define $p:Z\to X$ by: $\forall z\in Z$, $p_z=z_1$.\\
Define $q:Z\to Y$ by: $\forall z\in Z$, $q_z=z_2$.\\
We have: $\forall z\in Z$, $z=(z_1,z_2)=(p_z,q_z)$.\\
Let $s:=p\circ u$ and let $t:=q\circ u$.\\
Then we have: \qquad $\forall j\in\N$,\\
\phx\qquad $u_j\,\,=\,\,(\,p_{u_j}\,,\,q_{u_j}\,)\,\,=\,\,(\,(p\circ u)_j\,,\,(q\circ u)_j\,)\,\,=\,\,(\,s_j\,,\,t_j\,)\,\,=\,\,(s,t)_j$.\\
Then $u=(s,t)$. \quad
So, as $\,$ $u$ is Cauchy in $Z$ $\,$ and $\,$ $Z=X\times Y$, we get:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(s,t)$ is Cauchy in $X\times Y$.\\
Then, by HW\#3-2, \quad $s$ is Cauchy in $X$ \quad and \quad $t$ is Cauchy in $Y$.\\
So, since $X$ and $Y$ are both complete,\\
\phx\qquad $s$ is convergent in $X$ \quad and \quad $t$ is convergent in $Y$.\\
Choose $a\in X$ and $b\in Y$ s.t. \quad $s\to a$ in $X$ \quad and \quad $t\to b$ in $Y$.\\
Then, by a theorem from class, we have: \quad $(s,t)\to(a,b)$ in $X\times Y$.\\
So, since $u=(s,t)$ and $Z=X\times Y$, we get: \quad $\,u\to(a,b)$ in $Z$.\\
Then $u$ is convergent in $Z$, as desired. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
3-4. Let $p,q,r\in\N$, $A\in\R^{p\times q}$, $B\in\R^{q\times r}$. \qquad
Show: $|AB|\,\,\le\,\,|A|\cdot|B|$.
\vskip.1in\noindent
{\it Proof:} Since $|AB|\ge0$, we get $|AB|=\sqrt{|AB|^2}$.\\
Since $|A|\cdot|B|>0$, we get $|A|\cdot|B|=\sqrt{(|A|\cdot|B|)^2}$.\\
Want: $\sqrt{|AB|^2}\,\,\le\,\,\sqrt{(|A|\cdot|B|)^2}$.\\
Want: $0\,\,\le\,\,|AB|^2\,\,\le\,\,(|A|\cdot|B|)^2$.\\
We have $0\,\,\le\,\,|AB|^2$. \qquad Want: $|AB|^2\,\,\le\,\,(|A|\cdot|B|)^2$.\\
We have $\ds{|AB|^2=\sum_{i=1}^p\,\sum_{k=1}^r\,((AB)_{ik})^2=\sum_{i=1}^p\,\sum_{k=1}^r\,\left(\sum_{j=1}^q\,A_{ij}\cdot B_{jk}\right)^2}$.\\
Also, $\ds{(|A|\cdot|B|)^2=|A|^2\cdot|B|^2=\left(\sum_{i=1}^p\,\sum_{\ell=1}^q\,A_{i\ell}^2\right)\cdot\left(\sum_{m=1}^q\,\sum_{k=1}^r\,B_{mk}\right)}$\\
\phx\qquad\quad $\ds{=\sum_{i=1}^p\,\sum_{\ell=1}^q\,\sum_{m=1}^q\,\sum_{k=1}^r\,A_{i\ell}^2\cdot B_{mk}^2=\sum_{i=1}^p\,\sum_{k=1}^r\,\left(\sum_{\ell=1}^q\,\sum_{m=1}^q\,A_{i\ell}^2\cdot B_{mk}^2\right)}$.\\
Want: \quad $\ds{\sum_{i=1}^p\,\sum_{k=1}^r\,\left(\sum_{j=1}^q\,A_{ij}\cdot B_{jk}\right)^2\,\,\le\,\,\sum_{i=1}^p\,\sum_{k=1}^r\,\left(\sum_{\ell=1}^q\,\sum_{m=1}^q\,A_{i\ell}^2\cdot B_{mk}^2\right)}$.\\
Want: $\forall i\in[1..p]$, $\forall k\in[1..r]$, \quad
$\ds{\left(\sum_{j=1}^q\,A_{ij}\cdot B_{jk}\right)^2\,\,\le\,\,\sum_{\ell=1}^q\,\sum_{m=1}^q\,A_{i\ell}^2\cdot B_{mk}^2}$.\\
Given $i\in[1..p]$ and $k\in[1..r]$.\\
Want: $\ds{\left(\sum_{j=1}^q\,A_{ij}\cdot B_{jk}\right)^2\,\,\le\,\,\sum_{\ell=1}^q\,\sum_{m=1}^q\,A_{i\ell}^2\cdot B_{mk}^2}$.\\
Let $v:=(A_{i1},\ldots,A_{iq})$ and let $w:=(B_{1k},\ldots,B_{qk})$.\\
Then $\ds{v\sbullet w=\sum_{j=1}^q\,A_{ij}\cdot B_{jk}}$. \qquad
Want: $\ds{(v\sbullet w)^2\,\,\le\,\,\sum_{\ell=1}^q\,\sum_{m=1}^q\,A_{i\ell}^2\cdot B_{mk}^2}$.\\
We have: \quad $\ds{|v|^2=\sum_{\ell=1}^q\,A_{i\ell}^2}$ \quad and \quad $\ds{|w|^2=\sum_{m=1}^q\,B_{mk}^2}$.\\
Then: \quad $\ds{|v|^2\cdot|w|^2=\left(\sum_{\ell=1}^q\,A_{i\ell}^2\right)\cdot\left(\sum_{m=1}^q\,B_{mk}^2\right)=\sum_{\ell=1}^q\,\sum_{m=1}^q\,(A_{i\ell}^2\cdot B_{mk}^2)}$.\\
Want: $(v\sbullet w)^2\,\,\le\,\,|v|^2\cdot|w|^2$.\\
By Cauchy-Schwarz, we have: $|v\sbullet w|\,\,\le\,\,|v|\cdot|w|$.\\
Since $0\,\,\le\,\,|v\sbullet w|\,\,\le\,\,|v|\cdot|w|$, \qquad we get: $\,$ $|v\sbullet w|^2\,\,\le\,\,(|v|\cdot|w|)^2$.\\
Then $(v\sbullet w)^2\,\,=\,\,|v\sbullet w|^2\,\,\le\,\,(|v|\cdot|w|)^2\,\,=\,\,|v|^2\cdot|w|^2$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
3-5. Let $V\in\ES$. \qquad Show:\\
\phx\qquad\qquad\qquad (a) $|\bullet|_V$ is Lipschitz-1 from $V$ to $\R$\\
\phx\qquad and \qquad (b) $\|\bullet\|_V$ is Lipschitz-$\sqrt{\#\scri_V}$ from $V$ to $\R$.
\vskip.1in\noindent
{\it Proof:}\\
\\
{\it Proof of (a):}\\
Let $f:=|\bullet|_V$. \quad Want: $f$ is Lipschitz-1 from $V$ to $\R$.\\
Want: \quad $\forall x,y\in V$, $\,$ $d_\R(f_x,f_y)\le1\cdot(d_V(x,y))$.\\
Given $x,y\in V$. \qquad Want: $d_\R(f_x,f_y)\le1\cdot(d_V(x,y))$.\\
Want: $|f_x\,-\,f_y|\,\,\le\,\,|x\,-\,y|$.\\
Want: $(\,\,f_x\,-\,f_y\,\,\le\,\,|\,x\,-\,y\,|\,\,)\,\,\&\,\,(\,\,f_y\,-\,f_x\,\,\le\,\,|\,x\,-\,y\,|\,\,)$.\\
We have: \quad $\forall z\in V$, $\,$ $f_z=(|\bullet|_V)_z=|z|$.\\
Want: $(\,\,|x|\,-\,|y|\,\,\le\,\,|\,x\,-\,y\,|\,\,)\,\,\&\,\,(\,\,|y|\,-\,|x|\,\,\le\,\,|\,x\,-\,y\,|\,\,)$.\\
We have: \quad $|x|\,\,=\,\,|\,x\,-\,y\,+\,y\,|\,\,\le\,\,|x\,-\,y|\,+\,|y|$.\\
Then: $|x|\,-\,|y|\,\,\le\,\,|\,x\,-\,y\,|$. \quad
Want: $|y|\,-\,|x|\,\,\le\,\,|\,x\,-\,y\,|$.\\
We have: \quad $|y|\,\,=\,\,|\,y\,-\,x\,+\,x\,|\,\,\le\,\,|\,y\,-\,x\,|\,+\,|x|$.\\
Then: \quad $|y|\,-\,|x|\,\,\le\,\,|\,y\,-\,x\,|\,\,=\,\,|\,x\,-\,y\,|$, \qquad as desired.\\
{\it End of proof of (a).}\\
\\
{\it Proof of (b):}\\
Let $L:=\sqrt{\#\scri_V}$ and let $f:=\|\bullet\|_V$.\\
Want: $f$ is Lipschitz-$L$ from $V$ to $\R$.\\
Want: \quad $\forall x,y\in V$, $\,$ $d_\R(f_x,f_y)\le L\cdot(d_V(x,y))$.\\
Given $x,y\in V$. \qquad Want: $d_\R(f_x,f_y)\le L\cdot(d_V(x,y))$.\\
Want: $|f_x\,-\,f_y|\,\,\le\,\,L\,\cdot\,|x\,-\,y|$.\\
By a theorem from class, we have:\\
\phx\qquad$\forall z\in V$, \qquad\qquad\quad $\,\,\,\|z\|\,\,\le\,\,L\,\cdot\,|z|$.\\
\phx\qquad\qquad\quad $\,\,$ Then: $\|\,x\,-\,y\,\|\,\,\le\,\,L\,\cdot\,|x\,-\,y|$.\\
Want: $|f_x\,-\,f_y|\,\,\le\,\,\|\,x\,-\,y\,\|$.\\
We have: \quad $\forall z\in V$, $\,$ $f_z=(\|\bullet\|_V)_z=\|z\|$.\\
Want: $(\,\,f_x\,-\,f_y\,\,\le\,\,\|\,x\,-\,y\,\|\,\,)\,\,\&\,\,(\,\,f_y\,-\,f_x\,\,\le\,\,\|\,x\,-\,y\,\|\,\,)$.\\
Want: $(\,\,\|x\|\,-\,\|y\|\,\,\le\,\,\|\,x\,-\,y\,\|\,\,)\,\,\&\,\,(\,\,\|y\|\,-\,\|x\|\,\,\le\,\,\|\,x\,-\,y\,\|\,\,)$.\\
We have: \quad $\|x\|\,\,=\,\,\|\,x\,-\,y\,+\,y\,\|\,\,\le\,\,\|\,x\,-\,y\,\|\,+\,\|y\|$.\\
Then: $\|x\|\,-\,\|y\|\,\,\le\,\,\|\,x\,-\,y\,\|$. \quad
Want: $\|y\|\,-\,\|x\|\,\,\le\,\,\|\,x\,-\,y\,\|$.\\
We have: \quad $\|y\|\,\,=\,\,\|\,y\,-\,x\,+\,x\,\|\,\,\le\,\,\|\,y\,-\,x\,\|\,+\,\|x\|$.\\
Then: \quad $\|y\|\,-\,\|x\|\,\,\le\,\,\|\,y\,-\,x\,\|\,\,=\,\,\|\,x\,-\,y\,\|$, \qquad as desired.\\
{\it End of proof of (b).} \qquad
QED
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 2: Due on Tuesday 11 February}
\vskip.1in\hrule\vskip.1in\noindent
2-1. Show: \quad $\forall\ell\in\N$, $\,$ $((\bullet)^\ell)'=\ell\cdot(\bullet)^{\ell-1}$.
\vskip.1in\noindent
{\it Proof:}
Let $S:=\{\ell\in\N\,|\,((\bullet)^\ell)'=\ell\cdot(\bullet)^{\ell-1}\}$. \quad
Want: $S=\N$.\\
By a result from class, $\forall m\in\R$, $(m\cdot(\bullet))'=C_\R^m$. \quad Then $(1\cdot(\bullet))'=C_\R^1$.\\
Then $((\bullet)^1)'=(1\cdot(\bullet))'=C_\R^1=(\bullet)^0=1\cdot(\bullet)^0=1\cdot(\bullet)^{1-1}$, so $1\in S$.\\
Want: $\forall\ell\in S$, $\ell+1\in S$. \qquad Given $\ell\in S$. \qquad Want: $\ell+1\in S$.\\
Since $\ell\in S$, we have $((\bullet)^\ell)'=\ell\cdot(\bullet)^{\ell-1}$. \quad Want: $((\bullet)^{\ell+1})'=(\ell+1)\cdot(\bullet)^\ell$.\\
Want: $\forall x\in\R$, $((\bullet)^{\ell+1})'_x=((\ell+1)\cdot(\bullet)^\ell)_x$.\\
Given $x\in\R$. \qquad Want: $((\bullet)^{\ell+1})'_x=((\ell+1)\cdot(\bullet)^\ell)_x$.\\
By the Product Rule, $((\bullet)\cdot(\bullet)^\ell)'_x\,\,=^*\,\,(\bullet)'_x\cdot((\bullet)^\ell)_x+(\bullet)_x\cdot((\bullet)^\ell)'_x$.\\
Recall: $(1\cdot(\bullet))'=C_\R^1$. \quad Then $(\bullet)'=C_\R^1$. \quad Then $(\bullet)'_x=(C_\R^1)_x=1$.\\
Recall: $((\bullet)^\ell)'=\ell\cdot(\bullet)^{\ell-1}$. \quad Then $((\bullet)^\ell)'_x=(\ell\cdot(\bullet)^{\ell-1})_x=\ell\cdot x^{\ell-1}$.\\
Then $\,\,((\bullet)\cdot(\bullet)^\ell)'_x\,\,=^*\,\,(\bullet)'_x\cdot((\bullet)^\ell)_x+(\bullet)_x\cdot((\bullet)^\ell)'_x$\\
\phx\qquad\qquad\qquad\qquad $=\,\,\,\,\,\,\,1\,\,\,\,\,\cdot\,\,\,x^\ell\,\,\,\,\,\,\,\,+\,\,x\,\,\,\cdot\,\,\ell\cdot x^{\ell-1}$\\
\phx\qquad\qquad\qquad\qquad $=\qquad\quad\,\,\,\,\,x^\ell\quad\,\,+\,\,\ell\cdot x\cdot x^{\ell-1}$\\
\phx\qquad\qquad\qquad\qquad $=\qquad\quad\,\,\,\,\,x^\ell\quad\,\,+\,\,\ell\cdot x^\ell\,\,\,\,=\,\,\,\,(1+\ell)\cdot x^\ell\,\,\,\,\ne\,\,\,\,\frownie$.\\
Then $\,\,((\bullet)\cdot(\bullet)^\ell)'_x\,\,\,=\,\,\,(1+\ell)\cdot x^\ell\,\,=\,\,(\ell+1)\cdot x^\ell\,\,=\,\,((\ell+1)\cdot(\bullet)^\ell)_x$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad as desired. \qquad\qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
2-2. Let $m\in\R$, $L:\R\to\R$. \qquad Assume: $\forall x\in\R,\,\,L_x=m\cdot x$.\\
Show: $(\,\,\forall x\in\R, \,\, L'_x=m\,\,)\,\,\&\,\,(\,\,\forall x\in\R, \,\, L''_x=0\,\,)$.
\vskip.1in\noindent
{\it Proof:}
By a result from class, $(m\cdot(\bullet))'=C_\R^m$.\\
We have: $\forall x\in\R$, $L_x=m\cdot x=(m\cdot(\bullet))_x$. \quad Then $L=m\cdot(\bullet)$.\\
Then $L'=(m\cdot(\bullet))'=C_\R^m$. \quad Then: $\forall x\in\R$, $L'_x=(C_\R^m)_x=m$.\\
Then: $\forall x\in\R$, $L'_x=m$. \quad Want: $\forall x\in\R, \,\, L''_x=0$.\\
As $L'=C_\R^m$, we get: $L''=(C_\R^m)'$. \quad By a result from class, $(C_\R^m)'=\bfzero$.\\
Then $L''=(C_\R^m)'=\bfzero$. \quad Then: $\forall x\in\R$, $L''_x=\bfzero_x=0$.\\
Then: $\forall x\in\R$, $L''_x=0$, as desired. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
2-3. Let $c\in\R$, $Q:\R\to\R$. \qquad Assume: $\forall x\in\R,\,\,Q_x=c\cdot x^2$.\\
Show: $(\,\,\forall x\in\R, \,\, Q'_x=2\cdot c\cdot x\,\,)\,\,\&\,\,(\,\,\forall x\in\R, \,\, Q''_x=2\cdot c\,\,)$.
\vskip.1in\noindent
{\it Proof:}
We have: $\forall x\in\R,\,\,Q_x=c\cdot x^2=(c\cdot(\bullet)^2)_x$. \quad Then $Q=c\cdot(\bullet)^2$.\\
Want: \quad (1) $\forall x\in\R, \,\, Q'_x=2\cdot c\cdot x$ \quad and \quad (2) $\forall x\in\R, \,\, Q''_x=2\cdot c$.\\
\\
{\it Proof of (1):}\\
Given $x\in\R$. \qquad Want: $Q'_x=2\cdot c\cdot x$.\\
By HW\#2-1, we have: \quad $((\bullet)^2)'=2\cdot(\bullet)^{2-1}$.\\
Then: \quad $((\bullet)^2)'_x=(2\cdot(\bullet)^{2-1})_x=2\cdot x^{2-1}=2\cdot x$.\\
By linearity of derivative, $(c\cdot(\bullet)^2)'_x\,\,=^*\,\,c\cdot((\bullet)^2)'_x$.\\
Then $Q'_x\,\,=\,\,(c\cdot(\bullet)^2)'_x\,\,=^*\,\,c\cdot((\bullet)^2)'_x\,\,=\,\,c\cdot2\cdot x\,\,\ne\,\,\frownie$.\\
Then $Q'_x=c\cdot 2\cdot x$. \qquad $Q'_x=c\cdot 2\cdot x=2\cdot c\cdot x$, as desired.\\
{\it End of proof of (1).}\\
\\
{\it Proof of (2):}\\
By (1), we have: $\forall x\in\R$, $Q'_x=2\cdot c\cdot x$.\\
Let $m:=2\cdot c$ \qquad We have: $\forall x\in\R$, $Q'_x=m\cdot x$.\\
Then: $\,\forall x\in\R$, $Q'_x=m\cdot x=(m\cdot(\bullet))_x$. \qquad Then $Q'=m\cdot(\bullet)$.\\
\phx\qquad\qquad\quad $\,$ Let $L:=m\cdot(\bullet)$. \qquad\qquad\qquad Then $Q'=L$.\\
We have $\forall x\in\R$, $L_x=(m\cdot(\bullet))_x=m\cdot x$.\\
Then, by HW\#2-2, we get: $\forall x\in\R$, $L'_x=m$.\\
Then: $\forall x\in\R$, $Q''_x=L'_x=m=2\cdot c$. \quad
Then: $\forall x\in\R$, $Q''_x=2\cdot c$.\\
{\it End of proof of (2).}
QED
\vskip.1in\hrule\vskip.1in\noindent
2-4. Let $\alpha,\beta:\R\drta\R$. \qquad Assume: $\forall x\in\R,\,\,\,\,\beta_x\,\,=^*\,\,\alpha_x$.\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\,\,\,\,\,\,$ Show: $\forall x\in\R,\,\,\,\,\beta'_x\,\,=^*\,\,\alpha'_x$.
\vskip.1in\noindent
{\it Proof:}
Given $x\in\R$. \qquad Want: $\beta'_x\,\,=^*\,\,\alpha'_x$.\\
Want: $(\,\,\alpha'_x\ne\frownie\,\,)\,\,\Rightarrow\,\,(\,\,\beta'_x=\alpha'_x\,\,)$.\\
Assume $\alpha'_x\ne\frownie$. \qquad Want: $\beta'_x=\alpha'_x$.\\
By a theorem from class, it suffices to show: \quad $\beta=\alpha$ near $x$ in $\R$.\\
Want: $\exists B\in\scrb_\R(x)$ s.t.~$\beta=\alpha$ on $B$.\\
Since $\alpha'_x\ne\frownie$, \quad we get $x\in\D_{\alpha'}$, \quad and so, by a theorem from class,\\
\phx\qquad we conclude: \qquad $\alpha$ is defined near $x$ in $\R$.\\
Choose $B\in\scrb_\R(x)$ s.t.~$B\subseteq\D_\alpha$.\\
Then $B\in\scrb_\R(x)$. \qquad Want: $\beta=\alpha$ on $B$.\\
Want: $\forall t\in B$, $\beta_t=\alpha_t$. \quad
Given $t\in B$. \quad Want: $\beta_t=\alpha_t$.\\
Since $t\in B\subseteq\D_{\alpha}$, we get $\alpha_t\ne\frownie$. \quad
Since $B\in\scrb_\R(x)$, we get $B\subseteq\R$.\\
Since $t\in B\subseteq\R$, by hypothesis, we have: \qquad $\beta_t\,\,=^*\,\,\alpha_t$.\\
Since $\beta_t\,\,=^*\,\,\alpha_t\,\,\ne\frownie$, \quad we conclude $\beta_t=\alpha_t$, \quad as desired. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
2-5. Let $a\in\R$. \qquad Define $T:\R\to\R$ by: $\,\,\,\forall h\in\R$, $\,\,T_h=h+a$.\\
\phx\qquad\qquad\qquad\qquad\qquad\quad Show: $\,\,\,T'\,=\,C_\R^1$.
\vskip.1in\noindent
{\it Proof:}
As $\forall h\in\R$, $\,$ $T_h=h+a=((\bullet)+C_\R^a)_h$, we get: $T=(\bullet)+C_\R^a$.\\
By a theorem from class, we have $\forall m\in\R$, $(m\cdot(\bullet))'=C_\R^m$.\\
Then $(1\cdot(\bullet))'=C_\R^1$. \qquad Then $(\bullet)'=(1\cdot(\bullet))'=C_\R^1$.\\
By a theorem from class, $(C_\R^a)'=\bfzero$.\\
Want: $\forall h\in\R$, $T'_h=(C_\R^1)_h$. \quad Given $h\in\R$. \quad Want: $T'_h=(C_\R^1)_h$.\\
We have $(\bullet)'_h=(C_\R^1)_h=1$ and $(C_\R^a)'_h=\bfzero_h=0$.\\
By linearity of the derivative, $(\,(\bullet)\,+\,C_\R^a\,)'_h\,\,=^*\,\,(\bullet)'_h\,+\,(C_\R^a)'_h$.\\
Then $T'_h\,\,=\,\,(\,(\bullet)\,+\,C_\R^a\,)'_h\,\,=^*\,\,(\bullet)'_h\,+\,(C_\R^a)'_h\,\,=\,\,1+0\,\,=\,\,1\,\,\ne\,\,\frownie$.\\
Then $T'_h\,\,=\,\,1\,\,=\,\,(C_\R^1)_h$, \quad as desired. \qquad
QED
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 1: Due on Tuesday 4 February}
\vskip.1in\hrule\vskip.1in\noindent
1-1. Let $f:\R\drta\R$ and let $p\in\D_f$.\\
\phx\quad $\,$ Show: \qquad\qquad $(\,\,f\hbox{ has a global strict-maximum at }p\,\,)$\\
\phx\quad\qquad\qquad\quad $\,$ $\Leftrightarrow\,\,\,\,(\,\,f_p^\T\hbox{ has a global strict-maximum at }0\,\,)$.
\vskip.1in\noindent
{\it Proof:}
Since $p\in\D_f$, we get $(f_p^\T)_0=0$. \quad
Let $g:=f_p^\T$. \quad Then $g_0=0$.\\
Since $p\in\D_f$, we get: $f_p\in\I_f$. \quad
Want: \qquad\qquad\qquad $(\,\,f\hbox{ has a global strict-maximum at }p\,\,)$\\
\phx\quad\qquad\qquad\quad $\,$ $\Leftrightarrow\,\,\,\,(\,\,g\hbox{ has a global strict-maximum at }0\,\,)$.\\
\\
{\it Proof of $\Rightarrow$:}\\
Assume: $f\hbox{ has a global strict-maximum at }p$\\
Want: $g\hbox{ has a global strict-maximum at }0$.\\
Want: $\forall h\in(\D_g)^\times_0$, $g_h<g_0$. \qquad
Given $h\in(\D_g)^\times_0$. \quad Want: $g_h<g_0$.\\
Since $h\in(\D_g)^\times_0\subseteq\D_g$, we get: $g_h\ne\frownie$.\\
We have: \quad $f_{p+h}-f_p=(f_p^\T)_h=g_h$.\\
Then $f_{p+h}-f_p=g_h\ne\frownie$, so $f_{p+h}\ne\frownie$, so $p+h\in\D_f$.\\
Since $h\in(\D_g)^\times_0\subseteq\R^\times_0$, we get: $h+p\in\R^\times_p$, so $h+p\ne p$.\\
Since $h+p\in\D_f$ and $h+p\ne p$, we get: $h+p\in(\D_f)^\times_p$.\\
As $f$ has a global strict-maximum at $p$, we get: $\forall x\in(\D_f)^\times_p$, $f_x<f_p$.\\
So, since $h+p\in(\D_f)^\times_p$, we see that: $f_{h+p}<f_p$. \quad
Then $f_{h+p}-f_p<0$.\\
Then $g_h=f_{h+p}-f_p<0=g_0$, as desired.\\
{\it End of proof of $\Rightarrow$.}\\
\\
{\it Proof of $\Leftarrow$:}\\
Assume: $g\hbox{ has a global strict-maximum at }0$.\\
Want: $f\hbox{ has a global strict-maximum at }p$\\
Want: \qquad $\forall x\in(\D_f)^\times_p$, \quad $f_x<f_p$.\\
Given $x\in(\D_f)^\times_p$. \quad Want: $f_x<f_p$.\\
Since $x\in(\D_f)^\times_p\subseteq\D_f$, we get: $f_x\in\I_f$. \quad
Since $p\in\D_f$, we get: $f_p\in\I_f$.\\
Since $f_x,f_p\in\I_f\subseteq\R$, we get $f_x-f_p\in\R$.  \quad Want: $f_x-f_p<0$.\\
We have: \quad $g_{x-p}=(f_p^\T)_{x-p}=f_{p+x-p}-f_p=f_x-f_p$.\\
Then $g_{x-p}=f_x-f_p\in\R$, so $g_{x-p}\ne\frownie$, so $x-p\in\D_g$.\\
Since $x\in(\D_f)^\times_p\subseteq\R^\times_p$, we get $x-p\in\R^\times_0$, so $x-p\ne0$.\\
Since $x-p\in\D_g$ and $x-p\ne0$, we get: $x-p\in(\D_g)^\times_0$.\\
As $g$ has a global strict-maximum at $0$, we get: $\forall h\in(\D_g)^\times_0$, $g_h<g_0$.\\
So, since $x-p\in(\D_g)^\times_0$, it follows that: $g_{x-p}<g_0$.\\
Then $f_x-f_p=g_{x-p}<g_0=0$, as desired.\\
{\it End of proof of $\Leftarrow$.}
QED
\vskip.1in\hrule\vskip.1in\noindent
1-2. Let $f:\R\drta\R$, let $c\in\R$, let $L\in\LINS_cf$ and let $\vep>0$.\\
\phx\quad $\,$ Show: \quad $L-\vep\cdot|\bullet|\,\,\le\,\,f_c^\T\,\,\le\,\,L+\vep\cdot|\bullet|$\quad near $0$ in $\R$.
\vskip.1in\noindent
{\it Proof:}
Want: $\exists B\in\scrb_\R(0)$ s.t.~$L-\vep\cdot|\bullet|\,\,\le\,\,f_c^\T\,\,\le\,\,L+\vep\cdot|\bullet|$ on $B$.\\
Since $L\in\LINS_cf$, we get: $f_c^\T-L\in\scrlito_1$.\\
Let $\rho:=f_c^\T-L$. \qquad Then $\rho\in\scrlito_1$.\\
Choose $\delta>0$ s.t., $\forall h\in\R$, \quad $(\,\,|h|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|\rho_h|<\vep\cdot|h|^1\,\,)$.\\
Let $B:=B_\R(0,\delta)$. \qquad Then $B\in\scrb_\R(0)$.\\
Want: $L-\vep\cdot|\bullet|\,\,\le\,\,f_c^\T\,\,\le\,\,L+\vep\cdot|\bullet|$ on $B$.\\
Want: \qquad $\forall h\in B$, \quad $(L-\vep\cdot|\bullet|)_h\,\,\le\,\,(f_c^\T)_h\,\,\le\,\,(L+\vep\cdot|\bullet|)_h$.\\
Given $h\in B$. \qquad Want: $(L-\vep\cdot|\bullet|)_h\,\,\le\,\,(f_c^\T)_h\,\,\le\,\,(L+\vep\cdot|\bullet|)_h$.\\
\phx\qquad\qquad\qquad\qquad Want: $L_h-\vep\cdot|h|\,\,\,\,\,\,\,\le\,(f_c^\T)_h\,\,\,\le\,\,\,\,L_h+\vep\cdot|h|$.\\
Let $s:=L_h$ and $t:=(f_c^\T)_h$ and $u:=\vep\cdot|h|$. \quad Want: $s-u<t<s+u$.\\
Since $h\in B=B_\R(0,\delta)$, we get: $d(h,0)<\delta$.\\
Then $|h|=|h-0|=d(h,0)<\delta$. \quad
Then, by choice of $\delta$, $|\rho_h|<\vep\cdot|h|^1$.\\
We have $\rho_h=(f_c^\T-L)_h=(f_c^\T)_h-L_h=t-s$.\\
Then $|t-s|=|\rho_h|<\vep\cdot|h|^1=\vep\cdot|h|=u$.\\
Then $|t-s|<u$, and so $s-u<t<s+u$, as desired. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
1-3. Let $f:\R\drta\R$, let $c\in\R$ and let $L\in\LINS_cf$.\\
\phx\quad $\,$ Assume: \quad $f$ has a local semi-maximum at $c$ in $\R$.\\
\phx\quad $\,$ Show: \qquad $L=\bfzero$.
\vskip.1in\noindent
{\it Proof:}
Since $L\in\LINS_cf$, we get $D_cf=L$ and $L\in\scrl$.\\
Since $L\in\scrl$, we get $L:\R\to\R$. \qquad Then $\D_L=\R$.\\
Since $1\in\R=\D_L$, we conclude that: $L_1\ne\frownie$.\\
So, since $L_1=(D_cf)_1=f'_c$, we get $f'_c\ne\frownie$. \quad Then $c\in\D_{f'}$.\\
Then $c\in\D_{f'}\subseteq D_f$. \quad Then $(f_c^\T)_0=0$.\\
Since $L\in\LINS_cf$, we get $f_c^\T-L\in\scrlito_1$.\\
Since $f$ has a local semi-maximum at $c$ in $\R$,\\
\phx\qquad it follows that $f_c^\T$ has a local semi-maximum at $0$ in $\R$.\\
Let $g:=f_c^\T$. \qquad Then $g_0=0$ and $g-L\in\scrlito_1$.\\
Also, $g$ has a local semi-maximum at $0$ in $\R$.\\
Choose $A\in\scrb_\R(0)$ s.t.~$g\le g_0$ on $A$. \quad Then $g\le0$ on $A$.\\
Since $g-L\in\scrlito_1$ and $L\in\scrl=\scrh_1$, choose $\delta>0$ s.t., $\forall h\in B_\R(0,\delta)$,\\
\phx\qquad\quad $[\,\,(\,\,L_h<0\,\,)\,\,\Rightarrow\,\,(\,\,g_h<0\,\,)\,\,]\,\,\&\,\,[\,\,(\,\,L_h>0\,\,)\,\,\Rightarrow\,\,(\,\,g_h>0\,\,)\,\,]$.\\
Choose $\gamma>0$ s.t.~$A=B_\R(0,\gamma)$. \qquad Let $\beta:=\min\{\gamma/2,\delta/2\}$.\\
Then $\beta\le\gamma/2<\gamma$ and $\beta\le\delta/2<\delta$.\\
Also, since $\gamma/2,\delta/2>0$, we conclude that $\beta>0$. \quad Then $|\beta|=\beta$.\\
Assume $L\ne\bfzero$. \qquad Want: Contradiction.\\
Since $L\in\scrl$, choose $m\in\R$ s.t.~$L=m\cdot(\bullet)$.\\
Since $L\ne\bfzero$, we get $m\ne0$. \quad Then $m\in\R^\times_0$.\\
Since $m\in\R^\times_0$, it follows that $|m|>0$. \quad
Let $h:=(m/|m|)\cdot\beta$.\\
Then $d(h,0)=|h-0|=|h|=(|m|/|m|)\cdot|\beta|=1\cdot\beta=\beta$.\\
Since $d(h,0)=\beta<\gamma$, we get $h\in B_\R(0,\gamma)$.\\
Since $d(h,0)=\beta<\delta$, we get: $h\in B_\R(0,\delta)$.\\
Since $h\in B_\R(0,\gamma)=A$ and since $g\le0$ on $A$, we conclude: $g_h\le0$.\\
Since $m\in\R_0^\times$, it follows that $m^2>0$.\\
We have $L_h=m\cdot h=m\cdot(m/|m|)\cdot(\delta/2)=(m^2\cdot\delta)/(2\cdot|m|)$.\\
So, since $m^2>0$ and $\delta>0$ and $|m|>0$, we get: $L_h>0$.\\
So, since $h\in B_\R(0,\delta)$, by choice of $\delta$, we conclude that $g_h>0$.\\
Then $0<g_h\le0$, so $0<0$. \qquad Contradiction. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
1-4. Let $\phi:\R\drta\R$, let $a>0$ and let $Q:=a\cdot(\bullet)^2$.\\
\phx\quad $\,$ Assume: \quad $\phi-Q\,\,\in\,\,\scrlito_2$.\\
\phx\quad $\,$ Show: \qquad $\phi$ has a local strict-minimum at $0$ in $\R$.
\vskip.1in\noindent
{\it Proof:}
We have $Q=a\cdot(\bullet)^2\in\scrh_2$.\\
Since $\phi-Q\in\scrlito_2$ and $Q\in\scrh_2$, choose $\delta>0$ s.t., $\forall h\in B_\R(0,\delta)$,\\
\phx\qquad\quad $[\,\,(\,\,Q_h<0\,\,)\,\,\Rightarrow\,\,(\,\,\phi_h<0\,\,)\,\,]\,\,\&\,\,[\,\,(\,\,Q_h>0\,\,)\,\,\Rightarrow\,\,(\,\,\phi_h>0\,\,)\,\,]$.\\
Want: $\exists B\in\scrb_\R(0)$ s.t.~$\phi>\phi_0$ on $B_0^\times$.\\
Let $B:=B_\R(0,\delta)$. \quad Then $B\in\scrb_\R(0)$. \quad Want: $\phi>\phi_0$ on $B_0^\times$.\\
Want: $\forall h\in B_0^\times$, $\phi_h\ge\phi_0$. \quad Given $h\in B_0^\times$. \quad Want: $\phi_h>\phi_0$.\\
Since $\phi-Q\in\scrlito_2\subseteq\scrlito_0=\CVZ$, we get: $(\phi-Q)_0=0$.\\
We have $Q_0=(a\cdot(\bullet))_0=a\cdot0^2=0$.\\
Then $\phi_0=\phi_0-0=\phi_0-Q_0=(\phi-Q)_0=0$.\\
Since $h\in\R_0^\times$, we get $h^2>0$.
\quad So, since $a>0$, we get $a\cdot h^2>0$.\\
Then $Q_h=(a\cdot(\bullet)^2)_h=a\cdot h^2>0$.\\
So, since $h\in B=B_\R(0,\delta)$, by choice of $\delta$, we get: $\phi_h>0$.\\
Then $\phi_h>0=\phi_0$, as desired. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
1-5. Let $f,g:\R\drta\R$ and let $c\in\R$.\\
\phx\quad $\,$ Assume: \quad $f=g$ near $c$ in $\R$.\\
\phx\quad $\,$ Show: \qquad $\LINS_cf\,\,\subseteq\,\,\LINS_cg$.\\
\vskip.1in\noindent
{\it Proof:}
Want: $\forall L\in\LINS_cf$, $L\in\LINS_cg$.\\
Given $L\in\LINS_cf$. \qquad Want: $L\in\LINS_cg$.\\
Know: $f_c^\T-L\in\scrlito_1$. \qquad Want: $g_c^\T-L\in\scrlito_1$.\\
Want: $f_c^\T-L=g_c^\T-L$ near $0$ in $\R$.\\
Want: $\exists A\in\scrb_\R(0)$ s.t.~$f_c^\T-L=g_c^\T-L$ on $A$.\\
Since $f=g$ near $c$ in $\R$, choose $B\in\scrb_\R(c)$ s.t.~$f=g$ on $B$.\\
Since $B\in\scrb_\R(c)$, we get $c\in B$.\\
So, \quad since $f=g$ on $B$, \quad we get $f_c=g_c$.\\
Choose $\delta>0$ s.t.~$B=B_\R(c,\delta)$.\\
Let $A:=B_\R(0,\delta)$. \qquad Then $A\in\scrb_\R(0)$.\\
Want: $f_c^\T-L=g_c^\T-L$ on $A$.\\
Want: \quad $\forall h\in A$, \quad $(f_c^\T-L)_h=(g_c^\T-L)_h$.\\
Given $h\in A$. \qquad Want: $(f_c^\T-L)_h=(g_c^\T-L)_h$.\\
Since $h\in A=B_\R(0,\delta)$, we get $d(h,0)<\delta$.\\
Since $c,h\in\R$, we get $(c+h)-c=h$ and $h=h-0$.\\
Then $d(c+h,c)=|(c+h)-c|=|h|=|h-0|=d(h,0)<\delta$.\\
As $d(c+h,c)<\delta$, we get: $c+h\in B_\R(c,\delta)$. \quad Then $c+h\in B_\R(c,\delta)=B$.\\
So, \quad since $f=g$ on $B$, \quad we get $f_{c+h}=g_{c+h}$.\\
Recall that $f_c=g_c$. \quad Then $(f_c^\T)_h=f_{c+h}-f_c=g_{c+h}-g_c=(g_c^\T)_h$.\\
Then $(f_c^\T-L)_h=(f_c^\T)_h-L_h=(g_c^\T)_h-L_h=(g_c^\T-L)_h$. \quad
QED
\vskip.1in\hrule\vskip.1in\hrule
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