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\begin{document}

\centerline{\bf Solutions for MATH 4604 (Advanced Calculus II)}
\centerline{\bf Spring 2019}
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 13: Due on Tuesday 30 April}
\vskip.1in\hrule\vskip.1in\noindent
13-1. Let \qquad $W\in\TNSR^+$, \qquad $\phi:\R\to W$.\\
Assume: \qquad $\,\forall k\in\scri_W$, \quad $0\in\dom[((\pi_k^W)\circ\phi)']$.\\
Show: \qquad $\,\,\,\,\,\,$ $0\in\dom[\phi']$.
\vskip.1in\noindent
{\it Proof:}
Let $\ds{x:=\sum_{k\in\scri_W}\,(((\pi_k^W)\circ\phi)'_0)\cdot\varepsilon_k^W}$. \qquad
Want: $\phi'_0=x$.\\
We have: \qquad $\forall k\in\scri_W$, \quad $x_k=((\pi_k^W)\circ\phi)'_0$.\\
Define $L\in\scrl_\R^W$ by $L(t)=tx$. \qquad
Then $\sl_L^{\R W}=L(1)=x$.\\
Also, $\phi'_0=\sl_{D_0\phi}^{\R W}$. \qquad
Want: $\sl_{D_0\phi}^{\R W}=\sl_L^{\R W}$.\\
Want: $D_0\phi=L$. \qquad
Want: $\phi_0^T-L\in\scrlito_1^{\R W}$.\\
Let $\psi:=\phi_0^T-L$. \qquad Want: $\psi\in\scrlito_1^{\R W}$.\\
By HW\#12-1, it suffices to show: \quad $\forall k\in\scri_W$, \quad $(\pi_k^W)\circ\psi\in\scrlito_1^{\R\R}$.\\
Given $k\in\scri_W$. \qquad
Want: $(\pi_k^W)\circ\psi\in\scrlito_1^{\R\R}$.\\
Let $\chi:=(\pi_k^W)\circ\phi$. \qquad
Define $M\in\scrl_\R^\R$ by $M(t)=tx_k$.\\
We have $\sl_M^{\R\R}=M(1)=1\cdot x_k=x_k=((\pi_k^W)\circ\phi)'_0=\chi'_0=\sl_{D_0\chi}^{\R\R}$.\\
Then $M=[\sl_M^{\R\R}]_\R^\R=[\sl_{D_0\chi}^{\R\R}]_\R^\R=D_0\chi$.\\
Since $D_0\chi=M$, we get $\chi_0^T-M\in\scrlito_1^{\R\R}$.\\
Want: $\chi_0^T-M=(\pi_k^W)\circ\psi$. \qquad
Want: $\forall h\in\R$, $(\chi_0^T-M)_h=((\pi_k^W)\circ\psi)_h$.\\
Given $h\in\R$. \qquad
Want: $(\chi_0^T-M)_h=((\pi_k^W)\circ\psi)_h$.\\
We have $\psi_h=(\phi_0^T-L)_h=\phi_{0+h}-\phi_0-L_h=\phi_h-\phi_0-hx$.\\
Then $\pi_k^W(\psi_h)=(\pi_k^W(\phi_h))-(\pi_k^W(\phi_0))-hx_k$.\\
Then: $(\chi_0^T-M)_h\,\,\,=\,\,\,\chi_{0+h}-\chi_0-M_h\,\,\,=\,\,\,\chi_h-\chi_0-M_h$\\
\phx\qquad\qquad\qquad\quad $\,\,\,\,\,\,=\,\,\,((\pi_k^W)\circ\phi)_h-((\pi_k^W)\circ\phi)_0-hx_k$\\
\phx\qquad\qquad\qquad\quad $\,\,\,\,\,\,=\,\,\,(\pi_k^W(\phi_h))-(\pi_k^W(\phi_0))-hx_k$\\
\phx\qquad\qquad\qquad\quad $\,\,\,\,\,\,=\,\,\,\pi_k^W(\psi_h)\,\,\,=\,\,\,((\pi_k^W)\circ\psi)_h$, \quad
as desired. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
13-2. Let \qquad $V,W\in\TNSR^+$, \qquad $f:V\drta W$, \qquad $q,u\in V$.\\
Assume: \qquad $\,\forall k\in\scri_W$, \quad $q\in\dom[\partial_u((\pi_k^W)\circ f)]$.\\
Show: \qquad $\,\,\,\,\,\,$ $q\in\dom[\partial_uf]$.
\vskip.1in\noindent
{\it Proof:}
Let $\phi:=f\circ(i_q^u)$. \quad
Then $(\partial_uf)_q=\phi'_0$. \quad
Want: $0\in\dom[\phi']$.\\
By HW\#13-1, it suffices to show:
$\forall k\in\scri_W$, $0\in\dom[((\pi_k^W)\circ\phi)']$.\\
Given $k\in\scri_W$. \qquad
Want: $0\in\dom[((\pi_k^W)\circ\phi)']$.\\
By assumption, $q\in\dom[\partial_u((\pi_k^W)\circ f)]$.\\
So, since $(\partial_u((\pi_k^W)\circ f))_q=((\pi_k^W)\circ f\circ(i_q^u))'_0$,\\
\phx\qquad we get: $0\in\dom[((\pi_k^W)\circ f\circ(i_q^u))']$.\\
Then $0\in\dom[((\pi_k^W)\circ f\circ(i_q^u))']=\dom[((\pi_k^W)\circ\phi)']$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
13-3. Let \qquad $V,W\in\TNSR^+$, \qquad $f:V\drta W$, \qquad $p\in V$.\\
Assume: \qquad $\,f''$ is continuous near $p$.\\
Show: \qquad $\,\,\,\,\,\,$ $\forall i,j\in\scri_V$, \quad $\partial_i\partial_jf$ is continuous near $p$.
\vskip.1in\noindent
{\it Proof:}
Given $i,j\in\scri_V$. \qquad
Want: $\partial_i\partial_jf$ is continuous near $p$.\\
Choose $B\in\scrb_V(p)$ s.t.~$f''$ is continuous on $B$.\\
Want: $\partial_i\partial_jf$ is continuous on $B$.\\
Want: $\forall p\in B$, $\partial_i\partial_jf$ is continuous at $p$.\\
Given $p\in B$. \qquad
Want: $\partial_i\partial_jf$ is continuous at $p$.\\
By Theorem 99.1, it suffices to show:\\
\phx\qquad $\forall k\in\scri_W$, \quad $\pi_k\circ(\partial_i\partial_jf)$ is continuous at $p$.\\
Given $\forall k\in\scri_W$. \qquad
Want: $\pi_k\circ(\partial_i\partial_jf)$ is continuous at $p$.\\
Since $f''$ is continuous on $B$, we conclude: $B\subseteq\dom[f'']$.\\
So, since $\im[f'']\subseteq W\otimes V\otimes V=\dom[\pi_{k\|i\|j}]$,\\
\phx\qquad we conclude that: \quad
$B\subseteq\dom[\pi_{k\|i\|j}\circ(f'')]$.\\
By Thoerem 102.17, we have: \qquad
$\pi_k\circ(\partial_i\partial_jf)\,\,\supseteq\,\,\pi_{k\|i\|j}\circ(f'')$.\\
So, since $B\subseteq\dom[\pi_{k\|i\|j}\circ(f'')]$, we get:\\
\phx\qquad\qquad $\pi_k\circ(\partial_i\partial_jf)\,\,=\,\,\pi_{k\|i\|j}\circ(f'')$ \qquad on $B$.\\
So, since $p\in B$, by the Recentering Lemma (Theorem 38.16), we get:\\
\phx\qquad\qquad $\pi_k\circ(\partial_i\partial_jf)\,\,=\,\,\pi_{k\|i\|j}\circ(f'')$ \qquad near $p$.\\
Since $f''$ is continuous on $B$ and $p\in B$,\\
\phx\qquad\qquad we see that $f''$ is continuous at $p$.\\
Then, by Theorem 99.1, $\pi_{k\|i\|j}\circ(f'')$ is continuous at $p$.\\
So, as \qquad $\pi_k\circ(\partial_i\partial_jf)\,\,=\,\,\pi_{k\|i\|j}\circ(f'')$ \qquad near $p$,\\
\phx\qquad\qquad it follows, from Theorem 65.7,\\
\phx\qquad\qquad that $\pi_k\circ(\partial_i\partial_jf)$ is continuous at $p$, \qquad as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
13-4. Let  \qquad$V,W\in\TNSR^+$, \qquad $f:V\drta W$, \qquad $p\in V$.\\
Assume: \qquad $\,\,\forall h,i,j\in\scri_V$,\\
\phantom{x}\qquad\qquad\qquad\qquad $\partial_h\partial_i\partial_jf$ is both defined near $p$ and bounded near $p$.\\
Show: \qquad $\,\,\,\,\,\,$ $f''$ is continuous near $p$.
\vskip.1in\noindent
{\it Proof:}
By Theorem 102.21, it suffices to show:\\
\phx\qquad $\forall i,j\in\scri_V$, \qquad $\partial_i\partial_jf$ is continuous near $p$.\\
Given $i,j\in\scri_V$. \qquad
Want: $\partial_i\partial_jf$ is continuous near $p$.\\
Let $g:=\partial_i\partial_jf$. \qquad Want: $g$ is continuous near $p$.\\
By Theorem 102.5, it suffices to show:\\
\phx\qquad $\forall h\in\scri_V$, \qquad $\partial_hg$ is defined near $p$ and bounded near $p$.\\
Given $h\in\scri_V$. \qquad
Want: $\partial_hg$ is defined near $p$ and bounded near $p$.\\
By assumption, $\partial_h\partial_i\partial_jf$ is defined near $p$ and bounded near $p$.\\
So since $\partial_hg=\partial_h\partial_i\partial_jf$, we get:\\
\phx\qquad $\partial_hg$ is defined near $p$ and bounded near $p$, \qquad as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
13-5. Let \qquad $V,W\in\TNSR^+$, \qquad $f:V\drta W$, \qquad $p\in V$.\\
Assume: \qquad $\,\,\forall h,i,j\in\scri_V$,\\
\phantom{x}\qquad\qquad\qquad\qquad $\partial_h\partial_i\partial_jf$ is both defined near $p$ and continuous at $p$.\\
Show: \qquad $\,\,\,\,\,\,$ $f'''$ is continuous at $p$.
\vskip.1in\noindent
{\it Proof:}
By Theorem 102.3, $\forall h,i,j\in\scri_V$, $\partial_h\partial_i\partial_jf$ is bounded near $p$.\\
Also, by assumption, $\forall h,i,j\in\scri_V$, $\partial_h\partial_i\partial_jf$ is defined near $p$.\\
Then, by HW\#13-4, $f''$ is continuous near $p$.\\
Then: \quad $f''$ is continuous at $p$ \quad and \quad $f''$ is defined near $p$.\\
\\
{\it Claim:} Let $h,i,j\in\scri_V$, $k\in\scri_W$.\\
\phx\qquad\quad Then: \quad $\partial_h(\pi_{k\|i\|j}\circ(f''))\,\,=\,\,\pi_k\circ(\partial_h\partial_i\partial_jf)$ \quad near $p$.\\
{\it Proof of Claim:}\\
Since $f''$ is defined near $p$, by Theorem 97.3,\\
\phx\qquad we see that $\partial_j(f')=(\partial_jf)'$ near $p$.\\
Since $f''$ is defined near $p$ and since $\im[f'']\subseteq W\otimes V\otimes V=\dom[\pi_{k\|i\|j}]$,\\
\phx\qquad we conclude that $\pi_{k\|i\|j}\circ(f'')$ is defined near $p$.\\
Let $\alpha:=f'$ and $\ell:=k\|i$. \qquad
By Theorem 102.1, $\pi_\ell\circ(\partial_j\alpha)\supseteq\pi_{\ell\|j}\circ(\alpha')$.\\
That is, $\pi_{k\|i}\circ(\partial_j(f'))\supseteq\pi_{k\|i\|j}\circ(f'')$.\\
So, since $\pi_{k\|i\|j}\circ(f'')$ is defined near $p$,\\
\phx\qquad it follows that $\pi_{k\|i}\circ(\partial_j(f'))=\pi_{k\|i\|j}\circ(f'')$ near $p$.\\
Since $f''$ is defined near $p$, by Theorem 94.4,\\
\phx\qquad we see that $(\partial_jf)'$ is defined near $p$.\\
Then, since $\im[(\partial_jf)']\subseteq W\otimes V=\dom[\pi_{k\|i}]$,\\
\phx\qquad we conclude that $\pi_{k\|i}\circ((\partial_jf)')$ is defined near $p$.\\
Let $\beta:=\partial_jf$. \qquad
By Theorem 102.1, $\pi_k\circ(\partial_i\beta)\supseteq\pi_{k\|i}\circ(\beta')$.\\
That is, $\pi_k\circ(\partial_i\partial_jf)\supseteq\pi_{k\|i}\circ((\partial_jf)')$.\\
So, since $\pi_{k\|i}\circ((\partial_jf)')$ is defined near $p$,\\
\phx\qquad it follows that $\pi_k\circ(\partial_i\partial_jf)=\pi_{k\|i}\circ((\partial_jf)')$ near $p$.\\
So, since $\partial_j(f')=(\partial_jf)'$ near $p$,\\
\phx\qquad it follows that $\pi_k\circ(\partial_i\partial_jf)=\pi_{k\|i}\circ(\partial_j(f'))$ near $p$.\\
So, since $\pi_{k\|i}\circ(\partial_j(f'))=\pi_{k\|i\|j}\circ(f'')$ near $p$.\\
\phx\qquad it follows that $\pi_k\circ(\partial_i\partial_jf)=\pi_{k\|i\|j}\circ(f'')$ near $p$.\\
Then: \quad $\partial_h(\pi_{k\|i\|j}\circ(f''))\,\,=\,\,\partial_h(\pi_k\circ(\partial_i\partial_jf))$ \quad near $p$.\\
Want: \quad $\partial_h(\pi_k\circ(\partial_i\partial_jf))\,\,=\,\,\pi_k\circ(\partial_h\partial_i\partial_jf)$ \quad near $p$.\\
By assumption, $\partial_h\partial_i\partial_jf$ is defined near $p$.\\
So, since $\im[\partial_h\partial_i\partial_jf]\subseteq W=\dom[\pi_k]$,\\
\phx\qquad we see that $\pi_k\circ(\partial_h\partial_i\partial_jf)$ is defined near $p$.\\
By Theorem 94.3, $\partial_h(\pi_k\circ(\partial_i\partial_jf))\supseteq\pi_k\circ(\partial_h\partial_i\partial_jf)$.\\
So, since $\pi_k\circ(\partial_h\partial_i\partial_jf)$ is defined near $p$, it follows that:\\
\phx\qquad $\partial_h(\pi_k\circ(\partial_i\partial_jf))\,\,=\,\,\pi_k\circ(\partial_h\partial_i\partial_jf)$ \quad near $p$, \quad as desired.\\
{\it End of proof of Claim.}\\
\\
Let $g:=f''$. \qquad
Want: $g'$ is continuous at $p$.\\
Since $f:V\drta W$ and $g=f''$, we get $g:V\drta W\otimes V\otimes V$.\\
By Theorem 102.7, it suffices to show:\\
\phx\qquad $\forall h\in\scri_V$, \quad $\partial_hg$ is defined near $p$ and continuous at $p$.\\
Given $h\in\scri_V$. \qquad Want: \quad (A) $\partial_hg$ is defined near $p$ \qquad and\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $\,\,\,$ (B) $\partial_hg$ is continuous at $p$.\\
\\
{\it Proof of (A):}\\
By HW\#13-2, it suffices to show:\\
\phx\qquad $\forall\ell\in\scri_{W\otimes V}$, \quad $\partial_h(\pi_\ell\circ g)$ is defined near $p$.\\
Given $\ell\in\scri_{W\otimes V\otimes V}$. \qquad
Want: $\partial_h(\pi_\ell\circ g)$ is defined near $p$.\\
Since $\ell\in\scri_{W\otimes V\otimes V}$, choose $i,j\in\scri_V$ and $k\in\scri_W$ s.t.~$\ell=k\|i\|j$.\\
By assumption, $\partial_h\partial_i\partial_jf$ is defined near $p$.\\
So, since $\im[\partial_h\partial_i\partial_jf]\subseteq W=\dom[\pi_k]$,\\
\phx\qquad we see that $\pi_k\circ(\partial_h\partial_i\partial_jf)$ is defined near $p$.\\
By the claim: \quad $\partial_h(\pi_{k\|i\|j}\circ(f''))\,\,=\,\,\pi_k\circ(\partial_h\partial_i\partial_jf)$ \quad near $p$.\\
So, since $\ell=k\|i\|j$ and $g=f''$, we see that:\\
\phx\qquad\qquad $\partial_h(\pi_\ell\circ g)\,\,=\,\,\pi_k\circ(\partial_h\partial_i\partial_jf)$ \quad near $p$.\\
So, since $\pi_k\circ(\partial_h\partial_i\partial_jf)$ is defined near $p$, we conclude:\\
\phx\qquad\qquad $\partial_h(\pi_\ell\circ g)$ is defined near $p$, \qquad as desired.\\
{\it End of proof of (A).}\\
\\
{\it Proof of (B):}\\
By Theorem 99.1, it suffices to show:\\
\phx\qquad $\forall\ell\in\scri_{W\otimes V\otimes V}$, \quad $\pi_\ell\circ(\partial_hg)$ is continuous at $p$.\\
Given $\ell\in\scri_{W\otimes V\otimes V}$. \qquad
Want: $\pi_\ell\circ(\partial_hg)$ is continuous at $p$.\\
By (A), $\partial_hg$ is defined near $p$. \qquad
Then $p\in\dom[\partial_hg]$.\\
So, since $\im[\partial_hg]\subseteq W\otimes V\otimes V=\dom[\pi_\ell]$,\\
\phx\qquad we see that $p\in\dom[\pi_\ell\circ(\partial_hg)]$.\\
By Theorem 94.3, $\partial_h(\pi_\ell\circ g)\supseteq\pi_\ell\circ(\partial_hg)$.\\
Then, by Theorem 44.13, it suffices to show:\\
\phx\qquad\qquad $\partial_h(\pi_\ell\circ g)$ is continuous at $p$.\\
Since $\ell\in\scri_{W\otimes V\otimes V}$, choose $i,j\in\scri_V$ and $k\in\scri_W$ s.t.~$\ell=k\|i\|j$.\\
By assumption, $\partial_h\partial_i\partial_jf$ is continuous at $p$.\\
Then, by Theorem 99.1,\\
\phx\qquad we see that $\pi_k\circ(\partial_h\partial_i\partial_jf)$ is continuous at $p$.\\
By the claim, \quad $\partial_h(\pi_{k\|i\|j}\circ(f'))\,\,=\,\,\pi_k\circ(\partial_h\partial_i\partial_jf)$ \quad near $p$.\\
So, since $\ell=k\|i\|j$ and $g=f''$, we see that:\\
\phx\qquad\qquad $\partial_h(\pi_\ell\circ g)\,\,=\,\,\pi_k\circ(\partial_h\partial_i\partial_jf)$ \quad near $p$.\\
So, since $\pi_k\circ(\partial_h\partial_i\partial_jf)$ is continuous at $p$, by Theorem 65.7, we have:\\
\phx\qquad\qquad $\partial_h(\pi_\ell\circ g)$ is continuous at $p$, \qquad as desired.\\
{\it End of proof of (B).}
QED
\vskip.1in\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 12: Due on Tuesday 23 April}
\vskip.1in\hrule\vskip.1in\noindent
12-1. Let $V,W\in\TNSR^+$, $f:V\drta W$, $k\in\N_0$.\\
Assume: \qquad $\forall m\in\scri_W$, \quad $(\pi_m^W)\circ f\in\scrlito_k^{V\R}$.\\
Show: \qquad $f\in\scrlito_k^{VW}$.
\vskip.1in\noindent
{\it Proof:}\\
{\it Claim:} $f_{0_V}=0_W$.\\
{\it Proof of Claim:}\\
Want: $\forall m\in\scri_W$, $\pi_m^W(f_{0_V})=0$.\\
Given $m\in\scri_W$. \qquad
Want: $\pi_m^W(f_{0_V})=0$.\\
Since $(\pi_m^W)\circ f\in\scrlito_k^{V\R}\subseteq\scrlito_0^{V\R}=\CVZ_V^\R$, we conclude: $((\pi_m^W)\circ f)_{0_V}=0$.\\
Then $\pi_m^W(f_{0_V})=((\pi_m^W)\circ f)_{0_V}=0$, as desired.\\
{\it End of proof of Claim.}\\
\\
Let $\ds{\phi:=\adj_{0_V}^{0_W}\left(\frac{f}{|\bullet|_V^k}\right)}$. \qquad
Then $\phi_{0_V}=0_W$.\\
By the claim, $f_{0_V}=0_W$, and it follows that: $\phi\cdot(|\bullet|_V^k)=f$.\\
Want: $f\in(\CVZ_V^W)\cdot(|\bullet|_V^k)$. \qquad
Want: $\phi\in\CVZ_V^W$.\\
Want: $\forall m\in\scri_W$, $(\pi_m^W)\circ\phi\in\CVZ_V^\R$.\\
Given $m\in\scri_W$. \qquad
Want: $(\pi_m^W)\circ\phi\in\CVZ_V^\R$.\\
By assumption, $(\pi_m^W)\circ f\in\scrlito_k^{V\R}$. \qquad
Then $(\pi_m^W)\circ f\in(\CVZ_V^\R)\cdot(|\bullet|_V^k)$.\\
Choose $\psi\in\CVZ_V^\R$ s.t.~$(\pi_m^W)\circ f=\psi\cdot(|\bullet|_V^k)$.\\
Want: $(\pi_m^W)\circ\phi=\psi$. \qquad
Want: $(\pi_m^W)\circ\phi=\psi$ on $V$.\\
Since $\psi\in\CVZ_V^\R$, it follows that $\psi_{0_V}=0$.\\
We have $((\pi_m^W)\circ\phi)_{0_V}=(\pi_m^W)(\phi_{0_V})=\pi_m^W(0_W)=0=\psi_0$.\\
Want: $(\pi_m^W)\circ\phi=\psi$ on $V_{0_V}^\times$. \quad
Want: $\forall q\in V_{0_V}^\times$, $((\pi_m^W)\circ\phi)_q=\psi_q$.\\
Given $q\in V_{0_V}^\times$. \quad
Want: $((\pi_m^W)\circ\phi)_q=\psi_q$. \quad
Want: $(\pi_m^W)(\phi_q)=\psi_q$.\\
Since $q\in V_{0_V}^\times$, we get $|q|_V\ne0$, and so $|q|_V^k\ne0$.\\
Recall: $\phi\cdot(|\bullet|_V^k)=f$ and $(\pi_m^W)\circ f=\psi\cdot(|\bullet|_V^k)$.\\
Then $(\pi_m^W(\phi_q))\cdot(|q|_V^k)=\pi_m^W(\phi_q\cdot(|q|_V^k))=\pi_m^W((\phi\cdot(|\bullet|_V^k))_q)=\pi_m^W(f_q)$\\
\phantom{x}\qquad\qquad\qquad $=((\pi_m^W)\circ f)_q=(\psi\cdot(|\bullet|_V^k))_q=\psi_q\cdot(|q|_V^k)$.\\
So, since $|q|_V^k\ne0$, dividing by $|q|_V^k$, we get: $(\pi_m^W(\phi_q))=\psi_q$.
QED
\vskip.1in\hrule\vskip.1in\noindent
12-2. Let $V,W,X\in\TNSR^+$, $k\in\N_0$.\\
Show: \qquad $(\scrlito_k^{WX})\,\,\circ\,\,(\scrbigo_1^{VW})\quad\subseteq\quad\scrlito_k^{VX}$ \qquad and\\
\phantom{x}\qquad\qquad $\,\,\,(\scrbigo_1^{WX})\,\,\circ\,\,(\scrlito_k^{VW})\quad\subseteq\quad\scrlito_k^{VX}$ \qquad and\\
\phantom{x}\qquad\qquad $\,\,(\scrbigo_k^{WX})\,\,\circ\,\,(\scrbigo_1^{VW})\quad\subseteq\quad\scrbigo_k^{VX}$.
\vskip.1in\noindent
{\it Proof:}
Exactly one of the following is true:\\
\phantom{x}\qquad (1) $k\ne0$ \qquad\qquad or \qquad\qquad (2) $k=0$.\\
\\
{\it Case (1):}\\
Since $k\ne0$, $\,\,\,(\scrlito_k^{WX})\,\,\circ\,\,(\scrbigo_1^{VW})\,\,\,\,\subseteq\,\,\,\,\scrlito_{k\cdot1}^{VX}\,\,\,\,=\,\,\,\,\scrlito_k^{VX}$ \qquad and\\
\phantom{x}\qquad\qquad\quad $\,\,\,\,(\scrbigo_1^{WX})\,\,\circ\,\,(\scrlito_k^{VW})\,\,\,\,\subseteq\,\,\,\,\scrlito_{1\cdot k}^{VX}\,\,\,\,=\,\,\,\,\scrlito_k^{VX}$.\\
Want: \qquad $\,\,\,\,\,(\scrbigo_k^{WX})\,\,\circ\,\,(\scrbigo_1^{VW})\,\,\,\,\subseteq\,\,\,\,\scrbigo_k^{VX}$.\\
Since $k\ne0$, $\,\,(\scrbigo_k^{WX})\,\,\circ\,\,(\scrbigo_1^{VW})\,\,\,\,\subseteq\,\,\,\,\scrbigo_{k\cdot1}^{VX}\,\,\,\,=\,\,\,\,\scrbigo_k^{VX}$.\\
{\it End of Case (1).}\\
\\
{\it Case (2):}\\
We have $\,(\scrlito_k^{WX})\,\,\circ\,\,(\scrbigo_1^{VW})\,\,\,\,=\,\,\,\,\scrlito_0^{WX}\,\,\circ\,\,(\scrbigo_1^{VW})$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad $\,\subseteq\,\,\,\,(\scrlito_0^{WX})\,\,\circ\,\,(\scrlito_0^{VW})$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad $\,\subseteq\,\,\,\,(\CVZ_W^X)\,\,\circ\,\,(\CVZ_V^W)$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad $\,\subseteq\,\,\,\,\CVZ_V^X\,\,\,\,=\,\,\,\,\scrlito_0^{VX}\,\,\,\,\subseteq\,\,\,\,\scrlito_k^{VX}$.\\
Also, \quad $\,\,\,\,(\scrbigo_1^{WX})\,\,\circ\,\,(\scrlito_k^{VW})\,\,\,\,=\,\,\,\,(\scrbigo_1^{WX})\,\,\circ\,\,(\scrlito_0^{VW})$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad $\,\subseteq\,\,\,\,(\scrlito_0^{WX})\,\,\circ\,\,(\scrlito_0^{VW})$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad $\,\subseteq\,\,\,\,(\CVZ_W^X)\,\,\circ\,\,(\CVZ_V^W)$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad $\,\subseteq\,\,\,\,\CVZ_V^X\,\,\,\,=\,\,\,\,\scrlito_0^{VX}\,\,\,\,\subseteq\,\,\,\,\scrlito_k^{VX}$.\\
Want: $\,\,\,\,\,\,(\scrbigo_k^{WX})\,\,\circ\,\,(\scrbigo_1^{VW})\,\,\,\,\subseteq\,\,\,\,\scrbigo_k^{VX}$.\\
We have $(\scrbigo_k^{WX})\,\,\circ\,\,(\scrbigo_1^{VW})\,\,\,\,=\,\,\,\,(\scrbigo_0^{WX})\,\,\circ\,\,(\scrbigo_1^{VW})$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad $\,=\,\,\,\,(\scrbigo_0^{WX})\,\,\circ\,\,(\scrlito_0^{VW})$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad $\,=\,\,\,\,(\BNZ_W^X)\,\,\circ\,\,(\CVZ_V^W)$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad $\,\subseteq\,\,\,\,\BNZ_V^X\,\,\,\,=\,\,\,\,\scrbigo_0^{VX}\,\,\,\,\subseteq\,\,\,\,\scrbigo_k^{VX}$.\\
{\it End of Case (2).}
QED
\vskip.1in\hrule\vskip.1in\noindent
12-3. Let $V,W\in\TNSR^+$, $f:V\drta W$, $k\in\N_0$.\\
Assume: $f_{0_V}=0_W$.\\
Assume: $\forall j\in\scri_V$, $\partial_jf\in\scrbigo_k^{VW}$. \qquad\qquad
Show: $f\in\scrbigo_{k+1}^{VW}$.
\vskip.1in\noindent
{\it Proof:}
By Theorem 99.2,
it suffices to show: $\forall m\in\scri_W$, $\pi_m^W\circ f\in\scrbigo_{k+1}^{V\R}$.\\
Given $m\in\scri_W$. \qquad
Want: $\pi_m^W\circ f\in\scrbigo_{k+1}^{V\R}$.\\
Let $g:=\pi_m^W\circ f$. \qquad
Want: $g\in\scrbigo_{k+1}^{V\R}$.\\
By Theorem 98.10, choose $\beta:\scri_V\to\scrbigo_1^{VV}$ s.t.\\
\phx\qquad\qquad\qquad $\ds{g\,\,\,=\,\,\,\sum_{j\in\scri_V}\,((\partial_jg)\circ\beta_j)\cdot\pi_j^V}$ \quad near $0_V$.\\
Want: \qquad $\forall j\in\scri_V$, \qquad $((\partial_jg)\circ\beta_j)\cdot\pi_j^V\,\,\in\,\,\scrbigo_{k+1}^{V\R}$.\\
Given $j\in\scri_V$. \qquad
Want: $((\partial_jg)\circ\beta_j)\cdot\pi_j^V\,\,\in\,\,\scrbigo_{k+1}^{V\R}$.\\
Since $\pi_j^V\in\scrl_V^W\subseteq\scrbigo_1^{V\R}$,
it suffices to show: $(\partial_jg)\circ\beta_j\,\,\in\,\,\scrbigo_k^{V\R}$.\\
So, since $\beta_j\in\scrbigo_1^{VV}$,
it suffices to show: $\partial_jg\,\,\in\,\,\scrbigo_k^{V\R}$.\\
So, since $\partial_jg=\partial_j((\pi_m^W)\circ f)\supseteq\pi_m^W\circ(\partial_jf)$,\\
\phx\qquad it suffices to show: $\pi_m^W\circ(\partial_jf)\,\,\in\,\,\scrbigo_k^{V\R}$.\\
By assumption, $\partial_jf\in\scrbigo_k^{VW}$.\\
So, by Theorem 99.2,
we have $\pi_m^W\circ(\partial_jf)\,\,\in\,\,\scrbigo_k^{V\R}$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
12-4. Let $V,W\in\TNSR^+$, $f:V\drta W$, $k\in\N_0$.\\
Assume: $f_{0_V}=0_W$. \qquad\qquad
Assume: $\forall j\in\scri_V$, $(\partial_jf)_{0_V}=0_W$.\\
Assume: $\forall i,j\in\scri_V$, $\partial_i\partial_jf\in\scrbigo_k^{VW}$. \qquad\qquad
Show: $f\in\scrbigo_{k+2}^{VW}$.
\vskip.1in\noindent
{\it Proof:}
Since $f_{0_V}=0_W$, by HW\#12-3,\\
\phx\quad it suffices to show: $\forall j\in\scri_V$, $\partial_jf\in\scrbigo_{k+1}^{VW}$.\\
Given $j\in\scri_V$. \qquad
Want: $\partial_jf\in\scrbigo_{k+1}^{VW}$.\\
Let $g:=\partial_jf$. \qquad
Want: $g\in\scrbigo_{k+1}^{VW}$.\\
Since $g_{0_V}=(\partial_jf)_{0_V}=0_W$, by HW\#12-3,\\
\phx\quad it suffices to show: $\forall i\in\scri_V$, $\partial_ig\in\scrbigo_k^{VW}$.\\
Given $i\in\scri_V$. \qquad
Want: $\partial_ig\in\scrbigo_k^{VW}$.\\
We have $\partial_ig=\partial_i\partial_jf\in\scrbigo_k^{VW}$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
12-5. Let $V,W\in\TNSR^+$, $f:V\drta W$.\\
Assume: $f_{0_V}=0_W$.\\
Assume: $\forall j\in\scri_V$, $\partial_jf$ is both defined near $0_V$ and continuous at $0_V$.\\
Show: $0_V\in\dom[f']$.
\vskip.1in\noindent
{\it Proof:}
Define $q:\scri_V\to W$ by $q_j=(\partial_jf)_{0_V}$.\\
Define $L\in\scrl_V^W$ by $\ds{L(x)=\sum_{j\in\scri_V}\,x_jq_j}$. \quad
Want: $f'_{0_V}=\sl_L^{VW}$.\\
Want: $D_{0_V}^{VW}f=L$. \qquad
Want: $f_{0_V}^T-L\in\scrlito_1^{VW}$.\\
Since $f_{0_V}=0_W$, we see that $f_{0_V}^T=f$. \qquad
Want: $f-L\in\scrlito_1^{VW}$.\\
By Theorem 99.6, it suffices to show: $\forall j\in\scri_V$, $\partial_j(f-L)\in\scrlito_0^{VW}$.\\
Given $j\in\scri_V$. \qquad
Want: $\partial_j(f-L)\in\scrlito_0^{VW}$.\\
\\
{\it Claim:} $\partial_jL=C_V^{q_j}$.\\
{\it Proof of Claim:}\\
Want: $\forall x\in V$, $(\partial_jL)_x=(C_V^{q_j})_x$.\\
Given $x\in V$. \qquad
Want: $(\partial_jL)_x=(C_V^{q_j})_x$. \qquad
Want: $(\partial_jL)_x=q_j$.\\
Let $u:=\varepsilon_j^V$. \qquad
Then $\partial_uL=\partial_jL$. \qquad
Want: $(\partial_uL)_x=q_j$.\\
Let $i:=i_x^u$. \qquad
Want: $(L\circ i)'_0=q_j$.\\
We have $\,\ds{L(u)=\sum_{i\in\scri_V}\,u_iq_i=\sum_{i\in\scri_V}\,(\varepsilon^V_j)_i\cdot q_i}$\\
\phx\qquad\qquad\qquad $\ds{=\left((\varepsilon^V_j)_j\cdot q_j\right)\,\,+\,\,\left(\sum_{i\in\scri_V\backslash\{j\}}\,(\varepsilon^V_j)_i\cdot q_i\right)}$\\
\phx\qquad\qquad\qquad $\ds{=\left(1\cdot q_j\right)\,\,+\,\,\left(\sum_{i\in\scri_V\backslash\{j\}}\,0\cdot q_i\right)=q_j+0=q_j}$\\
Then $L(u)=q_j$. \qquad Let $y:=L(x)$.\\
Then, $\forall t\in\R$, we have:\\
\phx \qquad\qquad $(L\circ i)_t=L(i_t)=L(x+tu)=(L(x))+t\cdot(L(u))=y+tq_j$.\\
Then: \qquad $\forall t\in\R$, \quad $(L\circ i)_t=y+tq_j$.\\
Then: \qquad $\forall h\in\R$, \quad $(L\circ i)^T_0(h)=(y+(0+h)\cdot q_j)-(y+0\cdot q_j)=hq_j$.\\
Then $(L\circ i)^T_0\in\scrl_\R^W$. \qquad
Then: $D_0(L\circ i)=L\circ i$.\\
Then $(L\circ i)'_0=\sl_{L\circ i}^{\R W}=(L\circ i)(1)=1\cdot q_j=q_j$, as desired.\\
{\it End of proof of Claim.}\\
\\
We have: $\partial_j(f-L)\,\,\supseteq\,\,(\partial_jf)\,-\,(\partial_jL)$.\\
So, by the claim, $\partial_j(f-L)\,\,\supseteq\,\,(\partial_jf)\,-\,(C_V^{q_j})$.\\
Want: $(\partial_jf)\,-\,(C_V^{q_j})\in\scrlito_0^{VW}$. \qquad
Want: $(\partial_jf)\,-\,(C_V^{q_j})\in\CVZ_V^W$.\\
By assumption, $\partial_jf$ is both defined near $0_V$ and continuous at $0_V$.\\
Also, $C_V^{q_j}$ is both defined near $0_V$ and continuous at $0_V$.\\
Then $(\partial_jf)\,-\,(C_V^{q_j})$ is both defined near $0_V$ and continuous at $0_V$.\\
Want: $((\partial_jf)\,-\,(C_V^{q_j}))_{0_V}=0_W$.\\
By definition of $q$, we have $q_j=(\partial_jf)_{0_V}$.\\
We compute $((\partial_jf)\,-\,(C_V^{q_j}))_{0_V}=((\partial_jf)_{0_V})-((C_V^{q_j})_{0_V})$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\quad $\,\,=q_j-q_j=0_W$, as desired.
QED
\vskip.1in\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 11: Due on Tuesday 16 April}
\vskip.1in\hrule\vskip.1in\noindent
11-1. Let $\,$ $V:=\R^2$, $\,$ $S:=V\otimes V$, $\,$ $f:\R\drta V$, $\,$ $g:V\drta\R$.\\
Let $p\in\R$. \qquad $\,\,$
Assume \quad $f'_p=(1,2)$ \quad and \quad $f''_p=(3,4)$.\\
Let $q:=f_p$. \qquad
Assume \quad $g'_q=(5,6)$ \quad and \quad $\ds{g''_q=\left[\begin{array}{cc}7&8\\9&0\end{array}\right]}$.\\
\phantom{x}\qquad\qquad\qquad $\,\,\,\,\,$ Compute: \quad $(g\circ f)'_p$ \quad and \quad $(g\circ f)''_p$.
\vskip.1in\noindent
{\it Solution:}\\
We have $(g\circ f)'_p\,\,\,=^*\,\,\,g'_q\,\,\sbullet_V\,\,f'_p$\\
\phantom{x}\qquad\qquad\qquad $\,\,\,\,=\,\,\,(5,6)\,\,\sbullet_V\,\,(1,2)\,\,\,=\,\,\,5\cdot1\,\,+\,\,6\cdot2\,\,\,=\,\,\,17\,\,\,\ne\,\,\,\frownie$.\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\quad Then $\,\,\,\,\,(g\circ f)'_p\,\,\,=\,\,\,17$.\\
We have: $\ds{(f'_p)^{\otimes2}\quad=\quad\left[\begin{array}{cc}1\cdot1&1\cdot2\\2\cdot1&2\cdot2\end{array}\right]\quad=\quad\left[\begin{array}{cc}1&2\\2&4\end{array}\right]}$.\\
Also, $(g\circ f)''_p\quad=^*\quad g''_q\,\,\sbullet_S\,\,((f'_p)^{\otimes2})\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,g'_q\,\,\sbullet_V\,\,f''_p$\\
\phantom{x}\qquad\qquad $\ds{=\quad\left[\begin{array}{cc}7&8\\9&0\end{array}\right]\,\,\sbullet_S\,\,\left[\begin{array}{cc}1&2\\2&4\end{array}\right]\,\,\,\,+\,\,\,\,(5,6)\,\,\sbullet_V\,\,(3,4)}$\\
\phantom{x}\qquad\qquad $=\quad7\cdot1\,\,+\,\,8\cdot2\,\,+\,\,9\cdot2\,\,+\,\,0\cdot4\,\,+\,\,5\cdot3\,\,+\,\,6\cdot4$\\
\phantom{x}\qquad\qquad $=\quad7+16+18+0+15+24\,\,\,\,=\,\,\,\,80\,\,\,\,\ne\frownie$.\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\quad Then $\,\,(g\circ f)''_p\,\,\,\,=\,\,\,\,80$.\\
\vskip.1in\hrule\vskip.1in\noindent
11-2. Let \quad $V:=\R^2$, \quad $S:=V\otimes V$, \quad $\ds{A:=\left[\begin{array}{cc}1&2\\2&4\end{array}\right]}$.\\
Show: \qquad\qquad $\,\,$ $\forall q\in V$, \quad $A\,\,\sbullet_S\,\,(q^{\otimes2})\,\,\ge\,\,0$.
\vskip.1in\noindent
{\it Proof:}
Given $q\in V$. \qquad
Want: $A\,\,\sbullet_S\,\,(q^{\otimes2})\,\,\ge\,\,0$.\\
Let $x:=q_1$ and $y:=q_2$. \qquad
Then $q=(x,y)$.\\
Then $\ds{q^{\otimes2}\quad=\quad\left[\begin{array}{cc}x\cdot x&x\cdot y\\y\cdot x&y\cdot y\end{array}\right]\quad=\quad\left[\begin{array}{cc}x^2&xy\\xy&y^2\end{array}\right]}$.\\
Then \quad $\ds{A\,\,\sbullet_S\,\,(q^{\otimes2})\quad=\quad\left[\begin{array}{cc}1&2\\2&4\end{array}\right]\,\,\sbullet_S\,\,\left[\begin{array}{cc}x^2&xy\\xy&y^2\end{array}\right]}$\\
\phantom{x}\qquad\qquad\qquad\qquad\quad $\ds{\,\,\,=\quad 1\cdot x^2\,\,+\,\,2\cdot xy\,\,+\,\,2\cdot xy\,\,+\,\,4\cdot y^2}$\\
\phantom{x}\qquad\qquad\qquad\qquad\quad $\ds{\,\,\,=\quad x^2\,\,+\,\,4xy\,\,+\,\,4y^2}$\\
\phantom{x}\qquad\qquad\qquad\qquad\quad $\,\,\,=\quad(x+2y)^2\quad\ge\quad0$, \quad as desired. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
11-3. Let $V:=\R^2$, $S:=V\otimes V$, $a,b,c\in\R$, $\ds{A:=\left[\begin{array}{cc}a&b\\b&c\end{array}\right]}$, $z:=0_2$.\\
Assume: \quad\qquad $\,\,$ $a>0$ \quad and \quad $ac-b^2>0$.\\
Show: \qquad\qquad $\,\,$ $\forall q\in V_z^\times$, \quad $A\,\,\sbullet_S\,\,(q^{\otimes2})\,\,>\,\,0$.
\vskip.1in\noindent
{\it Proof:}
Given $q\in V_z^\times$. \qquad
Want: $A\,\,\sbullet_S\,\,(q^{\otimes2})\,\,>\,\,0$.\\
Let $x:=q_1$ and $y:=q_2$. \qquad
Then $q=(x,y)$.\\
Then $\ds{q^{\otimes2}\quad=\quad\left[\begin{array}{cc}x\cdot x&x\cdot y\\y\cdot x&y\cdot y\end{array}\right]\quad=\quad\left[\begin{array}{cc}x^2&xy\\xy&y^2\end{array}\right]}$.\\
We have $\ds{A\,\,\sbullet_S\,\,(q^{\otimes2})\quad=\quad\left[\begin{array}{cc}a&b\\b&c\end{array}\right]\,\,\sbullet_S\,\,\left[\begin{array}{cc}x^2&xy\\xy&y^2\end{array}\right]}$\\
\phantom{x}\qquad\qquad\qquad\qquad\quad $\ds{\,\,\,\,\,=\quad a\cdot x^2\,\,+\,\,b\cdot xy\,\,+\,\,b\cdot xy\,\,+\,\,c\cdot y^2}$\\
\phantom{x}\qquad\qquad\qquad\qquad\quad $\ds{\,\,\,\,\,=\quad ax^2\,\,+\,\,2bxy\,\,+\,\,cy^2}$\\
\phantom{x}\qquad\qquad\qquad\qquad\quad $\ds{\,\,\,\,\,=\quad a\,\,\cdot\,\,\left(\,\,x^2\,\,+\,\,\frac{2by}{a}x\,\,\right)\,\,+\,\,cy^2}$\\
\phantom{x}\qquad\qquad\qquad\qquad\quad $\ds{\,\,\,\,\,=\quad a\,\,\cdot\,\,\left(\,\,x^2\,\,+\,\,\frac{2by}{a}x\,\,+\,\,\frac{b^2y^2}{a^2}\,\,\right)}$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\ds{+\,\,\frac{acy^2}{a}\,\,-\,\,\frac{b^2y^2}{a}}$\\
\phantom{x}\qquad\qquad\qquad\qquad\quad $\ds{\,\,\,\,\,=\quad a\,\,\cdot\,\,\left(\,\,x\,\,+\,\,\frac{by}{a}\,\,\right)^2\quad\,\,\,\,+\,\,\,\,\frac{ac-b^2}{a}\,\,\cdot\,\,y^2}$.\\
Since $a>0$ and $\ds{\left(x+\frac{by}{a}\right)^2\ge0}$,\\
\phantom{x}\qquad we see that $\ds{a\,\cdot\,\left(\,x\,+\,\frac{by}{a}\,\right)^2\quad\ge\quad0}$.\\
Since $a>0$ and $ac-b^2>0$ and $y^2\ge0$,\\
\phantom{x}\qquad we see that $\ds{\frac{ac-b^2}{a}\,\cdot\,y^2\quad\ge\quad0}$.\\
Then $\ds{A\,\,\sbullet_S\,\,(q^{\otimes2})\,\,=\,\,a\,\cdot\,\left(\,x\,+\,\frac{by}{a}\,\right)^2\,+\,\frac{ac-b^2}{a}\,\cdot\,y^2\,\,\ge\,\,0}$.\\
Want: $A\,\,\sbullet_S\,\,(q^{\otimes2})\,\,\ne\,\,0$.\\
Assume: $A\,\,\sbullet_S\,\,(q^{\otimes2})\,\,=\,\,0$. \quad
Want: Contradiction.\\
We have $\ds{a\,\cdot\,\left(\,x\,+\,\frac{by}{a}\,\right)^2\,+\,\frac{ac-b^2}{a}\,\cdot\,y^2\quad=\quad A\,\sbullet_S\,(q^{\otimes2})\quad=\quad0}$.\\
Then $\ds{\frac{ac-b^2}{a}\,\cdot\,y^2\quad=\quad-\left(\,a\,\cdot\,\left(\,x\,+\,\frac{by}{a}\,\right)^2\,\right)\quad\le0}$.\\
So, since $\ds{\frac{ac-b^2}{a}\,\cdot\,y^2\quad\ge\quad0}$, we see that $\ds{\frac{ac-b^2}{a}\,\cdot\,y^2\quad=\quad0}$.\\
So, since $ac-b^2>0$ and $a>0$, we see that $y^2=0$, and so $y=0$.\\
Then $\ds{A\,\,\sbullet_S\,\,(q^{\otimes2})\,\,=\,\,a\,\cdot\,\left(\,x\,+\,\frac{b\cdot0}{a}\,\right)^2\,+\,\frac{ac-b^2}{a}\,\cdot\,0^2\,\,=\,\,ax^2}$.\\
Then $ax^2\,\,=\,\,A\,\,\sbullet_S\,\,(q^{\otimes2})\,\,=\,\,0$, \qquad so $ax^2\,\,=\,\,0$.\\
So, since $a>0$, we get $x^2=0$, \qquad and so $x=0$.\\
Since $q\in V_z^\times$, \qquad we conclude that $q\ne z$.\\
Then $0_2=z\ne q=(x,y)=(0,0)=0_2$, \qquad so $0_2\ne0_2$.\\
Contradiction. \qquad\qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
11-4. Let $\,$ $V:=\R^2$, $\,$ $S:=V\otimes V$, $\,$ $f:\R\drta V$, $\,$ $g:V\drta\R$.\\
Let $p\in\R$. \qquad $\,\,$
Assume \quad $f'_p\ne(0,0)$ \quad and \quad $f''_p\ne\frownie$.\\
Let $q:=f_p$. \qquad
Assume \quad $g'_q=(0,0)$ \quad and \quad $g''_q=\left[\begin{array}{cc}1&2\\2&5\end{array}\right]$.\\
Show: \qquad\qquad\qquad\qquad\qquad\quad $(g\circ f)''_p\,\,>\,\,0$.
\vskip.1in\noindent
{\it Proof:}
We have: $(g\circ f)''_p\,\,=^*\,\,g''_q\,\sbullet_S\,((f'_p)^{\otimes2})\,\,+\,\,g'_q\,\sbullet_V\,f''_p\,\,\ne\,\,\frownie$,\\
\phantom{x}\qquad\qquad\quad so $\,\,\,\,(g\circ f)''_p\,\,=\,\,\,\,g''_q\,\sbullet_S\,((f'_p)^{\otimes2})\,\,+\,\,g'_q\,\sbullet_V\,f''_p$.\\
So, since $g'_q=(0,0)$, we get $(g\circ f)''_p\,\,=\,\,g''_q\,\sbullet_S\,((f'_p)^{\otimes2})$.\\
Let $a:=1$, $b:=2$, $c:=5$, $\ds{A:=\left[\begin{array}{cc}a&b\\b&c\end{array}\right]}$, $z:=0_2$.\\
We have $A=\left[\begin{array}{cc}1&2\\2&5\end{array}\right]=g''_q$ \quad and \quad $(g\circ f)''_p\,\,=\,\,g''_q\,\sbullet_S\,((f'_p)^{\otimes2})$.\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad Then \quad $\,\,(g\circ f)''_p\,\,=\,\,A\,\,\sbullet_S\,((f'_p)^{\otimes2})$.\\
We have $a>0$ and $ac-b^2=1\cdot5-2^2=1>0$.\\
Since $f''_p\ne\frownie$, we get $p\in\dom[f'']\subseteq\dom[f']$, and so $f'_p\in\im[f']$.\\
Then $f'_p\in\im[f']\subseteq V\otimes\R=V$.\\
So, since $f'_p\ne(0,0)=0_2=z$, we conclude that $f'_p\in V_z^\times$.\\
Then, by HW\#11-3, we have $A\,\,\sbullet_S\,\,((f'_p)^{\otimes2})\,\,>\,\,0$.\\
Then: \qquad $(g\circ f)''_p\,\,=\,\,A\,\sbullet_S\,((f'_p)^{\otimes2})\,\,>\,\,0$, \qquad as desired. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
11-5. Let $V,W\in\TNSR^+$, $f:V\drta W$, $i\in\scri_V$, $j\in\scri_W$, $W':=W\otimes V$.\\
Show: \qquad\qquad $(\pi_{j\|i}^{W'})\,\,\circ\,\,(f')\quad\subseteq\quad(\pi_j^W)\,\,\circ\,\,(\partial_if)$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall x\in V$, $((\pi_{j\|i}^{W'})\,\,\circ\,\,(f'))_x\,\,\,\,{}^*\!\!=\,\,((\pi_j^W)\,\,\circ\,\,(\partial_if))_x$.\\
Given $x\in V$. \qquad
Want: $((\pi_{j\|i}^{W'})\,\,\circ\,\,(f'))_x\,\,\,\,{}^*\!\!=\,\,((\pi_j^W)\,\,\circ\,\,(\partial_if))_x$.\\
Want: $(\pi_{j\|i}^{W'})(f'_x)\,\,\,\,{}^*\!\!=\,\,(\pi_j^W)((\partial_if)_x)$.\\
Want: $[\,\,(\pi_{j\|i}^{W'})(f'_x)\,\,\ne\frownie\,\,]\,\,\Rightarrow\,\,[\,\,(\pi_{j\|i}^{W'})(f'_x)\,\,=\,\,(\pi_j^W)((\partial_if)_x)\,\,]$.\\
Assume $(\pi_{j\|i}^{W'})(f'_x)\,\,\ne\frownie$. \qquad
Want: $(\pi_{j\|i}^{W'})(f'_x)\,\,=\,\,(\pi_j^W)((\partial_if)_x)$.\\
Let $A:=f'_x$. \qquad\qquad\qquad\quad $\,$
Want: $(\pi_{j\|i}^{W'})(A)\,\,=\,\,(\pi_j^W)((\partial_if)_x)$.\\
Since $(\pi_{j\|i}^{W'})(f'_x)\,\,\ne\frownie$, we conclude that $f'_x\ne\frownie$, and so $f'_x\in\im[f']$.\\
Then $A=f'_x\in\im[f']\subseteq W\otimes V$.\\
By HW\#9-2, $(f'_x)\,\,*_{\R VW}\,\,\varepsilon_i^V\,\,\,\,{}^*\!\!=\,\,(\partial_{\varepsilon_i}f)_x$.\\
So, since $A=f'_x$ and $\partial_{\varepsilon_i}f=\partial_if$, we get $A\,\,*_{\R VW}\,\,\varepsilon_i^V\,\,\,\,{}^*\!\!=\,\,(\partial_if)_x$.\\
Since $A\in W\otimes V$ and $\varepsilon_i^V\in V$, we get $A\,\,*_{\R VW}\,\,\varepsilon_i^V\,\,\in W$.\\
In particular, $A\,\,*_{\R VW}\,\,\varepsilon_i^V\,\,\ne\,\,\frownie$.\\
Then $\frownie\,\,\ne\,\,A\,\,*_{\R VW}\,\,\varepsilon_i^V\,\,\,\,{}^*\!\!=\,\,(\partial_if)_x$,
so $A\,\,*_{\R VW}\,\,\varepsilon_i^V\,\,=\,\,(\partial_if)_x$.\\
Then $(\pi_j^W)(A\,\,*_{\R VW}\,\,\varepsilon_i^V)\,\,=\,\,(\pi_j^W)((\partial_if)_x)$.\\
We have $(A\,\,*_{\R VW}\,\,\varepsilon_i^V)_j\,\,=\,\,(A\,\,*_{\R VW}\,\,\varepsilon_i^V)_{j\|\emptyset}$\\
\phantom{x}\qquad\qquad\quad $\ds{=\,\,\sum_{h\in\scri_V}\,(A_{j\|h})\,\,\cdot\,\,((\varepsilon_i^V)_{h\|\emptyset})}$\\
\phantom{x}\qquad\qquad\quad $\ds{=\,\,\sum_{h\in\scri_V}\,(A_{j\|h})\,\,\cdot\,\,((\varepsilon_i^V)_h)}$\\
\phantom{x}\qquad\qquad\quad $\ds{=\,\,\left(\,\,\sum_{h\in\scri_V\backslash\{i\}}\,(A_{j\|h})\,\,\cdot\,\,((\varepsilon_i^V)_h)\,\,\right)\,\,+\,\,\left(\,\,(A_{j\|i})\,\,\cdot\,\,((\varepsilon_i^V)_i\,\,\right)}$\\
\phantom{x}\qquad\qquad\quad $\ds{=\,\,\left(\,\,\sum_{h\in\scri_V\backslash\{i\}}\,(A_{j\|h})\,\,\cdot\,\,0\,\,\right)\,\,+\,\,\left(\,\,(A_{j\|i})\,\,\cdot\,\,1\,\,\right)}$\\
\phantom{x}\qquad\qquad\quad $=\,\,0\,\,+\,\,A_{j\|i}\,\,=\,\,A_{j\|i}$.\\
Then $(\pi_{j\|i}^{W'})(A)\,\,=\,\,A_{j\|i}\,\,=\,\,(A\,\,*_{\R VW}\,\,\varepsilon_i^V)_j$\\
\phantom{x}\qquad\qquad $=\,\,(\pi_j^W)(A\,\,*_{\R VW}\,\,\varepsilon_i^V)\,\,=\,\,(\pi_j^W)((\partial_if)_x)$, $\,\,$ as desired. \quad
QED
\vskip.1in\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 10: Due on Tuesday 9 April}
\vskip.1in\hrule\vskip.1in\noindent
10-1. Let $V,W,X\in\TNSR^+$, \quad $f:V\drta W$, \quad $g:W\drta X$, \quad $u\in V$.\\
Show: \qquad $\forall p\in V$, \quad $(\partial_u^{VX}(g\circ f))_p\,\,\,\,=^*\,\,\,\,(g'_{f_p})\,\,*_{\R VW}\,\,((\partial_u^{VW}f)_p)$.
\vskip.1in\noindent
{\it Proof:}
Given $p\in V$.
Want: $(\partial_u^{VX}(g\circ f))_p\,\,\,\,=^*\,\,\,\,(g'_{f_p})\,\,*_{\R VW}\,\,((\partial_u^{VW}f)_p)$.\\
Let \quad $i:=i_p^v$. \quad
Then $i_0=p+0\cdot v=p$. \quad
Also, $(\partial_u^{VW}f)_p=(f\circ i)'_0$.\\
Let $\phi:=f\circ i$. \quad
Then $\phi_0=(f\circ i)_0=f_{i_0}=f_p$. \quad
Also, $(\partial_u^{VW}f)_p=\phi'_0$.\\
By the Chain Rule, \quad $(g\circ\phi)'_0\,\,\,\,=^*\,\,\,\,(g'_{\phi_0})\,\,*_{\R VW}\,\,(\phi'_0)$.\\
Then $(\partial_u^{VX}(g\circ f))_p\,\,=\,\,(g\circ f\circ i)'_0\,\,=\,\,(g\circ\phi)'_0$\\
\phantom{x}\qquad\qquad $=^*\,\,\,\,(g'_{\phi_0})\,\,*_{\R VW}\,\,(\phi'_0)\,\,=\,\,(g'_{f_p})\,\,*_{\R VW}\,\,((\partial_u^{VW}f)_p)$. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
10-2. Let $S,V,W,Z\in\TNSR^+$, \quad $*\in\scrb_{VW}^Z$.\\
Let $f:S\drta V$, \quad $g:S\drta W$, \quad $u\in S$.\\
Show: \quad $\forall p\in S$, $\,\,$ $(\partial_u^{SZ}(f*g))_p\,\,=^*\,\,(\,((\partial^{SV}_uf)_p)*g_p\,)\,\,+\,\,(\,f_p*((\partial^{SW}_ug)_p)\,)$.
\vskip.1in\noindent
{\it Proof:}
Given $p\in V$.\\
Want: \quad $(\partial_u^{SZ}(f*g))_p\,\,=^*\,\,(\,((\partial^{SV}_uf)_p)*g_p\,)\,\,+\,\,(\,f_p*((\partial^{SW}_ug)_p)\,)$.\\
Let \quad $i:=i_p^u$. \quad
Then $i_0=p+0\cdot u=p$.\\
Also, \quad $(\partial_u^{SV}f)_p=(f\circ i)'_0$ \quad and \quad $(\partial_u^{SW}g)_p=(g\circ i)'_0$.\\
Let \quad $\phi:=f\circ i$ \quad and \quad $\psi:=g\circ i$ \quad and \quad $\chi:=(f*g)\circ i$.\\
Then $\,\,$ $(\partial_u^{SV}f)_p=\phi'_0$ $\,\,$ and $\,\,$ $(\partial_u^{SW}g)_p=\psi'_0$ $\,\,$ and $\,\,$ $(\partial_u^{SZ}(f*g))_p=\chi'_0$.\\
Also, \quad $\phi_0=f_{i_0}=f_p$ \quad and \quad $\psi_0=g_{i_0}=g_p$.\\
Want: \quad $\chi'_0\,\,=^*\,\,\phi'_0\,*\,\chi_0\,\,+\,\,\phi_0\,*\,\psi'_0$.\\
\\
{\it Claim:} $\chi\,\,=\,\,\phi*\psi$.\\
{\it Proof of Claim:}\\
Want: $\forall t\in\R$, $\chi_t\,\,=\,\,(\phi*\psi)_t$. \quad
Given $t\in\R$. \quad
Want: $\chi_t\,\,=\,\,(\phi*\psi)_t$.\\
Let $q:=i_t$. \qquad
Then \quad $\phi_t=(f\circ i)_t=f_q$, \quad $\psi_t=(g\circ i)_t=g_q$ \quad and\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad $\chi_t=((f*g)\circ i)_t=(f*g)_q$.\\
Then $\chi_t\,\,=\,\,(f*g)_q\,\,=\,\,f_q*g_q\,\,=\,\,\phi_t*\psi_t\,\,=\,\,(\phi*\psi)_t$, as desired.\\
{\it End of proof of Claim.}\\
\\
By the Claim and by the Product Rule, we have:\\
\phantom{x}\qquad $D_0^{\R Z}\chi\qquad=^*\qquad(D_0^{\R V}\phi)\,\,*\,\,\psi_0\quad+\quad\phi_0\,\,*(D_0^{\R W}\psi)$.\\
Then: $(D_0^{\R Z}\chi)(1)\,=^*\,((D_0^{\R V}\phi)(1))\,*\,\psi_0\,+\,\phi_0\,*\,((D_0^{\R Z}\psi)(1))$.\\
So, \qquad since $\chi'_0=\sl_\bullet^{\R Z}(D_0^{\R Z}\chi)=(D_0^{\R Z}\chi)(1)$,\\
\phantom{x}\quad and since $\phi'_0=\sl_\bullet^{\R V}(D_0^{\R V}\phi)=(D_0^{\R V}\phi)(1)$,\\
\phantom{x}\quad and since $\psi'_0=\sl_\bullet^{\R W}(D_0^{\R W}\psi)=(D_0^{\R W}\psi)(1)$,\\
we see that: $\chi'_0\,\,=^*\,\,\phi'_0\,*\,\chi_0\,\,+\,\,\phi_0\,*\,\psi'_0$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
10-3. Let $V,W,X\in\TNSR^+$, \quad $A\in X\otimes V\otimes W$, \quad $y\in V$, \quad $z\in W$.\\
Show: \qquad $(A\,\,*_{\R,W,X\otimes V}\,\,z)\,\,*_{\R V\!X}\,\,y\quad=\quad A\,\,*_{\R,V\otimes W,X}\,\,(y\otimes z)$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall k\in\scri_X$,\\
\phantom{x}\qquad\qquad $((A\,\,*_{\R,W,X\otimes V}\,\,z)\,\,*_{\R V\!X}\,\,y)_k\quad=\quad(A\,\,*_{\R,V\otimes W,X}\,\,(y\otimes z))_k$.
Given $k\in\scri_X$.\\
Want: \qquad $((A\,\,*_{\R,W,X\otimes V}\,\,z)\,\,*_{\R V\!X}\,\,y)_k\quad=\quad(A\,\,*_{\R,V\otimes W,X}\,\,(y\otimes z))_k$.
We compute:\\
\phantom{x}\quad$\ds{(A\,\,*_{\R,V\otimes W,X}\,\,(y\otimes z))_k\quad=\quad(A\,\,*_{\R,V\otimes W,X}\,\,(y\otimes z))_{k\|\emptyset}}$\\
\phantom{x}\qquad\qquad$\ds{=\quad\sum_{\ell\in\scri_{V\otimes W}}\,A_{k\|\ell}\,\,\,\cdot\,\,\,(y\otimes z)_{\ell\|\emptyset}}$\\
\phantom{x}\qquad\qquad$\ds{=\quad\sum_{\ell\in\scri_{V\otimes W}}\,A_{k\|\ell}\,\,\,\cdot\,\,\,(y\otimes z)_\ell}$\\
\phantom{x}\qquad\qquad$\ds{=\quad\sum_{i\in\scri_V}\,\,\sum_{j\in\scri_W}\quad(\,\,\,A_{k\|i\|j}\,\,\,\cdot\,\,\,(y\otimes z)_{i\|j}\,\,\,)}$\\
\phantom{x}\qquad\qquad$\ds{=\quad\sum_{i\in\scri_V}\,\,\sum_{j\in\scri_W}\quad(\,\,\,A_{k\|i\|j}\,\,\,\cdot\,\,\,(y\,\,*_{V\R W}\,\,z)_{i\|j}\,\,\,)}$\\
\phantom{x}\qquad\qquad$\ds{=\quad\sum_{i\in\scri_V}\,\,\sum_{j\in\scri_W}\quad\left(\,\,\,A_{k\|i\|j}\,\,\,\cdot\,\,\,\left(\,\,\,\sum_{m\in\scri_\R}\quad y_{i\|m}\,\,\,\cdot\,\,\,z_{m\|j}\,\,\,\right)\,\,\,\right)}$\\
\phantom{x}\qquad\qquad$\ds{=\quad\sum_{i\in\scri_V}\,\,\sum_{j\in\scri_W}\quad(\,\,\,A_{k\|i\|j}\,\,\,\cdot\,\,\,y_{i\|\emptyset}\,\,\,\cdot\,\,\,z_{\emptyset\|j}\,\,\,)}$\\
\phantom{x}\qquad\qquad$\ds{=\quad\sum_{i\in\scri_V}\,\,\sum_{j\in\scri_W}\quad(\,\,\,A_{k\|i\|j}\,\,\,\cdot\,\,\,y_i\,\,\,\cdot\,\,\,z_j\,\,\,)}$.\\
Then:\\
\phantom{x}\quad$((A\,\,*_{\R,W,X\otimes V}\,\,z)\,\,*_{\R V\!X}\,\,y)_k\quad=\quad((A\,\,*_{\R,W,X\otimes V}\,\,z)\,\,*_{\R V\!X}\,\,y)_{k\|\emptyset}$\\
\phantom{x}\qquad\qquad\qquad$\ds{=\quad\sum_{i\in\scri_V}\quad(A\,\,*_{\R,W,X\otimes V}\,\,z)_{k\|i}\,\,\,\cdot\,\,\,y_{i\|\emptyset}}$\\
\phantom{x}\qquad\qquad\qquad$\ds{=\quad\sum_{i\in\scri_V}\quad(A\,\,*_{\R,W,X\otimes V}\,\,z)_{k\|i}\,\,\,\cdot\,\,\,y_i}$\\
\phantom{x}\qquad\qquad\qquad$\ds{=\quad\sum_{i\in\scri_V}\quad(A\,\,*_{\R,W,X\otimes V}\,\,z)_{k\|i\|\emptyset}\,\,\,\cdot\,\,\,y_i}$\\
\phantom{x}\qquad\qquad\qquad$\ds{=\quad\sum_{i\in\scri_V}\quad\left(\,\,\sum_{j\in\scri_W}\,A_{k\|i\|j}\,\,\,\cdot\,\,\,z_{j\|\emptyset}\,\,\right)\,\,\,\cdot\,\,\,y_i}$\\
\phantom{x}\qquad\qquad\qquad$\ds{=\quad\sum_{i\in\scri_V}\quad\left(\,\,\sum_{j\in\scri_W}\,A_{k\|i\|j}\,\,\,\cdot\,\,\,z_j\,\,\right)\,\,\,\cdot\,\,\,y_i}$\\
\phantom{x}\qquad\qquad\qquad$\ds{=\quad\sum_{i\in\scri_V}\,\,\sum_{j\in\scri_W}\quad(\,\,\,A_{k\|i\|j}\,\,\,\cdot\,\,\,y_i\,\,\,\cdot\,\,\,z_j\,\,\,)}$\\
\phantom{x}\qquad\qquad\qquad$=\quad(A\,\,*_{\R,V\otimes W,X}\,\,(y\otimes z))_k$, \qquad as desired. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
10-4. Let $V,W,X\in\TNSR^+$, $\,\,$ $f:V\drta W$, $\,\,$ $g:W\drta X$, $\,\,$ $u,v\in V$.\\
Show: \quad $\partial_v^{VX}\partial_u^{VX}(g\circ f)\,\,\,\,\supseteq\,\,\,\,(g''\circ f)\,\,\,\,*_{\R,W\otimes W,X}\,\,\,\,((\partial_u^{VW}f)\otimes(\partial_v^{VW}f))$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad $+$ \quad $(g'\circ f)\,\,\,\,*_{\R W\!X}\,\,\,\,(\partial_v^{VW}\partial_u^{VW}f)$.
\vskip.1in\noindent
{\it Proof:}
Let $\chi:=\partial_v^{VX}\partial_u^{VX}(g\circ f)$.\\
Let $\phi:=(g''\circ f)\,\,\,\,*_{\R,W\otimes W,X}\,\,\,\,((\partial_u^{VW}f)\otimes(\partial_v^{VW}f))$.\\
Let $\psi:=(g'\circ f)\,\,\,\,*_{\R W\!X}\,\,\,\,(\partial_v^{VW}\partial_u^{VW}f)$.\\
Want: $\chi\supseteq\phi+\psi$.\\
By HW\#10-1, \qquad $\partial_u^{VX}(g\circ f)\,\,\,\,\supseteq\,\,\,\,(g'\circ f)\,\,*_{\R WX}\,\,(\partial_u^{VW}f)$.\\
Then \qquad $\partial_v^{VX}\partial_u^{VX}(g\circ f)\,\,\,\,\supseteq\,\,\,\,\partial_v^{VX}(\,\,(g'\circ f)\,\,*_{\R WX}\,\,(\partial_u^{VW}f)\,\,)$.\\
Recall: $\chi=\partial_v^{VX}\partial_u^{VX}(g\circ f)$. \qquad
Let $\alpha:=g'\circ f$ and $\kappa:=\partial_u^{VW}f$.\\
Then \qquad $\chi\,\,\,\,\supseteq\,\,\,\,\partial_v^{VX}(\,\,\alpha\,\,*_{\R WX}\,\,\kappa\,\,)$.\\
Let $*:=*_{\R WX}$. \qquad
Then \qquad $\chi\,\,\,\,\supseteq\,\,\,\,\partial_v^{VX}(\,\,\alpha*\kappa\,\,)$.\\
Want: \qquad $\partial_v^{VX}(\,\,\alpha*\kappa\,\,)\,\,\,\,\supseteq\,\,\,\,\phi+\psi$.\\
Let $Y:=X\otimes W$. \qquad
Then $*\in\scrb_{YW}^X$.\\
By HW\#10-2, \quad $\partial_v^{VX}(\,\,\alpha*\kappa\,\,)\,\,\,\,\supseteq\,\,\,\,(\,(\partial_v^{VY}\alpha)*\kappa\,)\,+\,(\,\alpha*(\partial_v^{VW}\kappa)\,)$.\\
Want: \qquad $(\partial_v^{VY}\alpha)*\kappa\,\,\supseteq\,\,\phi$ \qquad and \qquad $\alpha*(\partial_v^{VW}\kappa)\,\,=\,\,\psi$.\\
We have \quad $\alpha*(\partial_v^{VW}\kappa)\,\,\,\,=\,\,\,\,(g'\circ f)\,\,*_{\R WX}\,\,(\partial_v^{VW}\partial_u^{VW}f)\,\,\,\,=\,\,\,\,\psi$.\\
Want: \qquad $(\partial_v^{VY}\alpha)*\kappa\,\,\supseteq\,\,\phi$.\\
Want: \qquad $\forall q\in V$, \quad $((\partial_v^{VY}\alpha)*\kappa)_q\,\,=^*\,\,\phi_q$.\\
Given $q\in V$. \qquad
Want: $((\partial_v^{VY}\alpha)*\kappa)_q\,\,=^*\,\,\phi_q$.\\
By HW\#10-1, \qquad $\partial_u^{VY}(g'\circ f)\,\,\,\,\supseteq\,\,\,\,(g''\circ f)\,\,*_{\R WY}\,\,(\partial_u^{VW}f)$.\\
Then $(\partial_v^{VY}\alpha)_q\,\,=\,\,(\partial_v^{VY}(g'\circ f))_q\,\,=^*\,\,((g''\circ f)\,\,*_{\R WY}\,\,(\partial_u^{VW}f))_q$.\\
Let \quad $A:=(g''\circ f)_q$, \quad $y:=(\partial_u^{VW}f)_q$, \quad $z:=(\partial_v^{VW}f)_q$.\\
Then $y=(\partial_u^{VW}f)_q=\kappa_q$.\\
Also, $\phi_q\,\,=\,\,((g''\circ f)\,\,\,\,*_{\R,W\otimes W,X}\,\,\,\,((\partial_u^{VW}f)\otimes(\partial_v^{VW}f)))_q$\\
\phantom{x}\qquad\qquad\qquad $=\,\,\,\,A\,\,\,\,*_{\R,W\otimes W,X}\,\,\,\,(y\otimes z)$\\
By HW\#10-3 (with $V$ replaced by $W$),\\
\phantom{x}\qquad $(A\,\,*_{\R,W,X\otimes W}\,\,z)\,\,*_{\R VW}\,\,y\quad=\quad A\,\,*_{\R,W\otimes W,X}\,\,(y\otimes z)$.\\
Then $((\partial_v^{VY}\alpha)*\kappa)_q\,\,=\,\,\,\,(\partial_v^{VY}\alpha)_q\,\,*\,\,\kappa_q\,\,=\,\,(\partial_v^{VY}\alpha)_q\,\,*\,\,y$\\
\phantom{x}\qquad\qquad\qquad\qquad $\,\,\,=^*\,\,((g''\circ f)\,\,*_{\R WY}\,\,(\partial_u^{VW}f))_q\,\,*\,\,y$\\
\phantom{x}\qquad\qquad\qquad\qquad $\,\,\,=\,\,\,\,((g''\circ f)_q\,\,*_{\R WY}\,\,(\partial_u^{VW}f)_q)\,\,*_{\R WX}\,\,y$.\\
\phantom{x}\qquad\qquad\qquad\qquad $\,\,\,=\,\,\,\,(A\,\,*_{\R,W,X\otimes W}\,\,z)\,\,*_{\R WX}\,\,y$.\\
\phantom{x}\qquad\qquad\qquad\qquad $\,\,\,=\,\,\,\,A\,\,*_{\R,W\otimes W,X}\,\,(y\otimes z)\,\,\,=\,\,\,\,\phi_q$. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
10-5. Let $V:=\R^2$, \quad $W:=\R$, \quad $g:V\drta W$, \quad $z:=0_2$, \quad $i:=i_z^{\varepsilon_1^V}$.\\
Assume: $(g_z=0)\,\,\&\,\,((\partial_1^{VW}g)_z=0)$. \qquad
Show: $g\circ i\in\scrlito_1^{WW}$.
\vskip.1in\noindent
{\it Proof:}
We have $i_0=z+0\cdot\varepsilon_1^V=z$ and $(\partial_1^{VW}g)_z=(\partial_{\varepsilon_1^V}^{VW}g)_z=(g\circ i)'_0$.\\
Then \quad $(g\circ i)_0=g_{i_0}=g_z=0$ \quad and \quad $(g\circ i)'_0=(\partial_1^{VW}g)_z=0$.\\
Since \quad $(g\circ i)_0=0$ \quad and \quad $(g\circ i)'_0=0$, \qquad it follows,\\
\phantom{x}\qquad from HW\#8-1 (with $S$ and $T$ replaced by $W$, and $R$ by $g\circ i$),\\
that: \qquad $g\circ i\in\scrlito_1^{WW}$, \qquad as desired. \qquad
QED
\vskip.1in\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 9: Due on Tuesday 2 April}
\vskip.1in\hrule\vskip.1in\noindent
9-1. Let $V,W\in\TNSR^+$, \quad $\alpha:V\drta W$, \quad $\delta>0$, \quad $B:=B_V(0_V,\delta)$.\\
\phantom{xx}\quad Assume: $B\subseteq\dom[\alpha]$. \qquad\qquad
Assume: $\forall q\in B$, $|\alpha_q|_W\le|q|_V$.\\
\phantom{x}\qquad\qquad\qquad\qquad Show: $\alpha\in\scrbigo_1^{VW}$.
\vskip.1in\noindent
{\it Proof:}
Let $z:=0_V$.\\
Since $z\in B_V(z,\delta)=B_V(0_V,\delta)=B$, by assumption, $|\alpha_z|_W\le|z|_V$.\\
Then $|\alpha_z|_W\le|z|_V=|0_V|_V=0$, so $|\alpha_z|_W\le0$.\\
Then $0\le|\alpha_z|_W\le0$, so $|\alpha_z|_W=0$, so $\alpha_z=0_W$.\\
Then $\alpha_{0_V}=\alpha_z=0_W$, so $\alpha_{0_V}=0_W$. \qquad
Let $\ds{\sigma:=\adj_{0_V}^{0_W}\left(\frac{\alpha}{|\bullet|_V}\right)}$.\\
As $\alpha_{0_V}=0_W$, we get: $(\,\,\dom[\alpha]=\dom[\sigma]\,\,)\,\,\&\,\,(\,\,\alpha=\sigma\cdot(|\bullet|_V)\,\,)$.\\
Want: $\alpha\in(\BNZ_V^W)\cdot(|\bullet|_V)$. \qquad
Want: $\sigma\in\BNZ_V^W$.\\
We have $B\in\dom[\alpha]=\dom[\sigma]$ and $B=B(0_V,\delta)\in\scrb_V(0_V)$.\\
Then $\sigma\in\DNZ_V^W$. \qquad
By definition of $\sigma$, we have $\sigma_{0_V}=0_W$.\\
Want: $\sigma_*(B)$ is bounded in $W$. \qquad\qquad
Let $C:=B_W(0_W,2)$.\\
Since $C\in\scrb_W$, it suffices to show: $\sigma_*(B)\subseteq C$.\\
Want: $\forall q\in B\cap(\dom[\sigma])$, $\sigma_q\in C$.\\
Given $q\in B\cap(\dom[\sigma])$. \qquad
Want: $\sigma_q\in C$.\\
Exactly one of the following is true:\\
\phantom{x}\qquad (1) $q=0_V$ \qquad\qquad or \qquad\qquad (2) $q\ne0_V$.\\
\\
{\it Case (1):}\\
We have $\sigma_q=\sigma_{0_V}=0_W\in B_W(0,2)=C$, as desired.\\
{\it End of Case (1).}\\
\\
{\it Case (2):}\\
By definition of $\sigma$, $\ds{\sigma_q=\frac{\alpha_q}{|q|_V}}$. \qquad
Then $\ds{|\sigma_q|_W=\frac{|\alpha_q|_W}{|q|_V}}$.\\
Since $q\in B\cap(\dom[\sigma])\subseteq B$, by assumption, $|\alpha_q|_W\le|q|_V$.\\
Then $\ds{|\sigma_q-0_W|_W=|\sigma_q|_W=\frac{|\alpha_q|_W}{|q|_V}\le1<2}$, so $\sigma_q\in B(0_W,2)$.\\
Then $\sigma_q\in B(0_W,1)=C$, as desired.
{\it End of Case (2).} \qquad\qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
9-2. Let $S,T\in\TNSR^+$, \quad $f:S\drta T$, \quad $x,v\in S$.\\
\phantom{xx}\quad Show: \qquad$(\partial_v^{ST}f)_x\quad=^*\quad(f'_x)\,\,*_{\R ST}\,\,v$.
\vskip.1in\noindent
{\it Proof:}
Want: $(\,\,(f'_x)\,\,*_{\R ST}\,\,v\ne\frownie\,\,)\,\,\Rightarrow\,\,(\,\,(\partial_v^{ST}f)_x\,\,=\,\,(f'_x)\,\,*_{\R ST}\,\,v\,\,)$.
Assume: $(f'_x)\,\,*_{\R ST}\,\,v\ne\frownie$. \qquad
Want: $(\partial_v^{ST}f)_x\,\,=\,\,(f'_x)\,\,*_{\R ST}\,\,v$.\\
Since $(f'_x)\,\,*_{\R ST}\,\,v\ne\frownie$, we see that $f'_x\ne\frownie$.\\
Then $\sl_\bullet^{ST}(D_x^{ST}f)=f'_x\ne\frownie$, so $D_x^{ST}f\ne\frownie$, so $D_x^{ST}f\in\scrl_S^T$.\\
Let $L:=D_x^{ST}f$. \qquad
Then $L\in\scrl_S^T$.\\
We have $\sl_L^{ST}=\sl_\bullet^{ST}(L)=\sl_\bullet^{ST}(D_x^{ST}f)=f'_x$.\\
Since $\sl_L^{ST}=f'_x$, by Theorem 85.6, we get $L_v\,\,=\,\,(f'_x)\,\,*_{\R ST}\,\,v$.\\
By HW\#8-4, $(\partial_v^{ST}f)_x\,\,=^*\,\,(D_x^{ST}f)_v$.\\
Then $(\partial_v^{ST}f)_x\,\,=^*\,\,(D_x^{ST}f)_v\,\,=\,\,L_v\,\,=(f'_x)\,\,*_{\R ST}\,\,v\,\,\ne\frownie$.\\
Then $(\partial_v^{ST}f)_x\,\,=\,\,(f'_x)\,\,*_{\R ST}\,\,v$, as desired. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
9-3. Let $V:=\R^2$, \quad $W:=\R$, \quad $f:V\drta W$, \quad $z:=0_2$.\\
\phantom{xx}\quad Assume: \qquad $(f'\hbox{ is defined near }z)\quad\&\quad(f_z=0)$.\\
\phantom{xx}\quad Let $g:=\partial_1^{VW}f$, \quad $h:=\partial_2^{VW}f$, \quad $i:=i_z^{\varepsilon_1^V}$.\\
\phantom{xx}\quad Show: $\exists\sigma\in\scrbigo_1^{WW}$, $\exists\tau\in\scrbigo_1^{VV}$ s.t.\\
\phantom{xx}\qquad\qquad $f\,\,=\,\,(g\circ i\circ\sigma\circ\pi_1^V)\cdot\pi_1^V\,\,+\,\,(h\circ\tau)\cdot\pi_2^V$ \qquad near $z$.
\vskip.1in\noindent
{\it Proof:}
Since $f'$ is defined near $z$, choose $C\in\scrb_V(z)$ s.t.~$C\subseteq\dom[f']$.\\
Since $C\in\scrb_V(z)$, choose $r>0$ s.t.~$C=B_V(z,r)$.\\
Let $\delta:=r/\sqrt{2}$ and $J:=(-\delta;\delta)$. \qquad
Then $J^2\subseteq B_V(z,r)$.\\
Then $J^2\subseteq B_V(z,r)=C\subseteq\dom[f]$.\\
Define $S:J\to2^\R$ and $T:J^2\to2^\R$ by\\
\phantom{x}\qquad $S_x=\{\alpha\in[0|x]~\hbox{s.t.}~f|_{(0,0)}^{(x,0)}\,\,=\,\,((\partial_1^{VW}f)(\alpha,0))\cdot x\}$ \qquad\qquad and\\
\phantom{x}\qquad $T_x=\{\beta\in[0|y]~\hbox{s.t.}~f|_{(x,0)}^{(x,y)}\,\,=\,\,((\partial_2^{VW}f)(x,\beta))\cdot y\}$.\\

\noindent
{\it Claim 1:} $\forall x\in J$, we have $\emptyset\ne S_x\subseteq J$.\\
{\it Proof of Claim 1:}\\
Given $x\in J$. \qquad
Want: $\emptyset\ne S_x\subseteq J$.\\
By HW\#8-5, we have $\emptyset\ne S_x$. \qquad
Want: $S_x\subseteq J$.\\
By definition of $S_x$, we have $S_x\subseteq[0|x]$.\\
We have $0\in(-\delta;\delta)=J$ and $x\in J$.\\
So, since $J$ is an interval, we get $[0|x]\subseteq J$.\\
Then $S_x\subseteq[0|x]\subseteq J$, as desired.\\
{\it End of proof of Claim 1.}\\

\noindent
{\it Claim 2:} $\forall x,y\in J$, we have $\emptyset\ne T_{(x,y)}\subseteq J$.\\
{\it Proof of Claim 2:}\\
Given $x,y\in J$. \qquad
Want: $\emptyset\ne T_{(x,y)}\subseteq J$.\\
By Theorem 90.13, p.~218, we have $\emptyset\ne T_{(x,y)}$. \qquad
Want: $T_{(x,y)}\subseteq J$.\\
By definition of $T_{(x,y)}$, we have $T_{(x,y)}\subseteq[0|y]$.\\
We have $0\in(-\delta;\delta)=J$ and $y\in J$.\\
So, since $J$ is an interval, we get $[0|y]\subseteq J$.\\
Then $T_{(x,y)}\subseteq[0|y]\subseteq J$, as desired.\\
{\it End of proof of Claim 2.}\\

\noindent
Define $\sigma:J\drta\R$ by $\sigma_x=\CH(S_x)$.\\
By Claim 1, \qquad $\forall x\in J$, we have $\emptyset\ne S_x\subseteq J$.\\
Then \quad $\sigma:J\to J$ \quad and, \quad $\forall x\in J$, $\sigma_x\in S_x\subseteq J$.\\
Define $\mu:J^2\drta\R$ by $\mu_{(x,y)}=\CH(T_{(x,y)})$.\\
By Claim 2, \qquad $\forall x,y\in J$, we have $\emptyset\ne T_{(x,y)}\subseteq J$.\\
Then \quad $\mu:J^2\to J$ \quad and, \quad $\forall x,y\in J$, $\mu_{(x,y)}\in T_{(x,y)}\subseteq J$.\\
Define $\tau:J^2\to J^2$ by $\tau_{(x,y)}=(x,\mu_{(x,y)})$.\\

\noindent
{\it Claim 3:} $\sigma\in\scrbigo_1^{WW}$.\\
{\it Proof of Claim 3:}\\
Since $J=B_\R(0,\delta)$ and $J=\dom[\sigma]$, by HW\#9-1,\\
\phantom{x}\qquad it suffices to prove: \qquad $\forall x\in J$, \quad $|\sigma_x|_W\le|x|_W$.\\
Given $x\in J$. \qquad
Want: $|\sigma_x|_W\le|x|_W$.\\
By definition of $S_x$, we have $S_x\subseteq[0|x]$.\\
Then $\sigma_x\in S_x\subseteq[0|x]$.\\
Then $\sigma_x\in[0|x]$, so $|\sigma_x|\le|x|$.\\
Then $|\sigma_x|_W=|\sigma_x|\le|x|=|x|_W$, as desired.\\
{\it End of proof of Claim 3.}\\

\noindent
{\it Claim 4:} $\tau\in\scrbigo_1^{VV}$.\\
{\it Proof of Claim 4:}\\
Let $B:=B_V(z,\delta)$. \qquad
Then $B\subseteq J^2=\dom[\tau]$.\\
By HW\#9-1, it suffices to prove: \qquad $\forall q\in B$, \quad $|\tau_q|_V\le|q|_V$.\\
Given $q\in B$. \qquad
Want: $|\tau_q|_V\le|q|_V$.\\
Let $x:=q_1$, $y:=q_2$. \qquad
Then $q=(x,y)$ and $|q|_V^2=x^2+y^2$.\\
Since $(x,y)=q\in B\subseteq J^2$, we see that $x,y\in J$.\\
By definition of $T_{(x,y)}$, we have $T_{(x,y)}\subseteq[0|y]$.\\
Then $\mu_{(x,y)}\in T_{(x,y)}\subseteq[0|y]$.\\
Then $\mu_{(x,y)}\in[0|y]$, so $|\mu_{(x,y)}|\le|y|$, so $(\mu_{(x,y)})^2\le y^2$.\\
Then $|\tau_q|=|\tau_{(x,y)}|_V^2=|(x,\mu_{(x,y)})|_V^2=x^2+(\mu_{(x,y)})^2\le x^2+y^2=|q|_V^2$.\\
Since $0\le|\tau_q|_V^2\le|q|_V^2$, we get $|\tau_q|_V\le|q|_V$, as desired.\\
{\it End of proof of Claim 4.}\\

\noindent
By Claim 3 and Claim 4, we have $\sigma\in\scrbigo_1^{WW}$ and $\sigma\in\scrbigo_1^{WW}$.\\
Want: $f\,\,=\,\,(g\circ i\circ\sigma\circ\pi_1^V)\cdot\pi_1^V\,\,+\,\,(h\circ\tau)\cdot\pi_2^V$ \qquad near $z$.\\
Since $B_V(0,\delta)\in\scrb_V(z)$, it suffices to prove:\\
\phantom{x}\qquad\qquad $f\,\,=\,\,(g\circ i\circ\sigma\circ\pi_1^V)\cdot\pi_1^V\,\,+\,\,(h\circ\tau)\cdot\pi_2^V$ \qquad on $B_V(0,\delta)$.\\
Want: $\forall q\in B_V(0,\delta)$, \quad $f_q\,\,=\,\,((g\circ i\circ\sigma\circ\pi_1^V)\cdot\pi_1^V\,\,+\,\,(h\circ\tau)\cdot\pi_2^V)_q$.\\
Given $q\in B_V(0,\delta)$.
Want: $f_q\,\,=\,\,((g\circ i\circ\sigma\circ\pi_1^V)\cdot\pi_1^V\,\,+\,\,(h\circ\tau)\cdot\pi_2^V)_q$.\\
Let $B:=B_V(z,\delta)$. \qquad
Let $x:=q_1$, $y:=q_2$. \qquad
Then $q=(x,y)$.\\
Then $\pi_1^V(q)=x$ and $\pi_2^V(q)=y$.\\
Since $(x,y)=q\in B\subseteq J^2$, we see that $x,y\in J$.\\
Since $i=i_z^{\varepsilon_1^V}$, we have: $\forall t\in\R$, $i(t)=z+t\varepsilon_1^V=(0,0)+t(1,0)=(t,0)$.\\
Then $(i\circ\sigma\circ\pi_1^V)_q=(i\circ\sigma)_x=i(\sigma_x)=(\sigma_x,0)$.\\
Want: $f_{(x,y)}\,\,=\,\,(g(\sigma_x,0))\cdot x\,\,+\,\,(h(\tau_{(x,y)}))\cdot y$.\\
Since $\sigma_x\in S_x$, by definition of $S_x$,\\
\phantom{x}\qquad\qquad we get $f|_{(0,0)}^{(x,0)}\,\,=\,\,((\partial_1^{VW}f)(\sigma_x,0))\cdot x$.\\
Then $f|_{(0,0)}^{(x,0)}\,\,=\,\,((\partial_1^{VW}f)(\sigma_x,0))\cdot x\,\,=\,\,(g(\sigma_x,0))\cdot x$.\\
Since $\mu_{(x,y)}\in T_{(x,y)}$, by definition of $T_{(x,y)}$,\\
\phantom{x}\qquad\qquad we get $f|_{(x,0)}^{(x,y)}\,\,=\,\,((\partial_1^{VW}f)(x,\mu_{(x,y)}))\cdot y$.\\
Then $f|_{(x,0)}^{(x,y)}\,\,=\,\,((\partial_1^{VW}f)(x,\mu_{(x,y)}))\cdot y\,\,=\,\,(h(x,\mu_{x,y}))\cdot y\,\,=\,\,(h(\tau_{x,y}))\cdot y$.\\
Since $f_{(0,0)}=f_z=0$, we get $f_{(x,y)}=f|_{(0,0)}^{(x,y)}$.\\
Then $f_{(x,y)}\,\,=\,\,f|_{((0,0)}^{(x,y)}\,\,=\,\,(\,f|_{(0,0)}^{(x,0)}\,)\,\,+\,\,(\,f|_{(x,0)}^{(x,y)}\,)$\\
\phantom{xx}\qquad\qquad $\,\,=\,\,(g(\sigma_x,0))\cdot x\,\,+\,\,(h(\tau_{(x,y)})\cdot y$, \quad as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
9-4. Let $V:=\R^2$, \quad $W:=\R$, \quad $h:V\drta W$, \quad $z:=0_2$.\\
\phantom{xx}\quad Assume: $z\in\dom[h']$. \quad
Assume: $h_z=(\partial_1^{VW}h)_z=(\partial_2^{VW}h)_z=0$.\\
\phantom{xx}\qquad\qquad\qquad\qquad Show: \qquad $h\in\scrlito_1^{VW}$.
\vskip.1in\noindent
{\it Proof:}
We have $h':V\drta W\otimes V$, so $\dom[h']\subseteq V$ and $\im[h']\subseteq W\otimes V$.\\
Since $z\in\dom[h']$, we get $h'_z\in\im[h']$.\\
Then $h'_z\in\im[h']\subseteq W\otimes V=\R\otimes\R^2=\R^2$.\\
Then $(h'_z)_1\in\R$ and $(h'_z)_2\in\R$. \qquad
Then $(h'_z)_1\ne\frownie\ne(h'_z)_2$.\\
We have $(\partial_1^{VW}h)_z\,\,=^*\,\,(h'_z)_1\,\,\ne\frownie$,\qquad so $(\partial_1^{VW}h)_z\,\,=\,\,(h'_z)_1$.\\
Also, $(\partial_2^{VW}h)_z\,\,=^*\,\,(h'_z)_2\,\,\ne\frownie$,\qquad so $(\partial_2^{VW}h)_z\,\,=\,\,(h'_z)_2$.\\
Then $(h'_z)_1=(\partial_1^{VW}h)_z=0$ and $(h'_z)_2=(\partial_2^{VW}h)_z=0$.\\
Then $h'_{0_V}=h'_z=(0,0)=0_{W\otimes V}$. \qquad
Also, $h_{0_V}=h_z=0=0_W$.\\
Then, by HW\#8-1, we get $h\in\scrlito_1^{VW}$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
9-5. Let $V:=\R^2$, \quad $W:=\R$, \quad $f:V\drta W$, \quad $z:=0_2$.\\
\phantom{xx}\quad Let $\alpha:=(\partial_1\partial_1f)_z$, \qquad $\beta:=(\partial_1\partial_2f)_z$, \qquad $\delta:=(\partial_2\partial_2f)_z$,\\
\phantom{xx}\qquad\qquad $\lambda:=(\partial_1f)_z$, \qquad $\mu:=(\partial_2f)_z$, \qquad $\rho:=f_z$.\\
Assume $\alpha\ne\frownie$, \quad $\beta\ne\frownie$, \quad $\delta\ne\frownie$, \quad $\lambda\ne\frownie$, \quad $\mu\ne\frownie$, \quad $\rho\ne\frownie$.\\
Define \quad $C\in\scrc_V^W$, \quad $L\in\scrl_V^W$, \quad $Q\in\scrq_V^W$ \quad by\\
\phantom{xx}\qquad\qquad $C(x,y)\,\,=\,\,\rho$, \qquad $L(x,y)\,\,=\,\,(\lambda,\mu)\,\,\sbullet_V\,\,(x,y)$,\\
\phantom{xx}\qquad\qquad\qquad $\ds{P(x,y)\,\,=\,\,\left[\begin{array}{cc}\alpha&\beta\\\beta&\delta\end{array}\right]\,\,\sbullet_{V\otimes V}\,\,((x,y)^{\otimes2})}$.\\
Let $\ds{R:=f-\left(C+L+\frac{P}{2!}\right)}$.\\
Show: $(\partial_1\partial_1R)_z=(\partial_1\partial_2R)_z=(\partial_2\partial_2R)_z=(\partial_1R)_z=(\partial_2R)_z=R_z=0$.
\vskip.1in\noindent
{\it Proof:}
We have: $\forall x,y\in\R$,\\
\phantom{x} $C_{(x,y)}=\rho$, \quad $(\partial_1C)_{(x,y)}=0$, \quad $(\partial_2C)_{(x,y)}=0$,\\
\phantom{x}\qquad $(\partial_1\partial_1C)_{(x,y)}=0$, \quad $(\partial_1\partial_2C)_{(x,y)}=0$, \quad $(\partial_2\partial_2C)_{(x,y)}=0$,\\
\phantom{x} $L_{(x,y)}=\lambda x+\mu y$, \quad $(\partial_1L)_{(x,y)}=\lambda$, \quad $(\partial_2L)_{(x,y)}=\mu$,\\
\phantom{x}\qquad $(\partial_1\partial_1L)_{(x,y)}=0$, \quad $(\partial_1\partial_2L)_{(x,y)}=0$, \quad $(\partial_2\partial_2L)_{(x,y)}=0$,\\
\phantom{x} $P_{(x,y)}=\alpha x^2+2\beta xy+\delta y^2$,\\
\phantom{x}\qquad $(\partial_1P)_{(x,y)}=2\alpha x+2\beta y$, \quad $(\partial_2P)_{(x,y)}=2\beta x+2\delta y$,\\
\phantom{x}\qquad $(\partial_1\partial_1P)_{(x,y)}=2\alpha$, \quad $(\partial_1\partial_2P)_{(x,y)}=2\beta$, \quad $(\partial_2\partial_2P)_{(x,y)}=2\delta$.\\
Then, since $z=0_2=(0,0)$, we get\\
\phantom{x} $C_z=\rho$, \quad $(\partial_1C)_z=0$, \quad $(\partial_2C)_z=0$,\\
\phantom{x}\qquad $(\partial_1\partial_1C)_z=0$, \quad $(\partial_1\partial_2C)_z=0$, \quad $(\partial_2\partial_2C)_z=0$,\\
\phantom{x} $L_z=0$, \quad $(\partial_1L)_z=\lambda$, \quad $(\partial_2L)_z=\mu$,\\
\phantom{x}\qquad $(\partial_1\partial_1L)_z=0$, \quad $(\partial_1\partial_2L)_z=0$, \quad $(\partial_2\partial_2L)_z=0$,\\
\phantom{x} $P_z=0$,\\
\phantom{x}\qquad $(\partial_1P)_z=0$, \quad $(\partial_2P)_z=z$,\\
\phantom{x}\qquad $(\partial_1\partial_1P)_z=2\alpha$, \quad $(\partial_1\partial_2P)_z=2\beta$, \quad $(\partial_2\partial_2P)_z=2\delta$.\\
Then\\
\phantom{x}\qquad $R_z\,\,=^*\,\,f_z-(\rho+0+0)\,\,=\,\,\rho-\rho\,\,=\,\,0\,\,\ne\,\,\frownie$,\\
\phantom{x}\qquad $(\partial_1R)_z\,\,=^*\,\,(\partial_1f)_z-(0+\lambda+0)\,\,=\,\,\lambda-\lambda\,\,=\,\,0\,\,\ne\,\,\frownie$,\\
\phantom{x}\qquad $(\partial_2R)_z\,\,=^*\,\,(\partial_2f)_z-(0+\mu+0)\,\,=\,\,\mu-\mu\,\,=\,\,0\,\,\ne\,\,\frownie$,\\
\phantom{x}\qquad $(\partial_1\partial_1R)_z\,\,=^*\,\,(\partial_1\partial_1f)_z-(0+0+(2\alpha/2))\,\,=\,\,\alpha-\alpha\,\,=\,\,0\,\,\ne\,\,\frownie$,\\
\phantom{x}\qquad $(\partial_1\partial_2R)_z\,\,=^*\,\,(\partial_1\partial_2f)_z-(0+0+(2\beta/2))\,\,=\,\,\beta-\beta\,\,=\,\,0\,\,\ne\,\,\frownie$,\\
\phantom{x}\qquad $(\partial_2\partial_2R)_z\,\,=^*\,\,(\partial_2\partial_2f)_z-(0+0+(2\delta/2))\,\,=\,\,\delta-\delta\,\,=\,\,0\,\,\ne\,\,\frownie$.\\
Then $(\partial_1\partial_1R)_z=(\partial_1\partial_2R)_z=(\partial_2\partial_2R)_z=(\partial_1R)_z=(\partial_2R)_z=R_z=0$,\\
\phantom{x}\qquad\qquad\qquad as desired. \qquad
QED
\vskip.1in\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 8: Due on Tuesday 26 March}
\vskip.1in\hrule\vskip.1in\noindent
8-1. Let $S,T\in\TNSR^+$, \quad $R:S\drta T$.\\
\phantom{x}\qquad Assume $(\,\,R_{0_S}=0_T\,\,)\,\,\&\,\,(\,\,R'_{0_S}=0_{T\otimes S}\,\,)$. \qquad
Show: $R\in\scrlito_1^{ST}$.
\vskip.1in\noindent
{\it Proof:}
Let $L:=D_{0_S}R$ and $A:=R'_{0_S}$.
Then $L=\Lin_A^{ST}$ and $A=0_{T\otimes S}$.\\
Then $L=\Lin_{0_{T\otimes S}}^{ST}=\bfzero_S^T$. \quad
Then $L\ne\frownie$.\\
Since $D_{0_S}R=L\ne\frownie$, it follows that $R_{0_S}^T-L\in\scrlito_1^{ST}$.\\
Since $R_{0_S}=0_T$, it follows that $R_{0_S}^T=R$.\\
Then $R=R-\bfzero_S^T=R_{0_S}^T-\bfzero_S^T=R_{0_S}^T-L\in\scrlito_1^{ST}$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
8-2. Let $S,T\in\TNSR^+$, \quad $f:S\drta T$.\\
\phantom{x}\qquad Let $p\in\dom[f']$, \quad $L:=\Lin_{f'_p}^{ST}$, \quad $R:=f^T_p-L$.\\
\phantom{x}\qquad Show: $(\,\,R_{0_S}=0_T\,\,)\,\,\&\,\,(\,\,R'_{0_S}=0_{T\otimes S}\,\,)$.
\vskip.1in\noindent
{\it Proof:}
We have $R_{0_S}=(f_p^T-L)(0_S)=[f_p^T(0_S)]-[L(0_S)]$.\\
Since $p\in\dom[f']\subseteq\dom[f]$, we get $f_p^T(0_S)=0_T$.\\
Since $L\in\scrl_S^T$, we get $L(0_S)=0_T$.\\
Then $R_{0_S}=R(0_S)=(f_p^T-L)(0_S)=[f_p^T(0_S)]-[L(0_S)]=0_T-0_T=0_T$.\\
Want: $R'_{0_S}=0_{T\otimes S}$.\\
We have $D_{0_S}^{ST}(f_p^T)=D_{p+0_S}^{ST}f=D_p^{ST}f=\Lin_{f'_p}^{ST}=L$.\\
Also, since $L\in\scrl_S^T$, we get $D_{0_S}^{ST}L=L$.\\
Then $D_{0_S}^{ST}R\,\,=\,\,D_{0_S}^{ST}(f_p^T-L)\,\,=^*\,\,[D_{0_S}^{ST}(f_p^T)]-[D_{0_S}^{ST}(L)]$\\
\phantom{xx}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $=\,\,L-L\,\,=\,\,\bfzero_S^T$.\\
Since $D_{0_S}R\,\,=^*\,\,\bfzero_S^T\,\,\ne\,\,\frownie$, it follows that $D_{0_S}R\,\,=\,\,\bfzero_S^T$.\\
Then $R-{0_S}=\sl_\bullet^{ST}(D_{0_S}R)=\sl_\bullet^{ST}(\bfzero_S^T)=0_{T\otimes S}$.
QED
\vskip.1in\hrule\vskip.1in\noindent
8-3. Let $S\in\TNSR^+$, \quad $p,v\in S$, \quad $i:=i_p^v$. \qquad
Show: $i'=C_\R^v$.
\vskip.1in\noindent
{\it Proof:}
We have: $\forall t\in\R$, $i(t)=i_p^v(t)=p+tv$.\\
Want: $\forall t\in\R$, \quad $i'(t)=C_\R^v(t)$.\\
Given $t\in\R$. \qquad
Want: $i'(t)=C_\R^v(t)$. \qquad
Want $i'_t=v$.\\
Define $L\in\scrl_\R^S$ by $L(h)=hv$. \qquad
Then $\sl_\bullet^{\R S}(L)=\sl_L^{\R S}=L(1)=v$.\\
We have: $\forall h\in\R$, \quad $i_t^T(h)=[i(t+h)]-[i(t)]$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad $=[p+(t+h)v]-[p+tv]=hv=L(h)$.\\
Then $i_t^T=L$, so $i_t^T-L=L-L=\bfzero_\R^S\in\scrlito_1^{\R S}$.\\
Since $i_t^T-L\in\scrlito_1^{\R S}$, it follows that $L=D_t^{\R S}i$.\\
Then $i'_t=\sl_\bullet^{\R S}(D_t^{\R S}i)=\sl_\bullet^{\R S}(L)=v$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
8-4. Let $S,T\in\TNSR^+$, \quad $f:S\drta T$, \quad $p,v\in S$.\\
\phantom{x}\qquad Show: $(\partial_v^{ST}f)_p\,\,=^*\,\,(D_p^{ST}f)_v$.
\vskip.1in\noindent
{\it Proof:}
Let $i:=i_p^v$ and let $g:=f\circ i$.\\
Then $(\partial_v^{ST}f)_p=(f\circ i)'_0=g'_0$.\\
Also, we have: $i_0=i(0)=i_p^v(0)=p+0\cdot v=p$.\\
By the Linearization Chain Rule,\\
\phantom{x}\qquad we have $D_0^{\R T}(f\circ i)\,\,=^*\,\,(D_{i_0}^{ST}f)\circ(D_0^{\R S}i)$.\\
Let $L:=D_0^{\R S}i$ and let $M:=D_0^{\R T}g$.\\
Then $M\,\,=\,\,D_0^{\R T}(f\circ i)\,\,=^*\,\,(D_{i_0}^{ST}f)\circ(D_0^{\R S}i)\,\,=\,\,(D_p^{ST}f)\circ L$.\\
Then $M(1)\,\,=^*\,\,((D_p^{ST}f)\circ L)(1)\,\,=\,\,(D_p^{ST}f)(L(1))$.\\
We have $g'_0=\sl_\bullet^{\R T}(D_0^{\R T}g)=\sl_\bullet^{\R T}(M)=\sl_M^{\R T}=M(1)$.\\
By HW\#8-3, $i'=C_\R^v$. \qquad
Then $i'_0=i'(0)=C_\R^v(0)=v$.\\
We have $i'_0=\sl_\bullet^{\R S}(D_0^{\R S}i)=\sl_\bullet^{\R S}(L)=\sl_L^{\R S}=L(1)$.\\
Then $(\partial_v^{ST}f)_p\,\,=\,\,g'_0\,\,=\,\,M(1)\,\,=^*\,\,(D_p^{ST})(L(1))$\\
\phantom{x}\qquad\qquad\qquad $=\,\,D_p^{ST}(i'_0)\,\,=\,\,D_p^{ST}(v)$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
8-5. Let $V:=\R^2$, \quad $W\in\R$, \quad $R:V\drta W$,\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad $\delta>0$, \quad $J:=(-\delta;\delta)$, \quad $x\in J$.\\
\phantom{x}\qquad Assume: \quad $J^2\subseteq\dom[R']$.\\
\phantom{x}\qquad Show: \quad $\exists\alpha\in[0|x]$ \quad s.t. \quad $R|_{(0,0)}^{(x,0)}\,\,=\,\,((\partial_1^{VW}R)(\alpha,0))\cdot x$.
\vskip.1in\noindent
{\it Proof:}
Let $i:=i_{(0,0)}^{(1,0)}$. \qquad
By HW\#8-3, we have $i'=C_\R^{(1,0)}$.\\
Also: $\forall t\in\R$, $i_t=i(t)=(0,0)+t\cdot(1,0)=(t,0)$.\\
Let $f:=R\circ i$. \quad
Then $f_x=R(i_x)=R(x,0)$ and $f_0=R(i_0)=R(0,0)$.\\
\\
{\it Claim:} $J\subseteq\dom[f']$.\\
{\it Proof of Claim:}\\
Want: $\forall t\in J$, $t\in\dom[f']$.\\
Given $t\in J$. \qquad
Want: $t\in\dom[f']$.\\
Since $t\in\R=\dom[C_\R^{(0,1)}=i'$ and $i_t=(t,0)\in J^2\subseteq\dom[R']$,\\
\phantom{x}\qquad we conclude that $t\in\dom[(R\circ i)']$.\\
Then $t\in\dom[(R\circ i)']=\dom[f']$, as desired.\\
{\it End of proof of Claim.}\\
\\
Since $x\in J$, by the Claim and the MVT,\\
\phantom{x}\qquad choose $\alpha\in[0|x]$ s.t.~$f_x-f_0=(f'_\alpha)\cdot(x-0)$.\\
Want: $R|_{(0,0)}^{(x,0)}\,\,=\,\,((\partial_1^{VW}R)(\alpha,0))\cdot x$.\\
We have $R|_{(0,0)}^{(x,0)}=[R(x,0)]-[R(0,0)]=f_x-f_0=(f'_\alpha)\cdot(x-0)=(f'_\alpha)\cdot x$.\\
Want: $f'_\alpha=(\partial_1^{VW}R)(\alpha,0)$.\\
We have $D_\alpha^{\R W}f\,\,=\,\,D_\alpha^{\R W}(R\circ i)\,\,=^*\,\,(D_{i(\alpha)}^{VW}R)\circ(D_\alpha^{\R V}i)$.\\
Also, $f'_\alpha=(D_\alpha^{\R W}f)(1)$ and $i'_\alpha=(D_\alpha^{\R V}i)(1)$. \qquad
Let $u:=\varepsilon_1^V$.\\
Then $(D_\alpha^{\R V}i)(1)=i'_\alpha=i'(\alpha)=C_\R^{(1,0)}(\alpha)=(1,0)=\varepsilon_1^V=u$.\\
Then $f'_\alpha\,\,=\,\,(D_\alpha^{\R W}f)(1)\,\,=^*\,\,((D_{i(\alpha)}^{VW}R)\circ(D_\alpha^{\R V}i))(1)\,\,=\,\,(D_{i(\alpha)}^{VW}R)(u)$.\\
Since $0,x\in J$ and since $J$ is an interval, we get $[0|x]\subseteq J$.\\
Then $\alpha\in[0|x]\subseteq J$. \quad
So, as $0\in J$, we get $(\alpha,0)\in J^2$.\\
Then $i(\alpha)=(\alpha,0)\in J^2\subseteq\dom[R']$, so $D_{i(\alpha)}^{VW}R\in\scrl_V^W$.\\
Then $(D_{i(\alpha)}^{VW}R)(u)\in W$, so $(D_{i(\alpha)}^{VW}R)(u)\ne\frownie$.\\
Since $f'_\alpha\,\,=^*\,\,(D_{i(\alpha)}^{VW}R)(u)\,\,\ne\,\,\frownie$, we conclude that $f'_\alpha\,\,=\,\,(D_{i(\alpha)}^{VW}R)(u)$.\\
Since $u=\varepsilon_1^V$, we get $\partial_u^{VW}R=\partial_1^{VW}R$.\\
We have $(\partial_u^{VW}R)(i(\alpha))\,\,=^*\,\,(D_{i(\alpha)}^{VW}R)_u\,\,=\,\,(D_{i(\alpha)}^{VW}R)(u)$.\\
Then $(\partial_u^{VW}R)(i(\alpha))\,\,=^*\,\,(D_{i(\alpha)}^{VW}R)(u)\,\,\ne\,\,\frownie$,\\
\phantom{x}\qquad so $(\partial_u^{VW}R)(i(\alpha))\,\,=\,\,(D_{i(\alpha)}^{VW}R)(u)$.\\
Then $f'_\alpha=(D_{i(\alpha)}^{VW}R)(u)=(\partial_u^{VW}R)(\alpha,0)=(\partial_1^{VW}R)(\alpha,0)$.
QED
\vskip.1in\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 7: Due on Tuesday 12 March}
\vskip.1in\hrule\vskip.1in\noindent
7-1. Let $V,W\in\TNSR$ and let $L,M\in\scrl_V^W$.\\
\phantom{x}\qquad\qquad\qquad Assume: $[L]_V^W=[M]_V^W$. \qquad\qquad
Show: $L=M$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall x\in V$, $L_x=M_x$.\\
Given $x\in V$. \qquad
Want: $L_x=M_x$.\\
Let $A:=\sl_L^{VW}$. \qquad
By Theorem 85.6, $L_x\,\,=\,\,A\,\,*_{\R VW}\,\,x$.\\
Let $B:=\sl_M^{VW}$. \qquad
By Theorem 85.6, $M_x\,\,=\,\,B\,\,*_{\R VW}\,\,x$.\\
We have $A=\sl_L^{VW}=[L]_V^W=[M]_V^W=\sl_M^{VW}=B$, so $A=B$.\\
Then $L_x\,\,=\,\,A\,\,*_{\R VW}\,\,x\,\,=\,\,B\,\,*_{\R VW}\,\,x\,\,=\,\,M_x$.
QED
\vskip.1in\hrule\vskip.1in\noindent
7-2. Let $V,W\in\TNSR$ and let $A\in W\otimes V$.\\
\phantom{x}\qquad\qquad\qquad Show: $\exists L\in\scrl_V^W$ s.t.~$[L]_V^W=A$.
\vskip.1in\noindent
{\it Proof:}
Define $L\in\scrl_V^W$ by $L_x\,\,=\,\,A\,\,*_{\R VW}\,\,x$. \quad
Want: $[L]_V^W=A$.\\
We have $[L]_V^W=\sl_L^{VW}=(\sl_\bullet^{VW})(L)$.\\
Want: $(\sl_\bullet^{VW})(L)=A$. \qquad\qquad
Want: $L=(\sl_\bullet^{VW})^{-1}(A)$. \\
Let $M:=(\sl_\bullet^{VW})^{-1}(A)$. \qquad
Want: $L=M$.\\
Want: $\forall x\in V$, $L_x=M_x$. \qquad
Given $x\in V$. \qquad
Want: $L_x=M_x$.\\
Let $B:=\sl_M^{VW}$. \qquad\qquad
By Theorem 85.6, \quad $M_x\,\,=\,\,B\,\,*_{\R VW}\,\,x$.\\
Since $M=(\sl_\bullet^{VW})^{-1}(A)$, we get $(\sl_\bullet^{VW})(M)=A$.\\
Then $B=\sl_M^{VW}=(\sl_\bullet^{VW})(M)=A$.\\
Then $L_x\,\,=\,\,A\,\,*_{\R VW}\,\,x\,\,=\,\,B\,\,*_{\R VW}\,\,x\,\,=\,\,M_x$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
7-3. Let $I$ be a finite set and let $z\in\R^I$.\\
\phantom{x}\qquad\qquad\qquad Show: \qquad $\ds{\sum_{j\in I}\,|z_j|\quad\le\quad\sqrt{\#\,I}\,\cdot\,\sqrt{\sum_{j\in I}\,z_j^2}}$.
\vskip.1in\noindent
{\it Proof:}
Define $x\in\R^I$ by $x_j=1$. \qquad\qquad
Then, $\forall j\in I$, $x_j^2=1$.\\
Then \qquad $\ds{\sum_{j\in I}\,x_j^2\,\,=\,\,\#I}$, \qquad so \qquad
$\ds{\sqrt{\sum_{j\in I}\,x_j^2}\,\,=\,\,\sqrt{\#I}}$.\\
Define $y\in\R^I$ by $y_j=|z_j|$. \qquad\qquad
Then, $\forall j\in I$, $y_j^2=z_j^2$.\\
Then \qquad $\ds{\sum_{j\in I}\,y_j^2\,\,=\,\,\sum_{j\in I}\,z_j^2}$, \qquad so \qquad
$\ds{\sqrt{\sum_{j\in I}\,y_j^2}\,\,=\,\,\sqrt{\sum_{j\in I}\,z_j^2}}$.\\
We have: \qquad $\forall j\in I$, $x_jy_j=1\cdot|z_j|=|z_j|$.\\
Then \qquad $\ds{\sum_{j\in I}\,x_jy_j\,\,=\,\,\sum_{j\in I}\,|z_j|}$.\\
By Theorem 84.11, \qquad $\ds{\sum_{j\in I}\,x_jy_j\,\,\le\,\,\sqrt{\sum_{j\in I}\,x_j^2}\,\cdot\,\sqrt{\sum_{j\in I}\,y_j^2}}$.\\
Then $\ds{\sum_{j\in I}\,|z_j|\,\,=\,\,\sum_{j\in I}\,x_jy_j\,\,\le\,\,
\sqrt{\sum_{j\in I}\,x_j^2}\,\cdot\,\sqrt{\sum_{j\in I}\,y_j^2}\,\,=\,\,\sqrt{\#\,I}\,\cdot\,\sqrt{\sum_{j\in I}\,z_j^2}}$,\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
7-4. Let $U,V,W\in\TNSR$ and let $B\in\scrb_{UV}^W$.\\
\phantom{x}\qquad\qquad\qquad Show: $\exists C\ge0$ s.t., $\forall p\in U$, $\forall q\in V$,\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad $|B(p,q)|_W\quad\le\quad C\,\cdot|p|_U\,\,\cdot|q|_V$.
\vskip.1in\noindent
{\it Proof:}
Let $K:=\max\{\,|\,B(\varepsilon_i^U,\varepsilon_j^V)\,|_W\,\,\hbox{s.t.}\,\,i\in\scri_U,\,\,j\in\scri_V\}$.\\
Then $K\ge0$. \qquad
Let $C:=K\cdot\sqrt{\#\scri_U}\cdot\sqrt{\#\scri_V}$. \qquad
Then $C\ge0$.\\
Want: $\forall p\in U$, $\forall q\in V$,
\qquad $|B(p,q)|_W\,\,\le\,\,C\,\cdot|p|_U\,\,\cdot|q|_V$.\\
Given $p\in U$, $q\in V$. \qquad
Want: $|B(p,q)|_W\,\,\le\,\,C\,\cdot|p|_U\,\,\cdot|q|_V$.\\
Since $\ds{p=\sum_{i\in\scri_U}\,p_i\varepsilon_i^U}$ and $\ds{q=\sum_{j\in\scri_V}\,q_j\varepsilon_j^V}$,
we get\\
\phantom{x}\qquad\qquad\qquad $\ds{B(p,q)\,\,=\,\,\sum_{i\in\scri_U}\,\sum_{j\in\scri_V}\,\bigg(\,p_i\cdot q_j\cdot B(\varepsilon_i^U,\varepsilon_j^V)\,\bigg)}$.\\
So, by subadditivity and absolute homogeneity of $|\bullet|_W$, we get\\
\phantom{x}\qquad\qquad\qquad $\ds{|\,B(p,q)\,|_W\,\,\le\,\,\sum_{i\in\scri_U}\,\sum_{j\in\scri_V}\,\bigg(\,|p_i|\cdot|q_j|\cdot|\,B(\varepsilon_i^U,\varepsilon_j^V)\,|_W\,\bigg)}$.\\
So, by the choice of $K$, we get\\
\phantom{x}\qquad\qquad\qquad $\ds{|\,B(p,q)\,|_W\,\,\le\,\,\sum_{i\in\scri_U}\,\sum_{j\in\scri_V}\,\bigg(\,|p_i|\cdot|q_j|\cdot K\,\bigg)}$.\\
So, as $\ds{K\cdot\left(\sum_{i\in\scri_U}\,|p_i|\right)\cdot\left(\sum_{j\in\scri_V}\,|q_j|\right)\,\,=\,\,\sum_{i\in\scri_U}\,\sum_{j\in\scri_V}\,\bigg(\,|p_i|\cdot|q_j|\cdot K\,\bigg)}$, we get
\phantom{x}\qquad\qquad\qquad $\ds{|\,B(p,q)\,|_W\,\,\le\,\,K\cdot\left(\sum_{i\in\scri_U}\,|p_i|\right)\cdot\left(\sum_{j\in\scri_V}\,|q_j|\right)}$.\\
By HW\#7-3, we get $\ds{\sum_{i\in\scri_U}\,|p_i|\le\sqrt{\#\scri_U}\cdot\sqrt{\sum_{i\in\scri_U}\,p_i^2}}$.\\
Also, by HW\#7-3, we get $\ds{\sum_{j\in\scri_V}\,|q_i|\le\sqrt{\#\scri_V}\cdot\sqrt{\sum_{j\in\scri_V}\,q_j^2}}$.\\
By definition of $|\bullet|_U$, we have $\ds{|p|_U=\sqrt{\sum_{i\in\scri_U}\,p_i^2}}$.\\
By definition of $|\bullet|_V$, we have $\ds{|q|_V=\sqrt{\sum_{j\in\scri_V}\,q_j^2}}$.\\
Then $\ds{\,\,\,|\,B(p,q)\,|_W\,\,\le\,\,K\cdot\left(\sum_{i\in\scri_U}\,|p_i|\right)\cdot\left(\sum_{j\in\scri_V}\,|q_j|\right)}$\\
\phantom{x}\qquad\qquad\qquad\qquad $\ds{\le\,\,K\cdot\sqrt{\#\scri_U}\cdot\sqrt{\sum_{i\in\scri_U}\,p_i^2}\cdot\sqrt{\#\scri_V}\cdot\sqrt{\sum_{j\in\scri_V}\,q_j^2}}$\\
\phantom{x}\qquad\qquad\qquad\qquad $\ds{\le\,\,K\cdot\sqrt{\#\scri_U}\cdot|p|_U\cdot\sqrt{\#\scri_V}\cdot|q|_V}$\\
\phantom{x}\qquad\qquad\qquad\qquad $\ds{=\,\,K\cdot\sqrt{\#\scri_U}\cdot\sqrt{\#\scri_V}\cdot|p|_U\cdot|q|_V}$\\
\phantom{x}\qquad\qquad\qquad\qquad $\ds{=\,\,C\cdot|p|_U\cdot|q|_V}$, \quad as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
7-5. Let $T,U,V,W\in\TNSR$ and let $*\in\scrb_{UV}^W$.\\
\phantom{x}\qquad\qquad\qquad Show: $\BNZ_{TU}\,\,*\,\,\BNZ_{TV}\subseteq\BNZ_{TW}$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall\alpha\in\BNZ_{TU}\,\,*\,\,\BNZ_{TV}$, $\alpha\in\BNZ_{TW}$.\\
Given $\alpha\in\BNZ_{TU}\,\,*\,\,\BNZ_{TV}$. \qquad
Want: $\alpha\in\BNZ_{TW}$.\\
Choose $\beta\in\BNZ_{TU}$ and $\gamma\in\BNZ_{TV}$ s.t.~$\alpha=\beta*\gamma$.\\
By HW\#7-4, choose $K\ge0$ s.t.~$\forall p\in U$, $\forall q\in V$,\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad $|p*q|_W\quad\le\quad K\,\cdot|p|_U\,\,\cdot|q|_V$.\\
Since $\beta\in\BNZ_{TU}\subseteq\DNZ_{TU}$, choose $B\in\scrb_T(0_T)$ s.t.~$B\subseteq\dom[\beta]$.\\
Since $\gamma\in\BNZ_{TV}\subseteq\DNZ_{TV}$, choose $C\in\scrb_T(0_T)$ s.t.~$C\subseteq\dom[\gamma]$.\\
Then $B\cap C\subseteq(\dom[\beta])\cap(\dom[\alpha])=\dom[\beta*\gamma]=\dom[\alpha]$.\\
So, since $B\cap C\in\{B,C\}\subseteq\scrb_T(0_T)$, we conclude that $\alpha\in\DNZ_{TW}$.\\
Want: $\exists A\in\scrb_T(0_T)$ s.t.~$\alpha_*(A)$ is bounded in $W$.\\
Since $\beta\in\BNZ_{TU}$, choose $D\in\scrb_T(0_T)$ s.t.~$\beta_*(D)$ is bounded in $U$.\\
Since $\beta_*(D)$ is bounded in $U$, choose $E\in\scrb_U$ s.t.~$\beta_*(D)\subseteq E$.\\
Since $E\in\scrb_U$, by the Superset Recentering Lemma,\\
\phantom{x}\qquad choose $F\in\scrb_U(0_U)$ s.t.~$E\subseteq F$.\\
Since $F\in\scrb_U(0_U)$, choose $r>0$ s.t.~$F=B_U(0_U,r)$.\\
Since $\gamma\in\BNZ_{TU}$, choose $X\in\scrb_T(0_T)$ s.t.~$\gamma_*(X)$ is bounded in $V$.\\
Since $\gamma_*(X)$ is bounded in $V$, choose $Y\in\scrb_V$ s.t.~$\gamma_*(X)\subseteq Y$.\\
Since $Y\in\scrb_V$, by the Superset Recentering Lemma,\\
\phantom{x}\qquad choose $Z\in\scrb_V(0_V)$ s.t.~$Y\subseteq Z$.\\
Since $Z\in\scrb_V(0_V)$, choose $s>0$ s.t.~$Z=B_V(0_V,s)$.\\
Let $A:=D\cap X$. \qquad
Then $A\in\{D,X\}\subseteq\scrb_T(0_T)$.\\
Want: $\alpha_*(A)$ is bounded in $W$. \qquad
Want: $\exists Q\in\scrb_W$ s.t.~$\alpha_*(A)\subseteq Q$.\\
Let $Q:=B_W(0_W,Krs+1)$. \qquad
Then $Q\in\scrb_W(0_W)\subseteq\scrb_W$.\\
Want: $\alpha_*(A)\subseteq Q$. \qquad
Want: $\forall w\in\alpha_*(A)$, $w\in Q$.\\
Given $w\in\alpha_*(A)$. \qquad
Want: $w\in Q$. \qquad
Want: $|w|_W<Krs+1$.\\
Since $w\in\alpha_*(A)$, choose $t\in(\dom[\alpha])\cap A$ s.t. $\alpha(t)=w$.\\
Since $t\in(\dom[\alpha])\cap A$, \quad we get \quad $t\in\dom[\alpha]$ \quad and \quad $t\in A$.\\
We have $t\in\dom[\alpha]=\dom[\beta*\gamma]=(\dom[\beta])\cap(\dom[\gamma])$,\\
\phantom{x}\qquad\qquad so \quad $t\in\dom[\beta]$ \quad and \quad $t\in\dom[\gamma]$.\\
Also, we have $t\in A=D\cap X$,\\
\phantom{x}\qquad\qquad so \quad $t\in D$ \quad and \quad $t\in X$.\\
Since $t\in\dom[\beta]$ and $t\in D$, we get $\beta(t)\in\beta_*(D)$.\\
So, since $\beta_*(D)\subseteq E\subseteq F=B_U(0_U,r)$,\\
\phantom{x}\qquad\qquad we get $\beta(t)\in B_U(0_U,r)$, \quad so $|\beta(t)|_U<r$.\\
Since $t\in\dom[\gamma]$ and $t\in X$, we get $\gamma(t)\in\gamma_*(X)$.\\
So, since $\gamma_*(X)\subseteq Y\subseteq Z=B_V(0_V,s)$,\\
\phantom{x}\qquad\qquad we get $\gamma(t)\in B_V(0_V,s)$, \quad so $|\gamma(t)|_V<s$.\\
Since $t\in\dom[\beta]$, we get $\beta(t)\in\im[\beta]$.\\
Since $\beta\in\BNZ_{TU}$, we get $\im[\beta]\subseteq U$.\\
Since $t\in\dom[\gamma]$, we get $\gamma(t)\in\im[\gamma]$.\\
Since $\gamma\in\BNZ_{TV}$, we get $\im[\gamma]\subseteq V$.\\
Since $\beta(t)\in\im[\beta]\subseteq U$ and $\gamma(t)\in\im[\gamma]\subseteq V$,\\
\phantom{x}\qquad it follows, from the choice of $K$,\\
\phantom{x}\qquad that $|(\beta(t))*(\gamma(t))|_W\le K\cdot|\beta(t)|_U\cdot|\gamma(t)|_V$.\\
We have $w=\alpha(t)=(\beta*\gamma)(t)=(\beta(t))*(\gamma(t))$.\\
Then $|w|_W=|(\beta(t))*(\gamma(t))|_W\le K\cdot|\beta(t)|_U\cdot|\gamma(t)|_V\le Krs$.\\
Then $|w|_W\le Krs<Krs+1$, as desired.
QED
\vskip.1in\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 6: Due on Tuesday 5 March}
\vskip.1in\hrule\vskip.1in\noindent
6-1. Let $T\in\TNSR$, \quad $v\in T$. \qquad
Show: $\ds{v=\sum_{j\in\scri_T}\,v_j\varepsilon_j^T}$.
\vskip.1in\noindent
{\it Proof:}
Let $\ds{w:=\sum_{j\in\scri_T}\,v_j\varepsilon_j^T}$. \qquad
Want $v=w$.\\
Want: $\forall i\in\scri_T$, $v_i=w_i$. \qquad
Given $i\in\scri_T$. \qquad
Want: $v_i=w_i$.\\
We have: \quad $\ds{w_i\quad=\quad\sum_{j\in\scri_T}\,v_j\cdot((\varepsilon_j^T)_i)}$\\
\phantom{x}\qquad\qquad\qquad\quad $\ds{\,\,=\quad\left(v_i\cdot((\varepsilon_i)_i)\right)\quad+\quad\left(\sum_{j\in\scri_T\backslash\{i\}}\,v_j\cdot((\varepsilon_j^T)_i)\right)}$\\
\phantom{x}\qquad\qquad\qquad\quad $\ds{\,\,=\quad\left(v_i\cdot1\right)\quad+\quad\left(\sum_{j\in\scri_T\backslash\{i\}}\,v_j\cdot0\right)\quad=\quad v_i+0}$.\\
Then $\ds{v_i=v_i+0=w_i}$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
6-2. Let $S,T\in\TNSR$, \quad $A\in T\otimes S$, \quad $L:=\Lin_A^{ST}$, \quad $i\in\scri_S$.\\
\phantom{x}\qquad\qquad Show: $\ds{L(\varepsilon_i^S)\,=\,\sum_{j\in\scri_T}\,A_{j\|i}\,\cdot\,\varepsilon_j^T}$.
\vskip.1in\noindent
{\it Proof:}
Let $v:=L(\varepsilon_i^S)$. \qquad
By HW\#6-1, we have: \quad $\ds{v=\sum_{j\in\scri_T}\,v_j\varepsilon_j^T}$.\\
By Theorem 83.5, p.~202, we have: \quad $\forall j\in\scri_T$, \quad $A_{j\|i}=(L(\varepsilon_i^S))_j$.\\
Then $\ds{L(\varepsilon_i^S)\,=\,v\,=\,\sum_{j\in\scri_T}\,v_j\varepsilon_j^T\,=\,\sum_{j\in\scri_T}\,(L(\varepsilon_i^S))_j\,\cdot\,\varepsilon_j^T}$\\
\phantom{x}\qquad\qquad\qquad\qquad $\ds{=\sum_{j\in\scri_T}\,A_{j\|i}\,\cdot\,\varepsilon_j^T}$, \qquad
as desired. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
6-3. Let $S,T\in\TNSR^+$, \quad $C\in\scrc_S^T$, \quad $p\in S$. \qquad
Show: $D_p^{ST}C=\bfzero_S^T$.
\vskip.1in\noindent
{\it Proof:}
Since $C\in\scrc_S^T$, choose $w\in T$ s.t.~$C=C_S^w$.\\
We have: \quad $\forall h\in S$, \quad $C_p^T(h)=[C(p+h)]\,-\,[C(p)]$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad $\,\,=[C_S^w(p+h)]\,-\,[C_S^w(p)]$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad $\,\,=\,\,w\,-\,w\,\,=\,\,0_T\,\,=\,\,\bfzero_S^T(h)$.\\
Then $C_p^T=\bfzero_S^T$. \qquad\qquad
Let $L:=\bfzero_S^T$.\\
Then $C_p^T-L\,\,=\,\,\bfzero_S^T\,-\,\bfzero_S^T\,\,=\,\,\bfzero_S^T\,\,\in\,\,\scrlito_1^{ST}$.\\
Then $L\,\,\in\,\,\LINS_pC$, \quad and so \quad $D_p^{ST}C\,\,=\,\,L$.\\
Then $D_p^{ST}C\,\,=\,\,L\,\,=\,\,\bfzero_S^T$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
6-4. Let $S,T\in\TNSR^+$, \quad $L\in\scrl_S^T$, \quad $p\in S$. \qquad
Show: $D_p^{ST}L=L$.
\vskip.1in\noindent
{\it Proof:}
We have: $\,\,\forall h\in S$, $\,\,L_p^T(h)=[L(p+h)]\,-\,[L(p)]$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\,=[L(p)]\,+\,[L(h)]\,-\,[L(p)]\,\,=\,\,L(h)$.\\
Then $L_p^T=L$. \qquad\qquad
Then $L_p^T-L\,\,=\,\,\bfzero_S^T\,\,\in\,\,\scrlito_1^{ST}$.\\
Then \quad $L\,\,\in\,\,\LINS_pL$, \quad and so \quad $D_p^{ST}L\,\,=\,\,L$, \quad as desired. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
6-5. Let $V,W\in\TNSR^+$, \quad $Q\in\scrq_V^W$, \quad $B\in\scrs\scrb_V^W$.\\
\phantom{x}\qquad Assume: $Q=B(\bullet,\bullet)$. \qquad
Show: $\forall x\in V$, $D_x^{VW}Q=2\cdot(B(x,\bullet))$.
\vskip.1in\noindent
{\it Proof:}
Given $x\in V$. \qquad
Want: $D_x^{VW}Q=2\cdot(B(x,\bullet))$.\\
We have: $\,\,\forall h\in S$,\\
\phantom{x}\qquad $Q_x^T(h)=(Q(x+h))\,-\,(Q(x))$\\
\phantom{x}\qquad\qquad\quad $=(B(x+h,x+h))\,-\,(B(x,x))$\\
\phantom{x}\qquad\qquad\quad $=(B(x,x))\,+(B(x,h))\,+(B(h,x))\,+\,(B(h,h))\,-\,(B(x,x))$\\
\phantom{x}\qquad\qquad\quad $=(B(x,x))\,-\,(B(x,x))\,+\,(B(x,h))\,+(B(x,h))\,+\,(B(h,h))$\\
\phantom{x}\qquad\qquad\quad $=2\cdot(B(x,h))\,+\,(B(h,h))$\\
\phantom{x}\qquad\qquad\quad $=(\,2\cdot(B(x,\bullet))+(B(\bullet,\bullet))\,)(h)$.\\
Then $Q_x^T=2\cdot(B(x,\bullet))+(B(\bullet,\bullet))$.\\
Let $L:=2\cdot(B(x,\bullet))$ and let $R:=B(\bullet,\bullet)$.\\
Then $Q_x^T=L+R$ and $L\in\scrl_V^W$ and $R\in\scrq_V^W$.\\
Since $L\in\scrl_V^W$, we get $L-L=\bfzero_S^T$, and so $L-L+R=\bfzero_S^T+R$.\\
Then $Q_x^T-L=L+R-L=L-L+R=\bfzero_V^W+R=R$.\\
Then $Q_x^T-L=R\in\scrq_V^W\subseteq\scrbigo_2^{VW}\subseteq\scrlito_1^{VW}$.\\
So, since $L\in\scrl_V^W$, we get $L\in\LINS_xQ$, and so $D_x^{VW}Q=L$.\\
Then $D_x^{VW}Q=L=2\cdot(B(x,\bullet))$, as desired.
QED
\vskip.1in\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 5: Due on Tuesday 26 February}
\vskip.1in\hrule\vskip.1in\noindent
5-1. Let $S:=\R^2$ and let $T:=\R$. \quad
Show: $\scrq_{ST}\subseteq\scrbigo_2^{ST}$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall Q\in\scrq_{ST}$, $Q\in\scrbigo_2^{ST}$.\\
Given $Q\in\scrq_{ST}$. \qquad
Want: $Q\in\scrbigo_2^{ST}$.\\
Since $Q\in\scrq_{ST}$, it follows that $\dom[Q]=S$.\\
Let $\ds{\alpha:=\adj_{0_S}^{0_T}\left(\frac{Q}{|\bullet|_S^2}\right)}$. \qquad
Then $\alpha(0_S)=0_T$.\\
Also, $\dom[\alpha]=((\dom[Q])_{0_S}^\times)_{0_S}^+=(\dom[Q])_{0_S}^+=S_{0_S}^+=S$.\\
Also, $\forall x\in S\backslash\{0_S\}$, we have $Q(x)=(\alpha(x))\cdot(|x|_S^2)$.\\
Since $Q\in\scrq_{ST}$, it follows that $Q(0_S)=0_T$.\\
Recall that $T=\R$. \qquad
Then $0_T=0$.\\
Then $\alpha(0_S)=0_T=0$ and $Q(0_S)=0_T=0$.\\
Then $Q(0_S)=0=0\cdot0^2=(\alpha(0_S))\cdot(|0_S|_S^2)$.\\
Then $\forall x\in S$, we have $Q(x)=(\alpha(x))\cdot(|x|_S^2)$. \quad
Then $Q=\alpha\cdot(|\bullet|_S^2)$.\\
Want: $Q\in(\BNZ_{ST})\cdot(|\bullet|_S^2)$. \qquad\qquad
Want: $\alpha\in\BNZ_{ST}$.\\
Since $\dom[\alpha]=S$, we see that $\alpha\in\DNZ_{ST}$.\\
Want: $\exists B\in\scrb_S(0_S)$ s.t.~$\alpha_*(B)$ is bounded in $T$.\\
Want: $\im[\alpha]$ is bounded in $T$.\\
Recall that $T=\R$. \qquad\qquad
Want: $\im[\alpha]$ is bounded in $\R$.\\
Want: $\exists r>0$ s.t.~$\im[\alpha]\subseteq B_\R(0,r)$.\\
Let $C:=\{v\in S~\hbox{s.t.}~|v|_S=1\}$.\\
Since $C$ is closed and bounded in $S$, we see that $C$ is compact.\\
Since $Q\in\scrq_{ST}$, it follows that $Q$ is continuous.\\
Then $Q|C$ is continuous.\\
So, since $C$ is compact, by the Extreme Value Theorem,\\
\phantom{x}\qquad we conclude: \qquad $\min(Q|C)\,\ne\,\frownie\,\ne\,\max(Q|C)$.\\
Let $K:=\min(Q|C)$ and $L:=\max(Q|C)$.\\
Then $(\,K,L\in\R\,)\,\&\,(\,K\le Q\le L\hbox{ on }C\,)$.\\
Let $r:=1+[\max\{|K|,|L|\}]$. \qquad
Then $r>0$ and $-r<K$ and $L<r$.\\
Want: $\im[\alpha]\subseteq B_\R(0,r)$.\\
Want: $\forall y\in\im[\alpha]$, $y\in B_\R(0,r)$.\\
Given $y\in\im[\alpha]$. \qquad
Want: $y\in B_\R(0,r)$.\\
Since $y\in\im[\alpha]$, choose $x\in\dom[\alpha]$ s.t.~$y=\alpha(x)$.\\
Exactly one of the following is true:\\
\phantom{x}\qquad (1) $x=0_S$ \qquad\qquad or \qquad\qquad (2) $x\ne0_S$.\\
\\
{\it Case (1):}\\
We have $y=\alpha(x)=\alpha(0_S)=0\in B_\R(0,r)$, as desired.\\
{\it End of Case (1).}\\
\\
{\it Case (2):}\\
Since $x\ne0_S$, by definition of $\alpha$, we get: $\ds{\alpha(x)=\frac{Q(x)}{|x|_S^2}}$.\\
Also, since $x\ne0_S$, we get $|x|_S>0$.\\
Let $b:=|x|_S$. \qquad
Then $b>0$, so $\ds{\frac{b}{|b|}=1}$.\\
Let $\ds{u:=\frac{x}{b}}$. \qquad
Then $\ds{|u|_S=\frac{|x|_S}{|b|}=\frac{b}{|b|}=1}$.\\
Then $u\in C$, so, since $K\le Q\le L$ on $C$,
we get: \quad $K\le Q(u)\le L$.\\
So, since $-r<K$ and $L<r$, we get: \quad $-r<Q(u)<r$.\\
Then $Q(u)\in(-r;r)=B_\R(0,r)$.\\
Since $Q\in\scrq_{ST}$, we have $\ds{Q\left(\frac{x}{b}\right)=\frac{Q(x)}{b^2}}$.\\
Then $\ds{Q(u)=Q\left(\frac{x}{b}\right)=\frac{Q(x)}{b^2}=\frac{Q(x)}{|x|_S^2}=\alpha(x)}$.\\
Then $y=\alpha(x)=Q(u)\in B_\R(0,r)$, as desired.\\
{\it End of Case (2).}
QED
\vskip.1in\hrule\vskip.1in\noindent
5-2. Let $S,T\in\TNSR^+$. \quad
Show: $(\BNZ_{ST})\cdot(\CVZ_{S\R})\subseteq\CVZ_{ST}$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall\gamma\in(\BNZ_{ST})\cdot(\CVZ_{S\R})$, \quad $\gamma\in\CVZ_{ST}$.\\
Given $\gamma\in(\BNZ_{ST})\cdot(\CVZ_{S\R})$. \qquad
Want: $\gamma\in\CVZ_{ST}$.\\
Since $\gamma\in(\BNZ_{ST})\cdot(\CVZ_{S\R})$,\\
\phantom{x}\qquad choose $\alpha\in\BNZ_{ST}$, $\beta\in\CVZ_{S\R}$ s.t.~$\gamma=\alpha\cdot\beta$.\\
Since $\alpha\in\BNZ_{ST}\subseteq\DNZ_{ST}$ and since $\beta\in\CVZ_{S\R}\subseteq\DNZ_{S\R}$,\\
\phantom{x}\qquad choose $P,Q\in\scrb_S(0_S)$ s.t.~$P\subseteq\dom[\alpha]$ and $Q\subseteq\dom[\beta]$.\\
Then $P\cap Q\in\{P,Q\}\subseteq\scrb_S(0_S)$.\\
So, since $P\cap Q\subseteq(\dom[\alpha])\cap(\dom[\beta])=\dom[\alpha\cdot\beta]=\dom[\gamma]$,\\
\phantom{x}\qquad we conclude that $\gamma\in\DNZ_{ST}$.\\
Want: $\gamma(0_S)=0_T$ and $\gamma$ is continuous at $0_S$.\\
Since $P\in\scrb_S(0_S)$, we get $0_S\in P$.\\
Since $\alpha\in\DNZ_{ST}$, we get $\alpha:S\drta T$.\\
So, since $0_S\in P\subseteq\dom[\alpha]$, we get $\alpha(0_S)\in T$, so $[\alpha(0_S)]\cdot0=0_T$.\\
Since $\beta\in\CVZ_{S\R}$, we get $\beta(0_S)=0$.\\
Then $\gamma(0_S)=(\alpha\cdot\beta)(0_S)=[\alpha(0_S)]\cdot[\beta(0_S)]=[\alpha(0_S)]\cdot0=0_T$.\\
Want: $\gamma$ is continuous at $0_S$.\\
Want: $\forall\varepsilon>0$, $\exists\delta>0$ s.t., $\forall x\in\dom[\gamma]$,\\
\phantom{x}\qquad\qquad\qquad $[\,\,|x-0_S|_S<\delta\,\,]\,\,\Rightarrow\,\,[\,\,|[\gamma(x)]-[\gamma(0_S)]|_T<\varepsilon\,\,]$.\\
Given $\varepsilon>0$. \qquad
Want: $\exists\delta>0$ s.t., $\forall x\in\dom[\gamma]$,\\
\phantom{x}\qquad\qquad\qquad $[\,\,|x-0_S|_S<\delta\,\,]\,\,\Rightarrow\,\,[\,\,|[\gamma(x)]-[\gamma(0_S)]|_T<\varepsilon\,\,]$.\\
Since $\alpha\in\BNZ_{ST}$, choose $A\in\scrb_S(0_S)$ s.t.~$\alpha_*(A)$ is bounded in $T$.\\
Since $\alpha_*(A)$ is bounded in $T$,
\phantom{x}\qquad choose $Y\in\scrb_T$ s.t.~$\alpha_*(A)\subseteq Y$.\\
By the Superset Recentering Lemma (Theorem 38.17),\\
\phantom{x}\qquad choose $Z\in\scrb_T(0_T)$ s.t.~$Y\subseteq Z$.\\
Choose $r>0$ s.t.~$Z=B_T(0_T,r)$.\\
Since $\beta\in\CVZ_{S\R}$, it follows that $\beta$ is continuous at $0_S$,\\
\phantom{x}\qquad so choose $\eta>0$ s.t., $\forall x\in\dom[\beta]$,\\
\phantom{x}\qquad\qquad\qquad $[\,\,|x-0_S|_S<\eta\,\,]\,\,\Rightarrow\,\,[\,\,|[\beta(x)]-[\beta(0)]|<\varepsilon/r\,\,]$.\\
Let $B:=B_S(0_S,\eta)$. \qquad
Then $A,B\in\scrb_S(0_S)$. \qquad
Let $Q:=A\cap B$.\\
Then $Q\in\{A,B\}\subseteq\scrb_S(0_S)$. \qquad
Choose $\delta>0$ s.t.~$Q=B_S(0_S,\delta)$.\\
Want: $\forall x\in\dom[\gamma]$, $(\,\,[\,\,|x-0_S|_S<\delta\,\,]\,\,\Rightarrow\,\,[\,\,|[\gamma(x)]-[\gamma(0_S)]|_T<\varepsilon\,\,]\,\,)$.\\
Given $x\in\dom[\gamma]$.\\
Want: $[\,\,|x-0_S|_S<\delta\,\,]\,\,\Rightarrow\,\,[\,\,|[\gamma(x)]-[\gamma(0_S)]|_T<\varepsilon\,\,]$.\\
Assume: $|x-0_S|_S<\delta$. \qquad
Want: $|[\gamma(x)]-[\gamma(0_S)]|_T<\varepsilon$.\\
We have $x\in\dom[\gamma]=\dom[\alpha\cdot\beta]=(\dom[\alpha])\cap(\dom[\beta])$.\\
Then $x\in(\dom[\alpha])\cap(\dom[\beta])\subseteq\dom[\beta]$.\\
Also, since $|x-0_S|_S<\delta$, we have $x\in B_S(0_S,\delta)$.\\
Then $x\in B_S(0_S,\delta)=Q=A\cap B$.\\
Then $x\in A\cap B\subseteq B=B_S(0_S,\eta)$, so $|x-0_S|_S<\eta$.\\
So, since $x\in\dom[\beta]$, by choice of $\eta$, we get: $|[\beta(x)]-[\beta(0_S)]|<\varepsilon/r$.\\
Since $\beta\in\CVZ_{S\R}$, we get $\beta(0_S)=0$. \qquad
Then $[\beta(x)]-[\beta(0_S)]=\beta(x)$.\\
Then $|\beta(x)|=|[\beta(x)]-[\beta(0_S)]|<\varepsilon/r$, so $|\beta(x)|<\varepsilon/r$.\\
Since $x\in(\dom[\alpha])\cap(\dom[\beta])\subseteq\dom[\alpha]$ and since $x\in A\cap B\subseteq A$,\\
\phantom{x}\qquad we conclude that $\alpha(x)\in\alpha_*(A)$.\\
Then $\alpha(x)\in\alpha_*(A)\subseteq Y\subseteq Z=B_T(0_T,r)$,
so $|\alpha(x)|_T<r$.\\
Since $0\le|\alpha(x)|_T<r$ and since $0\le|\beta(x)|<\varepsilon/r$,\\
\phantom{x}\qquad it follows that $|\alpha(x)|_T\cdot|\beta(x)|<r\cdot(\varepsilon/r)$.\\
Recall that $\gamma(0_S)=0_T$. \quad
We have $\gamma(x)=(\alpha\cdot\beta)(x)=[\alpha(x)]\cdot[\beta(x)]$.\\
Then $|[\gamma(x)]-[\gamma(0_S)]|_T=|[\gamma(x)]-0_T|_T=|\gamma(x)|_T=|[\alpha(x)]\cdot[\beta(x)]|_T$\\
\phantom{x}\qquad\qquad\qquad\qquad $\,\,=|\alpha(x)|_T\cdot|\beta(x)|<r\cdot(\varepsilon/r)=\varepsilon$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
5-3. Let $S\in\TNSR^+$, \quad $f,g:\R\drta S$, \quad $p\in\R$.\\
\phantom{x}\qquad Show: $(f\sbullet g)_p^T\,\,=\,\,(f_p^T)\sbullet(g_p)\,+\,(f_p)\sbullet(g_p^T)\,+\,(f_p^T)\sbullet(g_p^T)$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall h\in\R$,\\
\phantom{x}\qquad\qquad $(f\sbullet g)_p^T(h)=((f_p^T)\sbullet(g_p)+(f_p)\sbullet(g_p^T)+(f_p^T)\sbullet(g_p^T))(h)$.\\
Given $h\in\R$.\\
Want: $(f\sbullet g)_p^T(h)=((f_p^T)\sbullet(g_p)+(f_p)\sbullet(g_p^T)+(f_p^T)\sbullet(g_p^T))(h)$.\\
Let $A:=f_{p+h}$, $B:=g_{p+h}$.\\
Then $A-a=f_{p+h}-f_p=f_p^T(h)$.\\
Also, $B-b=g_{p+h}-g_p=g_p^T(h)$.\\
We have $A\sbullet B-a\sbullet b\,\,=\,\,a\sbullet(B-b)\,+\,b\sbullet(A-a)$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad $+\,(A-a)\sbullet(B-b)$.\\
Then
$(f\sbullet g)_p^T(h)=[(f\sbullet g)_{p+h}]-[(f\sbullet g)_p]=f_{p+h}\sbullet g_{p+h}-f_p\sbullet g_p$\\
\phantom{xxxx}$=A\sbullet B-a\sbullet b=(A-a)\sbullet b+a\sbullet(B-b)+(A-a)\sbullet(B-b)$\\
\phantom{xxxx}$=(f_p^T(h))\sbullet(g_p)+(f_p)\sbullet(g_p^T(h))+(f_p^T(h))\sbullet(g_p^T(h))$\\
\phantom{xxxx}$=((f_p^T)\sbullet(g_p)+(f_p)\sbullet(g_p^T)+(f_p^T)\sbullet(g_p^T))(h)$,\\
as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
5-4. Let $S,T,U\in\TNSR^+$, \quad $f:S\drta T$, \quad $g:T\drta U$.\\
\phantom{x}\qquad Let $p\in S$, \quad $q:=f_p$. \qquad
Show: $(f\circ g)_p^T=(g_q^T)\circ(f_p^T)$
\vskip.1in\noindent
{\it Proof:}
Want: $\forall h\in S$, $((g\circ f)_p^T)(h)=((g_q^T)\circ(f_p^T))(h)$.\\
Given $h\in S$. \qquad
Want: $((g\circ f)_p^T)(h)=((g_q^T)\circ(f_p^T))(h)$.\\
We compute $((g\circ f)_p^T)(h)=[(g\circ f)(p+h)]-[(g\circ f)(p)]$\\
\phantom{x}\qquad\qquad\qquad\qquad\quad $=[g(f_{p+h})]-[g(f_p)]=[g(f_{p+h})]-[g(q)]$.\\
Exactly one of the following is true:\\
\phantom{x}\qquad (1) $q=\frownie$ \qquad\qquad or \qquad\qquad (2) $q\ne\frownie$.\\
\\
{\it Case (1):}\\
Since $q=\frownie$, we get:\\
\phantom{x}\qquad both \quad $[g(f_{p+h})]-[g(q)]=\frownie$ \quad and \quad $((g_q^T)\circ(f_p^T))(h)=\frownie$.\\
Then $((g\circ f)_p^T)(h)=[g(f_{p+h})]-[g(q)]=\frownie=((g_q^T)\circ(f_p^T))(h)$.\\
{\it End of Case (1).}\\
\\
{\it Case (2):}\\
Since $f_p=q\ne\frownie$, we get $f_p\in\im[f]\subseteq\R$.\\
Then $q=f_p\in T$, so $q-q=0_T$.\\
Let $k:=(f_p^T)(h)$. \qquad
Then $k=f_{p+h}-f_p=f_{p+h}-q$.\\
Then $q+k=q+(f_{p+h}-q)=f_{p+h}+(q-q)=f_{p+h}+0_T=f_{p+h}$.\\
Then $((g\circ f)_p^T)(h)=[g(f_{p+h})]-[g(q)]=[g(q+k)]-[g(q)]$\\
\phantom{x}\qquad\qquad\qquad\qquad $=g_q^T(k)=(g_q^T)((f_p^T)(h))=((g_q^T)\circ(f_p^T))(h)$,\\
as desired.
{\it End of Case (2).}
QED
\vskip.1in\hrule\vskip.1in\noindent
5-5. Let $S,T\in\TNSR$, \quad $i\in\scri_S$, \quad $j\in\scri_T$. \qquad\qquad
Show: $i\|j\in\scri_{S\otimes T}$.
\vskip.1in\noindent
{\it Proof:}
Choose $\sigma,\tau\in\N$ s.t.~$S\in\TNSR_\sigma$ and $T\in\TNSR_\tau$.\\
Choose $m\in\R^\sigma$ and $n\in\R^\tau$ s.t.~$S=\R^m$ and $T=\R^n$.\\
We have \quad $\,\,\,S=\R^m$, \qquad so \qquad $\scri_S=[m]$.\\
Also, \qquad\quad $T=\R^n$, \qquad $\,$so \qquad $\scri_S=[n]$.\\
Also, \quad $\,S\otimes T=\R^{m\|n}$, \quad $\,$so \quad $\scri_{S\otimes T}=[m\|n]$.\\
We have: \qquad $\,\,\,m=(m_1,\ldots,m_\sigma)$,\\
\phantom{x}\qquad and \qquad $\,\,\,\,\,n=(n_1,\ldots,n_\tau)$\\
\phantom{x}\qquad and \quad $\,\,\,m\|n=(m_1,\ldots,m_\sigma,n_1,\ldots,n_\tau)$.\\
Then: \qquad\quad $\,\,\,[m]=[1..m_1]\times\cdots\times[1..m_\sigma]$,\\
\phantom{x}\qquad and \qquad $\,\,[n]=[1..n_1]\times\cdots\times[1..n_\tau]$\\
\phantom{x}\qquad and \quad $[m\|n]=[1..m_1]\times\cdots\times[1..m_\sigma]\times[1..n_1]\times\cdots\times[1..n_\tau]$.\\
Then: \qquad $i\in\scri_S=[m]=[1..m_1]\times\cdots\times[1..m_\sigma]$\\
\phantom{x}\qquad and $\,\,j\in\scri_T=[n]=[1..n_1]\times\cdots\times[1..n_\tau]$,\\
\phantom{x}\qquad and so \quad $i\|j\,\,\in\,\,[1..m_1]\times\cdots\times[1..m_\sigma]\times[1..n_1]\times\cdots\times[1..n_\tau]$.\\
Then $\,\,i\|j\,\,\in\,\,[m\|n]\,\,=\,\,\scri_{S\otimes T}$, as desired.
QED
\vskip.1in\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 4: Due on Tuesday 19 February}
\vskip.1in\hrule\vskip.1in\noindent
4-1. Let $k\in\N_0$, $\alpha\in\scrbigo_k$ and $\ds{\lambda:=\adj_0^0\left(\frac{\alpha}{|\bullet|^k}\right)}$. \qquad
Show: $\lambda\in\BNZ$.
\vskip.1in\noindent
{\it Proof:}
Since $\alpha\in\scrbigo_k=(\BNZ)\cdot(|\bullet|^k)$,\\
\phantom{x}\qquad\qquad choose $\mu\in\BNZ$ s.t.~$\alpha=\mu\cdot(|\bullet|^k)$.\\
We have $\dom[\alpha]=(\dom[\mu])\cap(\dom[|\bullet|^k])$.\\
So, since $\dom[\mu]\subseteq\R=\dom[|\bullet|^k]$, we see that $\dom[\alpha]=\dom[\mu]$.\\
Since $\mu\in\BNZ\subseteq\DNZ$, choose $A\in\scrb_\R(0)$ s.t.~$A\subseteq\dom[\mu]$.\\
Then $A\subseteq\dom[\mu]=\dom[\alpha]$, \quad so $A\subseteq\dom[\alpha]$.\\
So, since $A\in\scrb_\R(0)$, we see that $\alpha\in\DNZ$.\\
We have $\ds{\dom[\lambda]=\left(\dom\left[\frac{\alpha}{|\bullet|^k}\right]\right)_0^+=((\dom[\alpha])_0^\times)_0^+=(\dom[\alpha])_0^+}$.\\
Then $A\subseteq\dom[\alpha]\subseteq(\dom[\alpha])_0^+=\dom[\lambda]$, \quad so $A\subseteq\dom[\lambda]$.\\
So, since $A\in\scrb_\R(0)$, we conclude that $\lambda\in\DNZ$.\\
Want: $\exists B\in\scrb_\R(0)$ s.t.~$\lambda_*(B)$ is bounded in $\R$.\\
Since $\mu\in\BNZ$, choose $C\in\scrb_\R(0)$ s.t.~$\mu_*(C)$ is bounded in $\R$.\\
Since $A,C\in\scrb_\R(0)$, we get $A\cap C\in\{A,C\}$.\\
Let $B:=A\cap C$. \qquad
Then $B=A\cap C\in\{A,C\}\subseteq\scrb_\R(0)$.\\
Want: $\lambda_*(B)$ is bounded in $\R$.\\
Since $\mu_*(C)$ is bounded in $\R$, choose $D\in\scrb_\R$ s.t.~$\mu_*(C)\subseteq D$.\\
By the Superset Recentering Lemma, choose $E\in\scrb_\R(0)$ s.t.~$D\subseteq E$.\\
Since $E\in\scrb_\R(0)$, it follows that $0\in E$.\\
Since $\mu_*(C)\subseteq D\subseteq E$ and since $0\in E$,\\
\phantom{x}\qquad\qquad we conclude that $(\mu_*(C))\cup\{0\}\subseteq E$.\\
So, as $E\in\scrb_\R$, it suffices to show: \quad $\lambda_*(B)\subseteq E$.\\
Want: $\forall y\in\lambda_*(B)$, $y\in E$.\\
Given $y\in\lambda_*(B)$. \qquad
Want: $y\in E$.\\
Since $y\in\lambda_*(B)$, choose $x\in B\cap(\dom[\lambda])$ s.t.~$y=\lambda_x$.\\
Exactly one of the following is true:\\
\phantom{x}\qquad (1) $x=0$ \qquad\qquad or \qquad\qquad (2) $x\ne0$.\\
\\
{\it Case (1):}\\
We have $\ds{y=\lambda_x=\lambda_0=\left(\adj_0^0\left(\frac{\alpha}{|\bullet|^k}\right)\right)_0=0\in E}$, as desired.\\
{\it End of Case (1).}\\
\\
{\it Case (2):}\\
Since $x\ne0$, we have $\ds{\lambda_x=\left(\adj_0^0\left(\frac{\alpha}{|\bullet|^k}\right)\right)_x=\left(\frac{\alpha}{|\bullet|^k}\right)_x=\frac{\alpha_x}{|x|^k}}$.\\
Since $\alpha=\mu\cdot(|\bullet|^k)$, we get: $\alpha_x=(\mu\cdot(|\bullet|^k))_x$.\\
Then $\alpha_x=(\mu\cdot(|\bullet|^k))_x=\mu_x\cdot(|x|^k)$.\\
Since $B\in\scrb_\R(0)$, we get $B\subseteq\R$. \qquad
Then $x\in B\subseteq\R$.\\
So,  since $x\ne0$, we get $\ds{\frac{|x|^k}{|x|^k}=1}$.\\
Then $\ds{\lambda_x=\frac{\alpha_x}{|x|^k}=\frac{\mu_x\cdot(|x|^k)}{|x|^k}=\mu_x}$.\\
We have $x\in B\cap(\dom[\lambda])\subseteq B=A\cap C$, so $x\in A$ and $x\in C$.\\
Since $x\in A\subseteq\dom[\mu]$ and since $x\in C$, we get $\mu_x\in\mu_*(C)$.\\
Then $y=\lambda_x=\mu_x\in\mu_*(C)\subseteq D\subseteq E$, as desired.\\
{\it End of Case (2).}
QED
\vskip.1in\hrule\vskip.1in\noindent
4-2. Let $f:\R\drta\R$ and let $U\in\R$.\\
Assume that $U$ is open in $\R$ and that $U\subseteq\dom[f']$.\\
Assume that $f$ is semi-increasing on $U$.\\
Let $T:=f'_*(U)$. \qquad
Show: $T\ge0$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall m\in T$, $m\ge0$.\\
Given $m\in T$. \qquad
Want: $m\ge0$.\\
Assume $m<0$. \qquad
Want: Contradiction.\\
Since $m\in T=f'_*(U)$, choose $p\in U\cap(\dom[f'])$ s.t.~$m=f'_p$.\\
Then $p\in U\cap(\dom[f])\subseteq U$, \qquad so $p\in U$.\\
Since $U$ is open in $\R$, we get: $U=\Int_\R U$.\\
Since $p\in U=\Int_\R U$, choose $B\in\scrb_\R(p)$ s.t.~$B\subseteq U$.\\
Since $B\in\scrb_\R(p)$, choose $r>0$ s.t.~$B=B_\R(p,r)$.\\
Since $f'_p=m<0$, by Theorem 71.5, choose $\delta>0$ s.t.\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad (1) $f<f_p$ on $(p;p+\delta)$ \qquad\qquad and\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad (2) $f>f_p$ on $(p-\delta;p)$.\\
Let $\gamma:=[1/2]\cdot[\min\{\delta,r\}]$. \qquad
Then $0<\gamma<\delta$ and $-r<\gamma<r$.\\
Let $q:=p+\gamma$. \qquad
Then $p<q<p+\delta$ and $p-r<q<p+r$.\\
Then $q\in(p;p+\delta)$ and $q\in(p-r;p+r)$.\\
We have $q\in(p-r;p+r)=B_\R(p,r)=B\subseteq U$.\\
Then $p,q\in U$ and $p\le q$.\\
So, since $f$ is semi-increasing on $U$, we get: $f_p\le f_q$.\\
Since $q\in(p;p+\delta)$, by (1), we get: $f_q<f_p$.\\
Then $f_q<f_p\le f_q$, so $f_q<f_q$, so $f_q\ne f_q$. \quad
Contradiction. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
4-3. Let $f:\R\to\R$ and let $J$ be an interval.\\
Assume that $f$ is c/d on $J$. \qquad
Let $I:=\Int_\R J$ and $T:=f'_*(I)$.\\
Assume $T\ge0$. \qquad
Show: $f$ is semi-increasing on $J$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall a,b\in J$, $[\,(\,a\le b\,)\,\Rightarrow\,(\,f_a\le f_b\,)\,]$.\\
Given $a,b\in\T$. \qquad
Want: $(\,a\le b\,)\,\Rightarrow\,(\,f_a\le f_b\,)$.\\
Assume: $a\le b$. \qquad
Want: $f_a\le f_b$.\\
Assume $f_a>f_b$. \qquad
Want: Contradiction.\\
Since $f_a>f_b$, we get $f_a\ne f_b$, and so $a\ne b$.\\
Since $a\ne b$ and $a\le b$, we get: $a<b$. \qquad
Then $b-a>0$.\\
Since $f_a>f_b$, we get: $f_b-f_a<0$.\\
So, since $b-a>0$, we get $\ds{\frac{f_b-f_a}{b-a}<0}$.\\
Then $\ds{\DQ_f(a,b)=\frac{f_b-f_a}{b-a}<0}$, so $\DQ_f(a,b)<0$.\\
By Theorem 72.22, we have $\DQ_f(a,b)\in f'_*(I)$.\\
Then $\DQ_f(a,b)\in f'_*(I)=T\ge0$, so $0\le\DQ_f(a,b)$.\\
Then $0\le\DQ_f(a,b)<0$, so $0<0$. \qquad
Contradiction. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
4-4. Let $f:\R\drta\R$ and let $p\in\R$.\\
\phantom{xx}\quad Show: $(f(p+\bullet))'=f'(p+\bullet)$.
\vskip.1in\noindent
{\it Proof:}
Let $g:=f(p+\bullet)$. \qquad
Want: $g'=f'(p+\bullet)$.\\
Want: $\forall q\in\R$, $g'(q)=f'(p+q)$.\\
Given $q\in\R$. \qquad
Want: $g'(q)=f'(p+q)$. \qquad
Let $R:=\{\sl_L\,|\,L\in\LINS_qg\}$, $S:=\{\sl_L\,|\,L\in\LINS_{p+q}f\}$.\\
Then $g'(q)=\UE\,R$ and $f'(p+q)=\UE\,S$. \qquad
Want: $R=S$.\\
Want: $\LINS_qg=\LINS_{p+q}f$.\\
Want: $\{L\in\scrl\,|\,g_q^T-L\in\scrlito_1\}=\{L\in\scrl\,|\,f_{p+q}^T-L\in\scrlito_1\}$.\\
It suffices to show: $g_q^T=f_{p+q}^T$.\\
Want: $\forall h\in\R$, $(g_q^T)(h)=(f_{p+q}^T)(h)$.\\
Given $h\in\R$. \qquad
Want: $(g_q^T)(h)=(f_{p+q}^T)(h)$.\\
Want: $[g(q+h)]-[g(q)]=[f(p+q+h)]-[f(p+q)]$.\\
We have \qquad\quad $\,\,\,\,\,\,g(q+h)=(f(p+\bullet))(q+h)=f(p+q+h)$\\
\phantom{x}\qquad and \qquad\qquad\quad $\,\,g(q)=(f(p+\bullet))(q)=f(p+q)$.\\
Then \qquad\qquad\quad $\,\,\,g(q+h)=f(p+q+h)$\\
\phantom{x}\qquad and \qquad\qquad\quad $\,\,\,g(q)=f(p+q)$,\\
\phantom{x}\qquad so $[g(q+h)]-[g(q)]=[f(p+q+h)]-[f(p+q)]$.
QED
\vskip.1in\hrule\vskip.1in\noindent
4-5. Let $f:\R\drta\R$ and let $p\in\dom[f]$.\\
Show: $[\,(\,f\hbox{ has a strict local minimum at }p\hbox{ in }\R\,)$\\
\phantom{x}\quad $\Leftrightarrow\,\,\,\,(\,f_p^T\hbox{ has a strict local minimum at }0\hbox{ in }\R\,)\,]$.
\vskip.1in\noindent
{\it Proof:}
Let $g:=f_p^T$.\\
Since $p\in\dom[f]$, we get $f_p\in\im[f]$.\\
Then $f_p\in\im[f]\subseteq\R$, so $f_p\in\R$.\\
\\
{\it Proof of $\Rightarrow$:}\\
Assume: $f\hbox{ has a strict local minimum at }p\hbox{ in }\R$.\\
Want: $f_p^T\hbox{ has a strict local minimum at }0\hbox{ in }\R$.\\
Want: $g\hbox{ has a strict local minimum at }0\hbox{ in }\R$.\\
Choose $B\in\scrb_\R(p)$ s.t.~$f>f_p$ on $B_p^\times$.\\
Since $B\in\scrb_\R(p)$, we get $B-p\in\scrb_\R(0)$.\\
Let $A:=B-p$. \qquad
Then $A\in\scrb_\R(0)$.\\
It therefore suffices to show: $g>g_0$ on $A_0^\times$.\\
Want: $\forall h\in A_0^\times$, $g_h>g_0$.\\
Given $h\in A_0^\times$. \qquad
Want: $g_h>g_0$.\\
Since $A=B-p$, it follows that $p+A=B$.\\
Since $h\in A_0^\times$, we get $h\in A$ and $h\ne0$.\\
Then $p+h\in p+A$ and $p+h\ne p$.\\
Then $p+h\in B$ and $p+h\ne p$, so $p+h\in B_p^\times$.\\
So, by the choice of $B$, it follows that $f_{p+h}>f_p$.\\
So, since $f_p\in\R$, we get: $f_{p+h}-f_p>f_p-f_p$.\\
Then $g_h=(f_p^T)_h=f_{p+h}-f_p>f_p-f_p=f_{p+0}-f_p=g_0$, as desired.\\
{\it End of proof of $\Rightarrow$.}\\
\\
{\it Proof of $\Leftarrow$:}\\
Assume: $f_p^T\hbox{ has a strict local minimum at }0\hbox{ in }\R$.\\
Want: $f\hbox{ has a strict local minimum at }p\hbox{ in }\R$.\\
Know: $g\hbox{ has a strict local minimum at }0\hbox{ in }\R$.\\
Choose $A\in\scrb_\R(0)$ s.t.~$g>g_0$ on $A_0^\times$.\\
Since $A\in\scrb_\R(0)$, we get $A+p\in\scrb_\R(p)$.\\
Let $B:=A+p$. \qquad
Then $B\in\scrb_\R(p)$.\\
It therefore suffices to show: $f>f_p$ on $B_p^\times$.\\
Want: $\forall x\in B_p^\times$, $f_x>f_p$.\\
Given $x\in B_p^\times$. \qquad
Want: $f_x>f_p$.\\
Since $x\in B_p^\times$, we get $x\in B$ and $x\ne p$.\\
Then $x-p\in B-p$ and $x-p\ne0$. \qquad
Let $h:=x-p$.\\
Then $h\in B-p$ and $h\ne0$.\\
Since $B=A+p$, it follows that $B-p=A$.\\
Then $h\in A$ and $h\ne0$. \qquad
Then $h\in A_0^\times$.\\
So, by the choice of $A$, it follows that $g_h>g_0$.\\
So, as $f_p\in\R$, we get: $g_h+f_p>g_0+f_p$.\\
We have $g_h=f_{p+h}-f_p$ and $g_0=f_{p+0}-f_p$,\\
\phantom{x}\qquad so, as $f_p\in\R$, we get $g_h+f_p=f_{p+h}$ and $g_0+f_p=f_{p+0}$.\\
Since $h=x-p$, we get $p+h=x$.\\
Then $f_x=f_{p+h}=g_h+f_p>g_0+f_p=f_{p+0}=f_p$, as desired.\\
{\it End of proof of $\Leftarrow$.} \quad
QED
\vskip.1in\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 3: Due on Tuesday 12 February}
\vskip.1in\hrule\vskip.1in\noindent
3-1. Define $\alpha:\R\drta\R$ by $\ds{\alpha_h=\frac{h^2}{9\cdot(3+h)}}$. \qquad
Show: $\alpha\in\scrbigo_2$.
\vskip.1in\noindent
{\it Proof:}
Let $\ds{\lambda:=\adj_0^0\left(\frac{\alpha}{|\bullet|^2}\right)}$.\\
Since $\alpha_0=0$, it follows, from Theorem 69.5, that $\alpha=\lambda\cdot(|\bullet|^2)$.\\
Want: $\alpha\in(\BNZ)\cdot(|\bullet|^2)$. \qquad
Want: $\lambda\in\BNZ$.\\
We have $\ds{\dom[\lambda]=\left(\dom\left[\frac{\alpha}{|\bullet|^k}\right]\right)_0^+=((\dom[\alpha])_0^\times)_0^+\supseteq\dom[\alpha]}$.\\
So, since $\alpha\in\DNZ$, it follows that $\lambda\in\DNZ$.\\
Want: $\exists B\in\scrb_\R(0)$ s.t.~$\lambda_*(B)$ is bounded in $\R$.\\
Let $B:=B_\R(0,1)$. \qquad Then $B\in\scrb_\R(0)$.\\
Want: $\lambda_*(B)$ is bounded in $\R$. \qquad
Want: $\lambda_*(B)\subseteq B_\R(0,100)$.\\
Want: $\forall y\in\lambda_*(B)$, $y\in B_\R(0,100)$.\\
Given $y\in\lambda_*(B)$. \qquad
Want $y\in B_\R(0,100)$.\\
Since $y\in\lambda_*(B)$, choose $h\in B$ s.t.~$y=\lambda_h$.\\
Exactly one of the following is true:\\
\phantom{x}\qquad (1) $h=0$ \qquad\qquad or \qquad\qquad (2) $h\ne0$.\\
\\
{\it Case (1):}\\
We have $\ds{y=\lambda_h=\lambda_0=\left(\adj_0^0\left(\frac{\alpha}{|\bullet|^2}\right)\right)_0=0\in B_\R(0,100)}$, as desired.\\
{\it End of Case (1).}\\
\\
{\it Case (2):}\\
We have $\ds{y=\lambda_h=\left(\adj_0^0\left(\frac{\alpha}{|\bullet|^2}\right)\right)_h=\left(\frac{\alpha}{|\bullet|^2}\right)_h=\frac{\alpha_h}{|h|^2}}$.\\
Then $\ds{|y|=\frac{|\alpha_h|}{|h|^2}}$. \quad
Also, $\ds{|\alpha_h|=\frac{|h|^2}{9\cdot|3+h|}}$. \quad
Then $\ds{|y|=\frac{1}{9\cdot|3+h|}}$.\\
Since $h\in B=B_\R(0,1)$, it follows that $d_\R(h,0)<1$.\\
Then $|h-0|=d_\R(h,0)<1$, so $-1<h<1$.\\
Then $3-1<3+h<3+1$, so $2<3+h<4$.\\
Since $3+h>2>0$, we get $|3+h|=3+h$ and $\ds{\frac{1}{3+h}<\frac{1}{2}}$.\\
Since $\ds{\frac{1}{3+h}<\frac{1}{2}}$, we see that $\ds{\frac{1}{9}\cdot\frac{1}{3+h}<\frac{1}{9}\cdot\frac{1}{2}}$.\\
Then $\ds{d_\R(y,0)=|y-0|=|y|=\frac{1}{9\cdot|3+h|}=\frac{1}{9\cdot(3+h)}}$\\
\phantom{x}\qquad\qquad\qquad $\ds{=\frac{1}{9}\cdot\frac{1}{3+h}<\frac{1}{9}\cdot\frac{1}{2}<100}$.\\
Then $y\in B_\R(0,100)$.
{\it End of Case (2).}
QED
\vskip.1in\hrule\vskip.1in\noindent
3-2. Define $r:\R\drta\R$ by $r_x=1/x$. \qquad
Show: $r'_3=-1/9$.
\vskip.1in\noindent
{\it Proof:}
Define $L\in\scrl$ by $L_h=-h/9$.\\
Then $\sl_L=L_1=-1/9$. \qquad
Want $r'_3=\sl_L$.\\
By Theorem 64.49, it suffices to show: $r_3^T-L\in\scrlito_1$.\\
Define $\alpha:\R\drta\R$ by $\ds{\alpha_h=\frac{h^2}{9\cdot(3+h)}}$.\\
By HW\#3-1, $\alpha\in\scrbigo_2$,
so, since $\scrbigo_2\subseteq\scrlito_1$, we see that $\alpha\in\scrlito_1$.\\
It therefore suffices to show: $r_3^T-L=\alpha$.\\
Want: $\forall h\in\R$, $(r_3^T-L)_h=\alpha_h$.\\
Given $h\in\R$. \qquad
Want: $(r_3^T-L)_h=\alpha_h$.\\
We compute: $(r_3^T-L)_h\,\,\,=\quad[(r_3^T)_h]-[L_h]\,\,\,=\,\,\,[r_{3+h}-r_3]-[-h/9]$\\
\phantom{xtl}\qquad\qquad\qquad\qquad\qquad $\ds{=\quad r_{3+h}-r_3+\frac{h}{9}\quad=\quad\frac{1}{3+h}-\frac{1}{3}+\frac{h}{9}}$\\
\phantom{xtl}\qquad\qquad\qquad\qquad\qquad $\ds{=\quad\frac{9}{9\cdot(3+h)}-\frac{3\cdot(3+h)}{9\cdot(3+h)}+\frac{h\cdot(3+h)}{9\cdot(3+h)}}$\\
\phantom{xtl}\qquad\qquad\qquad\qquad\qquad $\ds{=\quad\frac{9-9-3h+3h+h^2}{9\cdot(3+h)}\quad=\quad\frac{h^2}{9\cdot(3+h)}}$\\
\phantom{xtl}\qquad\qquad\qquad\qquad\qquad $=\quad\alpha_h$, \qquad\qquad as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
3-3. Define $L\in\scrl$ by $L_x=7x$. \qquad
Show: $L'=C_\R^7$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall x\in\R$, $L'_x=(C_\R^7)_x$.\\
Given $x\in\R$. \qquad
Want: $L'_x=(C_\R^7)_x$. \qquad
Want: $L'_x=7$.\\
We have $\sl_L=L_1=7\cdot1=7$. \qquad
Want: $L'_x=\sl_L$.\\
By Theorem 64.49, it suffices to show: $L_x^T-L\in\scrlito_1$.\\
Since $\bfzero\in\scrlito_1$, it suffices to show: $L_x^T-L=\bfzero$.\\
Want: $\forall h\in\R$, $(L_x^T-L)_h=\bfzero_h$.\\
Given $h\in\R$. \qquad
Want: $(L_x^T-L)_h=\bfzero_h$.\\
We have $(L_x^T-L)_h=[(L_x^T)_h]-L_h=[L_{x+h}-L_x]-L_h$\\
\phantom{x}\qquad\qquad\qquad $=[7\cdot(x+h)-7x]-7h=0=\bfzero_h$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
3-4. Let $a,b\in\R$. \qquad
Assume $a<b$. \qquad
Let $I:=(a;b)$, $J:=[a;b]$.\\
\phantom{x}\qquad\qquad Show: $\Int_\R J=I$ and $\Cl_\R I=J$.
\vskip.1in\noindent
{\it Proof:}
We wish to prove all of the following:\\
\phantom{x}\qquad (1) $\Int_\R J\subseteq I$ \qquad and \qquad (2) $\Int_\R J\supseteq I$ \qquad and\\
\phantom{x}\qquad (3) $\Cl_\R I\subseteq J$ \qquad $\,$and \qquad (4) $\Cl_\R I\supseteq J$.\\
\\
{\it Proof of (1):}\\
Want: $\forall x\in\Int_\R J$, $x\in I$. \qquad
Given $x\in\Int_\R J$. \qquad
Want: $x\in I$.\\
\\
{\it Claim A:} $x\ne a$.\\
{\it Proof of Claim A:}\\
Assume $x=a$. \qquad
Want: Contradiction.\\
Since $a=x\in\Int_\R J$, choose $U\in\scrb_\R(a)$ s.t.~$U\subseteq J$.\\
Since $U\in\scrb_\R(a)$, choose $r>0$ s.t.~$U=B_\R(a,r)$.\\
Let $y:=a-(r/2)$. \qquad
Then $y<a$.\\
Also, $d_\R(y,a)=|a-y|=r/2<r$, so $y\in B_\R(a,r)$.\\
Then $y\in B_\R(a,r)=U\subseteq J=[a;b]$, so $y\ge a$.\\
Then $y<a$ and $y\ge a$. \qquad
Contradiction.\\
{\it End of proof of Claim A.}\\
\\
{\it Claim B:} $x\ne b$.\\
{\it Proof of Claim B:}\\
Assume $x=b$. \qquad
Want: Contradiction.\\
Since $b=x\in\Int_\R J$, choose $U\in\scrb_\R(b)$ s.t.~$U\subseteq J$.\\
Since $U\in\scrb_\R(b)$, choose $r>0$ s.t.~$U=B_\R(b,r)$.\\
Let $y:=b+(r/2)$. \qquad
Then $y>b$.\\
Also, $d_\R(y,b)=|b-y|=r/2<r$, so $y\in B_\R(b,r)$.\\
Then $y\in B_\R(b,r)=U\subseteq J=[a;b]$, so $y\le b$.\\
Then $y>b$ and $y\le b$. \qquad
Contradiction.\\
{\it End of proof of Claim B.}\\
\\
Since $x\in J=[a;b]$, it follows, from Claim A and Claim B,\\
\phantom{x}\qquad\qquad\qquad that $x\in[a;b]\backslash\{a,b\}$.\\
Then $x\in [a;b]\backslash\{a,b\}=(a;b)=I$, as desired.\\
{\it End of proof of (1).}\\
\\
{\it Proof of (2):}\\
Want: $\forall x\in I$, $x\in\Int_\R J$. \qquad
Given $x\in I$. \qquad
Want: $x\in\Int_\R J$.\\
Want: $\exists U\in\scrb_\R(x)$ s.t.~$U\subseteq J$.\\
Since $x\in I=(a;b)$, we have $a<x<b$.\\
Then $x-a>0$ and $b-x>0$, and so $\min\{x-a,b-x\}>0$.\\
Let $r:=\min\{x-a,b-x\}$. \qquad
Then $r>0$ and $r\le x-a$ and $r\le b-x$.\\
Let $U:=B_\R(x,r)$. \qquad
Then $U\in\scrb_\R(x)$. \qquad
Want: $U\subseteq J$.\\
Want: $\forall y\in U$, $y\in J$.\\
Given $y\in U$. \qquad
Want: $y\in J$.\\
Since $y\in U=B_\R(x,r)$, we get $d_\R(y,x)<r$.\\
Then $|y-x|=d_\R(y,x)<r$, so $x-r<y<x+r$.\\
Since $r\le x-a$, we get $x-r\ge x-(x-a)$, and so $x-r\ge a$.\\
Since $r\le b-x$, we get $x+r\le x+(b-x)$, and so $x+r\le b$.\\
We have $a\le x-r<y$, so $a<y$. \qquad
Also, $y<x+r\le b$, so $y<b$.\\
Then $a<y<b$, so $y\in(a;b)$.\\
So, since $(a;b)=I$, we get $y\in I$, as desired.\\
{\it End of proof of (2).}\\
\\
{\it Proof of (3):}\\
Want: $\forall x\in\Cl_\R I$, $x\in J$. \qquad
Given $x\in\Cl_\R I$. \qquad
Want: $x\in J$.\\
Since $x\in\Cl_\R I$, choose $s\in I^\N$ s.t.~$s_\bullet\to x$ in $\R$.\\
We have: $\forall j\in\N$, $s_j\in I=(a;b)\subseteq[a;b]$, so $a\le s_j\le b$.\\
Since $s_\bullet\to x$ in $\R$ and since, $\forall j\in\N$, $s_j\le b$,\\
\phantom{x}\qquad it follows, from Theorem 47.1, that $x\le b$.\\
Since $s_\bullet\to x$ in $\R$ and since, $\forall j\in\N$, $s_j\ge a$,\\
\phantom{x}\qquad it follows, from Theorem 47.2, that $x\ge a$.\\
Since $a\le x\le b$, we get $x\in[a;b]$.\\
So, since $[a;b]=J$, we get $x\in J$, as desired.\\
{\it End of proof of (3).}\\
\\
{\it Proof of (4):}\\
Want: $\forall x\in J$, $x\in\Cl_\R I$. \qquad
Given $x\in J$. \qquad
Want: $x\in\Cl_\R I$.\\
Want: $\exists s\in I^\N$ s.t.~$s_\bullet\to x$ in $\R$.\\
Exactly one of the following is true:\\
\phantom{x}\qquad ($\alpha$) $x=a$ \qquad\qquad or \qquad\qquad ($\beta$) $x\ne a$.\\
\\
{\it Case ($\alpha$):}\\
Since $a<b$, we see that $b-a>0$. \qquad
Then $(b-a)/2>0$.\\
Let $z:=(b-a)/2$. \qquad
Then $z>0$.\\
Also, as $b-a>0$, we get $(b-a)/2<b-a$.\\
Then $z<b-a$, and so $a+z<b$.\\
Define $s\in\R^\N$ by $s_j=a+(z/j)$.\\
\\
{\it Claim X:} $s\in I^\N$.\\
{\it Proof of Claim X:}\\
We have $\dom[s]=\N$. \qquad Want: $\im[s]\subseteq I$.\\
Want: $\forall q\in\im[s]$, $q\in I$.\\
Given $q\in\im[s]$. \qquad
Want: $q\in I$.\\
Since $q\in\im[s]$, choose $j\in\N$ s.t.~$s_j=q$. \qquad
Want: $s_j\in I$.\\
We have $j>0$, so, as $z>0$,\\
\phantom{x}\qquad we get $z/j>0$, so $a+(z/j)>a+0$.\\
Then $s_j=a+(z/j)>a+0=a$, so $a<s_j$.\\
We have $j\ge1$, so, as $z>0$,\\
\phantom{x}\qquad we get $z/j\le z/1$, so $a+(z/j)\le a+(z/1)$.\\
Then $s_j=a+(z/j)\le a+(z/)=a+z<b$, so $s_j<b$.\\
Since $a<s_j<b$, we get $s_j\in(a;b)$. \qquad
Then $s_j\in(a;b)=I$.\\
{\it End of proof of Claim X.}\\
\\
By Claim X, $s\in I^\N$. \qquad Want: $s_\bullet\to x$ in $\R$.\\
By definition of Case ($\alpha$), we know that $x=a$.\\
Define $r\in\R^\N$ by $r_j=1/j$. \qquad
Then, by Theorem 39.7, $r_\bullet\to0$ in $\R$.\\
Then, by Theorem 40.2, $(z\cdot r)_\bullet\to z\cdot0$ in $\R$.\\
Let $c:=C_\N^a$. \qquad
Then, by Theorem 40.3, $c_\bullet\to a$ in $\R$.\\
Then, by Theorem 40.1, $(c+(z\cdot r))_\bullet\to a+(z\cdot0)$ in $\R$.\\
So, since $a+(z\cdot0)=a+0=a=x$, we get $(c+(z\cdot r))_\bullet\to x$ in $\R$.\\
It therefore suffices to show: $s=c+(z\cdot r)$.\\
Want: $\forall j\in\N$, $s_j=(c+(z\cdot r))_j$.\\
Given $j\in\N$. \qquad
Want: $s_j=(c+(z\cdot r))_j$.\\
We have $c_j=(C_\N^a)_j=a$.\\
Also, $(z\cdot r)_j=z\cdot(r_j)=z\cdot(1/j)=z/j$.\\
Then $s_j=a+(z/j)=c_j+[(z\cdot r)_j]=(c+(z\cdot r))_j$.\\
{\it End of Case ($\alpha$).}\\
\\
{\it Case ($\beta$):}\\
Since $x\in J=[a;b]$, we get $a\le x\le b$.\\
Since $a\le x$ and $x\ne a$, we get $a<x$. \qquad
Then $x-a>0$.\\
Then $(x-a)/2>0$. \qquad
Let $z:=(x-a)/2$. \qquad
Then $z>0$.\\
Also, as $x-a>0$, we get $(x-a)/2<x-a$.\\
Then $z<x-a$, and so $x-z>a$.\\
Define $s\in\R^\N$ by $s_j=x-(z/j)$.\\
\\
{\it Claim Y:} $s\in I^\N$.\\
{\it Proof of Claim Y:}\\
We have $\dom[s]=\N$. \qquad Want: $\im[s]\subseteq I$.\\
Want: $\forall q\in\im[s]$, $q\in I$.\\
Given $q\in\im[s]$. \qquad
Want: $q\in I$.\\
Since $q\in\im[s]$, choose $j\in\N$ s.t.~$s_j=q$. \qquad
Want: $s_j\in I$.\\
We have $j>0$, so, as $z>0$,\\
\phantom{x}\qquad we get $z/j>0$, so $x-(z/j)<x-0$.\\
Then $s_j=x-(z/j)<x-0=x\le b$, so $s_j<b$.\\
We have $j\ge1$, so, as $z>0$,\\
\phantom{x}\qquad we get $z/j\le z/1$, so $x-(z/j)\ge x-(z/1)$.\\
Then $s_j=x-(z/j)\ge x-(z/1)=x-z\ge a$, so $s_j>a$, so $a<s_j$.\\
Since $a<s_j<b$, we get $s_j\in(a;b)$. \qquad
Then $s_j\in(a;b)=I$.\\
{\it End of proof of Claim Y.}\\
\\
By Claim Y, $s\in I^\N$. \qquad Want: $s_\bullet\to x$ in $\R$.\\
Define $r\in\R^\N$ by $r_j=1/j$. \qquad
Then, by Theorem 39.7, $r_\bullet\to0$ in $\R$.\\
Then, by Theorem 40.2, $(-z\cdot r)_\bullet\to-z\cdot0$ in $\R$.\\
Let $c:=C_\N^x$. \qquad
Then, by Theorem 40.3, $c_\bullet\to x$ in $\R$.\\
Then, by Theorem 40.1, $(c+(-z\cdot r))_\bullet\to x-(z\cdot0)$ in $\R$.\\
So, since $x-(z\cdot0)=x-0=x$, we get $(c+(-z\cdot r))_\bullet\to x$ in $\R$.\\
It therefore suffices to show: $s=c+(-z\cdot r)$.\\
Want: $\forall j\in\N$, $s_j=(c+(-z\cdot r))_j$.\\
Given $j\in\N$. \qquad
Want: $s_j=(c+(-z\cdot r))_j$.\\
We have $c_j=(C_\N^x)_j=x$.\\
Also, $(-z\cdot r)_j=-z\cdot(r_j)=-z\cdot(1/j)=-z/j$.\\
Then $s_j=x-(z/j)=x+(-z/j)=c_j+[(-z\cdot r)_j]=(c+(-z\cdot r))_j$.\\
{\it End of Case ($\beta$).}\\
\\
{\it End of proof of (4).}
QED
\vskip.1in\hrule\vskip.1in\noindent
3-5. Find a continuous function $f:\R\to\R$ such that\\
\phantom{x}\qquad\qquad $[\,\,f(-2)=f(2)\,\,]\,\,\&\,\,[\,\,\forall x\in(-2;2),\,\,f'_x\ne0\,\,]$.
\vskip.1in\noindent
{\it Solution:}
Let $f:=|\bullet|$.
\vskip.1in\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 2: Due on Tuesday 5 February}
\vskip.1in\hrule\vskip.1in\noindent
2-1. Let $f:\R\drta\R$, $a\in\R$, $p\in\dom[f]$, $S\subseteq\R$ and $q:=f_p$.\\
Show: $(\,f>a\hbox{ on }S\,)\,\Leftrightarrow\,(\,f_p^T>a-q\hbox{ on }S-p\,)$.
\vskip.1in\noindent
{\it Proof:}
{\it Proof of $\Rightarrow$:}
Assume: $f>a\hbox{ on }S$. \quad
Want: $f_p^T>a-q\hbox{ on }S-p$.\\
Want: $\forall h\in S-p,\,f_p^T(h)>a-q$.\\
Given $h\in S-p$. \qquad
Want: $f_p^T(h)>a-q$.\\
Since $h\in S-p$, choose $x\in S$ s.t.~$h=x-p$. \qquad
Then $p+h=x$.\\
Since $x\in S$ and since $f>a\hbox{ on }S$, we get: $f_x>a$.\\
Then $f_x-q>a-q$.\\
Then $f_p^T(h)=f_{p+h}-f_p=f_x-q>a-q$, as desired.
{\it End of proof of $\Rightarrow$.}\\
\\
{\it Proof of $\Leftarrow$:}
Assume: $f_p^T>a-q\hbox{ on }S-p$. \quad
Want: $f>a\hbox{ on }S$.\\
Want: $\forall x\in S,\,f_x>a$.\\
Given $x\in S$. \qquad
Want: $f_x>a$.\\
Since $x\in S$, we get: $x-p\in S-p$.\\
Let $h:=x-p$. \qquad
Then $h\in S-p$ and $p+h=x$.\\
Since $h\in S-p$ and since $f_p^T>a-q\hbox{ on }S-p$, we get: $f_p^T(h)>a-q$.\\
Then $[f_p^T(h)]+q>a-q+q$. \qquad
Also, $-f_p+q=-q+q=0$.\\
Then $f_x=f_x-f_p+q=f_{p+h}-f_p+q=[f_p^T(h)]+q>a-q+q=a$,\\
\phantom{x}\qquad as desired. \quad
{\it End of proof of $\Leftarrow$.} \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
2-2. Show: $(\BNZ)\circ(\CVZ)\subseteq\BNZ$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall\alpha\in(\BNZ)\circ(\CVZ),\,\alpha\in\BNZ$.\\
Given $\alpha\in(\BNZ)\circ(\CVZ)$. \qquad
Want: $\alpha\in\BNZ$.\\
Choose $\beta\in\BNZ$, $\gamma\in\CVZ$ s.t.~$\alpha=\beta\circ\gamma$.\\
Since $\beta\in\BNZ$, we get $\beta\in\DNZ$,
so choose $A\in\scrb_\R(0)$ s.t.~$A\subseteq\dom[\beta]$.\\
Since $\gamma\in\CVZ$, we get $\gamma\in\DNZ$,
so choose $D\in\scrb_\R(0)$ s.t.~$D\subseteq\dom[\gamma]$.\\
Since $\gamma\in\CVZ$, we get: \qquad
$\gamma$ is continuous at $0$ \quad and \quad $\gamma_0=0$.\\
Since $\gamma$ is continuous at $0$ and $A\in\scrb_\R(0)=\scrb_\R(\gamma_0)$,\\
\phantom{x}\qquad choose $E\in\scrb_\R(0)$ s.t.~$\gamma_*(E)\subseteq A$.\\
Let $F:=D\cap E$. \qquad Then $F\in\scrb_\R(0)$.\\
\\
{\it Claim 1: $F\subseteq\dom[\alpha]$}.\\
{\it Proof of Claim 1:}\\
Want: $\forall x\in F$, $x\in\dom[\alpha]$.\\
Given $x\in F$. \qquad
Want: $x\in\dom[\alpha]$.\\
We have $x\in F\subseteq D\subseteq\dom[\gamma]$. \qquad
Also, $x\in F\subseteq E$.\\
Since $x\in\dom[\gamma]$ and since $x\in E$,
we get: $\gamma(x)\in\gamma_*(E)$.\\
Then $\gamma(x)\in\gamma_*(E)\subseteq A\subseteq\dom[\beta]$,
so $\beta(\gamma(x))\ne\frownie$.\\
Then $\alpha(x)=(\beta\circ\gamma)(x)=\beta(\gamma(x))\ne\frownie$,
so $x\in\dom[\alpha]$, as desired.\\
{\it End of Proof of Claim 1.}\\
\\
Since $F\in\scrb_\R(0)$, by Claim 1, we conclude that $\alpha\in\DNZ$.\\
Want: $\exists Z\in\scrb_\R(0)$ s.t. $\alpha_*(Z)$ is bounded in $\R$.\\
Since $\beta\in\BNZ$, choose $B\in\scrb_\R(0)$ s.t.~$\beta_*(B)$ is bounded in $\R$.\\
Since $\gamma$ is continuous at $0$ and $B\in\scrb_\R(0)=\scrb_\R(\gamma_0)$,\\
\phantom{x}\qquad choose $Z\in\scrb_\R(0)$ s.t.~$\gamma_*(Z)\subseteq B$.\\
Want: $\alpha_*(Z)$ is bounded in $\R$.\\
Since $\beta_*(B)$ is bounded in $\R$, it suffices to prove:
$\alpha_*(Z)\subseteq\beta_*(B)$.\\
Want: $\forall q\in\alpha_*(Z)$, $q\in\beta_*(B)$.\\
Given $q\in\alpha_*(Z)$. \qquad
Want: $q\in\beta_*(B)$.\\
Since $q\in\alpha_*(Z)$, choose $p\in Z\cap(\dom[\alpha])$ s.t.~$q=\alpha(p)$.\\
We have: $p\in Z$ and $p\in\dom[\alpha]$.\\
Since $p\in\dom[\alpha]$, we get: $\alpha(p)\ne\frownie$.\\
Also, $\beta(\gamma(p))=(\beta\circ\gamma)(p)=\alpha(p)$.\\
Then $\beta(\gamma(p))=\alpha(p)\ne\frownie$, so $\gamma(p)\in\dom[\beta]$.\\
Then $\gamma(p)\ne\frownie$, so $p\in\dom[\gamma]$.\\
Since $p\in\dom[\gamma]$ and since $p\in Z$, we conclude that $\gamma(p)\in\gamma_*(Z)$.\\
So, since $\gamma_*(Z)\subseteq B$, it follows that $\gamma(p)\in B$.\\
Since $\gamma(p)\in\dom[\beta]$ and since $\gamma(p)\in B$, we see that $\beta(\gamma(p))\in\beta_*(B)$.\\
Then $q=\alpha(p)=\beta(\gamma(p))\in\beta_*(B)$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
2-3. Let $j,k\in\N$.
Show: $\scrbigo_j\circ\scrlito_k\subseteq\scrlito_{jk}$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall\alpha\in\scrbigo_j\circ\scrlito_k,\,\alpha\in\scrlito_{jk}$.\\
Given $\alpha\in\alpha\in\scrbigo_j\circ\scrlito_k$. \qquad
Want: $\alpha\in\scrlito_{jk}$.\\
Choose $\beta\in\scrbigo_j$, $\gamma\in\scrlito_k$ s.t.~$\alpha=\beta\circ\gamma$.\\
Since $\beta\in\scrbigo_j=(\BNZ)\cdot(|\bullet|^j)$,
choose $\phi\in\BNZ$ s.t.~$\beta=\phi\cdot(|\bullet|^j)$.\\
Since $\gamma\in\scrlito_k=(\CVZ)\cdot(|\bullet|^k)$,
choose $\psi\in\CVZ$ s.t.~$\gamma=\psi\cdot(|\bullet|^k)$.\\
We have: $\forall x\in\R$, $\,\alpha(x)=(\beta\circ\gamma)(x)=\beta(\gamma(x))=(\phi\cdot(|\bullet|^j))(\gamma(x))$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad $=[\phi(\gamma(x)]\cdot[\,\,\,\,|\,\,\,\,\gamma(x)\,\,\,\,|^j\,\,\,\,]$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad $=[(\phi\circ\gamma)(x)]\cdot[\,\,\,\,|\,\,\,\,(\,\psi\,\cdot\,(|\bullet|^k)\,)(x)\,\,\,\,|^j]$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad $=[(\phi\circ\gamma)(x)]\cdot[\,\,\,\,|\,\,\,\,[\psi(x)]\,\cdot\,[|x|^k]\,\,\,\,|^j\,\,\,\,]$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad $=[(\phi\circ\gamma)(x)]\cdot[\,\,\,\,|\,\,\,\,\psi(x)\,\,\,\,|^j\,\,\,\,]\cdot[\,\,\,\,|\,\,\,\,|x|^k\,\,\,\,|^j\,\,\,\,]$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad $=[(\phi\circ\gamma)(x)]\cdot[|\psi(x)|^j]\cdot[|x|^{jk}]$\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad $=([\phi\circ\gamma]\cdot[|\psi^j|]\cdot[|\bullet|^{jk}])(x)$.\\
Then $\alpha=[\phi\circ\gamma]\cdot[|\psi^j|]\cdot[|\bullet|^{jk}]$.\\
We have $\phi\in\BNZ$ and $\gamma\in\scrlito_k\subseteq\scrlito_0=\CVZ$, so $\phi\circ\gamma\in(\BNZ)\circ(\CVZ)$.\\
By HW\#2-2, $(\BNZ)\circ(\CVZ)\subseteq\BNZ$. \qquad
Then $\phi\circ\gamma\in\BNZ$.\\
Since $\psi\in\CVZ$, we get $|\psi|\in|\CVZ|\subseteq\CVZ$.\\
So, since $j\in\N$, it follows that $|\psi|^j\in\CVZ$.\\
Then $[\phi\circ\gamma]\cdot[|\psi^j|]\in(\BNZ)\cdot(\CVZ)$.\\
By HW\#1-4, $(\BNZ)\cdot(\CVZ)\subseteq\CVZ$. \qquad
Then $[\phi\circ\gamma]\cdot[|\psi^j|]\in\CVZ$.\\
Then $\alpha=[\phi\circ\gamma]\cdot[|\psi^j|]\cdot[|\bullet|^{jk}]\in(\CVZ)\cdot(|\bullet|^{jk})=\scrlito_{jk}$.
QED
\vskip.1in\hrule\vskip.1in\noindent
2-4. Let $\phi\in\CVZ$, $\varepsilon>0$.
Show: $\exists B\in\scrb_\R(0)$ s.t.~$|\phi|<\varepsilon$ on $B$.
\vskip.1in\noindent
{\it Proof:}
Since $\phi\in\CVZ$, we know:\\
\phantom{x}\qquad\qquad $\phi\in\DNZ$ \quad and \quad $\phi_0=0$ \quad and \quad $\phi$ is continuous at $0$.\\
Let $C:=B_\R(0,\varepsilon)$. \qquad
Then $C\in\scrb_\R(0)=\scrb_\R(\phi_0)$.\\
So, since $\phi$ is continuous at $0$, choose $A\in\scrb_\R(0)$ s.t.~$\phi_*(A)\subseteq C$.\\
Since $\phi\in\DNZ$, choose $D\in\scrb_\R(0)$ s.t.~$D\subseteq\dom[\phi]$.\\
Since $A,D\in\scrb_\R(0)$, we get $A\cap D\in\{A,D\}$.\\
Let $B:=A\cap D$. \qquad
Then $B\in\{A,D\}\subseteq\scrb_\R(0)$.\\
Want: $|\phi|<\varepsilon$ on $B$. \qquad\qquad
Want: $\forall x\in B$, $(|\phi|)(x)<\varepsilon$.\\
Given $x\in B$. \qquad
Want: $(|\phi|)(x)<\varepsilon$.\\
We have $x\in B=A\cap D\subseteq D\subseteq\dom[\phi]$.\\
Also, $x\in B=A\cap D\subseteq A$.\\
Since $x\in A$ and $x\in\dom[\phi]$, we get $\phi(x)\in\phi_*(A)$.\\
Then $\phi(x)\in\phi_*(A)\subseteq C=B_\R(0,\varepsilon)$, so $|[\phi(x)]-0|<\varepsilon$.\\
Then $(|\phi|)(x)=|\phi(x)|=|[\phi(x)]-0|<\varepsilon$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
2-5. Let $\alpha\in\scrlito_1$, $L\in\scrl\backslash\{\bfzero\}$.
Show: $\exists B\in\scrb_\R(0)$ s.t.~$|\alpha|\le|L|$ on $B$.
\vskip.1in\noindent
{\it Proof:}
Since $L\in\scrl$, choose $m\in\R$ s.t.~$L=m\cdot\id_\R$.\\
Then, for all $x\in\R$, we have: $L_x=mx$.\\
Since $L\ne\bfzero$, we see that $m\ne0$, so $m\in\R_0^\times$, so $|m|>0$.\\
Let $\varepsilon:=|m|$. \qquad
Then $\varepsilon>0$.\\
Since $\alpha\in\scrlito_1=\CVZ\cdot(|\bullet|)$, choose
$\phi\in\CVZ$ s.t.~$\alpha=\phi\cdot(|\bullet|)$.\\
By HW\#2-4, choose $B\in\scrb_\R(0)$ s.t.~$|\phi|<\varepsilon$ on $B$.\\
Want: $|\alpha|\le|L|$ on $B$. \qquad
Want: $\forall x\in B$, $|\alpha|_x\le|L|_x$.\\
Given $x\in B$. \qquad
Want: $|\alpha|_x\le|L|_x$.\\
Since $x\in B$ and $|\phi|<\varepsilon$ on $B$,
we get: $|\phi|_x<\varepsilon$.\\
Since $|\phi|_x\le\varepsilon$ and $|x|\ge0$,
we conclude that $|\phi|_x\cdot|x|\le\varepsilon\cdot|x|$.\\
Since $\alpha=\phi\cdot(|\bullet|)$,
it follows that $\alpha_x=\phi_x\cdot|x|$, and so $|\alpha_x|=|\phi_x|\cdot|x|$.\\
Then $|\alpha|_x=|\alpha_x|=|\phi_x|\cdot|x|=|\phi|_x\cdot|x|\le\varepsilon\cdot|x|$\\
\phantom{x}\qquad\qquad $=|m|\cdot|x|=|mx|=|L_x|=|L|_x$, as desired.
QED
\vskip.1in\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 1: Due on Tuesday 29 January}
\vskip.1in\hrule\vskip.1in\noindent
1-1. Let $j\in\N_0$.
Show: $\scrlito_j\supseteq(\CVZ)\cdot(|\bullet|^j)$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall\alpha\in(\CVZ)\cdot(|\bullet|^j)$, $\alpha\in\scrlito_j$.\\
Given $\alpha\in(\CVZ)\cdot(|\bullet|^j)$. \qquad
Want: $\alpha\in\scrlito_j$. \qquad
Let $\ds{\beta:=\frac{\alpha}{|\bullet|^j}}$.\\
Want: $(\,\alpha\in\DNZ\,)\,\&\,(\,\alpha_0=0\,)\,\&\,(\,\beta\to0\hbox{ near }0\,)$.\\
Since $\alpha\in(\CVZ)\cdot(|\bullet|^j)$, choose $\phi\in\CVZ$ s.t.~$\alpha=\phi\cdot(|\bullet|^j)$.\\
Since $\dom[|\bullet|^j]=\R$, we get $|\bullet|^j\in\DNZ$.\\
Then $\alpha=\phi\cdot(|\bullet|^j)\in(\DNZ)\cdot(\DNZ)\subseteq\DNZ$.\\
Want: $(\,\alpha_0=0\,)\,\&\,(\,\beta\to0\hbox{ near }0\,)$.\\
Since $\phi\in\CVZ$, it follows that $\phi_0=0$.\\
Then $\alpha_0=(\phi\cdot|\bullet|^j)_0=(\phi_0)\cdot(|0|^j)=0\cdot0=0$.\\
Want: $\beta\to0$ near $0$.\\
\\
{\it Claim:} $\beta=\phi$ on $\R_0^\times$.\\
{\it Proof of claim:}\\
Want: $\forall x\in\R_0^\times$, $\beta_x=\phi_x$.\\
Given $x\in\R_0^\times$. \qquad
Want: $\beta_x=\phi_x$.\\
Since $x\in\R_0^\times$, it follows that $\ds{\frac{|x|^j}{|x|^j}=1}$.\\
We have: $\alpha_x=(\phi\cdot(|\bullet|^j))_x=(\phi_x)\cdot(|x|^j)$.\\
Then: $\ds{\beta_x=\left(\frac{\alpha}{|\bullet|^j}\right)_x=\frac{\alpha_x}{|x|^j}=\frac{(\phi_x)\cdot(|x|^j)}{|x|^j}=(\phi_x)\cdot1=\phi_x}$, as desired.\\
{\it End of proof of claim.}\\
\\
Since $\phi\in\CVZ$, it follows that $\phi\to\phi_0$ near $0$.\\
So, since $\phi_0=0$, we get: $\phi\to0$ near $0$.\\
So, since $\beta=\phi$ on $\R_0^\times$,\\
\phantom{x}\qquad\qquad we conclude that $\beta\to0$ near $0$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
1-2. Show: $\CVZ\subseteq\BNZ$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall\alpha\in\CVZ$, $\alpha\in\BNZ$.\\
Given $\alpha\in\CVZ$. \qquad
Want: $\alpha\in\BNZ$.\\
We have $\alpha\in\CVZ\subseteq\DNZ$.\\
Want: $\exists B\in\scrb_\R(0)$ s.t.~$\alpha_*(B)$ is bounded in $\R$.\\
Since $\alpha\in\CVZ$, it follows that $\alpha$ is continuous at $0$.\\
Choose $B\in\scrb_\R(0)$ s.t.~$\alpha_*(B)\subseteq B_\R(0,1)$.\\
Want: $\alpha_*(B)$ is bounded in $\R$.\\
Since $\alpha_*(B)\subseteq B_\R(0,1)$, it follows that $\alpha_*(B)$ is bounded in $\R$.
QED
\vskip.1in\hrule\vskip.1in\noindent
1-3. Show: $(\BNZ)\cdot(\BNZ)\subseteq\BNZ$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall\gamma\in(\BNZ)\cdot(\BNZ)$, \quad $\gamma\in\BNZ$.\\
Given $\gamma\in(\BNZ)\cdot(\BNZ)$. \qquad
Want: $\gamma\in\BNZ$.\\
Since $\gamma\in(\BNZ)\cdot(\BNZ)$, choose $\alpha,\beta\in\BNZ$ s.t.~$\gamma=\alpha\cdot\beta$.\\
Since $\alpha,\beta\in\BNZ\subseteq\DNZ$,\\
\phantom{x}\qquad choose $P,Q\in\scrb_\R(0)$ s.t.~$P\subseteq\dom[\alpha]$ and $Q\in\dom[\beta]$.\\
Then $P\cap Q\in\scrb_\R(0)$.\\
So, since $P\cap Q\subseteq(\dom[\alpha])\cap(\dom[\beta])=\dom[\alpha\cdot\beta]=\dom[\gamma]$,\\
\phantom{x}\qquad we conclude that $\gamma\in\DNZ$.\\
Want: $\exists C\in\scrb_\R(0)$ s.t.~$\gamma_*(C)$ is bounded in $\R$.\\
Since $\alpha,\beta\in\BNZ$,
choose $A,B\in\scrb_\R(0)$ s.t.\\
\phantom{x}\qquad $\alpha_*(A)$ and $\beta_*(B)$ are bounded in $\R$.\\
Let $C:=A\cap B$. \qquad
Then $C\in\scrb_\R(0)$.\\
Want: $\gamma_*(C)$ is bounded in $\R$.\\
Since $\alpha_*(A)$ and $\beta_*(B)$ are bounded in $\R$,\\
\phantom{x}\qquad choose $T,U\in\scrb_\R$ s.t.~$\alpha_*(A)\subseteq T$ and $\beta_*(B)\subseteq U$.\\
By the Superset Recentering Lemma (Theorem 38.17),\\
\phantom{x}\qquad choose $V,W\in\scrb_\R(0)$ s.t.~$T\subseteq V$ and $U\subseteq W$.\\
Choose $r,s>0$ s.t.~$V=B_\R(0,r)$ and $W=B_\R(0,s)$.\\
Want: $\gamma_*(C)\subseteq B_\R(0,rs)$.\\
Want: $\forall y\in\gamma_*(C)$, $y\in B_\R(0,rs)$.\\
Given $y\in\gamma_*(C)$. \qquad
Want: $y\in B_\R(0,rs)$. \qquad
Want: $|y|<rs$.\\
Since $y\in\gamma_*(C)$, choose $x\in C\cap(\dom[\gamma])$ s.t. $\gamma(x)=y$.\\
We have $x\in\dom[\gamma]=\dom[\alpha\cdot\beta]=(\dom[\alpha])\cap(\dom[\beta])$.\\
Since $x\in\dom[\alpha]$ and $x\in C=A\cap B\subseteq A$, we get: $\alpha(x)\in\alpha_*(A)$.\\
Since $x\in\dom[\beta]$ and $x\in C=A\cap B\subseteq B$, we get: $\beta(x)\in\beta_*(B)$.\\
Then $\alpha(x)\in\alpha_*(A)\subseteq T\subseteq V=B_\R(0,r)$, so $|\alpha(x)|<r$.\\
Also, $\beta(x)\in\beta_*(A)\subseteq U\subseteq W=B_\R(0,s)$, so $|\beta(x)|<s$.\\
Since $0\le|\alpha(x)|<r$ and $0\le|\beta(x)|<s$, we get $|\alpha(x)|\cdot|\beta(x)|<rs$.\\
We have $\gamma(x)=(\alpha\cdot\beta)(x)=[\alpha(x)]\cdot[\beta(x)]$, so $|\gamma(x)|=|\alpha(x)|\cdot|\beta(x)|$.\\
Then $|y|=|\gamma(x)|=|\alpha(x)|\cdot|\beta(x)|<rs$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
1-4. Show: $(\BNZ)\cdot(\CVZ)\subseteq\CVZ$.
\vskip.1in\noindent
{\it Proof:}
Want: $\forall\gamma\in(\BNZ)\cdot(\CVZ)$, \quad $\gamma\in\CVZ$.\\
Given $\gamma\in(\BNZ)\cdot(\CVZ)$. \qquad
Want: $\gamma\in\CVZ$.\\
Since $\gamma\in(\BNZ)\cdot(\CVZ)$, choose $\alpha\in\BNZ$, $\beta\in\CVZ$ s.t.~$\gamma=\alpha\cdot\beta$.\\
Since $\alpha\in\BNZ\subseteq\DNZ$ and since $\beta\in\CVZ\subseteq\DNZ$,\\
\phantom{x}\qquad choose $P,Q\in\scrb_\R(0)$ s.t.~$P\subseteq\dom[\alpha]$ and $Q\subseteq\dom[\beta]$.\\
Then $P\cap Q\in\scrb_\R(0)$.\\
So, since $P\cap Q\in(\dom[\alpha])\cap(\dom[\beta])=\dom[\alpha\cdot\beta]=\dom[\gamma]$,\\
\phantom{x}\qquad we conclude that $\gamma\in\DNZ$.\\
Want: $\gamma(0)=0$ and $\gamma$ is continuous at $0$.\\
Since $P\in\scrb_\R(0)$, we get $0\in P$.\\
Since $\alpha\in\DNZ$, we get $\alpha:\R\drta\R$.\\
So, since $0\in P\subseteq\dom[\alpha]$, we get $\alpha(0)\in\R$, so $[\alpha(0)]\cdot0=0$.\\
Since $\beta\in\CVZ$, we get $\beta(0)=0$.\\
Then $\gamma(0)=(\alpha\cdot\beta)(0)=[\alpha(0)]\cdot[\beta(0)]=[\alpha(0)]\cdot0=0$.\\
Want: $\gamma$ is continuous at $0$.\\
Want: $\forall\varepsilon>0$, $\exists\delta>0$ s.t., $\forall x\in\dom[\gamma]$,\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,|x-0|<\delta\,\,]\,\,\Rightarrow\,\,[\,\,|[\gamma(x)]-[\gamma(0)]|<\varepsilon\,\,]$.\\
Given $\varepsilon>0$. \qquad
Want: $\exists\delta>0$ s.t., $\forall x\in\dom[\gamma]$,\\
\phantom{x}\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,|x-0|<\delta\,\,]\,\,\Rightarrow\,\,[\,\,|[\gamma(x)]-[\gamma(0)]|<\varepsilon\,\,]$.\\
Since $\alpha\in\BNZ$, choose $A\in\scrb_\R(0)$ s.t.~$\alpha_*(A)$ is bounded in $\R$.\\
Since $\alpha_*(A)$ is bounded in $\R$,\\
\phantom{x}\qquad choose $T\in\scrb_\R$ s.t.~$\alpha_*(A)\subseteq T$.\\
By the Superset Recentering Lemma (Theorem 38.17),\\
\phantom{x}\qquad choose $U\in\scrb_\R(0)$ s.t.~$T\subseteq U$.\\
Choose $r>0$ s.t.~$U=B_\R(0,r)$.\\
Since $\beta\in\CVZ$, it follows that $\beta$ is continuous at $0$,\\
\phantom{x}\qquad so choose $\eta>0$ s.t., $\forall x\in\dom[\beta]$,\\
\phantom{x}\qquad\qquad\qquad\qquad $[\,\,|x-0|<\eta\,\,]\,\,\Rightarrow\,\,[\,\,|[\beta(x)]-[\beta(0)]|<\varepsilon/r\,\,]$.\\
Let $B:=B_\R(0,\eta)$. \qquad
Then $A,B\in\scrb_\R(0)$. \qquad
Let $Q:=A\cap B$.\\
Then $Q\in\scrb_\R(0)$. \qquad
Choose $\delta>0$ s.t.~$Q=B_\R(0,\delta)$.\\
Want: $\forall x\in\dom[\gamma]$, $(\,\,[\,\,|x-0|<\delta\,\,]\,\,\Rightarrow\,\,[\,\,|[\gamma(x)]-[\gamma(0)]|<\varepsilon\,\,]\,\,)$.\\
Given $x\in\dom[\gamma]$.\\
Want: $[\,\,|x-0|<\delta\,\,]\,\,\Rightarrow\,\,[\,\,|[\gamma(x)]-[\gamma(0)]|<\varepsilon\,\,]$.\\
Assume: $|x-0|<\delta$. \qquad
Want: $|[\gamma(x)]-[\gamma(0)]|<\varepsilon$.\\
We have $x\in\dom[\gamma]=\dom[\alpha\cdot\beta]=(\dom[\alpha])\cap(\dom[\beta])$.\\
Then $x\in(\dom[\alpha])\cap(\dom[\beta])\subseteq\dom[\beta]$.\\
Also, since $|x-0|<\delta$, we have $x\in B_\R(0,\delta)$.\\
Then $x\in B_\R(0,\delta)=Q=A\cap B$.\\
Then $x\in A\cap B\subseteq B=B_\R(0,\eta)$, so $|x-0|<\eta$.\\
So, since $x\in\dom[\beta]$, by choice of $\eta$, we get: $|[\beta(x)]-[\beta(0)]|<\varepsilon/r$.\\
Since $\beta\in\CVZ$, we get $\beta(0)=0$. \qquad
Then $[\beta(x)]-[\beta(0)]=\beta(x)$.\\
Then $|\beta(x)|=|[\beta(x)]-[\beta(0)]|<\varepsilon/r$, so $|\beta(x)|<\varepsilon/r$.\\
Since $x\in(\dom[\alpha])\cap(\dom[\beta])\subseteq\dom[\alpha]$ and since $x\in A\cap B\subseteq A$,\\
\phantom{x}\qquad we conclude that $\alpha(x)\in\alpha_*(A)$.\\
Then $\alpha(x)\in\alpha_*(A)\subseteq T\subseteq U=B_\R(0,r)$,
so $|\alpha(x)|<r$.\\
Since $0\le|\alpha(x)|<r$ and since $0\le|\beta(x)|<\varepsilon/r$,\\
\phantom{x}\qquad it follows that $|\alpha(x)|\cdot|\beta(x)|<r\cdot(\varepsilon/r)$.\\
Recall that $\gamma(0)=0$. \qquad
We have $\gamma(x)=(\alpha\cdot\beta)(x)=[\alpha(x)]\cdot[\beta(x)]$.\\
Then $|[\gamma(x)]-[\gamma(0)]|=|[\gamma(x)]-0|=|\gamma(x)|=|[\alpha(x)]\cdot[\beta(x)]|$\\
\phantom{x}\qquad\qquad\qquad\qquad\quad $\,\,=|\alpha(x)|\cdot|\beta(x)|<r\cdot(\varepsilon/r)=\varepsilon$, as desired.
QED
\vskip.1in\hrule\vskip.1in\noindent
1-5. Let $j\in\N_0$.
Show: $\scrlito_j\subseteq\scrbigo_j$.
\vskip.1in\noindent
{\it Proof:}
By HW\#1-2, $\CVZ\subseteq\BNZ$.\\
Then $(\CVZ)\cdot(|\bullet|^j)\subseteq(\BNZ)\cdot(|\bullet|^j)$.\\
Then $\scrlito_j\subseteq\scrbigo_j$, as desired.
QED
\vskip.1in\hrule\vskip.1in\hrule
\end{document}
