221c221
< Let $a \in L (E_1, E_2)$ and let $b \in L (E_2, E_3)$.
---
> Let $b \in L (E_1, E_2)$ and let $a \in L (E_2, E_3)$.
239c239,291
< No proof has been written yet, but this is just calculations.
---
> % No proof has been written yet, but this is just calculations.
> We mostly use the characterization of $\| a \|$
> (done in class, but stated here in two parts for convenience of writing)
> %
> \begin{equation}\label{Eq_24208_Inf}
> \| a \| = \inf \bigl( \bigl\{ M \in [0, \I)
>       \colon {\mbox{for all $\xi \in E$ we
>       have $\| a \xi \| \leq M \| \xi \|$}} \bigr\} \bigr)
> \end{equation}
> %
> and, for all $\xi \in E$,
> %
> \begin{equation}\label{Eq_24208_Bound}
> \| a \xi \| \leq \| a \| \| \xi \|.
> \end{equation}
> %
> Nonemptiness of the set in~(\ref{Eq_24208_Inf}) is the definition
> of boundedness for a linear map,
> and~(\ref{Eq_24208_Bound}) says this set contains its infimum.
> (It is necessary to specify $M \geq 0$ in~(\ref{Eq_24208_Inf}),
> since otherwise if $E = \{ 0 \}$ and $a \in L (E, F)$ is the
> zero operator, then $\| a \| = - \I$.)
> 
> It is obvious that if $a \in L (E, F)$ then $\| a \| \geq 0$.
> 
> Suppose $\| a \| = 0$.
> By~(\ref{Eq_24208_Bound}), for any $\xi \in E$ we have
> $\| a \xi \| \leq \| a \| \| \xi \| = 0$, so $a \xi = 0$.
> Since this holds for all $\xi \in E$, we get $a = 0$.
> 
> For subadditivity, let $a, b \in L (E, F)$.
> For all $\xi \in E$ we have, using subadditivity of the norm on~$F$
> at the second step and~(\ref{Eq_24208_Bound}) at the third step,
> \[
> \| (a + b) \xi \|
>  = \| a \xi + b \xi \|
>  \leq \| a \xi \| + \| b \xi \|
>  \leq \| a \| \| \xi \| + \| b \| \| \xi \|
>  = \bigl( \| a \| + \| b \| \bigr) \| \xi \|.
> \]
> Since this is true for all $\xi \in E$, by~(\ref{Eq_24208_Inf})
> we get $\| a + b \| \leq \| a \| + \| b \|$.
> 
> To prove homogeneity, for $\ld \in \C$ and $a \in L (E, F)$ we calculate:
> \[
> \| \ld a \|
>  = \sup_{\| \xi \| \leq 1 \|} \| (\ld a) \xi \|
>  = \sup_{\| \xi \| \leq 1 \|} \| \ld \cdot (a \xi) \|
>  = \sup_{\| \xi \| \leq 1 \|} | \ld | \| a \xi \|
>  = | \ld | \sup_{\| \xi \| \leq 1 \|} \| a \xi \|
>  = | \ld | \| a \|.
> \]
> This completes the solution.
241a294,316
> \begin{proof}[Alternate proof of homogeneity]
> For all $\xi \in E$ we have, using homogeneity of the norm on~$F$
> at the second step and~(\ref{Eq_24208_Bound}) at the third step,
> \[
> \| (\ld a) \xi \|
>  = \| \ld \cdot (a \xi) \|
>  = | \ld | \| a \xi \|
>  \leq | \ld | \| a \| \| \xi \|.
> \]
> Therefore, by~(\ref{Eq_24208_Inf}),
> %
> \begin{equation}\label{Eq_24208_ScInq}
> \| \ld a \| \leq | \ld | \| a \|.
> \end{equation}
> %
> If $\ld = 0$, nonnegativity of the norm on $L (E, F)$ gives
> $\| \ld a \| = 0 = | \ld | \| a \|$.
> Otherwise, apply (\ref{Eq_24208_ScInq}) with $\ld^{-1}$ in place of $\ld$
> and $\ld a$ in place of~$a$, getting
> $\| a \| \leq | \ld |^{- 1} \| \ld a \|$.
> Combining this with~(\ref{Eq_24208_ScInq})
> gives $\| \ld a \| = | \ld | \| a \|$.
> \end{proof}
271c346
<  \leq \limsup_{n \to I} \| a_n \| \| \xi \|
---
>  \leq \limsup_{n \to \I} \| a_n \| \| \xi \|
314c389
< Set $c = \inf \bigl( \{ f (\af) \colon \af \in S \bigr)$.
---
> Set $c = \inf \bigl( \{ f (\af) \colon \af \in S \bigr\} \bigr)$.
332c407
< \frac{1}{c} \geq \| \bt \|_2 \geq \max_{1 leq j \leq n} | \bt_j |,
---
> \frac{1}{c} \geq \| \bt \|_2 \geq \max_{1 \leq j \leq n} | \bt_j |,
369c444
< Let $0 < \af \leq 1$ and let $[a, b] \S \R$ be a compact interval.
---
> Let $\af \in (0, 1]$ and let $[a, b] \S \R$ be a compact interval.
410c485
< we have $c_1 \| f \|_{\af} \leq\| f \|'_{\af} \leq c_2 \| f \|_{\af}$.
---
> we have $c_1 \| f \|_{\af} \leq \| f \|'_{\af} \leq c_2 \| f \|_{\af}$.
433a509,536
> \begin{lem}\label{L_6Z02_Lip_7}
> Let $c = 1 + \max (1, \, (b - a)^{\af} )$.
> Then $\| f \|'_{\af} \leq c \| f \|_{\af}$ for all
> $f \in {\mathrm{Lip}}^{\af}$.
> \end{lem}
> 
> \begin{proof}
> Let $f \in {\mathrm{Lip}}^{\af}$.
> For any $t \in [a, b]$, we have $| t - a | \leq b - a$, so
> \[
> \begin{split}
> | f (t) |
>  & \leq | f (a) | + | f (t) - f (a) |
>    \leq | f (a) | + M_{\af, f} | t - a |^{\af}   \\
>  & \leq | f (a) | + M_{\af, f} (b - a )^{\af}
>    \leq \max (1, \, (b - a)^{\af} ) \| f \|_{\af}.
> \end{split}
> \]
> Therefore $\| f \|_{\I} \leq \max (1, \, (b - a)^{\af} ) \| f \|_{\af}$.
> So
> \[
> \| f \|'_{\af} = \| f \|_{\I} + M_{\af, f}
>   \leq \max (1, \, (b - a)^{\af} ) \| f \|_{\af} + \| f \|_{\af}
>   = c \| f \|_{\af}.
> \]
> This completes the proof.
> \end{proof}
> 
441,444c544,546
< If $f \in {\mathrm{Lip}}^{\af}$ then $M_{\af, f} < \I$
< by definition, and $\| f \|_{\I} < \I$ by Lemma~\ref{L_6Z02_Lip_1},
< so $\| f \|'_{\af} < \I$.
< Therefore $\| f \|_{\af} < \I$ by Lemma~\ref{L_6Z02_Lip_2}.
---
> If $f \in {\mathrm{Lip}}^{\af}$ then $M_{\af, f} < \I$ by definition,
> and then clearly $\| f \|_{\I} < \I$.
> Therefore $\| f \|_{\af}' < \I$ by Lemma~\ref{L_6Z02_Lip_7}.
446,447c548
< A calculation (which we omit) shows that for any functions
< $f, \, g \colon [a, b] \to \C$ and $\ld \in \C$ we have
---
> We claim that if $f, g \colon [a, b] \to \C$ and $\ld \in \C$, then
451a553,582
> For the first,
> \[
> \begin{split}
> M_{\af, f + g}
> & = \sup_{s \neq t}
>     \frac{ | f (s) + g (s) - f (t) - g (t) |}{| s - t |^{\af} }
>   \leq \sup_{s \neq t} \left(
>     \frac{ | f (s) - f (t) |}{| s - t |^{\af} }
>         + \frac{ | g (s) - g (t) |}{| s - t |^{\af} } \right)
> \\
> & \leq \sup_{s \neq t} \frac{ | f (s) - f (t) |}{| s - t |^{\af} }
>         + \sup_{s \neq t} \frac{ | g (s) - g (t) |}{| s - t |^{\af} }
>   = M_{\af, f} + M_{\af, g}.
> \end{split}
> \]
> For the second,
> \[
> \begin{split}
> M_{\af, \ld f}
> & = \sup_{s \neq t} \frac{ | \ld f (s) - \ld f (t) |}{| s - t |^{\af} }
>   = \sup_{s \neq t} | \ld |
>       \left( \frac{ | f (s) - f (t) |}{| s - t |^{\af} } \right)
> \\
> & = | \ld | \sup_{s \neq t}
>       \left( \frac{ | f (s) - f (t) |}{| s - t |^{\af} } \right)
>   = | \ld | \cdot M_{\af, f}.
> \end{split}
> \]
> The claim is proved.
> 
514c645,646
< \begin{align*}
---
> \[
> \begin{split}
521c653,654
< \end{align*}
---
> \end{split}
> \]
534a668,675
> \begin{rmk}\label{R_24208_No_CX_cpt}
> Proving Lemma~\ref{L_6Z02_Lip_4}
> without using completeness of $C ([a, b])$
> requires only one or two more sentences,
> since continuity of the limit function is not used.
> This change eliminates the need for Lemma~\ref{L_6Z02_Lip_1}.
> \end{rmk}
> 
547,572d687
< 
< \begin{lem}\label{L_6Z02_Lip_7}
< Let $c = 1 + \max (1, \, (b - a)^{\af} )$.
< Then $\| f \|'_{\af} \leq c \| f \|_{\af}$ for all
< $f \in {\mathrm{Lip}}^{\af}$.
< \end{lem}
< 
< \begin{proof}
< Let $f \in {\mathrm{Lip}}^{\af}$.
< For any $t \in [a, b]$, we have $| t - a | \leq b - a$, so
< \begin{align*}
< | f (t) |
<  & \leq | f (a) | + | f (t) - f (a) |
<    \leq | f (a) | + M_{\af, f} | t - a |^{\af}   \\
<  & \leq | f (a) | + M_{\af, f} (b - a )^{\af}
<    \leq \max (1, \, (b - a)^{\af} ) \| f \|_{\af}.
< \end{align*}
< Therefore $\| f \|_{\I} \leq \max (1, \, (b - a)^{\af} ) \| f \|_{\af}$.
< So
< \[
< \| f \|'_{\af} = \| f \|_{\I} + M_{\af, f}
<   \leq \max (1, \, (b - a)^{\af} ) \| f \|_{\af} + \| f \|_{\af}
<   = c \| f \|_{\af}.
< \]
< This completes the proof.
< \end{proof}
635,638c750,751
< Moreover,
< prove that without the linearity hypothesis,
< the statement is false,
< even for $X = Y = \R$.
---
> Moreover, prove that, without the linearity hypothesis,
> the statement is false, even for $E = F = \R$.
655,656c768
< projection $E \oplus F \to E$
< is bijective and \ct.
---
> projection $E \oplus F \to E$ is bijective and \ct.
678,679c790
< It is well known that $E \oplus F$ as described is a vector
< space.
---
> It is well known that $E \oplus F$ as described is a vector space.
682,683c793
< does in fact define a norm on $E \oplus F$;
< the proof is omitted.
---
> does in fact define a norm on $E \oplus F$; the proof is omitted.
700c810
< with $\xi_n \in \E$ and $\et_n \in F$ for all $n \in \N$.
---
> with $\xi_n \in E$ and $\et_n \in F$ for all $n \in \N$.
704,705c814
< Therefore $\xi = \limi{n} \xi_n$ and $\et = \limi{n} \et_n$
< exist.
---
> Therefore $\xi = \limi{n} \xi_n$ and $\et = \limi{n} \et_n$ exist.
758,761c867,868
< For such values of~$n$,
< we have $y_n = \frac{1}{x_n}$.
< Since the function $h (x) = \frac{1}{x}$
< is \ct{} at~$x_0$,
---
> For such values of~$n$, we have $y_n = \frac{1}{x_n}$.
> Since the function $h (t) = \frac{1}{t}$ is \ct{} at~$x$,
763,764c870
< Therefore $y = \frac{1}{x}$,
< whence $(x, y) \in G$.
---
> Therefore $y = \frac{1}{x}$, whence $(x, y) \in G$.
769,772c875,876
< so $\limi{n} (x_n, y_n) = (0, 0)$,
< which is in~$G$.
< Otherwise,
< we show that $\limi{n} y_n$ does not exist,
---
> so $\limi{n} (x_n, y_n) = (0, 0)$, which is in~$G$.
> Otherwise, we show that $\limi{n} y_n$ does not exist,
780,787c884,886
< Since there are infinitely many $n \in \N$
< such that $x_n \neq 0$,
< we can find $n \in \N$
< such that $x_n \neq 0$ and $n > \max (N, N_0)$.
< Then $n \geq N$
< and
< $| y_n | > \frac{1}{\ep} = M$,
< as desired.
---
> Since there are infinitely many $n \in \N$ such that $x_n \neq 0$,
> we can find $n \in \N$ such that $x_n \neq 0$ and $n > \max (N, N_0)$.
> Then $n \geq N$ and $| y_n | > \frac{1}{\ep} = M$, as desired.