Lec 18 - Analysis 1 - MATH 3380.tex

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\begin{document}

\bsc{Example 1: Page 195 \#16(a)}{

Prove that \uf{S}{*} \eql \lmti{n} ( sup \bk{\uw{s}{n + 1}, \uw{s}{n + 2}, \uw{s}{n + 3}...}) \eql lim sup \uw{s}{n}, s \mem \br

Want to show that, although we know \uf{s}{*} \eql \lmti{n} ( sup \bk{\uw{s}{n + 1}, \uw{s}{n + 2}, \uw{s}{n + 3}...}), \uf{s}{*} is in fact lim sup \uw{s}{n}.

Let \uw{t}{n} \eql sup \bk{\uw{s}{n + 1}, \uw{s}{n + 2}, \uw{s}{n + 3}...}

\sidenote{
\av{\uw{s}{n}} \lse M \fa n \mem \bn

$-$M \lse \uw{s}{n} \lse M \fa n \mem \bn
}

\bk{\uw{t}{n}} is a bounded, decreasing sequence and
\eqn{t_{n + 1} = sup \bk{s_{n + 2}, s_{n + 3}...} \lse t_n = sup \bk{\uw{s}{n + 1}, \uw{s}{n + 2}, \uw{s}{n + 3}, ...} \textrm{ \fa n \mem } \bn}

If U is a bounded set in \br and V \sbs U, then sup V \lse sup U.

So, sup U \mem \br exists. Therefore,

\bilist
\item u \lse sup U \fa u \mem U
\item \fa \ep \gr 0, exs \uw{u}{1} \mem U st sup U \ms \ep \ls \uw{u}{1}
\elist

Notice that for v \mem V, v \lse sup U.

So, sup V \lse sup U.

\textbf{Is \uw{t}{n} bounded?}

It's bounded below since $-$M \lse \uw{s}{n + 1} \lse \uw{t}{n} \fa n \mem \bn

By the monotonic convergence theorem,

\lmti{n} \uw{t}{n} \eql \uf{s}{*} exists.

From Theorem 4.4.11, conditions (a) and (b):

\balist
\item \fa \ep \gr 0, \exs N \mem \bn st \uw{s}{n} \ls \uf{s}{*} \ps \ep, \fa n \gre N
\item \fa \ep \gr 0, i \mem \bn, \exs j \gr i st \uw{s}{j} \gr \uf{s}{*} \ms \ep
\elist

Notice that \uf{s}{*} \lse \uw{t}{n} \fa n \mem \bn

(since \uw{t}{n} is decreasing and it has a limit)

Let n \mem \bn and notice that \uw{t}{n + k} \lse \uw{t}{n} \fa k \mem \bn

So, 

\lmti{k} \uw{t}{n + k} \lse \uw{t}{n} \fa n \mem \bn

So,

\uf{s}{*} \eql \lmti{k} \uw{t}{n + k}, therefore \uf{s}{*} \lse \uw{t}{n} \fa n \mem \bn

Thus,

\fa \ep \gr 0, \exs \uw{N}{1} \mem \bn st 
\eqn{|t_n - s^*| < \epsilon \textrm{ for n \gre N}_1}
\eqn{t_n - s^* < \epsilon \textrm{ for n \gre N}_1}
\eqn{t_n < s^* + \epsilon \textrm{ for n \gre N}_1}
\eqn{s_{n + 1} \lse t_n < s^* + \epsilon \textrm{ for n + 1 \gre N}_1 + 1}
So,
\eqn{s_M < s^* + \epsilon \textrm{ for M \gre N}_1 + 1}
So, \uf{s}{*} satisfies (a).

Define \uw{t}{n} \eql sup \bk{\uw{s}{n + 1}, \uw{s}{n + 2}, \uw{s}{n + 3}...}

Now for any \ep \gr 0,
\step{t_n \gre s^* > s^* - \frac{\epsilon}{2} \textrm{ \fa n \mem \bn}}{1}
Also, for any i \eql n \mem \bn, \exs \uw{s}{j} where j \gr i st
\step{t_n - \frac{\epsilon}{2} < s_j}{2}
Notice that \uw{t}{n} \ms \frc{\epsilon}{2} is no longer a least upper bound for the set.

From \bpth{1} and \bpth{2}, \uw{s}{j} \gr \uw{t}{n} \ms \frc{\epsilon}{2} \gr \uf{s}{*} \ms \ep

So, since \uf{s}{*} satisfies \bpth{a} and \bpth{b}, \uf{s}{*} is the lim sup \uw{s}{n}.
}

\bsc{Unbounded Sequences}{
S \eql \bk{all subsequential limits of \uw{s}{n}}

We know that S is not empty if \uw{s}{n} is bounded since every bounded sequence has a convergent subsequence.

But what if \uw{s}{n} is bounded? page 192

Case:
\bilist
\item \bk{\uw{s}{n}} is unbounded above.
	
	From the proof of Theorem 4.4.8, \exs a monotonic subsequence \bk{\dbs{s}{n}{k}} of \bk{\uw{s}{n}} st \lmti{k} \dbs{s}{n}{k} \eql \infy
	
	Although \infy is not a real number, we say that, if \uw{s}{n} is unbounded above, then lim sup \uw{s}{n} \eql \infy
\item \bk{\uw{s}{n}} is bounded above but unbounded below.
	
	\textbf{Subcase i:} \exs a subsequence \bk{\dbs{s}{n}{k}} st \lmti{k} \dbs{s}{n}{k} \eql s \mem \br. Then, set lim sup \uw{s}{n} \eql sup S
	
	\textbf{Subcase ii:} There is no subsequence \bk{\dbs{s}{n}{k}} st \lmti{k} \dbs{s}{n}{k} \eql s \mem \br (a finite number).
	
	Then, \lmti{n} \uw{s}{n} \eql $-\infty$
	
	and, lim sup \uw{s}{n} \eql $-\infty$, so sup S \eql $-\infty$
	
	which means, since lim inf \uw{s}{n} \lse lim sup \uw{s}{n}, lim inf \eql $-\infty$
	
	
	
\elist
}

\end{document}