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REAL ANALYSIS

UNIFORM CONTINUITY

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\item[Definition]
$f(x)$ is said to be uniformly continuous in a set
$s\Leftrightarrow$, given
$\varepsilon>0\exists\delta=\delta(\varepsilon)||f(x_1)-f(x_2)|<\varepsilon$
whenever $|x_1-x_2|<\delta$ and $x_1,x_2\in S$.

\item[Theorem]
Suppose $f(x)$ is continuous in $\left[a,b\right]$ relative to
$[a,b]$, then $f(x)$ is uniformly continuous in $[a,b]$.

\item[Note:]
False for open intervals. To prove uniform continuity it suffices
to prove the following: Given $\varepsilon>0\exists$ a finite
subdivision of $[a,b]|$ oscillation $f(x)<\epsilon\
x\in[x_{\nu-1}x_\nu]$

or: if $M_\nu,\ m_\nu$ are the upper and lower bounds of $f(x)$ in
$[x_{\nu-1}x_\nu]$, we have

$$M_\nu-m_\nu<\varepsilon\ \ \nu=1,2,\ldots,n$$

\item[First Proof]
(using Bisection)

Suppose the result is false. Then $\exists\varepsilon_0$ so that
for this $\varepsilon_0$, there is no subdivision of the required
type.

We call an interval a good interval if
$bd-\underline{bd}<\varepsilon_0$, and bad otherwise.

Subdivide $[a\ b]$ into two equal closed intervals
$\left[a,\frac{a+b}{2}\right],\ \left[\frac{a+2}{2},b\right]$.

At least one of these is bad.

We define $J_1$ to be the bad interval if there is only one, or
the left hand one if there is a choice. Now subdivide $J_1$ into
two equal intervals as before, and again define $J_2$  to be the
bad (or left hand) subinterval of $J_1$. Continue this process
(which cannot terminate). We obtain a sequence of bad intervals
$J_1=[a_1\ b_1],\ J_2=[a_2\ b_2]\ldots$ where
$|a_n-b_n|=\frac{b-a}{2^n}$.

Now $a_1\leq a_2\leq\ldots\leq b$

Hence $\exists l\in[a,b]|a_n\to l$; and $b_n\to L$ as
$n\to\infty$.

But $f(x)$ is continuous at $l$ relative to $[a\ b]$. Hence
$\exists\delta>0$, such that, if we write
$I_\delta=[l-\delta,l+\delta]\cap[a\ b]$ and $\displaystyle
m_\delta=\underline{bd}_{x\in I_\delta}f\
M_\delta=\overline{bd}_{x\in I_\delta}f(x)$, we have
$M_\delta-m_\delta<\varepsilon_0$.

But $\exists N|J_N$ is contained in $I_\delta$. This gives a
contradiction since $I_\delta$ is good, $J_N$ is bad.

\item[Second Proof]
Given $\varepsilon>0$. Consider any pint $x$ in $[a\ b]$.

$\exists\delta||f(y)-f(x)|<\frac{1}{2}\varepsilon$ whenever
$y\in(x-\delta,x+\delta)-1$.

Let us define $f(y)=f(a)$ for $y<a$ and $f(y)=f(b)$ for $y>b$.
Then (1) defines a covering of $[a\ b]$.

By the Heine Borel theorem we con find a finite covering subset
$S$ of open intervals.

$$I_\nu=(x_\nu\ y_\nu)\ \ \nu=1,\ldots,n\ \ x_\nu<y_\nu$$

If we take all the points $x_\nu\ y_\nu\ (\nu=1,\ldots,n$ and
select those which lie in $[a\ b]$, together with $a$ and $b$,
they define a finite subdivision of $[a\ b],\
a=t_0<t_1<\ldots<t_m=b$. Consider any point in one of these
intervals of the subdivision. Each interval of the subdivision is
covered by one interval of $S$ defined by 1. Therefore in this
interval, $[t_{\nu-1}\ t_nu]$ which is covered by $(x_\mu\
y)\mu)\{x_\mu<t_{\nu-1}<t_\nu<v_\mu\}\
|f(x)-f(y)|<\frac{1}{2}\varepsilon$ for all $x$ in the interval
therefore $|M_\nu-f(y)|<\frac{1}{2}\varepsilon$ and
$|m_\nu-f(y)|<\frac{1}{2}\varepsilon$ therefore
$$M_\nu-m_\nu<\varepsilon\ \ \nu=1,2,\ldots,n.$$

Hence the result.

\item[Third Proof]
Suppose false. Then $\exists \varepsilon_0>0|$ there is no
$\delta$ of the required type.

Choose 2 points $x_1y_1||x_1-y_1|\leq1$ and
$|f(x_1)-f(y_1)|\geq\varepsilon_0$

Choose 2 points $x_2y_2||x_2-y_2|\leq\frac{1}{2}$ and
$|f(x_2)-f(y_2)|\geq\varepsilon_0$

\hspace{1cm}  \vdots

Choose 2 points $x_ny_n||x_n-y_n|\leq\frac{1}{n}$ and
$|f(x_n)-f(y_n)|\geq\varepsilon_0$

The choice is always possible, for otherwise $\frac{1}{n}$ would
be a possible $\delta$.

The sequence $x_1,x_2,x_3\ldots$ is bounded since each $x_n\in[a\
b]$. Hence it contains a convergent subsequence
$x_{r_1},x_{r_2}\ldots\to l$ as $n\to\infty$ where $l\in[a\ b]$
 $y_{r_1},y_{r_2}\ldots\to l$ as $x_{r_1}-y_r\to0$ as
 $r\to\infty$.

 Since $f(x)$ is continuous $\exists$ an interval $I$ about
 $l$|for all $x$ in $I\cap[a\ b]=J,\
 |f(x)-f(l)|<\frac{1}{2}\varepsilon_0$.

 But $\exists N|x_N$ and $y_N\in J$

 $$|f(x_N)-f(l)|<\frac{1}{2}\varepsilon_0\
 |f(y_N)-f(l)|<\frac{1}{2}\varepsilon_0\Rightarrow|f(x_N)-f(Y_N)|<\varepsilon_0$$

 Which is a contradiction hence the result.

 \item[]
 We can use Uniform Continuity to prove that a continuous function
 is Riemann integrable. We can find a subdivision $\triangle$ such
 that each interval $M_\nu-m_\nu<\frac{\varepsilon}{b-a}$, since
 $f(x)$ is uniformly continuous. Then

 $$\sum_{\nu=1}^n\delta_\nu(M_\nu-m_\nu)<\sum_{\nu=1}^n\delta_\nu\frac{\varepsilon}{b-a}=\varepsilon$$

 Therefore

 $$S_{\triangle}-s_{\triangle}<\varepsilon.$$

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