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\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
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\lhead{\color{blue} \large Math 207}
\lfoot{\small   \copyright $\;$  Hidegkuti,  2016}
\cfoot{}
\chead{\Large Definitions, Theorems - Part 1}
\rfoot{\small Last revised: October 6, 2016}
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\begin{document}


Definition: \ A set $S$ is \textbf{bounded from above} if there exists a
real number $U$ such that for all $x$ in $S,$ \ $x\leq U$.\bigskip

Definition: \ A set $S$ is \textbf{bounded from below} if there exists a
real number $L$ such that for all $x$ in $S,$ \ $x\geq L$.\bigskip

Definition: \ A set $S$ is \textbf{bounded} if it is bounded from above and
from below.\bigskip

Axiom: \ \ (\textbf{Least Upper Bound Property}) \ Every non-empty, set $S$
of real numbers has the following property: if $S$ is bounded from above,
then there exists a least upper bound for $S$.\bigskip

The least upper bound is also called supremum. \ \ The least upper bound
property (also called the completeness property) is a very fundamental one:
it is actually the single axiom that distinguishes the set of rational
numbers from the set of real numbers. \ Rational numbers do not have this
property, but real numbers do. \ In calculus, the least upper bound property
is a key ingredient in proving very important theorems later.\bigskip

The following is the actual definition of a (finite) limit of a function $f$
at a number $c$. \ In this course, we will not use this definiton.\bigskip 

Definition: \ Suppose that $f$ is a function and $c$, $L$ are real numbers.
\ We say that $\lim\limits_{x\rightarrow c}f\left( x\right) =L$ if for all $%
\varepsilon >0$ there exists $\delta >0$ such that for all $x\not=c$ with $%
\left\vert x-c\right\vert <\delta $, we also have that ($f$ is defined and) $%
\left\vert f\left( x\right) -L\right\vert <\varepsilon $.\bigskip 

The following definition is the one we will use.\bigskip 

Definition: \ If the left-hand side limit and the right-hand side limit both
exist (and are finite) and are equal, we say that $\lim\limits_{x\rightarrow
c}f\left( x\right) =L$.\bigskip

Definition: (Continuity at a point) \ A function $y=f\left( x\right) $ is
continuous at a number $c$ of its domain if the two-sided limit exists and $%
\lim\limits_{x\rightarrow c}f\left( x\right) =f\left( c\right) $.\bigskip

Definition: \ (Continuity on an interval)

(\textit{Open Interval}) \ A function $y=f\left( x\right) $ is continuous on
an interval $\left( a,b\right) $ if it is continuous at every $c$ in $\left(
a,b\right) $.\newline
(\textit{Closed Interval}) \ A function $y=f\left( x\right) $ is continuous
on an interval $\left[ a,b\right] $ if it is continuous at every $c$ in $%
\left( a,b\right) $ and 
\begin{equation*}
\lim\limits_{x\rightarrow a^{+}}f\left( x\right) =f\left( a\right) \text{ \
\ or \ \ }\lim\limits_{x\rightarrow b^{-}}f\left( x\right) =f\left( b\right) 
\text{, \ respectively.}
\end{equation*}

End-points of the interval require only one-sided limits.\bigskip 

Another way to express continuity is to say that $\lim\limits_{h\rightarrow
0}f\left( x+h\right) =f\left( x\right) $. \ Another alternative statement of
continuity is $\lim\limits_{x\rightarrow c}f\left( x\right) =f\left(
\lim\limits_{x\rightarrow c}x\right) $, so that there is a commutativity
between taking the limit and taking the function values.\bigskip 

Theorem: \ Suppose that $f$ and $g$ are functions that are continuous at $%
x=c $. \ \ Then:

\qquad 1) \ $f+g$ is continuous at $c$.

\qquad 2) \ $fg$ is continuous at $c$

\qquad 3) \ $f-g$ is continuous at $c$

\qquad 4) \ If $g\left( c\right) \not=0,$ then $\dfrac{f}{g}$ is continuous
at $c$

\qquad 5) If $g$ is continuos at $c$ and $f$ is continuous at $g\left(
c\right) ,$ then $f\circ g$ is continuous at $c$.\bigskip \bigskip

These properties can be proved using the properties of limits and the
definitions of the functions $f+g$, $fg$, $f-g$, $\dfrac{f}{g}$ and $f\circ
g $.\bigskip

Theorem: \ (\textbf{The Intermediate Value Theorem}) \ If $f$ is continuous
on a closed interval $\left[ a,b\right] $ and if $f\left( a\right) <0$ and $%
f\left( b\right) >0$, then there exists $c$ in $\left( a,b\right) $ so that $%
f\left( c\right) =0$.\bigskip

Proof: \ Suppose that the conditions hold. \ Define $S=\left\{ x:a\leq x\leq
b\text{ and \ }f\left( x\right) <0\right\} $. \ This set is non-empty
because $a$ is an element of it. \ This set is also bounded from above,
because for all $x$ in $S$, $x\leq b$ and so $b$ is an upper bound for $S$.
\ \ By the least upper bound property, $S$ has a least upper bound. \ Let us
detone it by $c$. \ Since $c$ is the least upper bound for $S$ and $b$ is an
upper bound, we also have that $c\leq b$. \ Since $a$ is in $S$ and $c$ is
an upper bound, we also have that $a\leq c$. \ Thus $c$ is in the interval $%
\left[ a,b\right] $. \ We will prove that $f\left( c\right) =0$.\bigskip

We will prove that $f\left( c\right) =0$ by showing that $f\left( c\right) $
cannot be positive or negative. \ \bigskip

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Suppose first that $f\left( c\right) $ is positive. \ Since $f$ is
continuous at $c$, that means that $f$ is positive on some open interval
containing $c.$ \ That means that $c$ is not the least upper bound for $S,$
because any number in that interval, to the left of $c$ is also an upper
bound for $S.$ \ That is impossible and so $f\left( c\right) $ cannot be
positive.\bigskip

Suppose now that $f\left( c\right) $ is negative. \ Since $f$ is continuous
at $c$, that means that $f$ is negative on some open interval containing $c.$
\ That means that $c$ is not an upper bound for $S,$ because a number in
that interval, to the right of $c$ is also an element of $S.$ \ That is
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So $f\left( c\right) =0$ which completes our proof.$~~\blacksquare $\bigskip
\bigskip

Theorem: \ (\textbf{The Intermediate Value Theorem for Continuous Functions}%
) \ If $f$ is continuous on a closed interval $\left[ a,b\right] $ and if $%
y_{0}$ is any value between $f\left( a\right) $ and $f\left( b\right) $,
then $y_{0}=f\left( c\right) $ for some $c$ in $\left[ a,b\right] $.\bigskip

Proof: \ Let us assume that $f\left( a\right) <f\left( b\right) $ and let $%
y_{0}$ be a fixed number between $f\left( a\right) $ and $f\left( b\right) $%
. We define a new function $g\left( x\right) =f\left( x\right) -y_{0}$. \
Clearly, $g$ is continuous on $\left[ a,b\right] $ because it is the
difference of two continuous functions. \ Also, $g\left( a\right) $ is
negative and $g\left( b\right) $ is positive, because\medskip

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $f\left( a\right) <y_{0}$ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $y_{0}<f\left(
b\right) $\medskip

\ \ \ \ \ \ \ \ \ \ \ \ \ \ $f\left( a\right) -y_{0}<0$ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $0<f\left( b\right)
-y_{0}$\medskip

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $g\left( a\right) <0$ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $0<g\left(
b\right) $\medskip \medskip

Then, by the Intermediate Value Theorem, there exists $c$ in $\left(
a,b\right) $ with $g\left( c\right) =0$.

\begin{eqnarray*}
0 &=&g\left( c\right)  \\
0 &=&f\left( c\right) -y_{0} \\
y_{0} &=&f\left( c\right) 
\end{eqnarray*}%
and this completes our proof.$\blacksquare $

\pagebreak 

Definition: \ Suppose that $f$ is a function and $c$ is an interior point of
its domain. \ If the (two-sided) limit%
\begin{equation*}
\lim\limits_{h\rightarrow 0}\dfrac{f\left( c+h\right) -f\left( c\right) }{h}
\end{equation*}%
exists (and is finite) we say that $f$ is \textbf{differentiable at }$c$ and
denote this limit as $f^{\prime }\left( c\right) $.

\bigskip 

Theorem: \ If $f$ is differentiable at $a$, then it is continuous
there.\bigskip

Proof: Suppose that $f$ is differentiable at a number $a$. \ Then $f^{\prime
}\left( a\right) $ exists which means that $f\left( a\right) $ exists and
the limit $\lim\limits_{h\rightarrow 0}\dfrac{f\left( a+h\right) -f\left(
a\right) }{h}$ also exists and is finite. \ Let us start with the true
statement that $0=0\cdot f^{\prime }\left( a\right) $.%
\begin{eqnarray*}
0 &=&0\cdot f^{\prime }\left( a\right)  \\
0 &=&\lim\limits_{h\rightarrow 0}h\cdot \lim\limits_{h\rightarrow 0}\dfrac{%
f\left( a+h\right) -f\left( a\right) }{h}\text{ \ \ \ \ \ \ \ \ \ \ \ by the
product rule of limits} \\
0 &=&\lim\limits_{h\rightarrow 0}\left( h\cdot \dfrac{f\left( a+h\right)
-f\left( a\right) }{h}\right) \text{ \ \ \ \ \ \ \ \ \ \ \ cancel out }h \\
0 &=&\lim\limits_{h\rightarrow 0}\left( f\left( a+h\right) -f\left( a\right)
\right) \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ by the difference rule
of limits} \\
0 &=&\lim\limits_{h\rightarrow 0}f\left( a+h\right)
-\lim\limits_{h\rightarrow 0}f\left( a\right)  \\
\lim\limits_{h\rightarrow 0}f\left( a\right)  &=&\lim\limits_{h\rightarrow
0}f\left( a+h\right) \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ by the constant rule of limits} \\
f\left( a\right)  &=&\lim\limits_{h\rightarrow 0}f\left( a+h\right) 
\end{eqnarray*}%
and $f\left( a\right) =\lim\limits_{h\rightarrow 0}f\left( a+h\right) $
means that $f$ is continuous at $a$.\bigskip 

So, differentiability implies continuity. \ What about backwards? \ The
answer is no. \ Consider the function 
\begin{equation*}
f\left( x\right) =\left\vert x\right\vert =\left\{ 
\begin{array}{ccc}
x & \text{if} & x\geq 0 \\ 
-x & \text{if} & y<0%
\end{array}%
\right. 
\end{equation*}

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continuous at zero, it is not differentiable there. \ Recall that the
derivative is a two-sided limit. \ As $h$ approaches zero, it is negative
when we compute the left-limit and positive when we compute the right limit.%
\begin{eqnarray*}
\lim\limits_{h\rightarrow 0^{-}}\dfrac{f\left( 0+h\right) -f\left( 0\right) 
}{h} &=&\lim\limits_{h\rightarrow 0^{-}}\dfrac{f\left( h\right) -f\left(
0\right) }{h}=\lim\limits_{h\rightarrow 0^{-}}\dfrac{-h-0}{h}%
=\lim\limits_{h\rightarrow 0^{-}}-1=-1\text{ \ \ \ and} \\
\lim\limits_{h\rightarrow 0^{+}}\dfrac{f\left( 0+h\right) -f\left( 0\right) 
}{h} &=&\lim\limits_{h\rightarrow 0^{+}}\dfrac{f\left( h\right) -f\left(
0\right) }{h}=\lim\limits_{h\rightarrow 0^{+}}\dfrac{h-0}{h}%
=\lim\limits_{h\rightarrow 0^{+}}1=1
\end{eqnarray*}%
So the derivative is not defined at $x=0$.

\pagebreak

Example 1: \ Suppose that $f$ is a function defined as $\ f\left( x\right)
=\left\{ 
\begin{array}{ccc}
mx-10 & \text{if} & x<-2 \\ 
x^{2}+9x-8 & \text{if} & x\geq -2%
\end{array}%
\right. .$ \ Find the value of $m$ if we know that $f$ is continuous
everywhere.

\bigskip

Solution: \ If $x<-2$, then the function is continuous for all $x.$ \
Similarly, $f$ is also continuous on all $x$ with $x\geq -2$. \ The only
questionable point is at $x=-2.$ \ For a continuous function, we need the
left limit and the right limit to exist and have the same value. \ 
\begin{eqnarray*}
\lim_{x\rightarrow -2^{-}}f\left( x\right) &=&\lim_{x\rightarrow
-2^{+}}f\left( x\right) \\
\lim_{x\rightarrow -2^{-}}\left( mx-10\right) &=&\lim_{x\rightarrow
-2^{+}}\left( x^{2}+9x-8\right)
\end{eqnarray*}%
By the various properties of limits, this equation can be simplified as
follows:%
\begin{eqnarray*}
m\left( -2\right) -10 &=&\left( -2\right) ^{2}+9\left( -2\right) -8 \\
-2m-10 &=&-22 \\
-2m &=&-12 \\
m &=&6
\end{eqnarray*}%
And so $m=6$ is the value for which $f$ is continuous on the entire number
line.

\bigskip

\bigskip

\begin{center}
{\Large Practice Problems}\bigskip
\end{center}

\begin{enumerate}
\item Suppose that $f$ is a function defined as $\ f\left( x\right) =\left\{ 
\begin{array}{ccc}
mx-13 & \text{if} & x<-10 \\ 
x^{2}+5x-3 & \text{if} & x\geq -10%
\end{array}%
\right. .$ \ Find the value of $m$ if we know that $f$ is continuous
everywhere.

\item Suppose that $f$ is a function defined as $\ f\left( x\right) =\left\{ 
\begin{array}{ccc}
8x-4 & \text{if} & x\leq 4 \\ 
-2x+b & \text{if} & x>4%
\end{array}%
\right. .$ \ Find the value of $b$ if we know that $f$ is continuous
everywhere.

\item Suppose that $f$ is a function defined as $\ f\left( x\right) =\left\{ 
\begin{array}{ccc}
mx-11 & \text{if} & x<-6 \\ 
x^{2}+4x-5 & \text{if} & x\geq -6%
\end{array}%
\right. .$ \ Find the value of $m$ if we know that $f$ is continuous
everywhere.

\item Suppose that $f$ is a function defined as $\ f\left( x\right) =\left\{ 
\begin{array}{ccc}
2x+b & \text{if} & x<7 \\ 
\sqrt{x+2} & \text{if} & x\geq 7%
\end{array}%
\right. .$ \ Find the value of $b$ if we know that $f$ is continuous
everywhere.
\end{enumerate}

\bigskip

\bigskip

\begin{center}
{\Large Answers - Practice Problems\bigskip }
\end{center}

\bigskip

1.) \ $-6$ \ \ \ \ \ \ 2.) \ $36$\ \ \ \ \ \ \ 3.) \ $-3$ \ \ \ \ \ 4) \ $%
-11 $\bigskip \bigskip

\end{document}