Theoretical Remarks #4

   In Monday's 6/Jan/2014 post we mentioned a proof for the existence of an antiderivative for any function continuous on an interval.
   In today's post, i am going to supply an alternative proof for the same proposition. For the reader's convenience, i am repeating at this point the statement of the proposition:
 Proposition: Let a real function $f$, continuous on an interval $\Delta$ and let $a \in \Delta$ be a fixed point. Then the function $F(x)=\int_{a}^{x}f(t)dt$ is an antiderivative function of $f$ in $\Delta$. In other words:
$$
F'(x) = \big( \int_{a}^{x}f(t)dt \big)' = f(x)
$$
for all $x \in \Delta$.
Prooof: (alternative)
   It is sufficient to show that for any fixed point $x_{0} \in \Delta$ we have $F'(x_{0})=f(x_{0})$. Let $x_{0}, x_{0}+h \in \Delta$ with $h \neq 0$. Then we can compute
$$
\begin{array}{c}
F(x_{0}+h) - F(x_{0}) = \int_{a}^{x_{0}+h}f(t)dt - \int_{a}^{x_{0}}f(t)dt = \\
     \\
\bigg( \int_{a}^{x_{0}}f(t)dt + \int_{x_{0}}^{x_{0}+h}f(t)dt \bigg)- \int_{a}^{x_{0}}f(t)dt =
 \int_{x_{0}}^{x_{0}+h}f(t)dt
\end{array}
$$
and since $h \neq 0$, this implies that
\begin{equation} \label{diff}
\frac{F(x_{0}+h) - F(x_{0})}{h} = \frac{1}{h} \int_{x_{0}}^{x_{0}+h}f(t)dt
\end{equation}
In order to proceed, we will distinguish between two cases:

• $h > 0$   $\rightsquigarrow$  (I)
• $h < 0$   $\rightsquigarrow$  (II)

(I). $h > 0$: Since $[x_{0},x_{0}+h] \subseteq \Delta$, $f$ is continuous on $[x_{0},x_{0}+h]$ and the Extreme value theorem applies: there are numbers $c,d \in [x_{0},x_{0}+h]$ such that $f(c)=m$ and $f(d)=M$ are the absolute minimum and absolute maximum values respectively of $f$ in $[x_{0},x_{0}+h]$. Consequently
$$
\begin{array} {c}
mh \leq \int_{x_{0}}^{x_{0}+h}f(t)dt \leq Mh \Leftrightarrow f(c)h \leq \int_{x_{0}}^{x_{0}+h}f(t)dt \leq f(d)h \Leftrightarrow \\ \\
   \\
\Leftrightarrow f(c) \leq \frac{1}{h} \int_{x_{0}}^{x_{0}+h}f(t)dt \leq f(d) \stackrel{\eqref{diff}}{\Leftrightarrow} f(c) \leq \frac{F(x_{0}+h) - F(x_{0})}{h} \leq f(d)
\end{array}
$$

So we have concluded that
\begin{equation} \label{sand1}
 f(c) \leq \frac{F(x_{0}+h) - F(x_{0})}{h} \leq f(d)
\end{equation}
At this point, we have to observe the following thing: by the application of the extreme value theorem on the continuous function $f$ on the interval $[x_{0},x_{0}+h]$ it follows that both $c$ and $d$ depend in general on the value of $h > 0$. It is easy to see that their values are actually functions of the positive $h$: So we can write $c(h)$ and $d(h)$. Not much needs to be said about these functions; their behaviour may be complicated in general (for example, you can provide an argument to show that $c(h), \ d(h)$ need not even be continuous in general!). However we have:
\begin{equation} \label{concomplim1}
\begin{array}{c}
\lim_{h \rightarrow 0^{+}} c(h) = x & ,   &  \lim_{h \rightarrow 0^{+}} d(h) = x
\end{array}
\end{equation}
\eqref{concomplim1} can be proved as a simple application of the $(\varepsilon, \delta)$-definition of the limit. Readers are adviced to show that explicitly for practise!
   Taken that $f$ is continuous on $\Delta$ and thus on $[x_{0},x_{0}+h]$, \eqref{concomplim1} imply that
\begin{equation} \label{concomplim2}
\begin{array}{c}
\lim_{h \rightarrow 0^{+}} f(c) = \lim_{h \rightarrow 0^{+}} f(c(h)) = f(x) \\
    \\
\lim_{h \rightarrow 0^{+}} f(d) = \lim_{h \rightarrow 0^{+}} f(d(h)) = f(x)
\end{array}
\end{equation}
Now combining \eqref{sand1} together with \eqref{concomplim2} and applying the squeeze theorem from the right, we get
\begin{equation} \label{from the right}
\lim_{h \rightarrow 0^{+}} \frac{F(x_{0}+h) - F(x_{0})}{h} = f(x)
\end{equation}

(II). $h < 0$:  In this case $[x_{0}+h,x_{0}] \subseteq \Delta$ and we proceed again following exactly the same steps as before keeping however in mind that now $h < 0$. We leave the intermediate details to the reader. We finally end up in
 \begin{equation} \label{from the left}
\lim_{h \rightarrow 0^{-}} \frac{F(x_{0}+h) - F(x_{0})}{h} = f(x)
\end{equation}
   Combining \eqref{from the right}, \eqref{from the left} we get the result
$$
\lim_{h \rightarrow 0} \frac{F(x_{0}+h) - F(x_{0})}{h} = F'(x_{0}) = f(x_{0})
$$
which finally concludes the proof!

Theoretical Remarks #3

We come in today's post to supply a proof for a well known Calculus proposition:  In friday's 27/Dec/2013 post we mentioned (without proof), the following proposition:
Proposition: Let a real function $f$, continuous on an interval $\Delta$ and let $a \in \Delta$ be a fixed point. Then the function $F(x)=\int_{a}^{x}f(t)dt$ is an antiderivative function of $f$ in $\Delta$. In other words:
$$
F'(x) = \big( \int_{a}^{x}f(t)dt \big)' = f(x)
$$
for all $x \in \Delta$.

Proof: 
   It is sufficient to show that for any fixed point $x_{0} \in \Delta$ we have $F'(x_{0})=f(x_{0})$.

   Let us first study the difference quotient $\frac{F(x)-F(x_{0})}{x-x_{0}}$ whose limit at $x \rightarrow x_{0}$, $x \neq x_{0}$ defines the value of $F'(x_{0})$:
$$
\begin{array}{c}
\frac{F(x)-F(x_{0})}{x-x_{0}}= \frac{1}{x-x_{0}}\bigg( \int_{a}^{x}f(t)dt - \int_{a}^{x_{0}}f(t)dt \bigg) =  \\
    \\
= \frac{1}{x-x_{0}}\bigg(  \int_{x_{0}}^{a}f(t)dt + \int_{a}^{x}f(t)dt  \bigg) =  \frac{1}{x-x_{0}} \int_{x_{0}}^{x}f(t)dt
\end{array}
$$
thus
\begin{equation} \label{diff*}
\frac{F(x)-F(x_{0})}{x-x_{0}}=\frac{1}{x-x_{0}} \int_{x_{0}}^{x}f(t)dt
\end{equation}
and since
\begin{equation} \label{fun}
f(x_{0}) = \frac{1}{x-x_{0}}(x-x_{0})f(x_{0}) = \frac{1}{x-x_{0}}\int_{x_{0}}^{x}f(x_{0}) dt
\end{equation}
combining \eqref{diff*} and \eqref{fun}, we readily get the following relation:
\begin{equation} \label{diffun}
\frac{F(x)-F(x_{0})}{x-x_{0}}-f(x_{0}) = \frac{1}{x-x_{0}} \int_{x_{0}}^{x} \big( f(t) - f(x_{0}) \big) dt
\end{equation}
   Since $f$ is continuous at $x_{0} \in \Delta$, for any $\varepsilon > 0$ there is a $\delta > 0$ such that: for any $t \in \Delta$ with $|t-x_{0}| < \delta$ we will have $|f(t)-f(x_{0})| < \varepsilon$.

   Thus, for any $x \in \Delta$ with $0 < |x-x_{0}| < \delta$, using \eqref{diffun} we get:
$$
\begin{array}{c}
\bigg| \frac{F(x)-F(x_{0})}{x-x_{0}}-f(x_{0})  \bigg| = \frac{1}{|x-x_{0}|} \bigg| \int_{x_{0}}^{x} \big( f(t) - f(x_{0}) \big) dt \bigg|  \leq \\
      \\
\leq \frac{1}{|x-x_{0}|} \bigg| \int_{x_{0}}^{x} \big| f(t) - f(x_{0}) \big| dt \bigg| < \frac{1}{|x-x_{0}|} \big| \int_{x_{0}}^{x} \varepsilon dt \big| =  \\
    \\
= \frac{1}{|x-x_{0}|} \varepsilon |x-x_{0}| = \varepsilon
\end{array}
$$
But the above means -according to the $(\varepsilon, \delta)$ definition of the limit- that
$$
F'(x_{0}) = \lim_{x \rightarrow x_{0}} \bigg( \frac{F(x)-F(x_{0})}{x-x_{0}} \bigg) = f(x_{0})
$$
which finally concludes the proof.

Mathematics Workbook - UoC

Contains 30 review exercises, indicative of the prerequisite level for someone intending to study Mathematics in a top University: Mathematics Workbook - UoC 

Notes for an introduction to Pure Mathematics (in Greek)

A general introduction to undergraduate level mathematics, focusing the viewpoint of pure mathematics (not for the faint-hearted !!)  Σημειώσεις για μια εισαγωγή στα Καθαρά Μαθηματικά

 

Fermat's last theorem (in Greek)

A very nice and interesting paper, with an appendix focusing on the historical development of the topic.
Unfortunately the text is in Greek only ;)    -->  Το τελευταίο θεώρημα του Fermat
(From: University of Athens, School of Mathematics)