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:

$$

F'(x) = \big( \int_{a}^{x}f(t)dt \big)' = f(x)

$$

for all $x \in \Delta$.

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:

$$

\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}

$$

\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!

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}

\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!

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