Monday, January 6, 2014

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
\label{diff*}
\frac{F(x)-F(x_{0})}{x-x_{0}}=\frac{1}{x-x_{0}} \int_{x_{0}}^{x}f(t)dt

and since
\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

combining \eqref{diff*} and \eqref{fun}, we readily get the following relation:
\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

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.