Question 8.T.11: Suppose that f ∈ R(a, b), and let F : [a, b] → R be defined ...
Suppose that f ∈ \mathcal{R}(a, b) , and let F : [a, b] → \mathbb{R} be defined by
F (x) = \int_{a}^{x}{f (t) dt}.
Then
(i) F is continuous and satisfies a Lipschitz condition on [a, b] .
(ii) If f is continuous at c ∈ [a, b] , F is differentiable at c and
F^{\prime} (c) = f (c) .
(i) Since f is bounded, there is a number K such that
|f (x)| ≤ K for all x ∈ [a, b] .
If x, y ∈ [a, b] and x < y, then, from the properties of the integral, we have
|F (y) − F (x)| = \left| \int_{a}^{y}{f (t) dt} − \int_{a}^{x}{f (t) dt}\right|
= \left| \int_{x}^{y}{f (t) dt} \right|
≤ \int_{x}^{y}{\left|f (t)\right| dt}
≤ \int_{x}^{y}{Kdt} = K |y − x| ,
which means F is Lipschitz continuous on [a, b] .
(ii) Suppose f is continuous at c. Then, given any ε > 0, there is a δ > 0 such that
x ∈ [a, b] , |x − c| < δ ⇒ |f (x) − f (c)| < ε.
If t lies between c and x, then |t − c| < δ and therefore |f (t) − f (c)| <
ε. Noting that
f (c) = \frac{1}{x − c} \int_{c}^{x}{f (c) dt}, x ≠ c,
we can write
\left|\frac{F (x) − F (c)}{x − c} − f (c)\right| = \left|\frac{1}{x − c} \int_{c}^{x}{[f (t) − f (c)] dt}\right|
≤ \frac{1}{\left|x − c\right|} \left|\int_{c}^{x}{\left|f (t) − f (c)\right| dt}\right|
≤ \frac{1}{\left|x − c\right|} \left|\int_{c}^{x}{εdt}\right| = ε,
which confirms that \lim_{x→c} \frac{F (x) − F (c)}{x − c} = f (c) , as required.