3
$\begingroup$

Let $[a,b]\subseteq \mathbb{R}$ a non-degenerate interval and $f,g:[a,b]\rightarrow \mathbb{R}$ two continuous functions, differentiable at $(a,b)$ such that |f'(x)| \leq g'(x), \forall x\in(a,b). Prove that $|f(a) - f(b)|\leq g(b) - g(a)$.

I've already proven that |f'(x)| < g'(x) implies $|f(a) - f(b)|\leq g(b) - g(a)$. But I'm having some trouble to show the other one, with the weaker hypothesis.

  • 2
    How did you prove the weaker statement? Why does it not extend to the stronger statement you really want to show?2011-12-02

2 Answers 2

6

One solution for such problems is to add in some $\epsilon$:

$g(x) + \epsilon (x-a)$ satisfies your stronger hypothesis for every $\epsilon > 0$, so $|f(b) - f(a)| \leq g(b) - g(a) + \epsilon (b-a).$ As $\epsilon \to 0$, you obtain the desired result.

3

It's just Fundamental Theorem of Calculus and elementary inequality.

In fact for $x\leq y \in (a,b)$ then: $|f(y)-f(x)| =\left| \int_x^y f^\prime (t)\ \text{d} t\right| \leq \int_x^y |f^\prime (t)|\ \text{d} t\leq \int_x^y g^\prime (t)\ \text{d} t =g(y)-g(x)\; ;$ hence the inequality follows by continuity (it suffices to take the limits for $x\to a^+$ and $y\to b^-$).