0
$\begingroup$

Let $a < b$. Suppose that to every continuous function $f : [a, b] \rightarrow \mathbb{R}$ and every subinterval $[\alpha, \beta] \subset [a, b]$ there is associated a number $I_\alpha^\beta(f)$ satisfying

  • $I_\alpha^\beta(sf + tg) = sI_\alpha^\beta(f) + tI_\alpha^\beta(g)$ for all $s, t \in \mathbb{R}$
  • $I_\alpha^\beta(1) = \beta - \alpha$
  • $I_\alpha^\beta(f) = I_\alpha^\gamma(f) + I_\gamma^\beta(f)$ for all $\gamma \in [\alpha,\beta]$
  • $\left\lvert I_\alpha^\beta(f)\right\rvert \leq (\beta - \alpha)\sup_{x \in [\alpha, \beta]} \left\lvert f(x) \right\rvert$

How can I show that $(\beta - \alpha)\inf_{x \in [\alpha, \beta]}f(x) \leq I_\alpha^\beta(f) \leq (\beta - \alpha)\sup_{x \in [\alpha, \beta]}f(x)$? So far, I can only show the right inequality in the case where $\sup_{x \in [\alpha, \beta]}\left\lvert f(x) \right\rvert = \sup_{x \in [\alpha, \beta]} f(x) $, and the left inequality in the case where $\sup_{x \in [\alpha, \beta]}\left\lvert f(x) \right\rvert = \sup_{x \in [\alpha, \beta]} -f(x) $. How can I show both inequalities in both cases?

In the end, I want to show that $d/d\beta\left[I_\alpha^\beta(f)\right] = f(\beta)$, and once I figure out how to prove the above then I know how to prove the rest. However, if there is a different way to prove this then I would accept that as well.

  • 0
    Are you sure you have the right amount of assumptions? I don't think these are sufficient to ensure that $f \ge 0 \implies I(f) \ge 0$ (which is what we need here) – 2017-01-22
  • 0
    @OpenBall Sorry, I forgot to add the absolute value sign on the left hand side of the last axiom, does this make a difference now? – 2017-01-22

0 Answers 0