0
$\begingroup$

Let $f: [a, b] \to \mathbb R$ be integrable function and $F: [a,b] \to \mathbb R$, $F(x) = \int_a^x f(t) dt$

a) Show that if $f(x) \ge 0$ for all $x \in [a, b]$ then $F$ is increasing

b) If $F$ is increasing, can you conclude that $f(x) \ge 0$ for all $x \in [a, b]$?

So, I can write down the proof of fundamental theorem of calculus, and that will prove it, but is there a neater way to prove it, just for the increasing part?

$\endgroup$
  • $\begingroup$ The fundamental theorem of calculus supposes the function $f$ is continuous. Here it is only supposed integrable. $\endgroup$ – Bernard Feb 24 '17 at 11:29
  • $\begingroup$ @Bernard: you are not right. One part of the fundamental theorem of calculus reads as follows: if $f$ is Riemann-integrable, then $F$ is Lipschitz-continuous. $\endgroup$ – Fred Feb 24 '17 at 11:36
  • 1
    $\begingroup$ @Fred: I'm quite sorry, but the function $f(x)=1$ if $x\ne0$, $f(0)=0$ is Riemann-integrable, but certainly not continuous (whether Lipschitz- or not). More generally, any step function is integrable, and not continuous. Unless we don't have the same definition of (Lipschitz-)continuity (?). $\endgroup$ – Bernard Feb 24 '17 at 11:46
  • $\begingroup$ @Bernard: the function $F(x) = \int_a^x f(t) dt$ is Lip. -continuous, if $f$ is Riemann- integrable. What is your problem ? $\endgroup$ – Fred Feb 24 '17 at 11:49
  • 1
    $\begingroup$ The fundamental theorem asserts $F$ is differentiable if $f$ is continuous. My problem was mentioning the fundamental theorem about a possibly non-continuous function. $\endgroup$ – Bernard Feb 24 '17 at 11:53
2
$\begingroup$

a) If $a \le x <y \le b$, then $F(y)-F(x)=\int_x^y f(t) dt \ge 0$, hence $F(x) \le F(y)$

b) Let $a=0,b=1$ and

$f(0)=-1$ and $f(x)=0$ for $x \in (0,1]$. Then $F$ is constant, hence increasing, but $f(0)<0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.