2
$\begingroup$

Let $f:[0,1]\to\mathbb{R}$ be a Riemann integrable function. Define a function $F:[0,1]\to\mathbb{R}$ by $$F(x)=\int_0^xf.$$ Suppose that $F$ is differentiable at $c\in(0,1)$. Then is it necessarily true that $F'(c)=f(c)$ ? Note that it is true if $f$ is continuous at $c$.

$\endgroup$
  • 1
    $\begingroup$ for discontinuous $f$, the limit for $F'$ may not exist, but if it does, how can it be anything else? $\endgroup$ – gt6989b Aug 13 at 2:35
  • 2
    $\begingroup$ Suppose $f$ is continuous, then $F'(x) = f(x)$ everywhere in $[0,1]$. What happens if you change the value of $f$ at a single point $c$? $\endgroup$ – RRL Aug 13 at 2:38
  • $\begingroup$ This is pretty much a statement of the Fundamental Theorem of Calculus. $\endgroup$ – Jared Aug 13 at 2:41
  • $\begingroup$ @gt6989b Apparently it can. See the answer below. $\endgroup$ – ashpool Aug 13 at 2:44
  • 1
    $\begingroup$ You can actually relax the condition of continuity at $c$ a bit to get the same result, i.e. $F'(c)=f(c)$. $c$ needs to be a so called Lebesgue point of $f$ (see en.wikipedia.org/wiki/Lebesgue_point and en.wikipedia.org/wiki/…) $\endgroup$ – humanStampedist Aug 13 at 11:36
10
$\begingroup$

Just take any continuous function with its value changed at a single point. For example let $f:[0, 1] \to \mathbb R$ be any continuous function and define $\tilde{f}: [0, 1] \to \mathbb R$ by $$ \tilde{f}(x) = \begin{cases} f(x) + 1 & \text{if $x = 1/2$}, \\ f(x) & \text{if $x \neq 1/2$}. \end{cases}$$ Then since $f$ and $\tilde{f}$ differ by only a single point, they have the same indefinite integral $$F(x) = \int_0^x f(x) \, \text{d}x = \int_0^x \tilde{f}(x) \, \text{d}x.$$ However $F'(1/2) = f(1/2) \neq \tilde{f}(1/2)$.

| cite | improve this answer | |
$\endgroup$

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.