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Let $f$ be Riemann integrable on $[a, b]$, let $c\in(a, b)$, and let $\displaystyle F(x)=\int_a^x f(t)\ dt$, $a\le x\le b$. For the following statement, give either a proof or a counterexample:

If $f$ is differentiable at $c$, then $F'$ is continuous at $c$.

My attempt:

My hunch is that the statement is false. Since $f$ is differentiable at $c$, $f$ is continuous at $c$ and so by the first Fundamental Theorem of Calculus, $F'(c)=f(c)$. But $F'$ need not be equal to $f$ at other points.

Is this correct? If so, how do I find a counterexample? If not, how do I prove the statement?

StubbornAtom
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Siddhartha
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  • Riemann or Lebesgue integrable? – Momo Dec 03 '16 at 00:14
  • @Momo Riemann integrable – Siddhartha Dec 03 '16 at 00:17
  • I think that one would like to check the continuity of $F(x)$, rather than that of $F'(x)$ as, by the theorem, you get $F'(x)=f(x)$ in the interval including $c$, and as $f$ is differentiable, then it is continuous at that point. – Miguel Angel Alarcon Bustos Dec 03 '16 at 00:23
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    The existence of $f'(c)$ does not imply $F'$ exists in a neighborhood of $c.$ For example, $f$ could have a bunch of jump discontinuities near $c,$ and we know that at a jump discontinuity, $F'$ won't exist. – zhw. Dec 03 '16 at 00:26

1 Answers1

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Construct a counterexample where $F'$ fails to exist at infinitely many points $x_n \to c$ and is technically not continuous at $c$, but where $f'(c)$ exists.

For example, take $c= 0$ and $f:[0,1/2] \to \mathbb{R}$ such that $f(0) = 0$ and for $x \in (0,1/2]$ and $k = 1,2, \ldots$

$$f(x) = x^2 \begin{cases}\,\,\,\,\,1, \,\,\,\, x\in \left(\frac{1}{2k+1},\frac{1}{2k}\right] \\ -1, \,\,\,\, x\in \left(\frac{1}{2k+2},\frac{1}{2k+1}\right] \end{cases}$$

RRL
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  • So $F'$ fails to exist at each $1/m, m>2.$ Is that the point here? – zhw. Dec 03 '16 at 00:59
  • Why the alteration? All you seem to be saying is that $F'$ doesn't exist everywhere in some neighborhood of $0$. That' what you had before, Is that right? – zhw. Dec 03 '16 at 01:30
  • That's the extent of it. – RRL Dec 03 '16 at 01:32
  • I suppose it becomes a technicality as to the definition of continuity or how the derivative is defined at those points. – RRL Dec 03 '16 at 01:35
  • Does your function $f$ look like two parabolas one concave the other convex meeting at the origin, changing from convex to concave infinitely many times for every subinterval, and with jump discontinuities at every $1/m$ So the graph of $F$ looks like a jagged $xsin(1/x)$ graph that is not differentiable at every $1/m$? – helios321 Oct 01 '17 at 04:19
  • @helios321: Yes and $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = 0$ exists. But in one sense the limit of $F'$ does not exist. However with the domain restricted it does. This question came up a second time -- answered in different ways -- one similar to this. I'll see if I can find it. – RRL Oct 01 '17 at 04:24
  • @helios321: See here for the same question asked 3 days later. As you can see one answer is similar to mine stating that if $F'$ fails to exist in a "full neighborhood" of $c$ then we technically can't say it is continuous at $c$. – RRL Oct 01 '17 at 04:41