5

While answering this question it occurred to me that there might be a function $f$ with the following properties:

  • $f$ is Riemann integrable on closed interval $[a, b] $.
  • The function $F$ defined by $$F(x) =\int_{a} ^{x} f(t) \, dt$$ is differentiable on $[a, b] $ (with one sided derivatives being considered at end points).
  • There is a point $c\in(a, b) $ such that $F'$ is discontinuous at $c$ and $f'(c) $ exists.

I wasn't able to construct such an example myself. Do help in finding one such function or show that such a function does not exist.

Clearly such an $f$ must be discontinuous at some points in $[a, b] $ and further its discontinuities must be of essential type and not the jump kind. The point $c$ must be a limit point of the set of discontinuities of $f$. Also via measure theory we must have $F'=f$ almost everywhere. I wonder if these properties are of any help here.

Paramanand Singh
  • 87,309

1 Answers1

5

We have existence of $F'(x) = \lim_{h \to 0} \frac{1}{h}\int_x^{x+h} f(t) \, dt$ for all $x \in (a,b)$. Since $f'(c)$ exists, $f$ is continuous at $c$ and $F'(c) = f(c)$.

Consider $h > 0$. We have

$$|F'(x) - F'(c)| = \left| \lim_{h \to 0+} \frac{1}{h}\int_x^{x+h} f(t) \, dt - f(c)\right| = \left| \lim_{h \to 0+} \frac{1}{h}\int_x^{x+h} (f(t)-f(c)) \, dt \right| \\ = \lim_{h \to 0+} \left| \frac{1}{h}\int_x^{x+h} (f(t)-f(c)) \, dt \right| \leqslant \lim_{h \to 0+} \frac{1}{|h|}\int_x^{x+h} |f(t)-f(c)| \, dt $$

If $|x -c|< \delta$ then $|f(x) - f(c)| < \epsilon$. If $h$ is sufficiently small and $|x - c| < \delta/2$, we also have $|x+h - c| < \delta$ and $|f(t) - f(c)| < \epsilon$, whence

$$|F'(x) - F'(c)| < \lim_{h \to 0+}\frac{h}{|h|}\epsilon = \epsilon$$

Therefore $F'$ is continuous at $c$.

RRL
  • 90,707