Question about the relation between integration and differentiation (From Calculus by Apostol)

Solution 1:

You start your argument correctly that $f'(c) $ exists and hence $f$ is continuous at $c$ and therefore by FTC $A'(c) =f(c) $. But beyond that you can't conclude anything.

For continuity of $A'$ at $c$ you need to ensure that $A'$ exists in some neighborhood of $c$ and further that $A'(x) \to A'(c) $ as $x\to c$.

For a concrete example let $f(0)=0$ and $$f(x) =x^2((1/x)-\lfloor 1/x\rfloor)\, \forall x\in(0,1], f(-x) =f(x) \,\forall x\in(0,1]$$ It is easy to prove that $f$ is discontinuous at points $$x=\pm 1/2,\pm 1/3,\dots,\pm 1/n,\dots$$ and continuous at rest of the points in $[-1,1]$. Moreover each of its discontinuity is a jump discontinuity.

With some effort one can prove that the function $f$ defined above is Riemann integrable on $[-1,1]$ (more generally if the set $D$ of discontinuities of a bounded function has a finite number of limit points then the function is Riemann integrable).

The corresponding function $$A(x) =\int_{-1}^{x}f(t)\,dt$$ is continuous on $[-1,1]$ and differentiable at all points of $[-1,1]$ except $\pm 1/2,\pm 1/3,\dots, \pm 1/n,\dots$. At these points $f$ has a jump discontinuity so $A$ is not differentiable there.

Further check that $f'(0)=0$ and $A'(0)=f(0)=0$ but $A'$ does not exist in any neighborhood of type $(-h,h) $ (because of trouble points $\pm 1/n$) and hence $A'$ is discontinuous at $0$.

There does not exist a counter example where $A'$ exists in entire interval but not continuous at some point of that interval.

Solution 2:

Paramanand gives an explicit counterexample, which proves that your claim is false. But I want to add this further answer to expose the flaw in your reasoning.

Your argument appears to rely on the following missing assumption: if $g$ and $h$ are two functions defined on a set containing $c$, $g(c)=h(c)$, and $g$ is continuous at $c$, then $h$ is continuous at $c$. (If this were true, your conclusion would follow. In your case, $g=f$ and $h=F’$.)

This assumption is false. Here’s a very simple counterexample. Take $g(x)=0$ on the real line and let $h(x)=1$ for all $x$ except zero, where $h$ jumps to zero. Then $g(0)=0=h(0)$ and $g$ is continuous at zero, but $h$ is not continuous at zero.