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I got stuck while doing exercise of the Apostol's Calculus, the exercise 28 of Section 5.5.

Here's the question

Given a function $f$ such that the integral $A(x) = \int_a^xf(t)dt$ exists for each $x$ in an interval $[a, b]$. Let $c$ be a point in the open interval $(a, b)$. Consider the following ten statements about this $f$ and this A:

And there are five (a) ~ (e) statements on the left, and five ($\alpha$) ~ ($\epsilon$) statements on the right. The author asks the reader to decide the implicative relation from statements on the left to statements on the right. I thought I answered correctly but the solution at the end tells different. I don't know why this is wrong.

(d) $f'(c)$ exists. $\implies$ ($\epsilon$) $A'$ is continuous at c.

This is my argument: By the Example 7 of Section 4.4, the differentiability of $f$ at c implies the continuity of $f$ at c. Since $f$ is differentiable at c, $f$ is continuous at c, so that $A'$, which equals to $f$, should continuous at c.

But the solution at the end says (d) does not implies ($\epsilon$).

Sorry for the partializing the problem, it maybe tough to point out what is wrong.

Paramanand Singh
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J.Bo
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  • Your argument '...so that A..′ is flawed. How did you come to that conclusion? – SL_MathGuy Jan 21 '20 at 03:53
  • Since $A' = f$ and $f$ is continuous at c, $A'$ should be continuous at c. – J.Bo Jan 21 '20 at 03:58
  • How do you claim $A' = f $? Which theorem did you use? – SL_MathGuy Jan 21 '20 at 04:00
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    The fundamental theorem of calculus. – J.Bo Jan 21 '20 at 04:01
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    In order to use FTC,$f$ needs to be continuous on $[a,b]$ not just at one point. So,the theorem can't be applied in this case. – SL_MathGuy Jan 21 '20 at 04:02
  • You don't need any continuity on an interval. – MoonLightSyzygy Jan 21 '20 at 04:10
  • If we assume that $f'$ is continuous at c, then is it possible to use FTC? We have an interval around c. – J.Bo Jan 21 '20 at 04:11
  • What I am saying is that the fundamental theorem of calculus doesn't need continuity of $f$ on an interval, only integrability and continuity at the point $c$. With that it ensures that $A'(c)$ exists and it is equal to $f(c)$. What you were told above of FTC requiring $f$ continuous on $[a,b]$ is Wikipedia knowledge. Check your book instead. – MoonLightSyzygy Jan 21 '20 at 04:26
  • @MoonLightSyzygy I forgot $f'(c)$ should exists when I mentioned my counter-example. But, your argument '...the fundamental theorem of calculus doesn't need continuity of $f$ on an interval, only integrability and continuity...' doesn't make sense since every continuous function on a closed bounded interval is integrable. Moreover,when we talk about FTC ,it's about a continuous function on a closed bounded interval not on 'ANY' interval. – SL_MathGuy Jan 21 '20 at 04:48
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    @SL_MathGuy Read the FTC from a good book, like Apostol's Calculus, rather than Wikipedia. – MoonLightSyzygy Jan 21 '20 at 04:51
  • @MoonLightSyzygy It's clear you don't know the relationship between continuity of a function on a closed bounded interval & the integrability of the function. I don't intend to carry this conversation forward. If possible, please give a counter-example to justify your argument that the continuity of $f$ is not necessary for FTC so that I can be convinced. – SL_MathGuy Jan 21 '20 at 04:55
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    @SL_MathGuy Don't carry any conversation, just open the book. Page 202, in the Second Edition, Theorem 5.1, with the proper proof (because maybe the statement could be wrong if one doesn't check the proof too) on page starting at the end of page 203. – MoonLightSyzygy Jan 21 '20 at 04:57
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    @SL_MathGuy: you should have a look at https://math.stackexchange.com/a/1900844/72031 where I have discussed FTC in a more general setting. – Paramanand Singh Jan 21 '20 at 10:09

2 Answers2

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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.

Paramanand Singh
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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.

symplectomorphic
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