5
$\begingroup$

Given a function $f$ that is differentiable at a point $x_0$, if we define (using the Riemann integral)

$$F(x) = \int_a^x f$$

Can we necessarily say that $F^{\prime}(x)$ is continuous at $x_0$? Going back and forth between $f$ and $F$ confuses me a bit. I think that the Fundamental Theorem of Calculus gives us some relation between $F^{\prime}(x_0)$ and $f(x_0)$, but I'm not sure.

$\endgroup$
5
  • 2
    $\begingroup$ Basically, if you have some function $f$, the function $F(x) = \int_a^x f$ is "nicer" than $f$. If $f$ is integrable, then $F$ is continuous. If $f$ is continuous, then $F$ is differentiable. If $f$ is differentiable, $F$ is twice differentiable. $\endgroup$ Commented Dec 6, 2016 at 18:14
  • $\begingroup$ @MathematicsStudent1122: That is true if $f$ is continuous near $x_0$. But it was only assumed that $f$ is differentiable and thus continuous at $x_0$. $F$ is almost everywhere differentiable and $F' = f$ almost everywhere. But $F'$ may not exists on a neighborhood of $x_0$, so speaking of continuity is difficult. $\endgroup$
    – user251257
    Commented Dec 6, 2016 at 19:50
  • $\begingroup$ @user251257 The first fundamental theorem of calculus states that if $f$ is continuous at $x_0$ (in our case, $f$ is differentiable, so this is certainly true), then $F(x) = \int_a^x f$ is differentiable at $x_0$ and $F'(x_0) = f(x_0)$. This implies $F'$ is continuous at $x_0$. Continuity in a neighbourhood of $x_0$ is not assumed. $\endgroup$ Commented Dec 6, 2016 at 20:10
  • $\begingroup$ @MathematicsStudent1122: Just because $F'$ exists in $x_0$ doesn't make it continuous at $x_0$... $\endgroup$
    – user251257
    Commented Dec 6, 2016 at 20:12
  • $\begingroup$ @user251257 Never mind, you're right then. Do you have a counterexample? $\endgroup$ Commented Dec 6, 2016 at 20:14

3 Answers 3

4
$\begingroup$

You can't say $F'$ is continuous at $x_0$ because $F'$ may not exist in a full neighborhood of $x_0.$ Take the interval $[-1,1]$ with $x_0=0$ for example. Choose any sequences $a_n,b_n$ you like such that $1\ge b_1 > a_1 > b_2 > a_2 > \cdots \to 0^+.$ Define $f(x) = x^2$ on each $[a_n,b_n],$ $f=0$ everywhere else. Then $f$ is Riemann integrable on $[-1,1]$ and $f'(0)=0.$ But at each $a_n$ and $b_n,$ $f$ has a jump discontinuity, hence $F'$ does not exist at these points. No matter which neighborhood of $0$ you examine, there will be lots of points, namely in the tail ends of the sequences $a_n,b_n,$ where $F'$ doesn't exist.

$\endgroup$
3
  • 2
    $\begingroup$ When discussing the continuity of a function, it's taken for granted that we're discussing its continuity on its domain. Is $F'$ continuous on its domain? You haven't answered that. $\endgroup$ Commented Dec 7, 2016 at 3:40
  • 1
    $\begingroup$ Nice counter-example. +1 $\endgroup$
    – Paramanand Singh
    Commented Dec 7, 2016 at 8:47
  • $\begingroup$ I don't understand where the discontinuity occurs, could you help me? $\endgroup$
    – Wrloord
    Commented Dec 14, 2021 at 22:14
2
$\begingroup$

The statement is true in the sense, that $F'$ is continuous at $x_0$ on its domain, which need not to be the entire interval $(a, b)$. We may drop that $f$ is differentiable at $x_0$, we may also replace Riemann integrability with Lebesgue.

Claim:

Let $D = \{ x\in (a, b) \mid F \text{ is differentiable at } x \}$. For $x_n\in D$ with $x_n\to x_0$ it follows $F'(x_n) \to F'(x_0) = f(x_0)$.

Proof:

Notice that as $f$ is continuous at $x_0$ it follows $$F'(x_0) = \lim_{y\to x_0} \frac{1}{y - x_0} \int_{x_0}^y f(t) dt = f(x_0).$$

Fix $\varepsilon > 0$.

  1. As $f$ is continuous at $x_0$, a $\delta > 0$ exists such that for every $x\in [a, b] \cap (x_0 - \delta, x_0 + \delta)$ it follows $$|f(x) - f(x_0)| < \frac\varepsilon2.$$
  2. As $x_n\to x_0$, a $N> 0$ exists such that for every $n \ge N$ it follows $$|x_n - x_0| < \frac{\delta}{2}.$$
  3. As $F$ is differentiable at $x_n$, a $\eta_n > 0$ exists such that for every $y\in [a, b] \cap (x_n - \eta_n, x_n + \eta_n)$ with $y\ne x_n$ it follows $$ \left|\frac{F(y) - F(x_n)}{y - x_n} - F'(x_n) \right| < \frac\varepsilon2. $$

Putting these together, for every $y$ with $|y - x_n| < \min(\eta_n, \frac{\delta_n}{2})$ it follows \begin{align*} \left| F'(x_n) - f(x_0) \right| &\le \left| F'(x_n) - \frac{F(y) - F(x_n)}{y - x_n} \right| + \left| \frac{F(y) - F(x_n)}{y - x_n} - f(x_0) \right| \\ & \le \frac\varepsilon 2 + \left| \frac{1}{y - x_n}\int_{x_n}^y f(t) - f(x_0) dt \right| \\ & \le\frac\varepsilon 2 + \frac{1}{y - x_n}\int_{x_n}^y \underbrace{|f(t) - f(x_0)|}_{\le \varepsilon / 2} dt \\ &\le \epsilon, \end{align*} as $|t - x_0| \le |t - x_n| + |x_n - x_0| \le |y - x_n| + |x_n - x_0| < \delta$.

That is, $F'(x_n)$ converges to $f(x_0) = F'(x_0)$.

Notes:

If $f$ is differentiable at $x_0$, a similar estimation shows that $$\lim_{n\to\infty} \frac{F'(x_n) - F'(x_0)}{x_n - x_0} = f'(x_0)$$


Old Answer for reference only:

This rather an extensive comment than an answer.

Since $f$ is Riemann integrable, its set of continuity $$ C = \left\{ x\in[a,b] \mid \lim_{t\to x} f(t) = f(x) \right\} $$ has full measure, that is $|C| = b-a$. Thus, $F'$ exists on $C$ and agrees with $f$ on $C$. Trivially, $F'$ restricted onto $C$ is continuous.

That says nothing about the domain of $F'$, which might be larger than $C$, and whether $F'$ is continuous at $x_0$.

$\endgroup$
4
  • $\begingroup$ Does this mean if we take the definition of continuity as being only applicable where $F'$ is defined, that $F'$ is continuous? $\endgroup$
    – SenZen
    Commented Jan 12, 2021 at 13:00
  • $\begingroup$ I understand that, but just because there exist points in the neighbourhood of $c$ where $F'$ is not continuous, doesn't mean it's not continuous at $c$? $\endgroup$
    – SenZen
    Commented Jan 12, 2021 at 18:14
  • $\begingroup$ @user251256 Maybe you can construct counterexample? $\endgroup$
    – SenZen
    Commented Jan 12, 2021 at 18:16
  • 1
    $\begingroup$ @SenZen the statement made precise is in fact correct. I have updated the answer. $\endgroup$
    – user251257
    Commented Jan 13, 2021 at 21:35
1
$\begingroup$

As stated in the problem, $F'(x)$ would be not only continuous but differentiable at $x_0$.

The FTC allows you to conclude that for $x_0\in [a,x]$, $$ F'(x_0)=f(x_0) $$ Which means that $F'(x)$ in its region of validity has the properties $f$ has.

$\endgroup$
5
  • $\begingroup$ Please comment along with the downvote, so I can improve my answer $\endgroup$ Commented Dec 6, 2016 at 19:49
  • 1
    $\begingroup$ The problem is $F'$ need not exist in a full neighborhood of $x_0.$ $\endgroup$
    – zhw.
    Commented Dec 6, 2016 at 22:48
  • 1
    $\begingroup$ @zhw. That's not a problem with the definition of differentiability I learned. $x_0$ is an accumulation point of the domain of $F'$, and itself in that domain. That suffices to define differentiability of $F'$ at $x_0$, and indeed $F'$ is differentiable at $x_0$ under the given hypotheses, using that definition. I'm aware that some authors gratuitously define differentiability only at interior points of the domain of the function. Just saying that definitions differ. $\endgroup$ Commented Dec 13, 2016 at 21:52
  • $\begingroup$ @qbert It's a little more complicated, we only need to have $F'(x) = f(x)$ almost everywhere. $F'$ may exist at some point $x_1$ with $F'(x_1) \neq f(x_1)$. $\endgroup$ Commented Dec 13, 2016 at 21:55
  • $\begingroup$ @DanielFischer I also learned that differentiability was a point property, unlike holomorphicity in complex analysis, if this is what is meant by your response to zhw. And fair enough, I realized this after. Figured I would leave my answer up anyway (and I'm glad I did as I have learned from it). $\endgroup$ Commented Dec 13, 2016 at 22:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .