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Does this hold:

If the derivative of a function $f$ is continuous at point $c$ (i.e. $f'$ is continuous at $c$), then the function $f$ is continuous in an open interval around $c$?

My understanding is that the answer is yes, with the following arguments:

  1. in order for $f'$ to be continuous at $c$, the derivative needs to exist for all points in a non-empty neighborhood of $c$ (i.e. $\exists \delta >0. \forall x \in (c-\delta, c+\delta). f'(x)$ exists).
  2. because existence of derivative implies continuity, we can conclude that $f$ is also continuous for all points in the neighborhood of $c$ ($\forall x \in (c-\delta, c+\delta). f'(x)$ exists $ \implies \lim_{t \to x} f(t) = f(x) $).

That could be a follow up to my previous question.

Thanks!

S11n
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    Some things were stated very badly here; for example "$f'(c)$ is continuous" does not say $f'$ is continuous at $c$. I corrected a few things like that - now everything above is correct; the answer is yes, for the simple reason you give. I'm mystified by the two "no" Answers. – David C. Ullrich Jul 10 '21 at 10:56
  • Thank you for the correction! I also corrected the text in the parenthesis of the first sentence. – S11n Jul 10 '21 at 12:31
  • If you are confident about that, why not just promote your comment to an answer, so we can close this question? – S11n Jul 10 '21 at 12:36
  • @David C. Ullrich. It seems to me that the discussion triggered by the OP's question comes down to this: we define the derivative of a function $f(x)$ as the limit of $a$ to zero of $[f(x+a) - f(x)]/a$. Now is it the sufficient to calculate this limit for positive $a$ only, or should the (same !) limit also be reached when considering negative $a$ ? – M. Wind Jul 10 '21 at 17:51
  • @M. Wind: as far as I understand, the derivative is defined (page 160 of Apostol's Calculus Vol.1) as the limit you mentioned, and the limit has to have the same value for both negative and positive $a$ in order for a derivative to exist at point $x$. – S11n Jul 12 '21 at 11:03
  • @S11n: That definition seems sensible to me. Still I have some doubts. Consider the function $f(x) = (1-x^2)*\sqrt{(1-x^2)}$ on the real numbers. The domain is the closed interval (-1,1). It would seem somewhat odd that the derivative $f'(x) = -3x \sqrt{(1-x^2)}$ exists on this interval, but not in the end points $x = -1$ and $x = +1$. – M. Wind Jul 12 '21 at 18:12
  • @M. Wind: I think you made a couple of mistakes in your comment. But I do not see anything odd about what you probably wanted to say (i.e. that if a function is defined only on an open interval, the function is not differentiable on the end points, for example you can take 1/x function, and point 0). – S11n Jul 13 '21 at 14:53
  • @S11n: The most important thing to remember is: it is all just a matter of definition. There is nothing particularly "deep" or "profound" or "meaningful" on whether the derivative formally exists (or not) in the end points of an interval. This has zero practical meaning in applied mathematics, physics, engineering. – M. Wind Jul 13 '21 at 15:45

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No, why would continuity require existence in a neighborhood? If that were the case, functions defined in closed intervals could never be continuous at the ends of the interval. We can imagine a function that's some kind of nowhere continuous function to the left of $c$, but something continuously differentiable to the right. If we connect the two in a nice way, it will be differentiable at $c$ such that the derivative is continuous. Like this:

Let $D:\mathbb R\to\mathbb R$ be the Dirichlet function, which is 1 on the rationals and 0 on the irrationals. It's a nowhere differentiable function. Now define

$$f:\mathbb R\to\mathbb R,~x\mapsto\begin{cases}x^2D(x)&x<0\\x^2&x\geq0\end{cases}$$

You can prove that it's not differentiable on $(-\infty,0)$, but is differentiable on $[0,\infty)$. We also have that $f'(x)=2x$, which makes $f'$ continuous, particularly at $0$. But $f$ is not continuous for any $x<0$, so it's not continuous in any neighborhood of $0$.

Vercassivelaunos
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  • No, your $f'$ is certainly not continuous at the origin. – David C. Ullrich Jul 10 '21 at 10:54
  • @DavidC.Ullrich The function $[0,\infty)\to\mathbb R,~x\mapsto 2x$ is certainly continuous at $0$. Or do you claim that this is not what $f'$ looks like? – Vercassivelaunos Jul 10 '21 at 11:41
  • I don't know exactly what we mean by "looks like", but no, $f'$ is not simply $2x$. – David C. Ullrich Jul 10 '21 at 12:17
  • I don't think that function is even continuous at 0 from the left, because it is not defined. – S11n Jul 10 '21 at 12:33
  • @DavidC.Ullrich Instead of vaguely pointing out that what I say is wrong, could you elaborate how you come to the conclusion that it's wrong? The function $f$ is clearly not differentiable on $(-\infty,0)$ since it's discontinuous there, clearly differentiable with derivative $2x$ on $(0,\infty)$ since it's equal to $x\mapsto x^2$ on an open domain, and it's also differentiable in $0$ with derivative $0$ since $\left\vert\frac{f(x)-f(0)}{x-0}\right\vert=\left\vert\frac{f(x)}{x}\right\vert\leq\left\vert\frac{x^2}{x}\right\vert=\vert x\vert$, which converges to $0$ as $x\to0$. – Vercassivelaunos Jul 10 '21 at 14:36
  • @S11n Continuous doesn't mean continuous from the left and right. Those are only equivalent if there is a left and right in the first place. If there is nothing to the left of a point in the domain, then the function need not be continuous from the left to be continuous. – Vercassivelaunos Jul 10 '21 at 14:39
  • @Vercassivelaunos: I had Apostol's definition of continuity in mind, which he wrote on page 130 of his Calculus, Vol.1. – S11n Jul 12 '21 at 11:05
  • I don't have that book at hand, so I can't compare. But if your definition of continuity requires existence in a neighborhood in general, regardless of differentiability (which would be a very restrictive definition: functions on closed intervals could never be continuous), then yes, it's still required if the function happens to be a derivative function. – Vercassivelaunos Jul 12 '21 at 12:32