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Suppose we want to prove that the derivative of a function across an interval exists at $C$, but the derivative at $C$ cannot be found. We know the function must be continuous. Can we take the limit of derivative from the negative and positive direction of $C$ and show that if they are equal, the derivative at $C$ exists and is equal to the limit obtained? Is this a necessary and sufficient condition?

EDIT:

Sufficiency - If a function is a derivative along some interval, it does not have a removable singularity at $C$.

Necessity - There is no interval of a derivative of some function in which a jump or essential discontinuity occurs.

There are two cases in which the condition is met if this is a necessary and sufficient condition. One is where the derivative is continuous, the other is where there is a removable discontinuity in the derivative. Is the latter possible?

jg mr chapb
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  • no of course not. I'm asking about the limit of the derivative. Not the limit of $\frac{f(x+h) - f(x)}{h}$ – jg mr chapb Apr 21 '17 at 10:51
  • @gebra Are you asking about $C^1$ functions? If the limit of the derivative to exists at a point, it is then continuous at the point. If the derivative is discontinuous, then your original function was not continuously differentiable – Matt Apr 21 '17 at 10:54
  • Yes, it is sufficient (Im not sure if necessary) because the unique kind of discontinuities of the derivative of a real function are essential discontinuities. Said in other form: a derivative doesnt have jump discontinuities, hence if the lateral limits are defined then they are equal and the derivative at this point is this limit. – Masacroso Apr 21 '17 at 11:16
  • @n.m. It doesn't seem that way to me. Just to be clear on what not necessary means. Take some continuous function whose derivative approaches different values from different directions at $C$. Now you're saying there exists some function of this form such that at $C$, the derivative exists. – jg mr chapb Apr 21 '17 at 11:16
  • @Masacroso hmm, I think one of us is confused here. (Probably me) but your statement shows that it is necessary. It doesn't say anything about sufficiency – jg mr chapb Apr 21 '17 at 11:21
  • @gebra by sufficient I mean that $$\text{lateral limits of derivative exists}\implies \text{derivative exists}$$ and by necessary $$\text{lateral limits of derivative exists}\impliedby \text{derivative exists}$$ I see now that probably this last implication (necessity) is not true because a function can be differentiable at a unique point (at least for functions with domain in $\Bbb C$, not sure if this hold for functions in $\Bbb R$) so there is no laterals limits. Anyway Im not sure if this empty condition is a negation. – Masacroso Apr 21 '17 at 11:27
  • @gebra I misread the problem statement – n. m. could be an AI Apr 21 '17 at 11:33
  • A derivative cannot have jumps (Darboux's theorem). Thus if a derivative exists across an interval, and there are one-sided limits at point $x_0$, then (a) they are equal and (b) they are both equal to the value at $x_0$. – n. m. could be an AI Apr 21 '17 at 11:44
  • @Masacroso A derivative can exist and be discontinuous at every point of an interval. – n. m. could be an AI Apr 21 '17 at 11:52
  • @n.m. this says the opposite. – Masacroso Apr 21 '17 at 11:55
  • @Masacroso my mistake; I thought of a derivative that has many discontinuities (a dense set), but I see it cannot be discontinuous everywhere. – n. m. could be an AI Apr 21 '17 at 12:04
  • Anyway there are differentiable functions with discontinuous derivatives, e.g. $x^2 \sin \frac{1}{x}, x \ne 0; 0, x = 0$. – n. m. could be an AI Apr 21 '17 at 12:14
  • @n.m. yes. This is enough to see that being differentiable at a point doesnt imply that $f'$ is continuous at this point. – Masacroso Apr 21 '17 at 12:22
  • A derivative cannot have jumps but can have essential discontinuities. – n. m. could be an AI Apr 21 '17 at 12:29

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If the $\lim_{x\to c^+}f^\prime(x)=\ell$ then using the mean value theorem you have that $$\frac{f(x)-f(c)}{x-c}=f'(d_x),$$ where $c<d_x<x$ and so $d_x\to c$ as $x\to c^+$. Hence, there is $$\lim_{x\to c^+}\frac{f(x)-f(c)}{x-c}=\lim_{x\to c^+}f'(d_x)=\ell.$$ Same for the other side. So if the limits of the derivatives exist and are equal then $f$ is differentiable. The other implication is false because $f'$ could exist everywhere but not be continuous at $c$. The function $f(x)=x^2\sin{1/x}$ if $x\ne 0$ and $f(0)=0$ is differentiable everywhere but the derivative is discontinuous. Even worse you could take $f(x)=x^2$ if $x$ is rational and $f(x)=0$ otherwise. then $f$ is differentiable at $x=0$ and discontinuous at every other point.

EDIT: The original question has been modified and so my answer makes no sense anymore. Anyway the derivative of a function is Darboux continuous, so if it cannot have certain types of discontinuities. Darboux

Gio67
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  • wait, can you explain how this shows that the derivative exists at c? – jg mr chapb Apr 21 '17 at 12:56
  • also i don't really understand the quadratic thing but i'll accept that it is not a necessary condition – jg mr chapb Apr 21 '17 at 12:58
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    @Masacroso: You have to distinguish between the one-sided limits of ${f(x)-f(c)\over x-c}$ and of $f'(x)$. Note that for the function $f(x):=|x|$ the one-sided limits of $f'(x)$ exist, and are different. The intermediate theorem for derivatives does not apply here since $f$ is not differentiable at $0$. – Christian Blatter Apr 21 '17 at 15:46
  • yeah, and the method in my question predicts that $|x|$ doesn't have a derivative at 0. My question is where it actually works. I'm still trying to see how the mean value theorem can actually be used to approach this – jg mr chapb Apr 21 '17 at 17:14
  • the intermediate value theorem still applies in $[c,x]$ the hypothesis are that $f$ is continuous in $[c,x]$ and differentiable in $(c,x)$. If the limits are different, you get that there exist the left and right derivative at $c$ but they are different, which is the case for $f(x)=|x|$. – Gio67 Apr 21 '17 at 18:41
  • oh, I see. Thank you @Christian, I overlooked this situation. – Masacroso Apr 22 '17 at 04:35