Discontinuous derivative. [duplicate]












119












$begingroup$



This question already has an answer here:




  • How discontinuous can a derivative be?

    1 answer




Could someone give an example of a ‘very’ discontinuous derivative? I myself can only come up with examples where the derivative is discontinuous at only one point. I am assuming the function is real-valued and defined on a bounded interval.










share|cite|improve this question











$endgroup$



marked as duplicate by A. Pongrácz, Cesareo, Gibbs, mrtaurho, GNUSupporter 8964民主女神 地下教會 Feb 15 at 13:33


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 2




    $begingroup$
    $f(x)=|x|$ at $x=0$? Not sure what you mean by "very".
    $endgroup$
    – Ron Gordon
    Feb 1 '13 at 18:36






  • 7




    $begingroup$
    He probably wants a something discontinuous almost everywhere.
    $endgroup$
    – Git Gud
    Feb 1 '13 at 18:38






  • 14




    $begingroup$
    Related: math.stackexchange.com/questions/112067/…
    $endgroup$
    – Deven Ware
    Feb 1 '13 at 18:40






  • 7




    $begingroup$
    @RonGordon: could you elaborate what you meant by $f(x)=|x|$ at $x=0$? Sine it is continuous but not differentiable I don't quite see how it fits in this context. Thank you.
    $endgroup$
    – Leo
    Jun 19 '15 at 21:52








  • 3




    $begingroup$
    Haskell Curry's contribution had an error now edited out by Mark McClure. For a reference for Curry's statements see Andy Bruckner, Differentiation of Real Functions, AMS (1994) p. 34. Mark McClure's is an excellent and lucid prelude before diving into the deep end and tackling the material in Andy's book. The poster who decided to tell us about the Weierstrass function missed the point, although one might vaguely think that this function has a "discontinous derivative," ...so discontinuous that it doesn't even exist? (I have deleted the comments that led to the correction.)
    $endgroup$
    – B. S. Thomson
    Nov 30 '15 at 22:39


















119












$begingroup$



This question already has an answer here:




  • How discontinuous can a derivative be?

    1 answer




Could someone give an example of a ‘very’ discontinuous derivative? I myself can only come up with examples where the derivative is discontinuous at only one point. I am assuming the function is real-valued and defined on a bounded interval.










share|cite|improve this question











$endgroup$



marked as duplicate by A. Pongrácz, Cesareo, Gibbs, mrtaurho, GNUSupporter 8964民主女神 地下教會 Feb 15 at 13:33


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 2




    $begingroup$
    $f(x)=|x|$ at $x=0$? Not sure what you mean by "very".
    $endgroup$
    – Ron Gordon
    Feb 1 '13 at 18:36






  • 7




    $begingroup$
    He probably wants a something discontinuous almost everywhere.
    $endgroup$
    – Git Gud
    Feb 1 '13 at 18:38






  • 14




    $begingroup$
    Related: math.stackexchange.com/questions/112067/…
    $endgroup$
    – Deven Ware
    Feb 1 '13 at 18:40






  • 7




    $begingroup$
    @RonGordon: could you elaborate what you meant by $f(x)=|x|$ at $x=0$? Sine it is continuous but not differentiable I don't quite see how it fits in this context. Thank you.
    $endgroup$
    – Leo
    Jun 19 '15 at 21:52








  • 3




    $begingroup$
    Haskell Curry's contribution had an error now edited out by Mark McClure. For a reference for Curry's statements see Andy Bruckner, Differentiation of Real Functions, AMS (1994) p. 34. Mark McClure's is an excellent and lucid prelude before diving into the deep end and tackling the material in Andy's book. The poster who decided to tell us about the Weierstrass function missed the point, although one might vaguely think that this function has a "discontinous derivative," ...so discontinuous that it doesn't even exist? (I have deleted the comments that led to the correction.)
    $endgroup$
    – B. S. Thomson
    Nov 30 '15 at 22:39
















119












119








119


124



$begingroup$



This question already has an answer here:




  • How discontinuous can a derivative be?

    1 answer




Could someone give an example of a ‘very’ discontinuous derivative? I myself can only come up with examples where the derivative is discontinuous at only one point. I am assuming the function is real-valued and defined on a bounded interval.










share|cite|improve this question











$endgroup$





This question already has an answer here:




  • How discontinuous can a derivative be?

    1 answer




Could someone give an example of a ‘very’ discontinuous derivative? I myself can only come up with examples where the derivative is discontinuous at only one point. I am assuming the function is real-valued and defined on a bounded interval.





This question already has an answer here:




  • How discontinuous can a derivative be?

    1 answer








real-analysis derivatives examples-counterexamples






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 4 '18 at 1:35









the_fox

2,89021537




2,89021537










asked Feb 1 '13 at 18:35









user58273user58273

644363




644363




marked as duplicate by A. Pongrácz, Cesareo, Gibbs, mrtaurho, GNUSupporter 8964民主女神 地下教會 Feb 15 at 13:33


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by A. Pongrácz, Cesareo, Gibbs, mrtaurho, GNUSupporter 8964民主女神 地下教會 Feb 15 at 13:33


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 2




    $begingroup$
    $f(x)=|x|$ at $x=0$? Not sure what you mean by "very".
    $endgroup$
    – Ron Gordon
    Feb 1 '13 at 18:36






  • 7




    $begingroup$
    He probably wants a something discontinuous almost everywhere.
    $endgroup$
    – Git Gud
    Feb 1 '13 at 18:38






  • 14




    $begingroup$
    Related: math.stackexchange.com/questions/112067/…
    $endgroup$
    – Deven Ware
    Feb 1 '13 at 18:40






  • 7




    $begingroup$
    @RonGordon: could you elaborate what you meant by $f(x)=|x|$ at $x=0$? Sine it is continuous but not differentiable I don't quite see how it fits in this context. Thank you.
    $endgroup$
    – Leo
    Jun 19 '15 at 21:52








  • 3




    $begingroup$
    Haskell Curry's contribution had an error now edited out by Mark McClure. For a reference for Curry's statements see Andy Bruckner, Differentiation of Real Functions, AMS (1994) p. 34. Mark McClure's is an excellent and lucid prelude before diving into the deep end and tackling the material in Andy's book. The poster who decided to tell us about the Weierstrass function missed the point, although one might vaguely think that this function has a "discontinous derivative," ...so discontinuous that it doesn't even exist? (I have deleted the comments that led to the correction.)
    $endgroup$
    – B. S. Thomson
    Nov 30 '15 at 22:39
















  • 2




    $begingroup$
    $f(x)=|x|$ at $x=0$? Not sure what you mean by "very".
    $endgroup$
    – Ron Gordon
    Feb 1 '13 at 18:36






  • 7




    $begingroup$
    He probably wants a something discontinuous almost everywhere.
    $endgroup$
    – Git Gud
    Feb 1 '13 at 18:38






  • 14




    $begingroup$
    Related: math.stackexchange.com/questions/112067/…
    $endgroup$
    – Deven Ware
    Feb 1 '13 at 18:40






  • 7




    $begingroup$
    @RonGordon: could you elaborate what you meant by $f(x)=|x|$ at $x=0$? Sine it is continuous but not differentiable I don't quite see how it fits in this context. Thank you.
    $endgroup$
    – Leo
    Jun 19 '15 at 21:52








  • 3




    $begingroup$
    Haskell Curry's contribution had an error now edited out by Mark McClure. For a reference for Curry's statements see Andy Bruckner, Differentiation of Real Functions, AMS (1994) p. 34. Mark McClure's is an excellent and lucid prelude before diving into the deep end and tackling the material in Andy's book. The poster who decided to tell us about the Weierstrass function missed the point, although one might vaguely think that this function has a "discontinous derivative," ...so discontinuous that it doesn't even exist? (I have deleted the comments that led to the correction.)
    $endgroup$
    – B. S. Thomson
    Nov 30 '15 at 22:39










2




2




$begingroup$
$f(x)=|x|$ at $x=0$? Not sure what you mean by "very".
$endgroup$
– Ron Gordon
Feb 1 '13 at 18:36




$begingroup$
$f(x)=|x|$ at $x=0$? Not sure what you mean by "very".
$endgroup$
– Ron Gordon
Feb 1 '13 at 18:36




7




7




$begingroup$
He probably wants a something discontinuous almost everywhere.
$endgroup$
– Git Gud
Feb 1 '13 at 18:38




$begingroup$
He probably wants a something discontinuous almost everywhere.
$endgroup$
– Git Gud
Feb 1 '13 at 18:38




14




14




$begingroup$
Related: math.stackexchange.com/questions/112067/…
$endgroup$
– Deven Ware
Feb 1 '13 at 18:40




$begingroup$
Related: math.stackexchange.com/questions/112067/…
$endgroup$
– Deven Ware
Feb 1 '13 at 18:40




7




7




$begingroup$
@RonGordon: could you elaborate what you meant by $f(x)=|x|$ at $x=0$? Sine it is continuous but not differentiable I don't quite see how it fits in this context. Thank you.
$endgroup$
– Leo
Jun 19 '15 at 21:52






$begingroup$
@RonGordon: could you elaborate what you meant by $f(x)=|x|$ at $x=0$? Sine it is continuous but not differentiable I don't quite see how it fits in this context. Thank you.
$endgroup$
– Leo
Jun 19 '15 at 21:52






3




3




$begingroup$
Haskell Curry's contribution had an error now edited out by Mark McClure. For a reference for Curry's statements see Andy Bruckner, Differentiation of Real Functions, AMS (1994) p. 34. Mark McClure's is an excellent and lucid prelude before diving into the deep end and tackling the material in Andy's book. The poster who decided to tell us about the Weierstrass function missed the point, although one might vaguely think that this function has a "discontinous derivative," ...so discontinuous that it doesn't even exist? (I have deleted the comments that led to the correction.)
$endgroup$
– B. S. Thomson
Nov 30 '15 at 22:39






$begingroup$
Haskell Curry's contribution had an error now edited out by Mark McClure. For a reference for Curry's statements see Andy Bruckner, Differentiation of Real Functions, AMS (1994) p. 34. Mark McClure's is an excellent and lucid prelude before diving into the deep end and tackling the material in Andy's book. The poster who decided to tell us about the Weierstrass function missed the point, although one might vaguely think that this function has a "discontinous derivative," ...so discontinuous that it doesn't even exist? (I have deleted the comments that led to the correction.)
$endgroup$
– B. S. Thomson
Nov 30 '15 at 22:39












2 Answers
2






active

oldest

votes


















167












$begingroup$

Haskell's answer does a great job of outlining conditions that a derivative $f'$ must satisfy, which then limits us in our search for an example. From there we see the key question: can we provide a concrete example of an everywhere differentiable function whose derivative is discontinuous on a dense, full-measure set of $mathbb R$? Here's a closer look at the Volterra-type functions referred to in Haskell's answer, together with a little indication as to how it might be extended.



Basic example



The basic example of a differentiable function with discontinuous derivative is
$$
f(x) = begin{cases}
x^2 sin(1/x) &mbox{if } x neq 0 \
0 & mbox{if } x=0.
end{cases}
$$
The differentiation rules show that this function is differentiable away from the origin and the difference quotient can be used to show that it is differentiable at the origin with value $f'(0)=0$. A graph is illuminating as well as it shows how $pm x^2$ forms an envelope for the function forcing differentiablity.



enter image description here



The derivative of $f$ is
$$
f'(x) = begin{cases}
2 x sin left(frac{1}{x}right)-cos left(frac{1}{x}right)&mbox{if } x neq 0 \
0 & mbox{if } x=0,
end{cases}
$$
which is discontinuous at $x=0$. Its graph looks something like so



enter image description here



Two points



The next step is to modify this example to obtain a function that is everywhere differentiable with a derivative that is continuous on all of $mathbb R$, except for two points. To this end, consider
$$
f(x) = begin{cases}
x^2 (1-x)^2 sin left(frac{1}{pi x (1-x)}right)&mbox{if } 0<x<1 \
0 & mbox{else}.
end{cases}
$$
The graph of $f$ and its derivative look like so.



enter image description here



A cantor set of discontinuties



Now that we have a way to construct a differentiable function whose derivative is discontinuous exactly at the endpoints of an interval, it should be clear how to construct a differentiable function whose derivative is discontinous on a Cantor set constructed in the interval. For $ninmathbb N$ and $m=1,2,ldots,2^n$, let $I_{m,n}$ denote one of the $2^n$ intervals removed during the $n^{th}$ stage of construction of the Cantor set. Then let $f_{m,n}$ be scaled to have support $I_{m,n}$ and to have maximum value $4^{-n}$. The function
$$F(x) = sum_{n=0}^{infty} sum_{m=1}^{2^n} f_{m,n}(x)$$
will be everywhere differentiable but its derivative will be discontinuous on the given Cantor set. Assuming we do this with Cantors standard ternary set, we get a picture that looks something like so:



enter image description here



Of course, there's really a sequence of functions here and care needs to be taken to show that the limit is truly differentiable. Let
$$F_N(x) = sum_{n=1}^{N} sum_{m=1}^{2^n} f_{m,n}(x).$$
The standard theorem then states that, as long as $F_N$ converges and $F_N'$ converges uniformly, then the limit of $F_N(x)$ will be differentiable. This is guaranteed by the choice of $4^{-n}$ as the max for $f_{m,n}$.



Increasing the measure



Again, the last example refers to the standard Cantor ternary set but there's no reason this can't be done with any Cantor set. In particular, it can be done with a so-called fat Cantor set, which can have positive measure arbitrarily close to the measure of the interval containing it. We immediately produce an everywhere differentiable function whose derivative is discontinuous on a nowhere dense set of positive measure. (Of course, care must again be taken to scale the heights of the functions go to zero quickly enough to guarantee differentiability.)



Finally, we can fill the holes of the removed intervals with more Cantor sets (and their corresponding functions) in such a way that the union of all of them is of full measure. This allows us to construct an everywhere differentiable function with derivative that is discontinuous on the union of those Cantor sets, which is a set of full measure.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    It's fairly clear that the derivative will be discontinuous at the removed endpoints. How do you ensure it is actually discontinuous on the whole Cantor set?
    $endgroup$
    – dfeuer
    Oct 14 '13 at 22:46










  • $begingroup$
    @dfeuer Those endpoints actually are elements of the cantor set. For example $2/3$ is in every intersection of cantor pieces. So it suffices to show for all $N$ that the function has a discontinuous derivative at those such end points if and only it is discontinuous at all $c$ in the cantor set.
    $endgroup$
    – MadcowD
    Oct 27 '15 at 2:20












  • $begingroup$
    In your first two examples, you give functions that are differentiable, but that the limit of the derivative, in one of the branches, does not exist, and hence cannot be continuous. Is there an example where the limit does exist, and it's still discontinuous?
    $endgroup$
    – An old man in the sea.
    Aug 31 '16 at 16:19










  • $begingroup$
    @Anoldmaninthesea. Sounds like a good question for the main site. I believe the answer is no.
    $endgroup$
    – Mark McClure
    Aug 31 '16 at 16:22






  • 1




    $begingroup$
    Mark, I've just asked that. ;)
    $endgroup$
    – An old man in the sea.
    Aug 31 '16 at 16:39



















74












$begingroup$

I guess that you are looking for a continuous function $ f: mathbb{R} to mathbb{R} $ such that $ f $ is differentiable everywhere but $ f' $ is ‘as discontinuous as possible’.



We have the following theorem in real analysis.




Theorem 1 If $ f: mathbb{R} to mathbb{R} $ is differentiable everywhere, then the set of points in $ mathbb{R} $ where $ f' $ is continuous is non-empty. More precisely, the set of all such points is a dense $ G_{delta} $-subset of $ mathbb{R} $.




Note: A $ G_{delta} $-subset of $ mathbb{R} $ is just the intersection of a countable collection of open subsets of $ mathbb{R} $.



The proof of Theorem 1 is an application of the Baire Category Theorem, and it can be found in Munkres’ Topology. By this theorem, it is therefore impossible to find an $ f: mathbb{R} to mathbb{R} $ whose derivative exists but is discontinuous everywhere.



There is another theorem that provides a necessary and sufficient condition for a set $E$ to be the set of discontinuities of some derivative.




Theorem 2 A set $E$ is the set of discontinuities of some derivative if and only if $E$ is a meagre $ F_{sigma} $-subset of $ mathbb{R} $.




Note: An $ F_{sigma} $-subset of $ mathbb{R} $ is just the union of a countable collection of closed subsets of $ mathbb{R} $.



Let me end off with a non-trivial example to add to yours. Volterra’s Function is differentiable everywhere, but its derivative is discontinuous on a set of positive measure, not just at a single point.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hi Mark. Yes this is Theorem 2.2, p. 34 of Andy's book (although he states it for the continuity points which is equivalent). Mr. Curry originally misstated the theorem as a characterization of derivatives, rather than a characterization of the discontinuity set of derivatives. Since the topic is about the discontinuities of a derivative this page is a rather complete picture of the situation. Too bad the Weierstrass function intrudes though.
    $endgroup$
    – B. S. Thomson
    Nov 30 '15 at 4:34


















2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









167












$begingroup$

Haskell's answer does a great job of outlining conditions that a derivative $f'$ must satisfy, which then limits us in our search for an example. From there we see the key question: can we provide a concrete example of an everywhere differentiable function whose derivative is discontinuous on a dense, full-measure set of $mathbb R$? Here's a closer look at the Volterra-type functions referred to in Haskell's answer, together with a little indication as to how it might be extended.



Basic example



The basic example of a differentiable function with discontinuous derivative is
$$
f(x) = begin{cases}
x^2 sin(1/x) &mbox{if } x neq 0 \
0 & mbox{if } x=0.
end{cases}
$$
The differentiation rules show that this function is differentiable away from the origin and the difference quotient can be used to show that it is differentiable at the origin with value $f'(0)=0$. A graph is illuminating as well as it shows how $pm x^2$ forms an envelope for the function forcing differentiablity.



enter image description here



The derivative of $f$ is
$$
f'(x) = begin{cases}
2 x sin left(frac{1}{x}right)-cos left(frac{1}{x}right)&mbox{if } x neq 0 \
0 & mbox{if } x=0,
end{cases}
$$
which is discontinuous at $x=0$. Its graph looks something like so



enter image description here



Two points



The next step is to modify this example to obtain a function that is everywhere differentiable with a derivative that is continuous on all of $mathbb R$, except for two points. To this end, consider
$$
f(x) = begin{cases}
x^2 (1-x)^2 sin left(frac{1}{pi x (1-x)}right)&mbox{if } 0<x<1 \
0 & mbox{else}.
end{cases}
$$
The graph of $f$ and its derivative look like so.



enter image description here



A cantor set of discontinuties



Now that we have a way to construct a differentiable function whose derivative is discontinuous exactly at the endpoints of an interval, it should be clear how to construct a differentiable function whose derivative is discontinous on a Cantor set constructed in the interval. For $ninmathbb N$ and $m=1,2,ldots,2^n$, let $I_{m,n}$ denote one of the $2^n$ intervals removed during the $n^{th}$ stage of construction of the Cantor set. Then let $f_{m,n}$ be scaled to have support $I_{m,n}$ and to have maximum value $4^{-n}$. The function
$$F(x) = sum_{n=0}^{infty} sum_{m=1}^{2^n} f_{m,n}(x)$$
will be everywhere differentiable but its derivative will be discontinuous on the given Cantor set. Assuming we do this with Cantors standard ternary set, we get a picture that looks something like so:



enter image description here



Of course, there's really a sequence of functions here and care needs to be taken to show that the limit is truly differentiable. Let
$$F_N(x) = sum_{n=1}^{N} sum_{m=1}^{2^n} f_{m,n}(x).$$
The standard theorem then states that, as long as $F_N$ converges and $F_N'$ converges uniformly, then the limit of $F_N(x)$ will be differentiable. This is guaranteed by the choice of $4^{-n}$ as the max for $f_{m,n}$.



Increasing the measure



Again, the last example refers to the standard Cantor ternary set but there's no reason this can't be done with any Cantor set. In particular, it can be done with a so-called fat Cantor set, which can have positive measure arbitrarily close to the measure of the interval containing it. We immediately produce an everywhere differentiable function whose derivative is discontinuous on a nowhere dense set of positive measure. (Of course, care must again be taken to scale the heights of the functions go to zero quickly enough to guarantee differentiability.)



Finally, we can fill the holes of the removed intervals with more Cantor sets (and their corresponding functions) in such a way that the union of all of them is of full measure. This allows us to construct an everywhere differentiable function with derivative that is discontinuous on the union of those Cantor sets, which is a set of full measure.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    It's fairly clear that the derivative will be discontinuous at the removed endpoints. How do you ensure it is actually discontinuous on the whole Cantor set?
    $endgroup$
    – dfeuer
    Oct 14 '13 at 22:46










  • $begingroup$
    @dfeuer Those endpoints actually are elements of the cantor set. For example $2/3$ is in every intersection of cantor pieces. So it suffices to show for all $N$ that the function has a discontinuous derivative at those such end points if and only it is discontinuous at all $c$ in the cantor set.
    $endgroup$
    – MadcowD
    Oct 27 '15 at 2:20












  • $begingroup$
    In your first two examples, you give functions that are differentiable, but that the limit of the derivative, in one of the branches, does not exist, and hence cannot be continuous. Is there an example where the limit does exist, and it's still discontinuous?
    $endgroup$
    – An old man in the sea.
    Aug 31 '16 at 16:19










  • $begingroup$
    @Anoldmaninthesea. Sounds like a good question for the main site. I believe the answer is no.
    $endgroup$
    – Mark McClure
    Aug 31 '16 at 16:22






  • 1




    $begingroup$
    Mark, I've just asked that. ;)
    $endgroup$
    – An old man in the sea.
    Aug 31 '16 at 16:39
















167












$begingroup$

Haskell's answer does a great job of outlining conditions that a derivative $f'$ must satisfy, which then limits us in our search for an example. From there we see the key question: can we provide a concrete example of an everywhere differentiable function whose derivative is discontinuous on a dense, full-measure set of $mathbb R$? Here's a closer look at the Volterra-type functions referred to in Haskell's answer, together with a little indication as to how it might be extended.



Basic example



The basic example of a differentiable function with discontinuous derivative is
$$
f(x) = begin{cases}
x^2 sin(1/x) &mbox{if } x neq 0 \
0 & mbox{if } x=0.
end{cases}
$$
The differentiation rules show that this function is differentiable away from the origin and the difference quotient can be used to show that it is differentiable at the origin with value $f'(0)=0$. A graph is illuminating as well as it shows how $pm x^2$ forms an envelope for the function forcing differentiablity.



enter image description here



The derivative of $f$ is
$$
f'(x) = begin{cases}
2 x sin left(frac{1}{x}right)-cos left(frac{1}{x}right)&mbox{if } x neq 0 \
0 & mbox{if } x=0,
end{cases}
$$
which is discontinuous at $x=0$. Its graph looks something like so



enter image description here



Two points



The next step is to modify this example to obtain a function that is everywhere differentiable with a derivative that is continuous on all of $mathbb R$, except for two points. To this end, consider
$$
f(x) = begin{cases}
x^2 (1-x)^2 sin left(frac{1}{pi x (1-x)}right)&mbox{if } 0<x<1 \
0 & mbox{else}.
end{cases}
$$
The graph of $f$ and its derivative look like so.



enter image description here



A cantor set of discontinuties



Now that we have a way to construct a differentiable function whose derivative is discontinuous exactly at the endpoints of an interval, it should be clear how to construct a differentiable function whose derivative is discontinous on a Cantor set constructed in the interval. For $ninmathbb N$ and $m=1,2,ldots,2^n$, let $I_{m,n}$ denote one of the $2^n$ intervals removed during the $n^{th}$ stage of construction of the Cantor set. Then let $f_{m,n}$ be scaled to have support $I_{m,n}$ and to have maximum value $4^{-n}$. The function
$$F(x) = sum_{n=0}^{infty} sum_{m=1}^{2^n} f_{m,n}(x)$$
will be everywhere differentiable but its derivative will be discontinuous on the given Cantor set. Assuming we do this with Cantors standard ternary set, we get a picture that looks something like so:



enter image description here



Of course, there's really a sequence of functions here and care needs to be taken to show that the limit is truly differentiable. Let
$$F_N(x) = sum_{n=1}^{N} sum_{m=1}^{2^n} f_{m,n}(x).$$
The standard theorem then states that, as long as $F_N$ converges and $F_N'$ converges uniformly, then the limit of $F_N(x)$ will be differentiable. This is guaranteed by the choice of $4^{-n}$ as the max for $f_{m,n}$.



Increasing the measure



Again, the last example refers to the standard Cantor ternary set but there's no reason this can't be done with any Cantor set. In particular, it can be done with a so-called fat Cantor set, which can have positive measure arbitrarily close to the measure of the interval containing it. We immediately produce an everywhere differentiable function whose derivative is discontinuous on a nowhere dense set of positive measure. (Of course, care must again be taken to scale the heights of the functions go to zero quickly enough to guarantee differentiability.)



Finally, we can fill the holes of the removed intervals with more Cantor sets (and their corresponding functions) in such a way that the union of all of them is of full measure. This allows us to construct an everywhere differentiable function with derivative that is discontinuous on the union of those Cantor sets, which is a set of full measure.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    It's fairly clear that the derivative will be discontinuous at the removed endpoints. How do you ensure it is actually discontinuous on the whole Cantor set?
    $endgroup$
    – dfeuer
    Oct 14 '13 at 22:46










  • $begingroup$
    @dfeuer Those endpoints actually are elements of the cantor set. For example $2/3$ is in every intersection of cantor pieces. So it suffices to show for all $N$ that the function has a discontinuous derivative at those such end points if and only it is discontinuous at all $c$ in the cantor set.
    $endgroup$
    – MadcowD
    Oct 27 '15 at 2:20












  • $begingroup$
    In your first two examples, you give functions that are differentiable, but that the limit of the derivative, in one of the branches, does not exist, and hence cannot be continuous. Is there an example where the limit does exist, and it's still discontinuous?
    $endgroup$
    – An old man in the sea.
    Aug 31 '16 at 16:19










  • $begingroup$
    @Anoldmaninthesea. Sounds like a good question for the main site. I believe the answer is no.
    $endgroup$
    – Mark McClure
    Aug 31 '16 at 16:22






  • 1




    $begingroup$
    Mark, I've just asked that. ;)
    $endgroup$
    – An old man in the sea.
    Aug 31 '16 at 16:39














167












167








167





$begingroup$

Haskell's answer does a great job of outlining conditions that a derivative $f'$ must satisfy, which then limits us in our search for an example. From there we see the key question: can we provide a concrete example of an everywhere differentiable function whose derivative is discontinuous on a dense, full-measure set of $mathbb R$? Here's a closer look at the Volterra-type functions referred to in Haskell's answer, together with a little indication as to how it might be extended.



Basic example



The basic example of a differentiable function with discontinuous derivative is
$$
f(x) = begin{cases}
x^2 sin(1/x) &mbox{if } x neq 0 \
0 & mbox{if } x=0.
end{cases}
$$
The differentiation rules show that this function is differentiable away from the origin and the difference quotient can be used to show that it is differentiable at the origin with value $f'(0)=0$. A graph is illuminating as well as it shows how $pm x^2$ forms an envelope for the function forcing differentiablity.



enter image description here



The derivative of $f$ is
$$
f'(x) = begin{cases}
2 x sin left(frac{1}{x}right)-cos left(frac{1}{x}right)&mbox{if } x neq 0 \
0 & mbox{if } x=0,
end{cases}
$$
which is discontinuous at $x=0$. Its graph looks something like so



enter image description here



Two points



The next step is to modify this example to obtain a function that is everywhere differentiable with a derivative that is continuous on all of $mathbb R$, except for two points. To this end, consider
$$
f(x) = begin{cases}
x^2 (1-x)^2 sin left(frac{1}{pi x (1-x)}right)&mbox{if } 0<x<1 \
0 & mbox{else}.
end{cases}
$$
The graph of $f$ and its derivative look like so.



enter image description here



A cantor set of discontinuties



Now that we have a way to construct a differentiable function whose derivative is discontinuous exactly at the endpoints of an interval, it should be clear how to construct a differentiable function whose derivative is discontinous on a Cantor set constructed in the interval. For $ninmathbb N$ and $m=1,2,ldots,2^n$, let $I_{m,n}$ denote one of the $2^n$ intervals removed during the $n^{th}$ stage of construction of the Cantor set. Then let $f_{m,n}$ be scaled to have support $I_{m,n}$ and to have maximum value $4^{-n}$. The function
$$F(x) = sum_{n=0}^{infty} sum_{m=1}^{2^n} f_{m,n}(x)$$
will be everywhere differentiable but its derivative will be discontinuous on the given Cantor set. Assuming we do this with Cantors standard ternary set, we get a picture that looks something like so:



enter image description here



Of course, there's really a sequence of functions here and care needs to be taken to show that the limit is truly differentiable. Let
$$F_N(x) = sum_{n=1}^{N} sum_{m=1}^{2^n} f_{m,n}(x).$$
The standard theorem then states that, as long as $F_N$ converges and $F_N'$ converges uniformly, then the limit of $F_N(x)$ will be differentiable. This is guaranteed by the choice of $4^{-n}$ as the max for $f_{m,n}$.



Increasing the measure



Again, the last example refers to the standard Cantor ternary set but there's no reason this can't be done with any Cantor set. In particular, it can be done with a so-called fat Cantor set, which can have positive measure arbitrarily close to the measure of the interval containing it. We immediately produce an everywhere differentiable function whose derivative is discontinuous on a nowhere dense set of positive measure. (Of course, care must again be taken to scale the heights of the functions go to zero quickly enough to guarantee differentiability.)



Finally, we can fill the holes of the removed intervals with more Cantor sets (and their corresponding functions) in such a way that the union of all of them is of full measure. This allows us to construct an everywhere differentiable function with derivative that is discontinuous on the union of those Cantor sets, which is a set of full measure.






share|cite|improve this answer











$endgroup$



Haskell's answer does a great job of outlining conditions that a derivative $f'$ must satisfy, which then limits us in our search for an example. From there we see the key question: can we provide a concrete example of an everywhere differentiable function whose derivative is discontinuous on a dense, full-measure set of $mathbb R$? Here's a closer look at the Volterra-type functions referred to in Haskell's answer, together with a little indication as to how it might be extended.



Basic example



The basic example of a differentiable function with discontinuous derivative is
$$
f(x) = begin{cases}
x^2 sin(1/x) &mbox{if } x neq 0 \
0 & mbox{if } x=0.
end{cases}
$$
The differentiation rules show that this function is differentiable away from the origin and the difference quotient can be used to show that it is differentiable at the origin with value $f'(0)=0$. A graph is illuminating as well as it shows how $pm x^2$ forms an envelope for the function forcing differentiablity.



enter image description here



The derivative of $f$ is
$$
f'(x) = begin{cases}
2 x sin left(frac{1}{x}right)-cos left(frac{1}{x}right)&mbox{if } x neq 0 \
0 & mbox{if } x=0,
end{cases}
$$
which is discontinuous at $x=0$. Its graph looks something like so



enter image description here



Two points



The next step is to modify this example to obtain a function that is everywhere differentiable with a derivative that is continuous on all of $mathbb R$, except for two points. To this end, consider
$$
f(x) = begin{cases}
x^2 (1-x)^2 sin left(frac{1}{pi x (1-x)}right)&mbox{if } 0<x<1 \
0 & mbox{else}.
end{cases}
$$
The graph of $f$ and its derivative look like so.



enter image description here



A cantor set of discontinuties



Now that we have a way to construct a differentiable function whose derivative is discontinuous exactly at the endpoints of an interval, it should be clear how to construct a differentiable function whose derivative is discontinous on a Cantor set constructed in the interval. For $ninmathbb N$ and $m=1,2,ldots,2^n$, let $I_{m,n}$ denote one of the $2^n$ intervals removed during the $n^{th}$ stage of construction of the Cantor set. Then let $f_{m,n}$ be scaled to have support $I_{m,n}$ and to have maximum value $4^{-n}$. The function
$$F(x) = sum_{n=0}^{infty} sum_{m=1}^{2^n} f_{m,n}(x)$$
will be everywhere differentiable but its derivative will be discontinuous on the given Cantor set. Assuming we do this with Cantors standard ternary set, we get a picture that looks something like so:



enter image description here



Of course, there's really a sequence of functions here and care needs to be taken to show that the limit is truly differentiable. Let
$$F_N(x) = sum_{n=1}^{N} sum_{m=1}^{2^n} f_{m,n}(x).$$
The standard theorem then states that, as long as $F_N$ converges and $F_N'$ converges uniformly, then the limit of $F_N(x)$ will be differentiable. This is guaranteed by the choice of $4^{-n}$ as the max for $f_{m,n}$.



Increasing the measure



Again, the last example refers to the standard Cantor ternary set but there's no reason this can't be done with any Cantor set. In particular, it can be done with a so-called fat Cantor set, which can have positive measure arbitrarily close to the measure of the interval containing it. We immediately produce an everywhere differentiable function whose derivative is discontinuous on a nowhere dense set of positive measure. (Of course, care must again be taken to scale the heights of the functions go to zero quickly enough to guarantee differentiability.)



Finally, we can fill the holes of the removed intervals with more Cantor sets (and their corresponding functions) in such a way that the union of all of them is of full measure. This allows us to construct an everywhere differentiable function with derivative that is discontinuous on the union of those Cantor sets, which is a set of full measure.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited May 4 '18 at 0:40









ziggurism

11.2k12971




11.2k12971










answered Jun 18 '13 at 2:16









Mark McClureMark McClure

23.6k34471




23.6k34471








  • 2




    $begingroup$
    It's fairly clear that the derivative will be discontinuous at the removed endpoints. How do you ensure it is actually discontinuous on the whole Cantor set?
    $endgroup$
    – dfeuer
    Oct 14 '13 at 22:46










  • $begingroup$
    @dfeuer Those endpoints actually are elements of the cantor set. For example $2/3$ is in every intersection of cantor pieces. So it suffices to show for all $N$ that the function has a discontinuous derivative at those such end points if and only it is discontinuous at all $c$ in the cantor set.
    $endgroup$
    – MadcowD
    Oct 27 '15 at 2:20












  • $begingroup$
    In your first two examples, you give functions that are differentiable, but that the limit of the derivative, in one of the branches, does not exist, and hence cannot be continuous. Is there an example where the limit does exist, and it's still discontinuous?
    $endgroup$
    – An old man in the sea.
    Aug 31 '16 at 16:19










  • $begingroup$
    @Anoldmaninthesea. Sounds like a good question for the main site. I believe the answer is no.
    $endgroup$
    – Mark McClure
    Aug 31 '16 at 16:22






  • 1




    $begingroup$
    Mark, I've just asked that. ;)
    $endgroup$
    – An old man in the sea.
    Aug 31 '16 at 16:39














  • 2




    $begingroup$
    It's fairly clear that the derivative will be discontinuous at the removed endpoints. How do you ensure it is actually discontinuous on the whole Cantor set?
    $endgroup$
    – dfeuer
    Oct 14 '13 at 22:46










  • $begingroup$
    @dfeuer Those endpoints actually are elements of the cantor set. For example $2/3$ is in every intersection of cantor pieces. So it suffices to show for all $N$ that the function has a discontinuous derivative at those such end points if and only it is discontinuous at all $c$ in the cantor set.
    $endgroup$
    – MadcowD
    Oct 27 '15 at 2:20












  • $begingroup$
    In your first two examples, you give functions that are differentiable, but that the limit of the derivative, in one of the branches, does not exist, and hence cannot be continuous. Is there an example where the limit does exist, and it's still discontinuous?
    $endgroup$
    – An old man in the sea.
    Aug 31 '16 at 16:19










  • $begingroup$
    @Anoldmaninthesea. Sounds like a good question for the main site. I believe the answer is no.
    $endgroup$
    – Mark McClure
    Aug 31 '16 at 16:22






  • 1




    $begingroup$
    Mark, I've just asked that. ;)
    $endgroup$
    – An old man in the sea.
    Aug 31 '16 at 16:39








2




2




$begingroup$
It's fairly clear that the derivative will be discontinuous at the removed endpoints. How do you ensure it is actually discontinuous on the whole Cantor set?
$endgroup$
– dfeuer
Oct 14 '13 at 22:46




$begingroup$
It's fairly clear that the derivative will be discontinuous at the removed endpoints. How do you ensure it is actually discontinuous on the whole Cantor set?
$endgroup$
– dfeuer
Oct 14 '13 at 22:46












$begingroup$
@dfeuer Those endpoints actually are elements of the cantor set. For example $2/3$ is in every intersection of cantor pieces. So it suffices to show for all $N$ that the function has a discontinuous derivative at those such end points if and only it is discontinuous at all $c$ in the cantor set.
$endgroup$
– MadcowD
Oct 27 '15 at 2:20






$begingroup$
@dfeuer Those endpoints actually are elements of the cantor set. For example $2/3$ is in every intersection of cantor pieces. So it suffices to show for all $N$ that the function has a discontinuous derivative at those such end points if and only it is discontinuous at all $c$ in the cantor set.
$endgroup$
– MadcowD
Oct 27 '15 at 2:20














$begingroup$
In your first two examples, you give functions that are differentiable, but that the limit of the derivative, in one of the branches, does not exist, and hence cannot be continuous. Is there an example where the limit does exist, and it's still discontinuous?
$endgroup$
– An old man in the sea.
Aug 31 '16 at 16:19




$begingroup$
In your first two examples, you give functions that are differentiable, but that the limit of the derivative, in one of the branches, does not exist, and hence cannot be continuous. Is there an example where the limit does exist, and it's still discontinuous?
$endgroup$
– An old man in the sea.
Aug 31 '16 at 16:19












$begingroup$
@Anoldmaninthesea. Sounds like a good question for the main site. I believe the answer is no.
$endgroup$
– Mark McClure
Aug 31 '16 at 16:22




$begingroup$
@Anoldmaninthesea. Sounds like a good question for the main site. I believe the answer is no.
$endgroup$
– Mark McClure
Aug 31 '16 at 16:22




1




1




$begingroup$
Mark, I've just asked that. ;)
$endgroup$
– An old man in the sea.
Aug 31 '16 at 16:39




$begingroup$
Mark, I've just asked that. ;)
$endgroup$
– An old man in the sea.
Aug 31 '16 at 16:39











74












$begingroup$

I guess that you are looking for a continuous function $ f: mathbb{R} to mathbb{R} $ such that $ f $ is differentiable everywhere but $ f' $ is ‘as discontinuous as possible’.



We have the following theorem in real analysis.




Theorem 1 If $ f: mathbb{R} to mathbb{R} $ is differentiable everywhere, then the set of points in $ mathbb{R} $ where $ f' $ is continuous is non-empty. More precisely, the set of all such points is a dense $ G_{delta} $-subset of $ mathbb{R} $.




Note: A $ G_{delta} $-subset of $ mathbb{R} $ is just the intersection of a countable collection of open subsets of $ mathbb{R} $.



The proof of Theorem 1 is an application of the Baire Category Theorem, and it can be found in Munkres’ Topology. By this theorem, it is therefore impossible to find an $ f: mathbb{R} to mathbb{R} $ whose derivative exists but is discontinuous everywhere.



There is another theorem that provides a necessary and sufficient condition for a set $E$ to be the set of discontinuities of some derivative.




Theorem 2 A set $E$ is the set of discontinuities of some derivative if and only if $E$ is a meagre $ F_{sigma} $-subset of $ mathbb{R} $.




Note: An $ F_{sigma} $-subset of $ mathbb{R} $ is just the union of a countable collection of closed subsets of $ mathbb{R} $.



Let me end off with a non-trivial example to add to yours. Volterra’s Function is differentiable everywhere, but its derivative is discontinuous on a set of positive measure, not just at a single point.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hi Mark. Yes this is Theorem 2.2, p. 34 of Andy's book (although he states it for the continuity points which is equivalent). Mr. Curry originally misstated the theorem as a characterization of derivatives, rather than a characterization of the discontinuity set of derivatives. Since the topic is about the discontinuities of a derivative this page is a rather complete picture of the situation. Too bad the Weierstrass function intrudes though.
    $endgroup$
    – B. S. Thomson
    Nov 30 '15 at 4:34
















74












$begingroup$

I guess that you are looking for a continuous function $ f: mathbb{R} to mathbb{R} $ such that $ f $ is differentiable everywhere but $ f' $ is ‘as discontinuous as possible’.



We have the following theorem in real analysis.




Theorem 1 If $ f: mathbb{R} to mathbb{R} $ is differentiable everywhere, then the set of points in $ mathbb{R} $ where $ f' $ is continuous is non-empty. More precisely, the set of all such points is a dense $ G_{delta} $-subset of $ mathbb{R} $.




Note: A $ G_{delta} $-subset of $ mathbb{R} $ is just the intersection of a countable collection of open subsets of $ mathbb{R} $.



The proof of Theorem 1 is an application of the Baire Category Theorem, and it can be found in Munkres’ Topology. By this theorem, it is therefore impossible to find an $ f: mathbb{R} to mathbb{R} $ whose derivative exists but is discontinuous everywhere.



There is another theorem that provides a necessary and sufficient condition for a set $E$ to be the set of discontinuities of some derivative.




Theorem 2 A set $E$ is the set of discontinuities of some derivative if and only if $E$ is a meagre $ F_{sigma} $-subset of $ mathbb{R} $.




Note: An $ F_{sigma} $-subset of $ mathbb{R} $ is just the union of a countable collection of closed subsets of $ mathbb{R} $.



Let me end off with a non-trivial example to add to yours. Volterra’s Function is differentiable everywhere, but its derivative is discontinuous on a set of positive measure, not just at a single point.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hi Mark. Yes this is Theorem 2.2, p. 34 of Andy's book (although he states it for the continuity points which is equivalent). Mr. Curry originally misstated the theorem as a characterization of derivatives, rather than a characterization of the discontinuity set of derivatives. Since the topic is about the discontinuities of a derivative this page is a rather complete picture of the situation. Too bad the Weierstrass function intrudes though.
    $endgroup$
    – B. S. Thomson
    Nov 30 '15 at 4:34














74












74








74





$begingroup$

I guess that you are looking for a continuous function $ f: mathbb{R} to mathbb{R} $ such that $ f $ is differentiable everywhere but $ f' $ is ‘as discontinuous as possible’.



We have the following theorem in real analysis.




Theorem 1 If $ f: mathbb{R} to mathbb{R} $ is differentiable everywhere, then the set of points in $ mathbb{R} $ where $ f' $ is continuous is non-empty. More precisely, the set of all such points is a dense $ G_{delta} $-subset of $ mathbb{R} $.




Note: A $ G_{delta} $-subset of $ mathbb{R} $ is just the intersection of a countable collection of open subsets of $ mathbb{R} $.



The proof of Theorem 1 is an application of the Baire Category Theorem, and it can be found in Munkres’ Topology. By this theorem, it is therefore impossible to find an $ f: mathbb{R} to mathbb{R} $ whose derivative exists but is discontinuous everywhere.



There is another theorem that provides a necessary and sufficient condition for a set $E$ to be the set of discontinuities of some derivative.




Theorem 2 A set $E$ is the set of discontinuities of some derivative if and only if $E$ is a meagre $ F_{sigma} $-subset of $ mathbb{R} $.




Note: An $ F_{sigma} $-subset of $ mathbb{R} $ is just the union of a countable collection of closed subsets of $ mathbb{R} $.



Let me end off with a non-trivial example to add to yours. Volterra’s Function is differentiable everywhere, but its derivative is discontinuous on a set of positive measure, not just at a single point.






share|cite|improve this answer











$endgroup$



I guess that you are looking for a continuous function $ f: mathbb{R} to mathbb{R} $ such that $ f $ is differentiable everywhere but $ f' $ is ‘as discontinuous as possible’.



We have the following theorem in real analysis.




Theorem 1 If $ f: mathbb{R} to mathbb{R} $ is differentiable everywhere, then the set of points in $ mathbb{R} $ where $ f' $ is continuous is non-empty. More precisely, the set of all such points is a dense $ G_{delta} $-subset of $ mathbb{R} $.




Note: A $ G_{delta} $-subset of $ mathbb{R} $ is just the intersection of a countable collection of open subsets of $ mathbb{R} $.



The proof of Theorem 1 is an application of the Baire Category Theorem, and it can be found in Munkres’ Topology. By this theorem, it is therefore impossible to find an $ f: mathbb{R} to mathbb{R} $ whose derivative exists but is discontinuous everywhere.



There is another theorem that provides a necessary and sufficient condition for a set $E$ to be the set of discontinuities of some derivative.




Theorem 2 A set $E$ is the set of discontinuities of some derivative if and only if $E$ is a meagre $ F_{sigma} $-subset of $ mathbb{R} $.




Note: An $ F_{sigma} $-subset of $ mathbb{R} $ is just the union of a countable collection of closed subsets of $ mathbb{R} $.



Let me end off with a non-trivial example to add to yours. Volterra’s Function is differentiable everywhere, but its derivative is discontinuous on a set of positive measure, not just at a single point.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 30 '15 at 3:39









Mark McClure

23.6k34471




23.6k34471










answered Feb 1 '13 at 21:08









Haskell CurryHaskell Curry

15.2k3886




15.2k3886












  • $begingroup$
    Hi Mark. Yes this is Theorem 2.2, p. 34 of Andy's book (although he states it for the continuity points which is equivalent). Mr. Curry originally misstated the theorem as a characterization of derivatives, rather than a characterization of the discontinuity set of derivatives. Since the topic is about the discontinuities of a derivative this page is a rather complete picture of the situation. Too bad the Weierstrass function intrudes though.
    $endgroup$
    – B. S. Thomson
    Nov 30 '15 at 4:34


















  • $begingroup$
    Hi Mark. Yes this is Theorem 2.2, p. 34 of Andy's book (although he states it for the continuity points which is equivalent). Mr. Curry originally misstated the theorem as a characterization of derivatives, rather than a characterization of the discontinuity set of derivatives. Since the topic is about the discontinuities of a derivative this page is a rather complete picture of the situation. Too bad the Weierstrass function intrudes though.
    $endgroup$
    – B. S. Thomson
    Nov 30 '15 at 4:34
















$begingroup$
Hi Mark. Yes this is Theorem 2.2, p. 34 of Andy's book (although he states it for the continuity points which is equivalent). Mr. Curry originally misstated the theorem as a characterization of derivatives, rather than a characterization of the discontinuity set of derivatives. Since the topic is about the discontinuities of a derivative this page is a rather complete picture of the situation. Too bad the Weierstrass function intrudes though.
$endgroup$
– B. S. Thomson
Nov 30 '15 at 4:34




$begingroup$
Hi Mark. Yes this is Theorem 2.2, p. 34 of Andy's book (although he states it for the continuity points which is equivalent). Mr. Curry originally misstated the theorem as a characterization of derivatives, rather than a characterization of the discontinuity set of derivatives. Since the topic is about the discontinuities of a derivative this page is a rather complete picture of the situation. Too bad the Weierstrass function intrudes though.
$endgroup$
– B. S. Thomson
Nov 30 '15 at 4:34



Popular posts from this blog

Human spaceflight

Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

張江高科駅