Is there a continuous monotone function that fails to be differentiable on a dense subset of $mathbb{R}$?












0














My textbook is asking to construct such a function, but this thread seems to say that's not possible. What am I missing here? I want to construct a function that's not differentiable for $x=c$, for any $c$ in the interval.



I am referring to part B below: this problem.










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  • 2




    Nowhere differentiable is not the same as being non-differentiable on a dense set.
    – Xander Henderson
    Nov 29 at 14:00






  • 1




    Maybe one can do something similar to the Devil's staircase?
    – Arthur
    Nov 29 at 14:14










  • @Arthur The Cantor function is only non-differentiable on the Cantor set, which is not dense in $mathbb{R}$.
    – Xander Henderson
    Nov 29 at 14:20










  • @XanderHenderson Which is why I didn't post it as an answer. I don't even know if the approach is possible to adapt to a dense set.
    – Arthur
    Nov 29 at 14:23












  • @bob Can you tell from which book you found this excercise?
    – Shubham
    Nov 29 at 14:58
















0














My textbook is asking to construct such a function, but this thread seems to say that's not possible. What am I missing here? I want to construct a function that's not differentiable for $x=c$, for any $c$ in the interval.



I am referring to part B below: this problem.










share|cite|improve this question




















  • 2




    Nowhere differentiable is not the same as being non-differentiable on a dense set.
    – Xander Henderson
    Nov 29 at 14:00






  • 1




    Maybe one can do something similar to the Devil's staircase?
    – Arthur
    Nov 29 at 14:14










  • @Arthur The Cantor function is only non-differentiable on the Cantor set, which is not dense in $mathbb{R}$.
    – Xander Henderson
    Nov 29 at 14:20










  • @XanderHenderson Which is why I didn't post it as an answer. I don't even know if the approach is possible to adapt to a dense set.
    – Arthur
    Nov 29 at 14:23












  • @bob Can you tell from which book you found this excercise?
    – Shubham
    Nov 29 at 14:58














0












0








0


1





My textbook is asking to construct such a function, but this thread seems to say that's not possible. What am I missing here? I want to construct a function that's not differentiable for $x=c$, for any $c$ in the interval.



I am referring to part B below: this problem.










share|cite|improve this question















My textbook is asking to construct such a function, but this thread seems to say that's not possible. What am I missing here? I want to construct a function that's not differentiable for $x=c$, for any $c$ in the interval.



I am referring to part B below: this problem.







calculus real-analysis integration derivatives






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 at 14:02

























asked Nov 29 at 13:59









bob

1089




1089








  • 2




    Nowhere differentiable is not the same as being non-differentiable on a dense set.
    – Xander Henderson
    Nov 29 at 14:00






  • 1




    Maybe one can do something similar to the Devil's staircase?
    – Arthur
    Nov 29 at 14:14










  • @Arthur The Cantor function is only non-differentiable on the Cantor set, which is not dense in $mathbb{R}$.
    – Xander Henderson
    Nov 29 at 14:20










  • @XanderHenderson Which is why I didn't post it as an answer. I don't even know if the approach is possible to adapt to a dense set.
    – Arthur
    Nov 29 at 14:23












  • @bob Can you tell from which book you found this excercise?
    – Shubham
    Nov 29 at 14:58














  • 2




    Nowhere differentiable is not the same as being non-differentiable on a dense set.
    – Xander Henderson
    Nov 29 at 14:00






  • 1




    Maybe one can do something similar to the Devil's staircase?
    – Arthur
    Nov 29 at 14:14










  • @Arthur The Cantor function is only non-differentiable on the Cantor set, which is not dense in $mathbb{R}$.
    – Xander Henderson
    Nov 29 at 14:20










  • @XanderHenderson Which is why I didn't post it as an answer. I don't even know if the approach is possible to adapt to a dense set.
    – Arthur
    Nov 29 at 14:23












  • @bob Can you tell from which book you found this excercise?
    – Shubham
    Nov 29 at 14:58








2




2




Nowhere differentiable is not the same as being non-differentiable on a dense set.
– Xander Henderson
Nov 29 at 14:00




Nowhere differentiable is not the same as being non-differentiable on a dense set.
– Xander Henderson
Nov 29 at 14:00




1




1




Maybe one can do something similar to the Devil's staircase?
– Arthur
Nov 29 at 14:14




Maybe one can do something similar to the Devil's staircase?
– Arthur
Nov 29 at 14:14












@Arthur The Cantor function is only non-differentiable on the Cantor set, which is not dense in $mathbb{R}$.
– Xander Henderson
Nov 29 at 14:20




@Arthur The Cantor function is only non-differentiable on the Cantor set, which is not dense in $mathbb{R}$.
– Xander Henderson
Nov 29 at 14:20












@XanderHenderson Which is why I didn't post it as an answer. I don't even know if the approach is possible to adapt to a dense set.
– Arthur
Nov 29 at 14:23






@XanderHenderson Which is why I didn't post it as an answer. I don't even know if the approach is possible to adapt to a dense set.
– Arthur
Nov 29 at 14:23














@bob Can you tell from which book you found this excercise?
– Shubham
Nov 29 at 14:58




@bob Can you tell from which book you found this excercise?
– Shubham
Nov 29 at 14:58










1 Answer
1






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6














The question to which you have linked discusses the impossibility of constructing a function that is continuous, monotone, and nowhere differentiable. However, you are not being asked to construct such a function. You are being asked to construct a continuous, monotone function which is non-differentiable on a dense subset of $mathbb{R}$.



My goto example of such a function is a function which has a little "jump" at each rational number. To build such a function, start by defining
$$ f(x) = begin{cases} 1 & text{if $x ge 0$, and} \ 0 & text{otherwise.} end{cases} $$
This function has a jump discontinuity at zero.



Next, observe that the rational numbers are countable, which implies that there is a bijection $q : mathbb{N} to mathbb{Q}$ which provides an enumeration of $mathbb{Q}$. Let $q_n := q(n)$ for each $ninmathbb{N}$. At each $n$, we are going to define a function $f_n$ which has a jump discontinuity at $q_n$. In order to ensure that the function we get doesn't "blow up", we are going to build the jumps in such a way that they get small relatively quickly. Specifically, define
$$ f_n(x) := frac{f(x-q_n)}{2^n}. $$
If you prefer something more explicit, note that
$$ f_n(x) = begin{cases} 2^{-n} & text{if $x ge q_n$, and} \ 0 & text{otherwise.} end{cases} $$



Because we have chosen the jumps in such a way that they get small fast, we can sum all of these functions and get something that converges. So define
$$ g(x) = sum_{n=1}^{infty} f_n(x). $$
It is not too hard to show that this function converges pointwise and has a discontinuity at every rational number.



Finally, as per the hint in your homework, define
$$ G(x) = int_{a}^x g(x),mathrm{d}x. $$
As per the quoted text, $G$ is not differentiable at any rational number.



Note that I have left out quite a few details which you should fill in. You should make sure that you understand why $mathbb{Q}$ is countable and dense in $mathbb{R}$ (unless these are facts that you are allowed to take for granted).
You should also check that $g$ really is discontinuous at each rational, and you should make sure that you understand why the series defining $g$ actually converges. Finally, it might be a good idea to build $G$ a little more carefully so that it is more artfully restricted to the interval of interest, i.e. the interval $[a,b]$.



EDIT: As per the comment left below by Dave L. Renfro, these kinds of results can be pushed farther to obtain, for example, Lipschitz continuous monotone functions which are non-differentiable on a dense subset of $mathbb{R}$. The relevant discussion is contained in this answer to another question on MSE.






share|cite|improve this answer























  • Regarding how far such examples can be pushed that also satisfy an additional restriction (namely, being Lipschitz continuous), see my answer to Monotone Function, Derivative Limit Bounded, Differentiable - 2.
    – Dave L. Renfro
    Nov 29 at 15:09












  • @DaveL.Renfro Nice!
    – Xander Henderson
    Nov 29 at 18:28











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6














The question to which you have linked discusses the impossibility of constructing a function that is continuous, monotone, and nowhere differentiable. However, you are not being asked to construct such a function. You are being asked to construct a continuous, monotone function which is non-differentiable on a dense subset of $mathbb{R}$.



My goto example of such a function is a function which has a little "jump" at each rational number. To build such a function, start by defining
$$ f(x) = begin{cases} 1 & text{if $x ge 0$, and} \ 0 & text{otherwise.} end{cases} $$
This function has a jump discontinuity at zero.



Next, observe that the rational numbers are countable, which implies that there is a bijection $q : mathbb{N} to mathbb{Q}$ which provides an enumeration of $mathbb{Q}$. Let $q_n := q(n)$ for each $ninmathbb{N}$. At each $n$, we are going to define a function $f_n$ which has a jump discontinuity at $q_n$. In order to ensure that the function we get doesn't "blow up", we are going to build the jumps in such a way that they get small relatively quickly. Specifically, define
$$ f_n(x) := frac{f(x-q_n)}{2^n}. $$
If you prefer something more explicit, note that
$$ f_n(x) = begin{cases} 2^{-n} & text{if $x ge q_n$, and} \ 0 & text{otherwise.} end{cases} $$



Because we have chosen the jumps in such a way that they get small fast, we can sum all of these functions and get something that converges. So define
$$ g(x) = sum_{n=1}^{infty} f_n(x). $$
It is not too hard to show that this function converges pointwise and has a discontinuity at every rational number.



Finally, as per the hint in your homework, define
$$ G(x) = int_{a}^x g(x),mathrm{d}x. $$
As per the quoted text, $G$ is not differentiable at any rational number.



Note that I have left out quite a few details which you should fill in. You should make sure that you understand why $mathbb{Q}$ is countable and dense in $mathbb{R}$ (unless these are facts that you are allowed to take for granted).
You should also check that $g$ really is discontinuous at each rational, and you should make sure that you understand why the series defining $g$ actually converges. Finally, it might be a good idea to build $G$ a little more carefully so that it is more artfully restricted to the interval of interest, i.e. the interval $[a,b]$.



EDIT: As per the comment left below by Dave L. Renfro, these kinds of results can be pushed farther to obtain, for example, Lipschitz continuous monotone functions which are non-differentiable on a dense subset of $mathbb{R}$. The relevant discussion is contained in this answer to another question on MSE.






share|cite|improve this answer























  • Regarding how far such examples can be pushed that also satisfy an additional restriction (namely, being Lipschitz continuous), see my answer to Monotone Function, Derivative Limit Bounded, Differentiable - 2.
    – Dave L. Renfro
    Nov 29 at 15:09












  • @DaveL.Renfro Nice!
    – Xander Henderson
    Nov 29 at 18:28
















6














The question to which you have linked discusses the impossibility of constructing a function that is continuous, monotone, and nowhere differentiable. However, you are not being asked to construct such a function. You are being asked to construct a continuous, monotone function which is non-differentiable on a dense subset of $mathbb{R}$.



My goto example of such a function is a function which has a little "jump" at each rational number. To build such a function, start by defining
$$ f(x) = begin{cases} 1 & text{if $x ge 0$, and} \ 0 & text{otherwise.} end{cases} $$
This function has a jump discontinuity at zero.



Next, observe that the rational numbers are countable, which implies that there is a bijection $q : mathbb{N} to mathbb{Q}$ which provides an enumeration of $mathbb{Q}$. Let $q_n := q(n)$ for each $ninmathbb{N}$. At each $n$, we are going to define a function $f_n$ which has a jump discontinuity at $q_n$. In order to ensure that the function we get doesn't "blow up", we are going to build the jumps in such a way that they get small relatively quickly. Specifically, define
$$ f_n(x) := frac{f(x-q_n)}{2^n}. $$
If you prefer something more explicit, note that
$$ f_n(x) = begin{cases} 2^{-n} & text{if $x ge q_n$, and} \ 0 & text{otherwise.} end{cases} $$



Because we have chosen the jumps in such a way that they get small fast, we can sum all of these functions and get something that converges. So define
$$ g(x) = sum_{n=1}^{infty} f_n(x). $$
It is not too hard to show that this function converges pointwise and has a discontinuity at every rational number.



Finally, as per the hint in your homework, define
$$ G(x) = int_{a}^x g(x),mathrm{d}x. $$
As per the quoted text, $G$ is not differentiable at any rational number.



Note that I have left out quite a few details which you should fill in. You should make sure that you understand why $mathbb{Q}$ is countable and dense in $mathbb{R}$ (unless these are facts that you are allowed to take for granted).
You should also check that $g$ really is discontinuous at each rational, and you should make sure that you understand why the series defining $g$ actually converges. Finally, it might be a good idea to build $G$ a little more carefully so that it is more artfully restricted to the interval of interest, i.e. the interval $[a,b]$.



EDIT: As per the comment left below by Dave L. Renfro, these kinds of results can be pushed farther to obtain, for example, Lipschitz continuous monotone functions which are non-differentiable on a dense subset of $mathbb{R}$. The relevant discussion is contained in this answer to another question on MSE.






share|cite|improve this answer























  • Regarding how far such examples can be pushed that also satisfy an additional restriction (namely, being Lipschitz continuous), see my answer to Monotone Function, Derivative Limit Bounded, Differentiable - 2.
    – Dave L. Renfro
    Nov 29 at 15:09












  • @DaveL.Renfro Nice!
    – Xander Henderson
    Nov 29 at 18:28














6












6








6






The question to which you have linked discusses the impossibility of constructing a function that is continuous, monotone, and nowhere differentiable. However, you are not being asked to construct such a function. You are being asked to construct a continuous, monotone function which is non-differentiable on a dense subset of $mathbb{R}$.



My goto example of such a function is a function which has a little "jump" at each rational number. To build such a function, start by defining
$$ f(x) = begin{cases} 1 & text{if $x ge 0$, and} \ 0 & text{otherwise.} end{cases} $$
This function has a jump discontinuity at zero.



Next, observe that the rational numbers are countable, which implies that there is a bijection $q : mathbb{N} to mathbb{Q}$ which provides an enumeration of $mathbb{Q}$. Let $q_n := q(n)$ for each $ninmathbb{N}$. At each $n$, we are going to define a function $f_n$ which has a jump discontinuity at $q_n$. In order to ensure that the function we get doesn't "blow up", we are going to build the jumps in such a way that they get small relatively quickly. Specifically, define
$$ f_n(x) := frac{f(x-q_n)}{2^n}. $$
If you prefer something more explicit, note that
$$ f_n(x) = begin{cases} 2^{-n} & text{if $x ge q_n$, and} \ 0 & text{otherwise.} end{cases} $$



Because we have chosen the jumps in such a way that they get small fast, we can sum all of these functions and get something that converges. So define
$$ g(x) = sum_{n=1}^{infty} f_n(x). $$
It is not too hard to show that this function converges pointwise and has a discontinuity at every rational number.



Finally, as per the hint in your homework, define
$$ G(x) = int_{a}^x g(x),mathrm{d}x. $$
As per the quoted text, $G$ is not differentiable at any rational number.



Note that I have left out quite a few details which you should fill in. You should make sure that you understand why $mathbb{Q}$ is countable and dense in $mathbb{R}$ (unless these are facts that you are allowed to take for granted).
You should also check that $g$ really is discontinuous at each rational, and you should make sure that you understand why the series defining $g$ actually converges. Finally, it might be a good idea to build $G$ a little more carefully so that it is more artfully restricted to the interval of interest, i.e. the interval $[a,b]$.



EDIT: As per the comment left below by Dave L. Renfro, these kinds of results can be pushed farther to obtain, for example, Lipschitz continuous monotone functions which are non-differentiable on a dense subset of $mathbb{R}$. The relevant discussion is contained in this answer to another question on MSE.






share|cite|improve this answer














The question to which you have linked discusses the impossibility of constructing a function that is continuous, monotone, and nowhere differentiable. However, you are not being asked to construct such a function. You are being asked to construct a continuous, monotone function which is non-differentiable on a dense subset of $mathbb{R}$.



My goto example of such a function is a function which has a little "jump" at each rational number. To build such a function, start by defining
$$ f(x) = begin{cases} 1 & text{if $x ge 0$, and} \ 0 & text{otherwise.} end{cases} $$
This function has a jump discontinuity at zero.



Next, observe that the rational numbers are countable, which implies that there is a bijection $q : mathbb{N} to mathbb{Q}$ which provides an enumeration of $mathbb{Q}$. Let $q_n := q(n)$ for each $ninmathbb{N}$. At each $n$, we are going to define a function $f_n$ which has a jump discontinuity at $q_n$. In order to ensure that the function we get doesn't "blow up", we are going to build the jumps in such a way that they get small relatively quickly. Specifically, define
$$ f_n(x) := frac{f(x-q_n)}{2^n}. $$
If you prefer something more explicit, note that
$$ f_n(x) = begin{cases} 2^{-n} & text{if $x ge q_n$, and} \ 0 & text{otherwise.} end{cases} $$



Because we have chosen the jumps in such a way that they get small fast, we can sum all of these functions and get something that converges. So define
$$ g(x) = sum_{n=1}^{infty} f_n(x). $$
It is not too hard to show that this function converges pointwise and has a discontinuity at every rational number.



Finally, as per the hint in your homework, define
$$ G(x) = int_{a}^x g(x),mathrm{d}x. $$
As per the quoted text, $G$ is not differentiable at any rational number.



Note that I have left out quite a few details which you should fill in. You should make sure that you understand why $mathbb{Q}$ is countable and dense in $mathbb{R}$ (unless these are facts that you are allowed to take for granted).
You should also check that $g$ really is discontinuous at each rational, and you should make sure that you understand why the series defining $g$ actually converges. Finally, it might be a good idea to build $G$ a little more carefully so that it is more artfully restricted to the interval of interest, i.e. the interval $[a,b]$.



EDIT: As per the comment left below by Dave L. Renfro, these kinds of results can be pushed farther to obtain, for example, Lipschitz continuous monotone functions which are non-differentiable on a dense subset of $mathbb{R}$. The relevant discussion is contained in this answer to another question on MSE.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 at 18:30

























answered Nov 29 at 14:17









Xander Henderson

14.1k103554




14.1k103554












  • Regarding how far such examples can be pushed that also satisfy an additional restriction (namely, being Lipschitz continuous), see my answer to Monotone Function, Derivative Limit Bounded, Differentiable - 2.
    – Dave L. Renfro
    Nov 29 at 15:09












  • @DaveL.Renfro Nice!
    – Xander Henderson
    Nov 29 at 18:28


















  • Regarding how far such examples can be pushed that also satisfy an additional restriction (namely, being Lipschitz continuous), see my answer to Monotone Function, Derivative Limit Bounded, Differentiable - 2.
    – Dave L. Renfro
    Nov 29 at 15:09












  • @DaveL.Renfro Nice!
    – Xander Henderson
    Nov 29 at 18:28
















Regarding how far such examples can be pushed that also satisfy an additional restriction (namely, being Lipschitz continuous), see my answer to Monotone Function, Derivative Limit Bounded, Differentiable - 2.
– Dave L. Renfro
Nov 29 at 15:09






Regarding how far such examples can be pushed that also satisfy an additional restriction (namely, being Lipschitz continuous), see my answer to Monotone Function, Derivative Limit Bounded, Differentiable - 2.
– Dave L. Renfro
Nov 29 at 15:09














@DaveL.Renfro Nice!
– Xander Henderson
Nov 29 at 18:28




@DaveL.Renfro Nice!
– Xander Henderson
Nov 29 at 18:28


















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