Is there a continuous monotone function that fails to be differentiable on a dense subset of $mathbb{R}$?
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: .
calculus real-analysis integration derivatives
|
show 1 more comment
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: .
calculus real-analysis integration derivatives
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
|
show 1 more comment
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: .
calculus real-analysis integration derivatives
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: .
calculus real-analysis integration derivatives
calculus real-analysis integration derivatives
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
|
show 1 more comment
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
|
show 1 more comment
1 Answer
1
active
oldest
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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.
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
add a comment |
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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.
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
add a comment |
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.
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
add a comment |
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.
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.
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
add a comment |
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
add a comment |
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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