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A continuous, nowhere differentiable but invertible function?
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I am aware of a few example of continuous, nowhere differentiable functions. The most famous is perhaps the Weierstrass functions
$$W(t)=sum_k^infty a^kcosleft(b^k tright)$$
but there are other examples, like the van der Waerden functions, or the Faber functions. Most of these "look like" some variation of:
(Weierstrass functions from Wolfram)
Specifically, they are clearly not invertible. Since these functions are generally self-similar at many scales, this non-invertability would seem to hold essentially everywhere.
I'm wondering if it's possible to construct such a function which is invertible. Intuitively, maybe this would be "jittery" in the same way as the Weierstrass function, but if it had a slope which always increased, it would be invertible. Or perhaps there is at least an example in which the function is invertible over some segment of the range.
analysis continuity differential-topology
edited Jul 16 at 22:18
Mutantoe
52039
asked Jul 16 at 17:31
levitopher
1,1191125
 |Â
show 4 more comments
up vote
79
down vote
favorite
I am aware of a few example of continuous, nowhere differentiable functions. The most famous is perhaps the Weierstrass functions
$$W(t)=sum_k^infty a^kcosleft(b^k tright)$$
but there are other examples, like the van der Waerden functions, or the Faber functions. Most of these "look like" some variation of:
(Weierstrass functions from Wolfram)
Specifically, they are clearly not invertible. Since these functions are generally self-similar at many scales, this non-invertability would seem to hold essentially everywhere.
I'm wondering if it's possible to construct such a function which is invertible. Intuitively, maybe this would be "jittery" in the same way as the Weierstrass function, but if it had a slope which always increased, it would be invertible. Or perhaps there is at least an example in which the function is invertible over some segment of the range.
analysis continuity differential-topology
edited Jul 16 at 22:18
Mutantoe
52039
asked Jul 16 at 17:31
levitopher
1,1191125
6
Guys, high school level amateur here, why can I not differentiate the individual terms to t and sum them to get the answer as an infinite sum over k of a^k.b^k.sin(b^k.t) ?
– HarryH
Jul 16 at 19:58
14
@HarryH: You can't always differentiate a series (a sum with infinitely many terms) term by term.
– Hans Lundmark
Jul 16 at 20:12
1
Looking at the graph, why can't you just use W(x)+cx, with c>>1 they should become invertible?
– Aganju
Jul 17 at 0:57
16
@Aganju basically, you would need to choose $c=infty$, because the function drops down arbitrary steeply locally.
– leftaroundabout
Jul 17 at 5:27
2
You asked a great question - it's a great question because it has an answer which is both interesting and important.
– jwg
Jul 19 at 9:39
 |Â
show 4 more comments
up vote
79
down vote
favorite
up vote
79
down vote
favorite
I am aware of a few example of continuous, nowhere differentiable functions. The most famous is perhaps the Weierstrass functions
$$W(t)=sum_k^infty a^kcosleft(b^k tright)$$
but there are other examples, like the van der Waerden functions, or the Faber functions. Most of these "look like" some variation of:
(Weierstrass functions from Wolfram)
Specifically, they are clearly not invertible. Since these functions are generally self-similar at many scales, this non-invertability would seem to hold essentially everywhere.
I'm wondering if it's possible to construct such a function which is invertible. Intuitively, maybe this would be "jittery" in the same way as the Weierstrass function, but if it had a slope which always increased, it would be invertible. Or perhaps there is at least an example in which the function is invertible over some segment of the range.
analysis continuity differential-topology
edited Jul 16 at 22:18
Mutantoe
52039
asked Jul 16 at 17:31
levitopher
1,1191125
I am aware of a few example of continuous, nowhere differentiable functions. The most famous is perhaps the Weierstrass functions
$$W(t)=sum_k^infty a^kcosleft(b^k tright)$$
but there are other examples, like the van der Waerden functions, or the Faber functions. Most of these "look like" some variation of:
(Weierstrass functions from Wolfram)
Specifically, they are clearly not invertible. Since these functions are generally self-similar at many scales, this non-invertability would seem to hold essentially everywhere.
I'm wondering if it's possible to construct such a function which is invertible. Intuitively, maybe this would be "jittery" in the same way as the Weierstrass function, but if it had a slope which always increased, it would be invertible. Or perhaps there is at least an example in which the function is invertible over some segment of the range.
analysis continuity differential-topology
edited Jul 16 at 22:18
Mutantoe
52039
asked Jul 16 at 17:31
levitopher
1,1191125
edited Jul 16 at 22:18
Mutantoe
52039
edited Jul 16 at 22:18
Mutantoe
52039
edited Jul 16 at 22:18
Mutantoe
52039
52039
asked Jul 16 at 17:31
levitopher
1,1191125
asked Jul 16 at 17:31
levitopher
1,1191125
asked Jul 16 at 17:31
levitopher
1,1191125
1,1191125
6
Guys, high school level amateur here, why can I not differentiate the individual terms to t and sum them to get the answer as an infinite sum over k of a^k.b^k.sin(b^k.t) ?
– HarryH
Jul 16 at 19:58
14
@HarryH: You can't always differentiate a series (a sum with infinitely many terms) term by term.
– Hans Lundmark
Jul 16 at 20:12
1
Looking at the graph, why can't you just use W(x)+cx, with c>>1 they should become invertible?
– Aganju
Jul 17 at 0:57
16
@Aganju basically, you would need to choose $c=infty$, because the function drops down arbitrary steeply locally.
– leftaroundabout
Jul 17 at 5:27
2
You asked a great question - it's a great question because it has an answer which is both interesting and important.
– jwg
Jul 19 at 9:39
 |Â
show 4 more comments
6
Guys, high school level amateur here, why can I not differentiate the individual terms to t and sum them to get the answer as an infinite sum over k of a^k.b^k.sin(b^k.t) ?
– HarryH
Jul 16 at 19:58
14
@HarryH: You can't always differentiate a series (a sum with infinitely many terms) term by term.
– Hans Lundmark
Jul 16 at 20:12
1
Looking at the graph, why can't you just use W(x)+cx, with c>>1 they should become invertible?
– Aganju
Jul 17 at 0:57
16
@Aganju basically, you would need to choose $c=infty$, because the function drops down arbitrary steeply locally.
– leftaroundabout
Jul 17 at 5:27
2
You asked a great question - it's a great question because it has an answer which is both interesting and important.
– jwg
Jul 19 at 9:39
6
6
Guys, high school level amateur here, why can I not differentiate the individual terms to t and sum them to get the answer as an infinite sum over k of a^k.b^k.sin(b^k.t) ?
– HarryH
Jul 16 at 19:58
Guys, high school level amateur here, why can I not differentiate the individual terms to t and sum them to get the answer as an infinite sum over k of a^k.b^k.sin(b^k.t) ?
– HarryH
Jul 16 at 19:58
14
14
@HarryH: You can't always differentiate a series (a sum with infinitely many terms) term by term.
– Hans Lundmark
Jul 16 at 20:12
@HarryH: You can't always differentiate a series (a sum with infinitely many terms) term by term.
– Hans Lundmark
Jul 16 at 20:12
1
1
Looking at the graph, why can't you just use W(x)+cx, with c>>1 they should become invertible?
– Aganju
Jul 17 at 0:57
Looking at the graph, why can't you just use W(x)+cx, with c>>1 they should become invertible?
– Aganju
Jul 17 at 0:57
16
16
@Aganju basically, you would need to choose $c=infty$, because the function drops down arbitrary steeply locally.
– leftaroundabout
Jul 17 at 5:27
@Aganju basically, you would need to choose $c=infty$, because the function drops down arbitrary steeply locally.
– leftaroundabout
Jul 17 at 5:27
2
2
You asked a great question - it's a great question because it has an answer which is both interesting and important.
– jwg
Jul 19 at 9:39
You asked a great question - it's a great question because it has an answer which is both interesting and important.
– jwg
Jul 19 at 9:39
 |Â
show 4 more comments
4 Answers
4
active
oldest
votes
up vote
156
down vote
accepted
Interestingly, there are no such examples! For a continuous function $f : mathbbR rightarrow mathbbR$ to be invertible, it must be either monotone increasing or decreasing. A famous classical result in analysis, Lebesgue's Monotone Function Theorem, states that any monotone function on an open interval is differentiable almost everywhere. Hence, there are no continuous functions that are invertible and nowhere differentiable.
answered Jul 16 at 17:37
Alex Nolte
2,0981419
add a comment |Â
up vote
54
down vote
Invertible implies bijective by set theory, and bijective together with continuity implies strictly increasing or decreasing, which imply differentiability almost everywhere! (This is known as Lebesgue's monotone function theorem.)
answered Jul 16 at 17:36
Alfred Yerger
9,7952044
add a comment |Â
up vote
16
down vote
If $f:(a,b)toBbb R$ is continuous and injective it must be monotone, hence differentiable almost everywhere.
answered Jul 16 at 17:37
David C. Ullrich
54.3k33583
add a comment |Â
up vote
5
down vote
I will assume by invertible you mean injective, i.e. bijective on its image. As others have mentioned, there are no such examples among functions from $mathbbR$ to $mathbbR$.
However you might find such examples in other topological spaces.
An example $mathbbR rightarrow mathbbR^2$ is given by $g(t) = (W(t), t)$ with $W$ the Weierstrass function; it is obviously injective and continuous but nowhere differentiable, considering its projection to the first coordinate.
You can also build an example from $mathbbR^2 rightarrow mathbbR^2$ by considering $g(t_1, t_2) = (t_1, W(t_1) + t_2)$ or its inverse, given by $g^-1(x_1, x_2) = (x_1, x_2 - W(x_1))$.
An injection $mathbbR^2 rightarrow mathbbR$ cannot be continuous so you won't find a counterexample there.
answered Jul 19 at 11:56
Maxim
33328
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
156
down vote
accepted
Interestingly, there are no such examples! For a continuous function $f : mathbbR rightarrow mathbbR$ to be invertible, it must be either monotone increasing or decreasing. A famous classical result in analysis, Lebesgue's Monotone Function Theorem, states that any monotone function on an open interval is differentiable almost everywhere. Hence, there are no continuous functions that are invertible and nowhere differentiable.
answered Jul 16 at 17:37
Alex Nolte
2,0981419
add a comment |Â
up vote
156
down vote
accepted
Interestingly, there are no such examples! For a continuous function $f : mathbbR rightarrow mathbbR$ to be invertible, it must be either monotone increasing or decreasing. A famous classical result in analysis, Lebesgue's Monotone Function Theorem, states that any monotone function on an open interval is differentiable almost everywhere. Hence, there are no continuous functions that are invertible and nowhere differentiable.
answered Jul 16 at 17:37
Alex Nolte
2,0981419
add a comment |Â
up vote
156
down vote
accepted
up vote
156
down vote
accepted
Interestingly, there are no such examples! For a continuous function $f : mathbbR rightarrow mathbbR$ to be invertible, it must be either monotone increasing or decreasing. A famous classical result in analysis, Lebesgue's Monotone Function Theorem, states that any monotone function on an open interval is differentiable almost everywhere. Hence, there are no continuous functions that are invertible and nowhere differentiable.
answered Jul 16 at 17:37
Alex Nolte
2,0981419
Interestingly, there are no such examples! For a continuous function $f : mathbbR rightarrow mathbbR$ to be invertible, it must be either monotone increasing or decreasing. A famous classical result in analysis, Lebesgue's Monotone Function Theorem, states that any monotone function on an open interval is differentiable almost everywhere. Hence, there are no continuous functions that are invertible and nowhere differentiable.
answered Jul 16 at 17:37
Alex Nolte
2,0981419
answered Jul 16 at 17:37
Alex Nolte
2,0981419
answered Jul 16 at 17:37
Alex Nolte
2,0981419
answered Jul 16 at 17:37
Alex Nolte
2,0981419
2,0981419
add a comment |Â
add a comment |Â
up vote
54
down vote
Invertible implies bijective by set theory, and bijective together with continuity implies strictly increasing or decreasing, which imply differentiability almost everywhere! (This is known as Lebesgue's monotone function theorem.)
answered Jul 16 at 17:36
Alfred Yerger
9,7952044
add a comment |Â
up vote
54
down vote
Invertible implies bijective by set theory, and bijective together with continuity implies strictly increasing or decreasing, which imply differentiability almost everywhere! (This is known as Lebesgue's monotone function theorem.)
answered Jul 16 at 17:36
Alfred Yerger
9,7952044
add a comment |Â
up vote
54
down vote
up vote
54
down vote
Invertible implies bijective by set theory, and bijective together with continuity implies strictly increasing or decreasing, which imply differentiability almost everywhere! (This is known as Lebesgue's monotone function theorem.)
answered Jul 16 at 17:36
Alfred Yerger
9,7952044
Invertible implies bijective by set theory, and bijective together with continuity implies strictly increasing or decreasing, which imply differentiability almost everywhere! (This is known as Lebesgue's monotone function theorem.)
answered Jul 16 at 17:36
Alfred Yerger
9,7952044
edited Jul 16 at 17:55
answered Jul 16 at 17:36
Alfred Yerger
9,7952044
answered Jul 16 at 17:36
Alfred Yerger
9,7952044
answered Jul 16 at 17:36
Alfred Yerger
9,7952044
9,7952044
add a comment |Â
add a comment |Â
up vote
16
down vote
If $f:(a,b)toBbb R$ is continuous and injective it must be monotone, hence differentiable almost everywhere.
answered Jul 16 at 17:37
David C. Ullrich
54.3k33583
add a comment |Â
up vote
16
down vote
If $f:(a,b)toBbb R$ is continuous and injective it must be monotone, hence differentiable almost everywhere.
answered Jul 16 at 17:37
David C. Ullrich
54.3k33583
add a comment |Â
up vote
16
down vote
up vote
16
down vote
If $f:(a,b)toBbb R$ is continuous and injective it must be monotone, hence differentiable almost everywhere.
answered Jul 16 at 17:37
David C. Ullrich
54.3k33583
If $f:(a,b)toBbb R$ is continuous and injective it must be monotone, hence differentiable almost everywhere.
answered Jul 16 at 17:37
David C. Ullrich
54.3k33583
answered Jul 16 at 17:37
David C. Ullrich
54.3k33583
answered Jul 16 at 17:37
David C. Ullrich
54.3k33583
answered Jul 16 at 17:37
David C. Ullrich
54.3k33583
54.3k33583
add a comment |Â
add a comment |Â
up vote
5
down vote
I will assume by invertible you mean injective, i.e. bijective on its image. As others have mentioned, there are no such examples among functions from $mathbbR$ to $mathbbR$.
However you might find such examples in other topological spaces.
An example $mathbbR rightarrow mathbbR^2$ is given by $g(t) = (W(t), t)$ with $W$ the Weierstrass function; it is obviously injective and continuous but nowhere differentiable, considering its projection to the first coordinate.
You can also build an example from $mathbbR^2 rightarrow mathbbR^2$ by considering $g(t_1, t_2) = (t_1, W(t_1) + t_2)$ or its inverse, given by $g^-1(x_1, x_2) = (x_1, x_2 - W(x_1))$.
An injection $mathbbR^2 rightarrow mathbbR$ cannot be continuous so you won't find a counterexample there.
answered Jul 19 at 11:56
Maxim
33328
add a comment |Â
up vote
5
down vote
I will assume by invertible you mean injective, i.e. bijective on its image. As others have mentioned, there are no such examples among functions from $mathbbR$ to $mathbbR$.
However you might find such examples in other topological spaces.
An example $mathbbR rightarrow mathbbR^2$ is given by $g(t) = (W(t), t)$ with $W$ the Weierstrass function; it is obviously injective and continuous but nowhere differentiable, considering its projection to the first coordinate.
You can also build an example from $mathbbR^2 rightarrow mathbbR^2$ by considering $g(t_1, t_2) = (t_1, W(t_1) + t_2)$ or its inverse, given by $g^-1(x_1, x_2) = (x_1, x_2 - W(x_1))$.
An injection $mathbbR^2 rightarrow mathbbR$ cannot be continuous so you won't find a counterexample there.
answered Jul 19 at 11:56
Maxim
33328
add a comment |Â
up vote
5
down vote
up vote
5
down vote
I will assume by invertible you mean injective, i.e. bijective on its image. As others have mentioned, there are no such examples among functions from $mathbbR$ to $mathbbR$.
However you might find such examples in other topological spaces.
An example $mathbbR rightarrow mathbbR^2$ is given by $g(t) = (W(t), t)$ with $W$ the Weierstrass function; it is obviously injective and continuous but nowhere differentiable, considering its projection to the first coordinate.
You can also build an example from $mathbbR^2 rightarrow mathbbR^2$ by considering $g(t_1, t_2) = (t_1, W(t_1) + t_2)$ or its inverse, given by $g^-1(x_1, x_2) = (x_1, x_2 - W(x_1))$.
An injection $mathbbR^2 rightarrow mathbbR$ cannot be continuous so you won't find a counterexample there.
answered Jul 19 at 11:56
Maxim
33328
I will assume by invertible you mean injective, i.e. bijective on its image. As others have mentioned, there are no such examples among functions from $mathbbR$ to $mathbbR$.
However you might find such examples in other topological spaces.
An example $mathbbR rightarrow mathbbR^2$ is given by $g(t) = (W(t), t)$ with $W$ the Weierstrass function; it is obviously injective and continuous but nowhere differentiable, considering its projection to the first coordinate.
You can also build an example from $mathbbR^2 rightarrow mathbbR^2$ by considering $g(t_1, t_2) = (t_1, W(t_1) + t_2)$ or its inverse, given by $g^-1(x_1, x_2) = (x_1, x_2 - W(x_1))$.
An injection $mathbbR^2 rightarrow mathbbR$ cannot be continuous so you won't find a counterexample there.
answered Jul 19 at 11:56
Maxim
33328
answered Jul 19 at 11:56
Maxim
33328
answered Jul 19 at 11:56
Maxim
33328
answered Jul 19 at 11:56
Maxim
33328
33328
add a comment |Â
add a comment |Â
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6
Guys, high school level amateur here, why can I not differentiate the individual terms to t and sum them to get the answer as an infinite sum over k of a^k.b^k.sin(b^k.t) ?
– HarryH
Jul 16 at 19:58
14
@HarryH: You can't always differentiate a series (a sum with infinitely many terms) term by term.
– Hans Lundmark
Jul 16 at 20:12
1
Looking at the graph, why can't you just use W(x)+cx, with c>>1 they should become invertible?
– Aganju
Jul 17 at 0:57
16
@Aganju basically, you would need to choose $c=infty$, because the function drops down arbitrary steeply locally.
– leftaroundabout
Jul 17 at 5:27
2
You asked a great question - it's a great question because it has an answer which is both interesting and important.
– jwg
Jul 19 at 9:39