Where is a “fat Cantor staircase” differentiable?What is the Lebesgue mean of the fat Cantor set?Strictly...

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Where is a “fat Cantor staircase” differentiable?


What is the Lebesgue mean of the fat Cantor set?Strictly monotone real function: stationary point, non-differentiable pointFunctions differentiable on “small” setsThe Cantor staircase function and related thingsIncreasing Sequence of Step Functions Converging to Characteristic Function of Fat Cantor SetCharacterizing discontinuous derivativesMaking a dense set of full measure from Cantor like setsContinuous extension from Cantor set to unit intervalDifferentiable A.E. with Positive Derivative Implies Increasing






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15














$begingroup$


The Cantor staircase is an example of a continuous function which is differentiable (and indeed flat) almost everywhere, and yet nonconstant. It is differentiable precisely at the points that aren't in the Cantor set.





Now, repeat the construction with a fat Cantor set (such as the Smith–Volterra–Cantor set, shown here).





Where is this fat Cantor staircase differentiable? Naively, I would've expected it to be differentiable on the complement of the Cantor set again, just like the original staircase. But this contradicts the theorem that says monotone functions are differentiable almost everywhere. So the fat Cantor staircase must be differentiable on some points in the Cantor set! Where, precisely, in the Cantor set does this happen? What is the derivative there? And what does the integral of this function's derivative look like?



(Conjecture: it is differentiable on the "pseudo-interior" - the set of points in the fat Cantor set without successor or predecessor, not counting 0 and 1.)










share|cite|improve this question











$endgroup$
















  • $begingroup$
    "And what does the integral of this function's derivative look like?" wouldnt that be just the function itself?
    $endgroup$
    – supinf
    7 hours ago






  • 1




    $begingroup$
    @supinf Not necessarily. Take the original Cantor staircase, for example - its derivative is zero almost everywhere, so the integral of its derivative is constant!
    $endgroup$
    – Akiva Weinberger
    7 hours ago










  • $begingroup$
    To clarify, are you defining the "pseudo-interior" as $bigcap_{n=1}^infty text{int}(F_n)$ where the $(F_n)$ is the decreasing sequence of compact sets given in your construction of this fat cantor set?
    $endgroup$
    – Brian Moehring
    7 hours ago










  • $begingroup$
    @BrianMoehring Yeah
    $endgroup$
    – Akiva Weinberger
    7 hours ago










  • $begingroup$
    All right. And second question, would it be accurate to state that your "fat cantor staircase" is given by $$x mapsto frac{mu(C cap [0,x])}{mu(C)},$$ where $C$ is your fat cantor set?
    $endgroup$
    – Brian Moehring
    7 hours ago




















15














$begingroup$


The Cantor staircase is an example of a continuous function which is differentiable (and indeed flat) almost everywhere, and yet nonconstant. It is differentiable precisely at the points that aren't in the Cantor set.





Now, repeat the construction with a fat Cantor set (such as the Smith–Volterra–Cantor set, shown here).





Where is this fat Cantor staircase differentiable? Naively, I would've expected it to be differentiable on the complement of the Cantor set again, just like the original staircase. But this contradicts the theorem that says monotone functions are differentiable almost everywhere. So the fat Cantor staircase must be differentiable on some points in the Cantor set! Where, precisely, in the Cantor set does this happen? What is the derivative there? And what does the integral of this function's derivative look like?



(Conjecture: it is differentiable on the "pseudo-interior" - the set of points in the fat Cantor set without successor or predecessor, not counting 0 and 1.)










share|cite|improve this question











$endgroup$
















  • $begingroup$
    "And what does the integral of this function's derivative look like?" wouldnt that be just the function itself?
    $endgroup$
    – supinf
    7 hours ago






  • 1




    $begingroup$
    @supinf Not necessarily. Take the original Cantor staircase, for example - its derivative is zero almost everywhere, so the integral of its derivative is constant!
    $endgroup$
    – Akiva Weinberger
    7 hours ago










  • $begingroup$
    To clarify, are you defining the "pseudo-interior" as $bigcap_{n=1}^infty text{int}(F_n)$ where the $(F_n)$ is the decreasing sequence of compact sets given in your construction of this fat cantor set?
    $endgroup$
    – Brian Moehring
    7 hours ago










  • $begingroup$
    @BrianMoehring Yeah
    $endgroup$
    – Akiva Weinberger
    7 hours ago










  • $begingroup$
    All right. And second question, would it be accurate to state that your "fat cantor staircase" is given by $$x mapsto frac{mu(C cap [0,x])}{mu(C)},$$ where $C$ is your fat cantor set?
    $endgroup$
    – Brian Moehring
    7 hours ago
















15












15








15


6



$begingroup$


The Cantor staircase is an example of a continuous function which is differentiable (and indeed flat) almost everywhere, and yet nonconstant. It is differentiable precisely at the points that aren't in the Cantor set.





Now, repeat the construction with a fat Cantor set (such as the Smith–Volterra–Cantor set, shown here).





Where is this fat Cantor staircase differentiable? Naively, I would've expected it to be differentiable on the complement of the Cantor set again, just like the original staircase. But this contradicts the theorem that says monotone functions are differentiable almost everywhere. So the fat Cantor staircase must be differentiable on some points in the Cantor set! Where, precisely, in the Cantor set does this happen? What is the derivative there? And what does the integral of this function's derivative look like?



(Conjecture: it is differentiable on the "pseudo-interior" - the set of points in the fat Cantor set without successor or predecessor, not counting 0 and 1.)










share|cite|improve this question











$endgroup$




The Cantor staircase is an example of a continuous function which is differentiable (and indeed flat) almost everywhere, and yet nonconstant. It is differentiable precisely at the points that aren't in the Cantor set.





Now, repeat the construction with a fat Cantor set (such as the Smith–Volterra–Cantor set, shown here).





Where is this fat Cantor staircase differentiable? Naively, I would've expected it to be differentiable on the complement of the Cantor set again, just like the original staircase. But this contradicts the theorem that says monotone functions are differentiable almost everywhere. So the fat Cantor staircase must be differentiable on some points in the Cantor set! Where, precisely, in the Cantor set does this happen? What is the derivative there? And what does the integral of this function's derivative look like?



(Conjecture: it is differentiable on the "pseudo-interior" - the set of points in the fat Cantor set without successor or predecessor, not counting 0 and 1.)







real-analysis measure-theory cantor-set






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question



share|cite|improve this question








edited 8 hours ago







Akiva Weinberger

















asked 8 hours ago









Akiva WeinbergerAkiva Weinberger

14.3k1 gold badge28 silver badges70 bronze badges




14.3k1 gold badge28 silver badges70 bronze badges















  • $begingroup$
    "And what does the integral of this function's derivative look like?" wouldnt that be just the function itself?
    $endgroup$
    – supinf
    7 hours ago






  • 1




    $begingroup$
    @supinf Not necessarily. Take the original Cantor staircase, for example - its derivative is zero almost everywhere, so the integral of its derivative is constant!
    $endgroup$
    – Akiva Weinberger
    7 hours ago










  • $begingroup$
    To clarify, are you defining the "pseudo-interior" as $bigcap_{n=1}^infty text{int}(F_n)$ where the $(F_n)$ is the decreasing sequence of compact sets given in your construction of this fat cantor set?
    $endgroup$
    – Brian Moehring
    7 hours ago










  • $begingroup$
    @BrianMoehring Yeah
    $endgroup$
    – Akiva Weinberger
    7 hours ago










  • $begingroup$
    All right. And second question, would it be accurate to state that your "fat cantor staircase" is given by $$x mapsto frac{mu(C cap [0,x])}{mu(C)},$$ where $C$ is your fat cantor set?
    $endgroup$
    – Brian Moehring
    7 hours ago




















  • $begingroup$
    "And what does the integral of this function's derivative look like?" wouldnt that be just the function itself?
    $endgroup$
    – supinf
    7 hours ago






  • 1




    $begingroup$
    @supinf Not necessarily. Take the original Cantor staircase, for example - its derivative is zero almost everywhere, so the integral of its derivative is constant!
    $endgroup$
    – Akiva Weinberger
    7 hours ago










  • $begingroup$
    To clarify, are you defining the "pseudo-interior" as $bigcap_{n=1}^infty text{int}(F_n)$ where the $(F_n)$ is the decreasing sequence of compact sets given in your construction of this fat cantor set?
    $endgroup$
    – Brian Moehring
    7 hours ago










  • $begingroup$
    @BrianMoehring Yeah
    $endgroup$
    – Akiva Weinberger
    7 hours ago










  • $begingroup$
    All right. And second question, would it be accurate to state that your "fat cantor staircase" is given by $$x mapsto frac{mu(C cap [0,x])}{mu(C)},$$ where $C$ is your fat cantor set?
    $endgroup$
    – Brian Moehring
    7 hours ago


















$begingroup$
"And what does the integral of this function's derivative look like?" wouldnt that be just the function itself?
$endgroup$
– supinf
7 hours ago




$begingroup$
"And what does the integral of this function's derivative look like?" wouldnt that be just the function itself?
$endgroup$
– supinf
7 hours ago




1




1




$begingroup$
@supinf Not necessarily. Take the original Cantor staircase, for example - its derivative is zero almost everywhere, so the integral of its derivative is constant!
$endgroup$
– Akiva Weinberger
7 hours ago




$begingroup$
@supinf Not necessarily. Take the original Cantor staircase, for example - its derivative is zero almost everywhere, so the integral of its derivative is constant!
$endgroup$
– Akiva Weinberger
7 hours ago












$begingroup$
To clarify, are you defining the "pseudo-interior" as $bigcap_{n=1}^infty text{int}(F_n)$ where the $(F_n)$ is the decreasing sequence of compact sets given in your construction of this fat cantor set?
$endgroup$
– Brian Moehring
7 hours ago




$begingroup$
To clarify, are you defining the "pseudo-interior" as $bigcap_{n=1}^infty text{int}(F_n)$ where the $(F_n)$ is the decreasing sequence of compact sets given in your construction of this fat cantor set?
$endgroup$
– Brian Moehring
7 hours ago












$begingroup$
@BrianMoehring Yeah
$endgroup$
– Akiva Weinberger
7 hours ago




$begingroup$
@BrianMoehring Yeah
$endgroup$
– Akiva Weinberger
7 hours ago












$begingroup$
All right. And second question, would it be accurate to state that your "fat cantor staircase" is given by $$x mapsto frac{mu(C cap [0,x])}{mu(C)},$$ where $C$ is your fat cantor set?
$endgroup$
– Brian Moehring
7 hours ago






$begingroup$
All right. And second question, would it be accurate to state that your "fat cantor staircase" is given by $$x mapsto frac{mu(C cap [0,x])}{mu(C)},$$ where $C$ is your fat cantor set?
$endgroup$
– Brian Moehring
7 hours ago












1 Answer
1






active

oldest

votes


















3
















$begingroup$

First note that the "fat cantor staircase function" is given by
$$f(x) = frac{mu(Ccap [0,x])}{mu(C)}$$
so immediately from Lebesgue's density theorem,
$$f'(x) = begin{cases}1/mu(C) & text{ if } d_C(x) = 1 \ 0 & text{ if } d_C(x) = 0end{cases}$$
describes the derivative almost everywhere, where $d_C(x) = lim_{hto 0^+} frac{mu(C cap [x-h, x+h])}{2h}.$





It now is left to show that $d_C^{-1}(1)$ equals the "pseudo-interior" of $C.$ I believe this is true, but I'll have to tackle it later (if anyone wants to use this work to finish, feel free, as it may be tomorrow before I can come back to it)






share|cite|improve this answer










$endgroup$















  • $begingroup$
    My intuition is that $d_C$ won't be $1$ on the entire pseudo-interior--only at points that are not "too quickly approximated" by the endpoints. That will be almost all points of the pseudo-interior, but I don't see a simple description of which points it is.
    $endgroup$
    – Eric Wofsey
    6 hours ago










  • $begingroup$
    @supinf It can take other values, but the set on which it's neither 0 nor 1 has measure zero.
    $endgroup$
    – Brian Moehring
    5 hours ago













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3
















$begingroup$

First note that the "fat cantor staircase function" is given by
$$f(x) = frac{mu(Ccap [0,x])}{mu(C)}$$
so immediately from Lebesgue's density theorem,
$$f'(x) = begin{cases}1/mu(C) & text{ if } d_C(x) = 1 \ 0 & text{ if } d_C(x) = 0end{cases}$$
describes the derivative almost everywhere, where $d_C(x) = lim_{hto 0^+} frac{mu(C cap [x-h, x+h])}{2h}.$





It now is left to show that $d_C^{-1}(1)$ equals the "pseudo-interior" of $C.$ I believe this is true, but I'll have to tackle it later (if anyone wants to use this work to finish, feel free, as it may be tomorrow before I can come back to it)






share|cite|improve this answer










$endgroup$















  • $begingroup$
    My intuition is that $d_C$ won't be $1$ on the entire pseudo-interior--only at points that are not "too quickly approximated" by the endpoints. That will be almost all points of the pseudo-interior, but I don't see a simple description of which points it is.
    $endgroup$
    – Eric Wofsey
    6 hours ago










  • $begingroup$
    @supinf It can take other values, but the set on which it's neither 0 nor 1 has measure zero.
    $endgroup$
    – Brian Moehring
    5 hours ago
















3
















$begingroup$

First note that the "fat cantor staircase function" is given by
$$f(x) = frac{mu(Ccap [0,x])}{mu(C)}$$
so immediately from Lebesgue's density theorem,
$$f'(x) = begin{cases}1/mu(C) & text{ if } d_C(x) = 1 \ 0 & text{ if } d_C(x) = 0end{cases}$$
describes the derivative almost everywhere, where $d_C(x) = lim_{hto 0^+} frac{mu(C cap [x-h, x+h])}{2h}.$





It now is left to show that $d_C^{-1}(1)$ equals the "pseudo-interior" of $C.$ I believe this is true, but I'll have to tackle it later (if anyone wants to use this work to finish, feel free, as it may be tomorrow before I can come back to it)






share|cite|improve this answer










$endgroup$















  • $begingroup$
    My intuition is that $d_C$ won't be $1$ on the entire pseudo-interior--only at points that are not "too quickly approximated" by the endpoints. That will be almost all points of the pseudo-interior, but I don't see a simple description of which points it is.
    $endgroup$
    – Eric Wofsey
    6 hours ago










  • $begingroup$
    @supinf It can take other values, but the set on which it's neither 0 nor 1 has measure zero.
    $endgroup$
    – Brian Moehring
    5 hours ago














3














3










3







$begingroup$

First note that the "fat cantor staircase function" is given by
$$f(x) = frac{mu(Ccap [0,x])}{mu(C)}$$
so immediately from Lebesgue's density theorem,
$$f'(x) = begin{cases}1/mu(C) & text{ if } d_C(x) = 1 \ 0 & text{ if } d_C(x) = 0end{cases}$$
describes the derivative almost everywhere, where $d_C(x) = lim_{hto 0^+} frac{mu(C cap [x-h, x+h])}{2h}.$





It now is left to show that $d_C^{-1}(1)$ equals the "pseudo-interior" of $C.$ I believe this is true, but I'll have to tackle it later (if anyone wants to use this work to finish, feel free, as it may be tomorrow before I can come back to it)






share|cite|improve this answer










$endgroup$



First note that the "fat cantor staircase function" is given by
$$f(x) = frac{mu(Ccap [0,x])}{mu(C)}$$
so immediately from Lebesgue's density theorem,
$$f'(x) = begin{cases}1/mu(C) & text{ if } d_C(x) = 1 \ 0 & text{ if } d_C(x) = 0end{cases}$$
describes the derivative almost everywhere, where $d_C(x) = lim_{hto 0^+} frac{mu(C cap [x-h, x+h])}{2h}.$





It now is left to show that $d_C^{-1}(1)$ equals the "pseudo-interior" of $C.$ I believe this is true, but I'll have to tackle it later (if anyone wants to use this work to finish, feel free, as it may be tomorrow before I can come back to it)







share|cite|improve this answer













share|cite|improve this answer




share|cite|improve this answer



share|cite|improve this answer










answered 6 hours ago









Brian MoehringBrian Moehring

3,0273 silver badges14 bronze badges




3,0273 silver badges14 bronze badges















  • $begingroup$
    My intuition is that $d_C$ won't be $1$ on the entire pseudo-interior--only at points that are not "too quickly approximated" by the endpoints. That will be almost all points of the pseudo-interior, but I don't see a simple description of which points it is.
    $endgroup$
    – Eric Wofsey
    6 hours ago










  • $begingroup$
    @supinf It can take other values, but the set on which it's neither 0 nor 1 has measure zero.
    $endgroup$
    – Brian Moehring
    5 hours ago


















  • $begingroup$
    My intuition is that $d_C$ won't be $1$ on the entire pseudo-interior--only at points that are not "too quickly approximated" by the endpoints. That will be almost all points of the pseudo-interior, but I don't see a simple description of which points it is.
    $endgroup$
    – Eric Wofsey
    6 hours ago










  • $begingroup$
    @supinf It can take other values, but the set on which it's neither 0 nor 1 has measure zero.
    $endgroup$
    – Brian Moehring
    5 hours ago
















$begingroup$
My intuition is that $d_C$ won't be $1$ on the entire pseudo-interior--only at points that are not "too quickly approximated" by the endpoints. That will be almost all points of the pseudo-interior, but I don't see a simple description of which points it is.
$endgroup$
– Eric Wofsey
6 hours ago




$begingroup$
My intuition is that $d_C$ won't be $1$ on the entire pseudo-interior--only at points that are not "too quickly approximated" by the endpoints. That will be almost all points of the pseudo-interior, but I don't see a simple description of which points it is.
$endgroup$
– Eric Wofsey
6 hours ago












$begingroup$
@supinf It can take other values, but the set on which it's neither 0 nor 1 has measure zero.
$endgroup$
– Brian Moehring
5 hours ago




$begingroup$
@supinf It can take other values, but the set on which it's neither 0 nor 1 has measure zero.
$endgroup$
– Brian Moehring
5 hours ago



















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