What is a tube in $mathbb{R}^n$?
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Lebesgue's differentiation theorem states that if $x$ is a point in $mathbb{R}^n$ and $f:mathbb{R}^nrightarrowmathbb{R}$ is a Lebesgue integrable function, then the limit of $frac{int_B f dlambda}{lambda(B)}$ over all balls $B$ centered at $x$ as the diameter of $B$ goes to $0$ is equal almost everywhere to $f(x)$. But if you replace balls with other kinds of set with diameter going to $0$, this need not be true. For instance it need not be true if you replace balls with rectangles.
But I just came across a journal paper which shows that if you take the collection of all "tubes" in $mathbb{R}^n$ oriented in certain directions, then the Lebesgue differentiation holds true for this collection for $L^p$ functions with $p>1$. But my question is, what exactly is a tube in $mathbb{R}^n$ as the term is used in this paper? The paper doesn't provide any definition as far as I can tell.
Is it like a cylinder, or what?
geometry measure-theory definition lebesgue-integral lebesgue-measure
asked Nov 19 at 7:03
Keshav Srinivasan
2,53111440
add a comment |
up vote
0
down vote
favorite
Lebesgue's differentiation theorem states that if $x$ is a point in $mathbb{R}^n$ and $f:mathbb{R}^nrightarrowmathbb{R}$ is a Lebesgue integrable function, then the limit of $frac{int_B f dlambda}{lambda(B)}$ over all balls $B$ centered at $x$ as the diameter of $B$ goes to $0$ is equal almost everywhere to $f(x)$. But if you replace balls with other kinds of set with diameter going to $0$, this need not be true. For instance it need not be true if you replace balls with rectangles.
But I just came across a journal paper which shows that if you take the collection of all "tubes" in $mathbb{R}^n$ oriented in certain directions, then the Lebesgue differentiation holds true for this collection for $L^p$ functions with $p>1$. But my question is, what exactly is a tube in $mathbb{R}^n$ as the term is used in this paper? The paper doesn't provide any definition as far as I can tell.
Is it like a cylinder, or what?
geometry measure-theory definition lebesgue-integral lebesgue-measure
asked Nov 19 at 7:03
Keshav Srinivasan
2,53111440
Why the downvote?
– Keshav Srinivasan
Nov 24 at 2:33
Not sure why you were downvoted. But my upvote should compensate for it :-)
– AOrtiz
Nov 24 at 2:34
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Lebesgue's differentiation theorem states that if $x$ is a point in $mathbb{R}^n$ and $f:mathbb{R}^nrightarrowmathbb{R}$ is a Lebesgue integrable function, then the limit of $frac{int_B f dlambda}{lambda(B)}$ over all balls $B$ centered at $x$ as the diameter of $B$ goes to $0$ is equal almost everywhere to $f(x)$. But if you replace balls with other kinds of set with diameter going to $0$, this need not be true. For instance it need not be true if you replace balls with rectangles.
But I just came across a journal paper which shows that if you take the collection of all "tubes" in $mathbb{R}^n$ oriented in certain directions, then the Lebesgue differentiation holds true for this collection for $L^p$ functions with $p>1$. But my question is, what exactly is a tube in $mathbb{R}^n$ as the term is used in this paper? The paper doesn't provide any definition as far as I can tell.
Is it like a cylinder, or what?
geometry measure-theory definition lebesgue-integral lebesgue-measure
asked Nov 19 at 7:03
Keshav Srinivasan
2,53111440
Lebesgue's differentiation theorem states that if $x$ is a point in $mathbb{R}^n$ and $f:mathbb{R}^nrightarrowmathbb{R}$ is a Lebesgue integrable function, then the limit of $frac{int_B f dlambda}{lambda(B)}$ over all balls $B$ centered at $x$ as the diameter of $B$ goes to $0$ is equal almost everywhere to $f(x)$. But if you replace balls with other kinds of set with diameter going to $0$, this need not be true. For instance it need not be true if you replace balls with rectangles.
But I just came across a journal paper which shows that if you take the collection of all "tubes" in $mathbb{R}^n$ oriented in certain directions, then the Lebesgue differentiation holds true for this collection for $L^p$ functions with $p>1$. But my question is, what exactly is a tube in $mathbb{R}^n$ as the term is used in this paper? The paper doesn't provide any definition as far as I can tell.
Is it like a cylinder, or what?
geometry measure-theory definition lebesgue-integral lebesgue-measure
geometry measure-theory definition lebesgue-integral lebesgue-measure
asked Nov 19 at 7:03
Keshav Srinivasan
2,53111440
asked Nov 19 at 7:03
Keshav Srinivasan
2,53111440
asked Nov 19 at 7:03
Keshav Srinivasan
2,53111440
asked Nov 19 at 7:03
Keshav Srinivasan
2,53111440
asked Nov 19 at 7:03
Keshav Srinivasan
2,53111440
2,53111440
Why the downvote?
– Keshav Srinivasan
Nov 24 at 2:33
Not sure why you were downvoted. But my upvote should compensate for it :-)
– AOrtiz
Nov 24 at 2:34
add a comment |
Why the downvote?
– Keshav Srinivasan
Nov 24 at 2:33
Not sure why you were downvoted. But my upvote should compensate for it :-)
– AOrtiz
Nov 24 at 2:34
Why the downvote?
– Keshav Srinivasan
Nov 24 at 2:33
Why the downvote?
– Keshav Srinivasan
Nov 24 at 2:33
Not sure why you were downvoted. But my upvote should compensate for it :-)
– AOrtiz
Nov 24 at 2:34
Not sure why you were downvoted. But my upvote should compensate for it :-)
– AOrtiz
Nov 24 at 2:34
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Based on reference 19 in the paper you linked, on page 224 of the corresponding article (or page 11 in the .pdf), it looks like a tube is the same as an "oriented" cylinder, and is determined by a direction $gammain S^{n-1}$, and two parameters: a height, and a radius of the cylinder itself.
answered Nov 24 at 2:32
AOrtiz
10.4k21341
add a comment |
up vote
0
down vote
It should be a $n$ dimensional cylinder right? So pick a disk in $mathbb{R}^{n-1}$ then map it's point set, $(x_1,dots,x_{n-1},0)$ to $(x_1,dots,x_{n-1},z)$ where $z in mathbb{R}$.
For the usual space, $n=3$, you get a regular cylinder which has it's base disk as a subset of the xy-plane.
answered Nov 23 at 23:57
bkbowser
11
Did you read the paper? Is that in fact the set the paper is describing?
– Keshav Srinivasan
Nov 24 at 0:22
I read a little bit of it but I'm just guessing. Topologists will define shapes in that way; you get the usual object for low dimensions, and a higher dimensional analogue in higher dimensions.
– bkbowser
Nov 24 at 1:00
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Based on reference 19 in the paper you linked, on page 224 of the corresponding article (or page 11 in the .pdf), it looks like a tube is the same as an "oriented" cylinder, and is determined by a direction $gammain S^{n-1}$, and two parameters: a height, and a radius of the cylinder itself.
answered Nov 24 at 2:32
AOrtiz
10.4k21341
add a comment |
up vote
1
down vote
accepted
Based on reference 19 in the paper you linked, on page 224 of the corresponding article (or page 11 in the .pdf), it looks like a tube is the same as an "oriented" cylinder, and is determined by a direction $gammain S^{n-1}$, and two parameters: a height, and a radius of the cylinder itself.
answered Nov 24 at 2:32
AOrtiz
10.4k21341
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Based on reference 19 in the paper you linked, on page 224 of the corresponding article (or page 11 in the .pdf), it looks like a tube is the same as an "oriented" cylinder, and is determined by a direction $gammain S^{n-1}$, and two parameters: a height, and a radius of the cylinder itself.
answered Nov 24 at 2:32
AOrtiz
10.4k21341
Based on reference 19 in the paper you linked, on page 224 of the corresponding article (or page 11 in the .pdf), it looks like a tube is the same as an "oriented" cylinder, and is determined by a direction $gammain S^{n-1}$, and two parameters: a height, and a radius of the cylinder itself.
answered Nov 24 at 2:32
AOrtiz
10.4k21341
edited Nov 24 at 2:40
answered Nov 24 at 2:32
AOrtiz
10.4k21341
answered Nov 24 at 2:32
AOrtiz
10.4k21341
answered Nov 24 at 2:32
AOrtiz
10.4k21341
10.4k21341
add a comment |
add a comment |
up vote
0
down vote
It should be a $n$ dimensional cylinder right? So pick a disk in $mathbb{R}^{n-1}$ then map it's point set, $(x_1,dots,x_{n-1},0)$ to $(x_1,dots,x_{n-1},z)$ where $z in mathbb{R}$.
For the usual space, $n=3$, you get a regular cylinder which has it's base disk as a subset of the xy-plane.
answered Nov 23 at 23:57
bkbowser
11
Did you read the paper? Is that in fact the set the paper is describing?
– Keshav Srinivasan
Nov 24 at 0:22
I read a little bit of it but I'm just guessing. Topologists will define shapes in that way; you get the usual object for low dimensions, and a higher dimensional analogue in higher dimensions.
– bkbowser
Nov 24 at 1:00
add a comment |
up vote
0
down vote
It should be a $n$ dimensional cylinder right? So pick a disk in $mathbb{R}^{n-1}$ then map it's point set, $(x_1,dots,x_{n-1},0)$ to $(x_1,dots,x_{n-1},z)$ where $z in mathbb{R}$.
For the usual space, $n=3$, you get a regular cylinder which has it's base disk as a subset of the xy-plane.
answered Nov 23 at 23:57
bkbowser
11
Did you read the paper? Is that in fact the set the paper is describing?
– Keshav Srinivasan
Nov 24 at 0:22
I read a little bit of it but I'm just guessing. Topologists will define shapes in that way; you get the usual object for low dimensions, and a higher dimensional analogue in higher dimensions.
– bkbowser
Nov 24 at 1:00
add a comment |
up vote
0
down vote
up vote
0
down vote
It should be a $n$ dimensional cylinder right? So pick a disk in $mathbb{R}^{n-1}$ then map it's point set, $(x_1,dots,x_{n-1},0)$ to $(x_1,dots,x_{n-1},z)$ where $z in mathbb{R}$.
For the usual space, $n=3$, you get a regular cylinder which has it's base disk as a subset of the xy-plane.
answered Nov 23 at 23:57
bkbowser
11
It should be a $n$ dimensional cylinder right? So pick a disk in $mathbb{R}^{n-1}$ then map it's point set, $(x_1,dots,x_{n-1},0)$ to $(x_1,dots,x_{n-1},z)$ where $z in mathbb{R}$.
For the usual space, $n=3$, you get a regular cylinder which has it's base disk as a subset of the xy-plane.
answered Nov 23 at 23:57
bkbowser
11
answered Nov 23 at 23:57
bkbowser
11
answered Nov 23 at 23:57
bkbowser
11
answered Nov 23 at 23:57
bkbowser
11
11
Did you read the paper? Is that in fact the set the paper is describing?
– Keshav Srinivasan
Nov 24 at 0:22
I read a little bit of it but I'm just guessing. Topologists will define shapes in that way; you get the usual object for low dimensions, and a higher dimensional analogue in higher dimensions.
– bkbowser
Nov 24 at 1:00
add a comment |
Did you read the paper? Is that in fact the set the paper is describing?
– Keshav Srinivasan
Nov 24 at 0:22
I read a little bit of it but I'm just guessing. Topologists will define shapes in that way; you get the usual object for low dimensions, and a higher dimensional analogue in higher dimensions.
– bkbowser
Nov 24 at 1:00
Did you read the paper? Is that in fact the set the paper is describing?
– Keshav Srinivasan
Nov 24 at 0:22
Did you read the paper? Is that in fact the set the paper is describing?
– Keshav Srinivasan
Nov 24 at 0:22
I read a little bit of it but I'm just guessing. Topologists will define shapes in that way; you get the usual object for low dimensions, and a higher dimensional analogue in higher dimensions.
– bkbowser
Nov 24 at 1:00
I read a little bit of it but I'm just guessing. Topologists will define shapes in that way; you get the usual object for low dimensions, and a higher dimensional analogue in higher dimensions.
– bkbowser
Nov 24 at 1:00
add a comment |
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Why the downvote?
– Keshav Srinivasan
Nov 24 at 2:33
Not sure why you were downvoted. But my upvote should compensate for it :-)
– AOrtiz
Nov 24 at 2:34