Prove a set contains an interval centered at zero.
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Prove that if $E subset [0,1]$ has positive measure, then the set $E-E = {x-y : x,y in E}$ contains an interval centered around zero. Hint: consider the function $h(x)=textbf{1}_{-E} star textbf{1}_{E} $.
Ideas: use the continuity of h(x)?
real-analysis measure-theory
asked Mar 25 '14 at 22:12
user62108
814
add a comment |
up vote
1
down vote
favorite
Prove that if $E subset [0,1]$ has positive measure, then the set $E-E = {x-y : x,y in E}$ contains an interval centered around zero. Hint: consider the function $h(x)=textbf{1}_{-E} star textbf{1}_{E} $.
Ideas: use the continuity of h(x)?
real-analysis measure-theory
asked Mar 25 '14 at 22:12
user62108
814
Why would $h$ be continuous?
– Frank
Mar 25 '14 at 22:15
Yes, using the continuity of $h$ is a good idea. You need to say why it is continuous, though.
– Daniel Fischer♦
Mar 25 '14 at 22:18
I know why h(x) is continuous but I'd rather not write it out on here. It was part of a previous exercise. Essentially if $fin L^1$ and $g in L^{infty}$ then $f star g$ is continuous. So my idea is to take the pre-image of some set to get an interval around zero.
– user62108
Mar 25 '14 at 22:59
Just saying that $h$ is continuous because both functions belong to $L^1cap L^infty$ is sufficient then. Now to reach the conclusion, find an $x$ such that $h(x) > 0$.
– Daniel Fischer♦
Mar 25 '14 at 23:04
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Prove that if $E subset [0,1]$ has positive measure, then the set $E-E = {x-y : x,y in E}$ contains an interval centered around zero. Hint: consider the function $h(x)=textbf{1}_{-E} star textbf{1}_{E} $.
Ideas: use the continuity of h(x)?
real-analysis measure-theory
asked Mar 25 '14 at 22:12
user62108
814
Prove that if $E subset [0,1]$ has positive measure, then the set $E-E = {x-y : x,y in E}$ contains an interval centered around zero. Hint: consider the function $h(x)=textbf{1}_{-E} star textbf{1}_{E} $.
Ideas: use the continuity of h(x)?
real-analysis measure-theory
real-analysis measure-theory
asked Mar 25 '14 at 22:12
user62108
814
asked Mar 25 '14 at 22:12
user62108
814
asked Mar 25 '14 at 22:12
user62108
814
asked Mar 25 '14 at 22:12
user62108
814
asked Mar 25 '14 at 22:12
user62108
814
814
Why would $h$ be continuous?
– Frank
Mar 25 '14 at 22:15
Yes, using the continuity of $h$ is a good idea. You need to say why it is continuous, though.
– Daniel Fischer♦
Mar 25 '14 at 22:18
I know why h(x) is continuous but I'd rather not write it out on here. It was part of a previous exercise. Essentially if $fin L^1$ and $g in L^{infty}$ then $f star g$ is continuous. So my idea is to take the pre-image of some set to get an interval around zero.
– user62108
Mar 25 '14 at 22:59
Just saying that $h$ is continuous because both functions belong to $L^1cap L^infty$ is sufficient then. Now to reach the conclusion, find an $x$ such that $h(x) > 0$.
– Daniel Fischer♦
Mar 25 '14 at 23:04
add a comment |
Why would $h$ be continuous?
– Frank
Mar 25 '14 at 22:15
Yes, using the continuity of $h$ is a good idea. You need to say why it is continuous, though.
– Daniel Fischer♦
Mar 25 '14 at 22:18
I know why h(x) is continuous but I'd rather not write it out on here. It was part of a previous exercise. Essentially if $fin L^1$ and $g in L^{infty}$ then $f star g$ is continuous. So my idea is to take the pre-image of some set to get an interval around zero.
– user62108
Mar 25 '14 at 22:59
Just saying that $h$ is continuous because both functions belong to $L^1cap L^infty$ is sufficient then. Now to reach the conclusion, find an $x$ such that $h(x) > 0$.
– Daniel Fischer♦
Mar 25 '14 at 23:04
Why would $h$ be continuous?
– Frank
Mar 25 '14 at 22:15
Why would $h$ be continuous?
– Frank
Mar 25 '14 at 22:15
Yes, using the continuity of $h$ is a good idea. You need to say why it is continuous, though.
– Daniel Fischer♦
Mar 25 '14 at 22:18
Yes, using the continuity of $h$ is a good idea. You need to say why it is continuous, though.
– Daniel Fischer♦
Mar 25 '14 at 22:18
I know why h(x) is continuous but I'd rather not write it out on here. It was part of a previous exercise. Essentially if $fin L^1$ and $g in L^{infty}$ then $f star g$ is continuous. So my idea is to take the pre-image of some set to get an interval around zero.
– user62108
Mar 25 '14 at 22:59
I know why h(x) is continuous but I'd rather not write it out on here. It was part of a previous exercise. Essentially if $fin L^1$ and $g in L^{infty}$ then $f star g$ is continuous. So my idea is to take the pre-image of some set to get an interval around zero.
– user62108
Mar 25 '14 at 22:59
Just saying that $h$ is continuous because both functions belong to $L^1cap L^infty$ is sufficient then. Now to reach the conclusion, find an $x$ such that $h(x) > 0$.
– Daniel Fischer♦
Mar 25 '14 at 23:04
Just saying that $h$ is continuous because both functions belong to $L^1cap L^infty$ is sufficient then. Now to reach the conclusion, find an $x$ such that $h(x) > 0$.
– Daniel Fischer♦
Mar 25 '14 at 23:04
add a comment |
1 Answer
1
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$$h(x)= int 1_E(y)1_{-E}(x-y)dy= int 1_E(y)1_{E(x+E)}(y)dy=m(Ecap (x+E))$$
Prove that $h$ is continuous. Then observe that $h(0)=m(E)>0$, so $0in h^{-1}(0,infty)$ which is an open set. Therefore, there exists an open interval around $0$, say $(-varepsilon,varepsilon)$ contained in $h^{-1}(0,infty)$. Show that this is the desired interval.
edited Nov 18 at 1:34
user10354138
6,500623
answered Mar 26 '14 at 2:46
Dimitris
4,9881640
Why is this function continuous?
– Ángela Flores
May 3 at 5:02
In general if you take the convolution of a bounded function with an $L^1$ function you get a continuous function. See math.stackexchange.com/questions/570494/…
– Dimitris
May 4 at 4:32
Spoiler alert: to show that this is the desired interval: take any $pin (-epsilon,epsilon)$. Then $0<h(p)=m(Ecap p+E)$ so $Ecap p+E not = emptyset$ hence there are $e_1,e_2 in E$ with $e_1=p+e_2$...
– Robson
Nov 18 at 0:37
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$$h(x)= int 1_E(y)1_{-E}(x-y)dy= int 1_E(y)1_{E(x+E)}(y)dy=m(Ecap (x+E))$$
Prove that $h$ is continuous. Then observe that $h(0)=m(E)>0$, so $0in h^{-1}(0,infty)$ which is an open set. Therefore, there exists an open interval around $0$, say $(-varepsilon,varepsilon)$ contained in $h^{-1}(0,infty)$. Show that this is the desired interval.
edited Nov 18 at 1:34
user10354138
6,500623
answered Mar 26 '14 at 2:46
Dimitris
4,9881640
Why is this function continuous?
– Ángela Flores
May 3 at 5:02
In general if you take the convolution of a bounded function with an $L^1$ function you get a continuous function. See math.stackexchange.com/questions/570494/…
– Dimitris
May 4 at 4:32
Spoiler alert: to show that this is the desired interval: take any $pin (-epsilon,epsilon)$. Then $0<h(p)=m(Ecap p+E)$ so $Ecap p+E not = emptyset$ hence there are $e_1,e_2 in E$ with $e_1=p+e_2$...
– Robson
Nov 18 at 0:37
add a comment |
up vote
2
down vote
$$h(x)= int 1_E(y)1_{-E}(x-y)dy= int 1_E(y)1_{E(x+E)}(y)dy=m(Ecap (x+E))$$
Prove that $h$ is continuous. Then observe that $h(0)=m(E)>0$, so $0in h^{-1}(0,infty)$ which is an open set. Therefore, there exists an open interval around $0$, say $(-varepsilon,varepsilon)$ contained in $h^{-1}(0,infty)$. Show that this is the desired interval.
edited Nov 18 at 1:34
user10354138
6,500623
answered Mar 26 '14 at 2:46
Dimitris
4,9881640
Why is this function continuous?
– Ángela Flores
May 3 at 5:02
In general if you take the convolution of a bounded function with an $L^1$ function you get a continuous function. See math.stackexchange.com/questions/570494/…
– Dimitris
May 4 at 4:32
Spoiler alert: to show that this is the desired interval: take any $pin (-epsilon,epsilon)$. Then $0<h(p)=m(Ecap p+E)$ so $Ecap p+E not = emptyset$ hence there are $e_1,e_2 in E$ with $e_1=p+e_2$...
– Robson
Nov 18 at 0:37
add a comment |
up vote
2
down vote
up vote
2
down vote
$$h(x)= int 1_E(y)1_{-E}(x-y)dy= int 1_E(y)1_{E(x+E)}(y)dy=m(Ecap (x+E))$$
Prove that $h$ is continuous. Then observe that $h(0)=m(E)>0$, so $0in h^{-1}(0,infty)$ which is an open set. Therefore, there exists an open interval around $0$, say $(-varepsilon,varepsilon)$ contained in $h^{-1}(0,infty)$. Show that this is the desired interval.
edited Nov 18 at 1:34
user10354138
6,500623
answered Mar 26 '14 at 2:46
Dimitris
4,9881640
$$h(x)= int 1_E(y)1_{-E}(x-y)dy= int 1_E(y)1_{E(x+E)}(y)dy=m(Ecap (x+E))$$
Prove that $h$ is continuous. Then observe that $h(0)=m(E)>0$, so $0in h^{-1}(0,infty)$ which is an open set. Therefore, there exists an open interval around $0$, say $(-varepsilon,varepsilon)$ contained in $h^{-1}(0,infty)$. Show that this is the desired interval.
edited Nov 18 at 1:34
user10354138
6,500623
answered Mar 26 '14 at 2:46
Dimitris
4,9881640
edited Nov 18 at 1:34
user10354138
6,500623
edited Nov 18 at 1:34
user10354138
6,500623
edited Nov 18 at 1:34
user10354138
6,500623
6,500623
answered Mar 26 '14 at 2:46
Dimitris
4,9881640
answered Mar 26 '14 at 2:46
Dimitris
4,9881640
answered Mar 26 '14 at 2:46
Dimitris
4,9881640
4,9881640
Why is this function continuous?
– Ángela Flores
May 3 at 5:02
In general if you take the convolution of a bounded function with an $L^1$ function you get a continuous function. See math.stackexchange.com/questions/570494/…
– Dimitris
May 4 at 4:32
Spoiler alert: to show that this is the desired interval: take any $pin (-epsilon,epsilon)$. Then $0<h(p)=m(Ecap p+E)$ so $Ecap p+E not = emptyset$ hence there are $e_1,e_2 in E$ with $e_1=p+e_2$...
– Robson
Nov 18 at 0:37
add a comment |
Why is this function continuous?
– Ángela Flores
May 3 at 5:02
In general if you take the convolution of a bounded function with an $L^1$ function you get a continuous function. See math.stackexchange.com/questions/570494/…
– Dimitris
May 4 at 4:32
Spoiler alert: to show that this is the desired interval: take any $pin (-epsilon,epsilon)$. Then $0<h(p)=m(Ecap p+E)$ so $Ecap p+E not = emptyset$ hence there are $e_1,e_2 in E$ with $e_1=p+e_2$...
– Robson
Nov 18 at 0:37
Why is this function continuous?
– Ángela Flores
May 3 at 5:02
Why is this function continuous?
– Ángela Flores
May 3 at 5:02
In general if you take the convolution of a bounded function with an $L^1$ function you get a continuous function. See math.stackexchange.com/questions/570494/…
– Dimitris
May 4 at 4:32
In general if you take the convolution of a bounded function with an $L^1$ function you get a continuous function. See math.stackexchange.com/questions/570494/…
– Dimitris
May 4 at 4:32
Spoiler alert: to show that this is the desired interval: take any $pin (-epsilon,epsilon)$. Then $0<h(p)=m(Ecap p+E)$ so $Ecap p+E not = emptyset$ hence there are $e_1,e_2 in E$ with $e_1=p+e_2$...
– Robson
Nov 18 at 0:37
Spoiler alert: to show that this is the desired interval: take any $pin (-epsilon,epsilon)$. Then $0<h(p)=m(Ecap p+E)$ so $Ecap p+E not = emptyset$ hence there are $e_1,e_2 in E$ with $e_1=p+e_2$...
– Robson
Nov 18 at 0:37
add a comment |
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Why would $h$ be continuous?
– Frank
Mar 25 '14 at 22:15
Yes, using the continuity of $h$ is a good idea. You need to say why it is continuous, though.
– Daniel Fischer♦
Mar 25 '14 at 22:18
I know why h(x) is continuous but I'd rather not write it out on here. It was part of a previous exercise. Essentially if $fin L^1$ and $g in L^{infty}$ then $f star g$ is continuous. So my idea is to take the pre-image of some set to get an interval around zero.
– user62108
Mar 25 '14 at 22:59
Just saying that $h$ is continuous because both functions belong to $L^1cap L^infty$ is sufficient then. Now to reach the conclusion, find an $x$ such that $h(x) > 0$.
– Daniel Fischer♦
Mar 25 '14 at 23:04