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The size of the Hardy-Littlewood maximal function $f^*$ of a function $f: mathbb R^n to mathbb R$?
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Let $f^*(mathbf x) = sup frac1Q int_Q |f(mathbf y)|dmathbf y$.
- Why are there positive constants $c_1$ and $c_2$ such that $(7.7)$ holds? What does $(7.7)$ tell us?
- Why is $chi_E^*$ not integrable over $mathbb R^n$?
real-analysis measure-theory proof-explanation
asked Jul 29 at 21:28
user398843
387215
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Let $f^*(mathbf x) = sup frac1Q int_Q |f(mathbf y)|dmathbf y$.
- Why are there positive constants $c_1$ and $c_2$ such that $(7.7)$ holds? What does $(7.7)$ tell us?
- Why is $chi_E^*$ not integrable over $mathbb R^n$?
real-analysis measure-theory proof-explanation
asked Jul 29 at 21:28
user398843
387215
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f^*(mathbf x) = sup frac1Q int_Q |f(mathbf y)|dmathbf y$.
- Why are there positive constants $c_1$ and $c_2$ such that $(7.7)$ holds? What does $(7.7)$ tell us?
- Why is $chi_E^*$ not integrable over $mathbb R^n$?
real-analysis measure-theory proof-explanation
asked Jul 29 at 21:28
user398843
387215
Let $f^*(mathbf x) = sup frac1Q int_Q |f(mathbf y)|dmathbf y$.
- Why are there positive constants $c_1$ and $c_2$ such that $(7.7)$ holds? What does $(7.7)$ tell us?
- Why is $chi_E^*$ not integrable over $mathbb R^n$?
real-analysis measure-theory proof-explanation
asked Jul 29 at 21:28
user398843
387215
asked Jul 29 at 21:28
user398843
387215
asked Jul 29 at 21:28
user398843
387215
asked Jul 29 at 21:28
user398843
387215
387215
add a comment |Â
add a comment |Â
2 Answers
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up vote
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Since $chi_E^*(x) ge fracClvert x rvert^n$ for large $x$, then using the tranformation formula (or in this case integration in polar coordinates), we have $$int_lvert x rvert ge R chi_E^*(x) dx ge int_lvert x rvert ge R fracClvert x rvert^ndx = alpha_n C int_R^infty fracr^n-1drr^n = alpha_n C int^infty_R fracdrr = infty,$$ where $alpha_n$ is the surface measure of $S^n-1 = x $. This shows that $chi_E^*$ is not integrable.
To prove the inequality, if $E$ is bounded, then for $x$ large, $x not in E$, and the distance $textdist(x,E) sim lvert x rvert$ as $lvert x rvert to infty$ (do you see why this is? A picture may help). Now if the cube $Q^x$ contains $E$, it has side length greater than some constant times $textdist(x,E)$ (indeed, the cube centered at $x$ of side length $2textdist(x,E)/sqrt n$ touches $E$ but does not contain any of the interior of $E$). Thus we will have $lvert Q^x rvert sim textdist(x,E)^n sim lvert x rvert^n$.
answered Jul 29 at 21:54
User8128
10.2k1522
Why do we integrate $chi^*_E$ over $$?
– user398843
Jul 29 at 23:22
@user398843 Well $chi_E^* ge 0$ which means that $$int_lvert x rvert ge R chi_E^*(x) dx ge int_mathbb R^n chi_E^*(x)dx$$ so it suffices to prove that the latter quantity is $+infty$ to see that $chi_E^* notin L^1(mathbb R^n)$.
– User8128
Jul 30 at 0:34
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If $E$ is bounded, let $M=sup_yin E|y|$. If $|x|$ is much larger than $M$, then for a cube centered at $x$ to intersect $E$ at all, the cube must have side length at least $c|x|$ for some constant $c$ (this constant is somewhat annoying to write down explicitly; if we were dealing with a ball instead of a cube we could just say the radius must be at least $|x|-M$ which is greater than $|x|/2$ for $|x|$ large, but for a cube we need an additional constant relating the $ell^infty$ norm to the Euclidean norm). So for large $|x|$, we have $chi_E^*(x)leq frac^n$ since if $Q$ is any cube centered at $x$ which intersects $E$ we have $|Qcap E|leq|E|$ and $|Q|geq c^n|x|^n$. So we can take $c_2=1/c^n$.
On the other hand, though, there is another constant $c'$ such that a cube of side length $c'|x|$ centered at $x$ will contain all of $E$. (Again, if we were taking a ball instead of a cube we could take a ball of radius $|x|+M$ which is at most $2|x|$ for $|x|$ large.) This cube has measure $c'^n|x|^n$ and witnesses that $chi_E^*(x)geq frac^n$, so we can take $c_1=1/c'^n$.
In particular, if $|E|>0$, then this gives a constant $b>0$ such that $chi_E^*(x)>fracbx$ when $|x|$ is sufficiently large. Since the integral of $frac1x$ over $mathbbR^n$ diverges (even away from the singularity at the origin), this means that that the integral of $chi_E^*$ diverges too.
answered Jul 29 at 21:52
Eric Wofsey
162k12188298
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Since $chi_E^*(x) ge fracClvert x rvert^n$ for large $x$, then using the tranformation formula (or in this case integration in polar coordinates), we have $$int_lvert x rvert ge R chi_E^*(x) dx ge int_lvert x rvert ge R fracClvert x rvert^ndx = alpha_n C int_R^infty fracr^n-1drr^n = alpha_n C int^infty_R fracdrr = infty,$$ where $alpha_n$ is the surface measure of $S^n-1 = x $. This shows that $chi_E^*$ is not integrable.
To prove the inequality, if $E$ is bounded, then for $x$ large, $x not in E$, and the distance $textdist(x,E) sim lvert x rvert$ as $lvert x rvert to infty$ (do you see why this is? A picture may help). Now if the cube $Q^x$ contains $E$, it has side length greater than some constant times $textdist(x,E)$ (indeed, the cube centered at $x$ of side length $2textdist(x,E)/sqrt n$ touches $E$ but does not contain any of the interior of $E$). Thus we will have $lvert Q^x rvert sim textdist(x,E)^n sim lvert x rvert^n$.
answered Jul 29 at 21:54
User8128
10.2k1522
Why do we integrate $chi^*_E$ over $$?
– user398843
Jul 29 at 23:22
@user398843 Well $chi_E^* ge 0$ which means that $$int_lvert x rvert ge R chi_E^*(x) dx ge int_mathbb R^n chi_E^*(x)dx$$ so it suffices to prove that the latter quantity is $+infty$ to see that $chi_E^* notin L^1(mathbb R^n)$.
– User8128
Jul 30 at 0:34
add a comment |Â
up vote
2
down vote
Since $chi_E^*(x) ge fracClvert x rvert^n$ for large $x$, then using the tranformation formula (or in this case integration in polar coordinates), we have $$int_lvert x rvert ge R chi_E^*(x) dx ge int_lvert x rvert ge R fracClvert x rvert^ndx = alpha_n C int_R^infty fracr^n-1drr^n = alpha_n C int^infty_R fracdrr = infty,$$ where $alpha_n$ is the surface measure of $S^n-1 = x $. This shows that $chi_E^*$ is not integrable.
To prove the inequality, if $E$ is bounded, then for $x$ large, $x not in E$, and the distance $textdist(x,E) sim lvert x rvert$ as $lvert x rvert to infty$ (do you see why this is? A picture may help). Now if the cube $Q^x$ contains $E$, it has side length greater than some constant times $textdist(x,E)$ (indeed, the cube centered at $x$ of side length $2textdist(x,E)/sqrt n$ touches $E$ but does not contain any of the interior of $E$). Thus we will have $lvert Q^x rvert sim textdist(x,E)^n sim lvert x rvert^n$.
answered Jul 29 at 21:54
User8128
10.2k1522
Why do we integrate $chi^*_E$ over $$?
– user398843
Jul 29 at 23:22
@user398843 Well $chi_E^* ge 0$ which means that $$int_lvert x rvert ge R chi_E^*(x) dx ge int_mathbb R^n chi_E^*(x)dx$$ so it suffices to prove that the latter quantity is $+infty$ to see that $chi_E^* notin L^1(mathbb R^n)$.
– User8128
Jul 30 at 0:34
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Since $chi_E^*(x) ge fracClvert x rvert^n$ for large $x$, then using the tranformation formula (or in this case integration in polar coordinates), we have $$int_lvert x rvert ge R chi_E^*(x) dx ge int_lvert x rvert ge R fracClvert x rvert^ndx = alpha_n C int_R^infty fracr^n-1drr^n = alpha_n C int^infty_R fracdrr = infty,$$ where $alpha_n$ is the surface measure of $S^n-1 = x $. This shows that $chi_E^*$ is not integrable.
To prove the inequality, if $E$ is bounded, then for $x$ large, $x not in E$, and the distance $textdist(x,E) sim lvert x rvert$ as $lvert x rvert to infty$ (do you see why this is? A picture may help). Now if the cube $Q^x$ contains $E$, it has side length greater than some constant times $textdist(x,E)$ (indeed, the cube centered at $x$ of side length $2textdist(x,E)/sqrt n$ touches $E$ but does not contain any of the interior of $E$). Thus we will have $lvert Q^x rvert sim textdist(x,E)^n sim lvert x rvert^n$.
answered Jul 29 at 21:54
User8128
10.2k1522
Since $chi_E^*(x) ge fracClvert x rvert^n$ for large $x$, then using the tranformation formula (or in this case integration in polar coordinates), we have $$int_lvert x rvert ge R chi_E^*(x) dx ge int_lvert x rvert ge R fracClvert x rvert^ndx = alpha_n C int_R^infty fracr^n-1drr^n = alpha_n C int^infty_R fracdrr = infty,$$ where $alpha_n$ is the surface measure of $S^n-1 = x $. This shows that $chi_E^*$ is not integrable.
To prove the inequality, if $E$ is bounded, then for $x$ large, $x not in E$, and the distance $textdist(x,E) sim lvert x rvert$ as $lvert x rvert to infty$ (do you see why this is? A picture may help). Now if the cube $Q^x$ contains $E$, it has side length greater than some constant times $textdist(x,E)$ (indeed, the cube centered at $x$ of side length $2textdist(x,E)/sqrt n$ touches $E$ but does not contain any of the interior of $E$). Thus we will have $lvert Q^x rvert sim textdist(x,E)^n sim lvert x rvert^n$.
answered Jul 29 at 21:54
User8128
10.2k1522
answered Jul 29 at 21:54
User8128
10.2k1522
answered Jul 29 at 21:54
User8128
10.2k1522
answered Jul 29 at 21:54
User8128
10.2k1522
10.2k1522
Why do we integrate $chi^*_E$ over $$?
– user398843
Jul 29 at 23:22
@user398843 Well $chi_E^* ge 0$ which means that $$int_lvert x rvert ge R chi_E^*(x) dx ge int_mathbb R^n chi_E^*(x)dx$$ so it suffices to prove that the latter quantity is $+infty$ to see that $chi_E^* notin L^1(mathbb R^n)$.
– User8128
Jul 30 at 0:34
add a comment |Â
Why do we integrate $chi^*_E$ over $$?
– user398843
Jul 29 at 23:22
@user398843 Well $chi_E^* ge 0$ which means that $$int_lvert x rvert ge R chi_E^*(x) dx ge int_mathbb R^n chi_E^*(x)dx$$ so it suffices to prove that the latter quantity is $+infty$ to see that $chi_E^* notin L^1(mathbb R^n)$.
– User8128
Jul 30 at 0:34
Why do we integrate $chi^*_E$ over $$?
– user398843
Jul 29 at 23:22
Why do we integrate $chi^*_E$ over $$?
– user398843
Jul 29 at 23:22
@user398843 Well $chi_E^* ge 0$ which means that $$int_lvert x rvert ge R chi_E^*(x) dx ge int_mathbb R^n chi_E^*(x)dx$$ so it suffices to prove that the latter quantity is $+infty$ to see that $chi_E^* notin L^1(mathbb R^n)$.
– User8128
Jul 30 at 0:34
@user398843 Well $chi_E^* ge 0$ which means that $$int_lvert x rvert ge R chi_E^*(x) dx ge int_mathbb R^n chi_E^*(x)dx$$ so it suffices to prove that the latter quantity is $+infty$ to see that $chi_E^* notin L^1(mathbb R^n)$.
– User8128
Jul 30 at 0:34
add a comment |Â
up vote
0
down vote
If $E$ is bounded, let $M=sup_yin E|y|$. If $|x|$ is much larger than $M$, then for a cube centered at $x$ to intersect $E$ at all, the cube must have side length at least $c|x|$ for some constant $c$ (this constant is somewhat annoying to write down explicitly; if we were dealing with a ball instead of a cube we could just say the radius must be at least $|x|-M$ which is greater than $|x|/2$ for $|x|$ large, but for a cube we need an additional constant relating the $ell^infty$ norm to the Euclidean norm). So for large $|x|$, we have $chi_E^*(x)leq frac^n$ since if $Q$ is any cube centered at $x$ which intersects $E$ we have $|Qcap E|leq|E|$ and $|Q|geq c^n|x|^n$. So we can take $c_2=1/c^n$.
On the other hand, though, there is another constant $c'$ such that a cube of side length $c'|x|$ centered at $x$ will contain all of $E$. (Again, if we were taking a ball instead of a cube we could take a ball of radius $|x|+M$ which is at most $2|x|$ for $|x|$ large.) This cube has measure $c'^n|x|^n$ and witnesses that $chi_E^*(x)geq frac^n$, so we can take $c_1=1/c'^n$.
In particular, if $|E|>0$, then this gives a constant $b>0$ such that $chi_E^*(x)>fracbx$ when $|x|$ is sufficiently large. Since the integral of $frac1x$ over $mathbbR^n$ diverges (even away from the singularity at the origin), this means that that the integral of $chi_E^*$ diverges too.
answered Jul 29 at 21:52
Eric Wofsey
162k12188298
add a comment |Â
up vote
0
down vote
If $E$ is bounded, let $M=sup_yin E|y|$. If $|x|$ is much larger than $M$, then for a cube centered at $x$ to intersect $E$ at all, the cube must have side length at least $c|x|$ for some constant $c$ (this constant is somewhat annoying to write down explicitly; if we were dealing with a ball instead of a cube we could just say the radius must be at least $|x|-M$ which is greater than $|x|/2$ for $|x|$ large, but for a cube we need an additional constant relating the $ell^infty$ norm to the Euclidean norm). So for large $|x|$, we have $chi_E^*(x)leq frac^n$ since if $Q$ is any cube centered at $x$ which intersects $E$ we have $|Qcap E|leq|E|$ and $|Q|geq c^n|x|^n$. So we can take $c_2=1/c^n$.
On the other hand, though, there is another constant $c'$ such that a cube of side length $c'|x|$ centered at $x$ will contain all of $E$. (Again, if we were taking a ball instead of a cube we could take a ball of radius $|x|+M$ which is at most $2|x|$ for $|x|$ large.) This cube has measure $c'^n|x|^n$ and witnesses that $chi_E^*(x)geq frac^n$, so we can take $c_1=1/c'^n$.
In particular, if $|E|>0$, then this gives a constant $b>0$ such that $chi_E^*(x)>fracbx$ when $|x|$ is sufficiently large. Since the integral of $frac1x$ over $mathbbR^n$ diverges (even away from the singularity at the origin), this means that that the integral of $chi_E^*$ diverges too.
answered Jul 29 at 21:52
Eric Wofsey
162k12188298
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If $E$ is bounded, let $M=sup_yin E|y|$. If $|x|$ is much larger than $M$, then for a cube centered at $x$ to intersect $E$ at all, the cube must have side length at least $c|x|$ for some constant $c$ (this constant is somewhat annoying to write down explicitly; if we were dealing with a ball instead of a cube we could just say the radius must be at least $|x|-M$ which is greater than $|x|/2$ for $|x|$ large, but for a cube we need an additional constant relating the $ell^infty$ norm to the Euclidean norm). So for large $|x|$, we have $chi_E^*(x)leq frac^n$ since if $Q$ is any cube centered at $x$ which intersects $E$ we have $|Qcap E|leq|E|$ and $|Q|geq c^n|x|^n$. So we can take $c_2=1/c^n$.
On the other hand, though, there is another constant $c'$ such that a cube of side length $c'|x|$ centered at $x$ will contain all of $E$. (Again, if we were taking a ball instead of a cube we could take a ball of radius $|x|+M$ which is at most $2|x|$ for $|x|$ large.) This cube has measure $c'^n|x|^n$ and witnesses that $chi_E^*(x)geq frac^n$, so we can take $c_1=1/c'^n$.
In particular, if $|E|>0$, then this gives a constant $b>0$ such that $chi_E^*(x)>fracbx$ when $|x|$ is sufficiently large. Since the integral of $frac1x$ over $mathbbR^n$ diverges (even away from the singularity at the origin), this means that that the integral of $chi_E^*$ diverges too.
answered Jul 29 at 21:52
Eric Wofsey
162k12188298
If $E$ is bounded, let $M=sup_yin E|y|$. If $|x|$ is much larger than $M$, then for a cube centered at $x$ to intersect $E$ at all, the cube must have side length at least $c|x|$ for some constant $c$ (this constant is somewhat annoying to write down explicitly; if we were dealing with a ball instead of a cube we could just say the radius must be at least $|x|-M$ which is greater than $|x|/2$ for $|x|$ large, but for a cube we need an additional constant relating the $ell^infty$ norm to the Euclidean norm). So for large $|x|$, we have $chi_E^*(x)leq frac^n$ since if $Q$ is any cube centered at $x$ which intersects $E$ we have $|Qcap E|leq|E|$ and $|Q|geq c^n|x|^n$. So we can take $c_2=1/c^n$.
On the other hand, though, there is another constant $c'$ such that a cube of side length $c'|x|$ centered at $x$ will contain all of $E$. (Again, if we were taking a ball instead of a cube we could take a ball of radius $|x|+M$ which is at most $2|x|$ for $|x|$ large.) This cube has measure $c'^n|x|^n$ and witnesses that $chi_E^*(x)geq frac^n$, so we can take $c_1=1/c'^n$.
In particular, if $|E|>0$, then this gives a constant $b>0$ such that $chi_E^*(x)>fracbx$ when $|x|$ is sufficiently large. Since the integral of $frac1x$ over $mathbbR^n$ diverges (even away from the singularity at the origin), this means that that the integral of $chi_E^*$ diverges too.
answered Jul 29 at 21:52
Eric Wofsey
162k12188298
answered Jul 29 at 21:52
Eric Wofsey
162k12188298
answered Jul 29 at 21:52
Eric Wofsey
162k12188298
answered Jul 29 at 21:52
Eric Wofsey
162k12188298
162k12188298
add a comment |Â
add a comment |Â
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