Hardy-Littlewood maximal function $f^*$ is greater than $f$ for measurable $f$
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Suppose $f$ is a Lebesgue measurable function on $mathbb R$, and $forall xin mathbb R$, define the Hardy-Littlewood maximal function $f^*(x)=sup_{t>0}$$1over{2t}$$int_{x-t}^{x+t} |f|$
My question is how to show $|f(x)|le f^*(x)$ for almost every $x$?
For the case $f$ is integrable, this can be proved using Lebesgue's Differentiation Theorem. But I do not know how to extend this argument to general measurable functions.
real-analysis measure-theory lebesgue-measure
asked yesterday
RunningMeatball
435
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show 2 more comments
up vote
0
down vote
favorite
Suppose $f$ is a Lebesgue measurable function on $mathbb R$, and $forall xin mathbb R$, define the Hardy-Littlewood maximal function $f^*(x)=sup_{t>0}$$1over{2t}$$int_{x-t}^{x+t} |f|$
My question is how to show $|f(x)|le f^*(x)$ for almost every $x$?
For the case $f$ is integrable, this can be proved using Lebesgue's Differentiation Theorem. But I do not know how to extend this argument to general measurable functions.
real-analysis measure-theory lebesgue-measure
asked yesterday
RunningMeatball
435
if $f$ is not integrable, then $f^*$ is infinite
– mathworker21
yesterday
@mathworker21 I don't think it's quite that simple. For example if $f(t)=1/(1+|t|^{1/2})$ then $f$ is not integrable but it seem to me that $f^*$ is finite almost everywhere. What's true is that if $f$ is not integrable in some neighborhood of $x$ then $f^*(x)$ is infinite.
– David C. Ullrich
yesterday
@DavidC.Ullrich yea, I thought that's what he meant by integrable. It's very well known that you just need $f$ locally integrable for Lebesgue's differentiation theorem
– mathworker21
yesterday
What if you apply the (locally) integrable-case result to $f_n(x):=f(x)cdot 1_{{|f|le n}}$, and then let $ntoinfty$?
– John Dawkins
yesterday
2
The desired inequality holds at all points $x$ where $f^*(x) = infty$. Otherwise, if $f^*(x) < infty$, then $f$ is integrable in the neighbourhood of $x$, and hence in that neighborhood you can apply Lebesgue's differentiation theorem.
– Hayk
yesterday
|
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose $f$ is a Lebesgue measurable function on $mathbb R$, and $forall xin mathbb R$, define the Hardy-Littlewood maximal function $f^*(x)=sup_{t>0}$$1over{2t}$$int_{x-t}^{x+t} |f|$
My question is how to show $|f(x)|le f^*(x)$ for almost every $x$?
For the case $f$ is integrable, this can be proved using Lebesgue's Differentiation Theorem. But I do not know how to extend this argument to general measurable functions.
real-analysis measure-theory lebesgue-measure
asked yesterday
RunningMeatball
435
Suppose $f$ is a Lebesgue measurable function on $mathbb R$, and $forall xin mathbb R$, define the Hardy-Littlewood maximal function $f^*(x)=sup_{t>0}$$1over{2t}$$int_{x-t}^{x+t} |f|$
My question is how to show $|f(x)|le f^*(x)$ for almost every $x$?
For the case $f$ is integrable, this can be proved using Lebesgue's Differentiation Theorem. But I do not know how to extend this argument to general measurable functions.
real-analysis measure-theory lebesgue-measure
real-analysis measure-theory lebesgue-measure
asked yesterday
RunningMeatball
435
asked yesterday
RunningMeatball
435
asked yesterday
RunningMeatball
435
asked yesterday
RunningMeatball
435
asked yesterday
RunningMeatball
435
435
if $f$ is not integrable, then $f^*$ is infinite
– mathworker21
yesterday
@mathworker21 I don't think it's quite that simple. For example if $f(t)=1/(1+|t|^{1/2})$ then $f$ is not integrable but it seem to me that $f^*$ is finite almost everywhere. What's true is that if $f$ is not integrable in some neighborhood of $x$ then $f^*(x)$ is infinite.
– David C. Ullrich
yesterday
@DavidC.Ullrich yea, I thought that's what he meant by integrable. It's very well known that you just need $f$ locally integrable for Lebesgue's differentiation theorem
– mathworker21
yesterday
What if you apply the (locally) integrable-case result to $f_n(x):=f(x)cdot 1_{{|f|le n}}$, and then let $ntoinfty$?
– John Dawkins
yesterday
2
The desired inequality holds at all points $x$ where $f^*(x) = infty$. Otherwise, if $f^*(x) < infty$, then $f$ is integrable in the neighbourhood of $x$, and hence in that neighborhood you can apply Lebesgue's differentiation theorem.
– Hayk
yesterday
|
show 2 more comments
if $f$ is not integrable, then $f^*$ is infinite
– mathworker21
yesterday
@mathworker21 I don't think it's quite that simple. For example if $f(t)=1/(1+|t|^{1/2})$ then $f$ is not integrable but it seem to me that $f^*$ is finite almost everywhere. What's true is that if $f$ is not integrable in some neighborhood of $x$ then $f^*(x)$ is infinite.
– David C. Ullrich
yesterday
@DavidC.Ullrich yea, I thought that's what he meant by integrable. It's very well known that you just need $f$ locally integrable for Lebesgue's differentiation theorem
– mathworker21
yesterday
What if you apply the (locally) integrable-case result to $f_n(x):=f(x)cdot 1_{{|f|le n}}$, and then let $ntoinfty$?
– John Dawkins
yesterday
2
The desired inequality holds at all points $x$ where $f^*(x) = infty$. Otherwise, if $f^*(x) < infty$, then $f$ is integrable in the neighbourhood of $x$, and hence in that neighborhood you can apply Lebesgue's differentiation theorem.
– Hayk
yesterday
if $f$ is not integrable, then $f^*$ is infinite
– mathworker21
yesterday
if $f$ is not integrable, then $f^*$ is infinite
– mathworker21
yesterday
@mathworker21 I don't think it's quite that simple. For example if $f(t)=1/(1+|t|^{1/2})$ then $f$ is not integrable but it seem to me that $f^*$ is finite almost everywhere. What's true is that if $f$ is not integrable in some neighborhood of $x$ then $f^*(x)$ is infinite.
– David C. Ullrich
yesterday
@mathworker21 I don't think it's quite that simple. For example if $f(t)=1/(1+|t|^{1/2})$ then $f$ is not integrable but it seem to me that $f^*$ is finite almost everywhere. What's true is that if $f$ is not integrable in some neighborhood of $x$ then $f^*(x)$ is infinite.
– David C. Ullrich
yesterday
@DavidC.Ullrich yea, I thought that's what he meant by integrable. It's very well known that you just need $f$ locally integrable for Lebesgue's differentiation theorem
– mathworker21
yesterday
@DavidC.Ullrich yea, I thought that's what he meant by integrable. It's very well known that you just need $f$ locally integrable for Lebesgue's differentiation theorem
– mathworker21
yesterday
What if you apply the (locally) integrable-case result to $f_n(x):=f(x)cdot 1_{{|f|le n}}$, and then let $ntoinfty$?
– John Dawkins
yesterday
What if you apply the (locally) integrable-case result to $f_n(x):=f(x)cdot 1_{{|f|le n}}$, and then let $ntoinfty$?
– John Dawkins
yesterday
2
2
The desired inequality holds at all points $x$ where $f^*(x) = infty$. Otherwise, if $f^*(x) < infty$, then $f$ is integrable in the neighbourhood of $x$, and hence in that neighborhood you can apply Lebesgue's differentiation theorem.
– Hayk
yesterday
The desired inequality holds at all points $x$ where $f^*(x) = infty$. Otherwise, if $f^*(x) < infty$, then $f$ is integrable in the neighbourhood of $x$, and hence in that neighborhood you can apply Lebesgue's differentiation theorem.
– Hayk
yesterday
|
show 2 more comments
1 Answer
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Case 1: $f^*(x) = infty$ $forall x$. The inequality holds.
Case 2: $f^*(x) < infty$ for some $ x$.
We have $frac{1}{2t}int_{x-t}^{x+t} |f|<infty$ and thus $int_{x-t}^{x+t} |f|<infty$ $forall t>0$. For any compact $K subset mathbb R$, there exists $t$ such that $Ksubset (x-t,x+t)$, hence $$int_K |f| leint_{x-t}^{x+t} |f|<infty$$This shows $f$ is locally integrable. By Lebesgue's Differentiation Theorem, $$|f(x)|=lim_{tto 0}frac{1}{2t}int_{x-t}^{x+t}|f|lesup_{t>0}frac{1}{2t}int_{x-t}^{x+t}|f|=f^*(x)$$ for almost every $xin mathbb R$.
answered 20 hours ago
RunningMeatball
435
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Case 1: $f^*(x) = infty$ $forall x$. The inequality holds.
Case 2: $f^*(x) < infty$ for some $ x$.
We have $frac{1}{2t}int_{x-t}^{x+t} |f|<infty$ and thus $int_{x-t}^{x+t} |f|<infty$ $forall t>0$. For any compact $K subset mathbb R$, there exists $t$ such that $Ksubset (x-t,x+t)$, hence $$int_K |f| leint_{x-t}^{x+t} |f|<infty$$This shows $f$ is locally integrable. By Lebesgue's Differentiation Theorem, $$|f(x)|=lim_{tto 0}frac{1}{2t}int_{x-t}^{x+t}|f|lesup_{t>0}frac{1}{2t}int_{x-t}^{x+t}|f|=f^*(x)$$ for almost every $xin mathbb R$.
answered 20 hours ago
RunningMeatball
435
add a comment |
up vote
0
down vote
Case 1: $f^*(x) = infty$ $forall x$. The inequality holds.
Case 2: $f^*(x) < infty$ for some $ x$.
We have $frac{1}{2t}int_{x-t}^{x+t} |f|<infty$ and thus $int_{x-t}^{x+t} |f|<infty$ $forall t>0$. For any compact $K subset mathbb R$, there exists $t$ such that $Ksubset (x-t,x+t)$, hence $$int_K |f| leint_{x-t}^{x+t} |f|<infty$$This shows $f$ is locally integrable. By Lebesgue's Differentiation Theorem, $$|f(x)|=lim_{tto 0}frac{1}{2t}int_{x-t}^{x+t}|f|lesup_{t>0}frac{1}{2t}int_{x-t}^{x+t}|f|=f^*(x)$$ for almost every $xin mathbb R$.
answered 20 hours ago
RunningMeatball
435
add a comment |
up vote
0
down vote
up vote
0
down vote
Case 1: $f^*(x) = infty$ $forall x$. The inequality holds.
Case 2: $f^*(x) < infty$ for some $ x$.
We have $frac{1}{2t}int_{x-t}^{x+t} |f|<infty$ and thus $int_{x-t}^{x+t} |f|<infty$ $forall t>0$. For any compact $K subset mathbb R$, there exists $t$ such that $Ksubset (x-t,x+t)$, hence $$int_K |f| leint_{x-t}^{x+t} |f|<infty$$This shows $f$ is locally integrable. By Lebesgue's Differentiation Theorem, $$|f(x)|=lim_{tto 0}frac{1}{2t}int_{x-t}^{x+t}|f|lesup_{t>0}frac{1}{2t}int_{x-t}^{x+t}|f|=f^*(x)$$ for almost every $xin mathbb R$.
answered 20 hours ago
RunningMeatball
435
Case 1: $f^*(x) = infty$ $forall x$. The inequality holds.
Case 2: $f^*(x) < infty$ for some $ x$.
We have $frac{1}{2t}int_{x-t}^{x+t} |f|<infty$ and thus $int_{x-t}^{x+t} |f|<infty$ $forall t>0$. For any compact $K subset mathbb R$, there exists $t$ such that $Ksubset (x-t,x+t)$, hence $$int_K |f| leint_{x-t}^{x+t} |f|<infty$$This shows $f$ is locally integrable. By Lebesgue's Differentiation Theorem, $$|f(x)|=lim_{tto 0}frac{1}{2t}int_{x-t}^{x+t}|f|lesup_{t>0}frac{1}{2t}int_{x-t}^{x+t}|f|=f^*(x)$$ for almost every $xin mathbb R$.
answered 20 hours ago
RunningMeatball
435
answered 20 hours ago
RunningMeatball
435
answered 20 hours ago
RunningMeatball
435
answered 20 hours ago
RunningMeatball
435
435
add a comment |
add a comment |
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if $f$ is not integrable, then $f^*$ is infinite
– mathworker21
yesterday
@mathworker21 I don't think it's quite that simple. For example if $f(t)=1/(1+|t|^{1/2})$ then $f$ is not integrable but it seem to me that $f^*$ is finite almost everywhere. What's true is that if $f$ is not integrable in some neighborhood of $x$ then $f^*(x)$ is infinite.
– David C. Ullrich
yesterday
@DavidC.Ullrich yea, I thought that's what he meant by integrable. It's very well known that you just need $f$ locally integrable for Lebesgue's differentiation theorem
– mathworker21
yesterday
What if you apply the (locally) integrable-case result to $f_n(x):=f(x)cdot 1_{{|f|le n}}$, and then let $ntoinfty$?
– John Dawkins
yesterday
2
The desired inequality holds at all points $x$ where $f^*(x) = infty$. Otherwise, if $f^*(x) < infty$, then $f$ is integrable in the neighbourhood of $x$, and hence in that neighborhood you can apply Lebesgue's differentiation theorem.
– Hayk
yesterday