maximal function bounded below by original function

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Is is true that for a locally integrable function, we always have $Mf(x)geq |f(x)|$ a.e.? I think that is true but I can not find any reference for that.

harmonic-analysis

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asked Jul 16 at 3:19

Dong Li

702413

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are you asking if there is a specific $M$ where the bound holds for all $x in mathbbR$?
  – Dionel Jaime
  Jul 16 at 3:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you may be asking about the third basic property listed here but I am not quite sure I understand your question yet. en.wikipedia.org/wiki/Maximal_function#Basic_properties
  – Mason
  Jul 16 at 3:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mason Well that's extremely different ...
  – Dionel Jaime
  Jul 16 at 3:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mason I am familiar with that properties but I did wonder if the OP is true.
  – Dong Li
  Jul 16 at 3:44
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  @Mason Yes it is the centred maximal function.
  – Dong Li
  Jul 16 at 3:55
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

Is is true that for a locally integrable function, we always have $Mf(x)geq |f(x)|$ a.e.? I think that is true but I can not find any reference for that.

harmonic-analysis

share|cite|improve this question

asked Jul 16 at 3:19

Dong Li

702413

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are you asking if there is a specific $M$ where the bound holds for all $x in mathbbR$?
  – Dionel Jaime
  Jul 16 at 3:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you may be asking about the third basic property listed here but I am not quite sure I understand your question yet. en.wikipedia.org/wiki/Maximal_function#Basic_properties
  – Mason
  Jul 16 at 3:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mason Well that's extremely different ...
  – Dionel Jaime
  Jul 16 at 3:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mason I am familiar with that properties but I did wonder if the OP is true.
  – Dong Li
  Jul 16 at 3:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Mason Yes it is the centred maximal function.
  – Dong Li
  Jul 16 at 3:55
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Is is true that for a locally integrable function, we always have $Mf(x)geq |f(x)|$ a.e.? I think that is true but I can not find any reference for that.

harmonic-analysis

share|cite|improve this question

asked Jul 16 at 3:19

Dong Li

702413

Is is true that for a locally integrable function, we always have $Mf(x)geq |f(x)|$ a.e.? I think that is true but I can not find any reference for that.

harmonic-analysis

share|cite|improve this question

asked Jul 16 at 3:19

Dong Li

702413

share|cite|improve this question

share|cite|improve this question

asked Jul 16 at 3:19

Dong Li

702413

asked Jul 16 at 3:19

Dong Li

702413

asked Jul 16 at 3:19

Dong Li

702413

702413

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are you asking if there is a specific $M$ where the bound holds for all $x in mathbbR$?
  – Dionel Jaime
  Jul 16 at 3:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you may be asking about the third basic property listed here but I am not quite sure I understand your question yet. en.wikipedia.org/wiki/Maximal_function#Basic_properties
  – Mason
  Jul 16 at 3:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mason Well that's extremely different ...
  – Dionel Jaime
  Jul 16 at 3:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mason I am familiar with that properties but I did wonder if the OP is true.
  – Dong Li
  Jul 16 at 3:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Mason Yes it is the centred maximal function.
  – Dong Li
  Jul 16 at 3:55
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are you asking if there is a specific $M$ where the bound holds for all $x in mathbbR$?
  – Dionel Jaime
  Jul 16 at 3:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you may be asking about the third basic property listed here but I am not quite sure I understand your question yet. en.wikipedia.org/wiki/Maximal_function#Basic_properties
  – Mason
  Jul 16 at 3:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mason Well that's extremely different ...
  – Dionel Jaime
  Jul 16 at 3:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mason I am familiar with that properties but I did wonder if the OP is true.
  – Dong Li
  Jul 16 at 3:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Mason Yes it is the centred maximal function.
  – Dong Li
  Jul 16 at 3:55
  

  
  
  
  

Are you asking if there is a specific $M$ where the bound holds for all $x in mathbbR$?
– Dionel Jaime
Jul 16 at 3:25

Are you asking if there is a specific $M$ where the bound holds for all $x in mathbbR$?
– Dionel Jaime
Jul 16 at 3:25

I think you may be asking about the third basic property listed here but I am not quite sure I understand your question yet. en.wikipedia.org/wiki/Maximal_function#Basic_properties
– Mason
Jul 16 at 3:26

I think you may be asking about the third basic property listed here but I am not quite sure I understand your question yet. en.wikipedia.org/wiki/Maximal_function#Basic_properties
– Mason
Jul 16 at 3:26

@Mason Well that's extremely different ...
– Dionel Jaime
Jul 16 at 3:29

@Mason Well that's extremely different ...
– Dionel Jaime
Jul 16 at 3:29

@Mason I am familiar with that properties but I did wonder if the OP is true.
– Dong Li
Jul 16 at 3:44

@Mason I am familiar with that properties but I did wonder if the OP is true.
– Dong Li
Jul 16 at 3:44

1

1

@Mason Yes it is the centred maximal function.
– Dong Li
Jul 16 at 3:55

@Mason Yes it is the centred maximal function.
– Dong Li
Jul 16 at 3:55

add a comment | 

1 Answer
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If $x$ is a Lebesgue point of $f$, then $Mf(x) geq |f(x)| $. Indeed, we have
$$
tag1 Mf(x) = sup_xin B frac1 int_B |f(y)| dy geq limlimits_rto 0 frac1 int_B(x,r) |f(y)| dy = |f(x)|,
$$
where the equality follows by the definition of Lebesgue point. Since for integrable $f$ almost every point is a Lebesgue point, then $(1)$ holds true almost everywhere (where $f$ is defined and is integrable).

share|cite|improve this answer

answered Jul 16 at 4:56

Hayk

1,39129

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

If $x$ is a Lebesgue point of $f$, then $Mf(x) geq |f(x)| $. Indeed, we have
$$
tag1 Mf(x) = sup_xin B frac1 int_B |f(y)| dy geq limlimits_rto 0 frac1 int_B(x,r) |f(y)| dy = |f(x)|,
$$
where the equality follows by the definition of Lebesgue point. Since for integrable $f$ almost every point is a Lebesgue point, then $(1)$ holds true almost everywhere (where $f$ is defined and is integrable).

share|cite|improve this answer

answered Jul 16 at 4:56

Hayk

1,39129

add a comment | 

up vote
2
down vote



accepted

If $x$ is a Lebesgue point of $f$, then $Mf(x) geq |f(x)| $. Indeed, we have
$$
tag1 Mf(x) = sup_xin B frac1 int_B |f(y)| dy geq limlimits_rto 0 frac1 int_B(x,r) |f(y)| dy = |f(x)|,
$$
where the equality follows by the definition of Lebesgue point. Since for integrable $f$ almost every point is a Lebesgue point, then $(1)$ holds true almost everywhere (where $f$ is defined and is integrable).

share|cite|improve this answer

answered Jul 16 at 4:56

Hayk

1,39129

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

If $x$ is a Lebesgue point of $f$, then $Mf(x) geq |f(x)| $. Indeed, we have
$$
tag1 Mf(x) = sup_xin B frac1 int_B |f(y)| dy geq limlimits_rto 0 frac1 int_B(x,r) |f(y)| dy = |f(x)|,
$$
where the equality follows by the definition of Lebesgue point. Since for integrable $f$ almost every point is a Lebesgue point, then $(1)$ holds true almost everywhere (where $f$ is defined and is integrable).

share|cite|improve this answer

answered Jul 16 at 4:56

Hayk

1,39129

If $x$ is a Lebesgue point of $f$, then $Mf(x) geq |f(x)| $. Indeed, we have
$$
tag1 Mf(x) = sup_xin B frac1 int_B |f(y)| dy geq limlimits_rto 0 frac1 int_B(x,r) |f(y)| dy = |f(x)|,
$$
where the equality follows by the definition of Lebesgue point. Since for integrable $f$ almost every point is a Lebesgue point, then $(1)$ holds true almost everywhere (where $f$ is defined and is integrable).

share|cite|improve this answer

answered Jul 16 at 4:56

Hayk

1,39129

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 4:56

Hayk

1,39129

answered Jul 16 at 4:56

Hayk

1,39129

answered Jul 16 at 4:56

Hayk

1,39129

1,39129

add a comment | 

add a comment | 

 

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