Counterexamples to the inequality $int_Omega |f(x)|^p dmu(x) leq left(int_Omega |f(x)| dmu(x)right)^p$ with $pgeq1$ [closed]

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Let $(mu,Sigma,Omega)$ be a measure space. Since we have the inequality $sqrtfraca^2+b^22 geq fraca+b2$, one may natually think that the following is true:
$$
int_Omega |f(x)|^p dmu leq (int_Omega |f(x)| dmu)^p, pgeq1.
$$
But I don't feel it is true since if it was true, the proof of Hoder's inequality would become trivial. But I cannot find any counterexamples for it, either.

So can anyone produce me some counterexamples? I would also be glad to see if the inequality about integral above is true in some special cases, e.g. in $mathbbR$, as I know.

Why does the inequality fail to be true? I wish to have some explaination.

real-analysis inequality lebesgue-integral examples-counterexamples holder-inequality

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edited Jul 28 at 13:14

Did

242k23208441

asked Jul 28 at 12:31

Ma Joad

722216

closed as off-topic by Did, Shailesh, amWhy, choco_addicted, Leucippus Jul 30 at 2:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Shailesh, amWhy, choco_addicted, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  left( and right).
  – user578878
  Jul 28 at 12:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suppose $Omega=[0,1]$, $Sigma$ are the Borel sets, $mu$ Lebesgue measure, $p>1$ (for $p=1$ the inequality is true), and $f(x)=x^-1/p$. Then $int |f|<infty$, while $int |f|^p=infty$.
  – user578878
  Jul 28 at 12:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  "But I cannot find any counterexamples for it, either." OK. Please give at least two different cases you checked unsuccessfully.
  – Did
  Jul 28 at 13:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did Sorry that I typeset the direction of the inequality wrong. My bad. Sorry for the confusion.
  – Ma Joad
  Jul 28 at 13:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your comment seems to be unrelated to mine. Should we understand that, all things considered, you checked no example at all?
  – Did
  Jul 28 at 14:40
  

  
  
  
  

 | 
show 1 more comment

up vote
-1
down vote

favorite

Let $(mu,Sigma,Omega)$ be a measure space. Since we have the inequality $sqrtfraca^2+b^22 geq fraca+b2$, one may natually think that the following is true:
$$
int_Omega |f(x)|^p dmu leq (int_Omega |f(x)| dmu)^p, pgeq1.
$$
But I don't feel it is true since if it was true, the proof of Hoder's inequality would become trivial. But I cannot find any counterexamples for it, either.

So can anyone produce me some counterexamples? I would also be glad to see if the inequality about integral above is true in some special cases, e.g. in $mathbbR$, as I know.

Why does the inequality fail to be true? I wish to have some explaination.

real-analysis inequality lebesgue-integral examples-counterexamples holder-inequality

share|cite|improve this question

edited Jul 28 at 13:14

Did

242k23208441

asked Jul 28 at 12:31

Ma Joad

722216

closed as off-topic by Did, Shailesh, amWhy, choco_addicted, Leucippus Jul 30 at 2:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Shailesh, amWhy, choco_addicted, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  left( and right).
  – user578878
  Jul 28 at 12:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suppose $Omega=[0,1]$, $Sigma$ are the Borel sets, $mu$ Lebesgue measure, $p>1$ (for $p=1$ the inequality is true), and $f(x)=x^-1/p$. Then $int |f|<infty$, while $int |f|^p=infty$.
  – user578878
  Jul 28 at 12:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  "But I cannot find any counterexamples for it, either." OK. Please give at least two different cases you checked unsuccessfully.
  – Did
  Jul 28 at 13:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did Sorry that I typeset the direction of the inequality wrong. My bad. Sorry for the confusion.
  – Ma Joad
  Jul 28 at 13:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your comment seems to be unrelated to mine. Should we understand that, all things considered, you checked no example at all?
  – Did
  Jul 28 at 14:40
  

  
  
  
  

 | 
show 1 more comment

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

Let $(mu,Sigma,Omega)$ be a measure space. Since we have the inequality $sqrtfraca^2+b^22 geq fraca+b2$, one may natually think that the following is true:
$$
int_Omega |f(x)|^p dmu leq (int_Omega |f(x)| dmu)^p, pgeq1.
$$
But I don't feel it is true since if it was true, the proof of Hoder's inequality would become trivial. But I cannot find any counterexamples for it, either.

So can anyone produce me some counterexamples? I would also be glad to see if the inequality about integral above is true in some special cases, e.g. in $mathbbR$, as I know.

Why does the inequality fail to be true? I wish to have some explaination.

real-analysis inequality lebesgue-integral examples-counterexamples holder-inequality

share|cite|improve this question

edited Jul 28 at 13:14

Did

242k23208441

asked Jul 28 at 12:31

Ma Joad

722216

Let $(mu,Sigma,Omega)$ be a measure space. Since we have the inequality $sqrtfraca^2+b^22 geq fraca+b2$, one may natually think that the following is true:
$$
int_Omega |f(x)|^p dmu leq (int_Omega |f(x)| dmu)^p, pgeq1.
$$
But I don't feel it is true since if it was true, the proof of Hoder's inequality would become trivial. But I cannot find any counterexamples for it, either.

So can anyone produce me some counterexamples? I would also be glad to see if the inequality about integral above is true in some special cases, e.g. in $mathbbR$, as I know.

Why does the inequality fail to be true? I wish to have some explaination.

real-analysis inequality lebesgue-integral examples-counterexamples holder-inequality

share|cite|improve this question

edited Jul 28 at 13:14

Did

242k23208441

asked Jul 28 at 12:31

Ma Joad

722216

share|cite|improve this question

share|cite|improve this question

edited Jul 28 at 13:14

Did

242k23208441

edited Jul 28 at 13:14

Did

242k23208441

edited Jul 28 at 13:14

Did

242k23208441

242k23208441

asked Jul 28 at 12:31

Ma Joad

722216

asked Jul 28 at 12:31

Ma Joad

722216

asked Jul 28 at 12:31

Ma Joad

722216

722216

closed as off-topic by Did, Shailesh, amWhy, choco_addicted, Leucippus Jul 30 at 2:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Shailesh, amWhy, choco_addicted, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Did, Shailesh, amWhy, choco_addicted, Leucippus Jul 30 at 2:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Shailesh, amWhy, choco_addicted, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  left( and right).
  – user578878
  Jul 28 at 12:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suppose $Omega=[0,1]$, $Sigma$ are the Borel sets, $mu$ Lebesgue measure, $p>1$ (for $p=1$ the inequality is true), and $f(x)=x^-1/p$. Then $int |f|<infty$, while $int |f|^p=infty$.
  – user578878
  Jul 28 at 12:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  "But I cannot find any counterexamples for it, either." OK. Please give at least two different cases you checked unsuccessfully.
  – Did
  Jul 28 at 13:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did Sorry that I typeset the direction of the inequality wrong. My bad. Sorry for the confusion.
  – Ma Joad
  Jul 28 at 13:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your comment seems to be unrelated to mine. Should we understand that, all things considered, you checked no example at all?
  – Did
  Jul 28 at 14:40
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  left( and right).
  – user578878
  Jul 28 at 12:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suppose $Omega=[0,1]$, $Sigma$ are the Borel sets, $mu$ Lebesgue measure, $p>1$ (for $p=1$ the inequality is true), and $f(x)=x^-1/p$. Then $int |f|<infty$, while $int |f|^p=infty$.
  – user578878
  Jul 28 at 12:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  "But I cannot find any counterexamples for it, either." OK. Please give at least two different cases you checked unsuccessfully.
  – Did
  Jul 28 at 13:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did Sorry that I typeset the direction of the inequality wrong. My bad. Sorry for the confusion.
  – Ma Joad
  Jul 28 at 13:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your comment seems to be unrelated to mine. Should we understand that, all things considered, you checked no example at all?
  – Did
  Jul 28 at 14:40
  

  
  
  
  

1

1

left( and right).
– user578878
Jul 28 at 12:34

left( and right).
– user578878
Jul 28 at 12:34

Suppose $Omega=[0,1]$, $Sigma$ are the Borel sets, $mu$ Lebesgue measure, $p>1$ (for $p=1$ the inequality is true), and $f(x)=x^-1/p$. Then $int |f|<infty$, while $int |f|^p=infty$.
– user578878
Jul 28 at 12:37

Suppose $Omega=[0,1]$, $Sigma$ are the Borel sets, $mu$ Lebesgue measure, $p>1$ (for $p=1$ the inequality is true), and $f(x)=x^-1/p$. Then $int |f|<infty$, while $int |f|^p=infty$.
– user578878
Jul 28 at 12:37

"But I cannot find any counterexamples for it, either." OK. Please give at least two different cases you checked unsuccessfully.
– Did
Jul 28 at 13:15

"But I cannot find any counterexamples for it, either." OK. Please give at least two different cases you checked unsuccessfully.
– Did
Jul 28 at 13:15

@Did Sorry that I typeset the direction of the inequality wrong. My bad. Sorry for the confusion.
– Ma Joad
Jul 28 at 13:44

@Did Sorry that I typeset the direction of the inequality wrong. My bad. Sorry for the confusion.
– Ma Joad
Jul 28 at 13:44

Your comment seems to be unrelated to mine. Should we understand that, all things considered, you checked no example at all?
– Did
Jul 28 at 14:40

Your comment seems to be unrelated to mine. Should we understand that, all things considered, you checked no example at all?
– Did
Jul 28 at 14:40

 | 
show 1 more comment

1 Answer
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1
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If $f(x)=1$, then the left hand side of your inequality is $|Omega|$ (the measure of $Omega$) and the right hand side is $|Omega|^p$. So it will be wrong if $|Omega|<1$.

share|cite|improve this answer

answered Jul 28 at 12:36

Kusma

1,097111

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

If $f(x)=1$, then the left hand side of your inequality is $|Omega|$ (the measure of $Omega$) and the right hand side is $|Omega|^p$. So it will be wrong if $|Omega|<1$.

share|cite|improve this answer

answered Jul 28 at 12:36

Kusma

1,097111

add a comment | 

up vote
1
down vote

If $f(x)=1$, then the left hand side of your inequality is $|Omega|$ (the measure of $Omega$) and the right hand side is $|Omega|^p$. So it will be wrong if $|Omega|<1$.

share|cite|improve this answer

answered Jul 28 at 12:36

Kusma

1,097111

add a comment | 

up vote
1
down vote

up vote
1
down vote

If $f(x)=1$, then the left hand side of your inequality is $|Omega|$ (the measure of $Omega$) and the right hand side is $|Omega|^p$. So it will be wrong if $|Omega|<1$.

share|cite|improve this answer

answered Jul 28 at 12:36

Kusma

1,097111

If $f(x)=1$, then the left hand side of your inequality is $|Omega|$ (the measure of $Omega$) and the right hand side is $|Omega|^p$. So it will be wrong if $|Omega|<1$.

share|cite|improve this answer

answered Jul 28 at 12:36

Kusma

1,097111

share|cite|improve this answer

share|cite|improve this answer

answered Jul 28 at 12:36

Kusma

1,097111

answered Jul 28 at 12:36

Kusma

1,097111

answered Jul 28 at 12:36

Kusma

1,097111

1,097111

add a comment | 

add a comment |