Show that the set of $fin L^1$ so that $fnotin L^p$ is residual.

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Consider $L^p([0,1])$ with Lebesgue measure. Note that if $fin L^p$ with $p>1$ then $fin L^1$. Show that the set of $fin L^1$ so that $fnotin L^p$ is residual.

First of all, a set $Esubset X$ is define to be generic if its complement is of the first category.

If we consider the set $$A_N:=leftfin L^1: displaystyleint_Ifleq Nm(I)^1-frac1pmbox for all intervals $I$ right$$

How can I prove this set is nowhere dense?

real-analysis general-topology lp-spaces baire-category

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edited Jul 30 at 8:45

Davide Giraudo

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asked Jul 30 at 7:32

julios

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up vote
2
down vote

favorite

1

Consider $L^p([0,1])$ with Lebesgue measure. Note that if $fin L^p$ with $p>1$ then $fin L^1$. Show that the set of $fin L^1$ so that $fnotin L^p$ is residual.

First of all, a set $Esubset X$ is define to be generic if its complement is of the first category.

If we consider the set $$A_N:=leftfin L^1: displaystyleint_Ifleq Nm(I)^1-frac1pmbox for all intervals $I$ right$$

How can I prove this set is nowhere dense?

real-analysis general-topology lp-spaces baire-category

share|cite|improve this question

edited Jul 30 at 8:45

Davide Giraudo

121k15146249

asked Jul 30 at 7:32

julios

564

add a comment | 

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

Consider $L^p([0,1])$ with Lebesgue measure. Note that if $fin L^p$ with $p>1$ then $fin L^1$. Show that the set of $fin L^1$ so that $fnotin L^p$ is residual.

First of all, a set $Esubset X$ is define to be generic if its complement is of the first category.

If we consider the set $$A_N:=leftfin L^1: displaystyleint_Ifleq Nm(I)^1-frac1pmbox for all intervals $I$ right$$

How can I prove this set is nowhere dense?

real-analysis general-topology lp-spaces baire-category

share|cite|improve this question

edited Jul 30 at 8:45

Davide Giraudo

121k15146249

asked Jul 30 at 7:32

julios

564

Consider $L^p([0,1])$ with Lebesgue measure. Note that if $fin L^p$ with $p>1$ then $fin L^1$. Show that the set of $fin L^1$ so that $fnotin L^p$ is residual.

First of all, a set $Esubset X$ is define to be generic if its complement is of the first category.

If we consider the set $$A_N:=leftfin L^1: displaystyleint_Ifleq Nm(I)^1-frac1pmbox for all intervals $I$ right$$

How can I prove this set is nowhere dense?

real-analysis general-topology lp-spaces baire-category

share|cite|improve this question

edited Jul 30 at 8:45

Davide Giraudo

121k15146249

asked Jul 30 at 7:32

julios

564

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 8:45

Davide Giraudo

121k15146249

edited Jul 30 at 8:45

Davide Giraudo

121k15146249

edited Jul 30 at 8:45

Davide Giraudo

121k15146249

121k15146249

asked Jul 30 at 7:32

julios

564

asked Jul 30 at 7:32

julios

564

asked Jul 30 at 7:32

julios

564

564

add a comment | 

add a comment | 

2 Answers
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Hint: If $int |f|^p > N$, show there is $h in L^infty$ such that $int h f > N$. Conclude that $f$ is open. To show it is dense, consider adding to any $f in L^1$ some function with small $L^1$ norm but large $L^p$ norm.

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answered Jul 30 at 7:50

Robert Israel

303k22201440

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Proof of the fact that $A_N$ is nowhere dense: Fatou's Lemma shows that $A_N$ is closed in $L^1$; suppose $A_N$ has an interior point $g$. Then there exists $r>0$ such that $||f-g||_1 <r$ implies $f in A_N$. By triangle inequality we get $int_I |f| leq 2N m(I)^1-frac 1 p $ for any interval $I$ whenever $||f||_1 <r$. Take $I=(0,c)$ and $f(x)=ax^alpha$ where $alpha in (-1,-frac 1 p)$ and $0<a<r(1+alpha)$. Then $||f||_1 <r$, and $int_I |f| =int_0^c ax^alpha =a frac c^1+alpha alpha +1$ so $a frac c^1+alpha alpha +1 leq 2N c^1-frac 1 p $. Dividing by $c^1+alpha$ and letting $c to 0$ we get a contradiction.

share|cite|improve this answer

answered Jul 30 at 8:36

Kavi Rama Murthy

19.7k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer. I have a little doubt, when you divide by $c^1+alpha$ and let $cto 0$, you have $int_Iftoinfty$, and this is the contradiction?
  – julios
  Jul 31 at 0:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You get the contradiction $frac a alpha +1=0$.
  – Kavi Rama Murthy
  Jul 31 at 5:26
  

  
  
  
  

add a comment | 

Your Answer

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

Hint: If $int |f|^p > N$, show there is $h in L^infty$ such that $int h f > N$. Conclude that $f$ is open. To show it is dense, consider adding to any $f in L^1$ some function with small $L^1$ norm but large $L^p$ norm.

share|cite|improve this answer

answered Jul 30 at 7:50

Robert Israel

303k22201440

add a comment | 

up vote
1
down vote

Hint: If $int |f|^p > N$, show there is $h in L^infty$ such that $int h f > N$. Conclude that $f$ is open. To show it is dense, consider adding to any $f in L^1$ some function with small $L^1$ norm but large $L^p$ norm.

share|cite|improve this answer

answered Jul 30 at 7:50

Robert Israel

303k22201440

add a comment | 

up vote
1
down vote

up vote
1
down vote

Hint: If $int |f|^p > N$, show there is $h in L^infty$ such that $int h f > N$. Conclude that $f$ is open. To show it is dense, consider adding to any $f in L^1$ some function with small $L^1$ norm but large $L^p$ norm.

share|cite|improve this answer

answered Jul 30 at 7:50

Robert Israel

303k22201440

Hint: If $int |f|^p > N$, show there is $h in L^infty$ such that $int h f > N$. Conclude that $f$ is open. To show it is dense, consider adding to any $f in L^1$ some function with small $L^1$ norm but large $L^p$ norm.

share|cite|improve this answer

answered Jul 30 at 7:50

Robert Israel

303k22201440

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 7:50

Robert Israel

303k22201440

answered Jul 30 at 7:50

Robert Israel

303k22201440

answered Jul 30 at 7:50

Robert Israel

303k22201440

303k22201440

add a comment | 

add a comment | 

up vote
1
down vote

Proof of the fact that $A_N$ is nowhere dense: Fatou's Lemma shows that $A_N$ is closed in $L^1$; suppose $A_N$ has an interior point $g$. Then there exists $r>0$ such that $||f-g||_1 <r$ implies $f in A_N$. By triangle inequality we get $int_I |f| leq 2N m(I)^1-frac 1 p $ for any interval $I$ whenever $||f||_1 <r$. Take $I=(0,c)$ and $f(x)=ax^alpha$ where $alpha in (-1,-frac 1 p)$ and $0<a<r(1+alpha)$. Then $||f||_1 <r$, and $int_I |f| =int_0^c ax^alpha =a frac c^1+alpha alpha +1$ so $a frac c^1+alpha alpha +1 leq 2N c^1-frac 1 p $. Dividing by $c^1+alpha$ and letting $c to 0$ we get a contradiction.

share|cite|improve this answer

answered Jul 30 at 8:36

Kavi Rama Murthy

19.7k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer. I have a little doubt, when you divide by $c^1+alpha$ and let $cto 0$, you have $int_Iftoinfty$, and this is the contradiction?
  – julios
  Jul 31 at 0:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You get the contradiction $frac a alpha +1=0$.
  – Kavi Rama Murthy
  Jul 31 at 5:26
  

  
  
  
  

add a comment | 

up vote
1
down vote

Proof of the fact that $A_N$ is nowhere dense: Fatou's Lemma shows that $A_N$ is closed in $L^1$; suppose $A_N$ has an interior point $g$. Then there exists $r>0$ such that $||f-g||_1 <r$ implies $f in A_N$. By triangle inequality we get $int_I |f| leq 2N m(I)^1-frac 1 p $ for any interval $I$ whenever $||f||_1 <r$. Take $I=(0,c)$ and $f(x)=ax^alpha$ where $alpha in (-1,-frac 1 p)$ and $0<a<r(1+alpha)$. Then $||f||_1 <r$, and $int_I |f| =int_0^c ax^alpha =a frac c^1+alpha alpha +1$ so $a frac c^1+alpha alpha +1 leq 2N c^1-frac 1 p $. Dividing by $c^1+alpha$ and letting $c to 0$ we get a contradiction.

share|cite|improve this answer

answered Jul 30 at 8:36

Kavi Rama Murthy

19.7k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer. I have a little doubt, when you divide by $c^1+alpha$ and let $cto 0$, you have $int_Iftoinfty$, and this is the contradiction?
  – julios
  Jul 31 at 0:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You get the contradiction $frac a alpha +1=0$.
  – Kavi Rama Murthy
  Jul 31 at 5:26
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Proof of the fact that $A_N$ is nowhere dense: Fatou's Lemma shows that $A_N$ is closed in $L^1$; suppose $A_N$ has an interior point $g$. Then there exists $r>0$ such that $||f-g||_1 <r$ implies $f in A_N$. By triangle inequality we get $int_I |f| leq 2N m(I)^1-frac 1 p $ for any interval $I$ whenever $||f||_1 <r$. Take $I=(0,c)$ and $f(x)=ax^alpha$ where $alpha in (-1,-frac 1 p)$ and $0<a<r(1+alpha)$. Then $||f||_1 <r$, and $int_I |f| =int_0^c ax^alpha =a frac c^1+alpha alpha +1$ so $a frac c^1+alpha alpha +1 leq 2N c^1-frac 1 p $. Dividing by $c^1+alpha$ and letting $c to 0$ we get a contradiction.

share|cite|improve this answer

answered Jul 30 at 8:36

Kavi Rama Murthy

19.7k2829

Proof of the fact that $A_N$ is nowhere dense: Fatou's Lemma shows that $A_N$ is closed in $L^1$; suppose $A_N$ has an interior point $g$. Then there exists $r>0$ such that $||f-g||_1 <r$ implies $f in A_N$. By triangle inequality we get $int_I |f| leq 2N m(I)^1-frac 1 p $ for any interval $I$ whenever $||f||_1 <r$. Take $I=(0,c)$ and $f(x)=ax^alpha$ where $alpha in (-1,-frac 1 p)$ and $0<a<r(1+alpha)$. Then $||f||_1 <r$, and $int_I |f| =int_0^c ax^alpha =a frac c^1+alpha alpha +1$ so $a frac c^1+alpha alpha +1 leq 2N c^1-frac 1 p $. Dividing by $c^1+alpha$ and letting $c to 0$ we get a contradiction.

share|cite|improve this answer

answered Jul 30 at 8:36

Kavi Rama Murthy

19.7k2829

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 8:36

Kavi Rama Murthy

19.7k2829

answered Jul 30 at 8:36

Kavi Rama Murthy

19.7k2829

answered Jul 30 at 8:36

Kavi Rama Murthy

19.7k2829

19.7k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer. I have a little doubt, when you divide by $c^1+alpha$ and let $cto 0$, you have $int_Iftoinfty$, and this is the contradiction?
  – julios
  Jul 31 at 0:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You get the contradiction $frac a alpha +1=0$.
  – Kavi Rama Murthy
  Jul 31 at 5:26
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer. I have a little doubt, when you divide by $c^1+alpha$ and let $cto 0$, you have $int_Iftoinfty$, and this is the contradiction?
  – julios
  Jul 31 at 0:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You get the contradiction $frac a alpha +1=0$.
  – Kavi Rama Murthy
  Jul 31 at 5:26
  

  
  
  
  

Thanks for your answer. I have a little doubt, when you divide by $c^1+alpha$ and let $cto 0$, you have $int_Iftoinfty$, and this is the contradiction?
– julios
Jul 31 at 0:41

Thanks for your answer. I have a little doubt, when you divide by $c^1+alpha$ and let $cto 0$, you have $int_Iftoinfty$, and this is the contradiction?
– julios
Jul 31 at 0:41

You get the contradiction $frac a alpha +1=0$.
– Kavi Rama Murthy
Jul 31 at 5:26

You get the contradiction $frac a alpha +1=0$.
– Kavi Rama Murthy
Jul 31 at 5:26

add a comment | 

 

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