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Confusion caused by the name $L^infty$
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The notation $L^infty(mu)$(the linear normed space made up of essentially bounded measurable functions) in Functional Analysis gives people the impression that the space $L^infty(mu)$ is obtained by finding the "limit" of the sequence $L^1,L^2,...$, however I cannot explain how this works.
Let $(Omega,Sigma,mu)$ be a measure space. A function $f$ is defined to be in $L^infty(mu)$ if $f$ is almost everywhere equal to a bounded function. Therefore, with the norm
$$
|f|_infty=inf_EsubsetOmega, mu(E)=0sup_tin Omega-E |f(t)|
$$
the space $L^infty(mu)$ is a Linear normed space.
I wish to see whether or not $lim_p to infty L^p=L^infty$. I begin by a special case, $ell^p$ spaces.
It is clear that $ell^1 subset ell^p subset ell^q subset ell^infty,q>p$. Let's consider the sequence $ell^1,ell^2,ell^3,...$, so $lim ell^p=bigcup_i=0^infty ell^i$, which is a linear space. However, $bigcup_i=0^infty ell^i$ doesn't include some elements in $ell^infty$. For example, the sequence $(1,1,1,...)$ lies in $ell^infty$, but not in $bigcup_i=0^infty ell^i$, because $(1,1,1,...)$ doesn't belong to any $ell^p, p<infty$.
So How can I explain why the notation $L^infty$ or $ell^infty$ are used?
real-analysis functional-analysis
asked Jul 29 at 13:10
Ma Joad
722216
add a comment |Â
up vote
2
down vote
favorite
The notation $L^infty(mu)$(the linear normed space made up of essentially bounded measurable functions) in Functional Analysis gives people the impression that the space $L^infty(mu)$ is obtained by finding the "limit" of the sequence $L^1,L^2,...$, however I cannot explain how this works.
Let $(Omega,Sigma,mu)$ be a measure space. A function $f$ is defined to be in $L^infty(mu)$ if $f$ is almost everywhere equal to a bounded function. Therefore, with the norm
$$
|f|_infty=inf_EsubsetOmega, mu(E)=0sup_tin Omega-E |f(t)|
$$
the space $L^infty(mu)$ is a Linear normed space.
I wish to see whether or not $lim_p to infty L^p=L^infty$. I begin by a special case, $ell^p$ spaces.
It is clear that $ell^1 subset ell^p subset ell^q subset ell^infty,q>p$. Let's consider the sequence $ell^1,ell^2,ell^3,...$, so $lim ell^p=bigcup_i=0^infty ell^i$, which is a linear space. However, $bigcup_i=0^infty ell^i$ doesn't include some elements in $ell^infty$. For example, the sequence $(1,1,1,...)$ lies in $ell^infty$, but not in $bigcup_i=0^infty ell^i$, because $(1,1,1,...)$ doesn't belong to any $ell^p, p<infty$.
So How can I explain why the notation $L^infty$ or $ell^infty$ are used?
real-analysis functional-analysis
asked Jul 29 at 13:10
Ma Joad
722216
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The notation $L^infty(mu)$(the linear normed space made up of essentially bounded measurable functions) in Functional Analysis gives people the impression that the space $L^infty(mu)$ is obtained by finding the "limit" of the sequence $L^1,L^2,...$, however I cannot explain how this works.
Let $(Omega,Sigma,mu)$ be a measure space. A function $f$ is defined to be in $L^infty(mu)$ if $f$ is almost everywhere equal to a bounded function. Therefore, with the norm
$$
|f|_infty=inf_EsubsetOmega, mu(E)=0sup_tin Omega-E |f(t)|
$$
the space $L^infty(mu)$ is a Linear normed space.
I wish to see whether or not $lim_p to infty L^p=L^infty$. I begin by a special case, $ell^p$ spaces.
It is clear that $ell^1 subset ell^p subset ell^q subset ell^infty,q>p$. Let's consider the sequence $ell^1,ell^2,ell^3,...$, so $lim ell^p=bigcup_i=0^infty ell^i$, which is a linear space. However, $bigcup_i=0^infty ell^i$ doesn't include some elements in $ell^infty$. For example, the sequence $(1,1,1,...)$ lies in $ell^infty$, but not in $bigcup_i=0^infty ell^i$, because $(1,1,1,...)$ doesn't belong to any $ell^p, p<infty$.
So How can I explain why the notation $L^infty$ or $ell^infty$ are used?
real-analysis functional-analysis
asked Jul 29 at 13:10
Ma Joad
722216
The notation $L^infty(mu)$(the linear normed space made up of essentially bounded measurable functions) in Functional Analysis gives people the impression that the space $L^infty(mu)$ is obtained by finding the "limit" of the sequence $L^1,L^2,...$, however I cannot explain how this works.
Let $(Omega,Sigma,mu)$ be a measure space. A function $f$ is defined to be in $L^infty(mu)$ if $f$ is almost everywhere equal to a bounded function. Therefore, with the norm
$$
|f|_infty=inf_EsubsetOmega, mu(E)=0sup_tin Omega-E |f(t)|
$$
the space $L^infty(mu)$ is a Linear normed space.
I wish to see whether or not $lim_p to infty L^p=L^infty$. I begin by a special case, $ell^p$ spaces.
It is clear that $ell^1 subset ell^p subset ell^q subset ell^infty,q>p$. Let's consider the sequence $ell^1,ell^2,ell^3,...$, so $lim ell^p=bigcup_i=0^infty ell^i$, which is a linear space. However, $bigcup_i=0^infty ell^i$ doesn't include some elements in $ell^infty$. For example, the sequence $(1,1,1,...)$ lies in $ell^infty$, but not in $bigcup_i=0^infty ell^i$, because $(1,1,1,...)$ doesn't belong to any $ell^p, p<infty$.
So How can I explain why the notation $L^infty$ or $ell^infty$ are used?
real-analysis functional-analysis
asked Jul 29 at 13:10
Ma Joad
722216
asked Jul 29 at 13:10
Ma Joad
722216
asked Jul 29 at 13:10
Ma Joad
722216
asked Jul 29 at 13:10
Ma Joad
722216
722216
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
If $Omega$ is any measure space, and $fin L^p(Omega)cap L^infty(Omega)$ for some $pin [1,infty)$, then $fin L^q(Omega)$ for all $qin [p,infty]$ and
$$lim_qto infty|f|_L^q(Omega)=|f|_L^infty(Omega) $$
So the limit of the $L^q$ norms is the $L^infty$ norms. In this sense, the space $L^infty$ is the limit of the $L^q$ spaces as $qto infty$.
To prove this result one might start from the Riesz-Thorin interpolation inequality:
If $Omega$ is a measure space and $fin L^p(Omega)cap L^q(Omega)$, with $1leq p<qleqinfty$, then $fin L^r(Omega)$ for all $r$ with $pleq rleq q$ and
$$ |f|_L^r(Omega)leq |f|_L^p(Omega)^lambda|f|_L^q(Omega)^1-lambda$$
Where $lambdain [0,1]$ is such that
$$frac1r=fraclambdap+frac1-lambdaq $$
answered Jul 29 at 13:16
Lorenzo Quarisa
2,249314
Could you give me some hint on how to write the proof?
– Ma Joad
Jul 29 at 14:30
Prove that $liminf_qto infty|f|_qgeq |f|_infty$ and then $limsup_qto infty|f|_qleq |f|_infty$ separatedly. For the latter inequality ($limsup$) you might find the Riesz-Thorin interpolation inequality particularly helpful.
– Lorenzo Quarisa
Jul 29 at 16:32
add a comment |Â
up vote
1
down vote
Lorenzo Quarisa's answer contains the essential point: Think of the limit of the norms, not of the spaces. Let me write out how this works in the simple example of the $ell^p$ spaces, and let me begin with an even simpler (but incomplete) vector space included in all of the $ell^p$ spaces, namely the space $F$ of sequences with only finitely many non-zero terms.
Consider a sequence $xin F$, and let $n$ be the number of its non-zero components. Also, let $m$ be the largest of the absolute values of the terms $x_i$ in $x$. Then the $p$-norm is bounded above and below in terms of $m$:
$$
mleq Vert xVert=left(sum_i|x_i|^pright)^1/pleq n^1/pm.
$$
(The first inequality is because $m^p$ is one of the summands; the second is because there are only $n$ summands, each majorized by $m^p$.) If we let $ptoinfty$ while keeping $x$ and therefore $n$ fixed, we have $n^1/pto1$ and therefore $Vert xVertto m$. That is, the $p$-norm approaches the $infty$-norm, for each fixed $x$.
It is in this sense that the $infty$-norm is the limit of the $p$-norms. EDIT: The next sentence is wrong in the case of $ell^infty$; see correction at the end. As for the $ell^p$ spaces, notice that they are the completions of $F$ with respect to the $p$-norms. So the limiting behavior of the norms is ascribed (guilt by association?) to the spaces.
CORRECTION: The closure of $F$ in the $infty$-norm is not $ell^infty$ but the subspace $c_0$ of sequences that converge to zero. So to get this approach to work, one would need to double-dualize, which has no effect on $ell^p$ for $1<p<infty$ but sends $c_0$ to $ell^infty$.
answered Jul 29 at 15:59
Andreas Blass
47.5k348104
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
If $Omega$ is any measure space, and $fin L^p(Omega)cap L^infty(Omega)$ for some $pin [1,infty)$, then $fin L^q(Omega)$ for all $qin [p,infty]$ and
$$lim_qto infty|f|_L^q(Omega)=|f|_L^infty(Omega) $$
So the limit of the $L^q$ norms is the $L^infty$ norms. In this sense, the space $L^infty$ is the limit of the $L^q$ spaces as $qto infty$.
To prove this result one might start from the Riesz-Thorin interpolation inequality:
If $Omega$ is a measure space and $fin L^p(Omega)cap L^q(Omega)$, with $1leq p<qleqinfty$, then $fin L^r(Omega)$ for all $r$ with $pleq rleq q$ and
$$ |f|_L^r(Omega)leq |f|_L^p(Omega)^lambda|f|_L^q(Omega)^1-lambda$$
Where $lambdain [0,1]$ is such that
$$frac1r=fraclambdap+frac1-lambdaq $$
answered Jul 29 at 13:16
Lorenzo Quarisa
2,249314
Could you give me some hint on how to write the proof?
– Ma Joad
Jul 29 at 14:30
Prove that $liminf_qto infty|f|_qgeq |f|_infty$ and then $limsup_qto infty|f|_qleq |f|_infty$ separatedly. For the latter inequality ($limsup$) you might find the Riesz-Thorin interpolation inequality particularly helpful.
– Lorenzo Quarisa
Jul 29 at 16:32
add a comment |Â
up vote
2
down vote
If $Omega$ is any measure space, and $fin L^p(Omega)cap L^infty(Omega)$ for some $pin [1,infty)$, then $fin L^q(Omega)$ for all $qin [p,infty]$ and
$$lim_qto infty|f|_L^q(Omega)=|f|_L^infty(Omega) $$
So the limit of the $L^q$ norms is the $L^infty$ norms. In this sense, the space $L^infty$ is the limit of the $L^q$ spaces as $qto infty$.
To prove this result one might start from the Riesz-Thorin interpolation inequality:
If $Omega$ is a measure space and $fin L^p(Omega)cap L^q(Omega)$, with $1leq p<qleqinfty$, then $fin L^r(Omega)$ for all $r$ with $pleq rleq q$ and
$$ |f|_L^r(Omega)leq |f|_L^p(Omega)^lambda|f|_L^q(Omega)^1-lambda$$
Where $lambdain [0,1]$ is such that
$$frac1r=fraclambdap+frac1-lambdaq $$
answered Jul 29 at 13:16
Lorenzo Quarisa
2,249314
Could you give me some hint on how to write the proof?
– Ma Joad
Jul 29 at 14:30
Prove that $liminf_qto infty|f|_qgeq |f|_infty$ and then $limsup_qto infty|f|_qleq |f|_infty$ separatedly. For the latter inequality ($limsup$) you might find the Riesz-Thorin interpolation inequality particularly helpful.
– Lorenzo Quarisa
Jul 29 at 16:32
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If $Omega$ is any measure space, and $fin L^p(Omega)cap L^infty(Omega)$ for some $pin [1,infty)$, then $fin L^q(Omega)$ for all $qin [p,infty]$ and
$$lim_qto infty|f|_L^q(Omega)=|f|_L^infty(Omega) $$
So the limit of the $L^q$ norms is the $L^infty$ norms. In this sense, the space $L^infty$ is the limit of the $L^q$ spaces as $qto infty$.
To prove this result one might start from the Riesz-Thorin interpolation inequality:
If $Omega$ is a measure space and $fin L^p(Omega)cap L^q(Omega)$, with $1leq p<qleqinfty$, then $fin L^r(Omega)$ for all $r$ with $pleq rleq q$ and
$$ |f|_L^r(Omega)leq |f|_L^p(Omega)^lambda|f|_L^q(Omega)^1-lambda$$
Where $lambdain [0,1]$ is such that
$$frac1r=fraclambdap+frac1-lambdaq $$
answered Jul 29 at 13:16
Lorenzo Quarisa
2,249314
If $Omega$ is any measure space, and $fin L^p(Omega)cap L^infty(Omega)$ for some $pin [1,infty)$, then $fin L^q(Omega)$ for all $qin [p,infty]$ and
$$lim_qto infty|f|_L^q(Omega)=|f|_L^infty(Omega) $$
So the limit of the $L^q$ norms is the $L^infty$ norms. In this sense, the space $L^infty$ is the limit of the $L^q$ spaces as $qto infty$.
To prove this result one might start from the Riesz-Thorin interpolation inequality:
If $Omega$ is a measure space and $fin L^p(Omega)cap L^q(Omega)$, with $1leq p<qleqinfty$, then $fin L^r(Omega)$ for all $r$ with $pleq rleq q$ and
$$ |f|_L^r(Omega)leq |f|_L^p(Omega)^lambda|f|_L^q(Omega)^1-lambda$$
Where $lambdain [0,1]$ is such that
$$frac1r=fraclambdap+frac1-lambdaq $$
answered Jul 29 at 13:16
Lorenzo Quarisa
2,249314
edited Jul 29 at 13:22
answered Jul 29 at 13:16
Lorenzo Quarisa
2,249314
answered Jul 29 at 13:16
Lorenzo Quarisa
2,249314
answered Jul 29 at 13:16
Lorenzo Quarisa
2,249314
2,249314
Could you give me some hint on how to write the proof?
– Ma Joad
Jul 29 at 14:30
Prove that $liminf_qto infty|f|_qgeq |f|_infty$ and then $limsup_qto infty|f|_qleq |f|_infty$ separatedly. For the latter inequality ($limsup$) you might find the Riesz-Thorin interpolation inequality particularly helpful.
– Lorenzo Quarisa
Jul 29 at 16:32
add a comment |Â
Could you give me some hint on how to write the proof?
– Ma Joad
Jul 29 at 14:30
Prove that $liminf_qto infty|f|_qgeq |f|_infty$ and then $limsup_qto infty|f|_qleq |f|_infty$ separatedly. For the latter inequality ($limsup$) you might find the Riesz-Thorin interpolation inequality particularly helpful.
– Lorenzo Quarisa
Jul 29 at 16:32
Could you give me some hint on how to write the proof?
– Ma Joad
Jul 29 at 14:30
Could you give me some hint on how to write the proof?
– Ma Joad
Jul 29 at 14:30
Prove that $liminf_qto infty|f|_qgeq |f|_infty$ and then $limsup_qto infty|f|_qleq |f|_infty$ separatedly. For the latter inequality ($limsup$) you might find the Riesz-Thorin interpolation inequality particularly helpful.
– Lorenzo Quarisa
Jul 29 at 16:32
Prove that $liminf_qto infty|f|_qgeq |f|_infty$ and then $limsup_qto infty|f|_qleq |f|_infty$ separatedly. For the latter inequality ($limsup$) you might find the Riesz-Thorin interpolation inequality particularly helpful.
– Lorenzo Quarisa
Jul 29 at 16:32
add a comment |Â
up vote
1
down vote
Lorenzo Quarisa's answer contains the essential point: Think of the limit of the norms, not of the spaces. Let me write out how this works in the simple example of the $ell^p$ spaces, and let me begin with an even simpler (but incomplete) vector space included in all of the $ell^p$ spaces, namely the space $F$ of sequences with only finitely many non-zero terms.
Consider a sequence $xin F$, and let $n$ be the number of its non-zero components. Also, let $m$ be the largest of the absolute values of the terms $x_i$ in $x$. Then the $p$-norm is bounded above and below in terms of $m$:
$$
mleq Vert xVert=left(sum_i|x_i|^pright)^1/pleq n^1/pm.
$$
(The first inequality is because $m^p$ is one of the summands; the second is because there are only $n$ summands, each majorized by $m^p$.) If we let $ptoinfty$ while keeping $x$ and therefore $n$ fixed, we have $n^1/pto1$ and therefore $Vert xVertto m$. That is, the $p$-norm approaches the $infty$-norm, for each fixed $x$.
It is in this sense that the $infty$-norm is the limit of the $p$-norms. EDIT: The next sentence is wrong in the case of $ell^infty$; see correction at the end. As for the $ell^p$ spaces, notice that they are the completions of $F$ with respect to the $p$-norms. So the limiting behavior of the norms is ascribed (guilt by association?) to the spaces.
CORRECTION: The closure of $F$ in the $infty$-norm is not $ell^infty$ but the subspace $c_0$ of sequences that converge to zero. So to get this approach to work, one would need to double-dualize, which has no effect on $ell^p$ for $1<p<infty$ but sends $c_0$ to $ell^infty$.
answered Jul 29 at 15:59
Andreas Blass
47.5k348104
add a comment |Â
up vote
1
down vote
Lorenzo Quarisa's answer contains the essential point: Think of the limit of the norms, not of the spaces. Let me write out how this works in the simple example of the $ell^p$ spaces, and let me begin with an even simpler (but incomplete) vector space included in all of the $ell^p$ spaces, namely the space $F$ of sequences with only finitely many non-zero terms.
Consider a sequence $xin F$, and let $n$ be the number of its non-zero components. Also, let $m$ be the largest of the absolute values of the terms $x_i$ in $x$. Then the $p$-norm is bounded above and below in terms of $m$:
$$
mleq Vert xVert=left(sum_i|x_i|^pright)^1/pleq n^1/pm.
$$
(The first inequality is because $m^p$ is one of the summands; the second is because there are only $n$ summands, each majorized by $m^p$.) If we let $ptoinfty$ while keeping $x$ and therefore $n$ fixed, we have $n^1/pto1$ and therefore $Vert xVertto m$. That is, the $p$-norm approaches the $infty$-norm, for each fixed $x$.
It is in this sense that the $infty$-norm is the limit of the $p$-norms. EDIT: The next sentence is wrong in the case of $ell^infty$; see correction at the end. As for the $ell^p$ spaces, notice that they are the completions of $F$ with respect to the $p$-norms. So the limiting behavior of the norms is ascribed (guilt by association?) to the spaces.
CORRECTION: The closure of $F$ in the $infty$-norm is not $ell^infty$ but the subspace $c_0$ of sequences that converge to zero. So to get this approach to work, one would need to double-dualize, which has no effect on $ell^p$ for $1<p<infty$ but sends $c_0$ to $ell^infty$.
answered Jul 29 at 15:59
Andreas Blass
47.5k348104
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Lorenzo Quarisa's answer contains the essential point: Think of the limit of the norms, not of the spaces. Let me write out how this works in the simple example of the $ell^p$ spaces, and let me begin with an even simpler (but incomplete) vector space included in all of the $ell^p$ spaces, namely the space $F$ of sequences with only finitely many non-zero terms.
Consider a sequence $xin F$, and let $n$ be the number of its non-zero components. Also, let $m$ be the largest of the absolute values of the terms $x_i$ in $x$. Then the $p$-norm is bounded above and below in terms of $m$:
$$
mleq Vert xVert=left(sum_i|x_i|^pright)^1/pleq n^1/pm.
$$
(The first inequality is because $m^p$ is one of the summands; the second is because there are only $n$ summands, each majorized by $m^p$.) If we let $ptoinfty$ while keeping $x$ and therefore $n$ fixed, we have $n^1/pto1$ and therefore $Vert xVertto m$. That is, the $p$-norm approaches the $infty$-norm, for each fixed $x$.
It is in this sense that the $infty$-norm is the limit of the $p$-norms. EDIT: The next sentence is wrong in the case of $ell^infty$; see correction at the end. As for the $ell^p$ spaces, notice that they are the completions of $F$ with respect to the $p$-norms. So the limiting behavior of the norms is ascribed (guilt by association?) to the spaces.
CORRECTION: The closure of $F$ in the $infty$-norm is not $ell^infty$ but the subspace $c_0$ of sequences that converge to zero. So to get this approach to work, one would need to double-dualize, which has no effect on $ell^p$ for $1<p<infty$ but sends $c_0$ to $ell^infty$.
answered Jul 29 at 15:59
Andreas Blass
47.5k348104
Lorenzo Quarisa's answer contains the essential point: Think of the limit of the norms, not of the spaces. Let me write out how this works in the simple example of the $ell^p$ spaces, and let me begin with an even simpler (but incomplete) vector space included in all of the $ell^p$ spaces, namely the space $F$ of sequences with only finitely many non-zero terms.
Consider a sequence $xin F$, and let $n$ be the number of its non-zero components. Also, let $m$ be the largest of the absolute values of the terms $x_i$ in $x$. Then the $p$-norm is bounded above and below in terms of $m$:
$$
mleq Vert xVert=left(sum_i|x_i|^pright)^1/pleq n^1/pm.
$$
(The first inequality is because $m^p$ is one of the summands; the second is because there are only $n$ summands, each majorized by $m^p$.) If we let $ptoinfty$ while keeping $x$ and therefore $n$ fixed, we have $n^1/pto1$ and therefore $Vert xVertto m$. That is, the $p$-norm approaches the $infty$-norm, for each fixed $x$.
It is in this sense that the $infty$-norm is the limit of the $p$-norms. EDIT: The next sentence is wrong in the case of $ell^infty$; see correction at the end. As for the $ell^p$ spaces, notice that they are the completions of $F$ with respect to the $p$-norms. So the limiting behavior of the norms is ascribed (guilt by association?) to the spaces.
CORRECTION: The closure of $F$ in the $infty$-norm is not $ell^infty$ but the subspace $c_0$ of sequences that converge to zero. So to get this approach to work, one would need to double-dualize, which has no effect on $ell^p$ for $1<p<infty$ but sends $c_0$ to $ell^infty$.
answered Jul 29 at 15:59
Andreas Blass
47.5k348104
edited Jul 29 at 19:05
answered Jul 29 at 15:59
Andreas Blass
47.5k348104
answered Jul 29 at 15:59
Andreas Blass
47.5k348104
answered Jul 29 at 15:59
Andreas Blass
47.5k348104
47.5k348104
add a comment |Â
add a comment |Â
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