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Intuition about L^p spaces
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I have read somewhere the following very nice intuition about $L^p(mathbbR)$ spaces.
This graphic shows a lot of nice relations:
1) There is no inclusion between $L^p$ and $L^q$
2) $L^p$ is the dual of $L^q$ for $frac1p + frac1q = 1$, since the boxes have the same shape, but just rotated
3) $L^2$ is self-dual
4) If $p<r<q$ and $fin L^pcap L^q$, then $fin L^r$.
Furthermore, the projections on the two distinct dimensions give us other cases. So while the full two-dimensional picture is the most abstract case $L^p(X,mathcalA,mu)$. The first dimension shows the case of finite measure spaces, the second dimension shows the case of measure spaces which do not have sets of arbitrarily small nonzero measure.
Now perhaps this is just a nice coincidence, and not more. But, and I'm sorry if this question is too vague, is there perhaps any deeper reason or theory which explains this picture?
fa.functional-analysis
edited 6 hours ago
Gerald Edgar
27.7k273154
asked 9 hours ago
cccdi
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up vote
8
down vote
favorite
I have read somewhere the following very nice intuition about $L^p(mathbbR)$ spaces.
This graphic shows a lot of nice relations:
1) There is no inclusion between $L^p$ and $L^q$
2) $L^p$ is the dual of $L^q$ for $frac1p + frac1q = 1$, since the boxes have the same shape, but just rotated
3) $L^2$ is self-dual
4) If $p<r<q$ and $fin L^pcap L^q$, then $fin L^r$.
Furthermore, the projections on the two distinct dimensions give us other cases. So while the full two-dimensional picture is the most abstract case $L^p(X,mathcalA,mu)$. The first dimension shows the case of finite measure spaces, the second dimension shows the case of measure spaces which do not have sets of arbitrarily small nonzero measure.
Now perhaps this is just a nice coincidence, and not more. But, and I'm sorry if this question is too vague, is there perhaps any deeper reason or theory which explains this picture?
fa.functional-analysis
edited 6 hours ago
Gerald Edgar
27.7k273154
asked 9 hours ago
cccdi
441
New contributor
cccdi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Very roughly the picture you drew has some similarities with the picture I have of interpolation spaces in my head, especially with the real interpolation methods.
– Willie Wong
3 hours ago
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
I have read somewhere the following very nice intuition about $L^p(mathbbR)$ spaces.
This graphic shows a lot of nice relations:
1) There is no inclusion between $L^p$ and $L^q$
2) $L^p$ is the dual of $L^q$ for $frac1p + frac1q = 1$, since the boxes have the same shape, but just rotated
3) $L^2$ is self-dual
4) If $p<r<q$ and $fin L^pcap L^q$, then $fin L^r$.
Furthermore, the projections on the two distinct dimensions give us other cases. So while the full two-dimensional picture is the most abstract case $L^p(X,mathcalA,mu)$. The first dimension shows the case of finite measure spaces, the second dimension shows the case of measure spaces which do not have sets of arbitrarily small nonzero measure.
Now perhaps this is just a nice coincidence, and not more. But, and I'm sorry if this question is too vague, is there perhaps any deeper reason or theory which explains this picture?
fa.functional-analysis
edited 6 hours ago
Gerald Edgar
27.7k273154
asked 9 hours ago
cccdi
441
New contributor
cccdi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have read somewhere the following very nice intuition about $L^p(mathbbR)$ spaces.
This graphic shows a lot of nice relations:
1) There is no inclusion between $L^p$ and $L^q$
2) $L^p$ is the dual of $L^q$ for $frac1p + frac1q = 1$, since the boxes have the same shape, but just rotated
3) $L^2$ is self-dual
4) If $p<r<q$ and $fin L^pcap L^q$, then $fin L^r$.
Furthermore, the projections on the two distinct dimensions give us other cases. So while the full two-dimensional picture is the most abstract case $L^p(X,mathcalA,mu)$. The first dimension shows the case of finite measure spaces, the second dimension shows the case of measure spaces which do not have sets of arbitrarily small nonzero measure.
Now perhaps this is just a nice coincidence, and not more. But, and I'm sorry if this question is too vague, is there perhaps any deeper reason or theory which explains this picture?
fa.functional-analysis
fa.functional-analysis
edited 6 hours ago
Gerald Edgar
27.7k273154
asked 9 hours ago
cccdi
441
New contributor
cccdi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
Gerald Edgar
27.7k273154
asked 9 hours ago
cccdi
441
New contributor
cccdi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
Gerald Edgar
27.7k273154
edited 6 hours ago
Gerald Edgar
27.7k273154
edited 6 hours ago
Gerald Edgar
27.7k273154
27.7k273154
asked 9 hours ago
cccdi
441
New contributor
cccdi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
cccdi
441
asked 9 hours ago
cccdi
441
441
New contributor
cccdi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
cccdi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
cccdi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Very roughly the picture you drew has some similarities with the picture I have of interpolation spaces in my head, especially with the real interpolation methods.
– Willie Wong
3 hours ago
add a comment |Â
1
Very roughly the picture you drew has some similarities with the picture I have of interpolation spaces in my head, especially with the real interpolation methods.
– Willie Wong
3 hours ago
1
1
Very roughly the picture you drew has some similarities with the picture I have of interpolation spaces in my head, especially with the real interpolation methods.
– Willie Wong
3 hours ago
Very roughly the picture you drew has some similarities with the picture I have of interpolation spaces in my head, especially with the real interpolation methods.
– Willie Wong
3 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
Perhaps this is more basic than what you're asking, but for me, the intuition for the relationships between $L^p$ spaces (which your diagram encapsulates) is as follows:
$fin L^p$ tells us two things: $f$ decays fast enough at infinity, and $f$ doesn't blow up too fast at any finite point. As $p$ increases, the first requirement gets less strict, and the second gets more strict. (Also, if $X$ is bounded, the first requirement is vacuous, and if $X$ is discrete, the second is vacuous.) That tells you why the inclusions work the way they do. For duality, the boundedness of a linear functional on $L^p$ comes down to whether $fg$ is in $L^1$ for some $f$ and $g$--the slower $f$ decays, the faster $g$ needs to decay, and the faster $f$ blows up, the more we need $g$ to not blow up. The previous sentence is equally true if we switch $f$ and $g$, which convinces us that $L^p$ spaces should be reflexive, and by considering $f=g$, we see that the restrictions on $f$ and $g$ should balance at exactly $p=2$, so $L^2$ should be self-dual. More generally, if $fin L^p$, then $fcdot f^p-1$ is in $L^1$, so $f^p-1$ acts as a linear functional on $f$, and since $fin L^p$, we have $f^p-1 in L^q$ with $q = p/(p-1)$, or in other words, $frac 1 p + frac 1 q = 1$. (One can also infer the right relationship between $p$ and $q$ by scaling Holder's inequality, but I like this better.)
answered 55 mins ago
Stanley Snelson
1314
add a comment |Â
up vote
3
down vote
Not an answer, but there are other pictures for the $L^p$-spaces that may be helpful. Hans Georg Feichtinger is known for his diagrams for various functions spaces - here are a few:
oth from Feichtinger's talk "Function spaces for time-frequency analysis: the usefulness of a Banach Gelfand Triple", https://www.univie.ac.at/nuhag-php/dateien/talks/3397_Varan28JanFei.pdf If I remember correctly, $S^0$ is the Feichtinger algebra and $FL^1$ is the Fourier transform of $L^1$. $C^0$ is the space of continuous functions vanishing at infinity, "Schw" the space of Schwartz functions.
From Feichtinger's talk "The Vast Family of Fourier Standard Spaces (invariant function spaces between S0 and S′0)" https://www.univie.ac.at/nuhag-php/dateien/talks/3318_TrondheimMay17A.pdf
answered 5 hours ago
Dirk
7,06443264
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Perhaps this is more basic than what you're asking, but for me, the intuition for the relationships between $L^p$ spaces (which your diagram encapsulates) is as follows:
$fin L^p$ tells us two things: $f$ decays fast enough at infinity, and $f$ doesn't blow up too fast at any finite point. As $p$ increases, the first requirement gets less strict, and the second gets more strict. (Also, if $X$ is bounded, the first requirement is vacuous, and if $X$ is discrete, the second is vacuous.) That tells you why the inclusions work the way they do. For duality, the boundedness of a linear functional on $L^p$ comes down to whether $fg$ is in $L^1$ for some $f$ and $g$--the slower $f$ decays, the faster $g$ needs to decay, and the faster $f$ blows up, the more we need $g$ to not blow up. The previous sentence is equally true if we switch $f$ and $g$, which convinces us that $L^p$ spaces should be reflexive, and by considering $f=g$, we see that the restrictions on $f$ and $g$ should balance at exactly $p=2$, so $L^2$ should be self-dual. More generally, if $fin L^p$, then $fcdot f^p-1$ is in $L^1$, so $f^p-1$ acts as a linear functional on $f$, and since $fin L^p$, we have $f^p-1 in L^q$ with $q = p/(p-1)$, or in other words, $frac 1 p + frac 1 q = 1$. (One can also infer the right relationship between $p$ and $q$ by scaling Holder's inequality, but I like this better.)
answered 55 mins ago
Stanley Snelson
1314
add a comment |Â
up vote
4
down vote
Perhaps this is more basic than what you're asking, but for me, the intuition for the relationships between $L^p$ spaces (which your diagram encapsulates) is as follows:
$fin L^p$ tells us two things: $f$ decays fast enough at infinity, and $f$ doesn't blow up too fast at any finite point. As $p$ increases, the first requirement gets less strict, and the second gets more strict. (Also, if $X$ is bounded, the first requirement is vacuous, and if $X$ is discrete, the second is vacuous.) That tells you why the inclusions work the way they do. For duality, the boundedness of a linear functional on $L^p$ comes down to whether $fg$ is in $L^1$ for some $f$ and $g$--the slower $f$ decays, the faster $g$ needs to decay, and the faster $f$ blows up, the more we need $g$ to not blow up. The previous sentence is equally true if we switch $f$ and $g$, which convinces us that $L^p$ spaces should be reflexive, and by considering $f=g$, we see that the restrictions on $f$ and $g$ should balance at exactly $p=2$, so $L^2$ should be self-dual. More generally, if $fin L^p$, then $fcdot f^p-1$ is in $L^1$, so $f^p-1$ acts as a linear functional on $f$, and since $fin L^p$, we have $f^p-1 in L^q$ with $q = p/(p-1)$, or in other words, $frac 1 p + frac 1 q = 1$. (One can also infer the right relationship between $p$ and $q$ by scaling Holder's inequality, but I like this better.)
answered 55 mins ago
Stanley Snelson
1314
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Perhaps this is more basic than what you're asking, but for me, the intuition for the relationships between $L^p$ spaces (which your diagram encapsulates) is as follows:
$fin L^p$ tells us two things: $f$ decays fast enough at infinity, and $f$ doesn't blow up too fast at any finite point. As $p$ increases, the first requirement gets less strict, and the second gets more strict. (Also, if $X$ is bounded, the first requirement is vacuous, and if $X$ is discrete, the second is vacuous.) That tells you why the inclusions work the way they do. For duality, the boundedness of a linear functional on $L^p$ comes down to whether $fg$ is in $L^1$ for some $f$ and $g$--the slower $f$ decays, the faster $g$ needs to decay, and the faster $f$ blows up, the more we need $g$ to not blow up. The previous sentence is equally true if we switch $f$ and $g$, which convinces us that $L^p$ spaces should be reflexive, and by considering $f=g$, we see that the restrictions on $f$ and $g$ should balance at exactly $p=2$, so $L^2$ should be self-dual. More generally, if $fin L^p$, then $fcdot f^p-1$ is in $L^1$, so $f^p-1$ acts as a linear functional on $f$, and since $fin L^p$, we have $f^p-1 in L^q$ with $q = p/(p-1)$, or in other words, $frac 1 p + frac 1 q = 1$. (One can also infer the right relationship between $p$ and $q$ by scaling Holder's inequality, but I like this better.)
answered 55 mins ago
Stanley Snelson
1314
Perhaps this is more basic than what you're asking, but for me, the intuition for the relationships between $L^p$ spaces (which your diagram encapsulates) is as follows:
$fin L^p$ tells us two things: $f$ decays fast enough at infinity, and $f$ doesn't blow up too fast at any finite point. As $p$ increases, the first requirement gets less strict, and the second gets more strict. (Also, if $X$ is bounded, the first requirement is vacuous, and if $X$ is discrete, the second is vacuous.) That tells you why the inclusions work the way they do. For duality, the boundedness of a linear functional on $L^p$ comes down to whether $fg$ is in $L^1$ for some $f$ and $g$--the slower $f$ decays, the faster $g$ needs to decay, and the faster $f$ blows up, the more we need $g$ to not blow up. The previous sentence is equally true if we switch $f$ and $g$, which convinces us that $L^p$ spaces should be reflexive, and by considering $f=g$, we see that the restrictions on $f$ and $g$ should balance at exactly $p=2$, so $L^2$ should be self-dual. More generally, if $fin L^p$, then $fcdot f^p-1$ is in $L^1$, so $f^p-1$ acts as a linear functional on $f$, and since $fin L^p$, we have $f^p-1 in L^q$ with $q = p/(p-1)$, or in other words, $frac 1 p + frac 1 q = 1$. (One can also infer the right relationship between $p$ and $q$ by scaling Holder's inequality, but I like this better.)
answered 55 mins ago
Stanley Snelson
1314
answered 55 mins ago
Stanley Snelson
1314
answered 55 mins ago
Stanley Snelson
1314
answered 55 mins ago
Stanley Snelson
1314
1314
add a comment |Â
add a comment |Â
up vote
3
down vote
Not an answer, but there are other pictures for the $L^p$-spaces that may be helpful. Hans Georg Feichtinger is known for his diagrams for various functions spaces - here are a few:
oth from Feichtinger's talk "Function spaces for time-frequency analysis: the usefulness of a Banach Gelfand Triple", https://www.univie.ac.at/nuhag-php/dateien/talks/3397_Varan28JanFei.pdf If I remember correctly, $S^0$ is the Feichtinger algebra and $FL^1$ is the Fourier transform of $L^1$. $C^0$ is the space of continuous functions vanishing at infinity, "Schw" the space of Schwartz functions.
From Feichtinger's talk "The Vast Family of Fourier Standard Spaces (invariant function spaces between S0 and S′0)" https://www.univie.ac.at/nuhag-php/dateien/talks/3318_TrondheimMay17A.pdf
answered 5 hours ago
Dirk
7,06443264
add a comment |Â
up vote
3
down vote
Not an answer, but there are other pictures for the $L^p$-spaces that may be helpful. Hans Georg Feichtinger is known for his diagrams for various functions spaces - here are a few:
oth from Feichtinger's talk "Function spaces for time-frequency analysis: the usefulness of a Banach Gelfand Triple", https://www.univie.ac.at/nuhag-php/dateien/talks/3397_Varan28JanFei.pdf If I remember correctly, $S^0$ is the Feichtinger algebra and $FL^1$ is the Fourier transform of $L^1$. $C^0$ is the space of continuous functions vanishing at infinity, "Schw" the space of Schwartz functions.
From Feichtinger's talk "The Vast Family of Fourier Standard Spaces (invariant function spaces between S0 and S′0)" https://www.univie.ac.at/nuhag-php/dateien/talks/3318_TrondheimMay17A.pdf
answered 5 hours ago
Dirk
7,06443264
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Not an answer, but there are other pictures for the $L^p$-spaces that may be helpful. Hans Georg Feichtinger is known for his diagrams for various functions spaces - here are a few:
oth from Feichtinger's talk "Function spaces for time-frequency analysis: the usefulness of a Banach Gelfand Triple", https://www.univie.ac.at/nuhag-php/dateien/talks/3397_Varan28JanFei.pdf If I remember correctly, $S^0$ is the Feichtinger algebra and $FL^1$ is the Fourier transform of $L^1$. $C^0$ is the space of continuous functions vanishing at infinity, "Schw" the space of Schwartz functions.
From Feichtinger's talk "The Vast Family of Fourier Standard Spaces (invariant function spaces between S0 and S′0)" https://www.univie.ac.at/nuhag-php/dateien/talks/3318_TrondheimMay17A.pdf
answered 5 hours ago
Dirk
7,06443264
Not an answer, but there are other pictures for the $L^p$-spaces that may be helpful. Hans Georg Feichtinger is known for his diagrams for various functions spaces - here are a few:
oth from Feichtinger's talk "Function spaces for time-frequency analysis: the usefulness of a Banach Gelfand Triple", https://www.univie.ac.at/nuhag-php/dateien/talks/3397_Varan28JanFei.pdf If I remember correctly, $S^0$ is the Feichtinger algebra and $FL^1$ is the Fourier transform of $L^1$. $C^0$ is the space of continuous functions vanishing at infinity, "Schw" the space of Schwartz functions.
From Feichtinger's talk "The Vast Family of Fourier Standard Spaces (invariant function spaces between S0 and S′0)" https://www.univie.ac.at/nuhag-php/dateien/talks/3318_TrondheimMay17A.pdf
answered 5 hours ago
Dirk
7,06443264
answered 5 hours ago
Dirk
7,06443264
answered 5 hours ago
Dirk
7,06443264
answered 5 hours ago
Dirk
7,06443264
7,06443264
add a comment |Â
add a comment |Â
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1
Very roughly the picture you drew has some similarities with the picture I have of interpolation spaces in my head, especially with the real interpolation methods.
– Willie Wong
3 hours ago