Counting measure in integral.
$begingroup$
Let $mu$ counting measure on $X$ countable set.
This is correct?
$int_{X}|f|^pdmu=int_{bigcup_{kinmathbb{N}} x_k} |f|^pdmu=sum_{kinmathbb{N}}int_{x_k}|f|^pdmu=sum_{kinmathbb{N}}int |f|^pmathcal{X}_{x_k}dmu=sum_{kinmathbb{N}} |f(x_k)|^pint mathcal{X}_{x_k}=sum_{kinmathbb{N}} |f(x_)|^pmu(x_k)=sum_{kinmathbb{N}}|f(x_k)|^p$
Therefore ${L(X)}^p={l(X)}^p$
real-analysis measure-theory
asked Jan 8 at 8:22
eraldcoileraldcoil
395211
$endgroup$
add a comment |
$begingroup$
Let $mu$ counting measure on $X$ countable set.
This is correct?
$int_{X}|f|^pdmu=int_{bigcup_{kinmathbb{N}} x_k} |f|^pdmu=sum_{kinmathbb{N}}int_{x_k}|f|^pdmu=sum_{kinmathbb{N}}int |f|^pmathcal{X}_{x_k}dmu=sum_{kinmathbb{N}} |f(x_k)|^pint mathcal{X}_{x_k}=sum_{kinmathbb{N}} |f(x_)|^pmu(x_k)=sum_{kinmathbb{N}}|f(x_k)|^p$
Therefore ${L(X)}^p={l(X)}^p$
real-analysis measure-theory
asked Jan 8 at 8:22
eraldcoileraldcoil
395211
$endgroup$
$begingroup$
Yes it is correct. A precise statement is the following: the map $fin L(X)^{p} to (f(x_k))$ is an isometric isomorphism of $L(X)^{p}$ onto $ell (X)^{p}$.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 8:27
add a comment |
$begingroup$
Let $mu$ counting measure on $X$ countable set.
This is correct?
$int_{X}|f|^pdmu=int_{bigcup_{kinmathbb{N}} x_k} |f|^pdmu=sum_{kinmathbb{N}}int_{x_k}|f|^pdmu=sum_{kinmathbb{N}}int |f|^pmathcal{X}_{x_k}dmu=sum_{kinmathbb{N}} |f(x_k)|^pint mathcal{X}_{x_k}=sum_{kinmathbb{N}} |f(x_)|^pmu(x_k)=sum_{kinmathbb{N}}|f(x_k)|^p$
Therefore ${L(X)}^p={l(X)}^p$
real-analysis measure-theory
asked Jan 8 at 8:22
eraldcoileraldcoil
395211
$endgroup$
Let $mu$ counting measure on $X$ countable set.
This is correct?
$int_{X}|f|^pdmu=int_{bigcup_{kinmathbb{N}} x_k} |f|^pdmu=sum_{kinmathbb{N}}int_{x_k}|f|^pdmu=sum_{kinmathbb{N}}int |f|^pmathcal{X}_{x_k}dmu=sum_{kinmathbb{N}} |f(x_k)|^pint mathcal{X}_{x_k}=sum_{kinmathbb{N}} |f(x_)|^pmu(x_k)=sum_{kinmathbb{N}}|f(x_k)|^p$
Therefore ${L(X)}^p={l(X)}^p$
real-analysis measure-theory
real-analysis measure-theory
asked Jan 8 at 8:22
eraldcoileraldcoil
395211
asked Jan 8 at 8:22
eraldcoileraldcoil
395211
asked Jan 8 at 8:22
eraldcoileraldcoil
395211
asked Jan 8 at 8:22
eraldcoileraldcoil
395211
asked Jan 8 at 8:22
eraldcoileraldcoil
395211
395211
$begingroup$
Yes it is correct. A precise statement is the following: the map $fin L(X)^{p} to (f(x_k))$ is an isometric isomorphism of $L(X)^{p}$ onto $ell (X)^{p}$.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 8:27
add a comment |
$begingroup$
Yes it is correct. A precise statement is the following: the map $fin L(X)^{p} to (f(x_k))$ is an isometric isomorphism of $L(X)^{p}$ onto $ell (X)^{p}$.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 8:27
$begingroup$
Yes it is correct. A precise statement is the following: the map $fin L(X)^{p} to (f(x_k))$ is an isometric isomorphism of $L(X)^{p}$ onto $ell (X)^{p}$.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 8:27
$begingroup$
Yes it is correct. A precise statement is the following: the map $fin L(X)^{p} to (f(x_k))$ is an isometric isomorphism of $L(X)^{p}$ onto $ell (X)^{p}$.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 8:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The equalities are in principle correct provided that $X$ is not finite.
Not completely correct is the notation that is practicized.
Every $x_k$ in the equalities should be replaced by ${x_k}$.
Further IMV it is a bit overdone.
If $mu$ denotes the counting measure and $X$ is countable then you can write immediately:$$int_X|f|^pdmu=sum_{xin X}|f(x)|^p$$
There is no need to go for $mathbb N$ as index set. It is even wrong if $X$ is a finite set.
Where it concerns $L(X)^p=l(X)^p$ see the comment of Kavi to your question.
answered Jan 8 at 8:41
drhabdrhab
102k545136
$endgroup$
add a comment |
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1 Answer
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$begingroup$
The equalities are in principle correct provided that $X$ is not finite.
Not completely correct is the notation that is practicized.
Every $x_k$ in the equalities should be replaced by ${x_k}$.
Further IMV it is a bit overdone.
If $mu$ denotes the counting measure and $X$ is countable then you can write immediately:$$int_X|f|^pdmu=sum_{xin X}|f(x)|^p$$
There is no need to go for $mathbb N$ as index set. It is even wrong if $X$ is a finite set.
Where it concerns $L(X)^p=l(X)^p$ see the comment of Kavi to your question.
answered Jan 8 at 8:41
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
The equalities are in principle correct provided that $X$ is not finite.
Not completely correct is the notation that is practicized.
Every $x_k$ in the equalities should be replaced by ${x_k}$.
Further IMV it is a bit overdone.
If $mu$ denotes the counting measure and $X$ is countable then you can write immediately:$$int_X|f|^pdmu=sum_{xin X}|f(x)|^p$$
There is no need to go for $mathbb N$ as index set. It is even wrong if $X$ is a finite set.
Where it concerns $L(X)^p=l(X)^p$ see the comment of Kavi to your question.
answered Jan 8 at 8:41
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
The equalities are in principle correct provided that $X$ is not finite.
Not completely correct is the notation that is practicized.
Every $x_k$ in the equalities should be replaced by ${x_k}$.
Further IMV it is a bit overdone.
If $mu$ denotes the counting measure and $X$ is countable then you can write immediately:$$int_X|f|^pdmu=sum_{xin X}|f(x)|^p$$
There is no need to go for $mathbb N$ as index set. It is even wrong if $X$ is a finite set.
Where it concerns $L(X)^p=l(X)^p$ see the comment of Kavi to your question.
answered Jan 8 at 8:41
drhabdrhab
102k545136
$endgroup$
The equalities are in principle correct provided that $X$ is not finite.
Not completely correct is the notation that is practicized.
Every $x_k$ in the equalities should be replaced by ${x_k}$.
Further IMV it is a bit overdone.
If $mu$ denotes the counting measure and $X$ is countable then you can write immediately:$$int_X|f|^pdmu=sum_{xin X}|f(x)|^p$$
There is no need to go for $mathbb N$ as index set. It is even wrong if $X$ is a finite set.
Where it concerns $L(X)^p=l(X)^p$ see the comment of Kavi to your question.
answered Jan 8 at 8:41
drhabdrhab
102k545136
answered Jan 8 at 8:41
drhabdrhab
102k545136
answered Jan 8 at 8:41
drhabdrhab
102k545136
answered Jan 8 at 8:41
drhabdrhab
102k545136
102k545136
add a comment |
add a comment |
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$begingroup$
Yes it is correct. A precise statement is the following: the map $fin L(X)^{p} to (f(x_k))$ is an isometric isomorphism of $L(X)^{p}$ onto $ell (X)^{p}$.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 8:27