Isomorophisms on $L^p$ and on $l^p$?
I want to prove that $l^p$ is isomorphic to the infinite dierct sum of $l^p$, similarly for $L^p$. Every time I try to define an operator, I lose one of the properties that this operators must have like surjectivity or linearity !
Can you help me please !
functional-analysis
edited Dec 7 at 3:49
Andrews
339317
add a comment |
I want to prove that $l^p$ is isomorphic to the infinite dierct sum of $l^p$, similarly for $L^p$. Every time I try to define an operator, I lose one of the properties that this operators must have like surjectivity or linearity !
Can you help me please !
functional-analysis
edited Dec 7 at 3:49
Andrews
339317
What topology do you want on the infinite direct sum? Or are you talking about isomorphism as vector spaces, without regard to topologies?
– Mariano Suárez-Álvarez
Mar 8 '13 at 2:24
The infinite direct product is not separable, @julien.
– Mariano Suárez-Álvarez
Mar 8 '13 at 2:28
@MarianoSuárez-Alvarez Sure, good point. What are talking about, here? I always find this terminology confusing. Is it the $C_0$ sum?
– Julien
Mar 8 '13 at 2:43
In our case we want the $ ||(x_n)||$ to be finite, were each coordinate of $(x_n)$ is an elements in $lp$, which means that the sequence of real numbers $(||x_n||)$ must belong to $lp$
– user61965
Mar 8 '13 at 3:17
add a comment |
I want to prove that $l^p$ is isomorphic to the infinite dierct sum of $l^p$, similarly for $L^p$. Every time I try to define an operator, I lose one of the properties that this operators must have like surjectivity or linearity !
Can you help me please !
functional-analysis
edited Dec 7 at 3:49
Andrews
339317
I want to prove that $l^p$ is isomorphic to the infinite dierct sum of $l^p$, similarly for $L^p$. Every time I try to define an operator, I lose one of the properties that this operators must have like surjectivity or linearity !
Can you help me please !
functional-analysis
functional-analysis
edited Dec 7 at 3:49
Andrews
339317
edited Dec 7 at 3:49
Andrews
339317
edited Dec 7 at 3:49
Andrews
339317
edited Dec 7 at 3:49
Andrews
339317
edited Dec 7 at 3:49
Andrews
339317
339317
asked Mar 8 '13 at 2:10
user61965
What topology do you want on the infinite direct sum? Or are you talking about isomorphism as vector spaces, without regard to topologies?
– Mariano Suárez-Álvarez
Mar 8 '13 at 2:24
The infinite direct product is not separable, @julien.
– Mariano Suárez-Álvarez
Mar 8 '13 at 2:28
@MarianoSuárez-Alvarez Sure, good point. What are talking about, here? I always find this terminology confusing. Is it the $C_0$ sum?
– Julien
Mar 8 '13 at 2:43
In our case we want the $ ||(x_n)||$ to be finite, were each coordinate of $(x_n)$ is an elements in $lp$, which means that the sequence of real numbers $(||x_n||)$ must belong to $lp$
– user61965
Mar 8 '13 at 3:17
add a comment |
What topology do you want on the infinite direct sum? Or are you talking about isomorphism as vector spaces, without regard to topologies?
– Mariano Suárez-Álvarez
Mar 8 '13 at 2:24
The infinite direct product is not separable, @julien.
– Mariano Suárez-Álvarez
Mar 8 '13 at 2:28
@MarianoSuárez-Alvarez Sure, good point. What are talking about, here? I always find this terminology confusing. Is it the $C_0$ sum?
– Julien
Mar 8 '13 at 2:43
In our case we want the $ ||(x_n)||$ to be finite, were each coordinate of $(x_n)$ is an elements in $lp$, which means that the sequence of real numbers $(||x_n||)$ must belong to $lp$
– user61965
Mar 8 '13 at 3:17
What topology do you want on the infinite direct sum? Or are you talking about isomorphism as vector spaces, without regard to topologies?
– Mariano Suárez-Álvarez
Mar 8 '13 at 2:24
What topology do you want on the infinite direct sum? Or are you talking about isomorphism as vector spaces, without regard to topologies?
– Mariano Suárez-Álvarez
Mar 8 '13 at 2:24
The infinite direct product is not separable, @julien.
– Mariano Suárez-Álvarez
Mar 8 '13 at 2:28
The infinite direct product is not separable, @julien.
– Mariano Suárez-Álvarez
Mar 8 '13 at 2:28
@MarianoSuárez-Alvarez Sure, good point. What are talking about, here? I always find this terminology confusing. Is it the $C_0$ sum?
– Julien
Mar 8 '13 at 2:43
@MarianoSuárez-Alvarez Sure, good point. What are talking about, here? I always find this terminology confusing. Is it the $C_0$ sum?
– Julien
Mar 8 '13 at 2:43
In our case we want the $ ||(x_n)||$ to be finite, were each coordinate of $(x_n)$ is an elements in $lp$, which means that the sequence of real numbers $(||x_n||)$ must belong to $lp$
– user61965
Mar 8 '13 at 3:17
In our case we want the $ ||(x_n)||$ to be finite, were each coordinate of $(x_n)$ is an elements in $lp$, which means that the sequence of real numbers $(||x_n||)$ must belong to $lp$
– user61965
Mar 8 '13 at 3:17
add a comment |
1 Answer
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For $ell_p$, you want to consider breaking up the index set (presumably ${mathbb N} = {1,2,3,ldots}$ as the union of infinitely many infinite sets. For example, take a $2$-dimensional grid
$$ pmatrix{ 1 &2 &4 & 7 & ldots cr
3 &5 &8 & 12 & ldots cr
6 &9 &13 & 18 & ldots cr
10 & 14 & 19 & 25 & ldots cr
ldots &ldots & ldots & ldots & ldots cr}$$
and use the rows. Then $ell_p$ is the infinite direct sum of copies of $ell_p$ with the $p$-norm $|(X_1, X_2, ldots )|_p = left( sum_i |X_i|_p right)^p$, such that
$X = (x_1, x_2, x_3, ldots)$ corresponds to $(X_1, X_2, X_3, ldots)$ with
$X_1 = (x_1, x_2, x_4, x_7, ldots)$, $X_2 = (x_3, x_5, x_8, x_{12})$, ....
Similarly, for $L_p([0,1])$, break up the interval into countably many subintervals, say
$I_k = [a_k, a_{k+1}]$ where $a_n$ is an increasing sequence with $a_1 = 0$
and $lim_{n to infty} a_n = 1$.
answered Mar 8 '13 at 2:56
Robert Israel
317k23207458
add a comment |
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1 Answer
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1 Answer
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active
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active
oldest
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For $ell_p$, you want to consider breaking up the index set (presumably ${mathbb N} = {1,2,3,ldots}$ as the union of infinitely many infinite sets. For example, take a $2$-dimensional grid
$$ pmatrix{ 1 &2 &4 & 7 & ldots cr
3 &5 &8 & 12 & ldots cr
6 &9 &13 & 18 & ldots cr
10 & 14 & 19 & 25 & ldots cr
ldots &ldots & ldots & ldots & ldots cr}$$
and use the rows. Then $ell_p$ is the infinite direct sum of copies of $ell_p$ with the $p$-norm $|(X_1, X_2, ldots )|_p = left( sum_i |X_i|_p right)^p$, such that
$X = (x_1, x_2, x_3, ldots)$ corresponds to $(X_1, X_2, X_3, ldots)$ with
$X_1 = (x_1, x_2, x_4, x_7, ldots)$, $X_2 = (x_3, x_5, x_8, x_{12})$, ....
Similarly, for $L_p([0,1])$, break up the interval into countably many subintervals, say
$I_k = [a_k, a_{k+1}]$ where $a_n$ is an increasing sequence with $a_1 = 0$
and $lim_{n to infty} a_n = 1$.
answered Mar 8 '13 at 2:56
Robert Israel
317k23207458
add a comment |
For $ell_p$, you want to consider breaking up the index set (presumably ${mathbb N} = {1,2,3,ldots}$ as the union of infinitely many infinite sets. For example, take a $2$-dimensional grid
$$ pmatrix{ 1 &2 &4 & 7 & ldots cr
3 &5 &8 & 12 & ldots cr
6 &9 &13 & 18 & ldots cr
10 & 14 & 19 & 25 & ldots cr
ldots &ldots & ldots & ldots & ldots cr}$$
and use the rows. Then $ell_p$ is the infinite direct sum of copies of $ell_p$ with the $p$-norm $|(X_1, X_2, ldots )|_p = left( sum_i |X_i|_p right)^p$, such that
$X = (x_1, x_2, x_3, ldots)$ corresponds to $(X_1, X_2, X_3, ldots)$ with
$X_1 = (x_1, x_2, x_4, x_7, ldots)$, $X_2 = (x_3, x_5, x_8, x_{12})$, ....
Similarly, for $L_p([0,1])$, break up the interval into countably many subintervals, say
$I_k = [a_k, a_{k+1}]$ where $a_n$ is an increasing sequence with $a_1 = 0$
and $lim_{n to infty} a_n = 1$.
answered Mar 8 '13 at 2:56
Robert Israel
317k23207458
add a comment |
For $ell_p$, you want to consider breaking up the index set (presumably ${mathbb N} = {1,2,3,ldots}$ as the union of infinitely many infinite sets. For example, take a $2$-dimensional grid
$$ pmatrix{ 1 &2 &4 & 7 & ldots cr
3 &5 &8 & 12 & ldots cr
6 &9 &13 & 18 & ldots cr
10 & 14 & 19 & 25 & ldots cr
ldots &ldots & ldots & ldots & ldots cr}$$
and use the rows. Then $ell_p$ is the infinite direct sum of copies of $ell_p$ with the $p$-norm $|(X_1, X_2, ldots )|_p = left( sum_i |X_i|_p right)^p$, such that
$X = (x_1, x_2, x_3, ldots)$ corresponds to $(X_1, X_2, X_3, ldots)$ with
$X_1 = (x_1, x_2, x_4, x_7, ldots)$, $X_2 = (x_3, x_5, x_8, x_{12})$, ....
Similarly, for $L_p([0,1])$, break up the interval into countably many subintervals, say
$I_k = [a_k, a_{k+1}]$ where $a_n$ is an increasing sequence with $a_1 = 0$
and $lim_{n to infty} a_n = 1$.
answered Mar 8 '13 at 2:56
Robert Israel
317k23207458
For $ell_p$, you want to consider breaking up the index set (presumably ${mathbb N} = {1,2,3,ldots}$ as the union of infinitely many infinite sets. For example, take a $2$-dimensional grid
$$ pmatrix{ 1 &2 &4 & 7 & ldots cr
3 &5 &8 & 12 & ldots cr
6 &9 &13 & 18 & ldots cr
10 & 14 & 19 & 25 & ldots cr
ldots &ldots & ldots & ldots & ldots cr}$$
and use the rows. Then $ell_p$ is the infinite direct sum of copies of $ell_p$ with the $p$-norm $|(X_1, X_2, ldots )|_p = left( sum_i |X_i|_p right)^p$, such that
$X = (x_1, x_2, x_3, ldots)$ corresponds to $(X_1, X_2, X_3, ldots)$ with
$X_1 = (x_1, x_2, x_4, x_7, ldots)$, $X_2 = (x_3, x_5, x_8, x_{12})$, ....
Similarly, for $L_p([0,1])$, break up the interval into countably many subintervals, say
$I_k = [a_k, a_{k+1}]$ where $a_n$ is an increasing sequence with $a_1 = 0$
and $lim_{n to infty} a_n = 1$.
answered Mar 8 '13 at 2:56
Robert Israel
317k23207458
answered Mar 8 '13 at 2:56
Robert Israel
317k23207458
answered Mar 8 '13 at 2:56
Robert Israel
317k23207458
answered Mar 8 '13 at 2:56
Robert Israel
317k23207458
317k23207458
add a comment |
add a comment |
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What topology do you want on the infinite direct sum? Or are you talking about isomorphism as vector spaces, without regard to topologies?
– Mariano Suárez-Álvarez
Mar 8 '13 at 2:24
The infinite direct product is not separable, @julien.
– Mariano Suárez-Álvarez
Mar 8 '13 at 2:28
@MarianoSuárez-Alvarez Sure, good point. What are talking about, here? I always find this terminology confusing. Is it the $C_0$ sum?
– Julien
Mar 8 '13 at 2:43
In our case we want the $ ||(x_n)||$ to be finite, were each coordinate of $(x_n)$ is an elements in $lp$, which means that the sequence of real numbers $(||x_n||)$ must belong to $lp$
– user61965
Mar 8 '13 at 3:17