Is $c$ isomorphic to $c_0$?
I want to prove that $c(k)$ and $c_0(k)$ are isomorphic.
It is easy to find a linear map $u$ given by $u(lbrace x_krbrace)=lbrace x_k-lim x_krbrace in c_0(k)$, and show that this is a surjection. But I couldn't find the constants $alpha ,beta$ such that
$alpha |x_n|_{sup}leq|u(x_n)|_{sup}leqbeta|x_n|_{sup}$, here $k$ is in $mathbb{R}$ or $mathbb{C}$.
Any help is welcome.
real-analysis functional-analysis banach-spaces vector-space-isomorphism
edited Jan 5 at 22:22
Davide Giraudo
125k16150261
asked Sep 22 '15 at 1:27
sti9111sti9111
377317
add a comment |
I want to prove that $c(k)$ and $c_0(k)$ are isomorphic.
It is easy to find a linear map $u$ given by $u(lbrace x_krbrace)=lbrace x_k-lim x_krbrace in c_0(k)$, and show that this is a surjection. But I couldn't find the constants $alpha ,beta$ such that
$alpha |x_n|_{sup}leq|u(x_n)|_{sup}leqbeta|x_n|_{sup}$, here $k$ is in $mathbb{R}$ or $mathbb{C}$.
Any help is welcome.
real-analysis functional-analysis banach-spaces vector-space-isomorphism
edited Jan 5 at 22:22
Davide Giraudo
125k16150261
asked Sep 22 '15 at 1:27
sti9111sti9111
377317
Your map is not an injection; it sends any constant sequence to $0$. Also, your terminology could use some clarification: what is $k$? What does "isomorphic" mean, isomorphic as topological vector spaces?
– Eric Wofsey
Sep 22 '15 at 1:29
Thank! but is possible find some $u$ such that $alpha ||x_n||_{sup}leq ||u(x_n)||_{sup}leqbeta ||x_n||_{sup}$ ?
– sti9111
Sep 22 '15 at 1:31
@sti9111 - when someone asks you a question about what you mean, it is not helpful to simply restate what you've already asked without bothering to answer what was requested of you. We are not sure what you mean, so we can't properly answer you until you clarify it.
– Paul Sinclair
Sep 22 '15 at 1:44
@PaulSinclair: In fairness to sti9111, I edited my comment to add the second half of it a minute after posting it and they may not have noticed.
– Eric Wofsey
Sep 22 '15 at 1:53
My apologies then. Actually, I think I can address his question about the norms as is, and am currently writing it up.
– Paul Sinclair
Sep 22 '15 at 1:54
add a comment |
I want to prove that $c(k)$ and $c_0(k)$ are isomorphic.
It is easy to find a linear map $u$ given by $u(lbrace x_krbrace)=lbrace x_k-lim x_krbrace in c_0(k)$, and show that this is a surjection. But I couldn't find the constants $alpha ,beta$ such that
$alpha |x_n|_{sup}leq|u(x_n)|_{sup}leqbeta|x_n|_{sup}$, here $k$ is in $mathbb{R}$ or $mathbb{C}$.
Any help is welcome.
real-analysis functional-analysis banach-spaces vector-space-isomorphism
edited Jan 5 at 22:22
Davide Giraudo
125k16150261
asked Sep 22 '15 at 1:27
sti9111sti9111
377317
I want to prove that $c(k)$ and $c_0(k)$ are isomorphic.
It is easy to find a linear map $u$ given by $u(lbrace x_krbrace)=lbrace x_k-lim x_krbrace in c_0(k)$, and show that this is a surjection. But I couldn't find the constants $alpha ,beta$ such that
$alpha |x_n|_{sup}leq|u(x_n)|_{sup}leqbeta|x_n|_{sup}$, here $k$ is in $mathbb{R}$ or $mathbb{C}$.
Any help is welcome.
real-analysis functional-analysis banach-spaces vector-space-isomorphism
real-analysis functional-analysis banach-spaces vector-space-isomorphism
edited Jan 5 at 22:22
Davide Giraudo
125k16150261
asked Sep 22 '15 at 1:27
sti9111sti9111
377317
edited Jan 5 at 22:22
Davide Giraudo
125k16150261
asked Sep 22 '15 at 1:27
sti9111sti9111
377317
edited Jan 5 at 22:22
Davide Giraudo
125k16150261
edited Jan 5 at 22:22
Davide Giraudo
125k16150261
edited Jan 5 at 22:22
Davide Giraudo
125k16150261
125k16150261
asked Sep 22 '15 at 1:27
sti9111sti9111
377317
asked Sep 22 '15 at 1:27
sti9111sti9111
377317
asked Sep 22 '15 at 1:27
sti9111sti9111
377317
377317
Your map is not an injection; it sends any constant sequence to $0$. Also, your terminology could use some clarification: what is $k$? What does "isomorphic" mean, isomorphic as topological vector spaces?
– Eric Wofsey
Sep 22 '15 at 1:29
Thank! but is possible find some $u$ such that $alpha ||x_n||_{sup}leq ||u(x_n)||_{sup}leqbeta ||x_n||_{sup}$ ?
– sti9111
Sep 22 '15 at 1:31
@sti9111 - when someone asks you a question about what you mean, it is not helpful to simply restate what you've already asked without bothering to answer what was requested of you. We are not sure what you mean, so we can't properly answer you until you clarify it.
– Paul Sinclair
Sep 22 '15 at 1:44
@PaulSinclair: In fairness to sti9111, I edited my comment to add the second half of it a minute after posting it and they may not have noticed.
– Eric Wofsey
Sep 22 '15 at 1:53
My apologies then. Actually, I think I can address his question about the norms as is, and am currently writing it up.
– Paul Sinclair
Sep 22 '15 at 1:54
add a comment |
Your map is not an injection; it sends any constant sequence to $0$. Also, your terminology could use some clarification: what is $k$? What does "isomorphic" mean, isomorphic as topological vector spaces?
– Eric Wofsey
Sep 22 '15 at 1:29
Thank! but is possible find some $u$ such that $alpha ||x_n||_{sup}leq ||u(x_n)||_{sup}leqbeta ||x_n||_{sup}$ ?
– sti9111
Sep 22 '15 at 1:31
@sti9111 - when someone asks you a question about what you mean, it is not helpful to simply restate what you've already asked without bothering to answer what was requested of you. We are not sure what you mean, so we can't properly answer you until you clarify it.
– Paul Sinclair
Sep 22 '15 at 1:44
@PaulSinclair: In fairness to sti9111, I edited my comment to add the second half of it a minute after posting it and they may not have noticed.
– Eric Wofsey
Sep 22 '15 at 1:53
My apologies then. Actually, I think I can address his question about the norms as is, and am currently writing it up.
– Paul Sinclair
Sep 22 '15 at 1:54
Your map is not an injection; it sends any constant sequence to $0$. Also, your terminology could use some clarification: what is $k$? What does "isomorphic" mean, isomorphic as topological vector spaces?
– Eric Wofsey
Sep 22 '15 at 1:29
Your map is not an injection; it sends any constant sequence to $0$. Also, your terminology could use some clarification: what is $k$? What does "isomorphic" mean, isomorphic as topological vector spaces?
– Eric Wofsey
Sep 22 '15 at 1:29
Thank! but is possible find some $u$ such that $alpha ||x_n||_{sup}leq ||u(x_n)||_{sup}leqbeta ||x_n||_{sup}$ ?
– sti9111
Sep 22 '15 at 1:31
Thank! but is possible find some $u$ such that $alpha ||x_n||_{sup}leq ||u(x_n)||_{sup}leqbeta ||x_n||_{sup}$ ?
– sti9111
Sep 22 '15 at 1:31
@sti9111 - when someone asks you a question about what you mean, it is not helpful to simply restate what you've already asked without bothering to answer what was requested of you. We are not sure what you mean, so we can't properly answer you until you clarify it.
– Paul Sinclair
Sep 22 '15 at 1:44
@sti9111 - when someone asks you a question about what you mean, it is not helpful to simply restate what you've already asked without bothering to answer what was requested of you. We are not sure what you mean, so we can't properly answer you until you clarify it.
– Paul Sinclair
Sep 22 '15 at 1:44
@PaulSinclair: In fairness to sti9111, I edited my comment to add the second half of it a minute after posting it and they may not have noticed.
– Eric Wofsey
Sep 22 '15 at 1:53
@PaulSinclair: In fairness to sti9111, I edited my comment to add the second half of it a minute after posting it and they may not have noticed.
– Eric Wofsey
Sep 22 '15 at 1:53
My apologies then. Actually, I think I can address his question about the norms as is, and am currently writing it up.
– Paul Sinclair
Sep 22 '15 at 1:54
My apologies then. Actually, I think I can address his question about the norms as is, and am currently writing it up.
– Paul Sinclair
Sep 22 '15 at 1:54
add a comment |
1 Answer
1
active
oldest
votes
No, there is no $alpha > 0$ satisfying the inequality for all sequences ${x_n}$. To see, this consider the case when the ${x_n}$ are positive real numbers converging to 0 and $k$ is a positive real constant, and for convenience, let $x = |x_n|_{sup}$, then $| x_n + k |_{sup} = x + k$. So your inequality becomes for ${x_n + k}, alpha(x + k) le x$. Therefore $(alpha - 1)x +alpha k le 0$. By letting $k gg x$, we see that this would require $alpha le 0$.
edited Jan 5 at 22:23
Davide Giraudo
125k16150261
answered Sep 22 '15 at 2:01
Paul SinclairPaul Sinclair
19.3k21441
More simply, instead of taking a sequence of positive numbers converging to $0$, just let $x_n=0$ for all $n$ (i.e., you are just looking at any nonzero constant sequence).
– Eric Wofsey
Sep 22 '15 at 2:21
True - I was avoiding the $0$ sequence because I was thinking it would zero out the inequality, but I failed to notice that after adding $k$, that was no longer true.
– Paul Sinclair
Sep 22 '15 at 2:40
Thanks! this say that for $u$ like before the numbers $alpha , beta$ doesn´t exist, but is possible find some $u$ which is a surjection and this number exist?
– sti9111
Sep 22 '15 at 19:28
Well, since I still am not absolutely sure what $c$ and $c_0$ are, that is hard to say. I am guessing that $c$ is the set of all convergents sequences in $Bbb K$, and $c_0$ are sequences that convege to 0, both supplied with the sup norm. Is this correct?
– Paul Sinclair
Sep 22 '15 at 22:22
add a comment |
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No, there is no $alpha > 0$ satisfying the inequality for all sequences ${x_n}$. To see, this consider the case when the ${x_n}$ are positive real numbers converging to 0 and $k$ is a positive real constant, and for convenience, let $x = |x_n|_{sup}$, then $| x_n + k |_{sup} = x + k$. So your inequality becomes for ${x_n + k}, alpha(x + k) le x$. Therefore $(alpha - 1)x +alpha k le 0$. By letting $k gg x$, we see that this would require $alpha le 0$.
edited Jan 5 at 22:23
Davide Giraudo
125k16150261
answered Sep 22 '15 at 2:01
Paul SinclairPaul Sinclair
19.3k21441
More simply, instead of taking a sequence of positive numbers converging to $0$, just let $x_n=0$ for all $n$ (i.e., you are just looking at any nonzero constant sequence).
– Eric Wofsey
Sep 22 '15 at 2:21
True - I was avoiding the $0$ sequence because I was thinking it would zero out the inequality, but I failed to notice that after adding $k$, that was no longer true.
– Paul Sinclair
Sep 22 '15 at 2:40
Thanks! this say that for $u$ like before the numbers $alpha , beta$ doesn´t exist, but is possible find some $u$ which is a surjection and this number exist?
– sti9111
Sep 22 '15 at 19:28
Well, since I still am not absolutely sure what $c$ and $c_0$ are, that is hard to say. I am guessing that $c$ is the set of all convergents sequences in $Bbb K$, and $c_0$ are sequences that convege to 0, both supplied with the sup norm. Is this correct?
– Paul Sinclair
Sep 22 '15 at 22:22
add a comment |
No, there is no $alpha > 0$ satisfying the inequality for all sequences ${x_n}$. To see, this consider the case when the ${x_n}$ are positive real numbers converging to 0 and $k$ is a positive real constant, and for convenience, let $x = |x_n|_{sup}$, then $| x_n + k |_{sup} = x + k$. So your inequality becomes for ${x_n + k}, alpha(x + k) le x$. Therefore $(alpha - 1)x +alpha k le 0$. By letting $k gg x$, we see that this would require $alpha le 0$.
edited Jan 5 at 22:23
Davide Giraudo
125k16150261
answered Sep 22 '15 at 2:01
Paul SinclairPaul Sinclair
19.3k21441
More simply, instead of taking a sequence of positive numbers converging to $0$, just let $x_n=0$ for all $n$ (i.e., you are just looking at any nonzero constant sequence).
– Eric Wofsey
Sep 22 '15 at 2:21
True - I was avoiding the $0$ sequence because I was thinking it would zero out the inequality, but I failed to notice that after adding $k$, that was no longer true.
– Paul Sinclair
Sep 22 '15 at 2:40
Thanks! this say that for $u$ like before the numbers $alpha , beta$ doesn´t exist, but is possible find some $u$ which is a surjection and this number exist?
– sti9111
Sep 22 '15 at 19:28
Well, since I still am not absolutely sure what $c$ and $c_0$ are, that is hard to say. I am guessing that $c$ is the set of all convergents sequences in $Bbb K$, and $c_0$ are sequences that convege to 0, both supplied with the sup norm. Is this correct?
– Paul Sinclair
Sep 22 '15 at 22:22
add a comment |
No, there is no $alpha > 0$ satisfying the inequality for all sequences ${x_n}$. To see, this consider the case when the ${x_n}$ are positive real numbers converging to 0 and $k$ is a positive real constant, and for convenience, let $x = |x_n|_{sup}$, then $| x_n + k |_{sup} = x + k$. So your inequality becomes for ${x_n + k}, alpha(x + k) le x$. Therefore $(alpha - 1)x +alpha k le 0$. By letting $k gg x$, we see that this would require $alpha le 0$.
edited Jan 5 at 22:23
Davide Giraudo
125k16150261
answered Sep 22 '15 at 2:01
Paul SinclairPaul Sinclair
19.3k21441
No, there is no $alpha > 0$ satisfying the inequality for all sequences ${x_n}$. To see, this consider the case when the ${x_n}$ are positive real numbers converging to 0 and $k$ is a positive real constant, and for convenience, let $x = |x_n|_{sup}$, then $| x_n + k |_{sup} = x + k$. So your inequality becomes for ${x_n + k}, alpha(x + k) le x$. Therefore $(alpha - 1)x +alpha k le 0$. By letting $k gg x$, we see that this would require $alpha le 0$.
edited Jan 5 at 22:23
Davide Giraudo
125k16150261
answered Sep 22 '15 at 2:01
Paul SinclairPaul Sinclair
19.3k21441
edited Jan 5 at 22:23
Davide Giraudo
125k16150261
edited Jan 5 at 22:23
Davide Giraudo
125k16150261
edited Jan 5 at 22:23
Davide Giraudo
125k16150261
125k16150261
answered Sep 22 '15 at 2:01
Paul SinclairPaul Sinclair
19.3k21441
answered Sep 22 '15 at 2:01
Paul SinclairPaul Sinclair
19.3k21441
answered Sep 22 '15 at 2:01
Paul SinclairPaul Sinclair
19.3k21441
19.3k21441
More simply, instead of taking a sequence of positive numbers converging to $0$, just let $x_n=0$ for all $n$ (i.e., you are just looking at any nonzero constant sequence).
– Eric Wofsey
Sep 22 '15 at 2:21
True - I was avoiding the $0$ sequence because I was thinking it would zero out the inequality, but I failed to notice that after adding $k$, that was no longer true.
– Paul Sinclair
Sep 22 '15 at 2:40
Thanks! this say that for $u$ like before the numbers $alpha , beta$ doesn´t exist, but is possible find some $u$ which is a surjection and this number exist?
– sti9111
Sep 22 '15 at 19:28
Well, since I still am not absolutely sure what $c$ and $c_0$ are, that is hard to say. I am guessing that $c$ is the set of all convergents sequences in $Bbb K$, and $c_0$ are sequences that convege to 0, both supplied with the sup norm. Is this correct?
– Paul Sinclair
Sep 22 '15 at 22:22
add a comment |
More simply, instead of taking a sequence of positive numbers converging to $0$, just let $x_n=0$ for all $n$ (i.e., you are just looking at any nonzero constant sequence).
– Eric Wofsey
Sep 22 '15 at 2:21
True - I was avoiding the $0$ sequence because I was thinking it would zero out the inequality, but I failed to notice that after adding $k$, that was no longer true.
– Paul Sinclair
Sep 22 '15 at 2:40
Thanks! this say that for $u$ like before the numbers $alpha , beta$ doesn´t exist, but is possible find some $u$ which is a surjection and this number exist?
– sti9111
Sep 22 '15 at 19:28
Well, since I still am not absolutely sure what $c$ and $c_0$ are, that is hard to say. I am guessing that $c$ is the set of all convergents sequences in $Bbb K$, and $c_0$ are sequences that convege to 0, both supplied with the sup norm. Is this correct?
– Paul Sinclair
Sep 22 '15 at 22:22
More simply, instead of taking a sequence of positive numbers converging to $0$, just let $x_n=0$ for all $n$ (i.e., you are just looking at any nonzero constant sequence).
– Eric Wofsey
Sep 22 '15 at 2:21
More simply, instead of taking a sequence of positive numbers converging to $0$, just let $x_n=0$ for all $n$ (i.e., you are just looking at any nonzero constant sequence).
– Eric Wofsey
Sep 22 '15 at 2:21
True - I was avoiding the $0$ sequence because I was thinking it would zero out the inequality, but I failed to notice that after adding $k$, that was no longer true.
– Paul Sinclair
Sep 22 '15 at 2:40
True - I was avoiding the $0$ sequence because I was thinking it would zero out the inequality, but I failed to notice that after adding $k$, that was no longer true.
– Paul Sinclair
Sep 22 '15 at 2:40
Thanks! this say that for $u$ like before the numbers $alpha , beta$ doesn´t exist, but is possible find some $u$ which is a surjection and this number exist?
– sti9111
Sep 22 '15 at 19:28
Thanks! this say that for $u$ like before the numbers $alpha , beta$ doesn´t exist, but is possible find some $u$ which is a surjection and this number exist?
– sti9111
Sep 22 '15 at 19:28
Well, since I still am not absolutely sure what $c$ and $c_0$ are, that is hard to say. I am guessing that $c$ is the set of all convergents sequences in $Bbb K$, and $c_0$ are sequences that convege to 0, both supplied with the sup norm. Is this correct?
– Paul Sinclair
Sep 22 '15 at 22:22
Well, since I still am not absolutely sure what $c$ and $c_0$ are, that is hard to say. I am guessing that $c$ is the set of all convergents sequences in $Bbb K$, and $c_0$ are sequences that convege to 0, both supplied with the sup norm. Is this correct?
– Paul Sinclair
Sep 22 '15 at 22:22
add a comment |
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Your map is not an injection; it sends any constant sequence to $0$. Also, your terminology could use some clarification: what is $k$? What does "isomorphic" mean, isomorphic as topological vector spaces?
– Eric Wofsey
Sep 22 '15 at 1:29
Thank! but is possible find some $u$ such that $alpha ||x_n||_{sup}leq ||u(x_n)||_{sup}leqbeta ||x_n||_{sup}$ ?
– sti9111
Sep 22 '15 at 1:31
@sti9111 - when someone asks you a question about what you mean, it is not helpful to simply restate what you've already asked without bothering to answer what was requested of you. We are not sure what you mean, so we can't properly answer you until you clarify it.
– Paul Sinclair
Sep 22 '15 at 1:44
@PaulSinclair: In fairness to sti9111, I edited my comment to add the second half of it a minute after posting it and they may not have noticed.
– Eric Wofsey
Sep 22 '15 at 1:53
My apologies then. Actually, I think I can address his question about the norms as is, and am currently writing it up.
– Paul Sinclair
Sep 22 '15 at 1:54