Matrix transformation of bounded sequence
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$A=(a_{nk})_{n,k=1}^infty$ and $A_{n}(x)=∑_{k=1}^{infty}a_{nk}x_{k}$ for fixed $n$. So $A_{n}(x)$ can be thought as a matrix transformation for any sequence. Now let $A∈(l_{infty},l_{infty})$ that is $forall x in l_{infty}, A_{n}(x) in l_{infty}$. And $l_{infty}$ represents bounded sequences. From here I get $sup_{n}|A_{n}(x)|<infty$ or $|A_{n}(x)|<infty$, $forall n$. By this transformation we find a new sequence $A_{n}(x)=(A_1(x),A_2(x),...)$ and supremum of this sequence is less than infinity. In this case can I say $A_{n}(x)=∑_{k=1}^{∞}a_{nk}x_{k}<infty , forall n$?
linear-algebra sequences-and-series
edited Dec 22 '18 at 4:00
Shaun
9,380113684
asked Dec 30 '15 at 18:06
ruudvaanruudvaan
187
$endgroup$
add a comment |
$begingroup$
$A=(a_{nk})_{n,k=1}^infty$ and $A_{n}(x)=∑_{k=1}^{infty}a_{nk}x_{k}$ for fixed $n$. So $A_{n}(x)$ can be thought as a matrix transformation for any sequence. Now let $A∈(l_{infty},l_{infty})$ that is $forall x in l_{infty}, A_{n}(x) in l_{infty}$. And $l_{infty}$ represents bounded sequences. From here I get $sup_{n}|A_{n}(x)|<infty$ or $|A_{n}(x)|<infty$, $forall n$. By this transformation we find a new sequence $A_{n}(x)=(A_1(x),A_2(x),...)$ and supremum of this sequence is less than infinity. In this case can I say $A_{n}(x)=∑_{k=1}^{∞}a_{nk}x_{k}<infty , forall n$?
linear-algebra sequences-and-series
edited Dec 22 '18 at 4:00
Shaun
9,380113684
asked Dec 30 '15 at 18:06
ruudvaanruudvaan
187
$endgroup$
$begingroup$
Note that you always have $forall ninmathbb{N},|A_n(x)|_infty<+infty$ with your definition for $A,$ and that it is not the same assertion that $sup_n|A_n(x)|_infty<+infty.$
$endgroup$
– Balloon
Jan 1 '16 at 12:48
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can you please explain me why I always have ∀n∈N,|An(x)|∞<+∞ in this case?
$endgroup$
– ruudvaan
Jan 1 '16 at 13:21
$begingroup$
Because you said that $forall ninmathbb{N},$ you have that $A_n$ maps any sequence $xin l_infty$ to an other sequence $A_n(x)in l_infty.$ As $l_infty$ is the space of bounded sequences, then you have that for any sequence $yin l_infty,$ $|y|_infty=sup_n|y_n|<+infty.$ So it is for $A_n(x).$
$endgroup$
– Balloon
Jan 1 '16 at 13:43
$begingroup$
since ∀n∈N,|An(x)|∞<+∞ then ∀n∈N,An(x)<+∞ please correct me:)) thats all I am looking for.
$endgroup$
– ruudvaan
Jan 1 '16 at 13:55
$begingroup$
This last point is true, but it seems I have misunderstood something : it is not $A_n(x)in l_infty$ but $(A_n(x))_{ninmathbb{N}}in l_infty.$ But it does'nt change the conclusion which is that $|(A_n(x))_{ninmathbb{N}}|_infty<+infty$ and in particular each term of this sequence is smaller than $+infty.$
$endgroup$
– Balloon
Jan 1 '16 at 14:04
add a comment |
$begingroup$
$A=(a_{nk})_{n,k=1}^infty$ and $A_{n}(x)=∑_{k=1}^{infty}a_{nk}x_{k}$ for fixed $n$. So $A_{n}(x)$ can be thought as a matrix transformation for any sequence. Now let $A∈(l_{infty},l_{infty})$ that is $forall x in l_{infty}, A_{n}(x) in l_{infty}$. And $l_{infty}$ represents bounded sequences. From here I get $sup_{n}|A_{n}(x)|<infty$ or $|A_{n}(x)|<infty$, $forall n$. By this transformation we find a new sequence $A_{n}(x)=(A_1(x),A_2(x),...)$ and supremum of this sequence is less than infinity. In this case can I say $A_{n}(x)=∑_{k=1}^{∞}a_{nk}x_{k}<infty , forall n$?
linear-algebra sequences-and-series
edited Dec 22 '18 at 4:00
Shaun
9,380113684
asked Dec 30 '15 at 18:06
ruudvaanruudvaan
187
$endgroup$
$A=(a_{nk})_{n,k=1}^infty$ and $A_{n}(x)=∑_{k=1}^{infty}a_{nk}x_{k}$ for fixed $n$. So $A_{n}(x)$ can be thought as a matrix transformation for any sequence. Now let $A∈(l_{infty},l_{infty})$ that is $forall x in l_{infty}, A_{n}(x) in l_{infty}$. And $l_{infty}$ represents bounded sequences. From here I get $sup_{n}|A_{n}(x)|<infty$ or $|A_{n}(x)|<infty$, $forall n$. By this transformation we find a new sequence $A_{n}(x)=(A_1(x),A_2(x),...)$ and supremum of this sequence is less than infinity. In this case can I say $A_{n}(x)=∑_{k=1}^{∞}a_{nk}x_{k}<infty , forall n$?
linear-algebra sequences-and-series
linear-algebra sequences-and-series
edited Dec 22 '18 at 4:00
Shaun
9,380113684
asked Dec 30 '15 at 18:06
ruudvaanruudvaan
187
edited Dec 22 '18 at 4:00
Shaun
9,380113684
asked Dec 30 '15 at 18:06
ruudvaanruudvaan
187
edited Dec 22 '18 at 4:00
Shaun
9,380113684
edited Dec 22 '18 at 4:00
Shaun
9,380113684
edited Dec 22 '18 at 4:00
Shaun
9,380113684
9,380113684
asked Dec 30 '15 at 18:06
ruudvaanruudvaan
187
asked Dec 30 '15 at 18:06
ruudvaanruudvaan
187
asked Dec 30 '15 at 18:06
ruudvaanruudvaan
187
187
$begingroup$
Note that you always have $forall ninmathbb{N},|A_n(x)|_infty<+infty$ with your definition for $A,$ and that it is not the same assertion that $sup_n|A_n(x)|_infty<+infty.$
$endgroup$
– Balloon
Jan 1 '16 at 12:48
$begingroup$
can you please explain me why I always have ∀n∈N,|An(x)|∞<+∞ in this case?
$endgroup$
– ruudvaan
Jan 1 '16 at 13:21
$begingroup$
Because you said that $forall ninmathbb{N},$ you have that $A_n$ maps any sequence $xin l_infty$ to an other sequence $A_n(x)in l_infty.$ As $l_infty$ is the space of bounded sequences, then you have that for any sequence $yin l_infty,$ $|y|_infty=sup_n|y_n|<+infty.$ So it is for $A_n(x).$
$endgroup$
– Balloon
Jan 1 '16 at 13:43
$begingroup$
since ∀n∈N,|An(x)|∞<+∞ then ∀n∈N,An(x)<+∞ please correct me:)) thats all I am looking for.
$endgroup$
– ruudvaan
Jan 1 '16 at 13:55
$begingroup$
This last point is true, but it seems I have misunderstood something : it is not $A_n(x)in l_infty$ but $(A_n(x))_{ninmathbb{N}}in l_infty.$ But it does'nt change the conclusion which is that $|(A_n(x))_{ninmathbb{N}}|_infty<+infty$ and in particular each term of this sequence is smaller than $+infty.$
$endgroup$
– Balloon
Jan 1 '16 at 14:04
add a comment |
$begingroup$
Note that you always have $forall ninmathbb{N},|A_n(x)|_infty<+infty$ with your definition for $A,$ and that it is not the same assertion that $sup_n|A_n(x)|_infty<+infty.$
$endgroup$
– Balloon
Jan 1 '16 at 12:48
$begingroup$
can you please explain me why I always have ∀n∈N,|An(x)|∞<+∞ in this case?
$endgroup$
– ruudvaan
Jan 1 '16 at 13:21
$begingroup$
Because you said that $forall ninmathbb{N},$ you have that $A_n$ maps any sequence $xin l_infty$ to an other sequence $A_n(x)in l_infty.$ As $l_infty$ is the space of bounded sequences, then you have that for any sequence $yin l_infty,$ $|y|_infty=sup_n|y_n|<+infty.$ So it is for $A_n(x).$
$endgroup$
– Balloon
Jan 1 '16 at 13:43
$begingroup$
since ∀n∈N,|An(x)|∞<+∞ then ∀n∈N,An(x)<+∞ please correct me:)) thats all I am looking for.
$endgroup$
– ruudvaan
Jan 1 '16 at 13:55
$begingroup$
This last point is true, but it seems I have misunderstood something : it is not $A_n(x)in l_infty$ but $(A_n(x))_{ninmathbb{N}}in l_infty.$ But it does'nt change the conclusion which is that $|(A_n(x))_{ninmathbb{N}}|_infty<+infty$ and in particular each term of this sequence is smaller than $+infty.$
$endgroup$
– Balloon
Jan 1 '16 at 14:04
$begingroup$
Note that you always have $forall ninmathbb{N},|A_n(x)|_infty<+infty$ with your definition for $A,$ and that it is not the same assertion that $sup_n|A_n(x)|_infty<+infty.$
$endgroup$
– Balloon
Jan 1 '16 at 12:48
$begingroup$
Note that you always have $forall ninmathbb{N},|A_n(x)|_infty<+infty$ with your definition for $A,$ and that it is not the same assertion that $sup_n|A_n(x)|_infty<+infty.$
$endgroup$
– Balloon
Jan 1 '16 at 12:48
$begingroup$
can you please explain me why I always have ∀n∈N,|An(x)|∞<+∞ in this case?
$endgroup$
– ruudvaan
Jan 1 '16 at 13:21
$begingroup$
can you please explain me why I always have ∀n∈N,|An(x)|∞<+∞ in this case?
$endgroup$
– ruudvaan
Jan 1 '16 at 13:21
$begingroup$
Because you said that $forall ninmathbb{N},$ you have that $A_n$ maps any sequence $xin l_infty$ to an other sequence $A_n(x)in l_infty.$ As $l_infty$ is the space of bounded sequences, then you have that for any sequence $yin l_infty,$ $|y|_infty=sup_n|y_n|<+infty.$ So it is for $A_n(x).$
$endgroup$
– Balloon
Jan 1 '16 at 13:43
$begingroup$
Because you said that $forall ninmathbb{N},$ you have that $A_n$ maps any sequence $xin l_infty$ to an other sequence $A_n(x)in l_infty.$ As $l_infty$ is the space of bounded sequences, then you have that for any sequence $yin l_infty,$ $|y|_infty=sup_n|y_n|<+infty.$ So it is for $A_n(x).$
$endgroup$
– Balloon
Jan 1 '16 at 13:43
$begingroup$
since ∀n∈N,|An(x)|∞<+∞ then ∀n∈N,An(x)<+∞ please correct me:)) thats all I am looking for.
$endgroup$
– ruudvaan
Jan 1 '16 at 13:55
$begingroup$
since ∀n∈N,|An(x)|∞<+∞ then ∀n∈N,An(x)<+∞ please correct me:)) thats all I am looking for.
$endgroup$
– ruudvaan
Jan 1 '16 at 13:55
$begingroup$
This last point is true, but it seems I have misunderstood something : it is not $A_n(x)in l_infty$ but $(A_n(x))_{ninmathbb{N}}in l_infty.$ But it does'nt change the conclusion which is that $|(A_n(x))_{ninmathbb{N}}|_infty<+infty$ and in particular each term of this sequence is smaller than $+infty.$
$endgroup$
– Balloon
Jan 1 '16 at 14:04
$begingroup$
This last point is true, but it seems I have misunderstood something : it is not $A_n(x)in l_infty$ but $(A_n(x))_{ninmathbb{N}}in l_infty.$ But it does'nt change the conclusion which is that $|(A_n(x))_{ninmathbb{N}}|_infty<+infty$ and in particular each term of this sequence is smaller than $+infty.$
$endgroup$
– Balloon
Jan 1 '16 at 14:04
add a comment |
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$begingroup$
Note that you always have $forall ninmathbb{N},|A_n(x)|_infty<+infty$ with your definition for $A,$ and that it is not the same assertion that $sup_n|A_n(x)|_infty<+infty.$
$endgroup$
– Balloon
Jan 1 '16 at 12:48
$begingroup$
can you please explain me why I always have ∀n∈N,|An(x)|∞<+∞ in this case?
$endgroup$
– ruudvaan
Jan 1 '16 at 13:21
$begingroup$
Because you said that $forall ninmathbb{N},$ you have that $A_n$ maps any sequence $xin l_infty$ to an other sequence $A_n(x)in l_infty.$ As $l_infty$ is the space of bounded sequences, then you have that for any sequence $yin l_infty,$ $|y|_infty=sup_n|y_n|<+infty.$ So it is for $A_n(x).$
$endgroup$
– Balloon
Jan 1 '16 at 13:43
$begingroup$
since ∀n∈N,|An(x)|∞<+∞ then ∀n∈N,An(x)<+∞ please correct me:)) thats all I am looking for.
$endgroup$
– ruudvaan
Jan 1 '16 at 13:55
$begingroup$
This last point is true, but it seems I have misunderstood something : it is not $A_n(x)in l_infty$ but $(A_n(x))_{ninmathbb{N}}in l_infty.$ But it does'nt change the conclusion which is that $|(A_n(x))_{ninmathbb{N}}|_infty<+infty$ and in particular each term of this sequence is smaller than $+infty.$
$endgroup$
– Balloon
Jan 1 '16 at 14:04