$xin l_1$, show sequence $x ^{left( k right)} = left( x _{1} , . . . , x _{k} , 0 , . . . right)in l_1$...
$begingroup$
Given any (fixed) element $xin l_1$, show that the sequence $x ^{left( k right)} = left( x _{1} , . . . , x _{k} , 0 , . . . right)in l_1$ (i.e. the first $k$ terms of $x$ followed by all $0s$) converges to x in $l_1-norm$. Show that the same holds true in $l_2$, and gives an example showing that it fails in $l_ {infty}$.
$l_1-norm = sum_{n=1}^{infty}|x_n|$
$l_p$ is the collection of all real sequences $x=(x_n)$ for which $sum _{n=1}^{infty }|x_n|^p < {infty }$.
$l_{infty}$ is the collection of all bounded real sequences.
Actually, I did't understand this question. I didn't know what the limit x is.
So, I didn't know how to prove this problem.
real-analysis analysis metric-spaces norm normed-spaces
edited Jun 24 '17 at 6:32
Empty
8,12652661
asked Sep 17 '16 at 21:04
User90User90
38917
$endgroup$
add a comment |
$begingroup$
Given any (fixed) element $xin l_1$, show that the sequence $x ^{left( k right)} = left( x _{1} , . . . , x _{k} , 0 , . . . right)in l_1$ (i.e. the first $k$ terms of $x$ followed by all $0s$) converges to x in $l_1-norm$. Show that the same holds true in $l_2$, and gives an example showing that it fails in $l_ {infty}$.
$l_1-norm = sum_{n=1}^{infty}|x_n|$
$l_p$ is the collection of all real sequences $x=(x_n)$ for which $sum _{n=1}^{infty }|x_n|^p < {infty }$.
$l_{infty}$ is the collection of all bounded real sequences.
Actually, I did't understand this question. I didn't know what the limit x is.
So, I didn't know how to prove this problem.
real-analysis analysis metric-spaces norm normed-spaces
edited Jun 24 '17 at 6:32
Empty
8,12652661
asked Sep 17 '16 at 21:04
User90User90
38917
$endgroup$
1
$begingroup$
$x$ is an element of $l^1$, and ${x^{(k)}}$ is a sequence of elements in $l^1$, the question is asking to show that $|x^{(k)} - x|_1$ goes to zero as $k$ approaches infinity.
$endgroup$
– Xiao
Sep 17 '16 at 21:09
add a comment |
$begingroup$
Given any (fixed) element $xin l_1$, show that the sequence $x ^{left( k right)} = left( x _{1} , . . . , x _{k} , 0 , . . . right)in l_1$ (i.e. the first $k$ terms of $x$ followed by all $0s$) converges to x in $l_1-norm$. Show that the same holds true in $l_2$, and gives an example showing that it fails in $l_ {infty}$.
$l_1-norm = sum_{n=1}^{infty}|x_n|$
$l_p$ is the collection of all real sequences $x=(x_n)$ for which $sum _{n=1}^{infty }|x_n|^p < {infty }$.
$l_{infty}$ is the collection of all bounded real sequences.
Actually, I did't understand this question. I didn't know what the limit x is.
So, I didn't know how to prove this problem.
real-analysis analysis metric-spaces norm normed-spaces
edited Jun 24 '17 at 6:32
Empty
8,12652661
asked Sep 17 '16 at 21:04
User90User90
38917
$endgroup$
Given any (fixed) element $xin l_1$, show that the sequence $x ^{left( k right)} = left( x _{1} , . . . , x _{k} , 0 , . . . right)in l_1$ (i.e. the first $k$ terms of $x$ followed by all $0s$) converges to x in $l_1-norm$. Show that the same holds true in $l_2$, and gives an example showing that it fails in $l_ {infty}$.
$l_1-norm = sum_{n=1}^{infty}|x_n|$
$l_p$ is the collection of all real sequences $x=(x_n)$ for which $sum _{n=1}^{infty }|x_n|^p < {infty }$.
$l_{infty}$ is the collection of all bounded real sequences.
Actually, I did't understand this question. I didn't know what the limit x is.
So, I didn't know how to prove this problem.
real-analysis analysis metric-spaces norm normed-spaces
real-analysis analysis metric-spaces norm normed-spaces
edited Jun 24 '17 at 6:32
Empty
8,12652661
asked Sep 17 '16 at 21:04
User90User90
38917
edited Jun 24 '17 at 6:32
Empty
8,12652661
asked Sep 17 '16 at 21:04
User90User90
38917
edited Jun 24 '17 at 6:32
Empty
8,12652661
edited Jun 24 '17 at 6:32
Empty
8,12652661
edited Jun 24 '17 at 6:32
Empty
8,12652661
8,12652661
asked Sep 17 '16 at 21:04
User90User90
38917
asked Sep 17 '16 at 21:04
User90User90
38917
asked Sep 17 '16 at 21:04
User90User90
38917
38917
1
$begingroup$
$x$ is an element of $l^1$, and ${x^{(k)}}$ is a sequence of elements in $l^1$, the question is asking to show that $|x^{(k)} - x|_1$ goes to zero as $k$ approaches infinity.
$endgroup$
– Xiao
Sep 17 '16 at 21:09
add a comment |
1
$begingroup$
$x$ is an element of $l^1$, and ${x^{(k)}}$ is a sequence of elements in $l^1$, the question is asking to show that $|x^{(k)} - x|_1$ goes to zero as $k$ approaches infinity.
$endgroup$
– Xiao
Sep 17 '16 at 21:09
1
1
$begingroup$
$x$ is an element of $l^1$, and ${x^{(k)}}$ is a sequence of elements in $l^1$, the question is asking to show that $|x^{(k)} - x|_1$ goes to zero as $k$ approaches infinity.
$endgroup$
– Xiao
Sep 17 '16 at 21:09
$begingroup$
$x$ is an element of $l^1$, and ${x^{(k)}}$ is a sequence of elements in $l^1$, the question is asking to show that $|x^{(k)} - x|_1$ goes to zero as $k$ approaches infinity.
$endgroup$
– Xiao
Sep 17 '16 at 21:09
add a comment |
1 Answer
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$begingroup$
The question is asking you to prove that the sequence $x^{(k)}$ converges to $x$ in the sense that for every $epsilon > 0$, $||x-x^{(k)}||_{l_1} < epsilon$ for all $k$ past a certain point.
Note $||x-x^{(k)}||_{l_1} = sum_{n=k+1}^{infty} x_n$. But we are given $x in l_1$ so that $sum_{n=1}^{infty} x_n$ converges. What does this tell you about sums of the form $sum_{n=k+1}^{infty} x_n$ for large $k$?
answered Sep 17 '16 at 21:46
mathworker21mathworker21
8,9421928
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1 Answer
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$begingroup$
The question is asking you to prove that the sequence $x^{(k)}$ converges to $x$ in the sense that for every $epsilon > 0$, $||x-x^{(k)}||_{l_1} < epsilon$ for all $k$ past a certain point.
Note $||x-x^{(k)}||_{l_1} = sum_{n=k+1}^{infty} x_n$. But we are given $x in l_1$ so that $sum_{n=1}^{infty} x_n$ converges. What does this tell you about sums of the form $sum_{n=k+1}^{infty} x_n$ for large $k$?
answered Sep 17 '16 at 21:46
mathworker21mathworker21
8,9421928
$endgroup$
add a comment |
$begingroup$
The question is asking you to prove that the sequence $x^{(k)}$ converges to $x$ in the sense that for every $epsilon > 0$, $||x-x^{(k)}||_{l_1} < epsilon$ for all $k$ past a certain point.
Note $||x-x^{(k)}||_{l_1} = sum_{n=k+1}^{infty} x_n$. But we are given $x in l_1$ so that $sum_{n=1}^{infty} x_n$ converges. What does this tell you about sums of the form $sum_{n=k+1}^{infty} x_n$ for large $k$?
answered Sep 17 '16 at 21:46
mathworker21mathworker21
8,9421928
$endgroup$
add a comment |
$begingroup$
The question is asking you to prove that the sequence $x^{(k)}$ converges to $x$ in the sense that for every $epsilon > 0$, $||x-x^{(k)}||_{l_1} < epsilon$ for all $k$ past a certain point.
Note $||x-x^{(k)}||_{l_1} = sum_{n=k+1}^{infty} x_n$. But we are given $x in l_1$ so that $sum_{n=1}^{infty} x_n$ converges. What does this tell you about sums of the form $sum_{n=k+1}^{infty} x_n$ for large $k$?
answered Sep 17 '16 at 21:46
mathworker21mathworker21
8,9421928
$endgroup$
The question is asking you to prove that the sequence $x^{(k)}$ converges to $x$ in the sense that for every $epsilon > 0$, $||x-x^{(k)}||_{l_1} < epsilon$ for all $k$ past a certain point.
Note $||x-x^{(k)}||_{l_1} = sum_{n=k+1}^{infty} x_n$. But we are given $x in l_1$ so that $sum_{n=1}^{infty} x_n$ converges. What does this tell you about sums of the form $sum_{n=k+1}^{infty} x_n$ for large $k$?
answered Sep 17 '16 at 21:46
mathworker21mathworker21
8,9421928
edited Nov 29 '18 at 15:26
answered Sep 17 '16 at 21:46
mathworker21mathworker21
8,9421928
answered Sep 17 '16 at 21:46
mathworker21mathworker21
8,9421928
answered Sep 17 '16 at 21:46
mathworker21mathworker21
8,9421928
8,9421928
add a comment |
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1
$begingroup$
$x$ is an element of $l^1$, and ${x^{(k)}}$ is a sequence of elements in $l^1$, the question is asking to show that $|x^{(k)} - x|_1$ goes to zero as $k$ approaches infinity.
$endgroup$
– Xiao
Sep 17 '16 at 21:09