Mixing
Proof that $l^p$ with $1 leq p < infty$ is dense in $c_0$
Proof that $l^p$ with $1 leq p < infty$ is dense in $c_0$
My definition of $c_0$ is the following:
$$ c_0 = left{ x vert lim_{kto infty} x_k = 0right}.$$
I want to prove it like this: every element of $c_0$ can be written as limit of $l^p$.
First question, should I use the $p$ norm or the sup norm?
Second question, how should I start given that I think this statement is true?
general-topology functional-analysis lp-spaces
edited Nov 20 '18 at 21:38
Davide Giraudo
125k16150259
asked Nov 20 '18 at 21:24
qcc101
458113
add a comment |
Proof that $l^p$ with $1 leq p < infty$ is dense in $c_0$
My definition of $c_0$ is the following:
$$ c_0 = left{ x vert lim_{kto infty} x_k = 0right}.$$
I want to prove it like this: every element of $c_0$ can be written as limit of $l^p$.
First question, should I use the $p$ norm or the sup norm?
Second question, how should I start given that I think this statement is true?
general-topology functional-analysis lp-spaces
edited Nov 20 '18 at 21:38
Davide Giraudo
125k16150259
asked Nov 20 '18 at 21:24
qcc101
458113
add a comment |
Proof that $l^p$ with $1 leq p < infty$ is dense in $c_0$
My definition of $c_0$ is the following:
$$ c_0 = left{ x vert lim_{kto infty} x_k = 0right}.$$
I want to prove it like this: every element of $c_0$ can be written as limit of $l^p$.
First question, should I use the $p$ norm or the sup norm?
Second question, how should I start given that I think this statement is true?
general-topology functional-analysis lp-spaces
edited Nov 20 '18 at 21:38
Davide Giraudo
125k16150259
asked Nov 20 '18 at 21:24
qcc101
458113
Proof that $l^p$ with $1 leq p < infty$ is dense in $c_0$
My definition of $c_0$ is the following:
$$ c_0 = left{ x vert lim_{kto infty} x_k = 0right}.$$
I want to prove it like this: every element of $c_0$ can be written as limit of $l^p$.
First question, should I use the $p$ norm or the sup norm?
Second question, how should I start given that I think this statement is true?
general-topology functional-analysis lp-spaces
general-topology functional-analysis lp-spaces
edited Nov 20 '18 at 21:38
Davide Giraudo
125k16150259
asked Nov 20 '18 at 21:24
qcc101
458113
edited Nov 20 '18 at 21:38
Davide Giraudo
125k16150259
asked Nov 20 '18 at 21:24
qcc101
458113
edited Nov 20 '18 at 21:38
Davide Giraudo
125k16150259
edited Nov 20 '18 at 21:38
Davide Giraudo
125k16150259
edited Nov 20 '18 at 21:38
Davide Giraudo
125k16150259
125k16150259
asked Nov 20 '18 at 21:24
qcc101
458113
asked Nov 20 '18 at 21:24
qcc101
458113
asked Nov 20 '18 at 21:24
qcc101
458113
458113
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The space $c_0$ is usually considered as a subspace of $ell^infty$, and hence automatically inherits the sup-norm.
For any given $x = (x_1, x_2, ldots) in c_0$, you can truncate it to obtain an $ell^p$ sequence, i.e. you can consider
$$ hat x^{(n)} = (x_1, x_2, ldots, x_n , 0, 0 ,ldots) in ell^p. $$
Then
$$ Vert hat x^{(n)} - xVert_{infty} = sup_{j geq n+1} |x_j|, $$
which can evidently be made arbitrarily small by taking $n$ to be large, since $x_j to 0$ as $j to infty$.
answered Nov 20 '18 at 21:32
MisterRiemann
5,7841624
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The space $c_0$ is usually considered as a subspace of $ell^infty$, and hence automatically inherits the sup-norm.
For any given $x = (x_1, x_2, ldots) in c_0$, you can truncate it to obtain an $ell^p$ sequence, i.e. you can consider
$$ hat x^{(n)} = (x_1, x_2, ldots, x_n , 0, 0 ,ldots) in ell^p. $$
Then
$$ Vert hat x^{(n)} - xVert_{infty} = sup_{j geq n+1} |x_j|, $$
which can evidently be made arbitrarily small by taking $n$ to be large, since $x_j to 0$ as $j to infty$.
answered Nov 20 '18 at 21:32
MisterRiemann
5,7841624
add a comment |
The space $c_0$ is usually considered as a subspace of $ell^infty$, and hence automatically inherits the sup-norm.
For any given $x = (x_1, x_2, ldots) in c_0$, you can truncate it to obtain an $ell^p$ sequence, i.e. you can consider
$$ hat x^{(n)} = (x_1, x_2, ldots, x_n , 0, 0 ,ldots) in ell^p. $$
Then
$$ Vert hat x^{(n)} - xVert_{infty} = sup_{j geq n+1} |x_j|, $$
which can evidently be made arbitrarily small by taking $n$ to be large, since $x_j to 0$ as $j to infty$.
answered Nov 20 '18 at 21:32
MisterRiemann
5,7841624
add a comment |
The space $c_0$ is usually considered as a subspace of $ell^infty$, and hence automatically inherits the sup-norm.
For any given $x = (x_1, x_2, ldots) in c_0$, you can truncate it to obtain an $ell^p$ sequence, i.e. you can consider
$$ hat x^{(n)} = (x_1, x_2, ldots, x_n , 0, 0 ,ldots) in ell^p. $$
Then
$$ Vert hat x^{(n)} - xVert_{infty} = sup_{j geq n+1} |x_j|, $$
which can evidently be made arbitrarily small by taking $n$ to be large, since $x_j to 0$ as $j to infty$.
answered Nov 20 '18 at 21:32
MisterRiemann
5,7841624
The space $c_0$ is usually considered as a subspace of $ell^infty$, and hence automatically inherits the sup-norm.
For any given $x = (x_1, x_2, ldots) in c_0$, you can truncate it to obtain an $ell^p$ sequence, i.e. you can consider
$$ hat x^{(n)} = (x_1, x_2, ldots, x_n , 0, 0 ,ldots) in ell^p. $$
Then
$$ Vert hat x^{(n)} - xVert_{infty} = sup_{j geq n+1} |x_j|, $$
which can evidently be made arbitrarily small by taking $n$ to be large, since $x_j to 0$ as $j to infty$.
answered Nov 20 '18 at 21:32
MisterRiemann
5,7841624
answered Nov 20 '18 at 21:32
MisterRiemann
5,7841624
answered Nov 20 '18 at 21:32
MisterRiemann
5,7841624
answered Nov 20 '18 at 21:32
MisterRiemann
5,7841624
5,7841624
add a comment |
add a comment |
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