Density in $ell^2$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Dense subspace of $ell^2$How to construct an “explicit” element of $(ell^infty(mathbb N))^* setminus ell^1(mathbb N)$?I would like to show that $ell^1$ is separableShow that sequences in $ell^1$ containing finite number of non-zero elements are dense in $ell^1$.Show the continuous embedding $ ell^2 subseteq c_0. $Question about relationship between $ell^ p$ spaces and Fourier coefficients.$ell^infty(mathbbN)$ is not separableIs $c_00$ space dense in $ell^infty$Question about surjectivity of operator in $ell^2$The spaces $ell^p, ; 1 leq p < + infty$ are separable. On the other side, $ell^infty$ is not.
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Density in $ell^2$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Dense subspace of $ell^2$How to construct an “explicit” element of $(ell^infty(mathbb N))^* setminus ell^1(mathbb N)$?I would like to show that $ell^1$ is separableShow that sequences in $ell^1$ containing finite number of non-zero elements are dense in $ell^1$.Show the continuous embedding $ ell^2 subseteq c_0. $Question about relationship between $ell^ p$ spaces and Fourier coefficients.$ell^infty(mathbbN)$ is not separableIs $c_00$ space dense in $ell^infty$Question about surjectivity of operator in $ell^2$The spaces $ell^p, ; 1 leq p < + infty$ are separable. On the other side, $ell^infty$ is not.
$begingroup$
How can i prove that $c_00$ is dense in $(ell^2, | cdot |_ell^2)$?
Where with $c_00$ i mean
beginequation
c_00= ,existsoverlineninmathbbN, forall n>overlinen , x(n)=0.
endequation
functional-analysis
asked Apr 1 at 7:42
GiovanniGiovanni
459
$endgroup$
add a comment |
$begingroup$
How can i prove that $c_00$ is dense in $(ell^2, | cdot |_ell^2)$?
Where with $c_00$ i mean
beginequation
c_00= ,existsoverlineninmathbbN, forall n>overlinen , x(n)=0.
endequation
functional-analysis
asked Apr 1 at 7:42
GiovanniGiovanni
459
$endgroup$
add a comment |
$begingroup$
How can i prove that $c_00$ is dense in $(ell^2, | cdot |_ell^2)$?
Where with $c_00$ i mean
beginequation
c_00= ,existsoverlineninmathbbN, forall n>overlinen , x(n)=0.
endequation
functional-analysis
asked Apr 1 at 7:42
GiovanniGiovanni
459
$endgroup$
How can i prove that $c_00$ is dense in $(ell^2, | cdot |_ell^2)$?
Where with $c_00$ i mean
beginequation
c_00= ,existsoverlineninmathbbN, forall n>overlinen , x(n)=0.
endequation
functional-analysis
functional-analysis
asked Apr 1 at 7:42
GiovanniGiovanni
459
asked Apr 1 at 7:42
GiovanniGiovanni
459
edited Apr 1 at 7:55
user408856
asked Apr 1 at 7:42
GiovanniGiovanni
459
asked Apr 1 at 7:42
GiovanniGiovanni
459
asked Apr 1 at 7:42
GiovanniGiovanni
459
459
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $epsilon>0$ be given, choose $xinell^2$. We know that $sum_iinmathbbN|x_i|^2<infty$, so there exists an $N$ such that $sum_i=N^infty |x_i|^2<epsilon$. Can you know construct a sequence $yin c_00$ such that $|x-y|_ell^2<epsilon$? How about
$$
y_n=begincases x_nquad,nleq N\
0quad,else
endcases
$$
$endgroup$
$begingroup$
Ok, now i got it! i was missing the part of taking a sequence $y_n$ that converges to $x$ in the $| cdot |_ell^2$ norm. Observing the convergence is enough to conclude about the density, right?
$endgroup$
– Giovanni
Apr 1 at 8:14
$begingroup$
A question: the above argument is enough to conclude that the space $(c_00, | cdot |_{ell^2)$ is not a complete space?
$endgroup$
– Giovanni
Apr 1 at 8:16
$begingroup$
I don't think this argument shows that the space is complete. It only shows denseness.
$endgroup$
– user370967
Apr 1 at 8:56
2
$begingroup$
Well, we know that $ell^2setminus c_00$ is not empty, so take $x in ell^2setminus c_00$. By denseness there is a sequence $(x_n)_n in BbbN$ that converges to $x$ in the $ell^2$-norm. As convergent sequences are always Cauchy (whenever Cauchy makes sense), we have found a Cauchy sequence in $c_00$ that does not converge in $c_00$
$endgroup$
– Jonas Lenz
Apr 1 at 9:23
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $epsilon>0$ be given, choose $xinell^2$. We know that $sum_iinmathbbN|x_i|^2<infty$, so there exists an $N$ such that $sum_i=N^infty |x_i|^2<epsilon$. Can you know construct a sequence $yin c_00$ such that $|x-y|_ell^2<epsilon$? How about
$$
y_n=begincases x_nquad,nleq N\
0quad,else
endcases
$$
$endgroup$
$begingroup$
Ok, now i got it! i was missing the part of taking a sequence $y_n$ that converges to $x$ in the $| cdot |_ell^2$ norm. Observing the convergence is enough to conclude about the density, right?
$endgroup$
– Giovanni
Apr 1 at 8:14
$begingroup$
A question: the above argument is enough to conclude that the space $(c_00, | cdot |_{ell^2)$ is not a complete space?
$endgroup$
– Giovanni
Apr 1 at 8:16
$begingroup$
I don't think this argument shows that the space is complete. It only shows denseness.
$endgroup$
– user370967
Apr 1 at 8:56
2
$begingroup$
Well, we know that $ell^2setminus c_00$ is not empty, so take $x in ell^2setminus c_00$. By denseness there is a sequence $(x_n)_n in BbbN$ that converges to $x$ in the $ell^2$-norm. As convergent sequences are always Cauchy (whenever Cauchy makes sense), we have found a Cauchy sequence in $c_00$ that does not converge in $c_00$
$endgroup$
– Jonas Lenz
Apr 1 at 9:23
add a comment |
$begingroup$
Let $epsilon>0$ be given, choose $xinell^2$. We know that $sum_iinmathbbN|x_i|^2<infty$, so there exists an $N$ such that $sum_i=N^infty |x_i|^2<epsilon$. Can you know construct a sequence $yin c_00$ such that $|x-y|_ell^2<epsilon$? How about
$$
y_n=begincases x_nquad,nleq N\
0quad,else
endcases
$$
$endgroup$
$begingroup$
Ok, now i got it! i was missing the part of taking a sequence $y_n$ that converges to $x$ in the $| cdot |_ell^2$ norm. Observing the convergence is enough to conclude about the density, right?
$endgroup$
– Giovanni
Apr 1 at 8:14
$begingroup$
A question: the above argument is enough to conclude that the space $(c_00, | cdot |_{ell^2)$ is not a complete space?
$endgroup$
– Giovanni
Apr 1 at 8:16
$begingroup$
I don't think this argument shows that the space is complete. It only shows denseness.
$endgroup$
– user370967
Apr 1 at 8:56
2
$begingroup$
Well, we know that $ell^2setminus c_00$ is not empty, so take $x in ell^2setminus c_00$. By denseness there is a sequence $(x_n)_n in BbbN$ that converges to $x$ in the $ell^2$-norm. As convergent sequences are always Cauchy (whenever Cauchy makes sense), we have found a Cauchy sequence in $c_00$ that does not converge in $c_00$
$endgroup$
– Jonas Lenz
Apr 1 at 9:23
add a comment |
$begingroup$
Let $epsilon>0$ be given, choose $xinell^2$. We know that $sum_iinmathbbN|x_i|^2<infty$, so there exists an $N$ such that $sum_i=N^infty |x_i|^2<epsilon$. Can you know construct a sequence $yin c_00$ such that $|x-y|_ell^2<epsilon$? How about
$$
y_n=begincases x_nquad,nleq N\
0quad,else
endcases
$$
$endgroup$
Let $epsilon>0$ be given, choose $xinell^2$. We know that $sum_iinmathbbN|x_i|^2<infty$, so there exists an $N$ such that $sum_i=N^infty |x_i|^2<epsilon$. Can you know construct a sequence $yin c_00$ such that $|x-y|_ell^2<epsilon$? How about
$$
y_n=begincases x_nquad,nleq N\
0quad,else
endcases
$$
edited Apr 1 at 7:51
answered Apr 1 at 7:47
user408856
$begingroup$
Ok, now i got it! i was missing the part of taking a sequence $y_n$ that converges to $x$ in the $| cdot |_ell^2$ norm. Observing the convergence is enough to conclude about the density, right?
$endgroup$
– Giovanni
Apr 1 at 8:14
$begingroup$
A question: the above argument is enough to conclude that the space $(c_00, | cdot |_{ell^2)$ is not a complete space?
$endgroup$
– Giovanni
Apr 1 at 8:16
$begingroup$
I don't think this argument shows that the space is complete. It only shows denseness.
$endgroup$
– user370967
Apr 1 at 8:56
2
$begingroup$
Well, we know that $ell^2setminus c_00$ is not empty, so take $x in ell^2setminus c_00$. By denseness there is a sequence $(x_n)_n in BbbN$ that converges to $x$ in the $ell^2$-norm. As convergent sequences are always Cauchy (whenever Cauchy makes sense), we have found a Cauchy sequence in $c_00$ that does not converge in $c_00$
$endgroup$
– Jonas Lenz
Apr 1 at 9:23
add a comment |
$begingroup$
Ok, now i got it! i was missing the part of taking a sequence $y_n$ that converges to $x$ in the $| cdot |_ell^2$ norm. Observing the convergence is enough to conclude about the density, right?
$endgroup$
– Giovanni
Apr 1 at 8:14
$begingroup$
A question: the above argument is enough to conclude that the space $(c_00, | cdot |_{ell^2)$ is not a complete space?
$endgroup$
– Giovanni
Apr 1 at 8:16
$begingroup$
I don't think this argument shows that the space is complete. It only shows denseness.
$endgroup$
– user370967
Apr 1 at 8:56
2
$begingroup$
Well, we know that $ell^2setminus c_00$ is not empty, so take $x in ell^2setminus c_00$. By denseness there is a sequence $(x_n)_n in BbbN$ that converges to $x$ in the $ell^2$-norm. As convergent sequences are always Cauchy (whenever Cauchy makes sense), we have found a Cauchy sequence in $c_00$ that does not converge in $c_00$
$endgroup$
– Jonas Lenz
Apr 1 at 9:23
$begingroup$
Ok, now i got it! i was missing the part of taking a sequence $y_n$ that converges to $x$ in the $| cdot |_ell^2$ norm. Observing the convergence is enough to conclude about the density, right?
$endgroup$
– Giovanni
Apr 1 at 8:14
$begingroup$
Ok, now i got it! i was missing the part of taking a sequence $y_n$ that converges to $x$ in the $| cdot |_ell^2$ norm. Observing the convergence is enough to conclude about the density, right?
$endgroup$
– Giovanni
Apr 1 at 8:14
$begingroup$
A question: the above argument is enough to conclude that the space $(c_00, | cdot |_{ell^2)$ is not a complete space?
$endgroup$
– Giovanni
Apr 1 at 8:16
$begingroup$
A question: the above argument is enough to conclude that the space $(c_00, | cdot |_{ell^2)$ is not a complete space?
$endgroup$
– Giovanni
Apr 1 at 8:16
$begingroup$
I don't think this argument shows that the space is complete. It only shows denseness.
$endgroup$
– user370967
Apr 1 at 8:56
$begingroup$
I don't think this argument shows that the space is complete. It only shows denseness.
$endgroup$
– user370967
Apr 1 at 8:56
2
2
$begingroup$
Well, we know that $ell^2setminus c_00$ is not empty, so take $x in ell^2setminus c_00$. By denseness there is a sequence $(x_n)_n in BbbN$ that converges to $x$ in the $ell^2$-norm. As convergent sequences are always Cauchy (whenever Cauchy makes sense), we have found a Cauchy sequence in $c_00$ that does not converge in $c_00$
$endgroup$
– Jonas Lenz
Apr 1 at 9:23
$begingroup$
Well, we know that $ell^2setminus c_00$ is not empty, so take $x in ell^2setminus c_00$. By denseness there is a sequence $(x_n)_n in BbbN$ that converges to $x$ in the $ell^2$-norm. As convergent sequences are always Cauchy (whenever Cauchy makes sense), we have found a Cauchy sequence in $c_00$ that does not converge in $c_00$
$endgroup$
– Jonas Lenz
Apr 1 at 9:23
add a comment |
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