Mixing
Normed vector space & Schauder basis exercise
$begingroup$
I am taking a new course in functional analysis, so appreciate any help for this exercise. Let the set of sequences $l_p = {a ∈ R^N: sum^∞_{i=1} |a_i|^p< ∞}$. I wish to determine or show:
(i) $l_p$ is a vector subspace of $R^N$. For this, I just have to show closed under scalar multiplication and addition?
(ii) $l_p$ has no countable basis? Do I need to consider some linearly independent vector to prove this?
(iii) The map $||a||_p := (sum_i|a_i|^p
)^{1/p}$ defines a norm on $l_p$. The main thing here is I can use Minkowski’s inequality? And to prove it, start with
$|x + y|^p = |x + y| · |x + y|^{p−1} ≤ |x| · |x + y|^{p−1} + |y| · |x + y|^{p−1}$ followed by an application of Holder’s inequality?
(iv) The normed vector space $(l_p, || − ||_p)$ is not Banach? So I have to find a cauchy sequence with limit not in $l_p$ right?
(v) The “standard basis” elements $e_i$ form a topological spanning set of $l_p$ so that the algebraic linear span of the $e_i$’s is dense in $l_p$. In other words, I want to prove it is a Schauder basis right because the sum converge in $l_p$?
real-analysis linear-algebra general-topology functional-analysis measure-theory
edited Feb 6 at 20:17
daw
25.1k1745
asked Feb 1 at 9:20
HomaniacHomaniac
502312
$endgroup$
|
show 1 more comment
$begingroup$
I am taking a new course in functional analysis, so appreciate any help for this exercise. Let the set of sequences $l_p = {a ∈ R^N: sum^∞_{i=1} |a_i|^p< ∞}$. I wish to determine or show:
(i) $l_p$ is a vector subspace of $R^N$. For this, I just have to show closed under scalar multiplication and addition?
(ii) $l_p$ has no countable basis? Do I need to consider some linearly independent vector to prove this?
(iii) The map $||a||_p := (sum_i|a_i|^p
)^{1/p}$ defines a norm on $l_p$. The main thing here is I can use Minkowski’s inequality? And to prove it, start with
$|x + y|^p = |x + y| · |x + y|^{p−1} ≤ |x| · |x + y|^{p−1} + |y| · |x + y|^{p−1}$ followed by an application of Holder’s inequality?
(iv) The normed vector space $(l_p, || − ||_p)$ is not Banach? So I have to find a cauchy sequence with limit not in $l_p$ right?
(v) The “standard basis” elements $e_i$ form a topological spanning set of $l_p$ so that the algebraic linear span of the $e_i$’s is dense in $l_p$. In other words, I want to prove it is a Schauder basis right because the sum converge in $l_p$?
real-analysis linear-algebra general-topology functional-analysis measure-theory
edited Feb 6 at 20:17
daw
25.1k1745
asked Feb 1 at 9:20
HomaniacHomaniac
502312
$endgroup$
$begingroup$
My guess is that you wrote $p$ instead of $ell_p$ in several places. Besides, the usual notation is $ell^p$.
$endgroup$
– José Carlos Santos
Feb 1 at 9:26
$begingroup$
Are you taking $p>0$ 0r $pge 1?$
$endgroup$
– zhw.
Feb 3 at 22:30
$begingroup$
We are taking 1 ≤ p < ∞
$endgroup$
– Homaniac
Feb 4 at 4:52
$begingroup$
Do you want detailed proofs? It seems like you are on the right track.
$endgroup$
– Matematleta
Feb 4 at 16:28
$begingroup$
Yea I am hoping for detailed proofs or at least more details for each part, thank you :)
$endgroup$
– Homaniac
Feb 4 at 17:52
|
show 1 more comment
$begingroup$
I am taking a new course in functional analysis, so appreciate any help for this exercise. Let the set of sequences $l_p = {a ∈ R^N: sum^∞_{i=1} |a_i|^p< ∞}$. I wish to determine or show:
(i) $l_p$ is a vector subspace of $R^N$. For this, I just have to show closed under scalar multiplication and addition?
(ii) $l_p$ has no countable basis? Do I need to consider some linearly independent vector to prove this?
(iii) The map $||a||_p := (sum_i|a_i|^p
)^{1/p}$ defines a norm on $l_p$. The main thing here is I can use Minkowski’s inequality? And to prove it, start with
$|x + y|^p = |x + y| · |x + y|^{p−1} ≤ |x| · |x + y|^{p−1} + |y| · |x + y|^{p−1}$ followed by an application of Holder’s inequality?
(iv) The normed vector space $(l_p, || − ||_p)$ is not Banach? So I have to find a cauchy sequence with limit not in $l_p$ right?
(v) The “standard basis” elements $e_i$ form a topological spanning set of $l_p$ so that the algebraic linear span of the $e_i$’s is dense in $l_p$. In other words, I want to prove it is a Schauder basis right because the sum converge in $l_p$?
real-analysis linear-algebra general-topology functional-analysis measure-theory
edited Feb 6 at 20:17
daw
25.1k1745
asked Feb 1 at 9:20
HomaniacHomaniac
502312
$endgroup$
I am taking a new course in functional analysis, so appreciate any help for this exercise. Let the set of sequences $l_p = {a ∈ R^N: sum^∞_{i=1} |a_i|^p< ∞}$. I wish to determine or show:
(i) $l_p$ is a vector subspace of $R^N$. For this, I just have to show closed under scalar multiplication and addition?
(ii) $l_p$ has no countable basis? Do I need to consider some linearly independent vector to prove this?
(iii) The map $||a||_p := (sum_i|a_i|^p
)^{1/p}$ defines a norm on $l_p$. The main thing here is I can use Minkowski’s inequality? And to prove it, start with
$|x + y|^p = |x + y| · |x + y|^{p−1} ≤ |x| · |x + y|^{p−1} + |y| · |x + y|^{p−1}$ followed by an application of Holder’s inequality?
(iv) The normed vector space $(l_p, || − ||_p)$ is not Banach? So I have to find a cauchy sequence with limit not in $l_p$ right?
(v) The “standard basis” elements $e_i$ form a topological spanning set of $l_p$ so that the algebraic linear span of the $e_i$’s is dense in $l_p$. In other words, I want to prove it is a Schauder basis right because the sum converge in $l_p$?
real-analysis linear-algebra general-topology functional-analysis measure-theory
real-analysis linear-algebra general-topology functional-analysis measure-theory
edited Feb 6 at 20:17
daw
25.1k1745
asked Feb 1 at 9:20
HomaniacHomaniac
502312
edited Feb 6 at 20:17
daw
25.1k1745
asked Feb 1 at 9:20
HomaniacHomaniac
502312
edited Feb 6 at 20:17
daw
25.1k1745
edited Feb 6 at 20:17
daw
25.1k1745
edited Feb 6 at 20:17
daw
25.1k1745
25.1k1745
asked Feb 1 at 9:20
HomaniacHomaniac
502312
asked Feb 1 at 9:20
HomaniacHomaniac
502312
asked Feb 1 at 9:20
HomaniacHomaniac
502312
502312
$begingroup$
My guess is that you wrote $p$ instead of $ell_p$ in several places. Besides, the usual notation is $ell^p$.
$endgroup$
– José Carlos Santos
Feb 1 at 9:26
$begingroup$
Are you taking $p>0$ 0r $pge 1?$
$endgroup$
– zhw.
Feb 3 at 22:30
$begingroup$
We are taking 1 ≤ p < ∞
$endgroup$
– Homaniac
Feb 4 at 4:52
$begingroup$
Do you want detailed proofs? It seems like you are on the right track.
$endgroup$
– Matematleta
Feb 4 at 16:28
$begingroup$
Yea I am hoping for detailed proofs or at least more details for each part, thank you :)
$endgroup$
– Homaniac
Feb 4 at 17:52
|
show 1 more comment
$begingroup$
My guess is that you wrote $p$ instead of $ell_p$ in several places. Besides, the usual notation is $ell^p$.
$endgroup$
– José Carlos Santos
Feb 1 at 9:26
$begingroup$
Are you taking $p>0$ 0r $pge 1?$
$endgroup$
– zhw.
Feb 3 at 22:30
$begingroup$
We are taking 1 ≤ p < ∞
$endgroup$
– Homaniac
Feb 4 at 4:52
$begingroup$
Do you want detailed proofs? It seems like you are on the right track.
$endgroup$
– Matematleta
Feb 4 at 16:28
$begingroup$
Yea I am hoping for detailed proofs or at least more details for each part, thank you :)
$endgroup$
– Homaniac
Feb 4 at 17:52
$begingroup$
My guess is that you wrote $p$ instead of $ell_p$ in several places. Besides, the usual notation is $ell^p$.
$endgroup$
– José Carlos Santos
Feb 1 at 9:26
$begingroup$
My guess is that you wrote $p$ instead of $ell_p$ in several places. Besides, the usual notation is $ell^p$.
$endgroup$
– José Carlos Santos
Feb 1 at 9:26
$begingroup$
Are you taking $p>0$ 0r $pge 1?$
$endgroup$
– zhw.
Feb 3 at 22:30
$begingroup$
Are you taking $p>0$ 0r $pge 1?$
$endgroup$
– zhw.
Feb 3 at 22:30
$begingroup$
We are taking 1 ≤ p < ∞
$endgroup$
– Homaniac
Feb 4 at 4:52
$begingroup$
We are taking 1 ≤ p < ∞
$endgroup$
– Homaniac
Feb 4 at 4:52
$begingroup$
Do you want detailed proofs? It seems like you are on the right track.
$endgroup$
– Matematleta
Feb 4 at 16:28
$begingroup$
Do you want detailed proofs? It seems like you are on the right track.
$endgroup$
– Matematleta
Feb 4 at 16:28
$begingroup$
Yea I am hoping for detailed proofs or at least more details for each part, thank you :)
$endgroup$
– Homaniac
Feb 4 at 17:52
$begingroup$
Yea I am hoping for detailed proofs or at least more details for each part, thank you :)
$endgroup$
– Homaniac
Feb 4 at 17:52
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
$1). $ Suppose $a=sum^∞_{i-1} |a_i|^p< ∞$ and $b=sum^∞_{i-1} |b_i|^p< ∞.$ Then, the Minkowski inequality applies to show that, since $|a|_p=a^{1/p}<infty; |b|_p=b^{1/p}<infty$ so $|a+b|_ple |a|_p+|b|_p<infty$ which implies that $a+bin ell^p$. Now, if $alphain mathbb R$, then $|alpha a|_p=alpha^{1/p}|a|_p<infty,$ so $alpha ain ell^p.$
$2). $ It is enough to show that there is an uncountable linearly independent set of vectors in $ell^p$. The trick I know is the following: Since $|mathbb N|=|mathbb Q|$, we can assume that $l^p = {a ∈ R^{mathbb Q}: sum^∞_{i-1} |a_i|^p< ∞}, $ that is, functions from $mathbb Q$ to $mathbb R.$ Now define, for each $xin mathbb R, f_x:mathbb Qto mathbb R$ by $f_x(q)=chi_E(q)$, where $E={qin mathbb Q:q<x}.$ Then, ${f_x:xin mathbb R}$ is the required uncountable, linearly independent set.
As an aside, if you know the Baire category theorem, you can show that no infinite dimensional Banach space, $X$, can have a countable Hamel basis, for if ${e_1, e_2,cdots }$ is such a basis, then, letting $F_n=$span${e_1,cdots,e_n},$ we have $X=bigcup F_n$, a countable union of closed sets with empty interior, and this is a contradiction of Baire's theorem.
$3). $ The Minkowski inequality shows that the triangle inequality holds. And if $alphain mathbb R, $ it is very easy to show that $|alpha a|_p=|alpha|cdot |a|_p.$ Finally, if $|a|_p=0, $ then $left ( sum |a_i|^p right )^{1/p}=0Rightarrow a_i=0; i=1,2,cdots $ so $a=0.$
$4). ell^p$ is complete and so Banach. Here is a sketch: take a Cauchy sequence $(a_n)subseteq ell^p$ and show that each coordinate $(a^i_n)_i$ converges to a real number $a_i: i=1,2,cdots$ (because the coordinates themselves are Cauchy in $mathbb R$ and $mathbb R$ is complete.) Now, consider $a=(a_1,a_2,cdots)$ and show that $ain ell^p$ and that $|a-a_n|to 0$ as $nto infty.$
$5). $ You need to show that span${e_i:iin mathbb N}$ is dense in $ell^p$. So, let $a=(a_1,a_2,cdots )in ell^p$ and consider $a_n=(a_1,a_2,cdots, a_n,0,0,cdots)=sum^n_{i=1} a_ie_i.$ Then, $a-a_n=(0,0,cdots, n+1,n+2,cdots)=sum^{infty}_{i=n+1}a_iRightarrow |a-a_n|^ple sum^{infty}_{i=n+1}|a_i|^pto 0$ because $ain ell^p.$
answered Feb 5 at 17:40
MatematletaMatematleta
12.1k21020
$endgroup$
$begingroup$
For part 4, I have tried to look for more details for the full proof in math.stackexchange.com/questions/72198/… and math.stackexchange.com/questions/328479/… but am not sure.. isit correct somewhere?
$endgroup$
– Homaniac
Feb 5 at 21:41
$begingroup$
Wouldn't you learn more if you tried yourself? Here's a hint on how to start: the Cauchy sequence $(a_n)$ is bounded by some $C$. Write it as a matrix. Fix a column $k$ and consider $(a_n^k)$ for $n=1,2,3,..$ etc. This is a Cauchy sequence in $mathbb R$ so it converges to an $a_k$. So now you have a sequence $a=(a_k)$ and you want to show it's in $ell^p$ and $(a_n)$ converges to it. So fix an integer $N$, take $sum^N_{k=1}|a_n^k|^ple C$ and let $nto infty,$ to get $sum^N_{k=1}|a_k|^ple C$. Letting $Nto infty$ shows that $(a_k)in ell^p.$ To show $a_nto a$, use a similar argument.
$endgroup$
– Matematleta
Feb 5 at 22:56
1
$begingroup$
Thank you, to show $a_n → a$, fix $epsilon>0$, take $N$ such that $|a_n-a_m|_pleq epsilon$ for all $n,mgeq N$. Then $left(sum_{k=1}^N|x_m(k)-x_n(k)|^pright)^{1/p}leq |x_m-x_n|_pleq epsilon$, then letting $m → infty$, followed by $N$, we get $|a-a_n|_pleq epsilon$ for all $ngeq N$. Right :)
$endgroup$
– Homaniac
Feb 8 at 15:09
$begingroup$
Yes, that's correct.
$endgroup$
– Matematleta
Feb 8 at 18:11
add a comment |
$begingroup$
(i) And that it is non-empty, but that's trivial.
(ii) Assume that it had a countable basis and try to construct an element that cannot be written as a finite linear combination.
(iii) Yes, your attempt looks good.
(iv) It is Banach (assuming $p geq 1$). You need to show completeness.
(v) Yes, exactly. Cannot be a Hamel basis because of (ii).
answered Feb 1 at 11:15
KlausKlaus
2,955214
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
$1). $ Suppose $a=sum^∞_{i-1} |a_i|^p< ∞$ and $b=sum^∞_{i-1} |b_i|^p< ∞.$ Then, the Minkowski inequality applies to show that, since $|a|_p=a^{1/p}<infty; |b|_p=b^{1/p}<infty$ so $|a+b|_ple |a|_p+|b|_p<infty$ which implies that $a+bin ell^p$. Now, if $alphain mathbb R$, then $|alpha a|_p=alpha^{1/p}|a|_p<infty,$ so $alpha ain ell^p.$
$2). $ It is enough to show that there is an uncountable linearly independent set of vectors in $ell^p$. The trick I know is the following: Since $|mathbb N|=|mathbb Q|$, we can assume that $l^p = {a ∈ R^{mathbb Q}: sum^∞_{i-1} |a_i|^p< ∞}, $ that is, functions from $mathbb Q$ to $mathbb R.$ Now define, for each $xin mathbb R, f_x:mathbb Qto mathbb R$ by $f_x(q)=chi_E(q)$, where $E={qin mathbb Q:q<x}.$ Then, ${f_x:xin mathbb R}$ is the required uncountable, linearly independent set.
As an aside, if you know the Baire category theorem, you can show that no infinite dimensional Banach space, $X$, can have a countable Hamel basis, for if ${e_1, e_2,cdots }$ is such a basis, then, letting $F_n=$span${e_1,cdots,e_n},$ we have $X=bigcup F_n$, a countable union of closed sets with empty interior, and this is a contradiction of Baire's theorem.
$3). $ The Minkowski inequality shows that the triangle inequality holds. And if $alphain mathbb R, $ it is very easy to show that $|alpha a|_p=|alpha|cdot |a|_p.$ Finally, if $|a|_p=0, $ then $left ( sum |a_i|^p right )^{1/p}=0Rightarrow a_i=0; i=1,2,cdots $ so $a=0.$
$4). ell^p$ is complete and so Banach. Here is a sketch: take a Cauchy sequence $(a_n)subseteq ell^p$ and show that each coordinate $(a^i_n)_i$ converges to a real number $a_i: i=1,2,cdots$ (because the coordinates themselves are Cauchy in $mathbb R$ and $mathbb R$ is complete.) Now, consider $a=(a_1,a_2,cdots)$ and show that $ain ell^p$ and that $|a-a_n|to 0$ as $nto infty.$
$5). $ You need to show that span${e_i:iin mathbb N}$ is dense in $ell^p$. So, let $a=(a_1,a_2,cdots )in ell^p$ and consider $a_n=(a_1,a_2,cdots, a_n,0,0,cdots)=sum^n_{i=1} a_ie_i.$ Then, $a-a_n=(0,0,cdots, n+1,n+2,cdots)=sum^{infty}_{i=n+1}a_iRightarrow |a-a_n|^ple sum^{infty}_{i=n+1}|a_i|^pto 0$ because $ain ell^p.$
answered Feb 5 at 17:40
MatematletaMatematleta
12.1k21020
$endgroup$
$begingroup$
For part 4, I have tried to look for more details for the full proof in math.stackexchange.com/questions/72198/… and math.stackexchange.com/questions/328479/… but am not sure.. isit correct somewhere?
$endgroup$
– Homaniac
Feb 5 at 21:41
$begingroup$
Wouldn't you learn more if you tried yourself? Here's a hint on how to start: the Cauchy sequence $(a_n)$ is bounded by some $C$. Write it as a matrix. Fix a column $k$ and consider $(a_n^k)$ for $n=1,2,3,..$ etc. This is a Cauchy sequence in $mathbb R$ so it converges to an $a_k$. So now you have a sequence $a=(a_k)$ and you want to show it's in $ell^p$ and $(a_n)$ converges to it. So fix an integer $N$, take $sum^N_{k=1}|a_n^k|^ple C$ and let $nto infty,$ to get $sum^N_{k=1}|a_k|^ple C$. Letting $Nto infty$ shows that $(a_k)in ell^p.$ To show $a_nto a$, use a similar argument.
$endgroup$
– Matematleta
Feb 5 at 22:56
1
$begingroup$
Thank you, to show $a_n → a$, fix $epsilon>0$, take $N$ such that $|a_n-a_m|_pleq epsilon$ for all $n,mgeq N$. Then $left(sum_{k=1}^N|x_m(k)-x_n(k)|^pright)^{1/p}leq |x_m-x_n|_pleq epsilon$, then letting $m → infty$, followed by $N$, we get $|a-a_n|_pleq epsilon$ for all $ngeq N$. Right :)
$endgroup$
– Homaniac
Feb 8 at 15:09
$begingroup$
Yes, that's correct.
$endgroup$
– Matematleta
Feb 8 at 18:11
add a comment |
$begingroup$
$1). $ Suppose $a=sum^∞_{i-1} |a_i|^p< ∞$ and $b=sum^∞_{i-1} |b_i|^p< ∞.$ Then, the Minkowski inequality applies to show that, since $|a|_p=a^{1/p}<infty; |b|_p=b^{1/p}<infty$ so $|a+b|_ple |a|_p+|b|_p<infty$ which implies that $a+bin ell^p$. Now, if $alphain mathbb R$, then $|alpha a|_p=alpha^{1/p}|a|_p<infty,$ so $alpha ain ell^p.$
$2). $ It is enough to show that there is an uncountable linearly independent set of vectors in $ell^p$. The trick I know is the following: Since $|mathbb N|=|mathbb Q|$, we can assume that $l^p = {a ∈ R^{mathbb Q}: sum^∞_{i-1} |a_i|^p< ∞}, $ that is, functions from $mathbb Q$ to $mathbb R.$ Now define, for each $xin mathbb R, f_x:mathbb Qto mathbb R$ by $f_x(q)=chi_E(q)$, where $E={qin mathbb Q:q<x}.$ Then, ${f_x:xin mathbb R}$ is the required uncountable, linearly independent set.
As an aside, if you know the Baire category theorem, you can show that no infinite dimensional Banach space, $X$, can have a countable Hamel basis, for if ${e_1, e_2,cdots }$ is such a basis, then, letting $F_n=$span${e_1,cdots,e_n},$ we have $X=bigcup F_n$, a countable union of closed sets with empty interior, and this is a contradiction of Baire's theorem.
$3). $ The Minkowski inequality shows that the triangle inequality holds. And if $alphain mathbb R, $ it is very easy to show that $|alpha a|_p=|alpha|cdot |a|_p.$ Finally, if $|a|_p=0, $ then $left ( sum |a_i|^p right )^{1/p}=0Rightarrow a_i=0; i=1,2,cdots $ so $a=0.$
$4). ell^p$ is complete and so Banach. Here is a sketch: take a Cauchy sequence $(a_n)subseteq ell^p$ and show that each coordinate $(a^i_n)_i$ converges to a real number $a_i: i=1,2,cdots$ (because the coordinates themselves are Cauchy in $mathbb R$ and $mathbb R$ is complete.) Now, consider $a=(a_1,a_2,cdots)$ and show that $ain ell^p$ and that $|a-a_n|to 0$ as $nto infty.$
$5). $ You need to show that span${e_i:iin mathbb N}$ is dense in $ell^p$. So, let $a=(a_1,a_2,cdots )in ell^p$ and consider $a_n=(a_1,a_2,cdots, a_n,0,0,cdots)=sum^n_{i=1} a_ie_i.$ Then, $a-a_n=(0,0,cdots, n+1,n+2,cdots)=sum^{infty}_{i=n+1}a_iRightarrow |a-a_n|^ple sum^{infty}_{i=n+1}|a_i|^pto 0$ because $ain ell^p.$
answered Feb 5 at 17:40
MatematletaMatematleta
12.1k21020
$endgroup$
$begingroup$
For part 4, I have tried to look for more details for the full proof in math.stackexchange.com/questions/72198/… and math.stackexchange.com/questions/328479/… but am not sure.. isit correct somewhere?
$endgroup$
– Homaniac
Feb 5 at 21:41
$begingroup$
Wouldn't you learn more if you tried yourself? Here's a hint on how to start: the Cauchy sequence $(a_n)$ is bounded by some $C$. Write it as a matrix. Fix a column $k$ and consider $(a_n^k)$ for $n=1,2,3,..$ etc. This is a Cauchy sequence in $mathbb R$ so it converges to an $a_k$. So now you have a sequence $a=(a_k)$ and you want to show it's in $ell^p$ and $(a_n)$ converges to it. So fix an integer $N$, take $sum^N_{k=1}|a_n^k|^ple C$ and let $nto infty,$ to get $sum^N_{k=1}|a_k|^ple C$. Letting $Nto infty$ shows that $(a_k)in ell^p.$ To show $a_nto a$, use a similar argument.
$endgroup$
– Matematleta
Feb 5 at 22:56
1
$begingroup$
Thank you, to show $a_n → a$, fix $epsilon>0$, take $N$ such that $|a_n-a_m|_pleq epsilon$ for all $n,mgeq N$. Then $left(sum_{k=1}^N|x_m(k)-x_n(k)|^pright)^{1/p}leq |x_m-x_n|_pleq epsilon$, then letting $m → infty$, followed by $N$, we get $|a-a_n|_pleq epsilon$ for all $ngeq N$. Right :)
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– Homaniac
Feb 8 at 15:09
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Yes, that's correct.
$endgroup$
– Matematleta
Feb 8 at 18:11
add a comment |
$begingroup$
$1). $ Suppose $a=sum^∞_{i-1} |a_i|^p< ∞$ and $b=sum^∞_{i-1} |b_i|^p< ∞.$ Then, the Minkowski inequality applies to show that, since $|a|_p=a^{1/p}<infty; |b|_p=b^{1/p}<infty$ so $|a+b|_ple |a|_p+|b|_p<infty$ which implies that $a+bin ell^p$. Now, if $alphain mathbb R$, then $|alpha a|_p=alpha^{1/p}|a|_p<infty,$ so $alpha ain ell^p.$
$2). $ It is enough to show that there is an uncountable linearly independent set of vectors in $ell^p$. The trick I know is the following: Since $|mathbb N|=|mathbb Q|$, we can assume that $l^p = {a ∈ R^{mathbb Q}: sum^∞_{i-1} |a_i|^p< ∞}, $ that is, functions from $mathbb Q$ to $mathbb R.$ Now define, for each $xin mathbb R, f_x:mathbb Qto mathbb R$ by $f_x(q)=chi_E(q)$, where $E={qin mathbb Q:q<x}.$ Then, ${f_x:xin mathbb R}$ is the required uncountable, linearly independent set.
As an aside, if you know the Baire category theorem, you can show that no infinite dimensional Banach space, $X$, can have a countable Hamel basis, for if ${e_1, e_2,cdots }$ is such a basis, then, letting $F_n=$span${e_1,cdots,e_n},$ we have $X=bigcup F_n$, a countable union of closed sets with empty interior, and this is a contradiction of Baire's theorem.
$3). $ The Minkowski inequality shows that the triangle inequality holds. And if $alphain mathbb R, $ it is very easy to show that $|alpha a|_p=|alpha|cdot |a|_p.$ Finally, if $|a|_p=0, $ then $left ( sum |a_i|^p right )^{1/p}=0Rightarrow a_i=0; i=1,2,cdots $ so $a=0.$
$4). ell^p$ is complete and so Banach. Here is a sketch: take a Cauchy sequence $(a_n)subseteq ell^p$ and show that each coordinate $(a^i_n)_i$ converges to a real number $a_i: i=1,2,cdots$ (because the coordinates themselves are Cauchy in $mathbb R$ and $mathbb R$ is complete.) Now, consider $a=(a_1,a_2,cdots)$ and show that $ain ell^p$ and that $|a-a_n|to 0$ as $nto infty.$
$5). $ You need to show that span${e_i:iin mathbb N}$ is dense in $ell^p$. So, let $a=(a_1,a_2,cdots )in ell^p$ and consider $a_n=(a_1,a_2,cdots, a_n,0,0,cdots)=sum^n_{i=1} a_ie_i.$ Then, $a-a_n=(0,0,cdots, n+1,n+2,cdots)=sum^{infty}_{i=n+1}a_iRightarrow |a-a_n|^ple sum^{infty}_{i=n+1}|a_i|^pto 0$ because $ain ell^p.$
answered Feb 5 at 17:40
MatematletaMatematleta
12.1k21020
$endgroup$
$1). $ Suppose $a=sum^∞_{i-1} |a_i|^p< ∞$ and $b=sum^∞_{i-1} |b_i|^p< ∞.$ Then, the Minkowski inequality applies to show that, since $|a|_p=a^{1/p}<infty; |b|_p=b^{1/p}<infty$ so $|a+b|_ple |a|_p+|b|_p<infty$ which implies that $a+bin ell^p$. Now, if $alphain mathbb R$, then $|alpha a|_p=alpha^{1/p}|a|_p<infty,$ so $alpha ain ell^p.$
$2). $ It is enough to show that there is an uncountable linearly independent set of vectors in $ell^p$. The trick I know is the following: Since $|mathbb N|=|mathbb Q|$, we can assume that $l^p = {a ∈ R^{mathbb Q}: sum^∞_{i-1} |a_i|^p< ∞}, $ that is, functions from $mathbb Q$ to $mathbb R.$ Now define, for each $xin mathbb R, f_x:mathbb Qto mathbb R$ by $f_x(q)=chi_E(q)$, where $E={qin mathbb Q:q<x}.$ Then, ${f_x:xin mathbb R}$ is the required uncountable, linearly independent set.
As an aside, if you know the Baire category theorem, you can show that no infinite dimensional Banach space, $X$, can have a countable Hamel basis, for if ${e_1, e_2,cdots }$ is such a basis, then, letting $F_n=$span${e_1,cdots,e_n},$ we have $X=bigcup F_n$, a countable union of closed sets with empty interior, and this is a contradiction of Baire's theorem.
$3). $ The Minkowski inequality shows that the triangle inequality holds. And if $alphain mathbb R, $ it is very easy to show that $|alpha a|_p=|alpha|cdot |a|_p.$ Finally, if $|a|_p=0, $ then $left ( sum |a_i|^p right )^{1/p}=0Rightarrow a_i=0; i=1,2,cdots $ so $a=0.$
$4). ell^p$ is complete and so Banach. Here is a sketch: take a Cauchy sequence $(a_n)subseteq ell^p$ and show that each coordinate $(a^i_n)_i$ converges to a real number $a_i: i=1,2,cdots$ (because the coordinates themselves are Cauchy in $mathbb R$ and $mathbb R$ is complete.) Now, consider $a=(a_1,a_2,cdots)$ and show that $ain ell^p$ and that $|a-a_n|to 0$ as $nto infty.$
$5). $ You need to show that span${e_i:iin mathbb N}$ is dense in $ell^p$. So, let $a=(a_1,a_2,cdots )in ell^p$ and consider $a_n=(a_1,a_2,cdots, a_n,0,0,cdots)=sum^n_{i=1} a_ie_i.$ Then, $a-a_n=(0,0,cdots, n+1,n+2,cdots)=sum^{infty}_{i=n+1}a_iRightarrow |a-a_n|^ple sum^{infty}_{i=n+1}|a_i|^pto 0$ because $ain ell^p.$
answered Feb 5 at 17:40
MatematletaMatematleta
12.1k21020
edited Feb 5 at 21:19
answered Feb 5 at 17:40
MatematletaMatematleta
12.1k21020
answered Feb 5 at 17:40
MatematletaMatematleta
12.1k21020
answered Feb 5 at 17:40
MatematletaMatematleta
12.1k21020
12.1k21020
$begingroup$
For part 4, I have tried to look for more details for the full proof in math.stackexchange.com/questions/72198/… and math.stackexchange.com/questions/328479/… but am not sure.. isit correct somewhere?
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– Homaniac
Feb 5 at 21:41
$begingroup$
Wouldn't you learn more if you tried yourself? Here's a hint on how to start: the Cauchy sequence $(a_n)$ is bounded by some $C$. Write it as a matrix. Fix a column $k$ and consider $(a_n^k)$ for $n=1,2,3,..$ etc. This is a Cauchy sequence in $mathbb R$ so it converges to an $a_k$. So now you have a sequence $a=(a_k)$ and you want to show it's in $ell^p$ and $(a_n)$ converges to it. So fix an integer $N$, take $sum^N_{k=1}|a_n^k|^ple C$ and let $nto infty,$ to get $sum^N_{k=1}|a_k|^ple C$. Letting $Nto infty$ shows that $(a_k)in ell^p.$ To show $a_nto a$, use a similar argument.
$endgroup$
– Matematleta
Feb 5 at 22:56
1
$begingroup$
Thank you, to show $a_n → a$, fix $epsilon>0$, take $N$ such that $|a_n-a_m|_pleq epsilon$ for all $n,mgeq N$. Then $left(sum_{k=1}^N|x_m(k)-x_n(k)|^pright)^{1/p}leq |x_m-x_n|_pleq epsilon$, then letting $m → infty$, followed by $N$, we get $|a-a_n|_pleq epsilon$ for all $ngeq N$. Right :)
$endgroup$
– Homaniac
Feb 8 at 15:09
$begingroup$
Yes, that's correct.
$endgroup$
– Matematleta
Feb 8 at 18:11
add a comment |
$begingroup$
For part 4, I have tried to look for more details for the full proof in math.stackexchange.com/questions/72198/… and math.stackexchange.com/questions/328479/… but am not sure.. isit correct somewhere?
$endgroup$
– Homaniac
Feb 5 at 21:41
$begingroup$
Wouldn't you learn more if you tried yourself? Here's a hint on how to start: the Cauchy sequence $(a_n)$ is bounded by some $C$. Write it as a matrix. Fix a column $k$ and consider $(a_n^k)$ for $n=1,2,3,..$ etc. This is a Cauchy sequence in $mathbb R$ so it converges to an $a_k$. So now you have a sequence $a=(a_k)$ and you want to show it's in $ell^p$ and $(a_n)$ converges to it. So fix an integer $N$, take $sum^N_{k=1}|a_n^k|^ple C$ and let $nto infty,$ to get $sum^N_{k=1}|a_k|^ple C$. Letting $Nto infty$ shows that $(a_k)in ell^p.$ To show $a_nto a$, use a similar argument.
$endgroup$
– Matematleta
Feb 5 at 22:56
1
$begingroup$
Thank you, to show $a_n → a$, fix $epsilon>0$, take $N$ such that $|a_n-a_m|_pleq epsilon$ for all $n,mgeq N$. Then $left(sum_{k=1}^N|x_m(k)-x_n(k)|^pright)^{1/p}leq |x_m-x_n|_pleq epsilon$, then letting $m → infty$, followed by $N$, we get $|a-a_n|_pleq epsilon$ for all $ngeq N$. Right :)
$endgroup$
– Homaniac
Feb 8 at 15:09
$begingroup$
Yes, that's correct.
$endgroup$
– Matematleta
Feb 8 at 18:11
$begingroup$
For part 4, I have tried to look for more details for the full proof in math.stackexchange.com/questions/72198/… and math.stackexchange.com/questions/328479/… but am not sure.. isit correct somewhere?
$endgroup$
– Homaniac
Feb 5 at 21:41
$begingroup$
For part 4, I have tried to look for more details for the full proof in math.stackexchange.com/questions/72198/… and math.stackexchange.com/questions/328479/… but am not sure.. isit correct somewhere?
$endgroup$
– Homaniac
Feb 5 at 21:41
$begingroup$
Wouldn't you learn more if you tried yourself? Here's a hint on how to start: the Cauchy sequence $(a_n)$ is bounded by some $C$. Write it as a matrix. Fix a column $k$ and consider $(a_n^k)$ for $n=1,2,3,..$ etc. This is a Cauchy sequence in $mathbb R$ so it converges to an $a_k$. So now you have a sequence $a=(a_k)$ and you want to show it's in $ell^p$ and $(a_n)$ converges to it. So fix an integer $N$, take $sum^N_{k=1}|a_n^k|^ple C$ and let $nto infty,$ to get $sum^N_{k=1}|a_k|^ple C$. Letting $Nto infty$ shows that $(a_k)in ell^p.$ To show $a_nto a$, use a similar argument.
$endgroup$
– Matematleta
Feb 5 at 22:56
$begingroup$
Wouldn't you learn more if you tried yourself? Here's a hint on how to start: the Cauchy sequence $(a_n)$ is bounded by some $C$. Write it as a matrix. Fix a column $k$ and consider $(a_n^k)$ for $n=1,2,3,..$ etc. This is a Cauchy sequence in $mathbb R$ so it converges to an $a_k$. So now you have a sequence $a=(a_k)$ and you want to show it's in $ell^p$ and $(a_n)$ converges to it. So fix an integer $N$, take $sum^N_{k=1}|a_n^k|^ple C$ and let $nto infty,$ to get $sum^N_{k=1}|a_k|^ple C$. Letting $Nto infty$ shows that $(a_k)in ell^p.$ To show $a_nto a$, use a similar argument.
$endgroup$
– Matematleta
Feb 5 at 22:56
1
1
$begingroup$
Thank you, to show $a_n → a$, fix $epsilon>0$, take $N$ such that $|a_n-a_m|_pleq epsilon$ for all $n,mgeq N$. Then $left(sum_{k=1}^N|x_m(k)-x_n(k)|^pright)^{1/p}leq |x_m-x_n|_pleq epsilon$, then letting $m → infty$, followed by $N$, we get $|a-a_n|_pleq epsilon$ for all $ngeq N$. Right :)
$endgroup$
– Homaniac
Feb 8 at 15:09
$begingroup$
Thank you, to show $a_n → a$, fix $epsilon>0$, take $N$ such that $|a_n-a_m|_pleq epsilon$ for all $n,mgeq N$. Then $left(sum_{k=1}^N|x_m(k)-x_n(k)|^pright)^{1/p}leq |x_m-x_n|_pleq epsilon$, then letting $m → infty$, followed by $N$, we get $|a-a_n|_pleq epsilon$ for all $ngeq N$. Right :)
$endgroup$
– Homaniac
Feb 8 at 15:09
$begingroup$
Yes, that's correct.
$endgroup$
– Matematleta
Feb 8 at 18:11
$begingroup$
Yes, that's correct.
$endgroup$
– Matematleta
Feb 8 at 18:11
add a comment |
$begingroup$
(i) And that it is non-empty, but that's trivial.
(ii) Assume that it had a countable basis and try to construct an element that cannot be written as a finite linear combination.
(iii) Yes, your attempt looks good.
(iv) It is Banach (assuming $p geq 1$). You need to show completeness.
(v) Yes, exactly. Cannot be a Hamel basis because of (ii).
answered Feb 1 at 11:15
KlausKlaus
2,955214
$endgroup$
add a comment |
$begingroup$
(i) And that it is non-empty, but that's trivial.
(ii) Assume that it had a countable basis and try to construct an element that cannot be written as a finite linear combination.
(iii) Yes, your attempt looks good.
(iv) It is Banach (assuming $p geq 1$). You need to show completeness.
(v) Yes, exactly. Cannot be a Hamel basis because of (ii).
answered Feb 1 at 11:15
KlausKlaus
2,955214
$endgroup$
add a comment |
$begingroup$
(i) And that it is non-empty, but that's trivial.
(ii) Assume that it had a countable basis and try to construct an element that cannot be written as a finite linear combination.
(iii) Yes, your attempt looks good.
(iv) It is Banach (assuming $p geq 1$). You need to show completeness.
(v) Yes, exactly. Cannot be a Hamel basis because of (ii).
answered Feb 1 at 11:15
KlausKlaus
2,955214
$endgroup$
(i) And that it is non-empty, but that's trivial.
(ii) Assume that it had a countable basis and try to construct an element that cannot be written as a finite linear combination.
(iii) Yes, your attempt looks good.
(iv) It is Banach (assuming $p geq 1$). You need to show completeness.
(v) Yes, exactly. Cannot be a Hamel basis because of (ii).
answered Feb 1 at 11:15
KlausKlaus
2,955214
answered Feb 1 at 11:15
KlausKlaus
2,955214
answered Feb 1 at 11:15
KlausKlaus
2,955214
answered Feb 1 at 11:15
KlausKlaus
2,955214
2,955214
add a comment |
add a comment |
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$begingroup$
My guess is that you wrote $p$ instead of $ell_p$ in several places. Besides, the usual notation is $ell^p$.
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– José Carlos Santos
Feb 1 at 9:26
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Are you taking $p>0$ 0r $pge 1?$
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– zhw.
Feb 3 at 22:30
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We are taking 1 ≤ p < ∞
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– Homaniac
Feb 4 at 4:52
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Do you want detailed proofs? It seems like you are on the right track.
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– Matematleta
Feb 4 at 16:28
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Yea I am hoping for detailed proofs or at least more details for each part, thank you :)
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– Homaniac
Feb 4 at 17:52