The subspace $S$ defines an equivalence relation $f sim g$ to mean $f-g in S$. Show that $B/S$ is a Banach...
$begingroup$
Suppose $B$ is a Banach space and $S$ is a closed linear subspace of $B$ . The subspace $S$ defines an equivalence relation $f sim g$ to mean $f-g in S$ . If $B/S$ denotes the collection of these equivalence classes , then show that $B/S$ is a Banach space with norm $$vertvert f vertvert_{B/S} = inf (vertvert f' vert vert_B ,, , f'sim f)$$
I can show that $B/S$ is a normed vector space. I want to proceed by show that for each $f_n$ , there exist a decomposition $f_n=f_n' + h_n$ with $h_n in S $ and $f_n'$ form a cauchy sequence in $B$. If this has been proved , then let $f_n' to f$ $$vertvert f-f_n vertvert_{B/S} le vertvert f-f_n' vertvert_{B/S} + vertvert f_n'-f_n vertvert_{B/S} le vertvert f-f_n' vertvert_B to 0$$
But I have no idea how to do this , and I don't get the point how to use the condition $S$ is a 'closed' subspace.
real-analysis functional-analysis banach-spaces
asked Dec 7 '18 at 19:03
J.GuoJ.Guo
3209
$endgroup$
add a comment |
$begingroup$
Suppose $B$ is a Banach space and $S$ is a closed linear subspace of $B$ . The subspace $S$ defines an equivalence relation $f sim g$ to mean $f-g in S$ . If $B/S$ denotes the collection of these equivalence classes , then show that $B/S$ is a Banach space with norm $$vertvert f vertvert_{B/S} = inf (vertvert f' vert vert_B ,, , f'sim f)$$
I can show that $B/S$ is a normed vector space. I want to proceed by show that for each $f_n$ , there exist a decomposition $f_n=f_n' + h_n$ with $h_n in S $ and $f_n'$ form a cauchy sequence in $B$. If this has been proved , then let $f_n' to f$ $$vertvert f-f_n vertvert_{B/S} le vertvert f-f_n' vertvert_{B/S} + vertvert f_n'-f_n vertvert_{B/S} le vertvert f-f_n' vertvert_B to 0$$
But I have no idea how to do this , and I don't get the point how to use the condition $S$ is a 'closed' subspace.
real-analysis functional-analysis banach-spaces
asked Dec 7 '18 at 19:03
J.GuoJ.Guo
3209
$endgroup$
$begingroup$
The condition that $S$ is closed is to show that $|,cdot,|_{B/S}$ is a norm and not just a seminorm. It is not used to show the completeness of the quotient space
$endgroup$
– Federico
Dec 7 '18 at 19:56
add a comment |
$begingroup$
Suppose $B$ is a Banach space and $S$ is a closed linear subspace of $B$ . The subspace $S$ defines an equivalence relation $f sim g$ to mean $f-g in S$ . If $B/S$ denotes the collection of these equivalence classes , then show that $B/S$ is a Banach space with norm $$vertvert f vertvert_{B/S} = inf (vertvert f' vert vert_B ,, , f'sim f)$$
I can show that $B/S$ is a normed vector space. I want to proceed by show that for each $f_n$ , there exist a decomposition $f_n=f_n' + h_n$ with $h_n in S $ and $f_n'$ form a cauchy sequence in $B$. If this has been proved , then let $f_n' to f$ $$vertvert f-f_n vertvert_{B/S} le vertvert f-f_n' vertvert_{B/S} + vertvert f_n'-f_n vertvert_{B/S} le vertvert f-f_n' vertvert_B to 0$$
But I have no idea how to do this , and I don't get the point how to use the condition $S$ is a 'closed' subspace.
real-analysis functional-analysis banach-spaces
asked Dec 7 '18 at 19:03
J.GuoJ.Guo
3209
$endgroup$
Suppose $B$ is a Banach space and $S$ is a closed linear subspace of $B$ . The subspace $S$ defines an equivalence relation $f sim g$ to mean $f-g in S$ . If $B/S$ denotes the collection of these equivalence classes , then show that $B/S$ is a Banach space with norm $$vertvert f vertvert_{B/S} = inf (vertvert f' vert vert_B ,, , f'sim f)$$
I can show that $B/S$ is a normed vector space. I want to proceed by show that for each $f_n$ , there exist a decomposition $f_n=f_n' + h_n$ with $h_n in S $ and $f_n'$ form a cauchy sequence in $B$. If this has been proved , then let $f_n' to f$ $$vertvert f-f_n vertvert_{B/S} le vertvert f-f_n' vertvert_{B/S} + vertvert f_n'-f_n vertvert_{B/S} le vertvert f-f_n' vertvert_B to 0$$
But I have no idea how to do this , and I don't get the point how to use the condition $S$ is a 'closed' subspace.
real-analysis functional-analysis banach-spaces
real-analysis functional-analysis banach-spaces
asked Dec 7 '18 at 19:03
J.GuoJ.Guo
3209
asked Dec 7 '18 at 19:03
J.GuoJ.Guo
3209
asked Dec 7 '18 at 19:03
J.GuoJ.Guo
3209
asked Dec 7 '18 at 19:03
J.GuoJ.Guo
3209
asked Dec 7 '18 at 19:03
J.GuoJ.Guo
3209
3209
$begingroup$
The condition that $S$ is closed is to show that $|,cdot,|_{B/S}$ is a norm and not just a seminorm. It is not used to show the completeness of the quotient space
$endgroup$
– Federico
Dec 7 '18 at 19:56
add a comment |
$begingroup$
The condition that $S$ is closed is to show that $|,cdot,|_{B/S}$ is a norm and not just a seminorm. It is not used to show the completeness of the quotient space
$endgroup$
– Federico
Dec 7 '18 at 19:56
$begingroup$
The condition that $S$ is closed is to show that $|,cdot,|_{B/S}$ is a norm and not just a seminorm. It is not used to show the completeness of the quotient space
$endgroup$
– Federico
Dec 7 '18 at 19:56
$begingroup$
The condition that $S$ is closed is to show that $|,cdot,|_{B/S}$ is a norm and not just a seminorm. It is not used to show the completeness of the quotient space
$endgroup$
– Federico
Dec 7 '18 at 19:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Take $(f_n)_n$ which is Cauchy in $B/S$. By passing to a subsequence, we can assume that $sum_{n} |f_n-f_{n-1}|_{B/S}<infty$. Take a positive sequence $ainell^1$. Inductively we can find decompositions $f_n=f_n'+h_n$ with $h_nin S$ such that $|f'_n-f_{n-1}'|_Bleq|f_n-f_{n-1}|_{B/S}+a(n)$. But then
$$
sum_n |f'_n-f_{n-1}'|_B < infty
$$
which implies that $f_n'to f$ for some $fin B$.
Then you are done because $|f_n-f|_{B/S}leq |f_n-f|_B$.
answered Dec 7 '18 at 19:28
FedericoFederico
4,984514
$endgroup$
$begingroup$
What does $l^1$ denote ? And since $B/S$ might be infinite dimensional ， can we use the method of induction ?
$endgroup$
– J.Guo
Dec 7 '18 at 19:41
$begingroup$
$ainell^1$ is just a way of saying that $sum_n |a(n)|<infty$.
$endgroup$
– Federico
Dec 7 '18 at 19:45
$begingroup$
"And since B/S might be infinite dimensional ， can we use the method of induction ?" What do you mean? The induction is on $n$. Once you have decomposed $f_1,dots,f_{n-1}$, you decompose $f_n$ in the required way
$endgroup$
– Federico
Dec 7 '18 at 19:46
$begingroup$
Sorry , I misunderstand the meaning of $l^1$ before. But I still don't see the point . By definition I can prove $|f_n-f_{n-1}+h_n|_Bleq|f_n-f_{n-1}|_{B/S}+a(n)$ ,but how to show that $|f'_n-f_{n-1}'|_Bleq|f_n-f_{n-1}|_{B/S}+a(n)$
$endgroup$
– J.Guo
Dec 7 '18 at 19:51
2
$begingroup$
I wonder if this might be easier to understand decomposed into: (1) Apply the theorem that a normed v.s. $B$ is a Banach space iff whenever $sum_{n=1}^infty lVert x_n rVert < infty$ then $sum_{n=1}^infty x_n$ converges in $B$. (2) If $sum_{n=1}^infty lVert f_n rVert_{B/S} < infty$, choose $f_n' in f_n + S$ s.t. $lVert f_n' rVert_B < lVert f_n rVert_{B/S} + a(n)$. (3) Then $sum_{n=1}^infty lVert f_n' rVert_B < infty$ so $sum_{n=1}^infty f_n'$ converges in $B$, say to $f'$. (4) Show that $sum_{n=1}^infty (f_n+S)$ converges to $f'+S$ in $B/S$.
$endgroup$
– Daniel Schepler
Dec 7 '18 at 20:01
|
show 5 more comments
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1 Answer
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1 Answer
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$begingroup$
Take $(f_n)_n$ which is Cauchy in $B/S$. By passing to a subsequence, we can assume that $sum_{n} |f_n-f_{n-1}|_{B/S}<infty$. Take a positive sequence $ainell^1$. Inductively we can find decompositions $f_n=f_n'+h_n$ with $h_nin S$ such that $|f'_n-f_{n-1}'|_Bleq|f_n-f_{n-1}|_{B/S}+a(n)$. But then
$$
sum_n |f'_n-f_{n-1}'|_B < infty
$$
which implies that $f_n'to f$ for some $fin B$.
Then you are done because $|f_n-f|_{B/S}leq |f_n-f|_B$.
answered Dec 7 '18 at 19:28
FedericoFederico
4,984514
$endgroup$
$begingroup$
What does $l^1$ denote ? And since $B/S$ might be infinite dimensional ， can we use the method of induction ?
$endgroup$
– J.Guo
Dec 7 '18 at 19:41
$begingroup$
$ainell^1$ is just a way of saying that $sum_n |a(n)|<infty$.
$endgroup$
– Federico
Dec 7 '18 at 19:45
$begingroup$
"And since B/S might be infinite dimensional ， can we use the method of induction ?" What do you mean? The induction is on $n$. Once you have decomposed $f_1,dots,f_{n-1}$, you decompose $f_n$ in the required way
$endgroup$
– Federico
Dec 7 '18 at 19:46
$begingroup$
Sorry , I misunderstand the meaning of $l^1$ before. But I still don't see the point . By definition I can prove $|f_n-f_{n-1}+h_n|_Bleq|f_n-f_{n-1}|_{B/S}+a(n)$ ,but how to show that $|f'_n-f_{n-1}'|_Bleq|f_n-f_{n-1}|_{B/S}+a(n)$
$endgroup$
– J.Guo
Dec 7 '18 at 19:51
2
$begingroup$
I wonder if this might be easier to understand decomposed into: (1) Apply the theorem that a normed v.s. $B$ is a Banach space iff whenever $sum_{n=1}^infty lVert x_n rVert < infty$ then $sum_{n=1}^infty x_n$ converges in $B$. (2) If $sum_{n=1}^infty lVert f_n rVert_{B/S} < infty$, choose $f_n' in f_n + S$ s.t. $lVert f_n' rVert_B < lVert f_n rVert_{B/S} + a(n)$. (3) Then $sum_{n=1}^infty lVert f_n' rVert_B < infty$ so $sum_{n=1}^infty f_n'$ converges in $B$, say to $f'$. (4) Show that $sum_{n=1}^infty (f_n+S)$ converges to $f'+S$ in $B/S$.
$endgroup$
– Daniel Schepler
Dec 7 '18 at 20:01
|
show 5 more comments
$begingroup$
Take $(f_n)_n$ which is Cauchy in $B/S$. By passing to a subsequence, we can assume that $sum_{n} |f_n-f_{n-1}|_{B/S}<infty$. Take a positive sequence $ainell^1$. Inductively we can find decompositions $f_n=f_n'+h_n$ with $h_nin S$ such that $|f'_n-f_{n-1}'|_Bleq|f_n-f_{n-1}|_{B/S}+a(n)$. But then
$$
sum_n |f'_n-f_{n-1}'|_B < infty
$$
which implies that $f_n'to f$ for some $fin B$.
Then you are done because $|f_n-f|_{B/S}leq |f_n-f|_B$.
answered Dec 7 '18 at 19:28
FedericoFederico
4,984514
$endgroup$
$begingroup$
What does $l^1$ denote ? And since $B/S$ might be infinite dimensional ， can we use the method of induction ?
$endgroup$
– J.Guo
Dec 7 '18 at 19:41
$begingroup$
$ainell^1$ is just a way of saying that $sum_n |a(n)|<infty$.
$endgroup$
– Federico
Dec 7 '18 at 19:45
$begingroup$
"And since B/S might be infinite dimensional ， can we use the method of induction ?" What do you mean? The induction is on $n$. Once you have decomposed $f_1,dots,f_{n-1}$, you decompose $f_n$ in the required way
$endgroup$
– Federico
Dec 7 '18 at 19:46
$begingroup$
Sorry , I misunderstand the meaning of $l^1$ before. But I still don't see the point . By definition I can prove $|f_n-f_{n-1}+h_n|_Bleq|f_n-f_{n-1}|_{B/S}+a(n)$ ,but how to show that $|f'_n-f_{n-1}'|_Bleq|f_n-f_{n-1}|_{B/S}+a(n)$
$endgroup$
– J.Guo
Dec 7 '18 at 19:51
2
$begingroup$
I wonder if this might be easier to understand decomposed into: (1) Apply the theorem that a normed v.s. $B$ is a Banach space iff whenever $sum_{n=1}^infty lVert x_n rVert < infty$ then $sum_{n=1}^infty x_n$ converges in $B$. (2) If $sum_{n=1}^infty lVert f_n rVert_{B/S} < infty$, choose $f_n' in f_n + S$ s.t. $lVert f_n' rVert_B < lVert f_n rVert_{B/S} + a(n)$. (3) Then $sum_{n=1}^infty lVert f_n' rVert_B < infty$ so $sum_{n=1}^infty f_n'$ converges in $B$, say to $f'$. (4) Show that $sum_{n=1}^infty (f_n+S)$ converges to $f'+S$ in $B/S$.
$endgroup$
– Daniel Schepler
Dec 7 '18 at 20:01
|
show 5 more comments
$begingroup$
Take $(f_n)_n$ which is Cauchy in $B/S$. By passing to a subsequence, we can assume that $sum_{n} |f_n-f_{n-1}|_{B/S}<infty$. Take a positive sequence $ainell^1$. Inductively we can find decompositions $f_n=f_n'+h_n$ with $h_nin S$ such that $|f'_n-f_{n-1}'|_Bleq|f_n-f_{n-1}|_{B/S}+a(n)$. But then
$$
sum_n |f'_n-f_{n-1}'|_B < infty
$$
which implies that $f_n'to f$ for some $fin B$.
Then you are done because $|f_n-f|_{B/S}leq |f_n-f|_B$.
answered Dec 7 '18 at 19:28
FedericoFederico
4,984514
$endgroup$
Take $(f_n)_n$ which is Cauchy in $B/S$. By passing to a subsequence, we can assume that $sum_{n} |f_n-f_{n-1}|_{B/S}<infty$. Take a positive sequence $ainell^1$. Inductively we can find decompositions $f_n=f_n'+h_n$ with $h_nin S$ such that $|f'_n-f_{n-1}'|_Bleq|f_n-f_{n-1}|_{B/S}+a(n)$. But then
$$
sum_n |f'_n-f_{n-1}'|_B < infty
$$
which implies that $f_n'to f$ for some $fin B$.
Then you are done because $|f_n-f|_{B/S}leq |f_n-f|_B$.
answered Dec 7 '18 at 19:28
FedericoFederico
4,984514
edited Dec 7 '18 at 19:45
answered Dec 7 '18 at 19:28
FedericoFederico
4,984514
answered Dec 7 '18 at 19:28
FedericoFederico
4,984514
answered Dec 7 '18 at 19:28
FedericoFederico
4,984514
4,984514
$begingroup$
What does $l^1$ denote ? And since $B/S$ might be infinite dimensional ， can we use the method of induction ?
$endgroup$
– J.Guo
Dec 7 '18 at 19:41
$begingroup$
$ainell^1$ is just a way of saying that $sum_n |a(n)|<infty$.
$endgroup$
– Federico
Dec 7 '18 at 19:45
$begingroup$
"And since B/S might be infinite dimensional ， can we use the method of induction ?" What do you mean? The induction is on $n$. Once you have decomposed $f_1,dots,f_{n-1}$, you decompose $f_n$ in the required way
$endgroup$
– Federico
Dec 7 '18 at 19:46
$begingroup$
Sorry , I misunderstand the meaning of $l^1$ before. But I still don't see the point . By definition I can prove $|f_n-f_{n-1}+h_n|_Bleq|f_n-f_{n-1}|_{B/S}+a(n)$ ,but how to show that $|f'_n-f_{n-1}'|_Bleq|f_n-f_{n-1}|_{B/S}+a(n)$
$endgroup$
– J.Guo
Dec 7 '18 at 19:51
2
$begingroup$
I wonder if this might be easier to understand decomposed into: (1) Apply the theorem that a normed v.s. $B$ is a Banach space iff whenever $sum_{n=1}^infty lVert x_n rVert < infty$ then $sum_{n=1}^infty x_n$ converges in $B$. (2) If $sum_{n=1}^infty lVert f_n rVert_{B/S} < infty$, choose $f_n' in f_n + S$ s.t. $lVert f_n' rVert_B < lVert f_n rVert_{B/S} + a(n)$. (3) Then $sum_{n=1}^infty lVert f_n' rVert_B < infty$ so $sum_{n=1}^infty f_n'$ converges in $B$, say to $f'$. (4) Show that $sum_{n=1}^infty (f_n+S)$ converges to $f'+S$ in $B/S$.
$endgroup$
– Daniel Schepler
Dec 7 '18 at 20:01
|
show 5 more comments
$begingroup$
What does $l^1$ denote ? And since $B/S$ might be infinite dimensional ， can we use the method of induction ?
$endgroup$
– J.Guo
Dec 7 '18 at 19:41
$begingroup$
$ainell^1$ is just a way of saying that $sum_n |a(n)|<infty$.
$endgroup$
– Federico
Dec 7 '18 at 19:45
$begingroup$
"And since B/S might be infinite dimensional ， can we use the method of induction ?" What do you mean? The induction is on $n$. Once you have decomposed $f_1,dots,f_{n-1}$, you decompose $f_n$ in the required way
$endgroup$
– Federico
Dec 7 '18 at 19:46
$begingroup$
Sorry , I misunderstand the meaning of $l^1$ before. But I still don't see the point . By definition I can prove $|f_n-f_{n-1}+h_n|_Bleq|f_n-f_{n-1}|_{B/S}+a(n)$ ,but how to show that $|f'_n-f_{n-1}'|_Bleq|f_n-f_{n-1}|_{B/S}+a(n)$
$endgroup$
– J.Guo
Dec 7 '18 at 19:51
2
$begingroup$
I wonder if this might be easier to understand decomposed into: (1) Apply the theorem that a normed v.s. $B$ is a Banach space iff whenever $sum_{n=1}^infty lVert x_n rVert < infty$ then $sum_{n=1}^infty x_n$ converges in $B$. (2) If $sum_{n=1}^infty lVert f_n rVert_{B/S} < infty$, choose $f_n' in f_n + S$ s.t. $lVert f_n' rVert_B < lVert f_n rVert_{B/S} + a(n)$. (3) Then $sum_{n=1}^infty lVert f_n' rVert_B < infty$ so $sum_{n=1}^infty f_n'$ converges in $B$, say to $f'$. (4) Show that $sum_{n=1}^infty (f_n+S)$ converges to $f'+S$ in $B/S$.
$endgroup$
– Daniel Schepler
Dec 7 '18 at 20:01
$begingroup$
What does $l^1$ denote ? And since $B/S$ might be infinite dimensional ， can we use the method of induction ?
$endgroup$
– J.Guo
Dec 7 '18 at 19:41
$begingroup$
What does $l^1$ denote ? And since $B/S$ might be infinite dimensional ， can we use the method of induction ?
$endgroup$
– J.Guo
Dec 7 '18 at 19:41
$begingroup$
$ainell^1$ is just a way of saying that $sum_n |a(n)|<infty$.
$endgroup$
– Federico
Dec 7 '18 at 19:45
$begingroup$
$ainell^1$ is just a way of saying that $sum_n |a(n)|<infty$.
$endgroup$
– Federico
Dec 7 '18 at 19:45
$begingroup$
"And since B/S might be infinite dimensional ， can we use the method of induction ?" What do you mean? The induction is on $n$. Once you have decomposed $f_1,dots,f_{n-1}$, you decompose $f_n$ in the required way
$endgroup$
– Federico
Dec 7 '18 at 19:46
$begingroup$
"And since B/S might be infinite dimensional ， can we use the method of induction ?" What do you mean? The induction is on $n$. Once you have decomposed $f_1,dots,f_{n-1}$, you decompose $f_n$ in the required way
$endgroup$
– Federico
Dec 7 '18 at 19:46
$begingroup$
Sorry , I misunderstand the meaning of $l^1$ before. But I still don't see the point . By definition I can prove $|f_n-f_{n-1}+h_n|_Bleq|f_n-f_{n-1}|_{B/S}+a(n)$ ,but how to show that $|f'_n-f_{n-1}'|_Bleq|f_n-f_{n-1}|_{B/S}+a(n)$
$endgroup$
– J.Guo
Dec 7 '18 at 19:51
$begingroup$
Sorry , I misunderstand the meaning of $l^1$ before. But I still don't see the point . By definition I can prove $|f_n-f_{n-1}+h_n|_Bleq|f_n-f_{n-1}|_{B/S}+a(n)$ ,but how to show that $|f'_n-f_{n-1}'|_Bleq|f_n-f_{n-1}|_{B/S}+a(n)$
$endgroup$
– J.Guo
Dec 7 '18 at 19:51
2
2
$begingroup$
I wonder if this might be easier to understand decomposed into: (1) Apply the theorem that a normed v.s. $B$ is a Banach space iff whenever $sum_{n=1}^infty lVert x_n rVert < infty$ then $sum_{n=1}^infty x_n$ converges in $B$. (2) If $sum_{n=1}^infty lVert f_n rVert_{B/S} < infty$, choose $f_n' in f_n + S$ s.t. $lVert f_n' rVert_B < lVert f_n rVert_{B/S} + a(n)$. (3) Then $sum_{n=1}^infty lVert f_n' rVert_B < infty$ so $sum_{n=1}^infty f_n'$ converges in $B$, say to $f'$. (4) Show that $sum_{n=1}^infty (f_n+S)$ converges to $f'+S$ in $B/S$.
$endgroup$
– Daniel Schepler
Dec 7 '18 at 20:01
$begingroup$
I wonder if this might be easier to understand decomposed into: (1) Apply the theorem that a normed v.s. $B$ is a Banach space iff whenever $sum_{n=1}^infty lVert x_n rVert < infty$ then $sum_{n=1}^infty x_n$ converges in $B$. (2) If $sum_{n=1}^infty lVert f_n rVert_{B/S} < infty$, choose $f_n' in f_n + S$ s.t. $lVert f_n' rVert_B < lVert f_n rVert_{B/S} + a(n)$. (3) Then $sum_{n=1}^infty lVert f_n' rVert_B < infty$ so $sum_{n=1}^infty f_n'$ converges in $B$, say to $f'$. (4) Show that $sum_{n=1}^infty (f_n+S)$ converges to $f'+S$ in $B/S$.
$endgroup$
– Daniel Schepler
Dec 7 '18 at 20:01
|
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$begingroup$
The condition that $S$ is closed is to show that $|,cdot,|_{B/S}$ is a norm and not just a seminorm. It is not used to show the completeness of the quotient space
$endgroup$
– Federico
Dec 7 '18 at 19:56