Mixing
Fredholm operators on non-Banach spaces.
$begingroup$
Apparently Fredholm operators are usually (at least in Wikipedia and my functional analysis lecture) only defined as operators $T$ between two Banach spaces.
As far as I can see, the definition can be extended to operators between arbitrary normed spaces without any problems, although then the condition that $operatorname{im} T$ is closed is no longer independent of $ker T < infty$ and $operatorname{coker} T < infty$.
Is there some reason why this is not usually done?
Are Fredholm operators between non-Banach spaces so much less interesting/useful than those between Banach spaces?
If yes, why?
functional-analysis soft-question operator-theory banach-spaces
edited Feb 25 at 12:59
Andrews
7361318
asked Jan 18 at 18:01
0x5390x539
1,425518
$endgroup$
add a comment |
$begingroup$
Apparently Fredholm operators are usually (at least in Wikipedia and my functional analysis lecture) only defined as operators $T$ between two Banach spaces.
As far as I can see, the definition can be extended to operators between arbitrary normed spaces without any problems, although then the condition that $operatorname{im} T$ is closed is no longer independent of $ker T < infty$ and $operatorname{coker} T < infty$.
Is there some reason why this is not usually done?
Are Fredholm operators between non-Banach spaces so much less interesting/useful than those between Banach spaces?
If yes, why?
functional-analysis soft-question operator-theory banach-spaces
edited Feb 25 at 12:59
Andrews
7361318
asked Jan 18 at 18:01
0x5390x539
1,425518
$endgroup$
add a comment |
$begingroup$
Apparently Fredholm operators are usually (at least in Wikipedia and my functional analysis lecture) only defined as operators $T$ between two Banach spaces.
As far as I can see, the definition can be extended to operators between arbitrary normed spaces without any problems, although then the condition that $operatorname{im} T$ is closed is no longer independent of $ker T < infty$ and $operatorname{coker} T < infty$.
Is there some reason why this is not usually done?
Are Fredholm operators between non-Banach spaces so much less interesting/useful than those between Banach spaces?
If yes, why?
functional-analysis soft-question operator-theory banach-spaces
edited Feb 25 at 12:59
Andrews
7361318
asked Jan 18 at 18:01
0x5390x539
1,425518
$endgroup$
Apparently Fredholm operators are usually (at least in Wikipedia and my functional analysis lecture) only defined as operators $T$ between two Banach spaces.
As far as I can see, the definition can be extended to operators between arbitrary normed spaces without any problems, although then the condition that $operatorname{im} T$ is closed is no longer independent of $ker T < infty$ and $operatorname{coker} T < infty$.
Is there some reason why this is not usually done?
Are Fredholm operators between non-Banach spaces so much less interesting/useful than those between Banach spaces?
If yes, why?
functional-analysis soft-question operator-theory banach-spaces
functional-analysis soft-question operator-theory banach-spaces
edited Feb 25 at 12:59
Andrews
7361318
asked Jan 18 at 18:01
0x5390x539
1,425518
edited Feb 25 at 12:59
Andrews
7361318
asked Jan 18 at 18:01
0x5390x539
1,425518
edited Feb 25 at 12:59
Andrews
7361318
edited Feb 25 at 12:59
Andrews
7361318
edited Feb 25 at 12:59
Andrews
7361318
7361318
asked Jan 18 at 18:01
0x5390x539
1,425518
asked Jan 18 at 18:01
0x5390x539
1,425518
asked Jan 18 at 18:01
0x5390x539
1,425518
1,425518
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I don't know if you have already found your answer but... anyway. I think the answer could lie on the closed graph theorem. Think about this:
The closed graph theorem states: Let $(E, |cdot |_E)$ and $(F, |cdot |_F)$ be two Banach spaces. if
$$T:(E, |cdot |_E) to (F, |cdot |_F)$$
is a linear mapping and
$$Gamma := { (x,Tx) : x in E }$$ (its graph) is in $Etimes F$ closed, then $T in mathcal{L}(E,F)$ (space of linear continuous mappings between E and F). If you consider linear mappings like the derivative between incomplete spaces whose graph is closed, then $T:E to F$ could be (not necessarily) not continuous (There are many details you have here to consider. This statement is not so trivial since the continuity of a linear functional depends on the topology induced by the norm).
Consider the linear the problem (maybe looking for solutions to differential equations)
$$Tu = f.$$
Some methods consider solutions approaching them by sequences (really roughly speaking). So, let ${x_n}_{nin mathbb{N}}, {y_n}_{nin mathbb{N}}subset E$ be two sequences with $x_n, y_n to u$. If $T:E to F$ is not continuous then the limits of these sequences could be different i.e.
$$lim_{n to infty} T(x_n) = f_1 neq f_2 = lim_{n to infty} T(y_n).$$
So you would not be really able to say much about the "solution" you found approaching it by one specific sequence.
But if you have already found an answer, I would be entirely glad and grateful to read it.
answered Feb 25 at 11:23
rarcrarc
789
New contributor
rarc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The way I see it, Fredholm's theory is about existence; if some conditions are satisfied, then the given equation admits at least one solution. All existence results in analysis rely on completeness. These results never build a solution explicitly; rather, they prove that some approximation procedure is convergent, because it is Cauchy. Fredholm's theory is no exception.
answered Feb 25 at 11:40
Giuseppe NegroGiuseppe Negro
17.2k331125
$endgroup$
$begingroup$
From the point I see... It would be possible to extend the concept of Fredholm Operators to incomplete spaces, even though this would be meaningless as you pointed out.
$endgroup$
– rarc
Feb 25 at 11:47
1
$begingroup$
@rarc: Exactly; that would be like considering "contraction mappings" between incomplete metric spaces. These mappings need not have a fixed point, which is their main feature in the complete case. So, why should we care? The same applies with Fredholm.
$endgroup$
– Giuseppe Negro
Feb 25 at 11:52
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I don't know if you have already found your answer but... anyway. I think the answer could lie on the closed graph theorem. Think about this:
The closed graph theorem states: Let $(E, |cdot |_E)$ and $(F, |cdot |_F)$ be two Banach spaces. if
$$T:(E, |cdot |_E) to (F, |cdot |_F)$$
is a linear mapping and
$$Gamma := { (x,Tx) : x in E }$$ (its graph) is in $Etimes F$ closed, then $T in mathcal{L}(E,F)$ (space of linear continuous mappings between E and F). If you consider linear mappings like the derivative between incomplete spaces whose graph is closed, then $T:E to F$ could be (not necessarily) not continuous (There are many details you have here to consider. This statement is not so trivial since the continuity of a linear functional depends on the topology induced by the norm).
Consider the linear the problem (maybe looking for solutions to differential equations)
$$Tu = f.$$
Some methods consider solutions approaching them by sequences (really roughly speaking). So, let ${x_n}_{nin mathbb{N}}, {y_n}_{nin mathbb{N}}subset E$ be two sequences with $x_n, y_n to u$. If $T:E to F$ is not continuous then the limits of these sequences could be different i.e.
$$lim_{n to infty} T(x_n) = f_1 neq f_2 = lim_{n to infty} T(y_n).$$
So you would not be really able to say much about the "solution" you found approaching it by one specific sequence.
But if you have already found an answer, I would be entirely glad and grateful to read it.
answered Feb 25 at 11:23
rarcrarc
789
New contributor
rarc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I don't know if you have already found your answer but... anyway. I think the answer could lie on the closed graph theorem. Think about this:
The closed graph theorem states: Let $(E, |cdot |_E)$ and $(F, |cdot |_F)$ be two Banach spaces. if
$$T:(E, |cdot |_E) to (F, |cdot |_F)$$
is a linear mapping and
$$Gamma := { (x,Tx) : x in E }$$ (its graph) is in $Etimes F$ closed, then $T in mathcal{L}(E,F)$ (space of linear continuous mappings between E and F). If you consider linear mappings like the derivative between incomplete spaces whose graph is closed, then $T:E to F$ could be (not necessarily) not continuous (There are many details you have here to consider. This statement is not so trivial since the continuity of a linear functional depends on the topology induced by the norm).
Consider the linear the problem (maybe looking for solutions to differential equations)
$$Tu = f.$$
Some methods consider solutions approaching them by sequences (really roughly speaking). So, let ${x_n}_{nin mathbb{N}}, {y_n}_{nin mathbb{N}}subset E$ be two sequences with $x_n, y_n to u$. If $T:E to F$ is not continuous then the limits of these sequences could be different i.e.
$$lim_{n to infty} T(x_n) = f_1 neq f_2 = lim_{n to infty} T(y_n).$$
So you would not be really able to say much about the "solution" you found approaching it by one specific sequence.
But if you have already found an answer, I would be entirely glad and grateful to read it.
answered Feb 25 at 11:23
rarcrarc
789
New contributor
rarc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I don't know if you have already found your answer but... anyway. I think the answer could lie on the closed graph theorem. Think about this:
The closed graph theorem states: Let $(E, |cdot |_E)$ and $(F, |cdot |_F)$ be two Banach spaces. if
$$T:(E, |cdot |_E) to (F, |cdot |_F)$$
is a linear mapping and
$$Gamma := { (x,Tx) : x in E }$$ (its graph) is in $Etimes F$ closed, then $T in mathcal{L}(E,F)$ (space of linear continuous mappings between E and F). If you consider linear mappings like the derivative between incomplete spaces whose graph is closed, then $T:E to F$ could be (not necessarily) not continuous (There are many details you have here to consider. This statement is not so trivial since the continuity of a linear functional depends on the topology induced by the norm).
Consider the linear the problem (maybe looking for solutions to differential equations)
$$Tu = f.$$
Some methods consider solutions approaching them by sequences (really roughly speaking). So, let ${x_n}_{nin mathbb{N}}, {y_n}_{nin mathbb{N}}subset E$ be two sequences with $x_n, y_n to u$. If $T:E to F$ is not continuous then the limits of these sequences could be different i.e.
$$lim_{n to infty} T(x_n) = f_1 neq f_2 = lim_{n to infty} T(y_n).$$
So you would not be really able to say much about the "solution" you found approaching it by one specific sequence.
But if you have already found an answer, I would be entirely glad and grateful to read it.
answered Feb 25 at 11:23
rarcrarc
789
New contributor
rarc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I don't know if you have already found your answer but... anyway. I think the answer could lie on the closed graph theorem. Think about this:
The closed graph theorem states: Let $(E, |cdot |_E)$ and $(F, |cdot |_F)$ be two Banach spaces. if
$$T:(E, |cdot |_E) to (F, |cdot |_F)$$
is a linear mapping and
$$Gamma := { (x,Tx) : x in E }$$ (its graph) is in $Etimes F$ closed, then $T in mathcal{L}(E,F)$ (space of linear continuous mappings between E and F). If you consider linear mappings like the derivative between incomplete spaces whose graph is closed, then $T:E to F$ could be (not necessarily) not continuous (There are many details you have here to consider. This statement is not so trivial since the continuity of a linear functional depends on the topology induced by the norm).
Consider the linear the problem (maybe looking for solutions to differential equations)
$$Tu = f.$$
Some methods consider solutions approaching them by sequences (really roughly speaking). So, let ${x_n}_{nin mathbb{N}}, {y_n}_{nin mathbb{N}}subset E$ be two sequences with $x_n, y_n to u$. If $T:E to F$ is not continuous then the limits of these sequences could be different i.e.
$$lim_{n to infty} T(x_n) = f_1 neq f_2 = lim_{n to infty} T(y_n).$$
So you would not be really able to say much about the "solution" you found approaching it by one specific sequence.
But if you have already found an answer, I would be entirely glad and grateful to read it.
answered Feb 25 at 11:23
rarcrarc
789
New contributor
rarc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Feb 25 at 11:34
answered Feb 25 at 11:23
rarcrarc
789
New contributor
rarc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Feb 25 at 11:23
rarcrarc
789
answered Feb 25 at 11:23
rarcrarc
789
789
New contributor
rarc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
rarc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
rarc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
The way I see it, Fredholm's theory is about existence; if some conditions are satisfied, then the given equation admits at least one solution. All existence results in analysis rely on completeness. These results never build a solution explicitly; rather, they prove that some approximation procedure is convergent, because it is Cauchy. Fredholm's theory is no exception.
answered Feb 25 at 11:40
Giuseppe NegroGiuseppe Negro
17.2k331125
$endgroup$
$begingroup$
From the point I see... It would be possible to extend the concept of Fredholm Operators to incomplete spaces, even though this would be meaningless as you pointed out.
$endgroup$
– rarc
Feb 25 at 11:47
1
$begingroup$
@rarc: Exactly; that would be like considering "contraction mappings" between incomplete metric spaces. These mappings need not have a fixed point, which is their main feature in the complete case. So, why should we care? The same applies with Fredholm.
$endgroup$
– Giuseppe Negro
Feb 25 at 11:52
add a comment |
$begingroup$
The way I see it, Fredholm's theory is about existence; if some conditions are satisfied, then the given equation admits at least one solution. All existence results in analysis rely on completeness. These results never build a solution explicitly; rather, they prove that some approximation procedure is convergent, because it is Cauchy. Fredholm's theory is no exception.
answered Feb 25 at 11:40
Giuseppe NegroGiuseppe Negro
17.2k331125
$endgroup$
$begingroup$
From the point I see... It would be possible to extend the concept of Fredholm Operators to incomplete spaces, even though this would be meaningless as you pointed out.
$endgroup$
– rarc
Feb 25 at 11:47
1
$begingroup$
@rarc: Exactly; that would be like considering "contraction mappings" between incomplete metric spaces. These mappings need not have a fixed point, which is their main feature in the complete case. So, why should we care? The same applies with Fredholm.
$endgroup$
– Giuseppe Negro
Feb 25 at 11:52
add a comment |
$begingroup$
The way I see it, Fredholm's theory is about existence; if some conditions are satisfied, then the given equation admits at least one solution. All existence results in analysis rely on completeness. These results never build a solution explicitly; rather, they prove that some approximation procedure is convergent, because it is Cauchy. Fredholm's theory is no exception.
answered Feb 25 at 11:40
Giuseppe NegroGiuseppe Negro
17.2k331125
$endgroup$
The way I see it, Fredholm's theory is about existence; if some conditions are satisfied, then the given equation admits at least one solution. All existence results in analysis rely on completeness. These results never build a solution explicitly; rather, they prove that some approximation procedure is convergent, because it is Cauchy. Fredholm's theory is no exception.
answered Feb 25 at 11:40
Giuseppe NegroGiuseppe Negro
17.2k331125
answered Feb 25 at 11:40
Giuseppe NegroGiuseppe Negro
17.2k331125
answered Feb 25 at 11:40
Giuseppe NegroGiuseppe Negro
17.2k331125
answered Feb 25 at 11:40
Giuseppe NegroGiuseppe Negro
17.2k331125
17.2k331125
$begingroup$
From the point I see... It would be possible to extend the concept of Fredholm Operators to incomplete spaces, even though this would be meaningless as you pointed out.
$endgroup$
– rarc
Feb 25 at 11:47
1
$begingroup$
@rarc: Exactly; that would be like considering "contraction mappings" between incomplete metric spaces. These mappings need not have a fixed point, which is their main feature in the complete case. So, why should we care? The same applies with Fredholm.
$endgroup$
– Giuseppe Negro
Feb 25 at 11:52
add a comment |
$begingroup$
From the point I see... It would be possible to extend the concept of Fredholm Operators to incomplete spaces, even though this would be meaningless as you pointed out.
$endgroup$
– rarc
Feb 25 at 11:47
1
$begingroup$
@rarc: Exactly; that would be like considering "contraction mappings" between incomplete metric spaces. These mappings need not have a fixed point, which is their main feature in the complete case. So, why should we care? The same applies with Fredholm.
$endgroup$
– Giuseppe Negro
Feb 25 at 11:52
$begingroup$
From the point I see... It would be possible to extend the concept of Fredholm Operators to incomplete spaces, even though this would be meaningless as you pointed out.
$endgroup$
– rarc
Feb 25 at 11:47
$begingroup$
From the point I see... It would be possible to extend the concept of Fredholm Operators to incomplete spaces, even though this would be meaningless as you pointed out.
$endgroup$
– rarc
Feb 25 at 11:47
1
1
$begingroup$
@rarc: Exactly; that would be like considering "contraction mappings" between incomplete metric spaces. These mappings need not have a fixed point, which is their main feature in the complete case. So, why should we care? The same applies with Fredholm.
$endgroup$
– Giuseppe Negro
Feb 25 at 11:52
$begingroup$
@rarc: Exactly; that would be like considering "contraction mappings" between incomplete metric spaces. These mappings need not have a fixed point, which is their main feature in the complete case. So, why should we care? The same applies with Fredholm.
$endgroup$
– Giuseppe Negro
Feb 25 at 11:52
add a comment |
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