$C^{infty}(mathbb{R})$ as a Fréchet space
$begingroup$
I think I'm misunderstanding what is being asked of me in the following question:
The space $C^{infty}(mathbb{R})$ has a Fréchet space topology with respect to which $f_n rightarrow f$ iff $f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets ($k$, non-negative, is the order of derivative).
To show the backwards implication I need to show that convergence ($f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets) give me seminorms, there are countably many such semi norms, and these seminorms give me a complete Hausdorff TVS. My book gives me some facts to help me in this direction.
I'm confused by the forward implication though - if endow the space of $C^{infty}(mathbb{R})$ with a topology given by countable seminorms which makes $C^{infty}(mathbb{R})$ a complete Hausdorff TVS -- how do I know this implies $f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets? What if the seminorms are really strange and have nothing to do with the derivatives of $f$ over compact subsets?
real-analysis general-topology functional-analysis topological-vector-spaces
edited Nov 30 '18 at 17:37
Bernard
119k639112
asked Nov 30 '18 at 16:27
yoshiyoshi
1,181817
$endgroup$
add a comment |
$begingroup$
I think I'm misunderstanding what is being asked of me in the following question:
The space $C^{infty}(mathbb{R})$ has a Fréchet space topology with respect to which $f_n rightarrow f$ iff $f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets ($k$, non-negative, is the order of derivative).
To show the backwards implication I need to show that convergence ($f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets) give me seminorms, there are countably many such semi norms, and these seminorms give me a complete Hausdorff TVS. My book gives me some facts to help me in this direction.
I'm confused by the forward implication though - if endow the space of $C^{infty}(mathbb{R})$ with a topology given by countable seminorms which makes $C^{infty}(mathbb{R})$ a complete Hausdorff TVS -- how do I know this implies $f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets? What if the seminorms are really strange and have nothing to do with the derivatives of $f$ over compact subsets?
real-analysis general-topology functional-analysis topological-vector-spaces
edited Nov 30 '18 at 17:37
Bernard
119k639112
asked Nov 30 '18 at 16:27
yoshiyoshi
1,181817
$endgroup$
1
$begingroup$
The seminirms are $p_{n,k}(f):= sup_{[-k,k]} |f^{(n)}|$. The key is that any compact set is contained in some interval $[-k,k]$
$endgroup$
– Tito Eliatron
Nov 30 '18 at 16:58
$begingroup$
I see that, but that's for a particular choice of seminorms. Why should it be true for any arbitrary countable family of seminorms though?
$endgroup$
– yoshi
Nov 30 '18 at 17:21
1
$begingroup$
For fixed k,n,m, The set ${p_{n,k}(f,0)<m}$ is open, so for your Family of seminorms, you can put a p- Bail inside.
$endgroup$
– Tito Eliatron
Nov 30 '18 at 17:31
$begingroup$
Can you elaborate on this statement, what is a p-ball and why is it suffient for one p-ball to be inside?
$endgroup$
– yoshi
Dec 1 '18 at 15:20
add a comment |
$begingroup$
I think I'm misunderstanding what is being asked of me in the following question:
The space $C^{infty}(mathbb{R})$ has a Fréchet space topology with respect to which $f_n rightarrow f$ iff $f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets ($k$, non-negative, is the order of derivative).
To show the backwards implication I need to show that convergence ($f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets) give me seminorms, there are countably many such semi norms, and these seminorms give me a complete Hausdorff TVS. My book gives me some facts to help me in this direction.
I'm confused by the forward implication though - if endow the space of $C^{infty}(mathbb{R})$ with a topology given by countable seminorms which makes $C^{infty}(mathbb{R})$ a complete Hausdorff TVS -- how do I know this implies $f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets? What if the seminorms are really strange and have nothing to do with the derivatives of $f$ over compact subsets?
real-analysis general-topology functional-analysis topological-vector-spaces
edited Nov 30 '18 at 17:37
Bernard
119k639112
asked Nov 30 '18 at 16:27
yoshiyoshi
1,181817
$endgroup$
I think I'm misunderstanding what is being asked of me in the following question:
The space $C^{infty}(mathbb{R})$ has a Fréchet space topology with respect to which $f_n rightarrow f$ iff $f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets ($k$, non-negative, is the order of derivative).
To show the backwards implication I need to show that convergence ($f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets) give me seminorms, there are countably many such semi norms, and these seminorms give me a complete Hausdorff TVS. My book gives me some facts to help me in this direction.
I'm confused by the forward implication though - if endow the space of $C^{infty}(mathbb{R})$ with a topology given by countable seminorms which makes $C^{infty}(mathbb{R})$ a complete Hausdorff TVS -- how do I know this implies $f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets? What if the seminorms are really strange and have nothing to do with the derivatives of $f$ over compact subsets?
real-analysis general-topology functional-analysis topological-vector-spaces
real-analysis general-topology functional-analysis topological-vector-spaces
edited Nov 30 '18 at 17:37
Bernard
119k639112
asked Nov 30 '18 at 16:27
yoshiyoshi
1,181817
edited Nov 30 '18 at 17:37
Bernard
119k639112
asked Nov 30 '18 at 16:27
yoshiyoshi
1,181817
edited Nov 30 '18 at 17:37
Bernard
119k639112
edited Nov 30 '18 at 17:37
Bernard
119k639112
edited Nov 30 '18 at 17:37
Bernard
119k639112
119k639112
asked Nov 30 '18 at 16:27
yoshiyoshi
1,181817
asked Nov 30 '18 at 16:27
yoshiyoshi
1,181817
asked Nov 30 '18 at 16:27
yoshiyoshi
1,181817
1,181817
1
$begingroup$
The seminirms are $p_{n,k}(f):= sup_{[-k,k]} |f^{(n)}|$. The key is that any compact set is contained in some interval $[-k,k]$
$endgroup$
– Tito Eliatron
Nov 30 '18 at 16:58
$begingroup$
I see that, but that's for a particular choice of seminorms. Why should it be true for any arbitrary countable family of seminorms though?
$endgroup$
– yoshi
Nov 30 '18 at 17:21
1
$begingroup$
For fixed k,n,m, The set ${p_{n,k}(f,0)<m}$ is open, so for your Family of seminorms, you can put a p- Bail inside.
$endgroup$
– Tito Eliatron
Nov 30 '18 at 17:31
$begingroup$
Can you elaborate on this statement, what is a p-ball and why is it suffient for one p-ball to be inside?
$endgroup$
– yoshi
Dec 1 '18 at 15:20
add a comment |
1
$begingroup$
The seminirms are $p_{n,k}(f):= sup_{[-k,k]} |f^{(n)}|$. The key is that any compact set is contained in some interval $[-k,k]$
$endgroup$
– Tito Eliatron
Nov 30 '18 at 16:58
$begingroup$
I see that, but that's for a particular choice of seminorms. Why should it be true for any arbitrary countable family of seminorms though?
$endgroup$
– yoshi
Nov 30 '18 at 17:21
1
$begingroup$
For fixed k,n,m, The set ${p_{n,k}(f,0)<m}$ is open, so for your Family of seminorms, you can put a p- Bail inside.
$endgroup$
– Tito Eliatron
Nov 30 '18 at 17:31
$begingroup$
Can you elaborate on this statement, what is a p-ball and why is it suffient for one p-ball to be inside?
$endgroup$
– yoshi
Dec 1 '18 at 15:20
1
1
$begingroup$
The seminirms are $p_{n,k}(f):= sup_{[-k,k]} |f^{(n)}|$. The key is that any compact set is contained in some interval $[-k,k]$
$endgroup$
– Tito Eliatron
Nov 30 '18 at 16:58
$begingroup$
The seminirms are $p_{n,k}(f):= sup_{[-k,k]} |f^{(n)}|$. The key is that any compact set is contained in some interval $[-k,k]$
$endgroup$
– Tito Eliatron
Nov 30 '18 at 16:58
$begingroup$
I see that, but that's for a particular choice of seminorms. Why should it be true for any arbitrary countable family of seminorms though?
$endgroup$
– yoshi
Nov 30 '18 at 17:21
$begingroup$
I see that, but that's for a particular choice of seminorms. Why should it be true for any arbitrary countable family of seminorms though?
$endgroup$
– yoshi
Nov 30 '18 at 17:21
1
1
$begingroup$
For fixed k,n,m, The set ${p_{n,k}(f,0)<m}$ is open, so for your Family of seminorms, you can put a p- Bail inside.
$endgroup$
– Tito Eliatron
Nov 30 '18 at 17:31
$begingroup$
For fixed k,n,m, The set ${p_{n,k}(f,0)<m}$ is open, so for your Family of seminorms, you can put a p- Bail inside.
$endgroup$
– Tito Eliatron
Nov 30 '18 at 17:31
$begingroup$
Can you elaborate on this statement, what is a p-ball and why is it suffient for one p-ball to be inside?
$endgroup$
– yoshi
Dec 1 '18 at 15:20
$begingroup$
Can you elaborate on this statement, what is a p-ball and why is it suffient for one p-ball to be inside?
$endgroup$
– yoshi
Dec 1 '18 at 15:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The question doesn't exclude that there are different topologies on $C^infty(mathbb{R})$ such that $C^infty(mathbb{R})$ is a Frechet space with this topology. However, since the open mapping theorem is also true for Frechet spaces, we cannot have a finer (or weaker) topology (as the usual one) on $C^infty(mathbb{R})$ in which $C^infty(mathbb{R})$ is also a Frechet space and $f^{(k)}_n rightarrow f^{(k)}$ uniformly on compact sets for each $k in mathbb{N}$.
One can proof that $C^infty(mathbb{R})$ has dimension $2^{aleph_0}$ in the sense of Hamel basis. Thus you can take e.g. the Banach-space $L^1(mathbb{R})$ and transport the norm to $C^infty(mathbb{R})$. With these norm $C^infty(mathbb{R})$ is a Banach space. Since the topology of local convergence of all derivates is not normable, this weird topology is different than the usual one.
answered Nov 30 '18 at 20:33
p4schp4sch
4,770217
$endgroup$
$begingroup$
why does "no finer topology" follow from the open mapping theorem?
$endgroup$
– yoshi
Dec 1 '18 at 15:22
1
$begingroup$
If $tau_1$, $tau_2$ are topologies on $C^infty(mathbb{R})$ with $tau_1 subset tau_2$, i.e. $tau_2$ is finer than $tau_1$, then $mathrm{id} colon (C^infty(mathbb{R}),tau_2) rightarrow (C^infty(mathbb{R}),tau_1)$ the map is continuous. If both topologies are Frechet, then the open mapping theorem already implies that this map is open, i.e. the inverse map $mathrm{id} colon (C^infty(mathbb{R}),tau_1) rightarrow (C^infty(mathbb{R}),tau_2)$ is continuous and thus $tau_1 = tau_2$.
$endgroup$
– p4sch
Dec 1 '18 at 21:27
$begingroup$
I like this, but my only problem is that we haven't covered the open mapping theorem for Frechet spaces. Is there a way to adapt Tito Aliatron's hints into this? I'm looking through the proof of the open mapping theorem for Frechet spaces to see if I can do it too.
$endgroup$
– yoshi
Dec 1 '18 at 21:44
1
$begingroup$
I don't think that you can prove this much easier. It is a feature coming from the completeness of the space. You may have a look in Rudin's book on functional analysis. There you can find a proof of the open mapping theorem - that is Theorem 2.11. In fact, the idea of the proof is the same as in the case of banach spaces.
$endgroup$
– p4sch
Dec 2 '18 at 11:03
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The question doesn't exclude that there are different topologies on $C^infty(mathbb{R})$ such that $C^infty(mathbb{R})$ is a Frechet space with this topology. However, since the open mapping theorem is also true for Frechet spaces, we cannot have a finer (or weaker) topology (as the usual one) on $C^infty(mathbb{R})$ in which $C^infty(mathbb{R})$ is also a Frechet space and $f^{(k)}_n rightarrow f^{(k)}$ uniformly on compact sets for each $k in mathbb{N}$.
One can proof that $C^infty(mathbb{R})$ has dimension $2^{aleph_0}$ in the sense of Hamel basis. Thus you can take e.g. the Banach-space $L^1(mathbb{R})$ and transport the norm to $C^infty(mathbb{R})$. With these norm $C^infty(mathbb{R})$ is a Banach space. Since the topology of local convergence of all derivates is not normable, this weird topology is different than the usual one.
answered Nov 30 '18 at 20:33
p4schp4sch
4,770217
$endgroup$
$begingroup$
why does "no finer topology" follow from the open mapping theorem?
$endgroup$
– yoshi
Dec 1 '18 at 15:22
1
$begingroup$
If $tau_1$, $tau_2$ are topologies on $C^infty(mathbb{R})$ with $tau_1 subset tau_2$, i.e. $tau_2$ is finer than $tau_1$, then $mathrm{id} colon (C^infty(mathbb{R}),tau_2) rightarrow (C^infty(mathbb{R}),tau_1)$ the map is continuous. If both topologies are Frechet, then the open mapping theorem already implies that this map is open, i.e. the inverse map $mathrm{id} colon (C^infty(mathbb{R}),tau_1) rightarrow (C^infty(mathbb{R}),tau_2)$ is continuous and thus $tau_1 = tau_2$.
$endgroup$
– p4sch
Dec 1 '18 at 21:27
$begingroup$
I like this, but my only problem is that we haven't covered the open mapping theorem for Frechet spaces. Is there a way to adapt Tito Aliatron's hints into this? I'm looking through the proof of the open mapping theorem for Frechet spaces to see if I can do it too.
$endgroup$
– yoshi
Dec 1 '18 at 21:44
1
$begingroup$
I don't think that you can prove this much easier. It is a feature coming from the completeness of the space. You may have a look in Rudin's book on functional analysis. There you can find a proof of the open mapping theorem - that is Theorem 2.11. In fact, the idea of the proof is the same as in the case of banach spaces.
$endgroup$
– p4sch
Dec 2 '18 at 11:03
add a comment |
$begingroup$
The question doesn't exclude that there are different topologies on $C^infty(mathbb{R})$ such that $C^infty(mathbb{R})$ is a Frechet space with this topology. However, since the open mapping theorem is also true for Frechet spaces, we cannot have a finer (or weaker) topology (as the usual one) on $C^infty(mathbb{R})$ in which $C^infty(mathbb{R})$ is also a Frechet space and $f^{(k)}_n rightarrow f^{(k)}$ uniformly on compact sets for each $k in mathbb{N}$.
One can proof that $C^infty(mathbb{R})$ has dimension $2^{aleph_0}$ in the sense of Hamel basis. Thus you can take e.g. the Banach-space $L^1(mathbb{R})$ and transport the norm to $C^infty(mathbb{R})$. With these norm $C^infty(mathbb{R})$ is a Banach space. Since the topology of local convergence of all derivates is not normable, this weird topology is different than the usual one.
answered Nov 30 '18 at 20:33
p4schp4sch
4,770217
$endgroup$
$begingroup$
why does "no finer topology" follow from the open mapping theorem?
$endgroup$
– yoshi
Dec 1 '18 at 15:22
1
$begingroup$
If $tau_1$, $tau_2$ are topologies on $C^infty(mathbb{R})$ with $tau_1 subset tau_2$, i.e. $tau_2$ is finer than $tau_1$, then $mathrm{id} colon (C^infty(mathbb{R}),tau_2) rightarrow (C^infty(mathbb{R}),tau_1)$ the map is continuous. If both topologies are Frechet, then the open mapping theorem already implies that this map is open, i.e. the inverse map $mathrm{id} colon (C^infty(mathbb{R}),tau_1) rightarrow (C^infty(mathbb{R}),tau_2)$ is continuous and thus $tau_1 = tau_2$.
$endgroup$
– p4sch
Dec 1 '18 at 21:27
$begingroup$
I like this, but my only problem is that we haven't covered the open mapping theorem for Frechet spaces. Is there a way to adapt Tito Aliatron's hints into this? I'm looking through the proof of the open mapping theorem for Frechet spaces to see if I can do it too.
$endgroup$
– yoshi
Dec 1 '18 at 21:44
1
$begingroup$
I don't think that you can prove this much easier. It is a feature coming from the completeness of the space. You may have a look in Rudin's book on functional analysis. There you can find a proof of the open mapping theorem - that is Theorem 2.11. In fact, the idea of the proof is the same as in the case of banach spaces.
$endgroup$
– p4sch
Dec 2 '18 at 11:03
add a comment |
$begingroup$
The question doesn't exclude that there are different topologies on $C^infty(mathbb{R})$ such that $C^infty(mathbb{R})$ is a Frechet space with this topology. However, since the open mapping theorem is also true for Frechet spaces, we cannot have a finer (or weaker) topology (as the usual one) on $C^infty(mathbb{R})$ in which $C^infty(mathbb{R})$ is also a Frechet space and $f^{(k)}_n rightarrow f^{(k)}$ uniformly on compact sets for each $k in mathbb{N}$.
One can proof that $C^infty(mathbb{R})$ has dimension $2^{aleph_0}$ in the sense of Hamel basis. Thus you can take e.g. the Banach-space $L^1(mathbb{R})$ and transport the norm to $C^infty(mathbb{R})$. With these norm $C^infty(mathbb{R})$ is a Banach space. Since the topology of local convergence of all derivates is not normable, this weird topology is different than the usual one.
answered Nov 30 '18 at 20:33
p4schp4sch
4,770217
$endgroup$
The question doesn't exclude that there are different topologies on $C^infty(mathbb{R})$ such that $C^infty(mathbb{R})$ is a Frechet space with this topology. However, since the open mapping theorem is also true for Frechet spaces, we cannot have a finer (or weaker) topology (as the usual one) on $C^infty(mathbb{R})$ in which $C^infty(mathbb{R})$ is also a Frechet space and $f^{(k)}_n rightarrow f^{(k)}$ uniformly on compact sets for each $k in mathbb{N}$.
One can proof that $C^infty(mathbb{R})$ has dimension $2^{aleph_0}$ in the sense of Hamel basis. Thus you can take e.g. the Banach-space $L^1(mathbb{R})$ and transport the norm to $C^infty(mathbb{R})$. With these norm $C^infty(mathbb{R})$ is a Banach space. Since the topology of local convergence of all derivates is not normable, this weird topology is different than the usual one.
answered Nov 30 '18 at 20:33
p4schp4sch
4,770217
edited Dec 1 '18 at 7:17
answered Nov 30 '18 at 20:33
p4schp4sch
4,770217
answered Nov 30 '18 at 20:33
p4schp4sch
4,770217
answered Nov 30 '18 at 20:33
p4schp4sch
4,770217
4,770217
$begingroup$
why does "no finer topology" follow from the open mapping theorem?
$endgroup$
– yoshi
Dec 1 '18 at 15:22
1
$begingroup$
If $tau_1$, $tau_2$ are topologies on $C^infty(mathbb{R})$ with $tau_1 subset tau_2$, i.e. $tau_2$ is finer than $tau_1$, then $mathrm{id} colon (C^infty(mathbb{R}),tau_2) rightarrow (C^infty(mathbb{R}),tau_1)$ the map is continuous. If both topologies are Frechet, then the open mapping theorem already implies that this map is open, i.e. the inverse map $mathrm{id} colon (C^infty(mathbb{R}),tau_1) rightarrow (C^infty(mathbb{R}),tau_2)$ is continuous and thus $tau_1 = tau_2$.
$endgroup$
– p4sch
Dec 1 '18 at 21:27
$begingroup$
I like this, but my only problem is that we haven't covered the open mapping theorem for Frechet spaces. Is there a way to adapt Tito Aliatron's hints into this? I'm looking through the proof of the open mapping theorem for Frechet spaces to see if I can do it too.
$endgroup$
– yoshi
Dec 1 '18 at 21:44
1
$begingroup$
I don't think that you can prove this much easier. It is a feature coming from the completeness of the space. You may have a look in Rudin's book on functional analysis. There you can find a proof of the open mapping theorem - that is Theorem 2.11. In fact, the idea of the proof is the same as in the case of banach spaces.
$endgroup$
– p4sch
Dec 2 '18 at 11:03
add a comment |
$begingroup$
why does "no finer topology" follow from the open mapping theorem?
$endgroup$
– yoshi
Dec 1 '18 at 15:22
1
$begingroup$
If $tau_1$, $tau_2$ are topologies on $C^infty(mathbb{R})$ with $tau_1 subset tau_2$, i.e. $tau_2$ is finer than $tau_1$, then $mathrm{id} colon (C^infty(mathbb{R}),tau_2) rightarrow (C^infty(mathbb{R}),tau_1)$ the map is continuous. If both topologies are Frechet, then the open mapping theorem already implies that this map is open, i.e. the inverse map $mathrm{id} colon (C^infty(mathbb{R}),tau_1) rightarrow (C^infty(mathbb{R}),tau_2)$ is continuous and thus $tau_1 = tau_2$.
$endgroup$
– p4sch
Dec 1 '18 at 21:27
$begingroup$
I like this, but my only problem is that we haven't covered the open mapping theorem for Frechet spaces. Is there a way to adapt Tito Aliatron's hints into this? I'm looking through the proof of the open mapping theorem for Frechet spaces to see if I can do it too.
$endgroup$
– yoshi
Dec 1 '18 at 21:44
1
$begingroup$
I don't think that you can prove this much easier. It is a feature coming from the completeness of the space. You may have a look in Rudin's book on functional analysis. There you can find a proof of the open mapping theorem - that is Theorem 2.11. In fact, the idea of the proof is the same as in the case of banach spaces.
$endgroup$
– p4sch
Dec 2 '18 at 11:03
$begingroup$
why does "no finer topology" follow from the open mapping theorem?
$endgroup$
– yoshi
Dec 1 '18 at 15:22
$begingroup$
why does "no finer topology" follow from the open mapping theorem?
$endgroup$
– yoshi
Dec 1 '18 at 15:22
1
1
$begingroup$
If $tau_1$, $tau_2$ are topologies on $C^infty(mathbb{R})$ with $tau_1 subset tau_2$, i.e. $tau_2$ is finer than $tau_1$, then $mathrm{id} colon (C^infty(mathbb{R}),tau_2) rightarrow (C^infty(mathbb{R}),tau_1)$ the map is continuous. If both topologies are Frechet, then the open mapping theorem already implies that this map is open, i.e. the inverse map $mathrm{id} colon (C^infty(mathbb{R}),tau_1) rightarrow (C^infty(mathbb{R}),tau_2)$ is continuous and thus $tau_1 = tau_2$.
$endgroup$
– p4sch
Dec 1 '18 at 21:27
$begingroup$
If $tau_1$, $tau_2$ are topologies on $C^infty(mathbb{R})$ with $tau_1 subset tau_2$, i.e. $tau_2$ is finer than $tau_1$, then $mathrm{id} colon (C^infty(mathbb{R}),tau_2) rightarrow (C^infty(mathbb{R}),tau_1)$ the map is continuous. If both topologies are Frechet, then the open mapping theorem already implies that this map is open, i.e. the inverse map $mathrm{id} colon (C^infty(mathbb{R}),tau_1) rightarrow (C^infty(mathbb{R}),tau_2)$ is continuous and thus $tau_1 = tau_2$.
$endgroup$
– p4sch
Dec 1 '18 at 21:27
$begingroup$
I like this, but my only problem is that we haven't covered the open mapping theorem for Frechet spaces. Is there a way to adapt Tito Aliatron's hints into this? I'm looking through the proof of the open mapping theorem for Frechet spaces to see if I can do it too.
$endgroup$
– yoshi
Dec 1 '18 at 21:44
$begingroup$
I like this, but my only problem is that we haven't covered the open mapping theorem for Frechet spaces. Is there a way to adapt Tito Aliatron's hints into this? I'm looking through the proof of the open mapping theorem for Frechet spaces to see if I can do it too.
$endgroup$
– yoshi
Dec 1 '18 at 21:44
1
1
$begingroup$
I don't think that you can prove this much easier. It is a feature coming from the completeness of the space. You may have a look in Rudin's book on functional analysis. There you can find a proof of the open mapping theorem - that is Theorem 2.11. In fact, the idea of the proof is the same as in the case of banach spaces.
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– p4sch
Dec 2 '18 at 11:03
$begingroup$
I don't think that you can prove this much easier. It is a feature coming from the completeness of the space. You may have a look in Rudin's book on functional analysis. There you can find a proof of the open mapping theorem - that is Theorem 2.11. In fact, the idea of the proof is the same as in the case of banach spaces.
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– p4sch
Dec 2 '18 at 11:03
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1
$begingroup$
The seminirms are $p_{n,k}(f):= sup_{[-k,k]} |f^{(n)}|$. The key is that any compact set is contained in some interval $[-k,k]$
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– Tito Eliatron
Nov 30 '18 at 16:58
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I see that, but that's for a particular choice of seminorms. Why should it be true for any arbitrary countable family of seminorms though?
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– yoshi
Nov 30 '18 at 17:21
1
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For fixed k,n,m, The set ${p_{n,k}(f,0)<m}$ is open, so for your Family of seminorms, you can put a p- Bail inside.
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– Tito Eliatron
Nov 30 '18 at 17:31
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Can you elaborate on this statement, what is a p-ball and why is it suffient for one p-ball to be inside?
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– yoshi
Dec 1 '18 at 15:20