$C^{infty}(mathbb{R})$ as a Fréchet space
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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 2 days ago
Bernard
116k637108
asked 2 days ago
yoshi
1,148817
add a comment |
up vote
1
down vote
favorite
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 2 days ago
Bernard
116k637108
asked 2 days ago
yoshi
1,148817
1
The seminirms are $p_{n,k}(f):= sup_{[-k,k]} |f^{(n)}|$. The key is that any compact set is contained in some interval $[-k,k]$
– Tito Eliatron
2 days ago
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?
– yoshi
2 days ago
1
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.
– Tito Eliatron
2 days ago
Can you elaborate on this statement, what is a p-ball and why is it suffient for one p-ball to be inside?
– yoshi
yesterday
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 2 days ago
Bernard
116k637108
asked 2 days ago
yoshi
1,148817
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 2 days ago
Bernard
116k637108
asked 2 days ago
yoshi
1,148817
edited 2 days ago
Bernard
116k637108
asked 2 days ago
yoshi
1,148817
edited 2 days ago
Bernard
116k637108
edited 2 days ago
Bernard
116k637108
edited 2 days ago
Bernard
116k637108
116k637108
asked 2 days ago
yoshi
1,148817
asked 2 days ago
yoshi
1,148817
asked 2 days ago
yoshi
1,148817
1,148817
1
The seminirms are $p_{n,k}(f):= sup_{[-k,k]} |f^{(n)}|$. The key is that any compact set is contained in some interval $[-k,k]$
– Tito Eliatron
2 days ago
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?
– yoshi
2 days ago
1
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.
– Tito Eliatron
2 days ago
Can you elaborate on this statement, what is a p-ball and why is it suffient for one p-ball to be inside?
– yoshi
yesterday
add a comment |
1
The seminirms are $p_{n,k}(f):= sup_{[-k,k]} |f^{(n)}|$. The key is that any compact set is contained in some interval $[-k,k]$
– Tito Eliatron
2 days ago
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?
– yoshi
2 days ago
1
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.
– Tito Eliatron
2 days ago
Can you elaborate on this statement, what is a p-ball and why is it suffient for one p-ball to be inside?
– yoshi
yesterday
1
1
The seminirms are $p_{n,k}(f):= sup_{[-k,k]} |f^{(n)}|$. The key is that any compact set is contained in some interval $[-k,k]$
– Tito Eliatron
2 days ago
The seminirms are $p_{n,k}(f):= sup_{[-k,k]} |f^{(n)}|$. The key is that any compact set is contained in some interval $[-k,k]$
– Tito Eliatron
2 days ago
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?
– yoshi
2 days ago
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?
– yoshi
2 days ago
1
1
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.
– Tito Eliatron
2 days ago
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.
– Tito Eliatron
2 days ago
Can you elaborate on this statement, what is a p-ball and why is it suffient for one p-ball to be inside?
– yoshi
yesterday
Can you elaborate on this statement, what is a p-ball and why is it suffient for one p-ball to be inside?
– yoshi
yesterday
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
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 2 days ago
p4sch
4,130216
why does "no finer topology" follow from the open mapping theorem?
– yoshi
yesterday
1
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 $tau_1 = tau_2$.
– p4sch
yesterday
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.
– yoshi
yesterday
1
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.
– p4sch
12 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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 2 days ago
p4sch
4,130216
why does "no finer topology" follow from the open mapping theorem?
– yoshi
yesterday
1
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 $tau_1 = tau_2$.
– p4sch
yesterday
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.
– yoshi
yesterday
1
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.
– p4sch
12 hours ago
add a comment |
up vote
1
down vote
accepted
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 2 days ago
p4sch
4,130216
why does "no finer topology" follow from the open mapping theorem?
– yoshi
yesterday
1
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 $tau_1 = tau_2$.
– p4sch
yesterday
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.
– yoshi
yesterday
1
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.
– p4sch
12 hours ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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 2 days ago
p4sch
4,130216
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 2 days ago
p4sch
4,130216
edited yesterday
answered 2 days ago
p4sch
4,130216
answered 2 days ago
p4sch
4,130216
answered 2 days ago
p4sch
4,130216
4,130216
why does "no finer topology" follow from the open mapping theorem?
– yoshi
yesterday
1
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 $tau_1 = tau_2$.
– p4sch
yesterday
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.
– yoshi
yesterday
1
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.
– p4sch
12 hours ago
add a comment |
why does "no finer topology" follow from the open mapping theorem?
– yoshi
yesterday
1
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 $tau_1 = tau_2$.
– p4sch
yesterday
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.
– yoshi
yesterday
1
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.
– p4sch
12 hours ago
why does "no finer topology" follow from the open mapping theorem?
– yoshi
yesterday
why does "no finer topology" follow from the open mapping theorem?
– yoshi
yesterday
1
1
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 $tau_1 = tau_2$.
– p4sch
yesterday
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 $tau_1 = tau_2$.
– p4sch
yesterday
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.
– yoshi
yesterday
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.
– yoshi
yesterday
1
1
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.
– p4sch
12 hours ago
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.
– p4sch
12 hours ago
add a comment |
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1
The seminirms are $p_{n,k}(f):= sup_{[-k,k]} |f^{(n)}|$. The key is that any compact set is contained in some interval $[-k,k]$
– Tito Eliatron
2 days ago
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?
– yoshi
2 days ago
1
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.
– Tito Eliatron
2 days ago
Can you elaborate on this statement, what is a p-ball and why is it suffient for one p-ball to be inside?
– yoshi
yesterday