Is there no norm in $C^infty ([a,b])$?
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Does anyone knows a reference, which proves the following:
Let $a,bin mathbbR$ with $a<b$. There is no norm in the space $C^infty([a,b])$, which makes it a Banach space.
functional-analysis reference-request
edited Nov 9 '17 at 1:04
Ooker
299318
asked Sep 3 '14 at 16:12
Tomás
15.7k31974
 |Â
show 4 more comments
up vote
16
down vote
favorite
Does anyone knows a reference, which proves the following:
Let $a,bin mathbbR$ with $a<b$. There is no norm in the space $C^infty([a,b])$, which makes it a Banach space.
functional-analysis reference-request
edited Nov 9 '17 at 1:04
Ooker
299318
asked Sep 3 '14 at 16:12
Tomás
15.7k31974
3
That's not true, there are norms making it a Banach space since there are Banach spaces of the same dimension ($2^aleph_0$). There is no "reasonable" norm making it a Banach space, "reasonable" meaning inducing a topology that is comparable with the standard Fréchet space topology (open mapping theorem).
– Daniel Fischer♦
Sep 3 '14 at 16:15
@DanielFischer, I see. I read in plenty of places, that the claim above were true. So I just lost my time... Do you know any reference for your reply?
– Tomás
Sep 3 '14 at 16:20
Maybe we can use the norm $|f| = sum_n frac12^nminf^(n)(x)$?
– Petite Etincelle
Sep 3 '14 at 16:24
Here is one place, where I found this claim @DanielFischer. There are others.
– Tomás
Sep 3 '14 at 16:29
2
@LiuGang, it does not seem to be a norm. For example, $|lambda f|neq lambda |f|$.
– Tomás
Sep 3 '14 at 16:35
 |Â
show 4 more comments
up vote
16
down vote
favorite
up vote
16
down vote
favorite
Does anyone knows a reference, which proves the following:
Let $a,bin mathbbR$ with $a<b$. There is no norm in the space $C^infty([a,b])$, which makes it a Banach space.
functional-analysis reference-request
edited Nov 9 '17 at 1:04
Ooker
299318
asked Sep 3 '14 at 16:12
Tomás
15.7k31974
Does anyone knows a reference, which proves the following:
Let $a,bin mathbbR$ with $a<b$. There is no norm in the space $C^infty([a,b])$, which makes it a Banach space.
functional-analysis reference-request
edited Nov 9 '17 at 1:04
Ooker
299318
asked Sep 3 '14 at 16:12
Tomás
15.7k31974
edited Nov 9 '17 at 1:04
Ooker
299318
edited Nov 9 '17 at 1:04
Ooker
299318
edited Nov 9 '17 at 1:04
Ooker
299318
299318
asked Sep 3 '14 at 16:12
Tomás
15.7k31974
asked Sep 3 '14 at 16:12
Tomás
15.7k31974
asked Sep 3 '14 at 16:12
Tomás
15.7k31974
15.7k31974
3
That's not true, there are norms making it a Banach space since there are Banach spaces of the same dimension ($2^aleph_0$). There is no "reasonable" norm making it a Banach space, "reasonable" meaning inducing a topology that is comparable with the standard Fréchet space topology (open mapping theorem).
– Daniel Fischer♦
Sep 3 '14 at 16:15
@DanielFischer, I see. I read in plenty of places, that the claim above were true. So I just lost my time... Do you know any reference for your reply?
– Tomás
Sep 3 '14 at 16:20
Maybe we can use the norm $|f| = sum_n frac12^nminf^(n)(x)$?
– Petite Etincelle
Sep 3 '14 at 16:24
Here is one place, where I found this claim @DanielFischer. There are others.
– Tomás
Sep 3 '14 at 16:29
2
@LiuGang, it does not seem to be a norm. For example, $|lambda f|neq lambda |f|$.
– Tomás
Sep 3 '14 at 16:35
 |Â
show 4 more comments
3
That's not true, there are norms making it a Banach space since there are Banach spaces of the same dimension ($2^aleph_0$). There is no "reasonable" norm making it a Banach space, "reasonable" meaning inducing a topology that is comparable with the standard Fréchet space topology (open mapping theorem).
– Daniel Fischer♦
Sep 3 '14 at 16:15
@DanielFischer, I see. I read in plenty of places, that the claim above were true. So I just lost my time... Do you know any reference for your reply?
– Tomás
Sep 3 '14 at 16:20
Maybe we can use the norm $|f| = sum_n frac12^nminf^(n)(x)$?
– Petite Etincelle
Sep 3 '14 at 16:24
Here is one place, where I found this claim @DanielFischer. There are others.
– Tomás
Sep 3 '14 at 16:29
2
@LiuGang, it does not seem to be a norm. For example, $|lambda f|neq lambda |f|$.
– Tomás
Sep 3 '14 at 16:35
3
3
That's not true, there are norms making it a Banach space since there are Banach spaces of the same dimension ($2^aleph_0$). There is no "reasonable" norm making it a Banach space, "reasonable" meaning inducing a topology that is comparable with the standard Fréchet space topology (open mapping theorem).
– Daniel Fischer♦
Sep 3 '14 at 16:15
That's not true, there are norms making it a Banach space since there are Banach spaces of the same dimension ($2^aleph_0$). There is no "reasonable" norm making it a Banach space, "reasonable" meaning inducing a topology that is comparable with the standard Fréchet space topology (open mapping theorem).
– Daniel Fischer♦
Sep 3 '14 at 16:15
@DanielFischer, I see. I read in plenty of places, that the claim above were true. So I just lost my time... Do you know any reference for your reply?
– Tomás
Sep 3 '14 at 16:20
@DanielFischer, I see. I read in plenty of places, that the claim above were true. So I just lost my time... Do you know any reference for your reply?
– Tomás
Sep 3 '14 at 16:20
Maybe we can use the norm $|f| = sum_n frac12^nminf^(n)(x)$?
– Petite Etincelle
Sep 3 '14 at 16:24
Maybe we can use the norm $|f| = sum_n frac12^nminf^(n)(x)$?
– Petite Etincelle
Sep 3 '14 at 16:24
Here is one place, where I found this claim @DanielFischer. There are others.
– Tomás
Sep 3 '14 at 16:29
Here is one place, where I found this claim @DanielFischer. There are others.
– Tomás
Sep 3 '14 at 16:29
2
2
@LiuGang, it does not seem to be a norm. For example, $|lambda f|neq lambda |f|$.
– Tomás
Sep 3 '14 at 16:35
@LiuGang, it does not seem to be a norm. For example, $|lambda f|neq lambda |f|$.
– Tomás
Sep 3 '14 at 16:35
 |Â
show 4 more comments
2 Answers
2
active
oldest
votes
up vote
22
down vote
accepted
Literally, the claim is wrong, since the space has dimension $2^aleph_0$, and there are Banach spaces with the same dimension (e.g. the $ell^p(mathbbN)$ spaces). Since the rationals are dense in $[a,b]$, there is a linear injection of $C^infty([a,b])$ into $mathbbR^mathbbQcap [a,b]$, and the latter space has cardinality
$$operatornamecard(mathbbR)^aleph_0 = left(2^aleph_0right)^aleph_0 = 2^aleph_0,$$
so $dim C^infty([a,b]) leqslant 2^aleph_0$. On the other hand, the functions $tmapsto e^ct,, cinmathbbR$ are linearly independent, so the dimension is at least $2^aleph_0$.
What is meant, even if it is not said, when that claim is made, is that the space $C^infty([a,b])$ in its natural topology - the usual Fréchet space topology induced by the seminorms $lVert frVert_k = sup lvert f^(k)(t)rvert : tin [a,b]$ for $kinmathbbN$ - is not normable.
An easy way to see that is to note that $C^infty([a,b])$ is a Fréchet-Montel space, that is, every closed and bounded subset is compact. That is a repeated application of the Ascoli-Arzelàtheorem; the boundedness of $f^(k+1) : fin B$ implies the equicontinuity of $ f^(k) : fin B$.
But a normed space has the Montel-Heine-Borel property if and only if it is finite-dimensional.
answered Sep 3 '14 at 17:34
Daniel Fischer♦
172k16156277
2
Nice proof, thank you very much. Maybe the following can save others a few minutes: To see that $left(e^ctright)_cinmathbbR$ is linearly independent, assume $sum_i=1^nalpha_ie^c_itequiv0$ on $left[a,bright]$ with all $alpha_ineq0$ and $c_1<dots<c_n$. Using the identity theorem (for holomorphic functions, say), this implies $sum_i=1^nalpha_ie^c_itequiv0$ on $mathbbR$. Division by $e^c_nt$ yields $$0equivsum_i=1^nalpha_ie^left(c_i-c_nright)txrightarrow[t to infty]alpha_n$$ in contradiction to $alpha_nneq0$.
– PhoemueX
Sep 3 '14 at 18:29
Am I wrong in think (intuitively), that any norm in $C^infty([a,b])$, which makes it a Banach space, will have to involve somehow, $f$ and all it's derivatives? I mean something like this $$|f|=H(f,f',f'',...,f^(n),...),$$ where $H$ is some function? If the question is to vague, I will delete it.
– Tomás
Sep 4 '14 at 12:53
I think that you cannot construct a norm making it into a Banach space explicitly. Finding Hamel bases of $C^infty([a,b])$ and $ell^p$ (or some other Banach space of dimension $2^aleph_0$) seems hopeless, and apart from transporting the norm via a linear isomorphism, I don't know how one would make it. But abstractly, or hypothetically, I think you're right, if it involved only finitely many derivatives, I don't see how $C^infty([a,b])$ could not be dense in $C^k([a,b])$ for a sufficiently large $k$.
– Daniel Fischer♦
Sep 4 '14 at 13:01
What is $2^aleph_0$ ? And why the dimension is $2^aleph_0$?
– lanse7pty
May 11 '16 at 1:59
@lanse7pty $aleph_0$ is the cardinality of $mathbbN$, hence $2^aleph_0$ is the cardinality of the power set of $mathbbN$, which also is the cardinality of $mathbbR$. The dimension is $2^aleph_0$ because a) the whole space has cardinality $2^aleph_0$, so the dimension is $leqslant 2^aleph_0$, and b) one can find linearly independent families in the space which have cardinality $2^aleph_0$, hence the dimension is $geqslant 2^aleph_0$.
– Daniel Fischer♦
May 11 '16 at 8:13
add a comment |Â
up vote
0
down vote
Assume there exists a norm generating the topology. Consider the open unit ball $B$ around $0$. Since the seminorms $f mapsto sup ||f^n||$ define the same topology there exists
an $epsilon > 0$ and $n$ so that
$$f^(i) subset B$$
The set $ $ is an open neighborhood of $0$ so there exists a $delta>0 $ so that $delta B subset $. We conclude from the above that
$$ sup subset $$ and therefore
$$sup ||f^(n+1)|| le frac1epsilon cdot delta max ( sup ||f^(i)|| , i=1,ldots n)$$
for all $f$.
It's not too hard to see that such an inequality is not possible. For instance consider $f$ with $f^(n)(x) = frac1M sin (Mx)$ for large $M$.
Note that there exists a metric that defines the topology given by
$d(f,g) = rho(f-g)$ where
$$rho(f) = sum_n=1^infty frac12^n cdot frac$$
answered Sep 9 '14 at 8:23
orangeskid
28.1k31746
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
22
down vote
accepted
Literally, the claim is wrong, since the space has dimension $2^aleph_0$, and there are Banach spaces with the same dimension (e.g. the $ell^p(mathbbN)$ spaces). Since the rationals are dense in $[a,b]$, there is a linear injection of $C^infty([a,b])$ into $mathbbR^mathbbQcap [a,b]$, and the latter space has cardinality
$$operatornamecard(mathbbR)^aleph_0 = left(2^aleph_0right)^aleph_0 = 2^aleph_0,$$
so $dim C^infty([a,b]) leqslant 2^aleph_0$. On the other hand, the functions $tmapsto e^ct,, cinmathbbR$ are linearly independent, so the dimension is at least $2^aleph_0$.
What is meant, even if it is not said, when that claim is made, is that the space $C^infty([a,b])$ in its natural topology - the usual Fréchet space topology induced by the seminorms $lVert frVert_k = sup lvert f^(k)(t)rvert : tin [a,b]$ for $kinmathbbN$ - is not normable.
An easy way to see that is to note that $C^infty([a,b])$ is a Fréchet-Montel space, that is, every closed and bounded subset is compact. That is a repeated application of the Ascoli-Arzelàtheorem; the boundedness of $f^(k+1) : fin B$ implies the equicontinuity of $ f^(k) : fin B$.
But a normed space has the Montel-Heine-Borel property if and only if it is finite-dimensional.
answered Sep 3 '14 at 17:34
Daniel Fischer♦
172k16156277
2
Nice proof, thank you very much. Maybe the following can save others a few minutes: To see that $left(e^ctright)_cinmathbbR$ is linearly independent, assume $sum_i=1^nalpha_ie^c_itequiv0$ on $left[a,bright]$ with all $alpha_ineq0$ and $c_1<dots<c_n$. Using the identity theorem (for holomorphic functions, say), this implies $sum_i=1^nalpha_ie^c_itequiv0$ on $mathbbR$. Division by $e^c_nt$ yields $$0equivsum_i=1^nalpha_ie^left(c_i-c_nright)txrightarrow[t to infty]alpha_n$$ in contradiction to $alpha_nneq0$.
– PhoemueX
Sep 3 '14 at 18:29
Am I wrong in think (intuitively), that any norm in $C^infty([a,b])$, which makes it a Banach space, will have to involve somehow, $f$ and all it's derivatives? I mean something like this $$|f|=H(f,f',f'',...,f^(n),...),$$ where $H$ is some function? If the question is to vague, I will delete it.
– Tomás
Sep 4 '14 at 12:53
I think that you cannot construct a norm making it into a Banach space explicitly. Finding Hamel bases of $C^infty([a,b])$ and $ell^p$ (or some other Banach space of dimension $2^aleph_0$) seems hopeless, and apart from transporting the norm via a linear isomorphism, I don't know how one would make it. But abstractly, or hypothetically, I think you're right, if it involved only finitely many derivatives, I don't see how $C^infty([a,b])$ could not be dense in $C^k([a,b])$ for a sufficiently large $k$.
– Daniel Fischer♦
Sep 4 '14 at 13:01
What is $2^aleph_0$ ? And why the dimension is $2^aleph_0$?
– lanse7pty
May 11 '16 at 1:59
@lanse7pty $aleph_0$ is the cardinality of $mathbbN$, hence $2^aleph_0$ is the cardinality of the power set of $mathbbN$, which also is the cardinality of $mathbbR$. The dimension is $2^aleph_0$ because a) the whole space has cardinality $2^aleph_0$, so the dimension is $leqslant 2^aleph_0$, and b) one can find linearly independent families in the space which have cardinality $2^aleph_0$, hence the dimension is $geqslant 2^aleph_0$.
– Daniel Fischer♦
May 11 '16 at 8:13
add a comment |Â
up vote
22
down vote
accepted
Literally, the claim is wrong, since the space has dimension $2^aleph_0$, and there are Banach spaces with the same dimension (e.g. the $ell^p(mathbbN)$ spaces). Since the rationals are dense in $[a,b]$, there is a linear injection of $C^infty([a,b])$ into $mathbbR^mathbbQcap [a,b]$, and the latter space has cardinality
$$operatornamecard(mathbbR)^aleph_0 = left(2^aleph_0right)^aleph_0 = 2^aleph_0,$$
so $dim C^infty([a,b]) leqslant 2^aleph_0$. On the other hand, the functions $tmapsto e^ct,, cinmathbbR$ are linearly independent, so the dimension is at least $2^aleph_0$.
What is meant, even if it is not said, when that claim is made, is that the space $C^infty([a,b])$ in its natural topology - the usual Fréchet space topology induced by the seminorms $lVert frVert_k = sup lvert f^(k)(t)rvert : tin [a,b]$ for $kinmathbbN$ - is not normable.
An easy way to see that is to note that $C^infty([a,b])$ is a Fréchet-Montel space, that is, every closed and bounded subset is compact. That is a repeated application of the Ascoli-Arzelàtheorem; the boundedness of $f^(k+1) : fin B$ implies the equicontinuity of $ f^(k) : fin B$.
But a normed space has the Montel-Heine-Borel property if and only if it is finite-dimensional.
answered Sep 3 '14 at 17:34
Daniel Fischer♦
172k16156277
2
Nice proof, thank you very much. Maybe the following can save others a few minutes: To see that $left(e^ctright)_cinmathbbR$ is linearly independent, assume $sum_i=1^nalpha_ie^c_itequiv0$ on $left[a,bright]$ with all $alpha_ineq0$ and $c_1<dots<c_n$. Using the identity theorem (for holomorphic functions, say), this implies $sum_i=1^nalpha_ie^c_itequiv0$ on $mathbbR$. Division by $e^c_nt$ yields $$0equivsum_i=1^nalpha_ie^left(c_i-c_nright)txrightarrow[t to infty]alpha_n$$ in contradiction to $alpha_nneq0$.
– PhoemueX
Sep 3 '14 at 18:29
Am I wrong in think (intuitively), that any norm in $C^infty([a,b])$, which makes it a Banach space, will have to involve somehow, $f$ and all it's derivatives? I mean something like this $$|f|=H(f,f',f'',...,f^(n),...),$$ where $H$ is some function? If the question is to vague, I will delete it.
– Tomás
Sep 4 '14 at 12:53
I think that you cannot construct a norm making it into a Banach space explicitly. Finding Hamel bases of $C^infty([a,b])$ and $ell^p$ (or some other Banach space of dimension $2^aleph_0$) seems hopeless, and apart from transporting the norm via a linear isomorphism, I don't know how one would make it. But abstractly, or hypothetically, I think you're right, if it involved only finitely many derivatives, I don't see how $C^infty([a,b])$ could not be dense in $C^k([a,b])$ for a sufficiently large $k$.
– Daniel Fischer♦
Sep 4 '14 at 13:01
What is $2^aleph_0$ ? And why the dimension is $2^aleph_0$?
– lanse7pty
May 11 '16 at 1:59
@lanse7pty $aleph_0$ is the cardinality of $mathbbN$, hence $2^aleph_0$ is the cardinality of the power set of $mathbbN$, which also is the cardinality of $mathbbR$. The dimension is $2^aleph_0$ because a) the whole space has cardinality $2^aleph_0$, so the dimension is $leqslant 2^aleph_0$, and b) one can find linearly independent families in the space which have cardinality $2^aleph_0$, hence the dimension is $geqslant 2^aleph_0$.
– Daniel Fischer♦
May 11 '16 at 8:13
add a comment |Â
up vote
22
down vote
accepted
up vote
22
down vote
accepted
Literally, the claim is wrong, since the space has dimension $2^aleph_0$, and there are Banach spaces with the same dimension (e.g. the $ell^p(mathbbN)$ spaces). Since the rationals are dense in $[a,b]$, there is a linear injection of $C^infty([a,b])$ into $mathbbR^mathbbQcap [a,b]$, and the latter space has cardinality
$$operatornamecard(mathbbR)^aleph_0 = left(2^aleph_0right)^aleph_0 = 2^aleph_0,$$
so $dim C^infty([a,b]) leqslant 2^aleph_0$. On the other hand, the functions $tmapsto e^ct,, cinmathbbR$ are linearly independent, so the dimension is at least $2^aleph_0$.
What is meant, even if it is not said, when that claim is made, is that the space $C^infty([a,b])$ in its natural topology - the usual Fréchet space topology induced by the seminorms $lVert frVert_k = sup lvert f^(k)(t)rvert : tin [a,b]$ for $kinmathbbN$ - is not normable.
An easy way to see that is to note that $C^infty([a,b])$ is a Fréchet-Montel space, that is, every closed and bounded subset is compact. That is a repeated application of the Ascoli-Arzelàtheorem; the boundedness of $f^(k+1) : fin B$ implies the equicontinuity of $ f^(k) : fin B$.
But a normed space has the Montel-Heine-Borel property if and only if it is finite-dimensional.
answered Sep 3 '14 at 17:34
Daniel Fischer♦
172k16156277
Literally, the claim is wrong, since the space has dimension $2^aleph_0$, and there are Banach spaces with the same dimension (e.g. the $ell^p(mathbbN)$ spaces). Since the rationals are dense in $[a,b]$, there is a linear injection of $C^infty([a,b])$ into $mathbbR^mathbbQcap [a,b]$, and the latter space has cardinality
$$operatornamecard(mathbbR)^aleph_0 = left(2^aleph_0right)^aleph_0 = 2^aleph_0,$$
so $dim C^infty([a,b]) leqslant 2^aleph_0$. On the other hand, the functions $tmapsto e^ct,, cinmathbbR$ are linearly independent, so the dimension is at least $2^aleph_0$.
What is meant, even if it is not said, when that claim is made, is that the space $C^infty([a,b])$ in its natural topology - the usual Fréchet space topology induced by the seminorms $lVert frVert_k = sup lvert f^(k)(t)rvert : tin [a,b]$ for $kinmathbbN$ - is not normable.
An easy way to see that is to note that $C^infty([a,b])$ is a Fréchet-Montel space, that is, every closed and bounded subset is compact. That is a repeated application of the Ascoli-Arzelàtheorem; the boundedness of $f^(k+1) : fin B$ implies the equicontinuity of $ f^(k) : fin B$.
But a normed space has the Montel-Heine-Borel property if and only if it is finite-dimensional.
answered Sep 3 '14 at 17:34
Daniel Fischer♦
172k16156277
answered Sep 3 '14 at 17:34
Daniel Fischer♦
172k16156277
answered Sep 3 '14 at 17:34
Daniel Fischer♦
172k16156277
answered Sep 3 '14 at 17:34
Daniel Fischer♦
172k16156277
172k16156277
2
Nice proof, thank you very much. Maybe the following can save others a few minutes: To see that $left(e^ctright)_cinmathbbR$ is linearly independent, assume $sum_i=1^nalpha_ie^c_itequiv0$ on $left[a,bright]$ with all $alpha_ineq0$ and $c_1<dots<c_n$. Using the identity theorem (for holomorphic functions, say), this implies $sum_i=1^nalpha_ie^c_itequiv0$ on $mathbbR$. Division by $e^c_nt$ yields $$0equivsum_i=1^nalpha_ie^left(c_i-c_nright)txrightarrow[t to infty]alpha_n$$ in contradiction to $alpha_nneq0$.
– PhoemueX
Sep 3 '14 at 18:29
Am I wrong in think (intuitively), that any norm in $C^infty([a,b])$, which makes it a Banach space, will have to involve somehow, $f$ and all it's derivatives? I mean something like this $$|f|=H(f,f',f'',...,f^(n),...),$$ where $H$ is some function? If the question is to vague, I will delete it.
– Tomás
Sep 4 '14 at 12:53
I think that you cannot construct a norm making it into a Banach space explicitly. Finding Hamel bases of $C^infty([a,b])$ and $ell^p$ (or some other Banach space of dimension $2^aleph_0$) seems hopeless, and apart from transporting the norm via a linear isomorphism, I don't know how one would make it. But abstractly, or hypothetically, I think you're right, if it involved only finitely many derivatives, I don't see how $C^infty([a,b])$ could not be dense in $C^k([a,b])$ for a sufficiently large $k$.
– Daniel Fischer♦
Sep 4 '14 at 13:01
What is $2^aleph_0$ ? And why the dimension is $2^aleph_0$?
– lanse7pty
May 11 '16 at 1:59
@lanse7pty $aleph_0$ is the cardinality of $mathbbN$, hence $2^aleph_0$ is the cardinality of the power set of $mathbbN$, which also is the cardinality of $mathbbR$. The dimension is $2^aleph_0$ because a) the whole space has cardinality $2^aleph_0$, so the dimension is $leqslant 2^aleph_0$, and b) one can find linearly independent families in the space which have cardinality $2^aleph_0$, hence the dimension is $geqslant 2^aleph_0$.
– Daniel Fischer♦
May 11 '16 at 8:13
add a comment |Â
2
Nice proof, thank you very much. Maybe the following can save others a few minutes: To see that $left(e^ctright)_cinmathbbR$ is linearly independent, assume $sum_i=1^nalpha_ie^c_itequiv0$ on $left[a,bright]$ with all $alpha_ineq0$ and $c_1<dots<c_n$. Using the identity theorem (for holomorphic functions, say), this implies $sum_i=1^nalpha_ie^c_itequiv0$ on $mathbbR$. Division by $e^c_nt$ yields $$0equivsum_i=1^nalpha_ie^left(c_i-c_nright)txrightarrow[t to infty]alpha_n$$ in contradiction to $alpha_nneq0$.
– PhoemueX
Sep 3 '14 at 18:29
Am I wrong in think (intuitively), that any norm in $C^infty([a,b])$, which makes it a Banach space, will have to involve somehow, $f$ and all it's derivatives? I mean something like this $$|f|=H(f,f',f'',...,f^(n),...),$$ where $H$ is some function? If the question is to vague, I will delete it.
– Tomás
Sep 4 '14 at 12:53
I think that you cannot construct a norm making it into a Banach space explicitly. Finding Hamel bases of $C^infty([a,b])$ and $ell^p$ (or some other Banach space of dimension $2^aleph_0$) seems hopeless, and apart from transporting the norm via a linear isomorphism, I don't know how one would make it. But abstractly, or hypothetically, I think you're right, if it involved only finitely many derivatives, I don't see how $C^infty([a,b])$ could not be dense in $C^k([a,b])$ for a sufficiently large $k$.
– Daniel Fischer♦
Sep 4 '14 at 13:01
What is $2^aleph_0$ ? And why the dimension is $2^aleph_0$?
– lanse7pty
May 11 '16 at 1:59
@lanse7pty $aleph_0$ is the cardinality of $mathbbN$, hence $2^aleph_0$ is the cardinality of the power set of $mathbbN$, which also is the cardinality of $mathbbR$. The dimension is $2^aleph_0$ because a) the whole space has cardinality $2^aleph_0$, so the dimension is $leqslant 2^aleph_0$, and b) one can find linearly independent families in the space which have cardinality $2^aleph_0$, hence the dimension is $geqslant 2^aleph_0$.
– Daniel Fischer♦
May 11 '16 at 8:13
2
2
Nice proof, thank you very much. Maybe the following can save others a few minutes: To see that $left(e^ctright)_cinmathbbR$ is linearly independent, assume $sum_i=1^nalpha_ie^c_itequiv0$ on $left[a,bright]$ with all $alpha_ineq0$ and $c_1<dots<c_n$. Using the identity theorem (for holomorphic functions, say), this implies $sum_i=1^nalpha_ie^c_itequiv0$ on $mathbbR$. Division by $e^c_nt$ yields $$0equivsum_i=1^nalpha_ie^left(c_i-c_nright)txrightarrow[t to infty]alpha_n$$ in contradiction to $alpha_nneq0$.
– PhoemueX
Sep 3 '14 at 18:29
Nice proof, thank you very much. Maybe the following can save others a few minutes: To see that $left(e^ctright)_cinmathbbR$ is linearly independent, assume $sum_i=1^nalpha_ie^c_itequiv0$ on $left[a,bright]$ with all $alpha_ineq0$ and $c_1<dots<c_n$. Using the identity theorem (for holomorphic functions, say), this implies $sum_i=1^nalpha_ie^c_itequiv0$ on $mathbbR$. Division by $e^c_nt$ yields $$0equivsum_i=1^nalpha_ie^left(c_i-c_nright)txrightarrow[t to infty]alpha_n$$ in contradiction to $alpha_nneq0$.
– PhoemueX
Sep 3 '14 at 18:29
Am I wrong in think (intuitively), that any norm in $C^infty([a,b])$, which makes it a Banach space, will have to involve somehow, $f$ and all it's derivatives? I mean something like this $$|f|=H(f,f',f'',...,f^(n),...),$$ where $H$ is some function? If the question is to vague, I will delete it.
– Tomás
Sep 4 '14 at 12:53
Am I wrong in think (intuitively), that any norm in $C^infty([a,b])$, which makes it a Banach space, will have to involve somehow, $f$ and all it's derivatives? I mean something like this $$|f|=H(f,f',f'',...,f^(n),...),$$ where $H$ is some function? If the question is to vague, I will delete it.
– Tomás
Sep 4 '14 at 12:53
I think that you cannot construct a norm making it into a Banach space explicitly. Finding Hamel bases of $C^infty([a,b])$ and $ell^p$ (or some other Banach space of dimension $2^aleph_0$) seems hopeless, and apart from transporting the norm via a linear isomorphism, I don't know how one would make it. But abstractly, or hypothetically, I think you're right, if it involved only finitely many derivatives, I don't see how $C^infty([a,b])$ could not be dense in $C^k([a,b])$ for a sufficiently large $k$.
– Daniel Fischer♦
Sep 4 '14 at 13:01
I think that you cannot construct a norm making it into a Banach space explicitly. Finding Hamel bases of $C^infty([a,b])$ and $ell^p$ (or some other Banach space of dimension $2^aleph_0$) seems hopeless, and apart from transporting the norm via a linear isomorphism, I don't know how one would make it. But abstractly, or hypothetically, I think you're right, if it involved only finitely many derivatives, I don't see how $C^infty([a,b])$ could not be dense in $C^k([a,b])$ for a sufficiently large $k$.
– Daniel Fischer♦
Sep 4 '14 at 13:01
What is $2^aleph_0$ ? And why the dimension is $2^aleph_0$?
– lanse7pty
May 11 '16 at 1:59
What is $2^aleph_0$ ? And why the dimension is $2^aleph_0$?
– lanse7pty
May 11 '16 at 1:59
@lanse7pty $aleph_0$ is the cardinality of $mathbbN$, hence $2^aleph_0$ is the cardinality of the power set of $mathbbN$, which also is the cardinality of $mathbbR$. The dimension is $2^aleph_0$ because a) the whole space has cardinality $2^aleph_0$, so the dimension is $leqslant 2^aleph_0$, and b) one can find linearly independent families in the space which have cardinality $2^aleph_0$, hence the dimension is $geqslant 2^aleph_0$.
– Daniel Fischer♦
May 11 '16 at 8:13
@lanse7pty $aleph_0$ is the cardinality of $mathbbN$, hence $2^aleph_0$ is the cardinality of the power set of $mathbbN$, which also is the cardinality of $mathbbR$. The dimension is $2^aleph_0$ because a) the whole space has cardinality $2^aleph_0$, so the dimension is $leqslant 2^aleph_0$, and b) one can find linearly independent families in the space which have cardinality $2^aleph_0$, hence the dimension is $geqslant 2^aleph_0$.
– Daniel Fischer♦
May 11 '16 at 8:13
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Assume there exists a norm generating the topology. Consider the open unit ball $B$ around $0$. Since the seminorms $f mapsto sup ||f^n||$ define the same topology there exists
an $epsilon > 0$ and $n$ so that
$$f^(i) subset B$$
The set $ $ is an open neighborhood of $0$ so there exists a $delta>0 $ so that $delta B subset $. We conclude from the above that
$$ sup subset $$ and therefore
$$sup ||f^(n+1)|| le frac1epsilon cdot delta max ( sup ||f^(i)|| , i=1,ldots n)$$
for all $f$.
It's not too hard to see that such an inequality is not possible. For instance consider $f$ with $f^(n)(x) = frac1M sin (Mx)$ for large $M$.
Note that there exists a metric that defines the topology given by
$d(f,g) = rho(f-g)$ where
$$rho(f) = sum_n=1^infty frac12^n cdot frac$$
answered Sep 9 '14 at 8:23
orangeskid
28.1k31746
add a comment |Â
up vote
0
down vote
Assume there exists a norm generating the topology. Consider the open unit ball $B$ around $0$. Since the seminorms $f mapsto sup ||f^n||$ define the same topology there exists
an $epsilon > 0$ and $n$ so that
$$f^(i) subset B$$
The set $ $ is an open neighborhood of $0$ so there exists a $delta>0 $ so that $delta B subset $. We conclude from the above that
$$ sup subset $$ and therefore
$$sup ||f^(n+1)|| le frac1epsilon cdot delta max ( sup ||f^(i)|| , i=1,ldots n)$$
for all $f$.
It's not too hard to see that such an inequality is not possible. For instance consider $f$ with $f^(n)(x) = frac1M sin (Mx)$ for large $M$.
Note that there exists a metric that defines the topology given by
$d(f,g) = rho(f-g)$ where
$$rho(f) = sum_n=1^infty frac12^n cdot frac$$
answered Sep 9 '14 at 8:23
orangeskid
28.1k31746
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Assume there exists a norm generating the topology. Consider the open unit ball $B$ around $0$. Since the seminorms $f mapsto sup ||f^n||$ define the same topology there exists
an $epsilon > 0$ and $n$ so that
$$f^(i) subset B$$
The set $ $ is an open neighborhood of $0$ so there exists a $delta>0 $ so that $delta B subset $. We conclude from the above that
$$ sup subset $$ and therefore
$$sup ||f^(n+1)|| le frac1epsilon cdot delta max ( sup ||f^(i)|| , i=1,ldots n)$$
for all $f$.
It's not too hard to see that such an inequality is not possible. For instance consider $f$ with $f^(n)(x) = frac1M sin (Mx)$ for large $M$.
Note that there exists a metric that defines the topology given by
$d(f,g) = rho(f-g)$ where
$$rho(f) = sum_n=1^infty frac12^n cdot frac$$
answered Sep 9 '14 at 8:23
orangeskid
28.1k31746
Assume there exists a norm generating the topology. Consider the open unit ball $B$ around $0$. Since the seminorms $f mapsto sup ||f^n||$ define the same topology there exists
an $epsilon > 0$ and $n$ so that
$$f^(i) subset B$$
The set $ $ is an open neighborhood of $0$ so there exists a $delta>0 $ so that $delta B subset $. We conclude from the above that
$$ sup subset $$ and therefore
$$sup ||f^(n+1)|| le frac1epsilon cdot delta max ( sup ||f^(i)|| , i=1,ldots n)$$
for all $f$.
It's not too hard to see that such an inequality is not possible. For instance consider $f$ with $f^(n)(x) = frac1M sin (Mx)$ for large $M$.
Note that there exists a metric that defines the topology given by
$d(f,g) = rho(f-g)$ where
$$rho(f) = sum_n=1^infty frac12^n cdot frac$$
answered Sep 9 '14 at 8:23
orangeskid
28.1k31746
edited Sep 9 '14 at 10:38
answered Sep 9 '14 at 8:23
orangeskid
28.1k31746
answered Sep 9 '14 at 8:23
orangeskid
28.1k31746
answered Sep 9 '14 at 8:23
orangeskid
28.1k31746
28.1k31746
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3
That's not true, there are norms making it a Banach space since there are Banach spaces of the same dimension ($2^aleph_0$). There is no "reasonable" norm making it a Banach space, "reasonable" meaning inducing a topology that is comparable with the standard Fréchet space topology (open mapping theorem).
– Daniel Fischer♦
Sep 3 '14 at 16:15
@DanielFischer, I see. I read in plenty of places, that the claim above were true. So I just lost my time... Do you know any reference for your reply?
– Tomás
Sep 3 '14 at 16:20
Maybe we can use the norm $|f| = sum_n frac12^nminf^(n)(x)$?
– Petite Etincelle
Sep 3 '14 at 16:24
Here is one place, where I found this claim @DanielFischer. There are others.
– Tomás
Sep 3 '14 at 16:29
2
@LiuGang, it does not seem to be a norm. For example, $|lambda f|neq lambda |f|$.
– Tomás
Sep 3 '14 at 16:35