Hilbert Space and Banach SpaceAre all Banach spaces also Hilbert spaces?Least norm in convex set in Banach spaceIs this space a banach space?Orthogonal complement in pre-hilbert spaceThis is Banach space?Are all Banach spaces also Hilbert spaces?Functions allowed in Hilbert and Banach spaces.Orthonormal basis for Hilbert spaceThe Hahn-Banach Theorem for Hilbert SpaceIs the space of maps between Hilbert spaces that have at most polynomial growth of order m a separable Banach space?May I know if there are some separate Banach spaces of maps between Hilbert spaces that are “richer” than Hilbert-Schmidt space?
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Hilbert Space and Banach Space
Are all Banach spaces also Hilbert spaces?Least norm in convex set in Banach spaceIs this space a banach space?Orthogonal complement in pre-hilbert spaceThis is Banach space?Are all Banach spaces also Hilbert spaces?Functions allowed in Hilbert and Banach spaces.Orthonormal basis for Hilbert spaceThe Hahn-Banach Theorem for Hilbert SpaceIs the space of maps between Hilbert spaces that have at most polynomial growth of order m a separable Banach space?May I know if there are some separate Banach spaces of maps between Hilbert spaces that are “richer” than Hilbert-Schmidt space?
$begingroup$
I just would like to ask which of the two is true.
Every Hilbert space is a Banach space or every Banach space is a Hilbert space.
Thanks.
real-analysis
asked 3 hours ago
LostinspaceLostinspace
406
$endgroup$
add a comment |
$begingroup$
I just would like to ask which of the two is true.
Every Hilbert space is a Banach space or every Banach space is a Hilbert space.
Thanks.
real-analysis
asked 3 hours ago
LostinspaceLostinspace
406
$endgroup$
2
$begingroup$
Have you checked the definitions?
$endgroup$
– avs
3 hours ago
$begingroup$
All that I know is that a Banach space is a complete normed vectors and a Hilbert space is the inner products of such these normed vectors. Am I right?
$endgroup$
– Lostinspace
3 hours ago
$begingroup$
Every Hilbert space is Banach.
$endgroup$
– Berci
3 hours ago
$begingroup$
The first is true by definition of a Hilbert space
$endgroup$
– Fakemistake
3 hours ago
add a comment |
$begingroup$
I just would like to ask which of the two is true.
Every Hilbert space is a Banach space or every Banach space is a Hilbert space.
Thanks.
real-analysis
asked 3 hours ago
LostinspaceLostinspace
406
$endgroup$
I just would like to ask which of the two is true.
Every Hilbert space is a Banach space or every Banach space is a Hilbert space.
Thanks.
real-analysis
real-analysis
asked 3 hours ago
LostinspaceLostinspace
406
asked 3 hours ago
LostinspaceLostinspace
406
asked 3 hours ago
LostinspaceLostinspace
406
asked 3 hours ago
LostinspaceLostinspace
406
asked 3 hours ago
LostinspaceLostinspace
406
406
2
$begingroup$
Have you checked the definitions?
$endgroup$
– avs
3 hours ago
$begingroup$
All that I know is that a Banach space is a complete normed vectors and a Hilbert space is the inner products of such these normed vectors. Am I right?
$endgroup$
– Lostinspace
3 hours ago
$begingroup$
Every Hilbert space is Banach.
$endgroup$
– Berci
3 hours ago
$begingroup$
The first is true by definition of a Hilbert space
$endgroup$
– Fakemistake
3 hours ago
add a comment |
2
$begingroup$
Have you checked the definitions?
$endgroup$
– avs
3 hours ago
$begingroup$
All that I know is that a Banach space is a complete normed vectors and a Hilbert space is the inner products of such these normed vectors. Am I right?
$endgroup$
– Lostinspace
3 hours ago
$begingroup$
Every Hilbert space is Banach.
$endgroup$
– Berci
3 hours ago
$begingroup$
The first is true by definition of a Hilbert space
$endgroup$
– Fakemistake
3 hours ago
2
2
$begingroup$
Have you checked the definitions?
$endgroup$
– avs
3 hours ago
$begingroup$
Have you checked the definitions?
$endgroup$
– avs
3 hours ago
$begingroup$
All that I know is that a Banach space is a complete normed vectors and a Hilbert space is the inner products of such these normed vectors. Am I right?
$endgroup$
– Lostinspace
3 hours ago
$begingroup$
All that I know is that a Banach space is a complete normed vectors and a Hilbert space is the inner products of such these normed vectors. Am I right?
$endgroup$
– Lostinspace
3 hours ago
$begingroup$
Every Hilbert space is Banach.
$endgroup$
– Berci
3 hours ago
$begingroup$
Every Hilbert space is Banach.
$endgroup$
– Berci
3 hours ago
$begingroup$
The first is true by definition of a Hilbert space
$endgroup$
– Fakemistake
3 hours ago
$begingroup$
The first is true by definition of a Hilbert space
$endgroup$
– Fakemistake
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A Hilbert space is necessarily a Banach space.
The converse isn't true : there are Banach spaces that cannot be transformed into Hilbert spaces. I.e., the class of Banach spaces is strictly larger than the class of Hilbert spaces. Indeed,
the norm in an Hilbert space must satisfy the parallelogram law which is not necessarily verified by the norm in a Banach space. See https://math.stackexchange.com/q/692529
said in a different way, there is a unique way to obtain a dot product by first defining a norm : it is through the polarization identity (Are all Banach spaces also Hilbert spaces?) ; having built this associated (would be) dot product in a Banach space, it happens that... it is not a dot product... thus such a Banach space cannot be considered as a Hilbert space.
Remark : in fact, parallelogram law and polarization identity are equivalent.
The first reference recalls the famous example of $C([0,1])$, the (Banach) space of real or complex continuous functions on interval $[0,1]$ with the uniform norm ($|f|_infty:=sup_x in [0,1]|f(x)|$).
... Whereas the same space $C([0,1])$ can be endowed with an Hilbert space structure with the ($L^2$) norm $|...|_2$ defined by $|f|_2^2:=int_0^1 f(t)^2dt$ and associated dot product $(f|g):=int_0^1 f(t)g(t)dt$.
Remark : historically, these two types of spaces (Hilbert vs. Banach) and their interplay have been set-up in a clear-cut manner in the 1920-1930s in particular with respect to the necessity to include or not the topological property of completeness.
answered 3 hours ago
Jean MarieJean Marie
31.8k42355
$endgroup$
$begingroup$
You switched it up in the first sentence, every Hilbert space is a Banach space.
$endgroup$
– Jannik Pitt
1 hour ago
$begingroup$
@Jannik Pitt Corrected. Thank you very much.
$endgroup$
– Jean Marie
1 hour ago
add a comment |
$begingroup$
A Hilbert space is a vector space with an inner product that is a complete metric space with respect to the distance function induced by the inner product.
A Banach space is a vector space with a norm that is a complete metric space with respect to the distance function induced by the norm.
Any inner product space is a normed vector space. (The norm of a vector is the square root of the inner product of the vector with itself.)
Any Hilbert space is a Banach space.
answered 3 hours ago
J. W. TannerJ. W. Tanner
5,2101520
$endgroup$
$begingroup$
Cf. this diagram
$endgroup$
– J. W. Tanner
3 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
A Hilbert space is necessarily a Banach space.
The converse isn't true : there are Banach spaces that cannot be transformed into Hilbert spaces. I.e., the class of Banach spaces is strictly larger than the class of Hilbert spaces. Indeed,
the norm in an Hilbert space must satisfy the parallelogram law which is not necessarily verified by the norm in a Banach space. See https://math.stackexchange.com/q/692529
said in a different way, there is a unique way to obtain a dot product by first defining a norm : it is through the polarization identity (Are all Banach spaces also Hilbert spaces?) ; having built this associated (would be) dot product in a Banach space, it happens that... it is not a dot product... thus such a Banach space cannot be considered as a Hilbert space.
Remark : in fact, parallelogram law and polarization identity are equivalent.
The first reference recalls the famous example of $C([0,1])$, the (Banach) space of real or complex continuous functions on interval $[0,1]$ with the uniform norm ($|f|_infty:=sup_x in [0,1]|f(x)|$).
... Whereas the same space $C([0,1])$ can be endowed with an Hilbert space structure with the ($L^2$) norm $|...|_2$ defined by $|f|_2^2:=int_0^1 f(t)^2dt$ and associated dot product $(f|g):=int_0^1 f(t)g(t)dt$.
Remark : historically, these two types of spaces (Hilbert vs. Banach) and their interplay have been set-up in a clear-cut manner in the 1920-1930s in particular with respect to the necessity to include or not the topological property of completeness.
answered 3 hours ago
Jean MarieJean Marie
31.8k42355
$endgroup$
$begingroup$
You switched it up in the first sentence, every Hilbert space is a Banach space.
$endgroup$
– Jannik Pitt
1 hour ago
$begingroup$
@Jannik Pitt Corrected. Thank you very much.
$endgroup$
– Jean Marie
1 hour ago
add a comment |
$begingroup$
A Hilbert space is necessarily a Banach space.
The converse isn't true : there are Banach spaces that cannot be transformed into Hilbert spaces. I.e., the class of Banach spaces is strictly larger than the class of Hilbert spaces. Indeed,
the norm in an Hilbert space must satisfy the parallelogram law which is not necessarily verified by the norm in a Banach space. See https://math.stackexchange.com/q/692529
said in a different way, there is a unique way to obtain a dot product by first defining a norm : it is through the polarization identity (Are all Banach spaces also Hilbert spaces?) ; having built this associated (would be) dot product in a Banach space, it happens that... it is not a dot product... thus such a Banach space cannot be considered as a Hilbert space.
Remark : in fact, parallelogram law and polarization identity are equivalent.
The first reference recalls the famous example of $C([0,1])$, the (Banach) space of real or complex continuous functions on interval $[0,1]$ with the uniform norm ($|f|_infty:=sup_x in [0,1]|f(x)|$).
... Whereas the same space $C([0,1])$ can be endowed with an Hilbert space structure with the ($L^2$) norm $|...|_2$ defined by $|f|_2^2:=int_0^1 f(t)^2dt$ and associated dot product $(f|g):=int_0^1 f(t)g(t)dt$.
Remark : historically, these two types of spaces (Hilbert vs. Banach) and their interplay have been set-up in a clear-cut manner in the 1920-1930s in particular with respect to the necessity to include or not the topological property of completeness.
answered 3 hours ago
Jean MarieJean Marie
31.8k42355
$endgroup$
$begingroup$
You switched it up in the first sentence, every Hilbert space is a Banach space.
$endgroup$
– Jannik Pitt
1 hour ago
$begingroup$
@Jannik Pitt Corrected. Thank you very much.
$endgroup$
– Jean Marie
1 hour ago
add a comment |
$begingroup$
A Hilbert space is necessarily a Banach space.
The converse isn't true : there are Banach spaces that cannot be transformed into Hilbert spaces. I.e., the class of Banach spaces is strictly larger than the class of Hilbert spaces. Indeed,
the norm in an Hilbert space must satisfy the parallelogram law which is not necessarily verified by the norm in a Banach space. See https://math.stackexchange.com/q/692529
said in a different way, there is a unique way to obtain a dot product by first defining a norm : it is through the polarization identity (Are all Banach spaces also Hilbert spaces?) ; having built this associated (would be) dot product in a Banach space, it happens that... it is not a dot product... thus such a Banach space cannot be considered as a Hilbert space.
Remark : in fact, parallelogram law and polarization identity are equivalent.
The first reference recalls the famous example of $C([0,1])$, the (Banach) space of real or complex continuous functions on interval $[0,1]$ with the uniform norm ($|f|_infty:=sup_x in [0,1]|f(x)|$).
... Whereas the same space $C([0,1])$ can be endowed with an Hilbert space structure with the ($L^2$) norm $|...|_2$ defined by $|f|_2^2:=int_0^1 f(t)^2dt$ and associated dot product $(f|g):=int_0^1 f(t)g(t)dt$.
Remark : historically, these two types of spaces (Hilbert vs. Banach) and their interplay have been set-up in a clear-cut manner in the 1920-1930s in particular with respect to the necessity to include or not the topological property of completeness.
answered 3 hours ago
Jean MarieJean Marie
31.8k42355
$endgroup$
A Hilbert space is necessarily a Banach space.
The converse isn't true : there are Banach spaces that cannot be transformed into Hilbert spaces. I.e., the class of Banach spaces is strictly larger than the class of Hilbert spaces. Indeed,
the norm in an Hilbert space must satisfy the parallelogram law which is not necessarily verified by the norm in a Banach space. See https://math.stackexchange.com/q/692529
said in a different way, there is a unique way to obtain a dot product by first defining a norm : it is through the polarization identity (Are all Banach spaces also Hilbert spaces?) ; having built this associated (would be) dot product in a Banach space, it happens that... it is not a dot product... thus such a Banach space cannot be considered as a Hilbert space.
Remark : in fact, parallelogram law and polarization identity are equivalent.
The first reference recalls the famous example of $C([0,1])$, the (Banach) space of real or complex continuous functions on interval $[0,1]$ with the uniform norm ($|f|_infty:=sup_x in [0,1]|f(x)|$).
... Whereas the same space $C([0,1])$ can be endowed with an Hilbert space structure with the ($L^2$) norm $|...|_2$ defined by $|f|_2^2:=int_0^1 f(t)^2dt$ and associated dot product $(f|g):=int_0^1 f(t)g(t)dt$.
Remark : historically, these two types of spaces (Hilbert vs. Banach) and their interplay have been set-up in a clear-cut manner in the 1920-1930s in particular with respect to the necessity to include or not the topological property of completeness.
answered 3 hours ago
Jean MarieJean Marie
31.8k42355
edited 1 hour ago
answered 3 hours ago
Jean MarieJean Marie
31.8k42355
answered 3 hours ago
Jean MarieJean Marie
31.8k42355
answered 3 hours ago
Jean MarieJean Marie
31.8k42355
31.8k42355
$begingroup$
You switched it up in the first sentence, every Hilbert space is a Banach space.
$endgroup$
– Jannik Pitt
1 hour ago
$begingroup$
@Jannik Pitt Corrected. Thank you very much.
$endgroup$
– Jean Marie
1 hour ago
add a comment |
$begingroup$
You switched it up in the first sentence, every Hilbert space is a Banach space.
$endgroup$
– Jannik Pitt
1 hour ago
$begingroup$
@Jannik Pitt Corrected. Thank you very much.
$endgroup$
– Jean Marie
1 hour ago
$begingroup$
You switched it up in the first sentence, every Hilbert space is a Banach space.
$endgroup$
– Jannik Pitt
1 hour ago
$begingroup$
You switched it up in the first sentence, every Hilbert space is a Banach space.
$endgroup$
– Jannik Pitt
1 hour ago
$begingroup$
@Jannik Pitt Corrected. Thank you very much.
$endgroup$
– Jean Marie
1 hour ago
$begingroup$
@Jannik Pitt Corrected. Thank you very much.
$endgroup$
– Jean Marie
1 hour ago
add a comment |
$begingroup$
A Hilbert space is a vector space with an inner product that is a complete metric space with respect to the distance function induced by the inner product.
A Banach space is a vector space with a norm that is a complete metric space with respect to the distance function induced by the norm.
Any inner product space is a normed vector space. (The norm of a vector is the square root of the inner product of the vector with itself.)
Any Hilbert space is a Banach space.
answered 3 hours ago
J. W. TannerJ. W. Tanner
5,2101520
$endgroup$
$begingroup$
Cf. this diagram
$endgroup$
– J. W. Tanner
3 hours ago
add a comment |
$begingroup$
A Hilbert space is a vector space with an inner product that is a complete metric space with respect to the distance function induced by the inner product.
A Banach space is a vector space with a norm that is a complete metric space with respect to the distance function induced by the norm.
Any inner product space is a normed vector space. (The norm of a vector is the square root of the inner product of the vector with itself.)
Any Hilbert space is a Banach space.
answered 3 hours ago
J. W. TannerJ. W. Tanner
5,2101520
$endgroup$
$begingroup$
Cf. this diagram
$endgroup$
– J. W. Tanner
3 hours ago
add a comment |
$begingroup$
A Hilbert space is a vector space with an inner product that is a complete metric space with respect to the distance function induced by the inner product.
A Banach space is a vector space with a norm that is a complete metric space with respect to the distance function induced by the norm.
Any inner product space is a normed vector space. (The norm of a vector is the square root of the inner product of the vector with itself.)
Any Hilbert space is a Banach space.
answered 3 hours ago
J. W. TannerJ. W. Tanner
5,2101520
$endgroup$
A Hilbert space is a vector space with an inner product that is a complete metric space with respect to the distance function induced by the inner product.
A Banach space is a vector space with a norm that is a complete metric space with respect to the distance function induced by the norm.
Any inner product space is a normed vector space. (The norm of a vector is the square root of the inner product of the vector with itself.)
Any Hilbert space is a Banach space.
answered 3 hours ago
J. W. TannerJ. W. Tanner
5,2101520
answered 3 hours ago
J. W. TannerJ. W. Tanner
5,2101520
answered 3 hours ago
J. W. TannerJ. W. Tanner
5,2101520
answered 3 hours ago
J. W. TannerJ. W. Tanner
5,2101520
5,2101520
$begingroup$
Cf. this diagram
$endgroup$
– J. W. Tanner
3 hours ago
add a comment |
$begingroup$
Cf. this diagram
$endgroup$
– J. W. Tanner
3 hours ago
$begingroup$
Cf. this diagram
$endgroup$
– J. W. Tanner
3 hours ago
$begingroup$
Cf. this diagram
$endgroup$
– J. W. Tanner
3 hours ago
add a comment |
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2
$begingroup$
Have you checked the definitions?
$endgroup$
– avs
3 hours ago
$begingroup$
All that I know is that a Banach space is a complete normed vectors and a Hilbert space is the inner products of such these normed vectors. Am I right?
$endgroup$
– Lostinspace
3 hours ago
$begingroup$
Every Hilbert space is Banach.
$endgroup$
– Berci
3 hours ago
$begingroup$
The first is true by definition of a Hilbert space
$endgroup$
– Fakemistake
3 hours ago