Check if map maps to Hilbert space
I am supposed to find a condition s.th. $A(x)=sum_{n=1}^inftyalpha_nlangle x,e_nrangle e_n$ maps from E to E. Where E is an Hilbert space with a complete ON-sequence $(e_n)^infty_{n=0}$, $(alpha_n)_{n=1}^inftyinmathbb{C}$.
What are the requirements for an element to belong to a general Hilbert space?
Can a Hilbert space contain an element with infinite norm?
And if the series does not converge, does that mean that it is undefined and hence cant be an element?
Would it be a problem if the series does not converge, and if so why?
functional-analysis hilbert-spaces
add a comment |
I am supposed to find a condition s.th. $A(x)=sum_{n=1}^inftyalpha_nlangle x,e_nrangle e_n$ maps from E to E. Where E is an Hilbert space with a complete ON-sequence $(e_n)^infty_{n=0}$, $(alpha_n)_{n=1}^inftyinmathbb{C}$.
What are the requirements for an element to belong to a general Hilbert space?
Can a Hilbert space contain an element with infinite norm?
And if the series does not converge, does that mean that it is undefined and hence cant be an element?
Would it be a problem if the series does not converge, and if so why?
functional-analysis hilbert-spaces
The exact condition is that $sum_n|alpha_n|^2|langle x,e_nrangle|^2 < infty$ for all $xin E$. A sufficient (and, I think, also necessary) condition is that $(alpha_n)_ninell^infty$.
– amsmath
Oct 16 '17 at 2:24
Thanks, yes saw a theorem that states that $sum_{n=1}^inftyalpha_ne_n<infty$, $(e_n)$ ON-sequence, if and only if $sum_{n=1}^infty|alpha_n|^2<infty$. But why does boundedness of the sum ensure that it is in E? Can't E contain unbounded elements?
– JoeDoe
Oct 16 '17 at 2:36
Is it becuase E is a vector space? And addition and multiplication becomes vaguely defined for infinite elements? But can't there be infinities of different orders, making the operations defined?
– JoeDoe
Oct 16 '17 at 2:39
How does infinite elements interfere with the definiton of hilbert spaces?
– JoeDoe
Oct 16 '17 at 2:40
If $E$ is a Hilbert space, then for any $fin E$ you have $|f| < infty$. That follows immediately from the definition of a norm.
– amsmath
Oct 16 '17 at 2:41
add a comment |
I am supposed to find a condition s.th. $A(x)=sum_{n=1}^inftyalpha_nlangle x,e_nrangle e_n$ maps from E to E. Where E is an Hilbert space with a complete ON-sequence $(e_n)^infty_{n=0}$, $(alpha_n)_{n=1}^inftyinmathbb{C}$.
What are the requirements for an element to belong to a general Hilbert space?
Can a Hilbert space contain an element with infinite norm?
And if the series does not converge, does that mean that it is undefined and hence cant be an element?
Would it be a problem if the series does not converge, and if so why?
functional-analysis hilbert-spaces
I am supposed to find a condition s.th. $A(x)=sum_{n=1}^inftyalpha_nlangle x,e_nrangle e_n$ maps from E to E. Where E is an Hilbert space with a complete ON-sequence $(e_n)^infty_{n=0}$, $(alpha_n)_{n=1}^inftyinmathbb{C}$.
What are the requirements for an element to belong to a general Hilbert space?
Can a Hilbert space contain an element with infinite norm?
And if the series does not converge, does that mean that it is undefined and hence cant be an element?
Would it be a problem if the series does not converge, and if so why?
functional-analysis hilbert-spaces
functional-analysis hilbert-spaces
edited Nov 29 at 0:16
asked Oct 16 '17 at 2:19
JoeDoe
103
103
The exact condition is that $sum_n|alpha_n|^2|langle x,e_nrangle|^2 < infty$ for all $xin E$. A sufficient (and, I think, also necessary) condition is that $(alpha_n)_ninell^infty$.
– amsmath
Oct 16 '17 at 2:24
Thanks, yes saw a theorem that states that $sum_{n=1}^inftyalpha_ne_n<infty$, $(e_n)$ ON-sequence, if and only if $sum_{n=1}^infty|alpha_n|^2<infty$. But why does boundedness of the sum ensure that it is in E? Can't E contain unbounded elements?
– JoeDoe
Oct 16 '17 at 2:36
Is it becuase E is a vector space? And addition and multiplication becomes vaguely defined for infinite elements? But can't there be infinities of different orders, making the operations defined?
– JoeDoe
Oct 16 '17 at 2:39
How does infinite elements interfere with the definiton of hilbert spaces?
– JoeDoe
Oct 16 '17 at 2:40
If $E$ is a Hilbert space, then for any $fin E$ you have $|f| < infty$. That follows immediately from the definition of a norm.
– amsmath
Oct 16 '17 at 2:41
add a comment |
The exact condition is that $sum_n|alpha_n|^2|langle x,e_nrangle|^2 < infty$ for all $xin E$. A sufficient (and, I think, also necessary) condition is that $(alpha_n)_ninell^infty$.
– amsmath
Oct 16 '17 at 2:24
Thanks, yes saw a theorem that states that $sum_{n=1}^inftyalpha_ne_n<infty$, $(e_n)$ ON-sequence, if and only if $sum_{n=1}^infty|alpha_n|^2<infty$. But why does boundedness of the sum ensure that it is in E? Can't E contain unbounded elements?
– JoeDoe
Oct 16 '17 at 2:36
Is it becuase E is a vector space? And addition and multiplication becomes vaguely defined for infinite elements? But can't there be infinities of different orders, making the operations defined?
– JoeDoe
Oct 16 '17 at 2:39
How does infinite elements interfere with the definiton of hilbert spaces?
– JoeDoe
Oct 16 '17 at 2:40
If $E$ is a Hilbert space, then for any $fin E$ you have $|f| < infty$. That follows immediately from the definition of a norm.
– amsmath
Oct 16 '17 at 2:41
The exact condition is that $sum_n|alpha_n|^2|langle x,e_nrangle|^2 < infty$ for all $xin E$. A sufficient (and, I think, also necessary) condition is that $(alpha_n)_ninell^infty$.
– amsmath
Oct 16 '17 at 2:24
The exact condition is that $sum_n|alpha_n|^2|langle x,e_nrangle|^2 < infty$ for all $xin E$. A sufficient (and, I think, also necessary) condition is that $(alpha_n)_ninell^infty$.
– amsmath
Oct 16 '17 at 2:24
Thanks, yes saw a theorem that states that $sum_{n=1}^inftyalpha_ne_n<infty$, $(e_n)$ ON-sequence, if and only if $sum_{n=1}^infty|alpha_n|^2<infty$. But why does boundedness of the sum ensure that it is in E? Can't E contain unbounded elements?
– JoeDoe
Oct 16 '17 at 2:36
Thanks, yes saw a theorem that states that $sum_{n=1}^inftyalpha_ne_n<infty$, $(e_n)$ ON-sequence, if and only if $sum_{n=1}^infty|alpha_n|^2<infty$. But why does boundedness of the sum ensure that it is in E? Can't E contain unbounded elements?
– JoeDoe
Oct 16 '17 at 2:36
Is it becuase E is a vector space? And addition and multiplication becomes vaguely defined for infinite elements? But can't there be infinities of different orders, making the operations defined?
– JoeDoe
Oct 16 '17 at 2:39
Is it becuase E is a vector space? And addition and multiplication becomes vaguely defined for infinite elements? But can't there be infinities of different orders, making the operations defined?
– JoeDoe
Oct 16 '17 at 2:39
How does infinite elements interfere with the definiton of hilbert spaces?
– JoeDoe
Oct 16 '17 at 2:40
How does infinite elements interfere with the definiton of hilbert spaces?
– JoeDoe
Oct 16 '17 at 2:40
If $E$ is a Hilbert space, then for any $fin E$ you have $|f| < infty$. That follows immediately from the definition of a norm.
– amsmath
Oct 16 '17 at 2:41
If $E$ is a Hilbert space, then for any $fin E$ you have $|f| < infty$. That follows immediately from the definition of a norm.
– amsmath
Oct 16 '17 at 2:41
add a comment |
2 Answers
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oldest
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By a general theorem in Hilbert space, your given E is isometrically isomorphic to $l^2 (mathbb{C})$, so that the first question: "What are the requirements for an element to belong to a general Hilbert space?" with the obvious answer: "An element belongs to a Hilbert space (or more generally to a Banach space) iff its norm is finite" has the tailored answer provided by @amsmath in the comments section:
[quote] The exact condition is that:
$$sum_n |alpha_n|^2 |langle x, e_nrangle|^2 <infty, forall xin E (1) $$ [/quote] and applies to the coefficients in the given basis expansion.
As for the other questions: "Can a Hilbert space contain an element with infinite norm?", the answer is "NO"; "And if the series does not converge, does that mean that it is undefined and hence cant be an element?", the answer is: "YES, for a series expansion of a vector with respect to an infinite basis, the convergence of this series in the norm is required by the finite-norm condition of the expanded vector and places boundedness conditions on the coefficients of this expansion, such as (1)".
Thank you, really cleared stuff up!
– JoeDoe
Oct 16 '17 at 9:50
add a comment |
As the other comments and answers clearly stat, the series only makes sense in the framework of Hilbert space if the coefficients are square-integrable. However, there are cases in which it is necessary to consider "generalized vectors", with infinite norm. The concept of rigged Hilbert spaces has been developed to deal with this kind of objects, especially in quantum mechanics and spectral theory.
The basic example is the plane wave function
$$e_xi(x)=exp(ixxi),qquad xi, xin mathbb R,$$
in the Hilbert space $L^2(mathbb R)$. It holds that $|e_xi|_{L^2(mathbb R)}=infty$, but $e_xi$ is a valid element of the rigging $(L^2(mathbb R), C^infty_c(mathbb R)$ ($C^infty_c$ denotes smooth functions with compact support), because the pairing
$$
int_{mathbb R} e_{xi}(x) phi(x), dx
$$
makes sense for all $phiin C^infty_c(mathbb R)$.
The rigging of a Hilbert space also introduces a weaker mode of convergence, so in this sense one may also have convergent orthonormal series and integrals even without square integrable coefficients. In the above example, this corresponds to convergence in the sense of distributions.
Thank you for the help!
– JoeDoe
Oct 16 '17 at 9:52
add a comment |
Your Answer
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2 Answers
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oldest
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2 Answers
2
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oldest
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By a general theorem in Hilbert space, your given E is isometrically isomorphic to $l^2 (mathbb{C})$, so that the first question: "What are the requirements for an element to belong to a general Hilbert space?" with the obvious answer: "An element belongs to a Hilbert space (or more generally to a Banach space) iff its norm is finite" has the tailored answer provided by @amsmath in the comments section:
[quote] The exact condition is that:
$$sum_n |alpha_n|^2 |langle x, e_nrangle|^2 <infty, forall xin E (1) $$ [/quote] and applies to the coefficients in the given basis expansion.
As for the other questions: "Can a Hilbert space contain an element with infinite norm?", the answer is "NO"; "And if the series does not converge, does that mean that it is undefined and hence cant be an element?", the answer is: "YES, for a series expansion of a vector with respect to an infinite basis, the convergence of this series in the norm is required by the finite-norm condition of the expanded vector and places boundedness conditions on the coefficients of this expansion, such as (1)".
Thank you, really cleared stuff up!
– JoeDoe
Oct 16 '17 at 9:50
add a comment |
By a general theorem in Hilbert space, your given E is isometrically isomorphic to $l^2 (mathbb{C})$, so that the first question: "What are the requirements for an element to belong to a general Hilbert space?" with the obvious answer: "An element belongs to a Hilbert space (or more generally to a Banach space) iff its norm is finite" has the tailored answer provided by @amsmath in the comments section:
[quote] The exact condition is that:
$$sum_n |alpha_n|^2 |langle x, e_nrangle|^2 <infty, forall xin E (1) $$ [/quote] and applies to the coefficients in the given basis expansion.
As for the other questions: "Can a Hilbert space contain an element with infinite norm?", the answer is "NO"; "And if the series does not converge, does that mean that it is undefined and hence cant be an element?", the answer is: "YES, for a series expansion of a vector with respect to an infinite basis, the convergence of this series in the norm is required by the finite-norm condition of the expanded vector and places boundedness conditions on the coefficients of this expansion, such as (1)".
Thank you, really cleared stuff up!
– JoeDoe
Oct 16 '17 at 9:50
add a comment |
By a general theorem in Hilbert space, your given E is isometrically isomorphic to $l^2 (mathbb{C})$, so that the first question: "What are the requirements for an element to belong to a general Hilbert space?" with the obvious answer: "An element belongs to a Hilbert space (or more generally to a Banach space) iff its norm is finite" has the tailored answer provided by @amsmath in the comments section:
[quote] The exact condition is that:
$$sum_n |alpha_n|^2 |langle x, e_nrangle|^2 <infty, forall xin E (1) $$ [/quote] and applies to the coefficients in the given basis expansion.
As for the other questions: "Can a Hilbert space contain an element with infinite norm?", the answer is "NO"; "And if the series does not converge, does that mean that it is undefined and hence cant be an element?", the answer is: "YES, for a series expansion of a vector with respect to an infinite basis, the convergence of this series in the norm is required by the finite-norm condition of the expanded vector and places boundedness conditions on the coefficients of this expansion, such as (1)".
By a general theorem in Hilbert space, your given E is isometrically isomorphic to $l^2 (mathbb{C})$, so that the first question: "What are the requirements for an element to belong to a general Hilbert space?" with the obvious answer: "An element belongs to a Hilbert space (or more generally to a Banach space) iff its norm is finite" has the tailored answer provided by @amsmath in the comments section:
[quote] The exact condition is that:
$$sum_n |alpha_n|^2 |langle x, e_nrangle|^2 <infty, forall xin E (1) $$ [/quote] and applies to the coefficients in the given basis expansion.
As for the other questions: "Can a Hilbert space contain an element with infinite norm?", the answer is "NO"; "And if the series does not converge, does that mean that it is undefined and hence cant be an element?", the answer is: "YES, for a series expansion of a vector with respect to an infinite basis, the convergence of this series in the norm is required by the finite-norm condition of the expanded vector and places boundedness conditions on the coefficients of this expansion, such as (1)".
answered Oct 16 '17 at 8:00
DanielC
429316
429316
Thank you, really cleared stuff up!
– JoeDoe
Oct 16 '17 at 9:50
add a comment |
Thank you, really cleared stuff up!
– JoeDoe
Oct 16 '17 at 9:50
Thank you, really cleared stuff up!
– JoeDoe
Oct 16 '17 at 9:50
Thank you, really cleared stuff up!
– JoeDoe
Oct 16 '17 at 9:50
add a comment |
As the other comments and answers clearly stat, the series only makes sense in the framework of Hilbert space if the coefficients are square-integrable. However, there are cases in which it is necessary to consider "generalized vectors", with infinite norm. The concept of rigged Hilbert spaces has been developed to deal with this kind of objects, especially in quantum mechanics and spectral theory.
The basic example is the plane wave function
$$e_xi(x)=exp(ixxi),qquad xi, xin mathbb R,$$
in the Hilbert space $L^2(mathbb R)$. It holds that $|e_xi|_{L^2(mathbb R)}=infty$, but $e_xi$ is a valid element of the rigging $(L^2(mathbb R), C^infty_c(mathbb R)$ ($C^infty_c$ denotes smooth functions with compact support), because the pairing
$$
int_{mathbb R} e_{xi}(x) phi(x), dx
$$
makes sense for all $phiin C^infty_c(mathbb R)$.
The rigging of a Hilbert space also introduces a weaker mode of convergence, so in this sense one may also have convergent orthonormal series and integrals even without square integrable coefficients. In the above example, this corresponds to convergence in the sense of distributions.
Thank you for the help!
– JoeDoe
Oct 16 '17 at 9:52
add a comment |
As the other comments and answers clearly stat, the series only makes sense in the framework of Hilbert space if the coefficients are square-integrable. However, there are cases in which it is necessary to consider "generalized vectors", with infinite norm. The concept of rigged Hilbert spaces has been developed to deal with this kind of objects, especially in quantum mechanics and spectral theory.
The basic example is the plane wave function
$$e_xi(x)=exp(ixxi),qquad xi, xin mathbb R,$$
in the Hilbert space $L^2(mathbb R)$. It holds that $|e_xi|_{L^2(mathbb R)}=infty$, but $e_xi$ is a valid element of the rigging $(L^2(mathbb R), C^infty_c(mathbb R)$ ($C^infty_c$ denotes smooth functions with compact support), because the pairing
$$
int_{mathbb R} e_{xi}(x) phi(x), dx
$$
makes sense for all $phiin C^infty_c(mathbb R)$.
The rigging of a Hilbert space also introduces a weaker mode of convergence, so in this sense one may also have convergent orthonormal series and integrals even without square integrable coefficients. In the above example, this corresponds to convergence in the sense of distributions.
Thank you for the help!
– JoeDoe
Oct 16 '17 at 9:52
add a comment |
As the other comments and answers clearly stat, the series only makes sense in the framework of Hilbert space if the coefficients are square-integrable. However, there are cases in which it is necessary to consider "generalized vectors", with infinite norm. The concept of rigged Hilbert spaces has been developed to deal with this kind of objects, especially in quantum mechanics and spectral theory.
The basic example is the plane wave function
$$e_xi(x)=exp(ixxi),qquad xi, xin mathbb R,$$
in the Hilbert space $L^2(mathbb R)$. It holds that $|e_xi|_{L^2(mathbb R)}=infty$, but $e_xi$ is a valid element of the rigging $(L^2(mathbb R), C^infty_c(mathbb R)$ ($C^infty_c$ denotes smooth functions with compact support), because the pairing
$$
int_{mathbb R} e_{xi}(x) phi(x), dx
$$
makes sense for all $phiin C^infty_c(mathbb R)$.
The rigging of a Hilbert space also introduces a weaker mode of convergence, so in this sense one may also have convergent orthonormal series and integrals even without square integrable coefficients. In the above example, this corresponds to convergence in the sense of distributions.
As the other comments and answers clearly stat, the series only makes sense in the framework of Hilbert space if the coefficients are square-integrable. However, there are cases in which it is necessary to consider "generalized vectors", with infinite norm. The concept of rigged Hilbert spaces has been developed to deal with this kind of objects, especially in quantum mechanics and spectral theory.
The basic example is the plane wave function
$$e_xi(x)=exp(ixxi),qquad xi, xin mathbb R,$$
in the Hilbert space $L^2(mathbb R)$. It holds that $|e_xi|_{L^2(mathbb R)}=infty$, but $e_xi$ is a valid element of the rigging $(L^2(mathbb R), C^infty_c(mathbb R)$ ($C^infty_c$ denotes smooth functions with compact support), because the pairing
$$
int_{mathbb R} e_{xi}(x) phi(x), dx
$$
makes sense for all $phiin C^infty_c(mathbb R)$.
The rigging of a Hilbert space also introduces a weaker mode of convergence, so in this sense one may also have convergent orthonormal series and integrals even without square integrable coefficients. In the above example, this corresponds to convergence in the sense of distributions.
answered Oct 16 '17 at 8:21
Giuseppe Negro
17.3k330122
17.3k330122
Thank you for the help!
– JoeDoe
Oct 16 '17 at 9:52
add a comment |
Thank you for the help!
– JoeDoe
Oct 16 '17 at 9:52
Thank you for the help!
– JoeDoe
Oct 16 '17 at 9:52
Thank you for the help!
– JoeDoe
Oct 16 '17 at 9:52
add a comment |
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The exact condition is that $sum_n|alpha_n|^2|langle x,e_nrangle|^2 < infty$ for all $xin E$. A sufficient (and, I think, also necessary) condition is that $(alpha_n)_ninell^infty$.
– amsmath
Oct 16 '17 at 2:24
Thanks, yes saw a theorem that states that $sum_{n=1}^inftyalpha_ne_n<infty$, $(e_n)$ ON-sequence, if and only if $sum_{n=1}^infty|alpha_n|^2<infty$. But why does boundedness of the sum ensure that it is in E? Can't E contain unbounded elements?
– JoeDoe
Oct 16 '17 at 2:36
Is it becuase E is a vector space? And addition and multiplication becomes vaguely defined for infinite elements? But can't there be infinities of different orders, making the operations defined?
– JoeDoe
Oct 16 '17 at 2:39
How does infinite elements interfere with the definiton of hilbert spaces?
– JoeDoe
Oct 16 '17 at 2:40
If $E$ is a Hilbert space, then for any $fin E$ you have $|f| < infty$. That follows immediately from the definition of a norm.
– amsmath
Oct 16 '17 at 2:41