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If $E$ is an infinite vector space, how to interprete $sum_ix_ie_i$?
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Let $E$ a vector space of infinite dimension and let $(e_i)_imathbb N$ a basis. How is interpreted $$sum_iinmathbb Nx_ie_i ?$$
Suppose now that $E$ is a Banach space with norm $|cdot |$. I suppose that $$sum_iinmathbb Nx_ie_i=lim_nto infty sum_i=1^n x_ie_i,$$
in the $|cdot |$ sense, i.e. $$forall varepsilon>0, exists N: forall nin mathbb N, ngeq Nimplies left|sum_iinmathbb Nx_ie_i-sum_i=1^nx_ie_iright|<varepsilon.$$
But since $E$ is a vector space, $sum_iinmathbb Nx_ie_iin E$ and thus exist (by definition of a vector space), but I guess it could happen that $$lim_nto infty sum_i=1^n x_ie_i$$ doesn't exist or is infinite, does it ? So how can we manage this case ?
linear-algebra functional-analysis vector-spaces summation
edited Jul 25 at 10:34
Sou
2,7062820
asked Jul 25 at 10:26
user330587
821310
add a comment |Â
up vote
6
down vote
favorite
Let $E$ a vector space of infinite dimension and let $(e_i)_imathbb N$ a basis. How is interpreted $$sum_iinmathbb Nx_ie_i ?$$
Suppose now that $E$ is a Banach space with norm $|cdot |$. I suppose that $$sum_iinmathbb Nx_ie_i=lim_nto infty sum_i=1^n x_ie_i,$$
in the $|cdot |$ sense, i.e. $$forall varepsilon>0, exists N: forall nin mathbb N, ngeq Nimplies left|sum_iinmathbb Nx_ie_i-sum_i=1^nx_ie_iright|<varepsilon.$$
But since $E$ is a vector space, $sum_iinmathbb Nx_ie_iin E$ and thus exist (by definition of a vector space), but I guess it could happen that $$lim_nto infty sum_i=1^n x_ie_i$$ doesn't exist or is infinite, does it ? So how can we manage this case ?
linear-algebra functional-analysis vector-spaces summation
edited Jul 25 at 10:34
Sou
2,7062820
asked Jul 25 at 10:26
user330587
821310
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Let $E$ a vector space of infinite dimension and let $(e_i)_imathbb N$ a basis. How is interpreted $$sum_iinmathbb Nx_ie_i ?$$
Suppose now that $E$ is a Banach space with norm $|cdot |$. I suppose that $$sum_iinmathbb Nx_ie_i=lim_nto infty sum_i=1^n x_ie_i,$$
in the $|cdot |$ sense, i.e. $$forall varepsilon>0, exists N: forall nin mathbb N, ngeq Nimplies left|sum_iinmathbb Nx_ie_i-sum_i=1^nx_ie_iright|<varepsilon.$$
But since $E$ is a vector space, $sum_iinmathbb Nx_ie_iin E$ and thus exist (by definition of a vector space), but I guess it could happen that $$lim_nto infty sum_i=1^n x_ie_i$$ doesn't exist or is infinite, does it ? So how can we manage this case ?
linear-algebra functional-analysis vector-spaces summation
edited Jul 25 at 10:34
Sou
2,7062820
asked Jul 25 at 10:26
user330587
821310
Let $E$ a vector space of infinite dimension and let $(e_i)_imathbb N$ a basis. How is interpreted $$sum_iinmathbb Nx_ie_i ?$$
Suppose now that $E$ is a Banach space with norm $|cdot |$. I suppose that $$sum_iinmathbb Nx_ie_i=lim_nto infty sum_i=1^n x_ie_i,$$
in the $|cdot |$ sense, i.e. $$forall varepsilon>0, exists N: forall nin mathbb N, ngeq Nimplies left|sum_iinmathbb Nx_ie_i-sum_i=1^nx_ie_iright|<varepsilon.$$
But since $E$ is a vector space, $sum_iinmathbb Nx_ie_iin E$ and thus exist (by definition of a vector space), but I guess it could happen that $$lim_nto infty sum_i=1^n x_ie_i$$ doesn't exist or is infinite, does it ? So how can we manage this case ?
linear-algebra functional-analysis vector-spaces summation
edited Jul 25 at 10:34
Sou
2,7062820
asked Jul 25 at 10:26
user330587
821310
edited Jul 25 at 10:34
Sou
2,7062820
edited Jul 25 at 10:34
Sou
2,7062820
edited Jul 25 at 10:34
Sou
2,7062820
2,7062820
asked Jul 25 at 10:26
user330587
821310
asked Jul 25 at 10:26
user330587
821310
asked Jul 25 at 10:26
user330587
821310
821310
add a comment |Â
add a comment |Â
4 Answers
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oldest
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up vote
1
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Answer for the case of normed linear spaces: if you are defining 'basis' the way you do in finite dimensional spaces (the so called Hamel basis) then you can form inifinite sums (in the case of normed linear space) only when you know that the series converges. There are examples where every element has unique expansion as a convergent inifinite sum in terms of a sequence $e_n$. For infinite dimensional spaces there are various types of bases: orthogonal bases, Schauder bases etc. In the case where the vector space has no norm/metric/topology you can only form finite sums.
answered Jul 25 at 10:31
Kavi Rama Murthy
20k2829
"you can form inifinite sums only when you know that the series converges". How do you define convergence of a series in a non metrisable vector space ?
– Surb
Jul 25 at 10:35
@Surb I have now included existence of a norm in the statement. The OP is clearly interested in Banach spaces so I thought of telling him about Schauder bases etc.
– Kavi Rama Murthy
Jul 25 at 10:41
add a comment |Â
up vote
0
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If $E=textSpane_imid iinmathbb N$ has infinite dimension, then $$E=bigoplus_iinmathbb N textSpane_i:=leftsum_i=1^n x_ie_imid x_ineq 0 text for a finite number of x_iright.$$
answered Jul 25 at 10:33
Surb
36.3k84274
add a comment |Â
up vote
0
down vote
For the second case, if $E$ is a Hilbert space and the collection $e_i$ is orthonormal (I.e., $left<e_i, e_jright> = delta_ij$), then $sum x_i e_i$ converges if and only if $sum |x_i|^2$ converges. By the triangle inequality, if the set $ $ is bounded, then the series converges if $sum |x_i|^2$ converges. So I guess the point is that if your coefficients are summable you can make the series converge by normalizing your basis elements.
edited Jul 25 at 10:50
Asaf Karagila
291k31402732
answered Jul 25 at 10:41
mheldman
54616
add a comment |Â
up vote
0
down vote
Another nonequivalent definition is:
Let $(v_n)_ninmathbbN$ be a sequence of vectors in a normed space $X$. We say that $sum_ninmathbbN v_n = v$ for some $v in X$ if for every $varepsilon > 0$ there exists a finite subset $F_0 subseteq mathbbN$ such that for every finite subset $F subseteq mathbbN$ with $F supseteq F_0$ we have
$$left|v - sum_nin Fv_nright| < varepsilon$$
It turns out that $sum_ninmathbbN v_n = v$ is equivalent to the statement that $$sum_n=1^infty v_sigma(n) = v$$
for all permutations $sigma : mathbbN to mathbbN$.
In this case we say that $sum_ninmathbbN v_n$ converges unconditionally to $v$.
answered Jul 25 at 11:44
mechanodroid
22.2k52041
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Answer for the case of normed linear spaces: if you are defining 'basis' the way you do in finite dimensional spaces (the so called Hamel basis) then you can form inifinite sums (in the case of normed linear space) only when you know that the series converges. There are examples where every element has unique expansion as a convergent inifinite sum in terms of a sequence $e_n$. For infinite dimensional spaces there are various types of bases: orthogonal bases, Schauder bases etc. In the case where the vector space has no norm/metric/topology you can only form finite sums.
answered Jul 25 at 10:31
Kavi Rama Murthy
20k2829
"you can form inifinite sums only when you know that the series converges". How do you define convergence of a series in a non metrisable vector space ?
– Surb
Jul 25 at 10:35
@Surb I have now included existence of a norm in the statement. The OP is clearly interested in Banach spaces so I thought of telling him about Schauder bases etc.
– Kavi Rama Murthy
Jul 25 at 10:41
add a comment |Â
up vote
1
down vote
Answer for the case of normed linear spaces: if you are defining 'basis' the way you do in finite dimensional spaces (the so called Hamel basis) then you can form inifinite sums (in the case of normed linear space) only when you know that the series converges. There are examples where every element has unique expansion as a convergent inifinite sum in terms of a sequence $e_n$. For infinite dimensional spaces there are various types of bases: orthogonal bases, Schauder bases etc. In the case where the vector space has no norm/metric/topology you can only form finite sums.
answered Jul 25 at 10:31
Kavi Rama Murthy
20k2829
"you can form inifinite sums only when you know that the series converges". How do you define convergence of a series in a non metrisable vector space ?
– Surb
Jul 25 at 10:35
@Surb I have now included existence of a norm in the statement. The OP is clearly interested in Banach spaces so I thought of telling him about Schauder bases etc.
– Kavi Rama Murthy
Jul 25 at 10:41
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Answer for the case of normed linear spaces: if you are defining 'basis' the way you do in finite dimensional spaces (the so called Hamel basis) then you can form inifinite sums (in the case of normed linear space) only when you know that the series converges. There are examples where every element has unique expansion as a convergent inifinite sum in terms of a sequence $e_n$. For infinite dimensional spaces there are various types of bases: orthogonal bases, Schauder bases etc. In the case where the vector space has no norm/metric/topology you can only form finite sums.
answered Jul 25 at 10:31
Kavi Rama Murthy
20k2829
Answer for the case of normed linear spaces: if you are defining 'basis' the way you do in finite dimensional spaces (the so called Hamel basis) then you can form inifinite sums (in the case of normed linear space) only when you know that the series converges. There are examples where every element has unique expansion as a convergent inifinite sum in terms of a sequence $e_n$. For infinite dimensional spaces there are various types of bases: orthogonal bases, Schauder bases etc. In the case where the vector space has no norm/metric/topology you can only form finite sums.
answered Jul 25 at 10:31
Kavi Rama Murthy
20k2829
edited Jul 25 at 10:39
answered Jul 25 at 10:31
Kavi Rama Murthy
20k2829
answered Jul 25 at 10:31
Kavi Rama Murthy
20k2829
answered Jul 25 at 10:31
Kavi Rama Murthy
20k2829
20k2829
"you can form inifinite sums only when you know that the series converges". How do you define convergence of a series in a non metrisable vector space ?
– Surb
Jul 25 at 10:35
@Surb I have now included existence of a norm in the statement. The OP is clearly interested in Banach spaces so I thought of telling him about Schauder bases etc.
– Kavi Rama Murthy
Jul 25 at 10:41
add a comment |Â
"you can form inifinite sums only when you know that the series converges". How do you define convergence of a series in a non metrisable vector space ?
– Surb
Jul 25 at 10:35
@Surb I have now included existence of a norm in the statement. The OP is clearly interested in Banach spaces so I thought of telling him about Schauder bases etc.
– Kavi Rama Murthy
Jul 25 at 10:41
"you can form inifinite sums only when you know that the series converges". How do you define convergence of a series in a non metrisable vector space ?
– Surb
Jul 25 at 10:35
"you can form inifinite sums only when you know that the series converges". How do you define convergence of a series in a non metrisable vector space ?
– Surb
Jul 25 at 10:35
@Surb I have now included existence of a norm in the statement. The OP is clearly interested in Banach spaces so I thought of telling him about Schauder bases etc.
– Kavi Rama Murthy
Jul 25 at 10:41
@Surb I have now included existence of a norm in the statement. The OP is clearly interested in Banach spaces so I thought of telling him about Schauder bases etc.
– Kavi Rama Murthy
Jul 25 at 10:41
add a comment |Â
up vote
0
down vote
If $E=textSpane_imid iinmathbb N$ has infinite dimension, then $$E=bigoplus_iinmathbb N textSpane_i:=leftsum_i=1^n x_ie_imid x_ineq 0 text for a finite number of x_iright.$$
answered Jul 25 at 10:33
Surb
36.3k84274
add a comment |Â
up vote
0
down vote
If $E=textSpane_imid iinmathbb N$ has infinite dimension, then $$E=bigoplus_iinmathbb N textSpane_i:=leftsum_i=1^n x_ie_imid x_ineq 0 text for a finite number of x_iright.$$
answered Jul 25 at 10:33
Surb
36.3k84274
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If $E=textSpane_imid iinmathbb N$ has infinite dimension, then $$E=bigoplus_iinmathbb N textSpane_i:=leftsum_i=1^n x_ie_imid x_ineq 0 text for a finite number of x_iright.$$
answered Jul 25 at 10:33
Surb
36.3k84274
If $E=textSpane_imid iinmathbb N$ has infinite dimension, then $$E=bigoplus_iinmathbb N textSpane_i:=leftsum_i=1^n x_ie_imid x_ineq 0 text for a finite number of x_iright.$$
answered Jul 25 at 10:33
Surb
36.3k84274
answered Jul 25 at 10:33
Surb
36.3k84274
answered Jul 25 at 10:33
Surb
36.3k84274
answered Jul 25 at 10:33
Surb
36.3k84274
36.3k84274
add a comment |Â
add a comment |Â
up vote
0
down vote
For the second case, if $E$ is a Hilbert space and the collection $e_i$ is orthonormal (I.e., $left<e_i, e_jright> = delta_ij$), then $sum x_i e_i$ converges if and only if $sum |x_i|^2$ converges. By the triangle inequality, if the set $ $ is bounded, then the series converges if $sum |x_i|^2$ converges. So I guess the point is that if your coefficients are summable you can make the series converge by normalizing your basis elements.
edited Jul 25 at 10:50
Asaf Karagila
291k31402732
answered Jul 25 at 10:41
mheldman
54616
add a comment |Â
up vote
0
down vote
For the second case, if $E$ is a Hilbert space and the collection $e_i$ is orthonormal (I.e., $left<e_i, e_jright> = delta_ij$), then $sum x_i e_i$ converges if and only if $sum |x_i|^2$ converges. By the triangle inequality, if the set $ $ is bounded, then the series converges if $sum |x_i|^2$ converges. So I guess the point is that if your coefficients are summable you can make the series converge by normalizing your basis elements.
edited Jul 25 at 10:50
Asaf Karagila
291k31402732
answered Jul 25 at 10:41
mheldman
54616
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For the second case, if $E$ is a Hilbert space and the collection $e_i$ is orthonormal (I.e., $left<e_i, e_jright> = delta_ij$), then $sum x_i e_i$ converges if and only if $sum |x_i|^2$ converges. By the triangle inequality, if the set $ $ is bounded, then the series converges if $sum |x_i|^2$ converges. So I guess the point is that if your coefficients are summable you can make the series converge by normalizing your basis elements.
edited Jul 25 at 10:50
Asaf Karagila
291k31402732
answered Jul 25 at 10:41
mheldman
54616
For the second case, if $E$ is a Hilbert space and the collection $e_i$ is orthonormal (I.e., $left<e_i, e_jright> = delta_ij$), then $sum x_i e_i$ converges if and only if $sum |x_i|^2$ converges. By the triangle inequality, if the set $ $ is bounded, then the series converges if $sum |x_i|^2$ converges. So I guess the point is that if your coefficients are summable you can make the series converge by normalizing your basis elements.
edited Jul 25 at 10:50
Asaf Karagila
291k31402732
answered Jul 25 at 10:41
mheldman
54616
edited Jul 25 at 10:50
Asaf Karagila
291k31402732
edited Jul 25 at 10:50
Asaf Karagila
291k31402732
edited Jul 25 at 10:50
Asaf Karagila
291k31402732
291k31402732
answered Jul 25 at 10:41
mheldman
54616
answered Jul 25 at 10:41
mheldman
54616
answered Jul 25 at 10:41
mheldman
54616
54616
add a comment |Â
add a comment |Â
up vote
0
down vote
Another nonequivalent definition is:
Let $(v_n)_ninmathbbN$ be a sequence of vectors in a normed space $X$. We say that $sum_ninmathbbN v_n = v$ for some $v in X$ if for every $varepsilon > 0$ there exists a finite subset $F_0 subseteq mathbbN$ such that for every finite subset $F subseteq mathbbN$ with $F supseteq F_0$ we have
$$left|v - sum_nin Fv_nright| < varepsilon$$
It turns out that $sum_ninmathbbN v_n = v$ is equivalent to the statement that $$sum_n=1^infty v_sigma(n) = v$$
for all permutations $sigma : mathbbN to mathbbN$.
In this case we say that $sum_ninmathbbN v_n$ converges unconditionally to $v$.
answered Jul 25 at 11:44
mechanodroid
22.2k52041
add a comment |Â
up vote
0
down vote
Another nonequivalent definition is:
Let $(v_n)_ninmathbbN$ be a sequence of vectors in a normed space $X$. We say that $sum_ninmathbbN v_n = v$ for some $v in X$ if for every $varepsilon > 0$ there exists a finite subset $F_0 subseteq mathbbN$ such that for every finite subset $F subseteq mathbbN$ with $F supseteq F_0$ we have
$$left|v - sum_nin Fv_nright| < varepsilon$$
It turns out that $sum_ninmathbbN v_n = v$ is equivalent to the statement that $$sum_n=1^infty v_sigma(n) = v$$
for all permutations $sigma : mathbbN to mathbbN$.
In this case we say that $sum_ninmathbbN v_n$ converges unconditionally to $v$.
answered Jul 25 at 11:44
mechanodroid
22.2k52041
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Another nonequivalent definition is:
Let $(v_n)_ninmathbbN$ be a sequence of vectors in a normed space $X$. We say that $sum_ninmathbbN v_n = v$ for some $v in X$ if for every $varepsilon > 0$ there exists a finite subset $F_0 subseteq mathbbN$ such that for every finite subset $F subseteq mathbbN$ with $F supseteq F_0$ we have
$$left|v - sum_nin Fv_nright| < varepsilon$$
It turns out that $sum_ninmathbbN v_n = v$ is equivalent to the statement that $$sum_n=1^infty v_sigma(n) = v$$
for all permutations $sigma : mathbbN to mathbbN$.
In this case we say that $sum_ninmathbbN v_n$ converges unconditionally to $v$.
answered Jul 25 at 11:44
mechanodroid
22.2k52041
Another nonequivalent definition is:
Let $(v_n)_ninmathbbN$ be a sequence of vectors in a normed space $X$. We say that $sum_ninmathbbN v_n = v$ for some $v in X$ if for every $varepsilon > 0$ there exists a finite subset $F_0 subseteq mathbbN$ such that for every finite subset $F subseteq mathbbN$ with $F supseteq F_0$ we have
$$left|v - sum_nin Fv_nright| < varepsilon$$
It turns out that $sum_ninmathbbN v_n = v$ is equivalent to the statement that $$sum_n=1^infty v_sigma(n) = v$$
for all permutations $sigma : mathbbN to mathbbN$.
In this case we say that $sum_ninmathbbN v_n$ converges unconditionally to $v$.
answered Jul 25 at 11:44
mechanodroid
22.2k52041
answered Jul 25 at 11:44
mechanodroid
22.2k52041
answered Jul 25 at 11:44
mechanodroid
22.2k52041
answered Jul 25 at 11:44
mechanodroid
22.2k52041
22.2k52041
add a comment |Â
add a comment |Â
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