$L(V,mathbb{R})$ with $|cdot|_1$ complete?
$begingroup$
Let $(V,|cdot|_V)$ and $(W,|cdot|_W)$ be normed vector spaces. By $L(V,W)$ we denote the space of bounded linear operators from $V$ to $W.$ If $W$ is complete, then $L(V,W)$ with the operator norm $|T| := sup left{|Tu|_W : |u|_V leq 1right}$ is a Banach space.
Now assume in addition that $gamma$ is probability measure on $V$ with $mathrm{supp}, gamma = V$ and with existing first moment.
My Question: Is $L(V,mathbb{R})$ closed subspace in $(mathrm{L}_1(V),|cdot|_1)$?
In this case $L(V,mathbb{R})$ with $|cdot|_1$ would also be a Banach space.
general-topology functional-analysis lp-spaces
edited Dec 31 '18 at 14:21
Davide Giraudo
127k17154268
asked Dec 30 '18 at 13:16
H17H17
313
$endgroup$
|
show 8 more comments
$begingroup$
Let $(V,|cdot|_V)$ and $(W,|cdot|_W)$ be normed vector spaces. By $L(V,W)$ we denote the space of bounded linear operators from $V$ to $W.$ If $W$ is complete, then $L(V,W)$ with the operator norm $|T| := sup left{|Tu|_W : |u|_V leq 1right}$ is a Banach space.
Now assume in addition that $gamma$ is probability measure on $V$ with $mathrm{supp}, gamma = V$ and with existing first moment.
My Question: Is $L(V,mathbb{R})$ closed subspace in $(mathrm{L}_1(V),|cdot|_1)$?
In this case $L(V,mathbb{R})$ with $|cdot|_1$ would also be a Banach space.
general-topology functional-analysis lp-spaces
edited Dec 31 '18 at 14:21
Davide Giraudo
127k17154268
asked Dec 30 '18 at 13:16
H17H17
313
$endgroup$
$begingroup$
do you define $Vert f Vert_1:=Vert f Vert + Vert f' Vert$? What is $L_1(V)$?
$endgroup$
– noctusraid
Dec 30 '18 at 13:22
$begingroup$
$mathrm{L}_1(V)$ denotes the usual Lebesgue space that contains all measurable $f : V to mathbb{R}$ such that $$ |f|_1 := int_V |f(v)| , gamma(mathrm{d}v) < infty$$
$endgroup$
– H17
Dec 30 '18 at 13:25
$begingroup$
@H17 The norm on $L(V,mathbb{R})$ is irrelevant, right?
$endgroup$
– mathworker21
Dec 30 '18 at 13:33
$begingroup$
@H17 Also, there is a technicality that is bothering me. $L_1(V)$ consists of equivalence classes of functions whereas $L(V,mathbb{R})$ consists of genuine functions. So, I don't really view $L(V,mathbb{R})$ as a subset of $L_1(V)$.
$endgroup$
– mathworker21
Dec 30 '18 at 13:35
1
$begingroup$
Actually the random variable (construction) is just the identity $x mapsto x.$ Random variables are very useful to work with (describing events etc ...). However, in probability theory the interesting object is actually the probability measure. Random variables are in fact rarely necessary.
$endgroup$
– H17
Dec 30 '18 at 15:30
|
show 8 more comments
$begingroup$
Let $(V,|cdot|_V)$ and $(W,|cdot|_W)$ be normed vector spaces. By $L(V,W)$ we denote the space of bounded linear operators from $V$ to $W.$ If $W$ is complete, then $L(V,W)$ with the operator norm $|T| := sup left{|Tu|_W : |u|_V leq 1right}$ is a Banach space.
Now assume in addition that $gamma$ is probability measure on $V$ with $mathrm{supp}, gamma = V$ and with existing first moment.
My Question: Is $L(V,mathbb{R})$ closed subspace in $(mathrm{L}_1(V),|cdot|_1)$?
In this case $L(V,mathbb{R})$ with $|cdot|_1$ would also be a Banach space.
general-topology functional-analysis lp-spaces
edited Dec 31 '18 at 14:21
Davide Giraudo
127k17154268
asked Dec 30 '18 at 13:16
H17H17
313
$endgroup$
Let $(V,|cdot|_V)$ and $(W,|cdot|_W)$ be normed vector spaces. By $L(V,W)$ we denote the space of bounded linear operators from $V$ to $W.$ If $W$ is complete, then $L(V,W)$ with the operator norm $|T| := sup left{|Tu|_W : |u|_V leq 1right}$ is a Banach space.
Now assume in addition that $gamma$ is probability measure on $V$ with $mathrm{supp}, gamma = V$ and with existing first moment.
My Question: Is $L(V,mathbb{R})$ closed subspace in $(mathrm{L}_1(V),|cdot|_1)$?
In this case $L(V,mathbb{R})$ with $|cdot|_1$ would also be a Banach space.
general-topology functional-analysis lp-spaces
general-topology functional-analysis lp-spaces
edited Dec 31 '18 at 14:21
Davide Giraudo
127k17154268
asked Dec 30 '18 at 13:16
H17H17
313
edited Dec 31 '18 at 14:21
Davide Giraudo
127k17154268
asked Dec 30 '18 at 13:16
H17H17
313
edited Dec 31 '18 at 14:21
Davide Giraudo
127k17154268
edited Dec 31 '18 at 14:21
Davide Giraudo
127k17154268
edited Dec 31 '18 at 14:21
Davide Giraudo
127k17154268
127k17154268
asked Dec 30 '18 at 13:16
H17H17
313
asked Dec 30 '18 at 13:16
H17H17
313
asked Dec 30 '18 at 13:16
H17H17
313
313
$begingroup$
do you define $Vert f Vert_1:=Vert f Vert + Vert f' Vert$? What is $L_1(V)$?
$endgroup$
– noctusraid
Dec 30 '18 at 13:22
$begingroup$
$mathrm{L}_1(V)$ denotes the usual Lebesgue space that contains all measurable $f : V to mathbb{R}$ such that $$ |f|_1 := int_V |f(v)| , gamma(mathrm{d}v) < infty$$
$endgroup$
– H17
Dec 30 '18 at 13:25
$begingroup$
@H17 The norm on $L(V,mathbb{R})$ is irrelevant, right?
$endgroup$
– mathworker21
Dec 30 '18 at 13:33
$begingroup$
@H17 Also, there is a technicality that is bothering me. $L_1(V)$ consists of equivalence classes of functions whereas $L(V,mathbb{R})$ consists of genuine functions. So, I don't really view $L(V,mathbb{R})$ as a subset of $L_1(V)$.
$endgroup$
– mathworker21
Dec 30 '18 at 13:35
1
$begingroup$
Actually the random variable (construction) is just the identity $x mapsto x.$ Random variables are very useful to work with (describing events etc ...). However, in probability theory the interesting object is actually the probability measure. Random variables are in fact rarely necessary.
$endgroup$
– H17
Dec 30 '18 at 15:30
|
show 8 more comments
$begingroup$
do you define $Vert f Vert_1:=Vert f Vert + Vert f' Vert$? What is $L_1(V)$?
$endgroup$
– noctusraid
Dec 30 '18 at 13:22
$begingroup$
$mathrm{L}_1(V)$ denotes the usual Lebesgue space that contains all measurable $f : V to mathbb{R}$ such that $$ |f|_1 := int_V |f(v)| , gamma(mathrm{d}v) < infty$$
$endgroup$
– H17
Dec 30 '18 at 13:25
$begingroup$
@H17 The norm on $L(V,mathbb{R})$ is irrelevant, right?
$endgroup$
– mathworker21
Dec 30 '18 at 13:33
$begingroup$
@H17 Also, there is a technicality that is bothering me. $L_1(V)$ consists of equivalence classes of functions whereas $L(V,mathbb{R})$ consists of genuine functions. So, I don't really view $L(V,mathbb{R})$ as a subset of $L_1(V)$.
$endgroup$
– mathworker21
Dec 30 '18 at 13:35
1
$begingroup$
Actually the random variable (construction) is just the identity $x mapsto x.$ Random variables are very useful to work with (describing events etc ...). However, in probability theory the interesting object is actually the probability measure. Random variables are in fact rarely necessary.
$endgroup$
– H17
Dec 30 '18 at 15:30
$begingroup$
do you define $Vert f Vert_1:=Vert f Vert + Vert f' Vert$? What is $L_1(V)$?
$endgroup$
– noctusraid
Dec 30 '18 at 13:22
$begingroup$
do you define $Vert f Vert_1:=Vert f Vert + Vert f' Vert$? What is $L_1(V)$?
$endgroup$
– noctusraid
Dec 30 '18 at 13:22
$begingroup$
$mathrm{L}_1(V)$ denotes the usual Lebesgue space that contains all measurable $f : V to mathbb{R}$ such that $$ |f|_1 := int_V |f(v)| , gamma(mathrm{d}v) < infty$$
$endgroup$
– H17
Dec 30 '18 at 13:25
$begingroup$
$mathrm{L}_1(V)$ denotes the usual Lebesgue space that contains all measurable $f : V to mathbb{R}$ such that $$ |f|_1 := int_V |f(v)| , gamma(mathrm{d}v) < infty$$
$endgroup$
– H17
Dec 30 '18 at 13:25
$begingroup$
@H17 The norm on $L(V,mathbb{R})$ is irrelevant, right?
$endgroup$
– mathworker21
Dec 30 '18 at 13:33
$begingroup$
@H17 The norm on $L(V,mathbb{R})$ is irrelevant, right?
$endgroup$
– mathworker21
Dec 30 '18 at 13:33
$begingroup$
@H17 Also, there is a technicality that is bothering me. $L_1(V)$ consists of equivalence classes of functions whereas $L(V,mathbb{R})$ consists of genuine functions. So, I don't really view $L(V,mathbb{R})$ as a subset of $L_1(V)$.
$endgroup$
– mathworker21
Dec 30 '18 at 13:35
$begingroup$
@H17 Also, there is a technicality that is bothering me. $L_1(V)$ consists of equivalence classes of functions whereas $L(V,mathbb{R})$ consists of genuine functions. So, I don't really view $L(V,mathbb{R})$ as a subset of $L_1(V)$.
$endgroup$
– mathworker21
Dec 30 '18 at 13:35
1
1
$begingroup$
Actually the random variable (construction) is just the identity $x mapsto x.$ Random variables are very useful to work with (describing events etc ...). However, in probability theory the interesting object is actually the probability measure. Random variables are in fact rarely necessary.
$endgroup$
– H17
Dec 30 '18 at 15:30
$begingroup$
Actually the random variable (construction) is just the identity $x mapsto x.$ Random variables are very useful to work with (describing events etc ...). However, in probability theory the interesting object is actually the probability measure. Random variables are in fact rarely necessary.
$endgroup$
– H17
Dec 30 '18 at 15:30
|
show 8 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The answer is no. Consider $V=l^2(mathbb{N}, mathbb{R})$ and let $e_i$ be the i-th basis vector of the usual Schauder basis. Set
$$ gamma = sum_{ngeq 1} 2^{-n} delta_{e_n}$$
And consider the following sequence of bounded linear functionals
$$l_m (x) = sum_{j=1}^m j cdot x_j $$
This forms a Cauchy sequence as for $m>k$ holds
$$ Vert l_m - l_k Vert = sum_{j=k+1}^m jcdot 2^{-j}$$
I leave it to you to you to show that this sequence does not converge to some bounded operator (assume it does and then do the computation for the norm to get a contradiction).
Added: I forgot to check that my probability measure admits the first moment (I did not even know how it was defined, thanks to H17 for telling me). In fact not only does the first moment exist, but all moments do. Indeed, for $kgeq 1$ we have
$$ int Vert x Vert^k gamma(dx) = sum_{ngeq 1} Vert e_n Vert^k cdot 2^{-n}
= sum_{ngeq 1} 2^{-n} = 1. $$
answered Dec 30 '18 at 14:23
Severin SchravenSeverin Schraven
6,5301935
$endgroup$
add a comment |
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$begingroup$
The answer is no. Consider $V=l^2(mathbb{N}, mathbb{R})$ and let $e_i$ be the i-th basis vector of the usual Schauder basis. Set
$$ gamma = sum_{ngeq 1} 2^{-n} delta_{e_n}$$
And consider the following sequence of bounded linear functionals
$$l_m (x) = sum_{j=1}^m j cdot x_j $$
This forms a Cauchy sequence as for $m>k$ holds
$$ Vert l_m - l_k Vert = sum_{j=k+1}^m jcdot 2^{-j}$$
I leave it to you to you to show that this sequence does not converge to some bounded operator (assume it does and then do the computation for the norm to get a contradiction).
Added: I forgot to check that my probability measure admits the first moment (I did not even know how it was defined, thanks to H17 for telling me). In fact not only does the first moment exist, but all moments do. Indeed, for $kgeq 1$ we have
$$ int Vert x Vert^k gamma(dx) = sum_{ngeq 1} Vert e_n Vert^k cdot 2^{-n}
= sum_{ngeq 1} 2^{-n} = 1. $$
answered Dec 30 '18 at 14:23
Severin SchravenSeverin Schraven
6,5301935
$endgroup$
add a comment |
$begingroup$
The answer is no. Consider $V=l^2(mathbb{N}, mathbb{R})$ and let $e_i$ be the i-th basis vector of the usual Schauder basis. Set
$$ gamma = sum_{ngeq 1} 2^{-n} delta_{e_n}$$
And consider the following sequence of bounded linear functionals
$$l_m (x) = sum_{j=1}^m j cdot x_j $$
This forms a Cauchy sequence as for $m>k$ holds
$$ Vert l_m - l_k Vert = sum_{j=k+1}^m jcdot 2^{-j}$$
I leave it to you to you to show that this sequence does not converge to some bounded operator (assume it does and then do the computation for the norm to get a contradiction).
Added: I forgot to check that my probability measure admits the first moment (I did not even know how it was defined, thanks to H17 for telling me). In fact not only does the first moment exist, but all moments do. Indeed, for $kgeq 1$ we have
$$ int Vert x Vert^k gamma(dx) = sum_{ngeq 1} Vert e_n Vert^k cdot 2^{-n}
= sum_{ngeq 1} 2^{-n} = 1. $$
answered Dec 30 '18 at 14:23
Severin SchravenSeverin Schraven
6,5301935
$endgroup$
add a comment |
$begingroup$
The answer is no. Consider $V=l^2(mathbb{N}, mathbb{R})$ and let $e_i$ be the i-th basis vector of the usual Schauder basis. Set
$$ gamma = sum_{ngeq 1} 2^{-n} delta_{e_n}$$
And consider the following sequence of bounded linear functionals
$$l_m (x) = sum_{j=1}^m j cdot x_j $$
This forms a Cauchy sequence as for $m>k$ holds
$$ Vert l_m - l_k Vert = sum_{j=k+1}^m jcdot 2^{-j}$$
I leave it to you to you to show that this sequence does not converge to some bounded operator (assume it does and then do the computation for the norm to get a contradiction).
Added: I forgot to check that my probability measure admits the first moment (I did not even know how it was defined, thanks to H17 for telling me). In fact not only does the first moment exist, but all moments do. Indeed, for $kgeq 1$ we have
$$ int Vert x Vert^k gamma(dx) = sum_{ngeq 1} Vert e_n Vert^k cdot 2^{-n}
= sum_{ngeq 1} 2^{-n} = 1. $$
answered Dec 30 '18 at 14:23
Severin SchravenSeverin Schraven
6,5301935
$endgroup$
The answer is no. Consider $V=l^2(mathbb{N}, mathbb{R})$ and let $e_i$ be the i-th basis vector of the usual Schauder basis. Set
$$ gamma = sum_{ngeq 1} 2^{-n} delta_{e_n}$$
And consider the following sequence of bounded linear functionals
$$l_m (x) = sum_{j=1}^m j cdot x_j $$
This forms a Cauchy sequence as for $m>k$ holds
$$ Vert l_m - l_k Vert = sum_{j=k+1}^m jcdot 2^{-j}$$
I leave it to you to you to show that this sequence does not converge to some bounded operator (assume it does and then do the computation for the norm to get a contradiction).
Added: I forgot to check that my probability measure admits the first moment (I did not even know how it was defined, thanks to H17 for telling me). In fact not only does the first moment exist, but all moments do. Indeed, for $kgeq 1$ we have
$$ int Vert x Vert^k gamma(dx) = sum_{ngeq 1} Vert e_n Vert^k cdot 2^{-n}
= sum_{ngeq 1} 2^{-n} = 1. $$
answered Dec 30 '18 at 14:23
Severin SchravenSeverin Schraven
6,5301935
edited Dec 30 '18 at 15:10
answered Dec 30 '18 at 14:23
Severin SchravenSeverin Schraven
6,5301935
answered Dec 30 '18 at 14:23
Severin SchravenSeverin Schraven
6,5301935
answered Dec 30 '18 at 14:23
Severin SchravenSeverin Schraven
6,5301935
6,5301935
add a comment |
add a comment |
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$begingroup$
do you define $Vert f Vert_1:=Vert f Vert + Vert f' Vert$? What is $L_1(V)$?
$endgroup$
– noctusraid
Dec 30 '18 at 13:22
$begingroup$
$mathrm{L}_1(V)$ denotes the usual Lebesgue space that contains all measurable $f : V to mathbb{R}$ such that $$ |f|_1 := int_V |f(v)| , gamma(mathrm{d}v) < infty$$
$endgroup$
– H17
Dec 30 '18 at 13:25
$begingroup$
@H17 The norm on $L(V,mathbb{R})$ is irrelevant, right?
$endgroup$
– mathworker21
Dec 30 '18 at 13:33
$begingroup$
@H17 Also, there is a technicality that is bothering me. $L_1(V)$ consists of equivalence classes of functions whereas $L(V,mathbb{R})$ consists of genuine functions. So, I don't really view $L(V,mathbb{R})$ as a subset of $L_1(V)$.
$endgroup$
– mathworker21
Dec 30 '18 at 13:35
1
$begingroup$
Actually the random variable (construction) is just the identity $x mapsto x.$ Random variables are very useful to work with (describing events etc ...). However, in probability theory the interesting object is actually the probability measure. Random variables are in fact rarely necessary.
$endgroup$
– H17
Dec 30 '18 at 15:30