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Relatively compactness of uniformly limited measures in variation w.r.t. weak convergence
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Let $f_n(x).dx$ be measures with density wrt Lebesgue measure, $f_n$ derivable, $f_n rightarrow f$ in uniform norm; $ n rightarrow infty$.
Assume that they are uniformly limited in variation, i.e. $exists M: int |Df_n|(dx)<M forall n$, where $Df$ is the distributional derivative of $f$.
This means $forall phi in C_c^1(mathbbR^n), int f(x)fracdelta phidelta x_i (x)dx=intphi(x)(D_i f)(dx)$.
Than there is a subsuccession $f_k_n n in mathbbN$ such that $f_k_n(x).dx$ converges weakly. How can i show it?
Original De Giorgi paper says it's an "easy" consequence of a theorem of de La Vallee Poussin.
Notice that i have very few confidence with this field, so please try to be as detailed as possible.
measure-theory weak-convergence bounded-variation
asked Jul 23 at 21:09
Andro
787
add a comment |Â
up vote
1
down vote
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Let $f_n(x).dx$ be measures with density wrt Lebesgue measure, $f_n$ derivable, $f_n rightarrow f$ in uniform norm; $ n rightarrow infty$.
Assume that they are uniformly limited in variation, i.e. $exists M: int |Df_n|(dx)<M forall n$, where $Df$ is the distributional derivative of $f$.
This means $forall phi in C_c^1(mathbbR^n), int f(x)fracdelta phidelta x_i (x)dx=intphi(x)(D_i f)(dx)$.
Than there is a subsuccession $f_k_n n in mathbbN$ such that $f_k_n(x).dx$ converges weakly. How can i show it?
Original De Giorgi paper says it's an "easy" consequence of a theorem of de La Vallee Poussin.
Notice that i have very few confidence with this field, so please try to be as detailed as possible.
measure-theory weak-convergence bounded-variation
asked Jul 23 at 21:09
Andro
787
The notions of Lebesgue measure, distributional derivative, and weak convergence are themselves not "elementary," so I'm not sure how you expect an answer to this question to be such.
– Math1000
Jul 23 at 22:09
You are right, i need answers to be detailful, with all implications needed, or with references.
– Andro
Jul 23 at 22:20
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f_n(x).dx$ be measures with density wrt Lebesgue measure, $f_n$ derivable, $f_n rightarrow f$ in uniform norm; $ n rightarrow infty$.
Assume that they are uniformly limited in variation, i.e. $exists M: int |Df_n|(dx)<M forall n$, where $Df$ is the distributional derivative of $f$.
This means $forall phi in C_c^1(mathbbR^n), int f(x)fracdelta phidelta x_i (x)dx=intphi(x)(D_i f)(dx)$.
Than there is a subsuccession $f_k_n n in mathbbN$ such that $f_k_n(x).dx$ converges weakly. How can i show it?
Original De Giorgi paper says it's an "easy" consequence of a theorem of de La Vallee Poussin.
Notice that i have very few confidence with this field, so please try to be as detailed as possible.
measure-theory weak-convergence bounded-variation
asked Jul 23 at 21:09
Andro
787
Let $f_n(x).dx$ be measures with density wrt Lebesgue measure, $f_n$ derivable, $f_n rightarrow f$ in uniform norm; $ n rightarrow infty$.
Assume that they are uniformly limited in variation, i.e. $exists M: int |Df_n|(dx)<M forall n$, where $Df$ is the distributional derivative of $f$.
This means $forall phi in C_c^1(mathbbR^n), int f(x)fracdelta phidelta x_i (x)dx=intphi(x)(D_i f)(dx)$.
Than there is a subsuccession $f_k_n n in mathbbN$ such that $f_k_n(x).dx$ converges weakly. How can i show it?
Original De Giorgi paper says it's an "easy" consequence of a theorem of de La Vallee Poussin.
Notice that i have very few confidence with this field, so please try to be as detailed as possible.
measure-theory weak-convergence bounded-variation
asked Jul 23 at 21:09
Andro
787
edited Jul 23 at 22:23
asked Jul 23 at 21:09
Andro
787
asked Jul 23 at 21:09
Andro
787
asked Jul 23 at 21:09
Andro
787
787
The notions of Lebesgue measure, distributional derivative, and weak convergence are themselves not "elementary," so I'm not sure how you expect an answer to this question to be such.
– Math1000
Jul 23 at 22:09
You are right, i need answers to be detailful, with all implications needed, or with references.
– Andro
Jul 23 at 22:20
add a comment |Â
The notions of Lebesgue measure, distributional derivative, and weak convergence are themselves not "elementary," so I'm not sure how you expect an answer to this question to be such.
– Math1000
Jul 23 at 22:09
You are right, i need answers to be detailful, with all implications needed, or with references.
– Andro
Jul 23 at 22:20
The notions of Lebesgue measure, distributional derivative, and weak convergence are themselves not "elementary," so I'm not sure how you expect an answer to this question to be such.
– Math1000
Jul 23 at 22:09
The notions of Lebesgue measure, distributional derivative, and weak convergence are themselves not "elementary," so I'm not sure how you expect an answer to this question to be such.
– Math1000
Jul 23 at 22:09
You are right, i need answers to be detailful, with all implications needed, or with references.
– Andro
Jul 23 at 22:20
You are right, i need answers to be detailful, with all implications needed, or with references.
– Andro
Jul 23 at 22:20
add a comment |Â
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The notions of Lebesgue measure, distributional derivative, and weak convergence are themselves not "elementary," so I'm not sure how you expect an answer to this question to be such.
– Math1000
Jul 23 at 22:09
You are right, i need answers to be detailful, with all implications needed, or with references.
– Andro
Jul 23 at 22:20