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Literature on relationship between strong and weak* (weak star) convergence
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I am trying to follow a proof in the paper Wasserstein Generative Adversarial Networks by Arjovsky et al. (proof A in supplementary material).
They show that the convergenc of the total variation distance of two probability distributions corresponds to convergence in a strong topology induced by the norm and the convergence of the Wasserstein-1 distance of two distributions corresponds to convergence in a weak* topology.
If I understand that correctly, the convergence of a sequence in a topology induced by the norm always converges strongly (see here). And convergence in a weak* topology is weak* convergence by definition.
As far as I understand this, strong convergence induces weak* convergence. I did also find that in some lecture notes (e. g. here or here), but not in any textbook or scientific article. So does anyone know of literature I could cite in my thesis concerning this?
Or a proof that strong convergence induces weak* convergence (easily) understandable by electrical engineers would be fine, too, I guess.
Thank you very much in advance!
general-topology functional-analysis reference-request weak-convergence weak-topology
edited Jan 31 at 23:57
Pedro
10.9k23475
asked Jan 31 at 11:14
Florian_1990Florian_1990
61
$endgroup$
add a comment |
$begingroup$
I am trying to follow a proof in the paper Wasserstein Generative Adversarial Networks by Arjovsky et al. (proof A in supplementary material).
They show that the convergenc of the total variation distance of two probability distributions corresponds to convergence in a strong topology induced by the norm and the convergence of the Wasserstein-1 distance of two distributions corresponds to convergence in a weak* topology.
If I understand that correctly, the convergence of a sequence in a topology induced by the norm always converges strongly (see here). And convergence in a weak* topology is weak* convergence by definition.
As far as I understand this, strong convergence induces weak* convergence. I did also find that in some lecture notes (e. g. here or here), but not in any textbook or scientific article. So does anyone know of literature I could cite in my thesis concerning this?
Or a proof that strong convergence induces weak* convergence (easily) understandable by electrical engineers would be fine, too, I guess.
Thank you very much in advance!
general-topology functional-analysis reference-request weak-convergence weak-topology
edited Jan 31 at 23:57
Pedro
10.9k23475
asked Jan 31 at 11:14
Florian_1990Florian_1990
61
$endgroup$
add a comment |
$begingroup$
I am trying to follow a proof in the paper Wasserstein Generative Adversarial Networks by Arjovsky et al. (proof A in supplementary material).
They show that the convergenc of the total variation distance of two probability distributions corresponds to convergence in a strong topology induced by the norm and the convergence of the Wasserstein-1 distance of two distributions corresponds to convergence in a weak* topology.
If I understand that correctly, the convergence of a sequence in a topology induced by the norm always converges strongly (see here). And convergence in a weak* topology is weak* convergence by definition.
As far as I understand this, strong convergence induces weak* convergence. I did also find that in some lecture notes (e. g. here or here), but not in any textbook or scientific article. So does anyone know of literature I could cite in my thesis concerning this?
Or a proof that strong convergence induces weak* convergence (easily) understandable by electrical engineers would be fine, too, I guess.
Thank you very much in advance!
general-topology functional-analysis reference-request weak-convergence weak-topology
edited Jan 31 at 23:57
Pedro
10.9k23475
asked Jan 31 at 11:14
Florian_1990Florian_1990
61
$endgroup$
I am trying to follow a proof in the paper Wasserstein Generative Adversarial Networks by Arjovsky et al. (proof A in supplementary material).
They show that the convergenc of the total variation distance of two probability distributions corresponds to convergence in a strong topology induced by the norm and the convergence of the Wasserstein-1 distance of two distributions corresponds to convergence in a weak* topology.
If I understand that correctly, the convergence of a sequence in a topology induced by the norm always converges strongly (see here). And convergence in a weak* topology is weak* convergence by definition.
As far as I understand this, strong convergence induces weak* convergence. I did also find that in some lecture notes (e. g. here or here), but not in any textbook or scientific article. So does anyone know of literature I could cite in my thesis concerning this?
Or a proof that strong convergence induces weak* convergence (easily) understandable by electrical engineers would be fine, too, I guess.
Thank you very much in advance!
general-topology functional-analysis reference-request weak-convergence weak-topology
general-topology functional-analysis reference-request weak-convergence weak-topology
edited Jan 31 at 23:57
Pedro
10.9k23475
asked Jan 31 at 11:14
Florian_1990Florian_1990
61
edited Jan 31 at 23:57
Pedro
10.9k23475
asked Jan 31 at 11:14
Florian_1990Florian_1990
61
edited Jan 31 at 23:57
Pedro
10.9k23475
edited Jan 31 at 23:57
Pedro
10.9k23475
edited Jan 31 at 23:57
Pedro
10.9k23475
10.9k23475
asked Jan 31 at 11:14
Florian_1990Florian_1990
61
asked Jan 31 at 11:14
Florian_1990Florian_1990
61
asked Jan 31 at 11:14
Florian_1990Florian_1990
61
61
add a comment |
add a comment |
2 Answers
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oldest
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welcome to MSE.
The fact that strong convergence implies weak*-convergence in the dual space is so basic, that most books don't bother mentioning it. What might be a good thing to do is to reference "Functional Analysis" 2nd edition by Walter Rudin which is a good book on functional analysis containing a good introduction to the weak and weak* topology.
As for the proof, this consists of two parts, showing that strong convergence implies weak convergence and noting that weak convergence implies weak*-convergence.
Let $X$ be a normed vector space and let $X^{*}$ denote its dual space. By definition $X^{*}$ is the space of all continuous linear functional, so for any $varphiin X^{*}$ the operator norm
$$|varphi|=sup{|varphi(x)|:xin X,|x|leq1}$$
is finite.
Now suppose $(x_{alpha})$ is a net in $X$ (or $(x_{n})$ is a sequence if you don't want to use nets) converging strongly to $xin X$. Then for all $varphiin X^{*}$ we find
$$lim_{alpha}|varphi(x_{alpha})-varphi(x)|=lim_{alpha}|varphi(x_{alpha}-x)|leqlim_{alpha}|varphi||x_{alpha}-x|=0.$$
As this holds for all $varphiin X^{*}$ by definition $(x_{alpha})$ weakly converges to $x$.
For the next part it is important to note that the weak* topology is not defined on $X$ but $X^{*}$. So we are considering strong convergence on $X^{*}$ with respect to the operator norm, and a net $(varphi_{alpha})$ in $X^{*}$ weakly converges to $varphi$ if for all $psiin (X^{*})^{*}$ we have $lim_{alpha}psi(varphi_{alpha})=psi(varphi)$. Recall that $(varphi_{alpha})$ converges to $varphi$ in the weak* topology if for all $xin X$ we have $lim_{alpha}varphi_{alpha}(x)=varphi(x)$. Note however that for $xin X$ the evaluation map $psi_{x}:X^{*}rightarrowmathbb{R},varphimapstovarphi(x)$ is a continuous linear functional on $X^{*}$ hence $psi_{x}in(X^{*})^{*}$. Therefore if $(varphi_{alpha})$ converges to $varphi$ in the weak topology it also converges in the weak* topology. (For a mathematical audience I would just write that the weak convergence implying weak* convergence is trivial.)
Combining these two results gives that strong convergence on $X^{*}$ implies weak* convergence.
Let me know if any of these steps is still unclear.
answered Jan 31 at 12:57
Floris ClaassensFloris Claassens
1,33229
$endgroup$
add a comment |
$begingroup$
As far as I understand this, strong convergence induces weak$^*$ convergence... does anyone know of literature I could cite in my thesis concerning this?
Yes: Proposition 3.13(ii) in Brezis book.
answered Jan 31 at 23:54
PedroPedro
10.9k23475
$endgroup$
add a comment |
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2 Answers
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$begingroup$
welcome to MSE.
The fact that strong convergence implies weak*-convergence in the dual space is so basic, that most books don't bother mentioning it. What might be a good thing to do is to reference "Functional Analysis" 2nd edition by Walter Rudin which is a good book on functional analysis containing a good introduction to the weak and weak* topology.
As for the proof, this consists of two parts, showing that strong convergence implies weak convergence and noting that weak convergence implies weak*-convergence.
Let $X$ be a normed vector space and let $X^{*}$ denote its dual space. By definition $X^{*}$ is the space of all continuous linear functional, so for any $varphiin X^{*}$ the operator norm
$$|varphi|=sup{|varphi(x)|:xin X,|x|leq1}$$
is finite.
Now suppose $(x_{alpha})$ is a net in $X$ (or $(x_{n})$ is a sequence if you don't want to use nets) converging strongly to $xin X$. Then for all $varphiin X^{*}$ we find
$$lim_{alpha}|varphi(x_{alpha})-varphi(x)|=lim_{alpha}|varphi(x_{alpha}-x)|leqlim_{alpha}|varphi||x_{alpha}-x|=0.$$
As this holds for all $varphiin X^{*}$ by definition $(x_{alpha})$ weakly converges to $x$.
For the next part it is important to note that the weak* topology is not defined on $X$ but $X^{*}$. So we are considering strong convergence on $X^{*}$ with respect to the operator norm, and a net $(varphi_{alpha})$ in $X^{*}$ weakly converges to $varphi$ if for all $psiin (X^{*})^{*}$ we have $lim_{alpha}psi(varphi_{alpha})=psi(varphi)$. Recall that $(varphi_{alpha})$ converges to $varphi$ in the weak* topology if for all $xin X$ we have $lim_{alpha}varphi_{alpha}(x)=varphi(x)$. Note however that for $xin X$ the evaluation map $psi_{x}:X^{*}rightarrowmathbb{R},varphimapstovarphi(x)$ is a continuous linear functional on $X^{*}$ hence $psi_{x}in(X^{*})^{*}$. Therefore if $(varphi_{alpha})$ converges to $varphi$ in the weak topology it also converges in the weak* topology. (For a mathematical audience I would just write that the weak convergence implying weak* convergence is trivial.)
Combining these two results gives that strong convergence on $X^{*}$ implies weak* convergence.
Let me know if any of these steps is still unclear.
answered Jan 31 at 12:57
Floris ClaassensFloris Claassens
1,33229
$endgroup$
add a comment |
$begingroup$
welcome to MSE.
The fact that strong convergence implies weak*-convergence in the dual space is so basic, that most books don't bother mentioning it. What might be a good thing to do is to reference "Functional Analysis" 2nd edition by Walter Rudin which is a good book on functional analysis containing a good introduction to the weak and weak* topology.
As for the proof, this consists of two parts, showing that strong convergence implies weak convergence and noting that weak convergence implies weak*-convergence.
Let $X$ be a normed vector space and let $X^{*}$ denote its dual space. By definition $X^{*}$ is the space of all continuous linear functional, so for any $varphiin X^{*}$ the operator norm
$$|varphi|=sup{|varphi(x)|:xin X,|x|leq1}$$
is finite.
Now suppose $(x_{alpha})$ is a net in $X$ (or $(x_{n})$ is a sequence if you don't want to use nets) converging strongly to $xin X$. Then for all $varphiin X^{*}$ we find
$$lim_{alpha}|varphi(x_{alpha})-varphi(x)|=lim_{alpha}|varphi(x_{alpha}-x)|leqlim_{alpha}|varphi||x_{alpha}-x|=0.$$
As this holds for all $varphiin X^{*}$ by definition $(x_{alpha})$ weakly converges to $x$.
For the next part it is important to note that the weak* topology is not defined on $X$ but $X^{*}$. So we are considering strong convergence on $X^{*}$ with respect to the operator norm, and a net $(varphi_{alpha})$ in $X^{*}$ weakly converges to $varphi$ if for all $psiin (X^{*})^{*}$ we have $lim_{alpha}psi(varphi_{alpha})=psi(varphi)$. Recall that $(varphi_{alpha})$ converges to $varphi$ in the weak* topology if for all $xin X$ we have $lim_{alpha}varphi_{alpha}(x)=varphi(x)$. Note however that for $xin X$ the evaluation map $psi_{x}:X^{*}rightarrowmathbb{R},varphimapstovarphi(x)$ is a continuous linear functional on $X^{*}$ hence $psi_{x}in(X^{*})^{*}$. Therefore if $(varphi_{alpha})$ converges to $varphi$ in the weak topology it also converges in the weak* topology. (For a mathematical audience I would just write that the weak convergence implying weak* convergence is trivial.)
Combining these two results gives that strong convergence on $X^{*}$ implies weak* convergence.
Let me know if any of these steps is still unclear.
answered Jan 31 at 12:57
Floris ClaassensFloris Claassens
1,33229
$endgroup$
add a comment |
$begingroup$
welcome to MSE.
The fact that strong convergence implies weak*-convergence in the dual space is so basic, that most books don't bother mentioning it. What might be a good thing to do is to reference "Functional Analysis" 2nd edition by Walter Rudin which is a good book on functional analysis containing a good introduction to the weak and weak* topology.
As for the proof, this consists of two parts, showing that strong convergence implies weak convergence and noting that weak convergence implies weak*-convergence.
Let $X$ be a normed vector space and let $X^{*}$ denote its dual space. By definition $X^{*}$ is the space of all continuous linear functional, so for any $varphiin X^{*}$ the operator norm
$$|varphi|=sup{|varphi(x)|:xin X,|x|leq1}$$
is finite.
Now suppose $(x_{alpha})$ is a net in $X$ (or $(x_{n})$ is a sequence if you don't want to use nets) converging strongly to $xin X$. Then for all $varphiin X^{*}$ we find
$$lim_{alpha}|varphi(x_{alpha})-varphi(x)|=lim_{alpha}|varphi(x_{alpha}-x)|leqlim_{alpha}|varphi||x_{alpha}-x|=0.$$
As this holds for all $varphiin X^{*}$ by definition $(x_{alpha})$ weakly converges to $x$.
For the next part it is important to note that the weak* topology is not defined on $X$ but $X^{*}$. So we are considering strong convergence on $X^{*}$ with respect to the operator norm, and a net $(varphi_{alpha})$ in $X^{*}$ weakly converges to $varphi$ if for all $psiin (X^{*})^{*}$ we have $lim_{alpha}psi(varphi_{alpha})=psi(varphi)$. Recall that $(varphi_{alpha})$ converges to $varphi$ in the weak* topology if for all $xin X$ we have $lim_{alpha}varphi_{alpha}(x)=varphi(x)$. Note however that for $xin X$ the evaluation map $psi_{x}:X^{*}rightarrowmathbb{R},varphimapstovarphi(x)$ is a continuous linear functional on $X^{*}$ hence $psi_{x}in(X^{*})^{*}$. Therefore if $(varphi_{alpha})$ converges to $varphi$ in the weak topology it also converges in the weak* topology. (For a mathematical audience I would just write that the weak convergence implying weak* convergence is trivial.)
Combining these two results gives that strong convergence on $X^{*}$ implies weak* convergence.
Let me know if any of these steps is still unclear.
answered Jan 31 at 12:57
Floris ClaassensFloris Claassens
1,33229
$endgroup$
welcome to MSE.
The fact that strong convergence implies weak*-convergence in the dual space is so basic, that most books don't bother mentioning it. What might be a good thing to do is to reference "Functional Analysis" 2nd edition by Walter Rudin which is a good book on functional analysis containing a good introduction to the weak and weak* topology.
As for the proof, this consists of two parts, showing that strong convergence implies weak convergence and noting that weak convergence implies weak*-convergence.
Let $X$ be a normed vector space and let $X^{*}$ denote its dual space. By definition $X^{*}$ is the space of all continuous linear functional, so for any $varphiin X^{*}$ the operator norm
$$|varphi|=sup{|varphi(x)|:xin X,|x|leq1}$$
is finite.
Now suppose $(x_{alpha})$ is a net in $X$ (or $(x_{n})$ is a sequence if you don't want to use nets) converging strongly to $xin X$. Then for all $varphiin X^{*}$ we find
$$lim_{alpha}|varphi(x_{alpha})-varphi(x)|=lim_{alpha}|varphi(x_{alpha}-x)|leqlim_{alpha}|varphi||x_{alpha}-x|=0.$$
As this holds for all $varphiin X^{*}$ by definition $(x_{alpha})$ weakly converges to $x$.
For the next part it is important to note that the weak* topology is not defined on $X$ but $X^{*}$. So we are considering strong convergence on $X^{*}$ with respect to the operator norm, and a net $(varphi_{alpha})$ in $X^{*}$ weakly converges to $varphi$ if for all $psiin (X^{*})^{*}$ we have $lim_{alpha}psi(varphi_{alpha})=psi(varphi)$. Recall that $(varphi_{alpha})$ converges to $varphi$ in the weak* topology if for all $xin X$ we have $lim_{alpha}varphi_{alpha}(x)=varphi(x)$. Note however that for $xin X$ the evaluation map $psi_{x}:X^{*}rightarrowmathbb{R},varphimapstovarphi(x)$ is a continuous linear functional on $X^{*}$ hence $psi_{x}in(X^{*})^{*}$. Therefore if $(varphi_{alpha})$ converges to $varphi$ in the weak topology it also converges in the weak* topology. (For a mathematical audience I would just write that the weak convergence implying weak* convergence is trivial.)
Combining these two results gives that strong convergence on $X^{*}$ implies weak* convergence.
Let me know if any of these steps is still unclear.
answered Jan 31 at 12:57
Floris ClaassensFloris Claassens
1,33229
answered Jan 31 at 12:57
Floris ClaassensFloris Claassens
1,33229
answered Jan 31 at 12:57
Floris ClaassensFloris Claassens
1,33229
answered Jan 31 at 12:57
Floris ClaassensFloris Claassens
1,33229
1,33229
add a comment |
add a comment |
$begingroup$
As far as I understand this, strong convergence induces weak$^*$ convergence... does anyone know of literature I could cite in my thesis concerning this?
Yes: Proposition 3.13(ii) in Brezis book.
answered Jan 31 at 23:54
PedroPedro
10.9k23475
$endgroup$
add a comment |
$begingroup$
As far as I understand this, strong convergence induces weak$^*$ convergence... does anyone know of literature I could cite in my thesis concerning this?
Yes: Proposition 3.13(ii) in Brezis book.
answered Jan 31 at 23:54
PedroPedro
10.9k23475
$endgroup$
add a comment |
$begingroup$
As far as I understand this, strong convergence induces weak$^*$ convergence... does anyone know of literature I could cite in my thesis concerning this?
Yes: Proposition 3.13(ii) in Brezis book.
answered Jan 31 at 23:54
PedroPedro
10.9k23475
$endgroup$
As far as I understand this, strong convergence induces weak$^*$ convergence... does anyone know of literature I could cite in my thesis concerning this?
Yes: Proposition 3.13(ii) in Brezis book.
answered Jan 31 at 23:54
PedroPedro
10.9k23475
answered Jan 31 at 23:54
PedroPedro
10.9k23475
answered Jan 31 at 23:54
PedroPedro
10.9k23475
answered Jan 31 at 23:54
PedroPedro
10.9k23475
10.9k23475
add a comment |
add a comment |
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