Constructive proof of Banach-Alaouglu
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Is there a constructive (i.e. not using Axiom of choice, and at most Axiom of dependent choice) proof of Banach-Alaoglu theorem in the case of separable Banach spaces. Even if it is needed assume that the dual is separable. Under even more assumptions - is there such a proof for infinite dimensional Banach spaces.
We know such proof exists for Hahn-Banach in the separable case.
functional-analysis banach-spaces separable-spaces
asked Jan 24 at 10:28
Stoyan ApostolovStoyan Apostolov
39229
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add a comment |
$begingroup$
Is there a constructive (i.e. not using Axiom of choice, and at most Axiom of dependent choice) proof of Banach-Alaoglu theorem in the case of separable Banach spaces. Even if it is needed assume that the dual is separable. Under even more assumptions - is there such a proof for infinite dimensional Banach spaces.
We know such proof exists for Hahn-Banach in the separable case.
functional-analysis banach-spaces separable-spaces
asked Jan 24 at 10:28
Stoyan ApostolovStoyan Apostolov
39229
$endgroup$
$begingroup$
Do you know a reference for Hahn-Banach in separable case, as you stated in the last sentence of the question?
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– Hayk
Jan 24 at 10:37
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I don't know too much about set-theoretic foundations, but I think the proof in Dieudonnè's Foundations of Modern Analysis II (Section 12.15 in my copy) works without axiom of choice (maybe you need countable choice?).
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– MaoWao
Jan 24 at 10:43
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As I recall the first proof published by Banach himself only dealt with the separable case, and was constructive (he proved sequential compactness directly). Later Alaouglu published the general theorem which relies on Tychonoff's theorem.
$endgroup$
– pitariver
Jan 24 at 10:54
add a comment |
$begingroup$
Is there a constructive (i.e. not using Axiom of choice, and at most Axiom of dependent choice) proof of Banach-Alaoglu theorem in the case of separable Banach spaces. Even if it is needed assume that the dual is separable. Under even more assumptions - is there such a proof for infinite dimensional Banach spaces.
We know such proof exists for Hahn-Banach in the separable case.
functional-analysis banach-spaces separable-spaces
asked Jan 24 at 10:28
Stoyan ApostolovStoyan Apostolov
39229
$endgroup$
Is there a constructive (i.e. not using Axiom of choice, and at most Axiom of dependent choice) proof of Banach-Alaoglu theorem in the case of separable Banach spaces. Even if it is needed assume that the dual is separable. Under even more assumptions - is there such a proof for infinite dimensional Banach spaces.
We know such proof exists for Hahn-Banach in the separable case.
functional-analysis banach-spaces separable-spaces
functional-analysis banach-spaces separable-spaces
asked Jan 24 at 10:28
Stoyan ApostolovStoyan Apostolov
39229
asked Jan 24 at 10:28
Stoyan ApostolovStoyan Apostolov
39229
asked Jan 24 at 10:28
Stoyan ApostolovStoyan Apostolov
39229
asked Jan 24 at 10:28
Stoyan ApostolovStoyan Apostolov
39229
asked Jan 24 at 10:28
Stoyan ApostolovStoyan Apostolov
39229
39229
$begingroup$
Do you know a reference for Hahn-Banach in separable case, as you stated in the last sentence of the question?
$endgroup$
– Hayk
Jan 24 at 10:37
$begingroup$
I don't know too much about set-theoretic foundations, but I think the proof in Dieudonnè's Foundations of Modern Analysis II (Section 12.15 in my copy) works without axiom of choice (maybe you need countable choice?).
$endgroup$
– MaoWao
Jan 24 at 10:43
$begingroup$
As I recall the first proof published by Banach himself only dealt with the separable case, and was constructive (he proved sequential compactness directly). Later Alaouglu published the general theorem which relies on Tychonoff's theorem.
$endgroup$
– pitariver
Jan 24 at 10:54
add a comment |
$begingroup$
Do you know a reference for Hahn-Banach in separable case, as you stated in the last sentence of the question?
$endgroup$
– Hayk
Jan 24 at 10:37
$begingroup$
I don't know too much about set-theoretic foundations, but I think the proof in Dieudonnè's Foundations of Modern Analysis II (Section 12.15 in my copy) works without axiom of choice (maybe you need countable choice?).
$endgroup$
– MaoWao
Jan 24 at 10:43
$begingroup$
As I recall the first proof published by Banach himself only dealt with the separable case, and was constructive (he proved sequential compactness directly). Later Alaouglu published the general theorem which relies on Tychonoff's theorem.
$endgroup$
– pitariver
Jan 24 at 10:54
$begingroup$
Do you know a reference for Hahn-Banach in separable case, as you stated in the last sentence of the question?
$endgroup$
– Hayk
Jan 24 at 10:37
$begingroup$
Do you know a reference for Hahn-Banach in separable case, as you stated in the last sentence of the question?
$endgroup$
– Hayk
Jan 24 at 10:37
$begingroup$
I don't know too much about set-theoretic foundations, but I think the proof in Dieudonnè's Foundations of Modern Analysis II (Section 12.15 in my copy) works without axiom of choice (maybe you need countable choice?).
$endgroup$
– MaoWao
Jan 24 at 10:43
$begingroup$
I don't know too much about set-theoretic foundations, but I think the proof in Dieudonnè's Foundations of Modern Analysis II (Section 12.15 in my copy) works without axiom of choice (maybe you need countable choice?).
$endgroup$
– MaoWao
Jan 24 at 10:43
$begingroup$
As I recall the first proof published by Banach himself only dealt with the separable case, and was constructive (he proved sequential compactness directly). Later Alaouglu published the general theorem which relies on Tychonoff's theorem.
$endgroup$
– pitariver
Jan 24 at 10:54
$begingroup$
As I recall the first proof published by Banach himself only dealt with the separable case, and was constructive (he proved sequential compactness directly). Later Alaouglu published the general theorem which relies on Tychonoff's theorem.
$endgroup$
– pitariver
Jan 24 at 10:54
add a comment |
1 Answer
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This is quite elementary. If $X$ is separable then the closed unit ball $B$ of $X^{*}$ is metrizable in the $weak^{*}$ topology. [ $d(f,g)=sumlimits_{k=1}^{infty} frac 1 {2^{i}} frac {|f(x_i)-g(x_i)|} {1+|f(x_i)-g(x_i)|}$ metrizes it if ${x_i}$ is dense in $X$]. Any sequence ${f_n}$ in $B$ has a subsequence ${f_n'}$ converging at each of the $x_j$'s (by a diagonal procedure). It is now fairly easy to see that if ${x_{i_l}}$ converges to some point $x$ then $f_n'(x_{i_l})$ is Cauchy. Call the limit $f(x)$. Obviously, $|f|leq 1$. It follows that $f_n'(x_{i_l})$ converges in the weak* topology. Thus, $B$ is sequentially compact and metrizable, hence compact.
answered Jan 24 at 11:58
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
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add a comment |
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1 Answer
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$begingroup$
This is quite elementary. If $X$ is separable then the closed unit ball $B$ of $X^{*}$ is metrizable in the $weak^{*}$ topology. [ $d(f,g)=sumlimits_{k=1}^{infty} frac 1 {2^{i}} frac {|f(x_i)-g(x_i)|} {1+|f(x_i)-g(x_i)|}$ metrizes it if ${x_i}$ is dense in $X$]. Any sequence ${f_n}$ in $B$ has a subsequence ${f_n'}$ converging at each of the $x_j$'s (by a diagonal procedure). It is now fairly easy to see that if ${x_{i_l}}$ converges to some point $x$ then $f_n'(x_{i_l})$ is Cauchy. Call the limit $f(x)$. Obviously, $|f|leq 1$. It follows that $f_n'(x_{i_l})$ converges in the weak* topology. Thus, $B$ is sequentially compact and metrizable, hence compact.
answered Jan 24 at 11:58
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
$endgroup$
add a comment |
$begingroup$
This is quite elementary. If $X$ is separable then the closed unit ball $B$ of $X^{*}$ is metrizable in the $weak^{*}$ topology. [ $d(f,g)=sumlimits_{k=1}^{infty} frac 1 {2^{i}} frac {|f(x_i)-g(x_i)|} {1+|f(x_i)-g(x_i)|}$ metrizes it if ${x_i}$ is dense in $X$]. Any sequence ${f_n}$ in $B$ has a subsequence ${f_n'}$ converging at each of the $x_j$'s (by a diagonal procedure). It is now fairly easy to see that if ${x_{i_l}}$ converges to some point $x$ then $f_n'(x_{i_l})$ is Cauchy. Call the limit $f(x)$. Obviously, $|f|leq 1$. It follows that $f_n'(x_{i_l})$ converges in the weak* topology. Thus, $B$ is sequentially compact and metrizable, hence compact.
answered Jan 24 at 11:58
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
$endgroup$
add a comment |
$begingroup$
This is quite elementary. If $X$ is separable then the closed unit ball $B$ of $X^{*}$ is metrizable in the $weak^{*}$ topology. [ $d(f,g)=sumlimits_{k=1}^{infty} frac 1 {2^{i}} frac {|f(x_i)-g(x_i)|} {1+|f(x_i)-g(x_i)|}$ metrizes it if ${x_i}$ is dense in $X$]. Any sequence ${f_n}$ in $B$ has a subsequence ${f_n'}$ converging at each of the $x_j$'s (by a diagonal procedure). It is now fairly easy to see that if ${x_{i_l}}$ converges to some point $x$ then $f_n'(x_{i_l})$ is Cauchy. Call the limit $f(x)$. Obviously, $|f|leq 1$. It follows that $f_n'(x_{i_l})$ converges in the weak* topology. Thus, $B$ is sequentially compact and metrizable, hence compact.
answered Jan 24 at 11:58
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
$endgroup$
This is quite elementary. If $X$ is separable then the closed unit ball $B$ of $X^{*}$ is metrizable in the $weak^{*}$ topology. [ $d(f,g)=sumlimits_{k=1}^{infty} frac 1 {2^{i}} frac {|f(x_i)-g(x_i)|} {1+|f(x_i)-g(x_i)|}$ metrizes it if ${x_i}$ is dense in $X$]. Any sequence ${f_n}$ in $B$ has a subsequence ${f_n'}$ converging at each of the $x_j$'s (by a diagonal procedure). It is now fairly easy to see that if ${x_{i_l}}$ converges to some point $x$ then $f_n'(x_{i_l})$ is Cauchy. Call the limit $f(x)$. Obviously, $|f|leq 1$. It follows that $f_n'(x_{i_l})$ converges in the weak* topology. Thus, $B$ is sequentially compact and metrizable, hence compact.
answered Jan 24 at 11:58
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
edited Jan 24 at 12:03
answered Jan 24 at 11:58
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
answered Jan 24 at 11:58
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
answered Jan 24 at 11:58
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
65.4k42766
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$begingroup$
Do you know a reference for Hahn-Banach in separable case, as you stated in the last sentence of the question?
$endgroup$
– Hayk
Jan 24 at 10:37
$begingroup$
I don't know too much about set-theoretic foundations, but I think the proof in Dieudonnè's Foundations of Modern Analysis II (Section 12.15 in my copy) works without axiom of choice (maybe you need countable choice?).
$endgroup$
– MaoWao
Jan 24 at 10:43
$begingroup$
As I recall the first proof published by Banach himself only dealt with the separable case, and was constructive (he proved sequential compactness directly). Later Alaouglu published the general theorem which relies on Tychonoff's theorem.
$endgroup$
– pitariver
Jan 24 at 10:54