Mixing
If B is a subalgebra of A, conclude that B closure is a subalgebra of A
$begingroup$
If B is a subalgebra of A, conclude that $bar{B}$ is a subalgebra of A.
This is from Real Analysis by N. L Carothers chapter 12 exercise 3. The purpose of this is to lead up to the Stone Weierstrass theorem. I found this question but I cannot understand the answer.
The first part of this question requires us to show
for f, g, h, k $in$ A $lvertlvert{fg - hk}lvertvert$ $leq$ $lvertlvert{f}lvertvert$$lvertlvert{g - k}lvertvert$ + $lvertlvert{k}lvertvert$$lvertlvert{f - h}lvertvert$. The second part required us to show that the multiplication operator is continuous.
Thanks again.
real-analysis sequences-and-series functional-analysis continuity algebras
asked Jan 30 at 22:50
MatthieuMatthieu
1007
$endgroup$
|
show 1 more comment
$begingroup$
If B is a subalgebra of A, conclude that $bar{B}$ is a subalgebra of A.
This is from Real Analysis by N. L Carothers chapter 12 exercise 3. The purpose of this is to lead up to the Stone Weierstrass theorem. I found this question but I cannot understand the answer.
The first part of this question requires us to show
for f, g, h, k $in$ A $lvertlvert{fg - hk}lvertvert$ $leq$ $lvertlvert{f}lvertvert$$lvertlvert{g - k}lvertvert$ + $lvertlvert{k}lvertvert$$lvertlvert{f - h}lvertvert$. The second part required us to show that the multiplication operator is continuous.
Thanks again.
real-analysis sequences-and-series functional-analysis continuity algebras
asked Jan 30 at 22:50
MatthieuMatthieu
1007
$endgroup$
$begingroup$
Do you not understand how to show these steps? Or do you not understand how they lead to an answer to your original question?
$endgroup$
– MaoWao
Jan 30 at 23:19
$begingroup$
@MaoWao I have proved up to continuity of multiplication. However, I do not understand how this leads to the answer to my question. Thanks.
$endgroup$
– Matthieu
Jan 31 at 6:57
$begingroup$
A map $Phi$ is continuous if and only if $Phi(overline{E})subset overline{Phi(E)}$ for all $E$. Now you can use the argument from the answer to the linked question to finish the proof.
$endgroup$
– MaoWao
Jan 31 at 9:39
$begingroup$
@MaoWao By showing continuity of the multiplication operation we have ensured that $bar{B}$ is closed under multiplication. Therefore, $bar{B}$ satisfies the properties of an algebra coupled with the fact $bar{B}$ $subseteq$ A this now makes it a subalgebra of A.
$endgroup$
– Matthieu
Jan 31 at 18:58
$begingroup$
One also has to show that $bar B$ is closed under addition and scalar multiplication. But maybe you have done this before. Anyway, the argument is very similar.
$endgroup$
– MaoWao
Feb 1 at 8:56
|
show 1 more comment
$begingroup$
If B is a subalgebra of A, conclude that $bar{B}$ is a subalgebra of A.
This is from Real Analysis by N. L Carothers chapter 12 exercise 3. The purpose of this is to lead up to the Stone Weierstrass theorem. I found this question but I cannot understand the answer.
The first part of this question requires us to show
for f, g, h, k $in$ A $lvertlvert{fg - hk}lvertvert$ $leq$ $lvertlvert{f}lvertvert$$lvertlvert{g - k}lvertvert$ + $lvertlvert{k}lvertvert$$lvertlvert{f - h}lvertvert$. The second part required us to show that the multiplication operator is continuous.
Thanks again.
real-analysis sequences-and-series functional-analysis continuity algebras
asked Jan 30 at 22:50
MatthieuMatthieu
1007
$endgroup$
If B is a subalgebra of A, conclude that $bar{B}$ is a subalgebra of A.
This is from Real Analysis by N. L Carothers chapter 12 exercise 3. The purpose of this is to lead up to the Stone Weierstrass theorem. I found this question but I cannot understand the answer.
The first part of this question requires us to show
for f, g, h, k $in$ A $lvertlvert{fg - hk}lvertvert$ $leq$ $lvertlvert{f}lvertvert$$lvertlvert{g - k}lvertvert$ + $lvertlvert{k}lvertvert$$lvertlvert{f - h}lvertvert$. The second part required us to show that the multiplication operator is continuous.
Thanks again.
real-analysis sequences-and-series functional-analysis continuity algebras
real-analysis sequences-and-series functional-analysis continuity algebras
asked Jan 30 at 22:50
MatthieuMatthieu
1007
asked Jan 30 at 22:50
MatthieuMatthieu
1007
asked Jan 30 at 22:50
MatthieuMatthieu
1007
asked Jan 30 at 22:50
MatthieuMatthieu
1007
asked Jan 30 at 22:50
MatthieuMatthieu
1007
1007
$begingroup$
Do you not understand how to show these steps? Or do you not understand how they lead to an answer to your original question?
$endgroup$
– MaoWao
Jan 30 at 23:19
$begingroup$
@MaoWao I have proved up to continuity of multiplication. However, I do not understand how this leads to the answer to my question. Thanks.
$endgroup$
– Matthieu
Jan 31 at 6:57
$begingroup$
A map $Phi$ is continuous if and only if $Phi(overline{E})subset overline{Phi(E)}$ for all $E$. Now you can use the argument from the answer to the linked question to finish the proof.
$endgroup$
– MaoWao
Jan 31 at 9:39
$begingroup$
@MaoWao By showing continuity of the multiplication operation we have ensured that $bar{B}$ is closed under multiplication. Therefore, $bar{B}$ satisfies the properties of an algebra coupled with the fact $bar{B}$ $subseteq$ A this now makes it a subalgebra of A.
$endgroup$
– Matthieu
Jan 31 at 18:58
$begingroup$
One also has to show that $bar B$ is closed under addition and scalar multiplication. But maybe you have done this before. Anyway, the argument is very similar.
$endgroup$
– MaoWao
Feb 1 at 8:56
|
show 1 more comment
$begingroup$
Do you not understand how to show these steps? Or do you not understand how they lead to an answer to your original question?
$endgroup$
– MaoWao
Jan 30 at 23:19
$begingroup$
@MaoWao I have proved up to continuity of multiplication. However, I do not understand how this leads to the answer to my question. Thanks.
$endgroup$
– Matthieu
Jan 31 at 6:57
$begingroup$
A map $Phi$ is continuous if and only if $Phi(overline{E})subset overline{Phi(E)}$ for all $E$. Now you can use the argument from the answer to the linked question to finish the proof.
$endgroup$
– MaoWao
Jan 31 at 9:39
$begingroup$
@MaoWao By showing continuity of the multiplication operation we have ensured that $bar{B}$ is closed under multiplication. Therefore, $bar{B}$ satisfies the properties of an algebra coupled with the fact $bar{B}$ $subseteq$ A this now makes it a subalgebra of A.
$endgroup$
– Matthieu
Jan 31 at 18:58
$begingroup$
One also has to show that $bar B$ is closed under addition and scalar multiplication. But maybe you have done this before. Anyway, the argument is very similar.
$endgroup$
– MaoWao
Feb 1 at 8:56
$begingroup$
Do you not understand how to show these steps? Or do you not understand how they lead to an answer to your original question?
$endgroup$
– MaoWao
Jan 30 at 23:19
$begingroup$
Do you not understand how to show these steps? Or do you not understand how they lead to an answer to your original question?
$endgroup$
– MaoWao
Jan 30 at 23:19
$begingroup$
@MaoWao I have proved up to continuity of multiplication. However, I do not understand how this leads to the answer to my question. Thanks.
$endgroup$
– Matthieu
Jan 31 at 6:57
$begingroup$
@MaoWao I have proved up to continuity of multiplication. However, I do not understand how this leads to the answer to my question. Thanks.
$endgroup$
– Matthieu
Jan 31 at 6:57
$begingroup$
A map $Phi$ is continuous if and only if $Phi(overline{E})subset overline{Phi(E)}$ for all $E$. Now you can use the argument from the answer to the linked question to finish the proof.
$endgroup$
– MaoWao
Jan 31 at 9:39
$begingroup$
A map $Phi$ is continuous if and only if $Phi(overline{E})subset overline{Phi(E)}$ for all $E$. Now you can use the argument from the answer to the linked question to finish the proof.
$endgroup$
– MaoWao
Jan 31 at 9:39
$begingroup$
@MaoWao By showing continuity of the multiplication operation we have ensured that $bar{B}$ is closed under multiplication. Therefore, $bar{B}$ satisfies the properties of an algebra coupled with the fact $bar{B}$ $subseteq$ A this now makes it a subalgebra of A.
$endgroup$
– Matthieu
Jan 31 at 18:58
$begingroup$
@MaoWao By showing continuity of the multiplication operation we have ensured that $bar{B}$ is closed under multiplication. Therefore, $bar{B}$ satisfies the properties of an algebra coupled with the fact $bar{B}$ $subseteq$ A this now makes it a subalgebra of A.
$endgroup$
– Matthieu
Jan 31 at 18:58
$begingroup$
One also has to show that $bar B$ is closed under addition and scalar multiplication. But maybe you have done this before. Anyway, the argument is very similar.
$endgroup$
– MaoWao
Feb 1 at 8:56
$begingroup$
One also has to show that $bar B$ is closed under addition and scalar multiplication. But maybe you have done this before. Anyway, the argument is very similar.
$endgroup$
– MaoWao
Feb 1 at 8:56
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
$fg-hk=f(g-k)+(f-h)k$. Using triangle inequality and the fact that $|xy|leq |x|y|$ we get $|fg-hk| leq |f(g-k)|+|(f-h)k| leq |f| |g-k|+|f-h| |k|$. To prove that the multiplication map $(f,g) to fg$ is continuous w ehave to show that $f_n to f$ adn $g_n to g$ together imply $f_ng_n to fg$. For this note that $ |f_ng_n-fg| leq |f_n| |g_n-g|+|f_n-f| |g|$. Also $|f_n| leq |f_n-f| +|f|<1+|f|$ for $n$ sufficiently large. Now can you complete the argument?
answered Jan 31 at 6:29
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
$begingroup$
Yes, this is what I got up to proving continuity of multiplication. However, I just do not see how continuity of multiplication coupled with your last inequality of ||$f_{n}$|| $leq$ 1 + $||f_{n}||$ leads to $bar{B}$ being a subset of A. I feel like I am missing something fundamental in my understanding. Thank you for the help.
$endgroup$
– Matthieu
Jan 31 at 6:50
$begingroup$
@MatteoLepur The fact that $overline {B}$ is a subset of $A$ requires no proof. By definition $overline {B}$ is the closure of $B$ in $A$, so it is a subset of $A$.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 7:14
$begingroup$
Therefore, the purpose of proving continuity of multiplication is to ensure that the set $bar{B}$ is closed under multiplication? Hence, we have that $bar{B}$ satisfies all the properties of being an algebra making it an subalgebra of A, the ambient space?
$endgroup$
– Matthieu
Jan 31 at 18:53
$begingroup$
@MatteoLepur You are right.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:11
$begingroup$
Awesome, thank you very much!
$endgroup$
– Matthieu
Feb 1 at 1:14
add a comment |
Your Answer
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1 Answer
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votes
1 Answer
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oldest
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$begingroup$
$fg-hk=f(g-k)+(f-h)k$. Using triangle inequality and the fact that $|xy|leq |x|y|$ we get $|fg-hk| leq |f(g-k)|+|(f-h)k| leq |f| |g-k|+|f-h| |k|$. To prove that the multiplication map $(f,g) to fg$ is continuous w ehave to show that $f_n to f$ adn $g_n to g$ together imply $f_ng_n to fg$. For this note that $ |f_ng_n-fg| leq |f_n| |g_n-g|+|f_n-f| |g|$. Also $|f_n| leq |f_n-f| +|f|<1+|f|$ for $n$ sufficiently large. Now can you complete the argument?
answered Jan 31 at 6:29
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
$begingroup$
Yes, this is what I got up to proving continuity of multiplication. However, I just do not see how continuity of multiplication coupled with your last inequality of ||$f_{n}$|| $leq$ 1 + $||f_{n}||$ leads to $bar{B}$ being a subset of A. I feel like I am missing something fundamental in my understanding. Thank you for the help.
$endgroup$
– Matthieu
Jan 31 at 6:50
$begingroup$
@MatteoLepur The fact that $overline {B}$ is a subset of $A$ requires no proof. By definition $overline {B}$ is the closure of $B$ in $A$, so it is a subset of $A$.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 7:14
$begingroup$
Therefore, the purpose of proving continuity of multiplication is to ensure that the set $bar{B}$ is closed under multiplication? Hence, we have that $bar{B}$ satisfies all the properties of being an algebra making it an subalgebra of A, the ambient space?
$endgroup$
– Matthieu
Jan 31 at 18:53
$begingroup$
@MatteoLepur You are right.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:11
$begingroup$
Awesome, thank you very much!
$endgroup$
– Matthieu
Feb 1 at 1:14
add a comment |
$begingroup$
$fg-hk=f(g-k)+(f-h)k$. Using triangle inequality and the fact that $|xy|leq |x|y|$ we get $|fg-hk| leq |f(g-k)|+|(f-h)k| leq |f| |g-k|+|f-h| |k|$. To prove that the multiplication map $(f,g) to fg$ is continuous w ehave to show that $f_n to f$ adn $g_n to g$ together imply $f_ng_n to fg$. For this note that $ |f_ng_n-fg| leq |f_n| |g_n-g|+|f_n-f| |g|$. Also $|f_n| leq |f_n-f| +|f|<1+|f|$ for $n$ sufficiently large. Now can you complete the argument?
answered Jan 31 at 6:29
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
$begingroup$
Yes, this is what I got up to proving continuity of multiplication. However, I just do not see how continuity of multiplication coupled with your last inequality of ||$f_{n}$|| $leq$ 1 + $||f_{n}||$ leads to $bar{B}$ being a subset of A. I feel like I am missing something fundamental in my understanding. Thank you for the help.
$endgroup$
– Matthieu
Jan 31 at 6:50
$begingroup$
@MatteoLepur The fact that $overline {B}$ is a subset of $A$ requires no proof. By definition $overline {B}$ is the closure of $B$ in $A$, so it is a subset of $A$.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 7:14
$begingroup$
Therefore, the purpose of proving continuity of multiplication is to ensure that the set $bar{B}$ is closed under multiplication? Hence, we have that $bar{B}$ satisfies all the properties of being an algebra making it an subalgebra of A, the ambient space?
$endgroup$
– Matthieu
Jan 31 at 18:53
$begingroup$
@MatteoLepur You are right.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:11
$begingroup$
Awesome, thank you very much!
$endgroup$
– Matthieu
Feb 1 at 1:14
add a comment |
$begingroup$
$fg-hk=f(g-k)+(f-h)k$. Using triangle inequality and the fact that $|xy|leq |x|y|$ we get $|fg-hk| leq |f(g-k)|+|(f-h)k| leq |f| |g-k|+|f-h| |k|$. To prove that the multiplication map $(f,g) to fg$ is continuous w ehave to show that $f_n to f$ adn $g_n to g$ together imply $f_ng_n to fg$. For this note that $ |f_ng_n-fg| leq |f_n| |g_n-g|+|f_n-f| |g|$. Also $|f_n| leq |f_n-f| +|f|<1+|f|$ for $n$ sufficiently large. Now can you complete the argument?
answered Jan 31 at 6:29
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
$fg-hk=f(g-k)+(f-h)k$. Using triangle inequality and the fact that $|xy|leq |x|y|$ we get $|fg-hk| leq |f(g-k)|+|(f-h)k| leq |f| |g-k|+|f-h| |k|$. To prove that the multiplication map $(f,g) to fg$ is continuous w ehave to show that $f_n to f$ adn $g_n to g$ together imply $f_ng_n to fg$. For this note that $ |f_ng_n-fg| leq |f_n| |g_n-g|+|f_n-f| |g|$. Also $|f_n| leq |f_n-f| +|f|<1+|f|$ for $n$ sufficiently large. Now can you complete the argument?
answered Jan 31 at 6:29
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
answered Jan 31 at 6:29
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
answered Jan 31 at 6:29
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
answered Jan 31 at 6:29
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
72.6k53170
$begingroup$
Yes, this is what I got up to proving continuity of multiplication. However, I just do not see how continuity of multiplication coupled with your last inequality of ||$f_{n}$|| $leq$ 1 + $||f_{n}||$ leads to $bar{B}$ being a subset of A. I feel like I am missing something fundamental in my understanding. Thank you for the help.
$endgroup$
– Matthieu
Jan 31 at 6:50
$begingroup$
@MatteoLepur The fact that $overline {B}$ is a subset of $A$ requires no proof. By definition $overline {B}$ is the closure of $B$ in $A$, so it is a subset of $A$.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 7:14
$begingroup$
Therefore, the purpose of proving continuity of multiplication is to ensure that the set $bar{B}$ is closed under multiplication? Hence, we have that $bar{B}$ satisfies all the properties of being an algebra making it an subalgebra of A, the ambient space?
$endgroup$
– Matthieu
Jan 31 at 18:53
$begingroup$
@MatteoLepur You are right.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:11
$begingroup$
Awesome, thank you very much!
$endgroup$
– Matthieu
Feb 1 at 1:14
add a comment |
$begingroup$
Yes, this is what I got up to proving continuity of multiplication. However, I just do not see how continuity of multiplication coupled with your last inequality of ||$f_{n}$|| $leq$ 1 + $||f_{n}||$ leads to $bar{B}$ being a subset of A. I feel like I am missing something fundamental in my understanding. Thank you for the help.
$endgroup$
– Matthieu
Jan 31 at 6:50
$begingroup$
@MatteoLepur The fact that $overline {B}$ is a subset of $A$ requires no proof. By definition $overline {B}$ is the closure of $B$ in $A$, so it is a subset of $A$.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 7:14
$begingroup$
Therefore, the purpose of proving continuity of multiplication is to ensure that the set $bar{B}$ is closed under multiplication? Hence, we have that $bar{B}$ satisfies all the properties of being an algebra making it an subalgebra of A, the ambient space?
$endgroup$
– Matthieu
Jan 31 at 18:53
$begingroup$
@MatteoLepur You are right.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:11
$begingroup$
Awesome, thank you very much!
$endgroup$
– Matthieu
Feb 1 at 1:14
$begingroup$
Yes, this is what I got up to proving continuity of multiplication. However, I just do not see how continuity of multiplication coupled with your last inequality of ||$f_{n}$|| $leq$ 1 + $||f_{n}||$ leads to $bar{B}$ being a subset of A. I feel like I am missing something fundamental in my understanding. Thank you for the help.
$endgroup$
– Matthieu
Jan 31 at 6:50
$begingroup$
Yes, this is what I got up to proving continuity of multiplication. However, I just do not see how continuity of multiplication coupled with your last inequality of ||$f_{n}$|| $leq$ 1 + $||f_{n}||$ leads to $bar{B}$ being a subset of A. I feel like I am missing something fundamental in my understanding. Thank you for the help.
$endgroup$
– Matthieu
Jan 31 at 6:50
$begingroup$
@MatteoLepur The fact that $overline {B}$ is a subset of $A$ requires no proof. By definition $overline {B}$ is the closure of $B$ in $A$, so it is a subset of $A$.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 7:14
$begingroup$
@MatteoLepur The fact that $overline {B}$ is a subset of $A$ requires no proof. By definition $overline {B}$ is the closure of $B$ in $A$, so it is a subset of $A$.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 7:14
$begingroup$
Therefore, the purpose of proving continuity of multiplication is to ensure that the set $bar{B}$ is closed under multiplication? Hence, we have that $bar{B}$ satisfies all the properties of being an algebra making it an subalgebra of A, the ambient space?
$endgroup$
– Matthieu
Jan 31 at 18:53
$begingroup$
Therefore, the purpose of proving continuity of multiplication is to ensure that the set $bar{B}$ is closed under multiplication? Hence, we have that $bar{B}$ satisfies all the properties of being an algebra making it an subalgebra of A, the ambient space?
$endgroup$
– Matthieu
Jan 31 at 18:53
$begingroup$
@MatteoLepur You are right.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:11
$begingroup$
@MatteoLepur You are right.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:11
$begingroup$
Awesome, thank you very much!
$endgroup$
– Matthieu
Feb 1 at 1:14
$begingroup$
Awesome, thank you very much!
$endgroup$
– Matthieu
Feb 1 at 1:14
add a comment |
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$begingroup$
Do you not understand how to show these steps? Or do you not understand how they lead to an answer to your original question?
$endgroup$
– MaoWao
Jan 30 at 23:19
$begingroup$
@MaoWao I have proved up to continuity of multiplication. However, I do not understand how this leads to the answer to my question. Thanks.
$endgroup$
– Matthieu
Jan 31 at 6:57
$begingroup$
A map $Phi$ is continuous if and only if $Phi(overline{E})subset overline{Phi(E)}$ for all $E$. Now you can use the argument from the answer to the linked question to finish the proof.
$endgroup$
– MaoWao
Jan 31 at 9:39
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@MaoWao By showing continuity of the multiplication operation we have ensured that $bar{B}$ is closed under multiplication. Therefore, $bar{B}$ satisfies the properties of an algebra coupled with the fact $bar{B}$ $subseteq$ A this now makes it a subalgebra of A.
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– Matthieu
Jan 31 at 18:58
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One also has to show that $bar B$ is closed under addition and scalar multiplication. But maybe you have done this before. Anyway, the argument is very similar.
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– MaoWao
Feb 1 at 8:56