Prove that F is a bounded linear functional and $||F|| _{X^*}=||w||_{infty}$. [closed]
$begingroup$
Let $(X,||cdot||)=(l^1,||cdot||_1)$, $w=(w_n)_{n geq 1} in l^{infty}$.
Define for all $x=(a_n)_{n geq 1}in l^{1}$,
$$F(x)=sum_{n geq 1}w_na_n.$$
Prove that $F$ is a bounded linear functional on $(X,||cdot||)$ and that
$||F|| _{X^*}=||w||_{infty}$.
sequences-and-series functional-analysis hilbert-spaces normed-spaces
asked Dec 9 '18 at 17:42
vladr10vladr10
102
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closed as off-topic by user10354138, Nosrati, Saad, mrtaurho, Lord_Farin Dec 23 '18 at 11:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user10354138, Nosrati, Saad, mrtaurho, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $(X,||cdot||)=(l^1,||cdot||_1)$, $w=(w_n)_{n geq 1} in l^{infty}$.
Define for all $x=(a_n)_{n geq 1}in l^{1}$,
$$F(x)=sum_{n geq 1}w_na_n.$$
Prove that $F$ is a bounded linear functional on $(X,||cdot||)$ and that
$||F|| _{X^*}=||w||_{infty}$.
sequences-and-series functional-analysis hilbert-spaces normed-spaces
asked Dec 9 '18 at 17:42
vladr10vladr10
102
$endgroup$
closed as off-topic by user10354138, Nosrati, Saad, mrtaurho, Lord_Farin Dec 23 '18 at 11:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user10354138, Nosrati, Saad, mrtaurho, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What exactly do you mean by $l^1$,$VertcdotVert_1$, and $(w_n)_{ngeq1}$?
$endgroup$
– R. Burton
Dec 9 '18 at 17:46
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@R.Burton These are classical functional analysis notations.
$endgroup$
– Rebellos
Dec 9 '18 at 17:49
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By $X^*$ you denote the dual space of $X$, right ?
$endgroup$
– Rebellos
Dec 9 '18 at 17:53
$begingroup$
Yes, that's right.
$endgroup$
– vladr10
Dec 9 '18 at 17:55
add a comment |
$begingroup$
Let $(X,||cdot||)=(l^1,||cdot||_1)$, $w=(w_n)_{n geq 1} in l^{infty}$.
Define for all $x=(a_n)_{n geq 1}in l^{1}$,
$$F(x)=sum_{n geq 1}w_na_n.$$
Prove that $F$ is a bounded linear functional on $(X,||cdot||)$ and that
$||F|| _{X^*}=||w||_{infty}$.
sequences-and-series functional-analysis hilbert-spaces normed-spaces
asked Dec 9 '18 at 17:42
vladr10vladr10
102
$endgroup$
Let $(X,||cdot||)=(l^1,||cdot||_1)$, $w=(w_n)_{n geq 1} in l^{infty}$.
Define for all $x=(a_n)_{n geq 1}in l^{1}$,
$$F(x)=sum_{n geq 1}w_na_n.$$
Prove that $F$ is a bounded linear functional on $(X,||cdot||)$ and that
$||F|| _{X^*}=||w||_{infty}$.
sequences-and-series functional-analysis hilbert-spaces normed-spaces
sequences-and-series functional-analysis hilbert-spaces normed-spaces
asked Dec 9 '18 at 17:42
vladr10vladr10
102
asked Dec 9 '18 at 17:42
vladr10vladr10
102
asked Dec 9 '18 at 17:42
vladr10vladr10
102
asked Dec 9 '18 at 17:42
vladr10vladr10
102
asked Dec 9 '18 at 17:42
vladr10vladr10
102
102
closed as off-topic by user10354138, Nosrati, Saad, mrtaurho, Lord_Farin Dec 23 '18 at 11:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user10354138, Nosrati, Saad, mrtaurho, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user10354138, Nosrati, Saad, mrtaurho, Lord_Farin Dec 23 '18 at 11:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user10354138, Nosrati, Saad, mrtaurho, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What exactly do you mean by $l^1$,$VertcdotVert_1$, and $(w_n)_{ngeq1}$?
$endgroup$
– R. Burton
Dec 9 '18 at 17:46
$begingroup$
@R.Burton These are classical functional analysis notations.
$endgroup$
– Rebellos
Dec 9 '18 at 17:49
$begingroup$
By $X^*$ you denote the dual space of $X$, right ?
$endgroup$
– Rebellos
Dec 9 '18 at 17:53
$begingroup$
Yes, that's right.
$endgroup$
– vladr10
Dec 9 '18 at 17:55
add a comment |
$begingroup$
What exactly do you mean by $l^1$,$VertcdotVert_1$, and $(w_n)_{ngeq1}$?
$endgroup$
– R. Burton
Dec 9 '18 at 17:46
$begingroup$
@R.Burton These are classical functional analysis notations.
$endgroup$
– Rebellos
Dec 9 '18 at 17:49
$begingroup$
By $X^*$ you denote the dual space of $X$, right ?
$endgroup$
– Rebellos
Dec 9 '18 at 17:53
$begingroup$
Yes, that's right.
$endgroup$
– vladr10
Dec 9 '18 at 17:55
$begingroup$
What exactly do you mean by $l^1$,$VertcdotVert_1$, and $(w_n)_{ngeq1}$?
$endgroup$
– R. Burton
Dec 9 '18 at 17:46
$begingroup$
What exactly do you mean by $l^1$,$VertcdotVert_1$, and $(w_n)_{ngeq1}$?
$endgroup$
– R. Burton
Dec 9 '18 at 17:46
$begingroup$
@R.Burton These are classical functional analysis notations.
$endgroup$
– Rebellos
Dec 9 '18 at 17:49
$begingroup$
@R.Burton These are classical functional analysis notations.
$endgroup$
– Rebellos
Dec 9 '18 at 17:49
$begingroup$
By $X^*$ you denote the dual space of $X$, right ?
$endgroup$
– Rebellos
Dec 9 '18 at 17:53
$begingroup$
By $X^*$ you denote the dual space of $X$, right ?
$endgroup$
– Rebellos
Dec 9 '18 at 17:53
$begingroup$
Yes, that's right.
$endgroup$
– vladr10
Dec 9 '18 at 17:55
$begingroup$
Yes, that's right.
$endgroup$
– vladr10
Dec 9 '18 at 17:55
add a comment |
1 Answer
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$begingroup$
For the case of linearity, take $x=(a_n) in ell^1$ and $y = (b_n) in ell^2$ and calculate $F(lambda x+y)$ with $lambda in mathbb R$. That is left as an exercise for you.
For boundedness :
$$|F(x)| = bigg|sum w_n a_n bigg| leq sum|w_na_n| leq |w_n|_inftysum |a_n| = |w_n|_infty|x|_1 $$
Thus, $F$ is a bounded linear operator with $|F| leq |w_n|_infty equiv |w|_infty$.
Now, the space $X^*$ aka the dual space of $X$ is the space of all the bounded linear functionals on $X$. But recall that a dual normed space is a Banach space When equipped with the norm : $||f||=sup{|f(x)|:||x||=1}.$ But $ell^1$ is a Banach space which means that $$|F|_{X^*} = sup{ |F(x)| : |x|_{X^*}=1} = |w_n|_infty equiv |w|_infty$$
answered Dec 9 '18 at 17:56
RebellosRebellos
14.5k31246
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the case of linearity, take $x=(a_n) in ell^1$ and $y = (b_n) in ell^2$ and calculate $F(lambda x+y)$ with $lambda in mathbb R$. That is left as an exercise for you.
For boundedness :
$$|F(x)| = bigg|sum w_n a_n bigg| leq sum|w_na_n| leq |w_n|_inftysum |a_n| = |w_n|_infty|x|_1 $$
Thus, $F$ is a bounded linear operator with $|F| leq |w_n|_infty equiv |w|_infty$.
Now, the space $X^*$ aka the dual space of $X$ is the space of all the bounded linear functionals on $X$. But recall that a dual normed space is a Banach space When equipped with the norm : $||f||=sup{|f(x)|:||x||=1}.$ But $ell^1$ is a Banach space which means that $$|F|_{X^*} = sup{ |F(x)| : |x|_{X^*}=1} = |w_n|_infty equiv |w|_infty$$
answered Dec 9 '18 at 17:56
RebellosRebellos
14.5k31246
$endgroup$
add a comment |
$begingroup$
For the case of linearity, take $x=(a_n) in ell^1$ and $y = (b_n) in ell^2$ and calculate $F(lambda x+y)$ with $lambda in mathbb R$. That is left as an exercise for you.
For boundedness :
$$|F(x)| = bigg|sum w_n a_n bigg| leq sum|w_na_n| leq |w_n|_inftysum |a_n| = |w_n|_infty|x|_1 $$
Thus, $F$ is a bounded linear operator with $|F| leq |w_n|_infty equiv |w|_infty$.
Now, the space $X^*$ aka the dual space of $X$ is the space of all the bounded linear functionals on $X$. But recall that a dual normed space is a Banach space When equipped with the norm : $||f||=sup{|f(x)|:||x||=1}.$ But $ell^1$ is a Banach space which means that $$|F|_{X^*} = sup{ |F(x)| : |x|_{X^*}=1} = |w_n|_infty equiv |w|_infty$$
answered Dec 9 '18 at 17:56
RebellosRebellos
14.5k31246
$endgroup$
add a comment |
$begingroup$
For the case of linearity, take $x=(a_n) in ell^1$ and $y = (b_n) in ell^2$ and calculate $F(lambda x+y)$ with $lambda in mathbb R$. That is left as an exercise for you.
For boundedness :
$$|F(x)| = bigg|sum w_n a_n bigg| leq sum|w_na_n| leq |w_n|_inftysum |a_n| = |w_n|_infty|x|_1 $$
Thus, $F$ is a bounded linear operator with $|F| leq |w_n|_infty equiv |w|_infty$.
Now, the space $X^*$ aka the dual space of $X$ is the space of all the bounded linear functionals on $X$. But recall that a dual normed space is a Banach space When equipped with the norm : $||f||=sup{|f(x)|:||x||=1}.$ But $ell^1$ is a Banach space which means that $$|F|_{X^*} = sup{ |F(x)| : |x|_{X^*}=1} = |w_n|_infty equiv |w|_infty$$
answered Dec 9 '18 at 17:56
RebellosRebellos
14.5k31246
$endgroup$
For the case of linearity, take $x=(a_n) in ell^1$ and $y = (b_n) in ell^2$ and calculate $F(lambda x+y)$ with $lambda in mathbb R$. That is left as an exercise for you.
For boundedness :
$$|F(x)| = bigg|sum w_n a_n bigg| leq sum|w_na_n| leq |w_n|_inftysum |a_n| = |w_n|_infty|x|_1 $$
Thus, $F$ is a bounded linear operator with $|F| leq |w_n|_infty equiv |w|_infty$.
Now, the space $X^*$ aka the dual space of $X$ is the space of all the bounded linear functionals on $X$. But recall that a dual normed space is a Banach space When equipped with the norm : $||f||=sup{|f(x)|:||x||=1}.$ But $ell^1$ is a Banach space which means that $$|F|_{X^*} = sup{ |F(x)| : |x|_{X^*}=1} = |w_n|_infty equiv |w|_infty$$
answered Dec 9 '18 at 17:56
RebellosRebellos
14.5k31246
edited Dec 9 '18 at 18:02
answered Dec 9 '18 at 17:56
RebellosRebellos
14.5k31246
answered Dec 9 '18 at 17:56
RebellosRebellos
14.5k31246
answered Dec 9 '18 at 17:56
RebellosRebellos
14.5k31246
14.5k31246
add a comment |
add a comment |
$begingroup$
What exactly do you mean by $l^1$,$VertcdotVert_1$, and $(w_n)_{ngeq1}$?
$endgroup$
– R. Burton
Dec 9 '18 at 17:46
$begingroup$
@R.Burton These are classical functional analysis notations.
$endgroup$
– Rebellos
Dec 9 '18 at 17:49
$begingroup$
By $X^*$ you denote the dual space of $X$, right ?
$endgroup$
– Rebellos
Dec 9 '18 at 17:53
$begingroup$
Yes, that's right.
$endgroup$
– vladr10
Dec 9 '18 at 17:55