Mixing
$B$ Banach $implies B^*$ Banach.
$begingroup$
Since we only need to check that $B^*$ is complete, we should prove Cauchy sequence $lbrace l_n rbrace$ converges.
In general idea, $$|(l-l_n)(f)| leq |(l-l_m)(f)| + |(l_m-l_n)(f)| leq |(l-l_m)(f)| + epsilon/2||f||$$
then let m be large enough. We have $$|(l-l_n)(f)| leq epsilon ||f||$$ And we finish the proof.
(This proof is in Stein's book Functional Analysis)
However, can I prove it in this way? I mean prove Cauchy sequence $lbrace l_n rbrace$ converges.
Can I use the definition of norm that $$||l-l_n||=underset{||f||=1}{sup}|(l-l_n)(f)|$$
and due to $|(l-l_n)(f)|$ converges to $0$ when $n$ goes to infinity, we have $l_n$ converges to $l$.
It seems to be wrong, but I don't know why.
functional-analysis banach-spaces dual-spaces
asked Jan 14 at 22:07
Busy ZhuBusy Zhu
143
$endgroup$
add a comment |
$begingroup$
Since we only need to check that $B^*$ is complete, we should prove Cauchy sequence $lbrace l_n rbrace$ converges.
In general idea, $$|(l-l_n)(f)| leq |(l-l_m)(f)| + |(l_m-l_n)(f)| leq |(l-l_m)(f)| + epsilon/2||f||$$
then let m be large enough. We have $$|(l-l_n)(f)| leq epsilon ||f||$$ And we finish the proof.
(This proof is in Stein's book Functional Analysis)
However, can I prove it in this way? I mean prove Cauchy sequence $lbrace l_n rbrace$ converges.
Can I use the definition of norm that $$||l-l_n||=underset{||f||=1}{sup}|(l-l_n)(f)|$$
and due to $|(l-l_n)(f)|$ converges to $0$ when $n$ goes to infinity, we have $l_n$ converges to $l$.
It seems to be wrong, but I don't know why.
functional-analysis banach-spaces dual-spaces
asked Jan 14 at 22:07
Busy ZhuBusy Zhu
143
$endgroup$
$begingroup$
The implication you're looking for is true but a bit strange to write down. If $X$ and $Y$ are normed vector spaces then the space of bounded, linear operators from $X$ to $Y$, $L(X,Y)$ is complete (and hence a Banach space) if and only if $Y$ is complete. $X^* = L(B, mathbb{R})$ and so is a Banach space since $mathbb{R}$ is complete. This doesn't require $X$ to be complete.
$endgroup$
– Rhys Steele
Jan 14 at 22:37
$begingroup$
Thank you for your reply! It's correct? But why does the book not use this simpler idea? I think the idea is probably wrong.
$endgroup$
– Busy Zhu
Jan 14 at 23:01
$begingroup$
If I'm honest I find your attempt confusing. Given a Cauchy sequence ${l_n}_{n geq 0}$ in $B^*$ you don't a priori have a good candidate for its limit $l$, so the $l$ you write down isn't defined here. Further, writing a bound like $|(l-l_n)(f)| leq |(l-l_m)(f)| + frac{varepsilon}{2} |f|$ isn't particularly helpful because you'll still be left to bound $|(l-l_m)(f)|$ in terms of $|f|$. For this proof to work you'd need to be able to pick $m$ independent of $f$ and if you could do that you'd just bound $|(l-l_n)(f)|$ for large enough $n$ straight away rather than doing this first.
$endgroup$
– Rhys Steele
Jan 14 at 23:05
$begingroup$
The usual technique is to check that for $f in B$, $(l_n(f))$ is a Cauchy sequence in $mathbb{R}$ and so converges to some number $L(f) in mathbb{R}$. You can then define a map $l: B to mathbb{R}$ by $l(f) = L(f)$. It's then not to hard to check that $l$ is linear by properties of limits. Then you check that $l$ in fact must be a bounded linear map and finally you show that $l_n to l$ in $B^*$ by sending $m to infty$ in $|l_n(f) - l_m(f)| leq varepsilon |f|$ for suitably large $m,n$.
$endgroup$
– Rhys Steele
Jan 14 at 23:10
add a comment |
$begingroup$
Since we only need to check that $B^*$ is complete, we should prove Cauchy sequence $lbrace l_n rbrace$ converges.
In general idea, $$|(l-l_n)(f)| leq |(l-l_m)(f)| + |(l_m-l_n)(f)| leq |(l-l_m)(f)| + epsilon/2||f||$$
then let m be large enough. We have $$|(l-l_n)(f)| leq epsilon ||f||$$ And we finish the proof.
(This proof is in Stein's book Functional Analysis)
However, can I prove it in this way? I mean prove Cauchy sequence $lbrace l_n rbrace$ converges.
Can I use the definition of norm that $$||l-l_n||=underset{||f||=1}{sup}|(l-l_n)(f)|$$
and due to $|(l-l_n)(f)|$ converges to $0$ when $n$ goes to infinity, we have $l_n$ converges to $l$.
It seems to be wrong, but I don't know why.
functional-analysis banach-spaces dual-spaces
asked Jan 14 at 22:07
Busy ZhuBusy Zhu
143
$endgroup$
Since we only need to check that $B^*$ is complete, we should prove Cauchy sequence $lbrace l_n rbrace$ converges.
In general idea, $$|(l-l_n)(f)| leq |(l-l_m)(f)| + |(l_m-l_n)(f)| leq |(l-l_m)(f)| + epsilon/2||f||$$
then let m be large enough. We have $$|(l-l_n)(f)| leq epsilon ||f||$$ And we finish the proof.
(This proof is in Stein's book Functional Analysis)
However, can I prove it in this way? I mean prove Cauchy sequence $lbrace l_n rbrace$ converges.
Can I use the definition of norm that $$||l-l_n||=underset{||f||=1}{sup}|(l-l_n)(f)|$$
and due to $|(l-l_n)(f)|$ converges to $0$ when $n$ goes to infinity, we have $l_n$ converges to $l$.
It seems to be wrong, but I don't know why.
functional-analysis banach-spaces dual-spaces
functional-analysis banach-spaces dual-spaces
asked Jan 14 at 22:07
Busy ZhuBusy Zhu
143
asked Jan 14 at 22:07
Busy ZhuBusy Zhu
143
edited Jan 14 at 23:09
Busy Zhu
asked Jan 14 at 22:07
Busy ZhuBusy Zhu
143
asked Jan 14 at 22:07
Busy ZhuBusy Zhu
143
asked Jan 14 at 22:07
Busy ZhuBusy Zhu
143
143
$begingroup$
The implication you're looking for is true but a bit strange to write down. If $X$ and $Y$ are normed vector spaces then the space of bounded, linear operators from $X$ to $Y$, $L(X,Y)$ is complete (and hence a Banach space) if and only if $Y$ is complete. $X^* = L(B, mathbb{R})$ and so is a Banach space since $mathbb{R}$ is complete. This doesn't require $X$ to be complete.
$endgroup$
– Rhys Steele
Jan 14 at 22:37
$begingroup$
Thank you for your reply! It's correct? But why does the book not use this simpler idea? I think the idea is probably wrong.
$endgroup$
– Busy Zhu
Jan 14 at 23:01
$begingroup$
If I'm honest I find your attempt confusing. Given a Cauchy sequence ${l_n}_{n geq 0}$ in $B^*$ you don't a priori have a good candidate for its limit $l$, so the $l$ you write down isn't defined here. Further, writing a bound like $|(l-l_n)(f)| leq |(l-l_m)(f)| + frac{varepsilon}{2} |f|$ isn't particularly helpful because you'll still be left to bound $|(l-l_m)(f)|$ in terms of $|f|$. For this proof to work you'd need to be able to pick $m$ independent of $f$ and if you could do that you'd just bound $|(l-l_n)(f)|$ for large enough $n$ straight away rather than doing this first.
$endgroup$
– Rhys Steele
Jan 14 at 23:05
$begingroup$
The usual technique is to check that for $f in B$, $(l_n(f))$ is a Cauchy sequence in $mathbb{R}$ and so converges to some number $L(f) in mathbb{R}$. You can then define a map $l: B to mathbb{R}$ by $l(f) = L(f)$. It's then not to hard to check that $l$ is linear by properties of limits. Then you check that $l$ in fact must be a bounded linear map and finally you show that $l_n to l$ in $B^*$ by sending $m to infty$ in $|l_n(f) - l_m(f)| leq varepsilon |f|$ for suitably large $m,n$.
$endgroup$
– Rhys Steele
Jan 14 at 23:10
add a comment |
$begingroup$
The implication you're looking for is true but a bit strange to write down. If $X$ and $Y$ are normed vector spaces then the space of bounded, linear operators from $X$ to $Y$, $L(X,Y)$ is complete (and hence a Banach space) if and only if $Y$ is complete. $X^* = L(B, mathbb{R})$ and so is a Banach space since $mathbb{R}$ is complete. This doesn't require $X$ to be complete.
$endgroup$
– Rhys Steele
Jan 14 at 22:37
$begingroup$
Thank you for your reply! It's correct? But why does the book not use this simpler idea? I think the idea is probably wrong.
$endgroup$
– Busy Zhu
Jan 14 at 23:01
$begingroup$
If I'm honest I find your attempt confusing. Given a Cauchy sequence ${l_n}_{n geq 0}$ in $B^*$ you don't a priori have a good candidate for its limit $l$, so the $l$ you write down isn't defined here. Further, writing a bound like $|(l-l_n)(f)| leq |(l-l_m)(f)| + frac{varepsilon}{2} |f|$ isn't particularly helpful because you'll still be left to bound $|(l-l_m)(f)|$ in terms of $|f|$. For this proof to work you'd need to be able to pick $m$ independent of $f$ and if you could do that you'd just bound $|(l-l_n)(f)|$ for large enough $n$ straight away rather than doing this first.
$endgroup$
– Rhys Steele
Jan 14 at 23:05
$begingroup$
The usual technique is to check that for $f in B$, $(l_n(f))$ is a Cauchy sequence in $mathbb{R}$ and so converges to some number $L(f) in mathbb{R}$. You can then define a map $l: B to mathbb{R}$ by $l(f) = L(f)$. It's then not to hard to check that $l$ is linear by properties of limits. Then you check that $l$ in fact must be a bounded linear map and finally you show that $l_n to l$ in $B^*$ by sending $m to infty$ in $|l_n(f) - l_m(f)| leq varepsilon |f|$ for suitably large $m,n$.
$endgroup$
– Rhys Steele
Jan 14 at 23:10
$begingroup$
The implication you're looking for is true but a bit strange to write down. If $X$ and $Y$ are normed vector spaces then the space of bounded, linear operators from $X$ to $Y$, $L(X,Y)$ is complete (and hence a Banach space) if and only if $Y$ is complete. $X^* = L(B, mathbb{R})$ and so is a Banach space since $mathbb{R}$ is complete. This doesn't require $X$ to be complete.
$endgroup$
– Rhys Steele
Jan 14 at 22:37
$begingroup$
The implication you're looking for is true but a bit strange to write down. If $X$ and $Y$ are normed vector spaces then the space of bounded, linear operators from $X$ to $Y$, $L(X,Y)$ is complete (and hence a Banach space) if and only if $Y$ is complete. $X^* = L(B, mathbb{R})$ and so is a Banach space since $mathbb{R}$ is complete. This doesn't require $X$ to be complete.
$endgroup$
– Rhys Steele
Jan 14 at 22:37
$begingroup$
Thank you for your reply! It's correct? But why does the book not use this simpler idea? I think the idea is probably wrong.
$endgroup$
– Busy Zhu
Jan 14 at 23:01
$begingroup$
Thank you for your reply! It's correct? But why does the book not use this simpler idea? I think the idea is probably wrong.
$endgroup$
– Busy Zhu
Jan 14 at 23:01
$begingroup$
If I'm honest I find your attempt confusing. Given a Cauchy sequence ${l_n}_{n geq 0}$ in $B^*$ you don't a priori have a good candidate for its limit $l$, so the $l$ you write down isn't defined here. Further, writing a bound like $|(l-l_n)(f)| leq |(l-l_m)(f)| + frac{varepsilon}{2} |f|$ isn't particularly helpful because you'll still be left to bound $|(l-l_m)(f)|$ in terms of $|f|$. For this proof to work you'd need to be able to pick $m$ independent of $f$ and if you could do that you'd just bound $|(l-l_n)(f)|$ for large enough $n$ straight away rather than doing this first.
$endgroup$
– Rhys Steele
Jan 14 at 23:05
$begingroup$
If I'm honest I find your attempt confusing. Given a Cauchy sequence ${l_n}_{n geq 0}$ in $B^*$ you don't a priori have a good candidate for its limit $l$, so the $l$ you write down isn't defined here. Further, writing a bound like $|(l-l_n)(f)| leq |(l-l_m)(f)| + frac{varepsilon}{2} |f|$ isn't particularly helpful because you'll still be left to bound $|(l-l_m)(f)|$ in terms of $|f|$. For this proof to work you'd need to be able to pick $m$ independent of $f$ and if you could do that you'd just bound $|(l-l_n)(f)|$ for large enough $n$ straight away rather than doing this first.
$endgroup$
– Rhys Steele
Jan 14 at 23:05
$begingroup$
The usual technique is to check that for $f in B$, $(l_n(f))$ is a Cauchy sequence in $mathbb{R}$ and so converges to some number $L(f) in mathbb{R}$. You can then define a map $l: B to mathbb{R}$ by $l(f) = L(f)$. It's then not to hard to check that $l$ is linear by properties of limits. Then you check that $l$ in fact must be a bounded linear map and finally you show that $l_n to l$ in $B^*$ by sending $m to infty$ in $|l_n(f) - l_m(f)| leq varepsilon |f|$ for suitably large $m,n$.
$endgroup$
– Rhys Steele
Jan 14 at 23:10
$begingroup$
The usual technique is to check that for $f in B$, $(l_n(f))$ is a Cauchy sequence in $mathbb{R}$ and so converges to some number $L(f) in mathbb{R}$. You can then define a map $l: B to mathbb{R}$ by $l(f) = L(f)$. It's then not to hard to check that $l$ is linear by properties of limits. Then you check that $l$ in fact must be a bounded linear map and finally you show that $l_n to l$ in $B^*$ by sending $m to infty$ in $|l_n(f) - l_m(f)| leq varepsilon |f|$ for suitably large $m,n$.
$endgroup$
– Rhys Steele
Jan 14 at 23:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint :
The space $B^*$ is the dual space of $B$, which is the space that contains all linear functionals such that $f : B to mathbb R$.
But, note that $mathbb R$ is a complete space with respect to its metric (norm), the absolute value. That means that $mathbb R$ is Banach.
answered Jan 14 at 22:26
RebellosRebellos
14.8k31248
$endgroup$
$begingroup$
Thank you for your reply. But it seems to not match with my question.
$endgroup$
– Busy Zhu
Jan 14 at 22:34
$begingroup$
@BusyZhu You want to prove that $B^*$ is a Banach space. All you need to take into account is that $mathbb R$ is complete, thus Banach. That means that if $x in B^*$ and is Cauchy, then it converges and the proof is really straightforward.
$endgroup$
– Rebellos
Jan 14 at 22:38
$begingroup$
thanks again! But actually what I want to know is the way I prove it is correct or not. The idea of using definition is straightforward but seems to be weird.
$endgroup$
– Busy Zhu
Jan 14 at 22:59
add a comment |
Your Answer
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Hint :
The space $B^*$ is the dual space of $B$, which is the space that contains all linear functionals such that $f : B to mathbb R$.
But, note that $mathbb R$ is a complete space with respect to its metric (norm), the absolute value. That means that $mathbb R$ is Banach.
answered Jan 14 at 22:26
RebellosRebellos
14.8k31248
$endgroup$
$begingroup$
Thank you for your reply. But it seems to not match with my question.
$endgroup$
– Busy Zhu
Jan 14 at 22:34
$begingroup$
@BusyZhu You want to prove that $B^*$ is a Banach space. All you need to take into account is that $mathbb R$ is complete, thus Banach. That means that if $x in B^*$ and is Cauchy, then it converges and the proof is really straightforward.
$endgroup$
– Rebellos
Jan 14 at 22:38
$begingroup$
thanks again! But actually what I want to know is the way I prove it is correct or not. The idea of using definition is straightforward but seems to be weird.
$endgroup$
– Busy Zhu
Jan 14 at 22:59
add a comment |
$begingroup$
Hint :
The space $B^*$ is the dual space of $B$, which is the space that contains all linear functionals such that $f : B to mathbb R$.
But, note that $mathbb R$ is a complete space with respect to its metric (norm), the absolute value. That means that $mathbb R$ is Banach.
answered Jan 14 at 22:26
RebellosRebellos
14.8k31248
$endgroup$
$begingroup$
Thank you for your reply. But it seems to not match with my question.
$endgroup$
– Busy Zhu
Jan 14 at 22:34
$begingroup$
@BusyZhu You want to prove that $B^*$ is a Banach space. All you need to take into account is that $mathbb R$ is complete, thus Banach. That means that if $x in B^*$ and is Cauchy, then it converges and the proof is really straightforward.
$endgroup$
– Rebellos
Jan 14 at 22:38
$begingroup$
thanks again! But actually what I want to know is the way I prove it is correct or not. The idea of using definition is straightforward but seems to be weird.
$endgroup$
– Busy Zhu
Jan 14 at 22:59
add a comment |
$begingroup$
Hint :
The space $B^*$ is the dual space of $B$, which is the space that contains all linear functionals such that $f : B to mathbb R$.
But, note that $mathbb R$ is a complete space with respect to its metric (norm), the absolute value. That means that $mathbb R$ is Banach.
answered Jan 14 at 22:26
RebellosRebellos
14.8k31248
$endgroup$
Hint :
The space $B^*$ is the dual space of $B$, which is the space that contains all linear functionals such that $f : B to mathbb R$.
But, note that $mathbb R$ is a complete space with respect to its metric (norm), the absolute value. That means that $mathbb R$ is Banach.
answered Jan 14 at 22:26
RebellosRebellos
14.8k31248
answered Jan 14 at 22:26
RebellosRebellos
14.8k31248
answered Jan 14 at 22:26
RebellosRebellos
14.8k31248
answered Jan 14 at 22:26
RebellosRebellos
14.8k31248
14.8k31248
$begingroup$
Thank you for your reply. But it seems to not match with my question.
$endgroup$
– Busy Zhu
Jan 14 at 22:34
$begingroup$
@BusyZhu You want to prove that $B^*$ is a Banach space. All you need to take into account is that $mathbb R$ is complete, thus Banach. That means that if $x in B^*$ and is Cauchy, then it converges and the proof is really straightforward.
$endgroup$
– Rebellos
Jan 14 at 22:38
$begingroup$
thanks again! But actually what I want to know is the way I prove it is correct or not. The idea of using definition is straightforward but seems to be weird.
$endgroup$
– Busy Zhu
Jan 14 at 22:59
add a comment |
$begingroup$
Thank you for your reply. But it seems to not match with my question.
$endgroup$
– Busy Zhu
Jan 14 at 22:34
$begingroup$
@BusyZhu You want to prove that $B^*$ is a Banach space. All you need to take into account is that $mathbb R$ is complete, thus Banach. That means that if $x in B^*$ and is Cauchy, then it converges and the proof is really straightforward.
$endgroup$
– Rebellos
Jan 14 at 22:38
$begingroup$
thanks again! But actually what I want to know is the way I prove it is correct or not. The idea of using definition is straightforward but seems to be weird.
$endgroup$
– Busy Zhu
Jan 14 at 22:59
$begingroup$
Thank you for your reply. But it seems to not match with my question.
$endgroup$
– Busy Zhu
Jan 14 at 22:34
$begingroup$
Thank you for your reply. But it seems to not match with my question.
$endgroup$
– Busy Zhu
Jan 14 at 22:34
$begingroup$
@BusyZhu You want to prove that $B^*$ is a Banach space. All you need to take into account is that $mathbb R$ is complete, thus Banach. That means that if $x in B^*$ and is Cauchy, then it converges and the proof is really straightforward.
$endgroup$
– Rebellos
Jan 14 at 22:38
$begingroup$
@BusyZhu You want to prove that $B^*$ is a Banach space. All you need to take into account is that $mathbb R$ is complete, thus Banach. That means that if $x in B^*$ and is Cauchy, then it converges and the proof is really straightforward.
$endgroup$
– Rebellos
Jan 14 at 22:38
$begingroup$
thanks again! But actually what I want to know is the way I prove it is correct or not. The idea of using definition is straightforward but seems to be weird.
$endgroup$
– Busy Zhu
Jan 14 at 22:59
$begingroup$
thanks again! But actually what I want to know is the way I prove it is correct or not. The idea of using definition is straightforward but seems to be weird.
$endgroup$
– Busy Zhu
Jan 14 at 22:59
add a comment |
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$begingroup$
The implication you're looking for is true but a bit strange to write down. If $X$ and $Y$ are normed vector spaces then the space of bounded, linear operators from $X$ to $Y$, $L(X,Y)$ is complete (and hence a Banach space) if and only if $Y$ is complete. $X^* = L(B, mathbb{R})$ and so is a Banach space since $mathbb{R}$ is complete. This doesn't require $X$ to be complete.
$endgroup$
– Rhys Steele
Jan 14 at 22:37
$begingroup$
Thank you for your reply! It's correct? But why does the book not use this simpler idea? I think the idea is probably wrong.
$endgroup$
– Busy Zhu
Jan 14 at 23:01
$begingroup$
If I'm honest I find your attempt confusing. Given a Cauchy sequence ${l_n}_{n geq 0}$ in $B^*$ you don't a priori have a good candidate for its limit $l$, so the $l$ you write down isn't defined here. Further, writing a bound like $|(l-l_n)(f)| leq |(l-l_m)(f)| + frac{varepsilon}{2} |f|$ isn't particularly helpful because you'll still be left to bound $|(l-l_m)(f)|$ in terms of $|f|$. For this proof to work you'd need to be able to pick $m$ independent of $f$ and if you could do that you'd just bound $|(l-l_n)(f)|$ for large enough $n$ straight away rather than doing this first.
$endgroup$
– Rhys Steele
Jan 14 at 23:05
$begingroup$
The usual technique is to check that for $f in B$, $(l_n(f))$ is a Cauchy sequence in $mathbb{R}$ and so converges to some number $L(f) in mathbb{R}$. You can then define a map $l: B to mathbb{R}$ by $l(f) = L(f)$. It's then not to hard to check that $l$ is linear by properties of limits. Then you check that $l$ in fact must be a bounded linear map and finally you show that $l_n to l$ in $B^*$ by sending $m to infty$ in $|l_n(f) - l_m(f)| leq varepsilon |f|$ for suitably large $m,n$.
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– Rhys Steele
Jan 14 at 23:10