Mixing
Prove that $C$ is Banach.
$begingroup$
$C={x_n lvert x_n converges } $
Let $x^n in C$ is Cauchy.
$rightarrow$ For $epsilon> 0$ there is N such that $n,m >N $ $$ lVert x^n -x^mrVert< frac {epsilon} {3} $$
we know that for every k $$lvert u^n -u^mrvert le sup_{ige 1} lvert x^n_i -x^m_irvert < frac {epsilon} {3} $$
So $u^n$ is Cauchy in $mathbb R$ which is Banach so $$u^n rightarrow u in mathbb R or lvert u^n -urvert < frac {epsilon} {3} $$
by this we can say that $lVert x^n -xrVert = sup lvert x^n_i -x_i rvert < frac {epsilon} {3} $ $ $ ($epsilon>0, nge Nin mathbb N$)
which means $x_n rightarrow x$
Now to show that $xin C$
$$lvert x-urvert le lvert x^n-xrvert+ lvert x^n-u^nrvert +lvert u^n-urvert <frac {epsilon} {3}+frac {epsilon} {3}+frac {epsilon} {3}=epsilon$$
This gives us $xrightarrow u$
So $xin C$
Is this Correct?
functional-analysis
asked Jan 5 at 23:05
HitmanHitman
1749
$endgroup$
add a comment |
$begingroup$
$C={x_n lvert x_n converges } $
Let $x^n in C$ is Cauchy.
$rightarrow$ For $epsilon> 0$ there is N such that $n,m >N $ $$ lVert x^n -x^mrVert< frac {epsilon} {3} $$
we know that for every k $$lvert u^n -u^mrvert le sup_{ige 1} lvert x^n_i -x^m_irvert < frac {epsilon} {3} $$
So $u^n$ is Cauchy in $mathbb R$ which is Banach so $$u^n rightarrow u in mathbb R or lvert u^n -urvert < frac {epsilon} {3} $$
by this we can say that $lVert x^n -xrVert = sup lvert x^n_i -x_i rvert < frac {epsilon} {3} $ $ $ ($epsilon>0, nge Nin mathbb N$)
which means $x_n rightarrow x$
Now to show that $xin C$
$$lvert x-urvert le lvert x^n-xrvert+ lvert x^n-u^nrvert +lvert u^n-urvert <frac {epsilon} {3}+frac {epsilon} {3}+frac {epsilon} {3}=epsilon$$
This gives us $xrightarrow u$
So $xin C$
Is this Correct?
functional-analysis
asked Jan 5 at 23:05
HitmanHitman
1749
$endgroup$
$begingroup$
You will not get a faster answer by reposting, even less if you do not explain, say, your notations.
$endgroup$
– Mindlack
Jan 5 at 23:09
1
$begingroup$
The second part seems flawed.. What is $u$ there?
$endgroup$
– Berci
Jan 5 at 23:10
$begingroup$
I have no idea what $C$ is, what your hypotheses on it are, what $u$ is, what $x$ is, what you mean by $x_n to x$, if $x$ is not necessarily in $C$, or what these $x^n_k$ are. So no, it isn't correct.
$endgroup$
– user3482749
Jan 5 at 23:10
$begingroup$
@Berci Please check now.
$endgroup$
– Hitman
Jan 5 at 23:30
$begingroup$
@Mindlack it is not a repost.
$endgroup$
– Hitman
Jan 5 at 23:30
add a comment |
$begingroup$
$C={x_n lvert x_n converges } $
Let $x^n in C$ is Cauchy.
$rightarrow$ For $epsilon> 0$ there is N such that $n,m >N $ $$ lVert x^n -x^mrVert< frac {epsilon} {3} $$
we know that for every k $$lvert u^n -u^mrvert le sup_{ige 1} lvert x^n_i -x^m_irvert < frac {epsilon} {3} $$
So $u^n$ is Cauchy in $mathbb R$ which is Banach so $$u^n rightarrow u in mathbb R or lvert u^n -urvert < frac {epsilon} {3} $$
by this we can say that $lVert x^n -xrVert = sup lvert x^n_i -x_i rvert < frac {epsilon} {3} $ $ $ ($epsilon>0, nge Nin mathbb N$)
which means $x_n rightarrow x$
Now to show that $xin C$
$$lvert x-urvert le lvert x^n-xrvert+ lvert x^n-u^nrvert +lvert u^n-urvert <frac {epsilon} {3}+frac {epsilon} {3}+frac {epsilon} {3}=epsilon$$
This gives us $xrightarrow u$
So $xin C$
Is this Correct?
functional-analysis
asked Jan 5 at 23:05
HitmanHitman
1749
$endgroup$
$C={x_n lvert x_n converges } $
Let $x^n in C$ is Cauchy.
$rightarrow$ For $epsilon> 0$ there is N such that $n,m >N $ $$ lVert x^n -x^mrVert< frac {epsilon} {3} $$
we know that for every k $$lvert u^n -u^mrvert le sup_{ige 1} lvert x^n_i -x^m_irvert < frac {epsilon} {3} $$
So $u^n$ is Cauchy in $mathbb R$ which is Banach so $$u^n rightarrow u in mathbb R or lvert u^n -urvert < frac {epsilon} {3} $$
by this we can say that $lVert x^n -xrVert = sup lvert x^n_i -x_i rvert < frac {epsilon} {3} $ $ $ ($epsilon>0, nge Nin mathbb N$)
which means $x_n rightarrow x$
Now to show that $xin C$
$$lvert x-urvert le lvert x^n-xrvert+ lvert x^n-u^nrvert +lvert u^n-urvert <frac {epsilon} {3}+frac {epsilon} {3}+frac {epsilon} {3}=epsilon$$
This gives us $xrightarrow u$
So $xin C$
Is this Correct?
functional-analysis
functional-analysis
asked Jan 5 at 23:05
HitmanHitman
1749
asked Jan 5 at 23:05
HitmanHitman
1749
edited Jan 6 at 0:22
Hitman
asked Jan 5 at 23:05
HitmanHitman
1749
asked Jan 5 at 23:05
HitmanHitman
1749
asked Jan 5 at 23:05
HitmanHitman
1749
1749
$begingroup$
You will not get a faster answer by reposting, even less if you do not explain, say, your notations.
$endgroup$
– Mindlack
Jan 5 at 23:09
1
$begingroup$
The second part seems flawed.. What is $u$ there?
$endgroup$
– Berci
Jan 5 at 23:10
$begingroup$
I have no idea what $C$ is, what your hypotheses on it are, what $u$ is, what $x$ is, what you mean by $x_n to x$, if $x$ is not necessarily in $C$, or what these $x^n_k$ are. So no, it isn't correct.
$endgroup$
– user3482749
Jan 5 at 23:10
$begingroup$
@Berci Please check now.
$endgroup$
– Hitman
Jan 5 at 23:30
$begingroup$
@Mindlack it is not a repost.
$endgroup$
– Hitman
Jan 5 at 23:30
add a comment |
$begingroup$
You will not get a faster answer by reposting, even less if you do not explain, say, your notations.
$endgroup$
– Mindlack
Jan 5 at 23:09
1
$begingroup$
The second part seems flawed.. What is $u$ there?
$endgroup$
– Berci
Jan 5 at 23:10
$begingroup$
I have no idea what $C$ is, what your hypotheses on it are, what $u$ is, what $x$ is, what you mean by $x_n to x$, if $x$ is not necessarily in $C$, or what these $x^n_k$ are. So no, it isn't correct.
$endgroup$
– user3482749
Jan 5 at 23:10
$begingroup$
@Berci Please check now.
$endgroup$
– Hitman
Jan 5 at 23:30
$begingroup$
@Mindlack it is not a repost.
$endgroup$
– Hitman
Jan 5 at 23:30
$begingroup$
You will not get a faster answer by reposting, even less if you do not explain, say, your notations.
$endgroup$
– Mindlack
Jan 5 at 23:09
$begingroup$
You will not get a faster answer by reposting, even less if you do not explain, say, your notations.
$endgroup$
– Mindlack
Jan 5 at 23:09
1
1
$begingroup$
The second part seems flawed.. What is $u$ there?
$endgroup$
– Berci
Jan 5 at 23:10
$begingroup$
The second part seems flawed.. What is $u$ there?
$endgroup$
– Berci
Jan 5 at 23:10
$begingroup$
I have no idea what $C$ is, what your hypotheses on it are, what $u$ is, what $x$ is, what you mean by $x_n to x$, if $x$ is not necessarily in $C$, or what these $x^n_k$ are. So no, it isn't correct.
$endgroup$
– user3482749
Jan 5 at 23:10
$begingroup$
I have no idea what $C$ is, what your hypotheses on it are, what $u$ is, what $x$ is, what you mean by $x_n to x$, if $x$ is not necessarily in $C$, or what these $x^n_k$ are. So no, it isn't correct.
$endgroup$
– user3482749
Jan 5 at 23:10
$begingroup$
@Berci Please check now.
$endgroup$
– Hitman
Jan 5 at 23:30
$begingroup$
@Berci Please check now.
$endgroup$
– Hitman
Jan 5 at 23:30
$begingroup$
@Mindlack it is not a repost.
$endgroup$
– Hitman
Jan 5 at 23:30
$begingroup$
@Mindlack it is not a repost.
$endgroup$
– Hitman
Jan 5 at 23:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The second part is not correct: $x$ should converge to a real number.
A hint for that: show that the real sequence $(x^n_n)$ is Cauchy, and show that $x=(lim x^n)$ converges to its limit.
answered Jan 5 at 23:39
BerciBerci
60.2k23672
$endgroup$
$begingroup$
Could you please check again. I have edited the question.
$endgroup$
– Hitman
Jan 5 at 23:48
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
The second part is not correct: $x$ should converge to a real number.
A hint for that: show that the real sequence $(x^n_n)$ is Cauchy, and show that $x=(lim x^n)$ converges to its limit.
answered Jan 5 at 23:39
BerciBerci
60.2k23672
$endgroup$
$begingroup$
Could you please check again. I have edited the question.
$endgroup$
– Hitman
Jan 5 at 23:48
add a comment |
$begingroup$
The second part is not correct: $x$ should converge to a real number.
A hint for that: show that the real sequence $(x^n_n)$ is Cauchy, and show that $x=(lim x^n)$ converges to its limit.
answered Jan 5 at 23:39
BerciBerci
60.2k23672
$endgroup$
$begingroup$
Could you please check again. I have edited the question.
$endgroup$
– Hitman
Jan 5 at 23:48
add a comment |
$begingroup$
The second part is not correct: $x$ should converge to a real number.
A hint for that: show that the real sequence $(x^n_n)$ is Cauchy, and show that $x=(lim x^n)$ converges to its limit.
answered Jan 5 at 23:39
BerciBerci
60.2k23672
$endgroup$
The second part is not correct: $x$ should converge to a real number.
A hint for that: show that the real sequence $(x^n_n)$ is Cauchy, and show that $x=(lim x^n)$ converges to its limit.
answered Jan 5 at 23:39
BerciBerci
60.2k23672
answered Jan 5 at 23:39
BerciBerci
60.2k23672
answered Jan 5 at 23:39
BerciBerci
60.2k23672
answered Jan 5 at 23:39
BerciBerci
60.2k23672
60.2k23672
$begingroup$
Could you please check again. I have edited the question.
$endgroup$
– Hitman
Jan 5 at 23:48
add a comment |
$begingroup$
Could you please check again. I have edited the question.
$endgroup$
– Hitman
Jan 5 at 23:48
$begingroup$
Could you please check again. I have edited the question.
$endgroup$
– Hitman
Jan 5 at 23:48
$begingroup$
Could you please check again. I have edited the question.
$endgroup$
– Hitman
Jan 5 at 23:48
add a comment |
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$begingroup$
You will not get a faster answer by reposting, even less if you do not explain, say, your notations.
$endgroup$
– Mindlack
Jan 5 at 23:09
1
$begingroup$
The second part seems flawed.. What is $u$ there?
$endgroup$
– Berci
Jan 5 at 23:10
$begingroup$
I have no idea what $C$ is, what your hypotheses on it are, what $u$ is, what $x$ is, what you mean by $x_n to x$, if $x$ is not necessarily in $C$, or what these $x^n_k$ are. So no, it isn't correct.
$endgroup$
– user3482749
Jan 5 at 23:10
$begingroup$
@Berci Please check now.
$endgroup$
– Hitman
Jan 5 at 23:30
$begingroup$
@Mindlack it is not a repost.
$endgroup$
– Hitman
Jan 5 at 23:30