Prove that $C_0$ is Banach.
$begingroup$
Let $x^n in C_0$ is Cauchy.
$rightarrow$ For $epsilon> 0$ there is N such that $n,m ge N $ $$ lVert x^m -x^nrVert< frac {epsilon} {2} $$
we know that for every k $$lvert x^m_k -x^n_krvert le sup_{ige 1} lvert x^m_i -x^n_irvert < frac {epsilon} {2} $$
So $x^n_k$ is Cauchy in $mathbb R$ which is Banach so $$x^n_k rightarrow x_k in mathbb R or lvert x^n_k -x_krvert le frac {epsilon} {2} $$
by this we can say that there is an N, $nge N$ such that $lVert x^n -xrVert = sup_{ige 1} lvert x^n_i -x_i lvert < frac {epsilon} {2} $ $ $
which means $x^n rightarrow x$
Since $x^m in c_0$,
there is some $N'$ such that $|x_i^m| < {1 over 2 } epsilon$ for $i ge N' (m=N)$
Now to show that $xin C_0$
$$lvert x_irvert le lvert x^m_i-x_irvert+ lvert x^m_irvert <frac {epsilon} {2}+frac {epsilon} {2}=epsilon for ige N' $$
This gives us $x_irightarrow 0$ for $ige N'$
So $xin C_0$
Is this Correct?
functional-analysis
asked Jan 5 at 21:53
HitmanHitman
1799
$endgroup$
add a comment |
$begingroup$
Let $x^n in C_0$ is Cauchy.
$rightarrow$ For $epsilon> 0$ there is N such that $n,m ge N $ $$ lVert x^m -x^nrVert< frac {epsilon} {2} $$
we know that for every k $$lvert x^m_k -x^n_krvert le sup_{ige 1} lvert x^m_i -x^n_irvert < frac {epsilon} {2} $$
So $x^n_k$ is Cauchy in $mathbb R$ which is Banach so $$x^n_k rightarrow x_k in mathbb R or lvert x^n_k -x_krvert le frac {epsilon} {2} $$
by this we can say that there is an N, $nge N$ such that $lVert x^n -xrVert = sup_{ige 1} lvert x^n_i -x_i lvert < frac {epsilon} {2} $ $ $
which means $x^n rightarrow x$
Since $x^m in c_0$,
there is some $N'$ such that $|x_i^m| < {1 over 2 } epsilon$ for $i ge N' (m=N)$
Now to show that $xin C_0$
$$lvert x_irvert le lvert x^m_i-x_irvert+ lvert x^m_irvert <frac {epsilon} {2}+frac {epsilon} {2}=epsilon for ige N' $$
This gives us $x_irightarrow 0$ for $ige N'$
So $xin C_0$
Is this Correct?
functional-analysis
asked Jan 5 at 21:53
HitmanHitman
1799
$endgroup$
add a comment |
$begingroup$
Let $x^n in C_0$ is Cauchy.
$rightarrow$ For $epsilon> 0$ there is N such that $n,m ge N $ $$ lVert x^m -x^nrVert< frac {epsilon} {2} $$
we know that for every k $$lvert x^m_k -x^n_krvert le sup_{ige 1} lvert x^m_i -x^n_irvert < frac {epsilon} {2} $$
So $x^n_k$ is Cauchy in $mathbb R$ which is Banach so $$x^n_k rightarrow x_k in mathbb R or lvert x^n_k -x_krvert le frac {epsilon} {2} $$
by this we can say that there is an N, $nge N$ such that $lVert x^n -xrVert = sup_{ige 1} lvert x^n_i -x_i lvert < frac {epsilon} {2} $ $ $
which means $x^n rightarrow x$
Since $x^m in c_0$,
there is some $N'$ such that $|x_i^m| < {1 over 2 } epsilon$ for $i ge N' (m=N)$
Now to show that $xin C_0$
$$lvert x_irvert le lvert x^m_i-x_irvert+ lvert x^m_irvert <frac {epsilon} {2}+frac {epsilon} {2}=epsilon for ige N' $$
This gives us $x_irightarrow 0$ for $ige N'$
So $xin C_0$
Is this Correct?
functional-analysis
asked Jan 5 at 21:53
HitmanHitman
1799
$endgroup$
Let $x^n in C_0$ is Cauchy.
$rightarrow$ For $epsilon> 0$ there is N such that $n,m ge N $ $$ lVert x^m -x^nrVert< frac {epsilon} {2} $$
we know that for every k $$lvert x^m_k -x^n_krvert le sup_{ige 1} lvert x^m_i -x^n_irvert < frac {epsilon} {2} $$
So $x^n_k$ is Cauchy in $mathbb R$ which is Banach so $$x^n_k rightarrow x_k in mathbb R or lvert x^n_k -x_krvert le frac {epsilon} {2} $$
by this we can say that there is an N, $nge N$ such that $lVert x^n -xrVert = sup_{ige 1} lvert x^n_i -x_i lvert < frac {epsilon} {2} $ $ $
which means $x^n rightarrow x$
Since $x^m in c_0$,
there is some $N'$ such that $|x_i^m| < {1 over 2 } epsilon$ for $i ge N' (m=N)$
Now to show that $xin C_0$
$$lvert x_irvert le lvert x^m_i-x_irvert+ lvert x^m_irvert <frac {epsilon} {2}+frac {epsilon} {2}=epsilon for ige N' $$
This gives us $x_irightarrow 0$ for $ige N'$
So $xin C_0$
Is this Correct?
functional-analysis
functional-analysis
asked Jan 5 at 21:53
HitmanHitman
1799
asked Jan 5 at 21:53
HitmanHitman
1799
edited Jan 6 at 6:26
Hitman
asked Jan 5 at 21:53
HitmanHitman
1799
asked Jan 5 at 21:53
HitmanHitman
1799
asked Jan 5 at 21:53
HitmanHitman
1799
1799
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The question has been modified based on this answer. The answer below is for the first version of the question.
First correction: $|a_n|<epsilon /2$ for all $n$ and $a_n to a$ does not imply $|a|<epsilon /2$. It implies $|a|leq epsilon /2$.
Second correction: in the last part $|x|$ has no meaning for a sequence $x$. Use coordinates. You should write $|x_i| leq |x_i^{n}|+|x_i^{n}-x_i|$. Then you should make the argument more rigorous as follows. First choose $n$ such that $|x_i^{n}-x_i|<epsilon /2$ for all $i$. Fixing this $n$ choose $k$ such that $|x_i^{n}|<epsilon /2$ for all $i geq k$. Now you get $|x_i| <epsilon$ for all $i geq k$.
answered Jan 5 at 23:53
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
$endgroup$
$begingroup$
Could you please check again? I have edited the question.
$endgroup$
– Hitman
Jan 6 at 0:19
$begingroup$
Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 0:38
$begingroup$
I think now it is okay.
$endgroup$
– Hitman
Jan 6 at 0:59
$begingroup$
what do you say?
$endgroup$
– Hitman
Jan 6 at 1:00
1
$begingroup$
In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 4:47
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The question has been modified based on this answer. The answer below is for the first version of the question.
First correction: $|a_n|<epsilon /2$ for all $n$ and $a_n to a$ does not imply $|a|<epsilon /2$. It implies $|a|leq epsilon /2$.
Second correction: in the last part $|x|$ has no meaning for a sequence $x$. Use coordinates. You should write $|x_i| leq |x_i^{n}|+|x_i^{n}-x_i|$. Then you should make the argument more rigorous as follows. First choose $n$ such that $|x_i^{n}-x_i|<epsilon /2$ for all $i$. Fixing this $n$ choose $k$ such that $|x_i^{n}|<epsilon /2$ for all $i geq k$. Now you get $|x_i| <epsilon$ for all $i geq k$.
answered Jan 5 at 23:53
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
$endgroup$
$begingroup$
Could you please check again? I have edited the question.
$endgroup$
– Hitman
Jan 6 at 0:19
$begingroup$
Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 0:38
$begingroup$
I think now it is okay.
$endgroup$
– Hitman
Jan 6 at 0:59
$begingroup$
what do you say?
$endgroup$
– Hitman
Jan 6 at 1:00
1
$begingroup$
In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 4:47
add a comment |
$begingroup$
The question has been modified based on this answer. The answer below is for the first version of the question.
First correction: $|a_n|<epsilon /2$ for all $n$ and $a_n to a$ does not imply $|a|<epsilon /2$. It implies $|a|leq epsilon /2$.
Second correction: in the last part $|x|$ has no meaning for a sequence $x$. Use coordinates. You should write $|x_i| leq |x_i^{n}|+|x_i^{n}-x_i|$. Then you should make the argument more rigorous as follows. First choose $n$ such that $|x_i^{n}-x_i|<epsilon /2$ for all $i$. Fixing this $n$ choose $k$ such that $|x_i^{n}|<epsilon /2$ for all $i geq k$. Now you get $|x_i| <epsilon$ for all $i geq k$.
answered Jan 5 at 23:53
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
$endgroup$
$begingroup$
Could you please check again? I have edited the question.
$endgroup$
– Hitman
Jan 6 at 0:19
$begingroup$
Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 0:38
$begingroup$
I think now it is okay.
$endgroup$
– Hitman
Jan 6 at 0:59
$begingroup$
what do you say?
$endgroup$
– Hitman
Jan 6 at 1:00
1
$begingroup$
In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 4:47
add a comment |
$begingroup$
The question has been modified based on this answer. The answer below is for the first version of the question.
First correction: $|a_n|<epsilon /2$ for all $n$ and $a_n to a$ does not imply $|a|<epsilon /2$. It implies $|a|leq epsilon /2$.
Second correction: in the last part $|x|$ has no meaning for a sequence $x$. Use coordinates. You should write $|x_i| leq |x_i^{n}|+|x_i^{n}-x_i|$. Then you should make the argument more rigorous as follows. First choose $n$ such that $|x_i^{n}-x_i|<epsilon /2$ for all $i$. Fixing this $n$ choose $k$ such that $|x_i^{n}|<epsilon /2$ for all $i geq k$. Now you get $|x_i| <epsilon$ for all $i geq k$.
answered Jan 5 at 23:53
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
$endgroup$
The question has been modified based on this answer. The answer below is for the first version of the question.
First correction: $|a_n|<epsilon /2$ for all $n$ and $a_n to a$ does not imply $|a|<epsilon /2$. It implies $|a|leq epsilon /2$.
Second correction: in the last part $|x|$ has no meaning for a sequence $x$. Use coordinates. You should write $|x_i| leq |x_i^{n}|+|x_i^{n}-x_i|$. Then you should make the argument more rigorous as follows. First choose $n$ such that $|x_i^{n}-x_i|<epsilon /2$ for all $i$. Fixing this $n$ choose $k$ such that $|x_i^{n}|<epsilon /2$ for all $i geq k$. Now you get $|x_i| <epsilon$ for all $i geq k$.
answered Jan 5 at 23:53
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
edited Jan 6 at 0:39
answered Jan 5 at 23:53
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
answered Jan 5 at 23:53
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
answered Jan 5 at 23:53
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
68.6k53169
$begingroup$
Could you please check again? I have edited the question.
$endgroup$
– Hitman
Jan 6 at 0:19
$begingroup$
Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 0:38
$begingroup$
I think now it is okay.
$endgroup$
– Hitman
Jan 6 at 0:59
$begingroup$
what do you say?
$endgroup$
– Hitman
Jan 6 at 1:00
1
$begingroup$
In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 4:47
add a comment |
$begingroup$
Could you please check again? I have edited the question.
$endgroup$
– Hitman
Jan 6 at 0:19
$begingroup$
Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 0:38
$begingroup$
I think now it is okay.
$endgroup$
– Hitman
Jan 6 at 0:59
$begingroup$
what do you say?
$endgroup$
– Hitman
Jan 6 at 1:00
1
$begingroup$
In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 4:47
$begingroup$
Could you please check again? I have edited the question.
$endgroup$
– Hitman
Jan 6 at 0:19
$begingroup$
Could you please check again? I have edited the question.
$endgroup$
– Hitman
Jan 6 at 0:19
$begingroup$
Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 0:38
$begingroup$
Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 0:38
$begingroup$
I think now it is okay.
$endgroup$
– Hitman
Jan 6 at 0:59
$begingroup$
I think now it is okay.
$endgroup$
– Hitman
Jan 6 at 0:59
$begingroup$
what do you say?
$endgroup$
– Hitman
Jan 6 at 1:00
$begingroup$
what do you say?
$endgroup$
– Hitman
Jan 6 at 1:00
1
1
$begingroup$
In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 4:47
$begingroup$
In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 4:47
add a comment |
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